Ci|0rtteU MttitJcraitg Sltbrarg
BOUGHT WITH THE INCOME OF THE
SAGE ENDOWMENT FUND
THE GIFT OF
HENRY W. SAGE
1891
Cornell University Library
arV18631
Graphical methods in applied mathematics
3 1924 031 246 386
olin.anx
Cornell University
Library
The original of tliis book is in
tine Cornell University Library.
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GRAPHICAL METHODS
IN APPLIED MATHEMATICS
MACMILLAN AND CO., Limited
LONDON ■ BOMBAY • CALCUTTA
MELBOURNE
THE MACMILLAN COMPANY
NEW YORK • BOSTON • CHICAGO
ATLANTA • SAN FRANCISCO
THE MACMILLAN CO. OF CANADA, Ltd.
TORONTO
GRAPHICAL METHODS IN
APPLIED MATHEMATICS
A COURSE OF WORK IN MENSURATION AND
STATICS FOR ENGINEERING AND
OTHER STUDENTS
BY
G. C. TURNER, B.Sc.(Lond.)
LECTURER IN MATHEMATICS IN THE.SCIBNOE DEPARTMENT OF G0LJ)SM1THS' COLLEGE
(university of LONDON)
MACMILLAN AND CO., LIMITED
ST. MARTIN'S STREET, LONDON
1909
First published in 1908, under the title of " Graphics.'
Beprinted as "Graphical Methods," 1909.
GLASGOW : PRINTED AT THB UNIVERSITY PKES6
BV ROBRRT MACLEHOSE AND CO. LTD.
PREFACE.
The importance of Graphics in modern mathematical training,
and its numerous uses in practical work, render unnecessary
any excuse for the publication of an elementary account of
some of its applications, provided these applications are chosen
with discretion and treated with clearness.
The author is hopeful that competent judges will consider
that the present book fulfils these requirements. It has not
been written with a view to any particular examination ; but the
easier parts will be found to meet the needs of secondary schools
and of candidates in military and naval examinations; while
students in technical colleges and candidates in the examinations
of the University of London will, it is believed, find most of
the chapters of definite use to them.
All sections and exercises marked with an asterisk should be
omitted in a first reading of the volume ; students who wish
further to curtail the course of work will find an easy First
Course mapped out on page ix.
Special attention is directed to the large number of concrete
examples, worked out in detail, which are supplied in the
various chapters. It is essential that the student should himself
work out the graphical constructions according to the instruc-
tions given, and afterwards compare his results with those
obtainable by measurement of the figures in the text. To
avoid the tendency to produce very small figures, which
characterise the work of almost all students, the instructions
supplied will be found to determine large drawings in nearly
all cases. An endeavour should be made so to construct the
diagrams that all lengths are correct to at least three numerical
figures ; it is hoped that this degree of accuracy has been attained
in the answers given at the end of the book. Owing to slight.
vi PREFACE.
perhaps very slight, errors in construction the final result,
obtained by measurement, will often be slightly incorrect in
the third figure..
The student is strongly urged not to confine himself to
graphical methods only in statics and mensuration. The
employment of 'calculation and graphics may be likened to
the use of our two hands; no matter how highly developed
one instrument may be, much more can be done with the two
conjointly than with one alone. Of necessity, in this book
analytical methods and calculations are only incidentally
touched upon, but students with a knowledge of Trigonometry
will see that even roughly drawn vector polygons can easily
be used for purposes of calculation.
This opportunity is gladly taken to acknowledge a debt
of gratitude to Prof. Henrici, F.E S., of the Central Technical
College, London to whom the author's first knowledge of the
true value of Graphics is due. His teaching showed Statics
and Dynamics not merely as a branch of somewhat unsatisfying
Mathematics, but as a real and interesting subject with important
applications. Those acquainted with Prof. Henrici's work and
lectures will appreciate the author's obligation to him.
Thanks are also due to Mr. E. F. Witchell of the Central
Technical and Goldsmiths' Colleges for reading most of the
proof sheets, suggesting improvements, and correcting some of
the answers; to Prof. E. A. Gregory and Mr. A. T. Simmons
for their unsparing trouble during the preparation of the MSS.
and while the book was passing through the press ; and finally
to the Senate of the University of London and the Controller of
H.M. Stationery Ofiice for permission to make use of problems
set in various University, Civil Service, Naval, Military, and
Board of Education examinations, The source of each such
problem, and the date when subsequent to 1902, has been
given after the question.
G. C. TUBNEE.
SiDcrrp, November, 1907.
PAGE
CONTENTS.
CHAPTER I.
Graphical Arithmetic,- 1
Miscellaneous Examples I., - 41
CHAPTER II.
Graphical Mensuration, 43
Miscellaneous Examples II., - 65
CHAPTER III.
Vectors and their Application to Velocities,
Accelerations and Mass-Centres, 69
Miscella/neous Examples III., 115
chapter iv.
Concurrent Forces, 119
Miscellaneous Examples IF., 166
CHAPTER V.
The Link Polygon, 173
Miscellaneous Examples V., 208
viii CONTENTS.
CHAPTER VI.
PAOE
Stress Diagrams, 210
Miscellaneous Examples VI., 252
■ CHAPTER VII.
Friction, 256
Miscellaneous Examples VII., - 281
CHAPTER VIII.
Moments, 284
Miscellaneovs Examples VIII, 305
CHAPTER IX.
Bending Moment and Shearing Force, 307
Miscellaneous Examples IX., 335
CHAPTER X.
Stress Diagrams (cmtinued), 337
Miscellaneous Examples X., - 353
CHAPTER XI.
Work, . 354
Miscellaneous Examples XL, 367
Appendix, Experiments on Moments, 368
INSTRUCTIONS TO THE STUDENT.
For the construction of the figures required in this book a set square
with a 3-inch side is useless. The side of the 45° set square should be at
least 6 or 7 inches.
The standard scale used should be flat on one side and bevelled on
the other and the scale divisions should reach to the edge. One edge
should be divided into fiftieths of an inch and the other into millimetres
or half-millimetres. Scales of this description can be obtained from
Messrs. Aston & Mander, Old Compton Street, London, W., and other
makers, at Is. 6d. each.
An angle is best set off or measured by means of its tangent or by a
scale of chords. If a protractor is used it should be a large semicircular
one of transparent material.
Hard chisel-pointed pencils should be used for all the constructions.
A EIEST COURSE OF STUDY.
Ch. I., pp. 1-17, 28-30, 33-35, 37-40.
Ch. II., pp. 43-48, 50, 53-59.
Ch. III., pp. 69-78, 84-92, 95-111.
Ch. IV., pp. 119-134, 137-162.
Ch. v., pp. 172-200.
Ch. VI., pp. 210-222, 2.36-245.
Ch. VII., pp. 256-264.
Ch. VIII., pp. 284-303.
Ch. XI., pp. 354-367.
CHAPTER I.
GRAPHICAL ARITHMETIC.
Scalar Quantities. In Mechanics and Physics, quantities
such as numbers, volumes, masses, time, temperature, dis-
placement, velocity and force are dealt with. Some of these
quantities are related to direction in space and cannot be defined
without reference to direction, others have no such relation to
space.
Mass, time, temperature, volume and number are examples
of quantities which are completely given when we know the
kind of quantity and how much there is of it; they are called
Scalar Quantities.
To specify the amount needs reference to some unit, a gramme,
a degree centigrade, a cubic centimetre, the number 1 . . . , so
that Scalar Quantities are specified by giving
(1) the unit quantity, (2) the number of units.
Vector Quantities. Those quantities which require for their
specification some reference to direction in space are called
Vector Quantities. Examples of these are displacement, velo
city, acceleration, force, ....
An hour differs from a minute only in amount, but the
pull of the earth on a book differs from the pull of a locomotive
on a train not only in amount but also in direction.
Time is a scalar quantity and force a vector quantity.
Quantities to Scale. The word scalar is used because these
quantities can be graphically represented to scale by lengths
(Latin scalae — a ladder —divided into equal parts by the rungs).
Thus, if we agree to represent unity by a length of 3 cms. then
the number 3 would be represented by a line 9 cms. long, and a
line of length 10'5 cms. would represent the number 3'5.
2 GRAPHICS.
In all cases of the representation of physical quantities by
lengths, the scale of the representation, i.e. the length repre-
senting the unit quantity, must be given either directly or by
implication.
Masses to Scale.
Example. To construci a scale of masses so that the mass
corresponding to any length, and the length corresponding to any
mass, can be read off at once.
The given line u (Fig. 1) represents 1 lb. mass. Transfer this
length to your drawing paper, by pricking through with needle
points or by the aid of dividers (having fine adjustment), and
mark the end points (left) and 1 (right). Mark off on this
line produced, lengths giving 2, 3, 4, ... 10 lbs., as follows :
In the figure 01 is (intentionally) not the same length as u.
(i) With dividers accurately adjusted to the length u, and with
the right-hand point as centre (marked 1 in figure) describe, a
Fig. 1
semicircle clockwise, pricking a slight mark at the point (marked
2 in figure) where the semicircle cuts the line. With 2 as
centre describe a semicircle contraclockwise, pricking through
at 3, and so on by alternate clock- and contraclockwise half
MASSES TO SCALE. 3
revolutions, pricking through points 4, 5, ... 10. With a properly
adjusted straight edge and set square draw fine sharp, short
lines perpendicular to 01 through the points marked.
(ii) Draw a straight line through making some angle
between 20° and 60° with 01, and mark off from inches
r, 2', 3', ... 10' along it.
Adjust a straight edge and set square with one edge of the
latter passing through 1 and I', so that when the set square
is moved parallel to itself along the straight edge to 10' it
still intersects 01 produced. Mark the points on 01 produced
when the set square passes through 2', 3', ... 10' by short, sharp,
fine lines.
These points so determined should coincide with the points
already marked on 01 produced ; why ?
We have now 01 representing 1 lb. mass, and 07 a mass of
7 lbs., etc.
(iii) To obtain, by method (ii), the division marks perpendicular
to 01.
Draw a fresh straight line and mark off 01 = m on it. Place
the inch scale and set square so that 1' 1 is perpendicular to 01
when the scale edge passes through 0, and mark off the points
2, 3, ... as before.
Example. To find the length which represents B-7 lbs.
(a) With a scale, adjusted at as before, mark the point
3-7" from along the scale, and with set square adjusted at
this point parallel to 3 3', mark a point on 01 produced 3-7,
then the length from to 3'7 represents 3'7 lbs.
(5) Produce 10 and I'O backwards through 0, and with an
inch scale mark tenths of inches along the latter up to 1 inch,
and from these points draw parallels to 3 3' cutting 01 produced
in points marked 0-1, 0-2, 0-3, ... 0-9. Then the distance
between 0-7 and 3 represents 3-7 lbs.
The final result in (b) is a scale of masses from which the
length corresponding to any mass between and 11 lbs. or the
mass corresponding to any length may be found.
4 GRAPHICS.
(1) With squared paper or straight edge ruled in mms. or 2 rams, find the
number represented by the line a, if (i) 1 cm. (ii) 2 cms. represents unity.
a
Via. 2.
(2) Make a scale for numbers from to 10 on squared paper, the length
representing unity being u.
Fio. 3.
(3) On a plan of a house J inch represents 3 feet. Draw a scale giving
feet and J feet. What length represents 7 ft. 6 in., and what length is
represented by 3 '2 ins., and oy the line a ?
Fro. 4.
(4) The areas of certain fields are represented by lengths to the seaie
of 6 cms. to an acre. Draw a scale giving 1 to 5 acres, tenths of an acre
and hundredths of an acre. Read off from your scale the area represented
by 17'3 cms. and the length which represents 4'25 acres.
Addition.
Example. The lines a, b, c, d, represent numbers to the scale of
I an, inch to unity. Find the sum of the numbers.
c -
C
Pio. 6.
Take a strip of paper with a straight edge and apply in turn to
the lines, marking with a fine sharp line the beginnings and ends
of the segments so that the segment OA is equal to a, AB is
equal to h and so on.
The edge, then, is marked OABCD as in Fig. 5.
In Fig. 5, OD is one-half the true length.
ADDITION. 5
Measure OD in half inches (or in inches and multiply mentally
by 2), this number of half inches is the required sum.
Notice that the order of addition is immaterial.
(5) A scale pan is suspended from the hook of a spring balance, and it is
loaded with small shot. The shot is put in by means of a small scoop.
The weight of shot added each time is given by the lines a, b, c, d, e, and
the line u represents 1 oz.
Pro. 6.
Find graphically the reading of the spring balance at each addition to
the load.
(Add the lengths as above and then draw the u scale along the straight
)
(6) A weight of shot given by the line in Fig. 7 is taken out of the
scale pan ; what is the reading of the balance ?
(7) What is the peri- ^
meter of the room of which
the accompanying figure is
the plan, drawn to a scale "s.
of 0-6" to lift. ^
Fig. 8.
Subtraction.
Example. The lines a, b, c, d, (Fig. 9) represent numbers to the
scale of r5 cms. to unity. Find the difference between the second
Tw/mker and the sum of the rest.
Add the lengths a + c + d as before and obtain OD on the
straight edge, cut off from D to the left DB=b, then OB is
6 GRAPHICS.
the length representing the required number. (In Fig. 9,
OJ, ... are half their true lengths.) Eead oiF the length of
OB on the 1'5 cm. scale.
O ABC D
Pio. 9.
Notice that since addition is performed as a continuous process
by adding lengths from left to right, subtraction must be
performed by setting off distances from right to left, if we wish
to measure our result from 0.
Example. Required the rmmier equal to the sum of the first cmd
third nv/mhers minus the sum of the second and fowth.
Mark, as before, on a straight edge, OA = a, AC=c, then to
the left, OD = d and DB = h. The point B comes to the left of
I 1 1 I I
BO DA C
Fio. 10.
the starting point (the origin), and the length OB, measured
to the left instead of the right, corresponds to the fact that the
required difference is negative. Measure OB on the proper
scale and prefix a negative sign to the number.
If distances to the right of represent positive numbers,
distances to the left must represent negative numbers.
Scale of Numbers. Such a line as BOD AC (Fig. 10) when
produced both ways represents numbers to the scale of 1-5 cms.
to unity. Every distance to the right of represents some
SUBTRACTION. 7
definite positive number, every distance to the left represents
some definite negative number; conversely, to every number
corresponds a definite point in the line.
(8) Find the sum of the numbers represented by a, b, c, d, the scale
being 0'4 inches to unity.
(9) Knd the algebraic sum corresponding toa + h-c + d.
( 10) Find the algebraic sum corresponding to a -h - c - d.
(11) Shew by actual measurement that
a + b-c + d^a + d-c + h,
and that a-b-c-d= -b~c + a-d.
Similar Triangles. The construction on page 3 depended
for its validity on a property of similar triangles, viz. the ratios
of the sides, taken in order, about the eaual angles are equal.
For triangles we can always ensure similarity by making them
equiangular. Crenerally, one figure is similar to another when
it is a copy of the second drawn to the same or a difierent
scale (in the first case the figures are congruent, i.e. identically
equal).
(12) Draw any triangle ABG and by the aid of the right angle of a set
square and a straight edge construct another triangle A-fi-fii, whose sides
are perpendicular to those of the first. Scale the sides and calculate the
'■^**°^ AB BG_ GA_
A^B^ B^Ci C^A{
(13) By aid of the 30° set squaTe construct .^jSjCj such that A^.2 is
turned clockwise through 30° from AB, and so on for the other sides.
Verify again that the triangles are similar.
A property of similar triangles often useful in graphical work
is that the ratio of their altitudes is equal to that of their bases.
(14) Verify this fact for the triangles drawn in the last exercise.
8 GRAPHICS.
The graphical constructions for multiplication, division, etc.,
depend on these properties of similar triangles. It should be
borne in mind that if the only object is to obtain the product,
quotient, root, or power of numbers, the graphical constructions
are but poor substitutes for abridged arithmetic, the slide rule
and logarithms ; it is only when in the course of other graphical
work it is found necessary to obtain, say, the product of two
numbers represented by lengths that the full advantage of the
methods becomes apparent.
Notation. To avoid circumlocutions and the constant repeti-
tion of ' the number represented by the length,' it is convenient
to use small letters a, b, c, ... for the lengths of lines, the numbers
represented by these lines being denoted by the corresponding
capitals A, B, C, — When the lengths a, b, ... are set off from
an origin or U, they will be lettered OJ, OB, ... or UA, UB, ....
The line representing unity is designated by u, unless some
measure in inches or centimetres is given.
Multiplication.
Example. The lengths a (5-98 cms.) amd b (8-84 cms.) represent
numbers to the scale 1-5" to unity. Find, (i) the length which gives
the product of the numbers, (ii) the product itself.
(i) Draw any two intersecting lines (Fig. 12). From the point
of intersection set off OU=l-b" and OB = h along one, and on
the other set off OA=a. Place a set square along AU and
move it parallel to itself until it passes through B, mark C
on OA where the set square cuts it.
OC is the required length, measure it by setting off the u scale
along OG and obtain the product.
OC OB
^^°°^- OA ^ W *°^' "®^"^ ^' ^' ^ ^^^ ^ *® numbers,
| = |, or C=A.B.
MULTIPLICATION.
Fig. 12.
10
GRAPHICS.
(ii) The construction given involves the transfer of lengths
from the given to the drawn intersecting lines. If the lines
a and b be already on the drawing paper this can be avoided.
JflG. 13.
From, one extremity of a draw u perpendicular to a (Fig. 13).
From one extremity of 6 draw a line c perpendicular to b, and
from the other a line perpendicular to the hypotenuse of the first
MULTIPLICATION. 11
right angle constructed. Then two similar triangles have been
drawn, the sides of the one being perpendicular to those of the
other.
Construct the u scale along c and measure c on that scale ; it
gives the product required.
Proof. From Fig. 13 ~ = ~, or ^ = 4, and .-. C=J.B.
° u B \
If the lines a and h are not parallel, draw u at one end of a
parallel to h, and complete the triangle ; then from the extremities
of h draw lines parallel to a and to the third side of the first
triangle.
The two triangles are similar (since they are drawn equi-
angular) and the line c giving the product is parallel to a.
For - = 1, or ah = cu, and A . B=0.
u
Measure c on the « scale and compare with the previous
results.
*(15) Vary the construction in the case where a is parallel to 6, by
drawing u at an angle of 60° with a.
(16) Draw two lines of lengths 7 "2 and .3 '9 cms. Let these represent
numbers to the scale of 0'7" to unity. Find the product by the methods
given. If unity be represented by 1"1", find the product of the new
numbers represented by the old lengths "'2 and 3 '9 cms.
12
GRAPHICS.
Multiplication on Squared Paper.
Example. 7/ u = 2", a = 8-38 cms , b = 6-82 cms., find the product
AxB.
Take a sheet of ordinary squared .paper (inches and tenths).
Mark off as indicated in Fig. 15, OA = a, OU=u, and UB = b.
Join OB and produce. Read off at once by the aid of the ruled
lines the length oi AG {AC being parallel to UB) on the 2 inch
scale ; it measures the product of A and B.
TL -7-
^ <E
-/(_
2 -/X-
1 X-
i r
-7 '<-
IS Z ;_
a/- 1-
Qf- \-
7 r
Z r
1 ^ i^
w
z: r
/
js ^ - i-
Y '
z ~ ^
/ r
r 1
Ai" 7 r X
-5 /O S >/l 1L 1-5A 2
J.
r
' t
7
t ^ (I _ _
^ 0-
1 ' .__
^ -j- A-
Flo. 15.
The side of each small square represents the number 0'05.
With a little practice a fifth of this, or the number 0-01, can be
estimated by the eye. To render the figure clearer in its
reduced size, the side of the smallest square shewn represents
0-1 and not 0-05.
THE STRAIGHT LINE. 13
Mark the points on OA and OA produced corresponding to
the numbers, 0'5, 1, r5, 2, 2-5, 3, 3-5 and 4, and to - 1 and - 2.
On the line through perpendicular to OA mark off the
points corresponding to the same numbers.
This method is exceedingly convenient when more than one
number has to be multiplied by the same factor. Any other
number being given by a length a,, we set off OA-^ = a-^, and then
read off the length of the corresponding perpendicular A-fi-^,
which is the product A^^x B.
(17) Read off the products of B and 2'7, 3'1 and 0-6.
(18) What number is represented by 6? Read oflf the products of this
number and 1'4, 2'3, 2'38and compare the results with those obtained by
actual multiplication.
(19) Multiply graphically 1-75 by 1-16, 2-35, 4-64, 3-88 and 5-26, using a
scale of 2" to unity.
(20) Multiply graphically 0-18 by 5-6, 2-4, 7'8, 6-9, using a scale of
1" to unity horizontally, and 10" to unity vertically. Read the products off
on the vertical scale.
Equation to a Straig^ht Line. Let any distance OM along
the line OA (Fig. 15) be x, and the corresponding perpendicular
distance PM be y. Then, wherever M may be along OA or OA
produced, y b ,
- = - always
= 1-34
= tangent of the angle OP makes with OM
(called the slope of the line).
The equation y = 1 '340; is called the equation to the straight line
OP. X and y are called the coordinates of the point P, and the
lines OA and its perpendicular through (lettered Ox and Oy in
Fig. 15) are called the axes of coordinates. PM or y is called the
ordinate, OM or x the abscissa of the point P. is the origin.
The straight line is called the graph of the corresponding
equation y=l-3ix, and any number of points on it could have
been obtained by giving x values 1, 2, 3, ... , and calculating the
corresponding values of y and marking* the points having these
* The points should be marked with a x , the two limbs of which must be fine
and sharp and intersect at the point.
14 GRAPHICS.
coordinates, or at once by drawing a straight line through
at a slope = 1 ■34.
Take any point A^ on OA produced to the left, then OA^
represents a negative number {A^. Read off what this number
is. Produce OB backwards through the origin and read off the
length on the 2" scale of the perpendicular line A^G^. This
number is the product B x A^. How does the figure shew that
the product is negative 1
(21) Find the product of 5 and -O'S, -1-5, -2-3.
(22) Find the product of --B and 0-8, 1-6, 2-3 directly from a figure,
(set off 6 downwards).
Since distances upwards along Oy represent positive numbers,
distances downwards along Oy produced must be considered
negative.
Note also that since — ^ = -, the equation -= 1 'Si represents
-X XX
the line POC2 produced indefinitely both ways.
(23) Draw the straight lines whose equations are
y=3x
y=x
y=\'5x
y=0'5a;
y=0-\x
i.e. draw lines through the origin the tangent of whose angles with the
axis of X are 3, 1, ... .
(24) In the third equation of the last exercise suppose x to have values
-2, - 1, 0, 1, 2, 3, in turn; calculate the corresponding values of y, and
shew that the points having these numbers as coordinates lie on the line
already drawn.
Notice that y = 0-lx, y = 0-Q\x, y = Q-QO\x, y = Q-QOQ\x are
successively nearer to the axis of x, and hence y = 0.x ov y =
must be the axis Qx itself. Similarly x=Q must be the axis of y.
(25) The lines o and 6 represent numbers to the scale u to unity.
Fio. 16.
Find by method (ii) the product of the corresponding numbers, con-
structing the u scale along c.
DIVISION. 15
(26) Find the product by method (i).
(27) Using paper divided into mras. or 2 mms., find the product of
1-74 and 082, 1-67, 2-31, 0-63, -1-31 and -2-36.
Different Scales. It is not necessary to use the same scale
horizontally and vertically. Suppose we wish to multiply 0'27
by 6-6 ; it would be better to represent the first number to a
scale 10 inches to unity, and the latter to a scale 1 inch to unity,
than to take the same scale and have lines differing greatly in
length. If the vertical scale be chosen as 10" to unity, then
the product must be read on that scale, for
GA OA
BU OV
and if OA and fZ be measured on the same scale so must CA
and BU.
Perform this multiplication graphically ; mark points V where
0U=\', and A where 0A = &-&' ; set up, perpendicular to OA,
UB^ 2-7", i.e. 0-27 of ten inches ; join OB and produce and read
off on the ten inch scale the length of A C.
(28) Multiply 0-037 by 8-1, 7-3, 5-6, 2-9 and 10-3.
A
Division. If C=A.B then — = — ; and, therefore, to find
1 B
C
the quotient -g, we have to make but a slight modification in
B
our construction for multiplication.
Example, a and b represent niimbers to the scale 2 cms. to unity.
Find the line representing the quotient = and the number itself.
a =3-92", b=l-34".
Notice that if we do not wish to find the line representing the
quotient, but only the number itself, we may take any length
whatsoever to represent unity. This follows from the fact that
a A
=- = ^ whatever the scale may be.
IB ^
(i) Draw any two intersecting lines (Fig. 17), from the point
of intersection set off 011=2 cms. along one, and 0A = a and
16
GRAPHICS.
0B = 1} along the other. Mark the point C on OU produced
where the line through A
parallel to BU cuts it. Then
OC is the required length c;
read the length on the 2 cm.
scale and the required quotient
2'93 is obtained.
Fio. 17.
(ii) Draw m (Fig. 18) perpendicular to 6 at its extremity. On
a construct a triangle similar to the one on 6 having its sides
Pio. 18.
DIVISION.
17
parallel to those of the first.
Then c being the side corre-
sponding to u we have
a h . A B n -^
- = -, i.e. -n = Y- or C=^-
c u 6 1 is
(iii) On squared paper
(mm.) take two axes at right
angles. Set oS0U=2 cms.
(Fig. 19) and 05 = 6 along
the axis Ox (the axis of x),
BA = a parallel to the axis
Oy (the axis of y). Finally,
UC, perpendicular to Ox,
cutting OA in C, is the
length representing the
quotient. Read this length
oflF on the 2 cm. scale and
obtain the quotient C
(29) Using squared paper find
the quotient corresponding to
d'
^ and 5 (Fig. 20),
where 5 cms. represents unity.
(30) Find by aid of squared
paper the quotients of 5 '6, 4 '7,
2-8, 1-8, -2-6 and -1-5 by 26.
(The negative os's must be set off
downwards. )
(31) What are the equations
to the sloping lines used in the
two previous exercises ?
ill
ill
Fig. 19.
T.G.
Fici. S
B
18
GRAPHICS.
(32) Find by method (ii) the quotient representing ~ where u represents
o
(1)6-
{2)6-
FiQ. 21.
unity. Change the length of u to 2" and
see that the quotient is the same, but
that the line representing it is altered.
(33) Find by direct graphical con-
struction the quotients of 2'8, 5*7, 4'5,
3-7, -21 and -1-8 by -33.
Combined Multiplication
and Division.
Example, a, b and c represent
three numbers wnd u represents
unity. Find the line which b.
A.B
itself.
C
and the nwnier
Fig. 28,
MULTIPLICATION AND DIVISION.
19
(i) Draw two intersecting lines and set off along one OU—u
(Fig. 22), OA=a, and along the other OB = h and OC=c. Mark
the point D on OA where BD, parallel to AO, cuts it, then OD is
the required length.
Proof.
OD OB , ^ A.B
^=^and2) = -^.
Construct the u scale along OD and read off the number D.
(ii) Multiply the ratio — by a set of numbers B.^, B^, B^, ... ,
unity being represented by 0-5 inches.
On squared paper mark two axes Ox and Oy (Fig. 23) ; set off
along Ox the distance c and draw CA == a parallel to Oy.
V
c
_
—
—
—
—
_
_
_
-
-
Si
--
--
—
--
—
--
—
—
—
—
—
r
•
/
/
/
/
D
'/
/
/
1
/
1
/
n,
/
1
V
,
/
;
/
i
[
/
/
1
'
1
J
/
1
1
/
1
1
/
i
1
\
•
1
/\
]
1
''
1
"
•
/
/
1
[
1
X
/
1
1
1
F
\
(
-^
F
\
'■„
_
_
i_
\ 1
Fig. 23.
Join OA and mark the points on it where the y ordinates through
^1, ^2, ^3, ... , cut it, viz. Dj, D^, Dg. Then B^D^, BJ}^, B^D^
are the required lengths. Read off the corresponding numbers.
20 GRAPHICS.
(34) Draw four lines of lengths 7'8, 2-6 and 3'1 cms. and 0-4 inch. If
the last represents unity, find the product of the numbers represented by
the third and the ratio of the first to the second.
Continued Multiplication.
EXAMPUE 1. FiThd a line representing the product A.B.C.D
where u represents unity, the numbers being given by the lines a, b, c, d.
Set off along any line OU=u (Fig. 24),
OA=a,
OB=h,
OC=c,
and along an intersecting line OB = d.
Mark X-^ where AX-^, parallel to UD, cuts OD produced.
„ Xg „ BX^ „ UX-^ „ OD.
„ Xg „ CXg „ UX2 „ OD produced.
Then OX^ is the length required. Measure OX^ on the u scale
and obtain the product A.B.C.D.
Proof. ^> = ^,
OD OU'
qx^oB
ox~oir
ox^_oc
oX~~ou'
Multiply these ratios together and obtain
OD.OX^.OX^
_ OA.OB.OC
OU^
or X^ = A.B.C.D.
Example 2. The scale being 1 inch to unity, find, on squared paper,
the continued product of the nwnbers represented by a,,h, c and d.
CONTINUED MULTIPLICATION.
b-
c-
d
u-
Fio. 24.
22 GRAPHICS.
Mark positions for U, A, G and D along Ox (Fig 25), and
set up UB^ = b, parallel to the axis of y.
Joini OB^ and produce. Mark the point A^ on O^j where the
ordinate at A. cuts it. Mark the point A^ where A-^A2, parallel
to Ox, cuts UB-^. Joini OA^ and produce and mark orj it G^
where the ordinate at G cuts it. Join" OC'g and mark Gg on
UB where C^Cg, parallel to Ox, cuts it. If OCg be joined, the
ordinate at D would not intersect it on the paper, so mark G^
on the ordinate a,t 2U and join 00^, marking Dg where the
ordinate at D cuts it. Then 2DD^ gives the product required.
Measure this on the u scale and write down the product.
. OA _0U OG _0U_ OP -20 U ,
AA~UB; CC~VA^ DDg UC,'
. OA.OG.OD 20m
■ " AA-^ . GG^ . DBf UB^ . UA^ . UG^
But A A^=UA 2, GG^=UGg;
. A.G.D 2.
DDg B'
:. A.B.G.D = '2DD^.
Do the multiplication again, taking A, B, G, on Ox and UB-^
vertically. When the construction lines go off the paper use
the 2m line instead of the u line.
Change of Scale. When the lengths a, b, c, etc., are long
compared with u, or when there are many multiplications to be
performed, the lines 00^ OD^ ... in (ii) become so steep that their
intersections with the verticals CCj DD^ ... will not be on the
paper. Similarly in (i) X^, X^, Xg ... get farther and farther
along OB, and the lines joining them to U become more and
more nearly parallel to OB.
When this is the case the triangles become ill-conditioned.
To avoid this difficulty the scale must be changed. Thus, in
(i) if OGXg becomes an ill-conditioned triangle in consequence of
!) being much larger than in Fig. 24, either halve OX^, or double
I The lines need not actually be drawn, it is sufficient to mark the points.
CHANGE OF SCALE
23
Pio. 26.
24 GRAPHICS.
OU, and proceed as before; the resulting length OX^ must now
be measured on the scale of ^u.
Should OGX^ be still an ill-conditioned triangle, take \ of
OX^ or quadruple OU; it still ill-conditioned take -^ of OX^ or
ten times U, and read the answer on ^ or j^ of the u scale.
A similar change can be made in method (ii) if necessary.
A B
(35) Find the value of — ^,— , where
a = 5-2",
6=2-7",
c=l-4",
M = 4 cms.
(36) Find, on squared paper (mm. ), the value of ^ x ^j . ^3 . -B3, where
u =1",
a =57 cms.,
c =6'8 cms.,
61=2-3 cms.,
62=5-6 cms.,
63=9-2 cms.
(37) Find the value of 6-8x27x1-9x3-4 gr^pjiieaUy.
(38) Find the value oiA.B.G.D.E, where
a=6-3 cms.,
5 = 5'1 cms.,
c=2-7 cms.,
d=3-8 cms.,
e=5-9 cms.,
and u=2".
Continued Product of Ratios.
Example. Find the line representing to the scale u to unUy ike
A P "R
roduct = ■ ^. =, the numbers lemg
given by the lines
a =9-9 cms.,
e=10-6 cms.,
6 = 5-52 cms.,
/= 13-05 cms.,
c = 6-3 cms.,
u= 1-65 inches.
(^=3-46 cms..
PRODUCT OF RATIOS.
25
Set off OA, OB, OC, OD, OE, OF from along any convenient
line and OCA along an intersecting one (Fig. 26).
Mark on OU the points Xj, X^, Zg, where AX-^ is parallel to
BU, CXg is parallel to DX^, EX^ is parallel to FX^.
Then OX^ gives the required product; measure this on the
u scale.
_ . OX, OA
Proof. — -J - —
ox^_qc ox^OE .
OU OF OZi OD OX^ OF'
qX^OA 00 OE A C_ E
Oil OD' OD' OF ^^ ^ B'D'E
(39) Draw lines of lengths 10-3, 7-8, 6-5, 43, 39, 27 cms., and find the
continued product of the ratios of the first to the second, the third to the
fourth, etc., if 0'7" represents unity.
(40) Find the continued product of the first four numbers in Ex. 39 and
the value of ( -=-^ J .
26
GRAPHICS.
Integral Powers (Positive and Negative). Since A^ means
1 1
Ax Ax Ax A and A~^
means — ; =
it is evident
A» Ax Ax A'
that the constructions already given cover the cases in which
numbers have to be raised to positive or negative integral
powers. It is, however, simpler to use the subjoined con-
struction.
Example. Given a (1-62") and the unit line u (3-08 cms.) to
construct Unes giving A^, A^, A*, ... and _, — ., -^, ....
A A^ A^
Draw any two lines intersecting at right angles. Set off along
these OU=u and OA=A.
Join UA, and draw AA2, A^A^. A^A^ ... so that each line is
perpendicular to the one drawn immediately before it, as in
Fig. 27. Also, draw UB^, B^B^,
BJB^ ... where the lines are
parallel to A A 2, A^A^,
Pia. 27
POWERS. 27
Construct the u scale along a straight-edged piece ot paper.
Measure 0^^, OA^, OA^ ... on the u scale; they are A^, A^,
A^ ....
Measure 0£j, OB^, OB^ ... on the same scale; they are
1 _L J_
A' A^' A^""
Proof. All the triangles drawn are similar, and hence
OA^_OA^_OA,_OA
OA^~OA~ OA~OU'
Multiplying the ratios, we get
OA^_fOA\^
OU~\OUj'
but U represents unity,
.-. A^ = A^.
Similarly, m = {mf' """^
.'. Ag = A% etc.
For the reciprocals we have
OB^OB^ _ OB^ _OBj_OU
OB,-
Multiplying together
OB,~OB,~OBj^ OU OA'
OB^_/qu\\
ou~\oa)'
•■• A = 24-
Similarly, ' Bg = -jg, etc.
For positive powers the construction stops aX A^, since A^
would not be on the paper. Take, then, ^ of OA^ and proceed
as before ; then
OA^ _0A^,
^OA, OA,'
OA.
AOA^
'5-10 OA,
28 GRAPHICS.
The succeeding intercepts must, therefore, be read on the
Jjyth u scale.
Again OB^ is too small to measure accurately, take 10 OB^
and read on the 10 m scale.
The points ... B^, Bg, B^, B^, U, A, A^, Ag, ... are points on
a curve called the equiangular spiral. The intermediate points
on this curve would give fractional and decimal powers, and
would thus enable one to find the values of such expressions
as A% A'^. It is not difficult to construct such a curve geo-
metrically.
Square Roots.
Example. Find the square root of A, given a and u.
Set off OA=a (Fig. 28) and OU=u in opposite senses along
Pio. 28.
a straight line. On UA describe a semicircle UCA, and measure
OC where OC is perpendicular to UA.
Then C = ^fA.
Proof. Since UA is a diameter of a circle and 00 a semi-
chord perpendicular to it ;
00^ = OU.OA;
:. O^^A or 0=s/A.
By repeating this process we can find rapidly A^, A^,
(41) Draw a line 6o cms. long. If unity be represented by 2", find the
powers of the number up to the 6*, and their reciprocals.
(42) Draw a line 6-5 cms. long. If unity be represented by 3", find the
square, cube and 4"' power of the number, and their reciprocals.
(43) Find the square and 4''> roots of the given numbers in (41) and (42).
SQUARES AND SQUARE ROOTS.
29
Powers by Squared Paper.
Example. Construct a curve which gives by inspection the squares
of all numbers, integral and decimal, from - 3 to +3.
Take two axes (Fig. 29) along the thick lines of the squared
paper, the axis of x horizontally and the axis of y vertically.
Mark the large divisions along the axis of x 0'5, 1, 1-5, 2, ...,
and those along the axis of y 1, 2, 3, Draw OF through 0,
and the point x=3, y=S. In Fig. 29 part only of the squared
paper and the curve is shewn.
T
y
s -3^f---
=====""===================:^^^^===
i:i:::::4:i::::: iiiii:::: 2:Z:t:::
=="=====================^-7^==^="
_A—/—i
^ — -. ,
3 \ \ ^ ^ Ll
^ y ;
V ^ 1-5
_ ^ ___ _ '^L_ _^ ^,^__
================2-===;^^=;^--===^===
T- — -7^-" L —
1 V j^ 1 ;
X ^'' Q 0-5 ^P 1 1-5 2 B 2-S
_,^" :: i ::i :■ : :
Fig. 29.
Mark any point ^^ on OP by a sharp short line perpendicular
to OP.
From ^1 go horizontally to A2 a point on the ordinate at 1,
put a straight edge along OA^t ^^^ mark the point A^ where
it cuts the ordinate through A^.
30 GRAPHICS.
Proceed similarly with points B^, Cj, ... , along OP, taking at
least twelve points. In Fig. 29, to save confusion, the construction
lines for only 2 points Ag and B^ have been shewn. With a
little care and practice the points A^, Bg, ... can be marked
accurately without actually drawing any construction lines
except OP, the ruled lines in the paper being a sufficient guide
for the eye.
Take points also on OP, produced backwards through the
origin, such as L^, and repeat the construction and find a
number of points like Z/g. Join all the points so obtained by a
smooth curve drawn by freehand ; see that it is a smooth curve
by looking along it, and smooth down any humps and irregu-
larities that appear on it.
The curve thus constructed is such that the ordinate for any
point on it represents the square of the number given by
the corresponding abscissa.
Proof. OAAg is similar to 01 A2, and OA = 2AA^^ as lengths ;
. AA^_OA_AAg
■■ 1^2 ~ Ol~AAj;
But AAj^ and OA represent the same number, viz. A, though
to a different scale ; hence A^ being the number given by AA^,
Ag = A^, i.e. AAg represents the square of the number A, the
scale being one half that on which A is measured.
A similar proof holds for negative numbers, and the construc-
tion shews that the square of a negative number is positive.
(44) Read off from the curve as accurately as possible the squares of 0"52,
0-68, 0-84, 1-75, 1-98, 2-24, 25, 2:85.
(45) Read off the square roots of 0-55, 0-85, 2-54, 3-8, 3-6, 4-54.
(46) Find the square's of the numbers given by lengths, 1, 1-8, 2'3, 4-6,
5, 6 '7 and 7 cm. if unity be represented by 2".
(47) Find the square roots of the numbers given by the lengths, 2, 3'8,
4-7, 6-9, 8-5, 10-3, 11-1 cms. if 2" represents unity.
Evidently the curve as drawn is not adapted for finding the
squares of numbers much greater than 3. To find the squares of
greater numbers, the scale of numbers along Oy must be made
still smaller than that along Ox. Thus for numbers from to
EQUATION TO CURVE OE SQUARES. 31
100 take 1 incK to represent 10 along Ox, but along Oy take 1 in'-h
to represent 1000. The construction is almost exactly the same ;
the line OP joining to the point for which x= 100, y= 100.
(48) Find graphically the squares of 2-7, 3'6, 7"7, 9'8, using a tenth scale
along the y axis. Find the square roots of 87, 73, 60, 31, 20 and 12.
(49) Construct a curve giving 7- of the squares of the numbers ranging
from -4 to +4. (The line OP must now go through the point (7, 1).)
Equation to Graph. The curve just constructed (p. 30) is
such that every ordinate like BB^ represents a number which is
the square of the number represented by the abscissa OB. If,
then, y and x are these numbers, y = x^, and since this equation
holds for all points on the curve it is called the equation to the
curve, and the curve is the graph of the equation.
It must be clearly understood that the equation y = x^ is only
true if, by y and x, we mean the numbers represented by the lines
and not the actual lengths of the lines themselves.
If y and x denote the lengths representing the number then
the equation y = x^ is not true.
Let u be the unit length along Oy, and 2m the unit length along
Ox, then, referring to Fig. 29, we have OA=x, AA^ = y, and
since ^^3 _ AA^
TA OT'
The last form of the relation brings us back to the original
equation ; for ^^ is the number represented by the length y, and
X ^
■^ is the number represented by the length x.
(J50) If the scale along y had been 1" to unity, and 5" to unity along x,
what would have been the equation connecting the lengths x and y of the
coordinates of any point on the curve, and what would be the relation
between corresponding numbers? What length would represent the
square of the number 3 ?
(51) If a be the unit of length along x, and h that along y, what is the
equation connecting the lengths x and y ? If as =2 '3" and 6 = 1 '5", what are
the lengths representing the squares of 1 and 3 ?
Another construction for the curve y = x^ follows from the
geometrical method explained on p. 26,
32
GRAPHICS.
Take two axes on squared paper (Fig. 30) ; let unity be repre-
sented by 1" along Ox and ^" along Oy.
Take any point A on Ox and
by the aid of set squares draw
AA-^ perpendicular to the line
joining A and - 4 (on Oy).
Mark on the ordinate at A
the point A^ where A^A^,
parallel to Ox, cuts it.
Eepeat this construction for
a number of points like A, and
join the points like A^ by a
smooth curve ; this is the
curve of squares.
Proof. As on p. 28,
OA^ = OAixO{-4:).
Taking all measurements
in inches, we have, if OA = x
and 0A-^ = y,
x^=2y
(x and y being in inches).
Hence, if the axis of y be
marked J" to unity, we have,
as nvmbers,
y = xK
Via. 30.
(52) Construct, by a similar method, a curve giving directly 1 '6 times the
square of numbers from - 3 to + 3.
Cubes and Cube Roots.
Example. From the owne y = x^ construct a cv/rve giving the cvhes
of nmnbers from - 2 to +2.
First construct the curve of squares, the origin being in the
centre of the squared paper, and the y scale ^ that of the x.
Take any point A^ (Fig. 31) on the curve, go horizontally to
r
\
K
10
*
V
»
\
8
*
\
\
\
6
h
\
\
^^
4
'•<
*
^
>
2
V
»
V
^
'*
X
1
2
3
■f
,'
^^
*■''
2
^*
'
'■*
y
.
,,-'
■4
,
•
CURVE OF CUBES.
33
A^ on the vertical through 1, mark A^ on AA-^ where OA^ cuts
it, then AA^ gives the cube of A.
Proceed similarly with a number of other points like Ay Join
all the points similar to A^ (for negative as well as positive a;'s) ;
this is the curve giving the cubes.
Proof.
Fio. 31.
AA^ = OA^ (as numbers)
= 1^, and^^ = ^3..
.'. OA^ — AAg (as numbers).
If, then, ;/ denote the number corresponding to any ordinate
AA^, and x the number for the corresponding abscissa, y — x%
is the equation to the curve.
Notice that for a negative number x, y is negative.
T.e.
34
GRAPHICS.
(53) Find the_cubes of 0-3, 0-5, 1-4, 1-7, 1-9, 2-1, 2-8 and 31, and the
values of 4^2, v'3-5, vf^.
(54) From the curve y = 3? construct the curve giving the fourth powers
of numbers from - 1 '8 to + 1 ■&.
(55) Read off from the curve y=x^, the values of
4/2, V5, VO, (1-22)S (l-85)<.
(56) Draw a curve by the above construction giving J of the cubes of
numbers from to 10. What is its equation ?
:-:;:-E--;:::::|:::!:::::::::::::::;-::--;;;:
:i:::::::=-:::::::|:::^:::::i::::::-::-
^^^^^^ N^
r %
L» 5s^l_I
L-_ A^-^Z5!__lir
:; ; : ::::::: t::::i::: ::;-:: :::: ;;t :: :::::::::;-:
1 A ^ 1 \L
U Tf--* r —1-
=2. $---[; % ' A---^ 3-
:!- = >; ;:::::: !::;:;_"- t- - - - -
-n •- A
^2
Pig. 32.
Curve giving- Reciprocals. Let 1 inch represent unity.
Take the origin at the centre of the squared paper. Mark any
point A-i (Fig. 32) on the unit line parallel to Ox, mark also A^
where OA^ cuts the unit line parallel to Oy. Go horizontally to
CURVE OF RECIPROCALS. 35
Ag, the point where A^A^ cuts the ordinate, through Ay Then
AA^ represents the reciprocal of the number A.
Eepeat this process for a number of. points like ^j on the
positive and the negative sides of Oy. Join all the points like
A^ by a smooth curve. This curve is such that the ordinate at
any point A^ gives the reciprocal of the corresponding abscissa
number.
Proof. ^^3 = 1^2 and AA.^ = Ol.
.'. AAg = jT-j (as numbers).
(57) Read off from the curve the reoiprooals of 0'35, 0'75, 1'2, 1'85, 2'15,
2-75, 3-84, -0-65 and -2-78.
(58) Using a oonstruction similar to that on page 32, draw the curve of
reciprocals. [Take OA =1 always, make 0( -4)=x, then OAi=y.'\
Equation to Curve of Reciprocals. Let y be any ordinate
number, and x the corresponding abscissa number; then evidently
from the construction
i = - or «« = 1,
1 X
which is the equation to the curve drawn.
Since -x.-y = xy=l,
we see that the two parts drawn by graphical construction are
really branches of the same curve.
Notice also that for very big x's the y'& are very small, and
vice versa, hence as we travel along x in the positive sense the
curve approaches nearer and nearer to the axis but never crosses
it. Similarly, for very large negative x's the curve gets very
near to the axis of x (negative side) but is below it.
(59) Construct the curve which gives J of the reciprocals of numbers.
What is its equation ?
(60) Construct the curve giving 1 '7 times the reciprocals of all numbers
from 1 to 100. What is its equation ?
36 GRAPHICS.
Curve of Reciprocals Squared. From the cmve y = -,
construct the ewrve y = —^ giving the squares of the reciprocals of
rmmbers.
The construction is very similar to the last. Put a straight
edge along OA-^ where -4j is any point on the curve •j;y=\,
mark A^ where the straight edge cuts the ordinate at 1, go
horizontally to A^ on the ordinate AA-^, then AA^ gives the
required reciprocal squared.
How does the construction shew that ( — ] is positive ?
(61) Oonatruet from the curve ^ = - the curves
y=^, and y=^,
(62) Construct from y = ^ the curves
y=:^ and y=^.
(6.3) Prom y= 3;^ construct y^=a^, or y=x^.
So far multiplication, division, etc., have referred to numbers,
represented by lengths. On page 1 it was pointed out that
a length may represent any other scalar quantity, the length
representing the unit quantity being given.
Areas to Scale. The product of two lengths a and h is
defined as the area of a rectangle having a and h as adjacent
sides
The product of two unit lengths is unit area. To represent
the product of two lengths by a line, we must first choose a line
to represent unit area. This line may be the unit of length, or,
if more convenient, some other length.
The methods, for finding the lines representing areas or volumes,
are exactly the same as for multiplying numbers together ; it is
only the interpretation that is different.
AREAS TO SCALE.
37
Example. Represent the product 0/ a x b by a line, imit airea
being represented by u, the unit of length.
Set off OU==u (Fig. 33), OA = a along any line, and UB = h
perpendicular to it, tben
AG.OU=OA. VB
or AC .u = a.b.
AC is the height of a rect-
angle having unit length as
base. Measure this on the
M scale ; it gives the number
of unit areas contained in
a. b.
Note that although the
same line u represents unit
area and unit length, it is not
correct to say that lengths "
•' ? fio. 33.
and areas are represented to
the same scale. They are different physical quantities, and all
we can say is that the same length represents the same number of
units of length as of units of area.
Example. Lines of lengths 7 and 15 cms. represent the sides of a
rectangular room to the scale of 1" to 10'. Find a line giving the floor
area, when the unit area is
(i) 10 sq. ft., (ii) sq. yds., (iii) 17 sq. ft.
(i) Draw, as in Fig. 33, 011=1", OA = 15 cms., UB = 7 cms.;
produce OB to cut AC, the perpendicular at A to OA, in C.
Then, as before, AG.OU=OA.UB.
This equation remains true whatever the scale on which we
measure the lengths. If, then, we measure on the tenth inch
scale, each tenth represents one ft. (for OU represents 10 ft.), and
AC represents the height of a rectangle of base 10 ft. and area
equal to the given floor, i.e. gives the floor area in 10 sq. ft.
(ii) Set off OU=Q-%" and measure AC in tenths of inches.
(iii) Setoff 0C/'= 1-7"
38 GRAPHICS.
Any Scale. Suppose the given lengths OA and UB repre-
sented lengths to the scale 1" to x ft. and we want to find the
area represented by OJ x UB in sq. yds.
X being a number we can always set oft a line representing
9
that number of feet, and - of this will be OU, and AC must be
measured on the scale (- j to 1 sq. yd.
Thus, if x=1' we must divide 1" into 7 equal parts (or if
more convenient 10" into 7 equal parts), 9 of these will equal
OU. Make the construction as before, and measure AC on the
scale of J-" to 1 sq. yd.
(64) Lines of lengths 2-3 and 47 inches represent, to the scale of 10 ems.
to 7 ft. , the sides of a rectangular room. Find by construction the floor
area in sq. yds.
(65) Lines of lengths 1'82 and 3-65 inches represent the altitude and
base of a rectangle to the scale of 1 inch to 350 cms. , find geometrically the
area in 100 sq. cms. '
*(66) Find a line representing the volume of a rectangular box, in ob. ins. ,
whose edges are 7, 15 and 17 cms. in length, unit volume being represented
by 0"1 inch.
*(67) Find graphically the volume of a rectangular room whose dimensions
are given by lines of 3'2, 5'3 and 6'7 cms., the scale being 1" to 10',
(1) in cb. yds ; (2) in 10 ob. ft.
Work done. The work done in lifting a body vertically
upwards is defined as the product of the weight of the body
and the vertical distance moved through. If the weight be
expressed in pounds and the distance in feet, the product is in
foot-pounds (ft. -lbs.). The work done in lifting a 1 lb. weight
vertically through 1 ft. is thus 1 ft. -lb., and is the unit of work.
Obviously the work done in lifting 10 lbs. through 1 ft. is the same
as that done in lifting 1 lb. through 10 ft. or 1 oz. through 160 ft.
Example, w represents the weight of a body to the scale n to a lb.
weight, a represents the vertical distance moved through, to the scale f to
a ft. ; find graphically the work done in ft.-lhs.
Notice that mx/ is the area representing a ft. -lb., and wxs
the area we wish to find in terms of m x/.
WORK DONE.
39
We can most conveniently do this by finding the rectangle
whose base is m or / and whose area is w x s.
(i) Set off OU=u (Fig. 34) and OW=w along one axis, and
OF=f axiA. OS=s along an intersecting one.
Draw WX parallel to US and measure OX on the / scale.
This gives the work done in ft. lbs.
/-
JT -
Proof.
Pio. 34.
OX ow
or OX.OU=OW.OS;
OS OU
.'. ox .u = w.s.
Hence OX is the altitude of a rectangle whose base is u and
whose area is w .s; and therefore OX represents the vertical
40
GRAPHICS.
distance through which 1 lb. weight must be raised in order that
the given work may be done. If, then, OX be measured on the
/ scale the number of units in OX will be the
number of ft -lbs. represented hj w . s.
(ii) Set off OF=f (Fig. 35) and OJF^w along
any line, OS=s along an intersecting line, and
draw JVX parallel to FS.
rp, OJF OF
^^^" ox=-os'
:. OF. 0X=0W.0S;
i.e. /. OX=w . 8.
Fig. 36.
OX therefore measures the weight which, lifted
vertically through 1 ft., requires an expenditure
of work represented by w . s. Hence measure OX
on the u scale ; it gives the number of ft. -lbs.
represented by w .s.
Pio. 36.
(68) If a is actually 6", which oonstniction, (i) or (ii),
would be moat convenient ?
(69) In the CG.s. system the unit of work is an erg=dynex centi-
metre. If d (Pig. 36) is the weight of a body in dynes and 1,000,000 dynes
is represented by u, find the work done in lifting the body through the
distance c.
MOMENT OF A FORCE. 41
*{70) The speed of a body is given by v (Fig. 37), where u represents a
foot per second. Find the time the body takes to go a distance D, repre-
sented "by d ; /represents a foot.
/-
Fig. 3Y.
(71) The weight of a body is given by a line 4" long, the lb. being
represented by 1'3 ems. If 1 ft. is represented by a line of length 4",
find the work done in lifting the body through a distance given by a line
of length 15".
(72) Find graphically in ft. -tons the work done in raising a body, weight
0"75 ton, through a distance of 23 '2 ft.
Moment of a Force. If a force be applied to the arm of a
lever, the turning moment or torque of the force about the axis
(fulcrum) of the lever is measured in magnitude by the product
of the force and the perpendicular on its line of action from the
axis. The geometrical representation of a moment is (like that
of work done) an area. The difference between the two products
we shall see later.
(73) A straight bar PQ, 12 ft. long, is hinged at Q, a, force of 13 lbs. Is
applied at P making an angle of 35° with PQ. If a force of 1 lb. be
represented by a line 1 cm. long and if 1 ft. be represented by O'l", find in
two ways a line which represents the moment about Q, and read oflf the
moment by scale.
MISCELLANEOUS EXAMPLES. I.
1. Draw the lines a, h and c of lengths 4-7, 3 9, and 5-2 cms. Find
lines representing :^, A . B, .4^, to the scale of 0-5" to unity, and the
numerical values of those quantities.
2. Find a line which represents the fraction ^ to the scale of 9 cms.
to unity.
3. Determine graphically the value of ^6 and Vq.
42 GRAPHICS.
4. Find a line whose length represents \/7'2 to the scale of 07 inch
to unity, and read off the value of the square root.
5. Construct the line whose equation is l-7y = 5'8a;, and from the line
read off the values of ?l2i^, 5:2^8 ^^^ 4-6x5-8
1-7 1-7 1-7
6. Construct geometrically the curve y='2-'lx^ and find the values of
2-7x4-lS 3-6^ X 2-7 and a/^.
V .&'7
7. If a line of length 7 '2 cms. represents unity, find the product of the
ratio ^ and C, where a = 3-48", 5 = 1-85" and c=l-62", 2-08", 3-55", 4'28"
in turn.
8. In constructing the curve of cubes, unity is represented along Ox by
2" and along Oy by 0'5". What is the relation between the lengths
X and y for any point on the curve ?
9. Find the product A . Bin three ways, where a = 8 '7 cms. , 6 = 4 '8 ems.
and u=l'62".
10. w -2-3" represents, to the scale 2 cms. to 1 lb., the weight of a body ;
/i=7'2 cms. represents, to the scale 1 cm. to 1', the vertical distance the
body is moved through ; find the work done in ft. -lbs.
11. Construct geometrically the curve xy = 3'2, and find the values of
3'2 times the reciprocals of 1'3, 2'7, 4'2 and 0'8.
12. By aid of a straight line divide 2-72, 0-85, 3-64, 1-88 in turn by 1-35.
*13. The volume of a pyramid being -j base area x height, find graphically
the volume in cubic feet when the base is a rectangle, the sides of the
rectangle being given by lines of 7 '2 and 3 '9 ems. and the height by a line
of 4'3 cms. , the scale being 2" to 1 foot.
14. To divide a set of numbers by 5 '44 use a straight line graph, taking
the vertical scale (for the numbers to be divided) as 1 quarter-inch for 10,
and the horizontal scale (for the quotients) as 1 quarter-inch for 1. Obtain
from your graph the quotients of 60 and 218 by 5 '44, and verify by
calculation with the tables. Explain why the graphical method gives the
result. (Military Entrance Examination, 1905.)
15. Find graphically the values of 2-38, 18-3, 47 "5 when multiplied by
0'763
5-47 ■
CHAPTER II.
GRAPHICAL MENSUEATION.
The chief problem studied in this chapter may be concisely
stated as follows : Given an area bownded by straight or curved lines, to
find a length which will represent to a given scale the magnitude of the
area.
The process for effecting this is called reducing the given area
to unit base. The required length is the altitude of a rectangle
whose base is the unit of length and whose area is equal to the
given area.
The Triangle. The area of a triangle being half the pro-
duct of the base and altitude, if we can find another triangle
of equal area having one side twice the unit of length the
altitude of this second triangle measures the area.
Method I. Transfer ABC (Fig. 38) to drawing paper. Draw
through A a line
parallel to BG ;
with B as centre,
describe an arc of
a circle of radius
2 inches cutting
AA-^^ at A-^ (or
put a scale at B
in such a position
that BA^ = 2").
By the aid of set
squares, draw CD
perpendicular to
BA,. Put the
1 FlQ. 3S.
inch scale along
CD and read off p, the number of square inches in ABC.
44 GRAPHICS.
Proof. Since AJ-i is parallel to BC, the area of ^iJC=area
oiA-^BC; and, since BA-^ = 2 inches,
the area of ABC=^x 2 xp=p (sq. ins.).
(1) Draw a triangle having sides 3"7, 2'8 and 4'3 inches, and find a line
giving its area in sq. ins. (In this case 2 inches is less than anyperpen-
dioular from a vertex to the opposite side, so take 4 inches for BA^ and
read p on the J" scale. )
(2) Draw a triangle having sides 9, 7'3, 5'6 cms. and find a line giving
the area in sq. cms. (Take BA^, 10 cms. and read^ in mms.)
Method II. Since the lengths of BAj^ and CD may be inter-
changed without altering the area (i.e. we may make CD = 2
inches or in general = 2m), if BDC be kept a right angle, the
line from A to the base, parallel to BD, measures the area.
Transfer ABC (Fig. 39) to drawing paper. With C as centre
describe an arc of a circle of radius 2 units. Place the set
squares, in contact along one edge, so that an edge of one going
through B is perpendicular to an edge of the other going through
C ; a position can easily be found for the set squares in which
these edges intersect on the arc at D (say). In this position BD
is the tangent to the arc at D and CD is the radius to the point
of contact.* Move the B set square, parallel to itself, until the
edge passes through A, and draw AE to cut the base in E.
Measure AE ( =p) ; it gives the area of ABC (3-23 sq. ins.).
Proof. From (Fig. 39) we see that the areas ABC and A^BG
are equal, and that the area of the latter is
iA^C X altitude = ^AE . 2m,
and hence AE measures the area in sq. units.
When does this construction fail 1 See that the difficulty can
be got over by taking u, Jm, ^u, ... instead of 2m.
(3) Repeat this measurement ■ by describing a semicircle on BC and
setting off a chord, 0D=2u, in it, and then proceed as before.
(4) Find the area in sq. inches of the triangle whose sides are 7-5, 6-3
and 4 "7 cms.
Method III. Transfer ABC (Fig. 40) to drawing paper. Set
off along BC, BD^%i. Mark the point E on AB where CE,
*D could be found with the right angle of one set square only if the corner
were perfect and not rounded by use.
TRIANGLE.
A A,
45
E --'C
Fig. 39.
parallel to AD, cuts it ; measure EF ( =p), where EF is perpen-
dicular to BO ; p gives the area (1'58 sq. ins.).
46
GRAPHICS.
Proof. Join DE ;
then (as areas) AEG=DEG, :. ABC = EEC + DEC = EBB,
and the last triangle has 2m for its base.
(5) Repeat the construction, taking BA and AG as bases. Is this
construction always possible ?
Rectangle. Parallelograms and rectangles can be treated by
the method given for quadrilaterals in the next section ; but the
following way is a little simpler.
Draw a rectangle ABCD whose height BA is 8-5 cms. and
base BG is 3-2 cm. To find its area in sq. inches, set off along
Fig. 41.
BA, BU=l" (Fig. 41) and draw AE parallel to UC. Measure
BE in inches, and the number so obtained is the area of ABGD
in sq. inches.
This is like the old construction of pp. 8 and 9 over again
and needs no further demonstration.
Quadrilateral. Transfer the quadrilateral ABCl) (Fig. 42)
to drawing paper. With B as centre, describe an arc of radius
2m (2"), and draw the tangent DE to it from D (or describe a
semicircle on BD, and set off BE=2u in it). From A and C
draw AAj^ and CG-^ parallel to BD, and measure A^C^ which
gives the area of ABGD (2-57 sq. inches).
Proof. Join BA-,^ and BG-^ ; then (as areas) ABD = A.^BD and
BDG=BDG„ :. ABGD-
2m and base A-fi-^.
A^BG-^, a triangle whose altitude is
QUADRILATERAL.
47
(6) The sides of n, quadrilateral, taken in the order A BOD A, are 3 "7,
2, 4 and 2'8 inches, the angle ABC is a right angle ; find the area (i) by
using the diagonal A O, (ii) by using BD.
(7) From a point in a field lengths are measured OA =35 f t. ,
OB=72 ft. and 00=51 ft., the angles AOB and BOG being 55° and 50°
respectively. Draw the figure OABG to scale (2 ems. to 10 ft. say).
Reduce the figure to unit base, and determine the area marked out on the
field by the contour OABG.
C
Re-entrant Quadrilateral. The construction already given
holds for a re-entrant quadrilateral. Transfer ABCD (Fig. 43)
to drawing paper and proceed exactly as before. In this case,
using the diagonal BD, A^ and Cj are on the same side of E.
Measure A-fi-^^ in inches, this gives the area in sq. inches (0-97).
Notice that A-fi-^^ is now the difference between A-J) and C^D
instead of the sum
Proof. Join BA■^ and CA-^; ABCD is now the difference between
the triangles ^52) and GBD, and BAD = BAJ), BCD = BC^D.
Hence BADC=BA-^D-BOjD = BA^C^,
a triangle of altitude BE ("2?*) and base A-^Cj^.
48
GRAPHICS.
* If C, in Fig. 42, be moved nearer and nearer to BD, BCD
gets smaller and smaller and vanishes when C is on BD. If be
moved still further, so that it crosses BD, the triangle becomes
negative and has to be subtracted from ABD. Corresponding to
this change of sign of the area, there is a change in the sense of
the boundary as determined by the order of the letters. In
Fig. 42 the boundary, in the order of the letters BCD, is described
FIO. 43.
clockwise, whereas, in Fig. 43, the boundary, taken in the same
order, is described contraclockwise. On changing the sense of
the boundary of an area, we must, therefore, change the sign of
the area The equation JBCD = ABD + BCD holds, therefore,
for both Figs. 42 and 43, and since BCD = - CBD, we have
ABCD = ABD-CBD,
where the three areas have the same sense to their boundaries.
CROSS QUADRILATERAL.
49
It is usual to consider an area, whose boundary is described
contraclockwise, as positive, one with a clockwise boundary as
negative. In Fig. 43 ABGD is a negative area, but BADC is
a positive one.
(8) Given BAD=m', AB=1-2, AD=6, SG=5 and CD=3-3 cms., find
the area in sq. inches.
* Gross Quadrilateral. If is taken on the other side of
AB or AD the figure is called a cross quadrilateral, and the
area is still the difference between ABD and CBD.
C
\2u
Fig. 44.
Transfer the annexed figure ABGD (Fig. 44) to paper and
reduce to unit base as before. See that A■fi-^ (0'65") still repre-
sents the area ABD - the area DOB. Hence the area A BOD is
that of the triangle AOD - the triangle BOO. On going round the
figure ABGD in the order of the letters from A back to ^, it is
seen that AOD is described clockwise and BGO contraclockwise.
60
GRAPHICS.
The triangle AOD being greater than BOO, the figure ABGD,
taken in the order of the letters, is negative, but A DOB is
positive.
(9) Given ^0=1-2, D0=2-l, CO=0i and 50=0-8 inches, and
OOB=110° ; find the area of the cross quadrilateral in sq. ins.
Polygons (including the quadrilateral as a particular case).
Transfer ABODEF (Fig. 45) to drawing paper. Produce
AB both ways. With set squares mark Cj on AB so that CC^
is parallel to DB ; mark D-^ so that DD^ is parallel to G-^E,
FlO. 46.
then on the other side mark F^ so that F-yF is parallel to AE ;
find the area of EF-J).^ by reducing to unit base; this is the
area of ABODEFA (19 sq. cms.).
In general, to reduce any polygon to a triangle having its
base on a side AB of the polygon, put a set square along the
line joining B to the next but one vertex D, and bring down the
omitted vertex C to C^ on AB by a parallel to BD.
Then put the set square along the line joining Cj to the next
but one vertex E, and bring down D to B^ on AB as before.
This process is to be continued until only one vertex is left
POLYGONS. 51
above AB. Should the construction lines become awkward, the
transformation can be transferred to the end A of AB, or a
new base line may be taken on another side of the polygon.
Proof. Since BCD = BG^D in area
the hexagon ^ BGDi/i^^ = the pentagon AC^BEFA.
Join D^E; then, since C^DE=C^D^E
the pentagon AC^DE FA = the quadrilateral AD^EFA.
Join F-^^E; then, since AEF=AEF^
the quadrilateral AD-^EFA = the triangle F-J)yE.
(10) Reduce the same polygon to a triangle having its base (i) on BG,
(ii) on AD.
(11) Reduce the re-entrant pentagon AEDCB (Fig. 46) to unit base.
AB='i-5", BC=2-S8", AE=6-0&', ED=S-8".
Example. To find the area between a polygon and one enclosed
mtfdn it.
Instead of reducing the two polygons separately, the mensur-
ation may be effected by a continuous process.
Transfer the figure ABC... (Fig. 47) to drawing paper.
Produce KH to cut FA at L, then ABCDEFA - HKJJH^the
re-entrant polygon ABCDEFGHIJKLA when G and L are
coincident.
Reduce this polygon to unit base as before (32-3 sq. ems.).
52
GRAPHICS.
For many-sided figures the process is tedious and errors may
easily accumulate unless very great care is taken. In such cases,
and for curved boundaries, the strip division method (p. 58) may
Fio. 47.
be used. The planimeter, where one is obtainable, is, however,
best for such cases.
(12) ABODEA is a small pentagonal field, the sides AB, BO and AE
were measured and found to be 136, 52 and 95 yds. long respectively, and
the angles ABC, BCD, DEA and EAB had magnitudes 75*, 70°, 60° and
50°. Draw a plan of the field to a scale of 1 cm. to 10 yds., reduce the area
to unit base and determine the area of the field in sq. yds.
Circular Arc. Draw a circular arc AB (Fig. 48) of radius
3 inches subtending 90° at the centre. Draw the tangent AB-i
to this at A. Set the dividers so that the distance apart of the
end points is about 1-5". Step off the arc from B towards A
with alternate clock- and contraclockwise sweeps, prick a point
on AB^ where the last semicircular sweep would cut it. From
CIRCULAR ARC.
53
this point make as many steps along AB^ as were taken for
the arc, mark B^ the end of the last step. Then AB^ is very
roughly the length of AB.
Adjust the dividers again so that the length of step is about
I" (roughly), and
repeat the operation ;
you will come to a
point B2 near B^ .
Which most nearly
gives the length of
the arc BA, BA-^ or
BA^, and why?
Adjust the dividers
again so that length
of step is about ^'
(roughly). Repeat
the operation again,
and find a point B^
on AB. Can you "'"' *"'
distinguish between B^ and Bg ?
AB^ is approximately the length of the arc AB.
Professor Eankine gave the following construction for finding
as a straight line
the length of a
circular arc AB.
Produce the
chord AB (Fig.
49) to G making
BO=^AB. Draw
the tangent at B,
and with C as
centre describe
*■'«• *»• the arc AD cut-
ting BD at D, then BD is the length of the arc AB approxi-
mately.
54
GRAPHICS.
For an angle of 90° the error is about 1 %, so that the method
should not be used for arcs greater than a quadrant.
(13) Draw a oireular arc of radius 4" and subtending 50° at the centre.
Measure the length of the arc in cms. by the two methods given.
(14) Draw the circular arc whose base is 10'5 cms. and arc length 11 '4 cms.
In this problem the length AB is given and therefore .4C is known.
The point £> is therefore given by the intersection of two circles. The
centre of the circular are AB is therefore found as the point of intersection
of the perpendicular at .8 to BD, and the perpendicular to AB at its
mid-point.
For the arc of a semicircle there is a method due to a Polish
Jesuit, Kochansky (circa 1685).
Draw a semicircle ABC (Fig.
50) of radius 3". Draw a tangent
at one end of the diameter (B) ;
set off BD so that BOD = 30°.
From D step off D.E = three
times the radius, then AE is the
length of the arc ACS nearly.^
Any arc greater than a semi-
circle can be now found by first
finding the length of the semi-
circle by the above construction
and the remainder by Professor
Eankine's construction.
Fig. 50.
(15) Find the length of a semicircular arc of 4-72" radius and compare
with twice the length of the corresponding quadrant given by Rankine's
rule and the length given by stepping oflf the arc along a tangent.
iThe theory of the construction gives a value of 7r=3-14153 instead of 3-14159.
CIRCULAR SECTOR.
SS
Area of Circular Sector. Draw a circular arc AB (Fig. 51)
of radius 3'5" subtending at the centre an angle of 70°. At any
point G of the arc draw a tangent, and step off CB^ equal to the
arc CB, and GA■^ equal to the arc CA.
Reduce the tri- ^■'-
angle A-^OB-^ to unit
base by one of the
given methods, and
measure the alti-
tude in inches. This
number is the area
of the sector AOB
in sq. inches.
Proof. If the arc
be supposed divided
up into a very great
number of small parts LM (LM as drawn is not very small, but
this is only because if it were very small the points L and M
would seem to the eye coincident). LM being very small, it is
very nearly straight, and its area is therefore approximately
^LM X the perpendicular (p) from on it.
In addition, if all the chords like LM are equal, the per-
pendiculars are all equal, and therefore
Area of sector AOB is approximately equal to
^p X (the sum of the steps LM)
= ^pl.LM. (Read : "the sum of all terms like LM")
As the number of steps increases indefinitely the length LM
diminishes without limit; but SLilf approaches and ultimately
becomes the arc AB, p becomes the radius r of the arc, while
^p'2LM becomes the area of the sector.
Hence, area of the sector = area of A-^OB^.
The formula for the area is : Area = ^rW,
where r is the radius, 9 the circular measure ^ of the angle, and rO
the length of the arc.
1 See Note A, p. 374.
56
GRAPHICS.
(16) Draw a circnlar sector of radius 8 cms. such that the base of the
segment is 9 cms. Find the area in eq. inches.
(17) Find the area of a, quadrant of a circle of radius 3 '4 inches.
Area of a Segment of a Circle.
(i) Segment less than a semicircle. Draw a sector AOBC
(Fig. 52) of angle 75° and radius 4" ; join AB, cutting oflf the
segment ABC. Set ofif the arc along the tangent at C, OB^ = CB
and CA^ = CA.
Join BBj and draw
OB^ parallel to it;
similarly, draw OA2
parallel to AAj^, then
A^A^B^^ has the
same area as ABC.
Reduce this quad-
rilateral to unit base,
and measure the area
in sq. inches.
Fig. 52.
* Proof. Since segment .(4 (7i?
= sector OACB-A OAB = A OA^B^ - A OAB
= a0^2-82 + a0^i^2 + aO^^^i -i-^2^i.Biii2 - A0v45
= A OA^B^ + aOAA^ + a OB^B - A OAB + A^^B^B^
.'. segment = A^^B^B^.
A simplification is effected by taking C at A, so that A^ is at
A and the quadrilateral reduces to a triangle AB-^B^. Draw the
figure and make the construction. The proof then simplifies to :
segment ^5a= sector AOBC- A OBA
= aA0B^-aA0B2-A OBB^
= A AOB^ - A AOB^ - A OBjB^
=aAB^B^.
Find the area of this triangle and compare it with the previous
result.
CIRCULAR SEGMENT.
57
(18) Take 4 OS = 60°, 0.4 = 3". Eind the area graphically, and com-
pare with the value of 4'5 j J - — - J.
(The formula giving the area \s' A = -^(a- sin 2a), where r is the radius
and .40^= 2a in circular measure.)
(ii) Segment greater than a semicircle. Draw a segment
ACB (Fig. 53) of a circle such that AOB= 150°, and repeat the
A Bj A, B
Fig. 53.
previous construction (p. 56). A crossed quadrilateral A^A^B^B^
is obtained. Reduce this to unit base.
Verify, by means of the formula in Ex. 18.
* Proof. This is very similar to the preceding one.
Segment ACB = sector OACB + aAOB
= A^A^B^B^ + AO^B^- OA^A.^ - OB^B^ + AOB
= a'^A^B^B^ + OA^B^ -OA^A- OBB^ + AOB
= A^A,B^B,.
For the same segment, draw the tangent at A and step off the arc
along it. The area is thus reduced to a triangle ^^j-^i- Find
the area of this triangle and compare it with the previous result.
(19) Reduce a semicircle of radius 7'3 cms. to unit base.
(20) Reduce a segment having three quadrants to its arc to unit base,
the radius being 3 •2".
Irregular and Curved Figures. The areas of figures
bounded by curves or many straight lines are best obtained by a
58
GRAPHICS.
planimeter. The method of strip division gives, with care, a
very good approximation.
Place a sheet of tracing paper over the accompanying figure
and draw its outline. Divide the distance between the two
O, Y
Xv
Plo. 54.
extreme points and Oj into 12 strips of equal width. To do
this, set a scale slantwise between two parallel lines drawn at
and Oj (the whole figure must lie between these parallels),
in such a position XY that there are 12 equal divisions between
X and Y.
MID-ORDINATE. SIMPSON'S RULE. 59
Mark the mid-points Xj, X^, ... of these divisions, and
through these points draw lines parallel to XO.
Take a long strip of paper with a straight edge and add up
graphically the segments of the mid-lines intercepted by the curve.
The given area is approximately that of a rectangle whose
base is the width of the strips and whose altitude is the length
on the long strip. Calculate this in sq. cms. (53 '2).
The approximation made is in the assumption that each strip
has the area of a rectangle whose height is the line through the
mid-point (the mid-ordinate) and of the given width, and this,
again, assumes that the little shaded areas outside the figure are
just balanced by the shaded areas inside.
A more accurate value of the area is obtained by using
Simpson's Rule, viz.
where the area is divided into an even number of equally wide
strips of width h; y„ is the length of the first, and y^ the
length of the second side of the first strip; similarly, 2/1 is
the length of the first, and y^ the length of the second side
of the second strip, and so on; shortly put, y^, y.^, yi,---y„ are
the lengths of the bounding straight lines of the strips. Since
the number of strips is even, n must be even. The rule only
holds for an even number of strips.
Apply the rule to determine the area of Fig. 54. Here,
taking the strips from bottom to top, yQ = 0, 2/1 is the length of
the first dotted line within the curve, 1/2 ^^^ second ; the other
y's are not shewn, but «/j2 = (w=12). Add the ordinates
according to the rule, and compare the result with that obtained
by the mid-ordinate rule.
Place the tracing paper over a sheet of squared mm. paper and
count up the number of large squares wholly within the curve,
and then the number of small ones between these squares and
the curves, estimating for any decimals of a small square. Find
the area by this means and compare with the previous results.
60
GRAPHICS.
(21) Obtain the area of the hexagon on p. 50 by the strip division method.
(22) Draw a semicircle of radius 3"1" on squared paper.
Find the area by
(1) The mid-ordinate rule. (2) Simpson's Rule.
(3) Counting the squares. (4) Calculating from formula : Area-.^Trr^.
* Another Method for Figures bounded by Curved Lines.
For these figures it is usual to assume that the boundary can be
divided into parabolic arcs. In most elementary text books on
Geometrical Conies, it
is shewn that the area
of any parabolic seg-
ment, such as AGB
(Fig. 55), is equal to ^
of the triangle having
as base the base of the
segment, and having
its vertex on the tan-
gent to the curve
parallel to the base.
In Fig. 55 ^ of triangle ACB = a,ve& of segment.
We may, therefore, choose any convenient point P on the
tangent, divide AP into three equal parts, produce to Q, where
i'^ = one of these parts, and join QB. Then the area of AQB is
the parabolic segment area
(23) Find the area of the circular segment of Exercise (16) by treating it
a3 a parabolic segment.
Any curved area such as ABODE A (Fig. 56) may be divided
into a number of approximately parabolic arcs, AB, BC,
Each arc must be curved in one way only (i.e. must not contain
a point of inflexion) and through considerably less than 180° ;
the procedure is as foUows. Start with the base BC say ; draw
a tangent parallel to BC and produce AB to cut it at iT; divide
BK into three equal parts and make BB^ equal to four of these ;
join CB-^. Then CBBi = given parabolic segment in area.
Again, produce B-^C to cut the tangent parallel to CJD, and take
the I point C^ on B-^C and join C^D. Then CC^D = area of corre-
FlG. 55.
PARABOLIC SEGMENTS.
61
Dv
>Ci
A,*-
Fig. 6(1.
spending segment. Proceed in this way round the curve and
reduce the curved area to the rectilinear one B^O-^D^E^A^B.
Reduce this to unit base in the usual way (6-9 s(j. ins.).
62 GRAPHICS.
This method is, however, rather tedious, and errors due to want
of parallelism may, unless very great care be taken, lead to
considerable final error. It is generally better to use the method
of strip division and either the mid-ordinate or Simpson's Eule
for such cases.
Volumes of Revolution. Any such volume may be found
by a double reduction. Suppose the given area in Fig. 57 to
revolve about the line XX; then it will generate a figure called
a volume of revolution. In particular, if a right-angled triangle
ABC revolve about its base AB, it will generate a cone ; if a
semicircle revolve about its diameter, it will generate a sphere;
a rectangle about its base will generate a cylinder ; any triangle
about one side will generate a spear-head volume ; a circle about
an exterior line in its plane will generate an anchor ring ; and a
rectangle about a line parallel to its base will generate a figure
like the rim of a fly-wheel.
The construction to be explained is one, therefore, of great
generality.
Make a tracing of the outline of the given figure.
Divide the area up into 10 equally- wide strips, parallel to XX,
and draw the mid-lines for each of these as in Fig. 57. Only
the mid-lines are shewn.
Draw a line YF above XX and parallel to it at a distance
a inches, given by a = — (that in Fig. 57 is purposely not at
the correct distance).
Project the end points, like AB of each mid-line up to A'F,
on YY, AA' and BB' being perpendicular to YY. Join A' and
B' to any fixed point on XX, cutting AB in A-^ and B-^.
Connect all the points like Aj^B^ by a curve. The area of this
curve is proportional to the volume If a planimeter is not
obtainable, add up all the mid-lines like A^B^ of the new figure
(called the First Ectuivalent Figure) by the straight-edged paper
method, and obtain the area approximately in square inches;
multiply this by 20 ; the result is the volume of revolution in
cubic inches-
VOLUME OF REVOLUTION.
63
* Proof. Suppose the area divided up, not into 10 only, but
into a very great number of equally thin strips, of which AB
may represent any one. The strip must be considered infinitely
thin, so that it is all at the same distance from XX. Let x be
Fig. 67.
the length AB, and h the width of the strip, then, however small
A may be, 2te (read : "the sum of all terms like hx") will be the
area.
Let y be the distance of AB from XX, then when the strip
revolves round XX, keeping always at the same distance y from
it, it will generate a very thin hollow cylinder. The height of
this cylinder is x, its circumference is 27ry and its thickness is h ;
its volume, therefore, will be 2'n-yxh,
64 GRAPHICS.
All the other strips into which the area has been divided, will,
on revolution about XX, also generate very thin hollow cylinders,
and the sum of all these hollow cylinders is the volume of
revolution of the original area. Hence the
volume of revolution = "S/iTryhx.
In this sum, x and y change from term to term, but 27rA is the
same, and is therefore a common factor to all terms in the sum.
Hence, if V is the volume,
V= iirh 'Sxy.
Let a be the distance of YY from XX, then, from the method
of constructing the First Equivalent Figure, we see that OA-^B-^
and OA'E are similar triangles, and therefore
y a a a!
hence xy = a. A-^B.^.
A similar equation holds for all the strips like AB, and hence
"Shxy = "ZahAjB^ ,
= a'EhA-^By
But hA-^B-^ is the area of one strip of the Equivalent Figure,
and IhA-^B-^ is therefore the whole area.
.'. "Shxy or hlxy^ax area of Equivalent Figure,
.'. '2-irh "Exy = lira x area,
i.e. V= 27ra x area of Equivalent Figure.
If, then, in our special ease
10"
a = -^, 2;ra = 20"
and V= 20 x area of Equivalent Figure,
and is in cubic inches if the area be in square inches.
(24) Find the volume generated by a right-angled triangle ABC in
revolving about its base AB, if AB=i", BO=Z"
*(25) Find the volume of a sphere of radius 1'73".
*(26) The coordinates of three points A, B, C are, in inches, (0'5, 0'6),
(1, 2'5), (3'5, 0). Find the volume generated by the revolution of ABC
about the axis of x. Take point {.3 "5, 0) as 0.
*(27) Draw a segment of a circle of base 4'4" and height 2'9". Find the
volume generated by revolving the segment about its base.
MISCELLANEOUS EXAMPLES.
6S
MISCELLANEOUS EXAMPLES. IL
1. Reduce an equilateral triangle of side 9-1 cms. to unit base u= 1" by
the three methods given, and compare these determinations of the area
with those obtained by (i) measuring the base and altitude and taking
half the product, (ii) by calculation from the formula.
Area=l.»^f
a being the length of a side in inches. (1 inch =2 -54 cms.)
2. The coordinates of four points A, B, G, D are (2-3, 0), (4-1, 2-1),
(1 -2, 4-9), (-0-8, 2-2) in inches. Mnd the areas of the quadrilaterals ABGD
and A OBB.
3. Find the area of the figure ^BOD^
(Fig. 58) in square inches, where AB=5•^
cms., BC=9-S cms., and AEZ) is a cir-
cular arc for which £!F=3 cms.
Fig. 58.
4. The coordinates of 5 points A, B, O, D, E are (M, 2'2), (4-9, 0-8),
• (7-3, 5-8), (5-1, 7-3), (0-8, 5'2) cms. Find the area of ABODE in sq. ins.
5. Find also the area of 402)^5.
[If O be the point of intersection oi AC and BE, then
the difference of the areas OCDE and ABO has to be
found ; the construction for effecting the reduction to
unit base is exactly as on p. 50.]
6, Reduce to unit base the area of the lens section
shewn (Fig. 59), AB=i", 0E=l-8", ED=l-5", (i) by
drawing the figure to scale on squared paper and counting
up the contained squares ; (ii) by reducing the segments
separately to unit base.
7, Draw a circular sector of radius 12 cms. and angle 150 degrees. Find
the area of the segment (i) by the construction given in the text, (ii) by
treating it as a parabolic segment, (iii) by dividing it up into eight strips
of equal width and adding the mid-lines.
66 GRAPHICS.
*8. The circular segment of question 7 revolves about its base ; find the
volume of the solid generated.
9. One side of a field is straight and of length 200 ft. At distances
increasing by 20 ft. from one end, the width is measured and found to be
70, 100, 100, 130, 137, 180, 150, 145, 100 ft. Find approximately the area
of the field.
10. Reduce to unit base the area A BCD, where .ilS=5'9, AD=S-1 cms.
and AB is perpendicular to AD, and BOD is » circular arc of 12'5 cms.
length, convexity outwards. (The arc may be drawn by reversing
Rankine's construction.)
11. The comers of a triangular field PQR are determined with reference
to a base line AB by the dimensions PAB=5T, PBA=94:°, QAB-W,
QBA = UV,EAB=130°,SBA=iT. 4 Sis 50 feet long. Drawadiagram
to a scale of an inch to 100 feet, and determine the area of the field.
(Military Entrance Examination, 1905.)
12. A triangle has sides of 3 '9, 3"2 and 4-2 inches. Draw the triangle,
measure each of the angles with a protractor and find the area.
13. Test or prove geometrically the accuracy of the following graphical
method of determining the area of a quadrilateral A BCD: "Join BD;
through C draw CE parallel to BD meeting AB, produced if necessary,
in E ; with centre E and radius equal to twice the unit of length, describe
a circle ; from A draw a tangent to this circle, to meet DX, which is
parallel to .45, in X. Then the number of units of length in j4X is the
number of units of area in A BCD." Data for the test figure : BD=2 in.,
AB=V6ia., BG=1-Sin., CD=l-6m., DA = l-3in.
(Military Entrance Examination, 1905.)
Fig. 60.
14. In a survey of a field ABODE, of which a sketch is given (Pig. 60),
the following measurements were made: ^5=84 yards, .4(7=173 yards,
^i)=175 yards, ^S= 130 yards, LBA0=4r, LCAD^Se', LDAE=9Cf.
Draw a plan to a scale of an inch to 30 yards, and find the area of the field
from your plan. (Naval and Engineer Cadets, 1904.)
MISCELLANEOUS EXAMPLES.
67
90°
90
30
Fia. 61.
15. Above (Fig. 61) is a rough plan of the city of Paris drawn to the
scale of 1 centimetre to the kilometre. Find the area of the city in square
kilometres by measuring any lines you like. (Naval Cadets, 1903. )
68
GRAPHICS.
PlO. 62.
16. In any way you please, find the area of the given figure (Pig. 62)
to the nearest square inch. State your method.
(Naval and Engineer Cadets, 1905. )
CHAPTER III.
VECTORS AND THEIE APPLICATION TO VELOCITIES,
ACCELERATIONS, AND MASS-CENTRES.
Displacement. If a point moves from to A (Fig. 63)
along some curved or straight path, the line drawn from to A
is the displacement of the point. This displacement is inde-
pendent of the actual path of the point, and depends only
on the relative positions of the new and the initial points.
Fig. 63.
To specify the displacement, there must be given not only
the magnitude of OA {e.g. 1'32 inches), but the direction or lie
of the line (e.g. the North-South line) ; and not only the direction
of the line, but the sense of the motion in that line (e.g. towards
the North).
The displacement OAj though equal to OA in magnitude and
direction is of the opposite sense.
Displacements may be represented by lines drawn to scale, if
the lines be placed in the proper directions and given the required
senses. The sense of the displacement is indicated by an arrow
head on the line.
To indicate that the line from to A involves direction and
sense as well as magnitude, it is convenient to print the letters
in block type; thus OA means the length OA in its proper
direction and with its correct sense.
10 GRAPHICS.
In writing it is extremely difficult to keep the distinction
between the block and the ordinary capital, and so^when writing,
it is better to use a bar over the letters ; thus, OJ (" Maxwell "
notation) means the same thing as OA.
Sum of Two Displacements. If the point moves irom
to ^ and then to B, the final displacement from is OB,
while the displacement from J is AB and that of J from is
OA. These f:.cts are symbolised by the equation
OB = OA + AB.
Such an equation does not mean that the length OB is the
sum of the lengths OA and AB ; but simply, that the final
position of the moving point is the same whether displaced
directly from to 5, or first to A and then from A to B.
Sense and Sign. If the second displacement brings the
point from ^ to (so that B is at 0), then the final displace-
ment is zero, and
OA + AO = 0, or 0A= -AO.
Hence, changing the sense of a displacement changes the sign
of its symbol. (See also p. 6.)
Example. A train travels due N. for 20 miles, then N.E. fm
10 miles, what is its displacement? Another train goes 10 miles
N.E. and then 20 miles N., show that its total displacement is the
same as that of the first train.
Eepresent 10 miles by 0'5 in.
(i) Set off 0A = 4:" (Fig. 64) vertically upwards and AB = 2"
making 45° with OA produced. Measure OB in y\j in. and
divide mentally by 2 ; this gives the magnitude of the displace-
ment (28). With a good protractor (the vernier protractor is
best), or by aid of a scale of chords, measure 6 (15°) ; then the
displacement of the train is 28 miles 15° to the East of North
(approximately). Measure also by the aid of p and a table of
sines.
Notice that OB itself with the arrow head gives the displace-
ment, but, if it is necessary to state the di.splacement in words,
ADDITION OF DISPLACEMENTS.
71
we must give not only its magnitude but also the direction and
sense compared with some standard direction and sense — in this
case the line drawn towards the North.
[It is perhaps as well to notice that, drawing
OA parallel to the bound edge of the paper
and towards the top of the page, is only the
conventional way of representing " towards
the North." The line OA will only represent
the true displacement when the book is placed
so that the arrow head on OA does point
due North.]
Fig. 64.
(ii) Set off OB^ = 2" at 45° with the N. line, then 4" due N. ;
arrive at B as before, for OB^BA is a parallelogram.
Order of Addition. The order in which two displacements
are added is immaterial ; as an equation
and the displacement BjB is equal to OA and OBj is equal to AB.
(1) A circus horse trots with uniform speed round a circus of radius
80 feet in 1 minute. Starting from the south position, give the displace-
ment in 15, 30, 45 and 60 seconds. Make the drawing to the scale 2 mms.
to a foot.
(2) A ring slides 5 ft. along a 7 ft. rod from E. to W. whilst the rod
moves parallel to itself 7 ft. S. E. Find the total displacement of the ring.
The rod now rotates through 70° about its West end in a clockwise sense,
the ring remaining in the same position relatively to the rod ; find the
total displacement of the ring, if it starts at the E. end of the rod.
72
GRAPHICS.
Addition of any number of Displacements.
Example. A point is displaced successively from to A, from
A to B, to C aTid to D, the displacements being given in magnitude,
direction and sense by OiAj, OiBi, OjCi, OjD^. Find the resultant
O^A.^ = i% 0i5i = 7-92, Oi(7i=10-5, O^Di = i-lcms.,
J^0iBi = 25°, 5iOiCi = 120° and C^O^I\ = 80°.
From any point draw OA (Fig. 65) equal and parallel to
OjA^; from A draw ^5 equal and parallel to O-^B-^; from B
Fio. 65.
draw 5(7 equal and parallel to OjCj ; and, finally, CD equal and
parallel to O^D^. Then OD is the sum of the displacements
OD = OA + AB + BC + CD = OjAj + O^B^ + OjCj + OjDj .
Measure OD and the angle AOD ; these measurements give
the displacement in magnitude, direction and sense.
(3) From the same point" O add the displacements in a, diflferent order,
e.g. find
0, Ai + OiCi + OiDi + OiBi
OiDi + 0]Bi + OiAi + OiCi ,
and shew that the same resultant displacement is obtained.
(4) Find the sum of the displacements
OiA,-OBi + OCi-ODi
and OiCi-OiBi-OiDi + OiA,.
Since OA=-AO, to subtract a displacement AO we have only to
change the sense and add.
VECTORS. 73
Relative Displacement. All displacements are relative, for
there is no point in space known to be fixed. The earth turns
on its axis, the axis moves round the sun, and the sun itself is
moving in space.
Example Given the displacement of two points A and B relative
to 0, to find the displacement of B relative to A.
In Fig. 66 OA and OB are the displacements relative to ;
then AB is the displacement of B relative to A.
But
AB = AO + OB;
.-. AB = OB-OA,
Fia. 66.
and AB is the difference of the displacements of A and B
relative to 0.
If the displacement of A relative to B, viz. BA, had been
required we should have had
BA = BO + OA = OA-OB.
Complete the parallelogram OABB^ then
AB = OB, =OB + BB,.
The displacement of B relative to A may therefore be regarded
as follows. Give to both A and B a common displacement
BBj = AO, making the total displacement of A zero ; then the
total displacement of B is the relative displacement required.
Vectors and Vector Quantities. Displacements are
examples of directed, or vector quantities. The magnitude of
the quantity — e.g. how many feet displaced — ^being represented
to scale by a length, then, if this length be placed in the proper
direction and given the proper sense, we have a complete repre-
sentation of the vector quantity. Velocities of 20 and 30 miles
an hour due N. and E. respectively are represented by lines of
74 GRAPHICS
lengths 2 and 3 inches respectively, if the lines be placed in the
given directions and arrow heads marked on them to give the
senses : the scale is 1 inch to 10 miles an hour.
To specify a vector, magnitude, direction and sense must be
given, and a vector is defined as a geometrical CLuantity {e.g. a
line) which has magnitude, direction and sense.
Position. A vector has no definite position, it may be con-
ceived as occupying any one of an infinite number of parallel
positions.
In Fig. 67 AB and CD are equal vectors, and we write
AB = CD.
C D
Fio. 67.
But by changing the sense of CD we have
AB=-CD, or AB + OD = 0.
The connection between sense in geometry and sign in algebra
was considered on p. 70.
Notation. In order to avoid the repetition of the word
vector, Greek letters will often be used to symbolise them, the
corresponding English letter denoting the magnitude ; a, b, c,
d, e denote the magnitudes of a, /8, y, S, e.i Block letters are often
used in books to denote vectors— A, B, . — this is the " Heavi-
side" notation. Owing to the difficulty of writing these, small
■ Greek letters are to be preferred — "Henrici" notation." When
the vector is denoted by the letters placed at its ends, block
letters will be used ; in writing, use the Maxwell notation.
Addition of Vectors. The process is exactly the same as
for displacements, the formal enunciation is : To add vectors,
place the first anywhere, at the end of the first place the
beginning of the second, at the end of the second the beginning
'For the pronunciation see Note B, p. 374.
ADDITION OF VECTORS. 75
of the third, and so on ; then the vector from the beginning of
the first to the end of the last is the sum of the given vectors.
The vector giving the sum is often called the resultant vector,
and in relation to this the vectors are called the components.
When the end of the last and the beginning of the first vector
coincide, the sum is zero and the vectors are said to cancel.
(5) Draw any five lines and give them senses. Denoting these lines by
a, |8, 7, S, e prove that
a + p + y + S + e = p + d + e + y + a = S + y + p + a + e.
(6) Find the vector sum of
a--p + y-S + e = y-d-p + e + a.
(7) The lengths of five vectors are 3-5, 2-6, 47, 6-2 and 7-8 cms.
re.spectively, and they point N., S.W., 20° S. of E., 25° E. of S. and E.
Find the resultant vector, taking care to give its magnitude, direction and
Order of Addition. The order in which vectors are added
is immaterial.
In Fig. 68,
a- = a + j3 + y + S,
but /^ + y = y + /3,
:. cr = a + y + P+S.
PiQ. 68.
By changing the order two at a time, any desired order may
be obtained, and hence the theorem is established.
(8) Draw a regular hexagon OABCDE, find the vector <r such that
a + (3 + 7+3 + e + (r=0, where o = OA, ;8=0B,
Average Velocity. When a body moves, its displacement,
from some standard point or origin, changes witli the time.
The total displacement in any time divided by the time is
defined as the average velocity for that time.
This average velocity is then measured by a displacement and
is therefore a vector quantity.
76
GRAPHICS.
Example. A train at 10 a.m. is 100 miles N.E. oj London, d
noon it is 80 miles 15° N. of E. WTiat is its average velocity ?
Draw OP^ (Fig. 69) and OP^ to scale, giving the displacements;
measure PjPg on same scale and find its direction. Bisect P^P,
at M. Then PiP2is the total displacement, and P^M gives the
average velocity in miles per hour.
It should be evident, that, as nothing is known of the motion
between 10 a.m. and noon, the displacement in one hour is not
necessarily ^P^P^.
The average velocity means only that velocity which, if it
remained constant, would give the actual displacement in the
given time.
The average speed of a point for any interval of time is
defined as the distance traversed divided by the time.
(9) A man walks from O towards the S.W. for 4 miles and arrives at
A in 65 minutes, he then walks for 3 miles W. to S in 40 minutes, then
6-5 miles due N. to C7 in 105 minutes. What are his average velocities
from Oto^jOtoB, OtoG, and what are his average speeds?
(10) A man walks round (clockwise) a rectangular field A BCD in
27 minutes. Starting at A he is at B (due E. of A), distant 200 yds.,
in 7 minutes ; at C, distant 160 yds. from B, in 16 minutes ; and at D in
21 minutes. What are his average velocities and speeds from A to B,
A to G, A to D and irom A oack to A^
VELOCITY.
77
Speed and Velocity. A point moves on a curve from P
(Fig. 70) to Pj in time t. Its displacement in that time is the
FP
chord PPj and its average velocity is — 1.
t
During this time it travels over the distance PPj (arc) and
— - — 1 is defined as Ihe average speed.
t
If we take t very small, then the arc, the chord, and the
tangent at P become indistinguishable
the one from the other at P. At the
limit, when Pj approaches nearer and
nearer and finally comes up to P, the
chord PjP produced becomes the tangent
at P, and the magnitude of the velocity
is the speed at P. The direction of
motion is therefore always tangential to
the path.
At every instant of the motion the
point is moving with a definite speed in
a definite direction.
The smaller the time interval, the
smaller will be the vector PPj, and the
more nearly will the average velocity be
the same as at the beginning or end of
the time interval. The problem of finding, in particular cases,
to what fixed value this average velocity tends, as the interval
of time is taken smaller and smaller without limit, belongs to
the Calculus and cannot be discussed here. This limiting value
of the average velocity is evidently the velocity at the instant
under consideration, and its magnitude is the speed of the point
at that instant.
A velocity is specified by giving the speed, the direction and
the sense of the motion. ,
Speed is constant only when the point passes over equal
distances in equal times, no matter how small the equal times may
Pio. 70.
78 GRAPHICS.
be, or the average speed is always the same whatever the time
interval.
Velocity is constant only when the speed is constant, and the
direction and sense of the motion remain unaltered.
A point moving along a curved path may have constant speed
but cannot have constant velocity.
Units. The unit of length being a foot, and the unit of time
a second, the unit of speed is a ft. per sec. often written 1 ft./sec.
Other units in common use are 1 mile per hour, or 1 m./hr., and
1 cm. per sec. or 1 cm./sec.
Definition. The Velocity of a point is its rate of displace-
ment and is measured by the displacement in unit time, or the
displacement that would have taken place if the velocity had
remained constant.
Velocity is, therefore, a vector quantity and can be represented
by a line vector ; the length of the vector represents to scale the
magnitude of the velocity (the speed), the direction and sense of
the motion being shown by the direction and sense of the vector.
Velocities are added or compounded like vectors since they
are measured by displacements.
Example. A skip is momng due W. at a speed o/ 15 miles an
hour ; a passenger runs across the deck from S. to N. at 7 miles an
hour, find the velocity of the passenger relative to the earth.
Fig. 71.
Set off a (Fig. 71) from right to left of length 15 cms. ; add
P of length 7 cms. drawn vertically upwards ; then y the sum
(y = a + /?) gives the magnitude, direction and sense of the velocity
required. Scale y in cms. and measure the angle 6.
RELATIVE VELOCITY. 79
(11) A ship sails N. relative to the water at 5 ft. per sec. whilst a
current takes it E. at 3 ft. per see. ; what is the velocity of the ship
relative to the earth ?
(12) A river current runs at 2 miles an hour ; in what direction should
a swimmer go, who can swim 2-5 miles an hour, in order to cross the river
perpendicular to the banks? [Draw the vector of the velocity of the
current ; through the beginning of this, draw a line perpendicular to it,
and with the other end as centre, describe a circular arc of radius 2 '5 to
cut the perpendicular. This construction gives the velocities, of the
swimmer relative to the water and to the land. There are two solutions,
giving the directions from both banks.]
(13) A boat can be rowed at 6 miles an hour in still water, a river
current flows at 3 miles an hour ; how should the head of the boat be
pointed if it be desired to cross the river at an angle of 45° up stream ?
(14) A train travels E. at 65 miles an hour, a shot is projected from the
train at an angle of 30° with the forward direction and at a speed of 200 miles
an hour relative to the train ; what is the velocity of the shot relative to the
earth?
(15) A ball is moving at 10 miles an hour S.W. and is struck by a bat
with a force which would, if the ball had been at rest, have given it a
speed of 8 miles an hour due S. ; with what velocity does the ball leave
the bat ?
* Relative Velocity. A point at any instant can only have
one definite velocity ; it is impossible to conceive it as moving in
two diiFerent directions, or with two different speeds, at one and
the same instant. Relative to other moving points it may have
all sorts of velocities.
* Example. A train moves 20° E. of N. at 50 miles an hour ;
another train is moving TV. at 22 '6 miles an hour, and a third is
travelling due S. at 35 '2 miles an how. What are the velocities of
the first and third relative to the second ?
Notice that the relative motions of two or more bodies are
unaffected by any motion common to them all ; thus, the relative
motions of trains, people, ships, etc., are quite independent of the
motion of the earth round the sun. We may, therefore, suppose
any common velocity given to the trains.
In Fig. 72, a represents the velocity of the first train, /8 and y
those of the second and third.
80 GRAPHICS.
Add - /3 to a, then the sum o-j gives the relative velocity of the
first to the second train, for - (i reduces the second train to rest,
and then a- (3 is the velocity of the first relative to a supposed
Pia. 72.
fixed point. Scale o-j and measure the angle it makes with the
E. line (61-8 miles an hour 50-3° N. of E.).
Similarly, add -/8 to y and obtain o-g the relative velocity
of the third to the second.
(16) Two cyclists meet on a road, one is going S. at 10 miles an hour, the
other N. at 12 miles an hour ; what are their relative velocities ?
(17) A cyclist travels N.W. at 12 miles an hour, the wind is due E. and
travels at 20 miles an hour ; what is the apparent direction and speed of the
wind to the cyclist? [Add to the wind velocity one of 12 miles an
hour S.E.]
(18) A train travels due W. at 40 miles an hour ; the smoke from its
funnel makes an angle of 157° with the forward motion ; the wind is
blowing from the N. , what is its speed ?
(19) In the last example if the speed of the wind had been 20 miles an
hour, what are the possiole directions whence it could have come ?
RELATIVE VELOCITY.
81
* Example. Two ships are 20 miles apart, one (A) is then due E.
of the other (B), and is steaming due N. at 18 miles an hour; B is
steaming E. at 15 miles an hour. Find when they mil be nearest one
another, and their distance apart at that time.
Suppose a speed of 18 miles an hour due S. is given to both,
then A is reduced to rest, and the relative motion of A and B
B 15 . A
Fio. 73.
is unaltered. Hence, add the vectors (Fig. 73) representing the
velocities 15 due E. and 18 due S. and obtain a resultant vector
BC(7). From A drop a perpendicular on BC, and measure its
length on the scale to which the distance AB was drawn ; it is
evidently the shortest distance the ships will be apart. BC is
the relative displacement in one hour ; in order to find the time
for the relative displacement BD, set oif BC^ = 10 cms. in any
direction to represent 1 hour ; then, if DDi be drawn parallel to
OCj, BDj^ gives the time in hours.
* Example. Find ly construction the actual positions of the two
steamers, in the last example, when at their shortest distance apart.
To do this, draw through D a line diie North cutting BA
T.G, P
82
GRAPHICS.
PlO. 74.
at B-^, then B^ is the required position of B. Through B^ draw
-Bj^i parallel to DA cutting the N. line at ^ in A^; then A-^ is
the required position of A.
(20) Two ships are 11 miles apart, and both are steaming direct towards
the same point distant 1 1 and 7 miles from them respectively. They both
travel at 14 miles an hour. Find their shortest distance apart and the
corresponding time.
* Total Acceleration. When the velocity of a body changes,
the motion is said to be accelerated. This acceleration may be
due to the velocity increasing or
diminishing (retardation), or to >■
the change in the direction of
the motion or to both. Thus, if
aj (Fig. 74) gives the velocity
at one instant and a^ at a subsequent one, the magnitude only
has changed, and the total acceleration is a^ - a^ and is negative.
If jSj and /Sg (Fig. 75) denote the velocities at two instants,
then the change in the velocity is y, where
/3i + r=^2 or 7 = ^2-^1-
•y, the change in the velocity, is
simply the velocity which must be
added on — as a vector — to the
initial velocity ^j to give the final
velocity /Sg. Change in velocity is
then a vector quantity and must be
represented by a vector.
Fig. 76.
The velocity has changed both in magnitude and in direction
and y is the total acceleration.
(21) A cyclist at noon is travelling N. at 12 miles per hour ; at 1 p.m. he
is travelling 10 miles an hour 75° E. of N. ; what is his total acceleration ?
* Average Acceleration. Dividing the total acceleration by
the time we get the average acceleration,
ACCELERATION. 83
' * Acceleration. Notice that both total and average accelera-
tions are vector quantities, and the latter gives the average
velocity added per unit of time.
If the acceleration is constant, i.e. if the same velocity be
added during equal intervals of time, no matter how small the
latter may be, then the velocity added per second is the
acceleration.
If the acceleration changes, then the value to v^hich the
average acceleration approximates, as the time interval becomes
smaller and smaller without limit, is the acceleration at the
instant (cf. Velocity, on p. 77).
* Definition. Acceleration is the rate of change of velocity,
and is measured by the velocity added per unit time, or the
velocity that would have been added if the acceleration had kept
constant.
Acceleration is therefore a vector quantity, and accelerations
are added (or compounded) as vectors.
It can be shewn by experiment that bodies falling freely under
the influence of gravity have a constant acceleration directed
towards the centre of the earth, and measured by a velocity of
32"2 ft. per second added per second, or 32'2 ft. per sec. per
sec. (at Greenwich). This acceleration is usually denoted by the
letter g.
(22) A train at noon is moving 35° E. of N. with a speed of 32 m./hr.
At 12 h. 35 m. p.m. it is moving with a speed of 27 m./hr. 50° E. of N.
What are its total and average accelerations during this time ?
(23) A point moves in a horizontal circle of radius 5 '3 ft. in the contra-
clockwise sense. When at the most northern point its speed is 11 '3 ft. /sec,
when at the W. point it is 12'7 ft./sec, and when at the S. and E. points
its speed is 14 '8 and 11 '3 ft./sec. If the time taken to move through each
quadrant be 1-2 minutes, find the average and total accelerations for
. 1, 2, 3 and 4 quadrants.
(24) Two trains are moving towards the same point in directions in-
clined at 70° with one another. One train is increasing speed at the rate
of 33 ft. per sec. per sec. ; the other is diminishing its speed at the rate
of 17 ft. per see. per sec. Find the acceleration of the first relative to the
second.
Components of a Vector. Finding the sum of a number of
vectors is a unique process ; i.e. one and only one resultant
84
GRAPHICS.
vector is obtained. The converse, i.e. finding the components
when the resultant is known, is not unique in general. The
components of a vector in two given directions are, however,
uniquely determined.
Example. Find the components of a in the two given directions.
From the ends of a (Fig. 76), draw lines parallel to the two
given directions, these determine two vectors Oj and a^ which are
Fia. 76.
the components. The construction can be done in two ways
as indicated, but the component vectors are the same in both
constructions.
(25) Draw any vector a and any three lines ; shew that any number of
components of a can be found in the three directions.
(26) a represents a velocity of 10 ft. per sec. due N. ; find the component
velocities N.W. and N.E.
(27) Find the components of a displacement 15 ft. E. in direction,
making angles 15° N. and 30° S. respectively with this line.
* (28) A falling stone has an acceleration of 32'2 ft. per sec. per sec.
vertically downwards ; find the components along and perpendicular to a
line making an angle of 60° with the horizontal.
* (29) A train has an acceleration of 5 ft. per sec. per sec. down an
incline of 1 in 6 (1 vertical 6 along the incline) ; find the component
accelerations horizontally and verticafly.
When only one component is spoken of, the other is supposed
to be at right angles to the first component.
VECTORS MULTIPLIED BY SCALARS.
85
(30) A ship journeys 50 miles 20° N. of W. ; what is its displacement
dueW.
*(31) A bead slides freely down a straight wire making 65° with the
horizontal ; what is the acceleration of the Dead down the wire ?
*(32) A cable car fails to grip • on a down incline of 10 in 73 ; if the
retardation due to friction be equivalent to a negative acceleration of
2'8 ft. per sec. per sec, what is the actual acceleration of the oar ?
*(33) If in question 32 a man jumps vip from his seat so that his body
has a vertical acceleration (relative to the car) of 1'9 feet per sec. per sec,
what is the real acceleration of his body ?
Multiplication of Vectors by Scalars. Multiplying a
vector by a number merely multiplies the length of that
vector, thus na means a vector n times as long as a.
Similarly multiplying a vector by any scalar quantity multiplies
its length by that quantity.
OJB (Fig. 77) is any triangle.
If A^B-^ be drawn parallel to the
base AB, cutting OA and OB pro-
duced in Aj^ and Bj, then we know
that OAB and OA^B^ are similar
triangles. Hence A-^B-^ is the same
multiple oi AB that OA-^ is of OA.
FlQ. 77.
If OA = a and OB = fi, be the two sides of the triangle,
then if OAi = wa and 0Bi = m/8
we have AB = j8-a and AiBj = m(/3-a),
so that from ^^^l = O^j - OA^,
we have the vector law,
m (/3 - a) = w;8 - ma, ( 1 )
(34) Establish the equation 7ia + np=n(a + p) by (i) adding to the sum
of n vectors a the svim of n vectors |8, and (ii) adding ;8 to o and then
adding n of these vectors together.
Centre of Mean Position. Given two points A and B,
the point M bisecting the line AB evidently occupies a mean
position with regard to A and B.
S6
GRAPHICS.
Choose any two origins and 0^ (Fig. 78), let OA = a and
OB = 13, and from 0^ draw a and add ;8 to it, then the sum
a + l3 = (r;
A
Pia. 78.
from set off Jo- and shew that the point M thus determined
is the mid-point of JB. (This is also obvious from OAB, for
AB=j8-a, and, therefore, the vector to the mid-point of ^-8 is
proving, incidentally, that the diagonals of parallelograms bisect
one another.)
Take any three points A, B, C (Fig. 79), and a point of
reference 0. Find the sum a + ^S+y (=o-) as indicated (away
from A, B, C). From set off -jO- and determine thus the
point M, the centre of mean position.
Instead of take other origins 0-^ and 0^ and shew by a
similar construction that the same point M is obtained. This
shews that M depends on A, B and C, and not on the origin
used to determine it. Draw the medians of ABC and see that
M is their point of concurrence.
(35) Draw a regular hexagon, take any origin (not the centre) and find
in a separate figure the sum of the six position vectors of the vertices. Set
off from the origin -j- of this sum, and thus find M, the centre of mean
position of the vertices.
Perform a siniilar construction when the origin is at the centre.
CENTRE OP MEAN POSITION.
87
(36) Draw a parallelogram and find the centre of mean position of the
vertices by taking the origin (i) outside, (ii) at the intersection of the
diagonals.
In each case the centre of mean position is the end point of the
vector, drawn from the origin, which is ^ of the sum of the position
vectors of the vertices.
(37) Take any five points and any origin, find -g of the sum of tlie
position vectors ; set this ofi' from the origin and determine thus the point
M. Choose another origin and show, by a similar construction, that the
same point M is determined.
Definition. Generally, if there are n points A,B,C, ... whose
position vectors with reference to some origin are a, (3, y, ... ,
a + /8 + 7+... .
then
is the position vector of a point M called
the centre of mean position.
88
GfeAPHICS.
Let a, ;S, 7, ... (Fig. 80) be the position vectors of the points
relative to 0, a^, /8j, yj, ... their position vectors relative to 0^,
and let p = 00i, then
OM = ° + ^ + y+- and 0,M' = ^^^±M2l±^.
But
Fio. 80.
.-. OiM' =
,_a + P + y+... -np
n
M' is at M.
om-p;
The centre of mean position is thus a point dependent only on the
relative position of the points themselves and not on the origin used
to determine it.
The points A, B, C, ... need not be in a plane.
(38) The coordinates of five points are (1, 1-2), (-1, 1-5), (2-1, IS),
(3'3, -1'4) and (-2, -3'4) ; find the centre of mean position. (Add the
vectors and divide by 5 ; set off this one-fifth vector from the origin, and
measure the coordinates of its end point M. )
CENTRE OF FIGURE. 89
Centre of Figure. In the case of a line, curve, area or
volume, the centre of mean position of all the points in it is
called the centre of figure or centroid. In many cases we can
determine the centre of figure by inspection.
A straight line. Choosing the origin at the point of bisection
we see that to every point P, having a position vector p, there is
a point P^, having - p for its vector, hence, '2p = 0, and is the
centre of mean position. Where is the centre of figure for
lines bounding a square, a rectangle, a parallelogram, a circle,
respectively, and why 1
Area of a Parallelogram. Taking the origin at the intersection
of the diagonals, to every point P or Q (Fig. 81) there is a corre-
sponding point Pj or Oi such that OP + OPj or OQ + OQj = 0, and,
therefore, the centre of mean position of figure is at this point
of intersection.
(39) Where is the centre of figure of the area of a rectangle, of a circle,
of a regular hexagon ?
(40) Shew that the centre of figure of a regular pentagonal area cannot
be proved to be at the centre of the circumscribing circle by this argument
alone.
(41) Mark the positions of any seven points on your drawing paper.
Find the centre of mean position of any four of them and then of the
remaining three, and mark these two points. Find the centre of mean
position of these two points, counting the first one four times and the last
three times. Is this final centre of mean position the same as would be
determined directly from the position vectors of the seven points ?
90 GRAPHICS.
Theorem on Centres of Mean Position. In finding the
centre of mean position of a system of points, any number oi
them may be replaced by their centre of mean position, if the
position vector of that point be multiplied by that number.
Let there be Wj points whose position vectors are a^, y8j, y^, ... .
ij j> ^2 " " " " '^2' Pi' y2' ••• •
» I) '^3 >> » V )) ""si Ps' "Ys! ••■ •
Then the position vector o- of the centre of mean position for
the whole 'm^ + 'm,2 + m^ points, is given by
(wil 4- wig + %) 0- = 2 (oj + a^ + ag)
= 2aj + Sag + 2og .
If o-j is the centre of mean position for the m-^ points, a-^ that
for the mg points and o-g that for the wig points, then
.'. (TOj + m2 + »lg)<7 = OTjCrj + TO2<''2 + ''*3"'B-
The nij points may therefore be treated as if concentrated at
their centre of mean position, provided the position vector of the
latter be counted m^ times, and so on for the other points.
Evidently the argument holds however many partial systems
of points we suppose the whole system divided up into.
Mass-Centres. Let there be n points having unit mass at
each point, then the centre of mean position of the points is
called the mass-centre of the masses. If aj , a, . . . are the position
vectors of the points, and o- that of the mass-centre, then
mo" = 2aj .
Divide the points up into groups, m-y points having mass-centre
o-j, wig points mass-centre o-^, ... then, since 7i = 2mj, we have
(miH-mg + mg-i- ...)cr = miO-j-t-TO20-2-i- (1)
This equation remains unaltered however the jWj points be
moved, provided that their mass-centre given by o-j remains
unaltered ; we may, therefore, suppose them to come together and
coincide at o-j, and so for the other partial systems.
MASS CENTRE.
91
At the end point of o-j we have now a mass m^ units, at o-j a
mass mg units.... Moreover, equation (1) remains true when
multiplied throughout by the same scalar quantity, and hence
»ij, m^, . . . may be taken as the masses at the points.
Hence, given a number of points M^, M^, ... having masses
%, m^, ... concentrated there, and having position vectors o-^,
o-^, ... the mass centre of the system is given by the position
vector o", where
Mass-Points or Particles. We
have thus arrived at the conception
of points with masses concentrated
at them ; such points are called
mass-points, and have exactly the
same meaning as the more usual
term particles.
To find the Mass-Centre (m.c.) of
a number of mass-points: Choose
any convepient origin 0, multiply
each position vector by the mass at
its end point, add all these mass-
vectors,' and divide the resultant
mass-vector by the sum of the
masses. Set off from the vector
so determined; its end point will
be the M.C. required.
Example. Find the Mass-centre
(m.c.) of three particles of masses 1, 2
and 3 grammes placed at the vertices
of an equilateral tricmgle.
Draw any equilateral triangle ABC (Fig. 82). Take any
origin 0. Draw a equal to OA, add 2/3 (where ^ = 0B), and
Sy (where 7 = 00). Then, if o- = a -t- 2/3 -)- 3y, set oif ^a- from 0,
and find M the M.C. required.
Fig. S2.
92
GRAPHICS,
(42) Find the M.c. of four particles of masses 1, 2, 3, 4 grammes, placed
at equal intervals round a circle of radius 3 inches.
(43) Find the m.c. of five particles of masses, 2, 3, 1-8, 3-3, and 4-7 lbs.
aoed in order at the vertices of a regular pentagon of 1 '5" side.
placed
(44) Take the m.c. as found in Ex. 43 as origin, repeat the construc-
tion, and show that the vector polygon is closed, i.e. that the origin is
the M.c.
Mass-Points in a Line. For two points, A and B, having
masses 3 and 2 lbs., we have, by takuig the origin in the line
(Fig. 83), 3Q^ ^ 2Qg ^ gQ]y[
If M be now taken as origin,
3MA + 2MB = 0,
Fia. 83.
and therefore M divides AB inversely as the masses. Hence
the simplest construction would be : Set off BA^ = 3", and AB^
parallel to BA^ and of length 2". Put a straight edge along
A^B.^, and mark the point M where it cuts AB.
(45) At A and B, distant apart 3", are masses given by lines of length
71 and 5 '6 cms. Find the M.c.
(46) If in Ex. 45 the masses at A and B are given by (i) the squares,
(ii) the cubes of the lengths, determine graphically in each case the position
of the M.c.
MASS POINTS IN A LINE. 93
(47) Draw any line and mark an origin and four points in the line.
Suppose masses of 2, 3, 1, and 4 grammes to be at the points. Construct
the position of the M.c, and measure its distance from O. Shew that
this position could have been calculated by multiplying each mass by its
distance from (positive if to the right, negative if to the left), adding
the products together, and dividing by 10, the sum of the masses.
Formula for the m.c. of Points in a Line. Let Xj, ajg, ^g, ...
be the distances of. a number of points in a line from an origin
in that line, and mj, mg, wig, ... the masses at those points.
Then, all the position vectors being parallel, they are added as
scalars, one sense giving a positive scalar, the opposite a negative
one. If, then, x denote the distance of the M.C. from 0, we have
xS.m-^ = mjXj + mjKg + . . . = "Sm^^
If this sum is negative, it shows that the M.c. lies on the
negative side of the origin.
(48) Find by calculation the position of the M.c. of masses 2'7, 3"6, 4'7
and 69 grms. situated at points in a line distant 11-7, 1'6, 1'2 and 9'3 cms.
from an origin O in the line.
(49) Calculate the position of the m.c. of masses 10, 5, 3, 8, and 1 lbs.
situated in a line, the position of the points from a fixed origin in the line
being 1, 5, -2, - 3, and 4 ft. respectively.
Graphical Construction.
Example. Masses given by lines m^, m^, m^ and m^ are concen-
trated at points Xj, Xj, Xg, X^ in a line, to construct the position of
the mass-centre.
Draw OTj, wij, ... and the distances Xj, X^... twice the size of
those in Fig. 84.
Through any point in the line draw Oy perpendicular to it,
set off from along this perpendicular OM^, OM^, OM^, 0M^
and OM equal to Mj, m^, m^, m^, and m-^ + m^ + m^ + m^ ( = 2wij).
OM should be found by the strip method of addition.
Through M^ draw M-^x^ parallel to MXy
„ M^ „ M^^ „ MX^.
„ M^ „ M^x^ „ MX,.
„ 'Mi „ M^x^ „ MX^.
94 GRAPHICS.
By the strip method find the sum of Ox-^ , Ox^ , Ox^ , Ox^ , and set off
OX equal to this sum, then X is the M.C. of the four mass-points. J ^
M
IVI3
■1^=*, ^3 X, \ X, X X3 X4
Fig. 84.
_ . OM OX.
^''''- mr-oi'
:. OM.Ox^ = OM^.OX^.
Similarly, OM . Ox^ = OM^ . OX^ ,
OM.Ox^=OM^.OX^,
OM.Ox^ = OM^.OX^;
:. by addition OM .Wx^^WM^.OX^.
The right-hand side of this equation is the sum of the products
of each mass and its distance from 0, the left-hand side is OX
multiplied by the sum of the masses, henpe OX is the distance
of the M,o. from 0,
MASS CENTRES OE FIGURES.
95
(50) Repeat the oonstruotion as above, drawing Oy through M-^ , M,^ and
Mi in turn. Notice that this simplifies the work, since only three
parallels have to be drawn.
(51) Masses given by lines of lengths 2'3, 5, 7'8, 8'5 eras., scale 1" to
5 lbs., are at points (in a straight line) whose distances apart are 0'5, 1, 17,
0-8 inches taken in order. Find the M.c. graphically, and test by calcu-
lation from the formula of p. 93.
Non-CoUinear Mass-Points. For points not in a line and
having masses given by lines, general constructions are given
(i) on page 112, (ii) on page 299.
A few simple cases may be treated by repeated constructions
similar to tbat on page 92.
(52) Find the m.c. of three masses given by the lines m^, m^, mj,
situated at A, B, G respectively, where AJ3=5% £0=1'5 and C.4 = 5"98
cms. First, find the M.o. G^ of m^ and Wig, and then the m.c. 0^ of m^ at
C, and mj + m^ at (?i .
(53) Masses given by lines of lengths 4'7, 2 '3 and 3 "8 ems. are situated at
points whose coordinates in inches are (1, 0), (2, 3), (3, - 1). Find the
mass-centre by construction, and give its coordinates.
Mass-Centre of a figure with an axis of Symmetry. If
any curved or broken zig-zag line has an axis of symmetry the mass
centre of the line must lie on that axis.
For to every point P distant MP from the axis of symmetry
there is a point P-^ at the same distance
on the other side of the axis (Fig. 85).
Choosing any origin 0, then, for two
such points,
OP + OP^ = OM+MP + 0M+ MP^
=:20M,
and as this holds for every pair of points,
the M.C. must lie on the axis.
When the axis of symmetry is per-
pendicular to the lines joining corre-
sponding points, it is called an axis of
right symmetry, otherwise it is an axis of skew symmetry. The
line joining the mid-points of opposite sides of a rectangle is an
axis of right symmetry, a similar line in a parallelogram is
fin axis of skew symmetry.
PiQ. 85.
96
GRAPHICS.
M.O. of any number of equal consecutive lines inscribed
in a circle. Draw a circle of radius 4" (Fig. 86), and set off in
it six chords, each of length 1-.5", forming an open polygon
A-iA^A^A^A^A^A^. Draw the axis of symmetry OA^, being
the centre of the circle, and join A^A^.
Bisect ^1^2 ** ^1 ^"•i ®^* off ^^ along the axis of symmetry
equal to OM^. Draw AB parallel to A-^A,j = \ the sum of the
sides. Join OB, and draw A^D parallel to OA, and DG parallel
A-^A^, as in Fig. 86, then the point G so determined is the
mass-centre of the six lines.
Proof. Draw through 0, OX parallel to A^A^, and through
A-^ and A^ lines parallel to OA and OX as in Fig. 86. Through
the mid-points M^, M^, M^ of the lines draw MjZ,, etc., parallel
to OA.
Then, by construction, OM-^X-^ is similar to A^A^C^ ;
. OM^^A^
■ ■ M^X^ A^G-, ''
M-^X^ = OM^.A^Ci.
.M,X„=OM,.AX\,
A-^A^.
In the same way A^A^
A^A^.
M^X^^OM^.AJO,.
MASS-CENTRE. OF A CIRCULAR ARC. 97
But A^A^ = A^Ag = A^J^ and OM^ = OM^=OM^;
:. by addition A^^A^iM^X^ + M^X^ + M^X^) = ^A^A^ . OM^ .
Now, for finding the mass-centre, we may suppose the mass of
each line concentrated at its mid-point, and since OA is the axis
of symmetry, the pairs of mass-points M-^, M^... have their mass-
centres on OA, and at distances M^X^, ... from OX.
But the distance y of the M.c. from is given by
A^A^{M,X^ + M,X, + M^,) = 3.^1^2 . y ;
:. 3A^A^.y = ^A,Aj.0M^; (i)
.-. AB.y=GD.OA,
y . GD
°^ 6a~ab'
and y = OG.
If there had been eight sides instead of six, equation (i) would
have been iA^A^ . y - iA^Ag . OM^,
if 2?i sides, nA^A^ . y = \A-^A^+i. OM-^,
or semi-perimeter . y = semi-closing chord . perpendicular
from centre on polygon,
or perimeter . y = closing chord . perpendicular, (ii)
(54) Find by this method the M.c. of any six sides of a regular heptagon.
Mass-Centre of a Circular Arc. The formula embodied
in equation (ii) is independent of the number of sides to the
polygon. When the number of sides becomes very large, the
sides themselves being very small, the polygon becomes nearly the
same as the circular arc, and the perpendicular Oilfj becomes nearly
the radius of the circle. The limiting case, when the number of
sides becomes infinitely large and their size infinitely small, is
the arc itself, and hence for any circular arc
perimeter . y = closing chord . radius.
Construction for the m.c. of a Circular Arc. Draw a
circular arc BCD (Fig. 87) of radius 4" and angle 135°, and its
axis of symmetry, OC. Construct the tangent at C, and step ofi'
CD = axQCD
T.O. G
98
GRAPHICS.
along it. Join OD, and draw DL and LG parallel and per-
pendicular to 00, then G is the mass-centre.
^ 7
Proof. This follows at once from the equation
y _ closing chord
perimeter
radius
From this result a formula for calculation can be deduced,
for if r = radius and 2a the angle BOD,
y_rsina
r ra '
. _ sin a
we have
which, in the case of a semicircular arc, becomes
_ 2r
IT
(55) Construct the M.c. of a semicircular arc, and compare the measured
- 2r
y with the calculated value —
w
(56) Construct the M.c. of a circular arc subtending 270° at the centre.
(57) Find the M.c. of the lines bounding a circular sector of angle 75°.
(58) Find the M.c. of a uniform U-rod formed of two parallel pieces, each
of length 6", connected by a semicircular piece of radius 3".
Mass-Centres of Areas. If mass be supposed distributed
uniformly over an area, the preceding processes enable us to find
the mass-centre (or centroid) in many cases.
The principles of general use are :
(i) If the area has an axis of symmetry, the M.O. must lie on
that axis.
MASS-CENTRES OP AREAS.
(ii) If the area can be divided into parts, for each of which
the M.c. can be seen by inspection or easily found, then the m.c.
of the whole is found by finding the M.c. of these points, each
point having a mass proportional to the corresponding area.
Example. To find the m.c. of a triangular lamina ABC.
Draw the median AM-^ (Fig. 88) ;
this bisects all lines parallel to BC,
i.e. the M.c.'s of all lines (or very
narrow strips) parallel to £C lie in
AMj^. Hence to a mass m at Pj
there is an equal mass m at P,
where PjOP is parallel to BC and
OPi + OP = 0. AMj^ is an axis of
skew symmetry; BC is called the
conjugate direction.
Draw a second median BM ; the point of intersection is
the M.C.
(59) Shew, from the property of the M.c, that the three medians meet
in a point.
(60) Shew that the triangle and three equal masses
placed at the vertices have the same axes of skew
symmetry. Hence prove that m.c. of the triangle
lies on the median M^A, J of the way up from base.
Example. Find graphically the
M.c. of the area in Fig. 89.
Draw the figure to scale.
Find the mass-centres ffj and G^
of the two rectangles as indicated.
Join G-fi2i ^'^^ divide G-fi^ iii *^^
ratio of 16 to 9.
100 GRAPHICS.
Quadrilateral.
Example. Find, graphically the m.c. of a quadrilateral.
Draw any quadrilateral J BCD (Fig. 90) and its diagonals AC
and BD intersecting in 0.
Cut off AE = CO. Find the M.C. of BED, it is the same point
as the M.C. of ABCD.
D
y^v — — c
Proof. Since AE = CO, the medians through D of CAD and
OED are coincident. But the M.C. of a triangle (Ex. 60) lies \ up
the median from the base, hence the M.c. of OED is that of CAD.
Similarly, the M.C. of DEB is that of CBA. Also, the masses
of AED and BEA, DOC and BOC, DEO and BEO are propor-
tional to their altitudes, which are proportional to DO and OB,
ABC BO BEG
"■ ACD~DO~DEff
hence the M.c. of ABCD is that of BED.
Example. Find the m.c. of the airea of the re-entrant quadrilateral
ABED (Fig. 90).
Set o^AE^^EO along EA produced, and find the M.C. of the
triangle DE^B ; it is the M.c. of the re-entrant quadrilateral.
Construct also the M.c. by finding separately the M.O.'s of
ADE and ABE.
* Example. Find the m.c. of a cross quadrilateral A BED.
Draw a cross quadrilateral from the re-entrant quadrilateral
(Fig. 90) by taking E on the other side of AD, and follow out
the construction just given ; the M.c. of the triangle DE-^B is the
M.c. of the cross quadrilateral.
MASS-CENTRE OF QUADRILATEtlAL.
101
Proof. The cross quadrilateral has an area ABE - ADE ;
the M.C. of ABE is that of E-^^BO and the M.C. of ADE is that
of E^DO. The areas added are in each case proportional to
the original areas and hence the M.C. of ABE - ADE is that of
E^BO - E^DO, i.e. is the m.c. of E^DB.
(61) Find the mass-centres of the areas of the following figures (Figs.
91-101).
The figures must be drawn full size according to the dimen-
sions given. The angle d in Fig. 101 is equal to. the same
lettered angle in Fig. 100.
+
It
?'
i
t — r-13"—*
■* - 0.82"*
Fig. 91.
4 3-6 V
t
1
1
i"
1
1
1
1
V
FlG. 92.
102
GRAPHICS.
. 1-B-
T] U-o-65^
-a-s"--
Fig. 95.
ni
<- 1.44-'-—*
« o-e/*
Fig. 96.
Fig. 98.
MASS-CENTRE EXERCISES.
103
/■5"\
3-37"
2-82"
Fro. 101.
104 GRAPHICS.
M.O. of Trapezoidal Area. Draw any trapezium ABGD
(Fig. 102) where £G and DA are the parallel sides.
Produce DA to A^ and BG to B-^ in opposite senses, so that
A A.^ = BC and GB^ = AD.
D M, A A,
Find the point of intersection G of the line A-^B-^ and the line
M^M^ joining the mid-points of the parallel sides AD and BG.
is the M.C. of the area.
Proof. The triangles DBG, ABD being of equal altitude have
areas proportional to their bases BG and AD. We may, there-
j>ri
fore, replace the two triangles by mass-points -— at D, B and C
AD
and —^ at A, D and B.
o
.'. in the line A-^D there is a mass given by — — .
o
RR AD + 2BC
But the M.c. must lie on M-^M^,
.'. G, the mass-centre, must divide M^M^ so that
GMi_ AD + 2BG _ BG+^AD M^A^
GM^ BG+2AD AD + ^BG~ M^B(
Notice that the distance of the mass-centre of a number of
points from a given line is unaltered by any movement of the
points parallel to that line.
(62) Find the m.c. of the trapezium for which BC=4-3, AD = Ql,
AB =5-S and Oi)=4'7 cms., (i) by this method and (ii) by the general
quadrilateral method.
(63) Divide the trapezium into a triangle and parallelogram, and find the
M.c. of the whole by finding in what ratio it divides the line joining the
M.o.'s of the triangle and parallelogram.
MASS-CENTRE 0? A CIRCULAR SECTOR.
lOS
M.C. of a Circular Sector.
Example. Construct the m.c. of a circular sector of radius 4
inches and angle 120°.
Draw the axis of symmetry 00 (Fig. 103).
Set off 0B■^ = §05 and draw the concentric arc ^iBy
Step off the arc O^B^ along the tangent Cj^g ; draw B^^B^
vertically to cut OB^, and B^G horizontally to cut 00 at G. G is
the M.c.
Proof. Suppose the sector divided into a great number of
very small sectors, of which OLM is a very enlarged copy, then
OLM is at its limit a triangle, LM being a tangent to the circle,
the M.C. of this triangle is at g, where Og — ^OL. The sector
OACB may therefore be replaced, as far as the M.c. is concerned,
by a circular arc of radius ^OA (p. 97).
The y for the circular sector is, therefore, given by
which, in the case of a semicircle, becomes
4r
106
GRAPHICS.
(64) Find the M.c. of a semicircle and compare your result with the
calculated position.
(65) Construct directly from the semicircle the position of the M.c. of
a quadrant of a circle, and from that the M.c.'s of i, J and J of a quadrant.
Negative Mass and Area.
Example. To find the m.c. of the area ABCDDiCjBjAi, given
that AB = 6", BC = 8", AA^ = 1", AjBj = 5", BjCj = 5".
Find the mass-centres G^ and ffg (^i^- ^^'^) °f *^® rectangles
ABCD and A-^B-JJ-^D-^. Join G1G2 and produce. Set oif along
parallel lines 6*2^2 proportional to ABCD, and ffjZ, proportional
to A^B-fi^D-^ both in the same sense. Where K-^K^ cuts G-fi^ is
the point G, the mass centre of the area.
Since the given area is the difference between two rectangles,
the smaller rectangle (a square) must be considered as a negative
area or as having negative mass, and hence G^K^, G^K^ instead
of being set off in opposite senses must have the same sense.
NEGATIVE MASS.
107
Proof. In order to justify the construction on p. 106 refer to
Fig. 105. Let M denote the mass of the whole area enclosed by
the outer boundary, and m the mass of the shaded area enclosed
by the inner curve.
Let Gj and G^ be the mass-
centres of the area between the
two curves (M-m) and of the
shaded area. Then, to find G the
M.C. of the whole area M, we have
to divide Gj^G^ at G so that
G^G_ M-m
G^G~ m ■
The Graphical construction for
effecting this division is indicated
in Fig. 105.
Fio. 106.
On the other hand, if G and G^ be known, G-^ can be deter-
mined.
Add unity to each side of the last equation j then since
we get
M-m _M
m m
G^G + GGi _ G^Gi _M
GG^ ~ GG^ ~ m
This shews that Gjjr has to be divided externally at G-^ in the
ratio — .
m
Hence, from G and G^, set off parallel lines, in the same sense,
representing the masses to scale. Join their end points and
produce the line to cut G^ in ffj, the M.O. required. This
was the construction made in the example.
Mass-Centre of a Segment of a Circle. Draw a circular
sector OACB (Fig. 106) of radius 4-5" and angle 150°. Construct
108
fiRAtHlCS.
the M.C. G, of this sector as before. Divide OM (Fig. 106)
into three equal parts and take OQ^ = §OJf. Through 0^ and ftj
draw parallels and set off GJi-^ =p, the perpendicular from M on
c
y^^^^
\
/
G ^\.
1 ,-"""^^
G, \y
K — ^~~^
^^ '
i — - A
o
Fio. 106.
OA and G^^ = \!iiQ arc AC* (In Fig. 106 f of these distances
are set off.) Join K^-^^ cutting 00 at 0, the M.C. of the segment.
Proof. The problem is to find the M.C. of the sector area and
the negative area of the triangle OAB.
The area of the sector = \ radius x arc (p. 55)
= 0Ax arc AC.
The area of the triangle = Oilf. MA
OA 0M\
^. ( . OA 0M\
= 0^.^; (since -^=—j.
area of sector arc AC
area of triangle p
Since the area of the triangle must be considered as negative,
p and the arc AC must be set off in the same sense.
(66) Construct the M.c. of a circular area of radius 3 inches having a
circular hole of radius 1" cut out, the distance apart of the centres being
1'5". (Areas are proportional to radii squared.)
* These distances must be set off very carefully, especially when the angle of
the sector is very small or nearly 180°.
MASS-CENTRE OF A SEGMENT OP A CIRCLE.
109
(67) Construct the M.c. of the area as above, the radii being if and r,
and the distance apart of the centres being c. Construct the ratio of the
squares as in Fig. 107.
-R
-r
■ c
Fig. 107.
(68) Construct the M.c. of the figure shewn (Fig. 108).
Fig. 108.
(69) Find the m.c. of a rectangle of sides 5 '64" and 7 '85" with a square of
side 2-83" cut out, the distance apart of their centres being 1 -5", and the
line of centres being a diagonal of the rectangle. (Use a similar con-
struction to that in Ex. 67 for getting lines proportional to areas. )
(70) Find the M.C. of a rectangle (5"64" x 4'85"), having a circular hole of
radius 1 '4" out out, the distance apart of the centres being 1 '5!'.
(71) Draw a rectangle of sides 4 and 3 inches. On the .S" side as diameter
describe a semicircle inside the rectangle; now suppose it cut away.
Find the M.c. of the remaining area.
110
GRAPHICS.
Vectors in a Plane. Parallel vectors are said to be like
vectors, they can all be expressed as multiples of any other
parallel vector.
Two non-parallel vectors are independent, i.e. one cannot be
expressed in terms of the others.
Draw any two non-parallel vectors a and /3 ; see tljat the sum or
diiferenoe of any multiples of these is the third side: of a triangle,
which can only be zero when a and /S have the same direction, or
the two multiples are zero. If, therefore, we have an equation
aa = 6/3, where a and ^ are independent, it can only be satisfied
when a = b = 0.
Similarly, if 1a + 3/3 = aa + h^,
then {7-a)a = {b-3)l3,
and a must be 7 and b must be 3 if the vectors are not parallel.
Scalar Equations for Mass-Centre. A number of masses
nij , nig , wig , ... are at points whose position vectors are pitP^tP^, ■■■■
Take any two axes through (Fig. 109), and let the
components of Pj parallel to the axes be a^ and yS^,
I) ^2 „ » Oj and ^21
SCALAR EQUATIONS FOR MASS-CENTRES. Ill
Then Pi = ai + ^i,
P2 = «2 + /^2'
Let p be the position vector of the mass-centre and a and j8
the components of p.
Then, for the mass-centre,
.'. (a + P) Swij = (jWjaj + mi/?! + ™2"2 + ™2/^2 + • • • )
by rearrangement of the terms.
But a, Oj, ftg, ... are like vectors, and so are P, p^, ^2) •• > ^^d
Oj and ySj are independent.
Hence a2?TCi = 2wiiai and ySSwij = Sm^jSj.
If, then, jCj is the length of a^, y^ of /Sj, ^ of a, etc.,
iS?ni = Smja;i and ySmi = 2miyi,
two scalar equations, each of which determines a line on which
the mass-centre must lie, the point of intersection of the two
lines being G the mass-centre.
Generally, it is convenient to take the axes perpendicular, and
in this case mj/^ is the mass at a point multiplied by the per-
pendicular distance of the point from the axis of x, and is called
the mass moment about Ox.
'Sm-^y.^ is then the sum of the mass moments about the axis of x.
The whole mass 2»ij, supposed concentrated at the mass
centre, is called the resultant mass.
We have then the theorem :
The sum of the mass moments about any line is equal to the
moment of the resultant mass.
Graphical Construction for m.c. The construction given
on p. 94 for the position of the M.c. of points in a line can be
applied to points in a plane, the construction being made for two
intersecting lines. A better method, however, is the link polygon
construction, given on p. 299.
112
GRAPHICS.
Example. Find the m.o. of masses 2, 3, 5, and 1 lis. at points
whose coordinates are (1, 1), (2, 3), (4, 1) and (3, 2).
Draw the axes of coordinates (Fig. 110), and mark the co-
ordinates of the points on each axis; draw a line bisecting the
angle between them, and set off along this line the masses to any
convenient scale
Y.
1
, .
! ;
- ■
:
:: :
Y*
I \
:::;.
m
;: :::
Y,
y^
y
3 '
m
1
!.: :: ^
::: :l |
1
n
■ -
i: ::
IHIUfHIf-
:: :::
O «, ^4 X^.
X,
Fig. 110.
In Fig. 110, 0X.^=1, OY^ = l and 0M.^ = 2. (on mass scale);
OZj = 2, Or^ = 3 and OM^ = 3 . . . . (OZ3 = 4 is not shewn.) The
suffixes indicate the order of the points in the example.
Join M (Oilf =11, the sum of the masses) to 1, 2, 4, 3 on the
X axis, and mark the points where parallels to these lines from
2, 3, 5, 1 on the mass axis respectively cut the z axis, a;,, asj, x^, x^.
Then
"^=U ., n^=..,.
MASS POINTS IN A PLANE. 113
Similarly ^ = ^ or 11*2 = 3. 2,
lla!3 = 5.4,
lla;^=1.2.
Hence ll{Xj^ + x^ + x^ + x^) = 2 .1 + 3. 2 + 5. 4 + 1 . 2 = 2???,ia;j.
Add by a straight edged strip x^, x^, x^ and x^, the sum is x.
Make a similar construction on the y axis and obtain y.
Mark the point whose coordinates are x and y ; it is the mass
centre, G.
(72) Masses are given by lines of length 5'1, 2'3, 1'5 and 2'15 cms. ; the
coordinates in inches of the points are (0, Tl), (2-3, 0), (3-2, 1-7), (2-2, 4-3).
Find the mass-centre by construction, and test by calculation from the
. , _ SmiKi _ Sm,«,
tormula a;=-=-^-^i V= ^ •
Graphical Construction for the iii.c. of any area,
irregular or otherwise. Transfer the figure given in Fig. Ill
to your drawing paper, and draw its axis of symmetry XY.
Divide it up into strips parallel to the base, and draw the first
equivalent figure as in the construction on p. 62, only take the
point in the base and a equal to the height of the figure.
Divide the height at G so that
XG_A^
XY~ A '
where A and A^ are lines proportional to the areas of the given
and the equivalent figure. G is the M.C.
* Proof. This is very similar to the proof on p. 63.
In Fig. Ill, ^5 is one of the very thin strips, parallel to the
base, into which the area is supposed to be divided. The mass
of each of these strips may be supposed concentrated at the
mid-point, i.e. in the axis of symmetry, XY.
If m is the mass of any one of these strips and «/ is the distance
from the base, then, if y is the distance of the m.c. from the base
y2m = 'Zmy.
But m is proportional to the area of the strip, hence if AB = x
and h is the thickness of the strip, m is proportional to hx, and
therefore y 2A.r = '2,hxy.
T.e, K
114
GRAPHICS.
Now (see Fig. 57)
and
- = ^, hence xy = ax,
a y
'. y . area of given Fig. = aShx
= a . area of first equivalent figure,
and
y=
11 J-
X
Fig. 111.
If the area has not an axis of symmetry, y only determines the
distance of the M.c. from XX. The process must therefore be
repeated for another line which intersects XX (preferably at
right angles to it), and the distance of the M.C. from this line
must be determined. From these two distances the M.C. can
be determined as the intersection of two lines.
Another method is given on p. 299 in connection with the
Link Polygon.
MISCELLANEOUS EXAMPLES. 115
MISCELLANEOUS EXAMPLES. III.
1. Draw a square A BOD of side 3 inches to represent a square field of
side 300 yards. A man starting at A walks round at 80 yards a minute,
another man starting at D at the same instant, and walking at 100 yards
a minute, begins to overtake him. Construct the relative displacements
at the end of the lat, 2nd, 3rd, 4th, 6th, 9th, 12th and 15th minutes.
2. Construct the minimum relative displacements when the men are on
adjacent sides.
3. A toy gun is pointed at an elevation of 45° ; on firing it begins to
recoil with a speed of 5 ft. per sec. (horizontally), the speed of the shot
relative to the gun is 30 ft. per sec, construct the true velocity of the
shot.
4. A sailing boat is going N.W. at 8 miles an hour, a, sailor moves
across the deck from the S.W. at 2'7 miles an hour, a current is flowing at
3-6 miles an hour 15° S. of E. ; what is the velocity of the sailor relative
to the current ?
5. Find the centre of mean position of five points whose coordinates
in cms. are (2-1, 3-3), (4-7, I'S), (2-6, 1-75), (1-95, 4-6), (0-75, 6-25).
6. Find the mass-centre of the above points supposing masses of 3, 4'1,
2'8, 7'3 and 4'6 grammes to be concentrated at the points, first by the
vector polygon method, secondly by the graphical construction of p. 113,
and finally by calculation.
7. Find the M.c. of the part of a circular area between two parallel
lines at distances 3'74 and 2'66 inches from the centre, the radius of the
circle being 4"5 inches. First treat it as the difference between two
segments and then by the strip division method.
8. Find the M.c. of a circular arc and its chord, the arc subtending an
angle of 135° at the centre of a circle of radius 3'7"-
9. Masses of 3, 8, 7, 6, 2, 4 grammes are placed at the vertices .4, B, ...
of a regular hexagon ; construct the position of the M.c.
10. Draw a circular arc of radius 3" and one of radius 2", the distances
apart of the centres being 2". Find the M.c. of the lens shaped area between
the two arcs by the method of strip division.
11. A horizontal wooden cylinder rests on the top of a rectangular block
of wood, the radius of the cylinder = width of block =2 '52 ft., the height
of the block is 7 "86 ft., and the length of the cylinder and depth of block
are the same. Find the m.c. (Treat as a circle on a rectangle. )
12. A steamer which is steaming in still water due S.E. at a speed of
14 knots enters a current flowing due W. at a speed of 2 knots. Deter-
mine in any way you please the actual velocity of the steamer when in the
current and the direction in which she will travel.
If it is desired to maintain a due S.E. course and to cover exactly as
much distance per hour in this direction as when in the still water, what
course would the steamer require to steer, and what must be the speed of
the ship in regard to still water? (Military Entrance, 1906.)
116
GRAPHICS.
13. A uniform iron girder has a cross-
section of the given form (Fig, 112).
Determine the position of the centre of
gravity of the section.
(Naval Cadets, 1904.)
;s
■ % wide
4--
Fio. 112.
14. Fig. 113 represents a figure formed of a rectangle and an isosceles
triangle. Find and mark the position of its centre of gravity.
(Naval and Engineer Cadets March, 1904.)
Fig. 113.
Fio. 114.
15. The letter T in the diagram (Fig. 114) is made of wire of uniform
thickness. Find its centre of gravity (M.c), stating your method.
(Naval Cadets, 1903.)
16. Fig. 115 represents a hexagon
frame, the length of each side being
2 inches ; equal masses of 4 pounds each
are placed at the four corners a, 6, d, e,
and a mass of 8 pounds is placed at the
comer c. The mass of each of the six
sides of the hexagon frame is IJ pounds.
Find the common centre of gravity (m.c.)
of the whole system.
(Military Entrance, 1905.)
Fja. 115,
MISCELLANEOUS EXAMPLES. 117
17. A cistern, without lid, -whose thickness may be neglected, measures
3 ft. 6 in. in height by 2 ft. 3 in. by 3 ft. 3 in. Find the position of its
centre of gravity.
18. A and B are two points 20 miles apart. At noon one man starts
from A to walk to B at the rate of 4 miles an hour, and at 2 p. m. another
man starts after him on a bicycle at 10 miles an hour. Draw a diagram
on your ruled paper to show how far they are apart at any given time,
and at what times they pass any given point between A and B. [Scale to
be 5 mile = l inch, and 1 hour = l inch.]
Also, find from your diagram or otherwise when and where the cyclist
overtakes the man walking. (Engineer Students, Navy, 1903.)
19. Two small spheres, of weights 4 ounces and 7 ounces, are placed so
that their centres are 5 inches apart. How far is their centre of gravity
from the centre of each? (Engineer Students, 1903.)
20. Given the centre of gravity of a body, and that of one of its parts,
explain how to find the centre of gravity of the remaining part.
ABDO is a rectangular lamina of uniform density; E is the middle
point of AB ; join DE ; find the perpendicular distances of the centre of
gravity of BGDE from the sides BC and CD. (B. of E. , II. )
21. A river which is 2 miles wide is flowing between parallel straight
banks at the rate of 4 miles an hour. A steamer starts from a point A on
one bank and 'steers a straight course at 7 miles an hour. Show on a
graph the distance above orT)elow A of her point of arrival at the other
bank, as a function of the inclination of her course to the direction of the
river. (Naval Cadets, 1905.)
22. Explain the method of determining the motion of one body relative
to another. To a passenger on a steamer going N. at twelve miles per hr. ,
the clouds appear to travel from the E., at 8 miles per hr. ; find their true
velocity.
Two steamers are at a given instant 10 miles apart in an B. and W.
line ; they are going towards each other, one N. E. at 20 miles per hr. , and
the other N.W. at 16 miles per hr. Eind how near they approach.
(Inter. Sci., 1901.)
23. Find the centre of gravity of a triangular frame formed of three
uniform bars of equal weight. Where must a mass equal to that of a
uniform triangular plate be fixed on the plate so that the mass oentro
of the whole may be at the middle of the line joining a vertex to the point
of bisection of the opposite side. (Inter. Sci., 1901.)
24. Explain the phrase "velocity of one body relatively to another
body."
25. Two level roads are inclined at an angle of 60°. Two motors, each
half-a-mile from the junction are being driven towards it at speeds 10
and 1 1 miles an hour. Find the velocity of the first motor relative to the
second, and the distance the motors are apart after 2 minutes 50 seconds.
(B. of E., IL, 1904.)
118 GRAPHICS.
26. A ship A that steams 23 knots sights another ship B to the N. at a
distance of 1 '4 sea miles, and steaming E. at 19 knots. In what direction
must A steam in order that her motion relative to B may be directly
towards 5? and in what time does she reach B1 A knot is a speed of 1 sea
inile per hour.
Shew that if A does not know S's speed she can deduce it from her own
by steering in such a direction as to keep due S. of B. If this direction
is 78° E. of N. , and the other data are as already given, what is ^'s speed ?
(Military Entrance, 1906.)
27. Eind the position of the centre of gravity of a circular sector. Find
the distance of the centre of gravity of a circular segment from its chord.
(B. of E;, II., 1906.)
28. Ox and Oy are two lines at right angles ; P and Q are points
moving from O to a; and from O to y, with speeds 7 and 12 respectively.
At the same instant OP =8 and 0Q = 5. Find the velocity (speed) with
which they are separating from each other, and explain whether or not the
velocity of separation is their relative velocity. (B. of E., II., 1903.)
29. Give an instance of a moving body that is at rest relatively to
another moving body. State how the relative velocity of one point with
respect to another point can be found.
Two points A and B are moving with equal speeds and opposite senses
round a given circle ; at the instant that the arc between them is a
quadrant, find the relative velocity of A with respect to B.
(B. of E., n., 1905.)
CHAPTER IV.
CONCURRENT FORCES.
EXFEBIMENTS.
ExPT. I. Lay an envelope or sheet of paper on a fairly smooth table.
Push it by a pencil parallel to the shorter edge (i) near one corner,
(ii) near the middle. Notice that the motion is quite different in the two
cases, even though the push is otherwise the same. Does the effect of a
force on a body depend on its line of action (axis) ? Think of other simple
experiments illustrating this point.
For the remaining experiments the following apparatus is
necessary : A vertically fixed drawing board and paper ; light
freely-running pulleys which can be clamped round the board in
any desired positions ; a set of weights from 5 to 1000 grammes ;
one or two scale pans of known weight ; some light, tough, stiff
cardboard ; strong, black, fine thread ; some small polished steel
rings, about the size of a threepenny piece ; some thin, strong
wire (for making little hooks) ; and a spring balance.
ExPT. II. Fasten, by means of loops, two threads to one of the steel
rings, and attach 100 gramme-weights to the other ends. Put the threads
over two pulleys as indicated in Fig. 116, and let the whole come to rest.
Take another piece of thread in the hands, and, stretching tightly, see if
the two threads are in a straight Une.
The function of the pulleys is only to change the directions of
the forces due to the weights; any effect due to their having
friction may be minimised by good lubrication.
What are the forces acting on the ring, neglecting its weight, in magni-
tude, direction and sense ? Draw lines representing these in magnitude,
direction and sense ; these are the vectors of the forces. Add these
120
GRAPHICS.
vectors. What is their sum ? Why may the weight of the ring be left
out of consideration ? See if equilibrium is possible with different weights
Wi and W^.
Fig. 116.
What is the pull at B on the part BA ? What is the pull at B on the
part between B and the ring ? In what respect do these pulls differ ?
BxPT. III. Replace the ring by a- piece of cardboard and attach the
threads by wire hooks passing through two holes punched in the card.
See that the sum of the veotprs of the forces is again zero. Mark on the
card, points A, B, C, in a, line with the threads. Punch holes at these
points, and insert the lower hook in turn through each of these holes.
Is equilibrium still maintained ? Does it matter at what point in its axis
a force may be supposed applied to a rigid body ?
(1) Pour hooks A, B, C, D are connected together by three strings AB,
BG, CD, Fig. 117. Weights Wi = V)a grammes, 1^2=50 grammes, #3=150
grammes are attached by long strings to B, C and D, and the whole is
EXPERIMENTS.
121
suspended from the hook A fixed to a beam or wall. If iWi , M.^ and Jfg
are points in AB, BO and CD, what is the pull at M^ on O, M^ on B, andi
at M^Qn A"!
Verify at M^ by inserting a spring balance there.
ExPT. IV. Attach three threads by loops to one of the rings and sus-
pend weights 200, 150, and 100 grammes as indicated (Fig. 118). Let the
ring take up its position of equilibrium. Mark two points on the drawing
f Sii
•■B
■ C
■■>D
iWa
Fig. 117.
150
Pig. lis.
paper under each thread. A set square does very well for this purpose, if
placed approximately perpendicular to the plane of the board, but a small
cube or right prism is better. Indicate the sense of each pull on the ring.
Remove the drawing paper and draw to scale the vectors, o, /3 and y,
of the forces acting on the ring, and find their sum.
Are the lines of action of the forces concurrent ?
Eepeat the experiment and drawing for other weights.
Is equilibrium possible with 50, 60 and 150 grammes ?
122
GRAPHICS.
ExPT. V. Use three pulleys and four threads with attached weights
of 80, 120, 200 and 120 grammes. Mark the axes and the senses of the
pulls on the ring as before, and find the sum of the vectors.
Perform a similar experiment and construction using five
different weights. Notice in each ease whether the forces are
concurrent or not.
ExPT. VI. Draw a triangle ABO (Fig. 119) on cardboard having sides
3-4, 3-6 and 5-2 inches, and give the boundary a clockwise sense. Draw
concurrent lines parallel to these sides, indicating the senses as in Kg.
119. Punch holes on these three Unes and cut away the card as indicated
Fig. 119.
by dotted lines. Fix the card, with one axis vertically downwards, on the
drawing board by pins. Adjust the threads, hooked through the holes,
so as to lie over the lines and attach weights proportional to the corre-
sponding sides of the vector triangle. Remove the pins and notice if the
card remains in position.
Perform a similar experiment starting (i) with a quadrilateral
of sides 5, 3, 2 and 6 inches, the senses being the same way round,
(ii) with a pentagon ; the axes must be concurrent in both cases.
ExPT. VII. Attach three weights to a card (one thread hanging verti-
cally and two passing over pulleys). The card will take up some position
of equilibrium. See if the axes of the forces are concurrent. Attach four
EXPERIMENTS.
m
weights to the card, and after the card has taken up its position of
equilibrium see if the axes are concurrent.
Perform a similar experiment with five weights.
Example. A student repeating Expt. IV. has a vertical pvll of
40 grms. weight on the ring, the two parts of the left-hand thread
supporting 25 grms. weight make an angle of 45° mth one another.
What was the third weight used, and what was the angle between the
two parts of its thread ?
Set off OA = i" (Fig. 120) vertically downwards, 05 = 2-5" at
an angle of 45° with OA. Join AB. Then OAB is thfe vector
triangle of the forces, and AB = 2-85" gives the third pull on the
ring of magnitude 28'5
grms. weight (nearly). The
direction is given by the
angle OAB and the sense
is from A to B.
Measure OAB (a) with
a protractor, (&) by a scale
of chords, (c) by means of
the tangent of the angle
{i.e. measure p on the 2-inch
scale, p being the perpen-
dicular to AB drawn at 2"
from A, and find the angle
from the table of tangents).
See that these three results
are approximately 38'7°
each.
Scale of Forces
lo
grammes wt.
Fio. i20.
124
GUlAtHlCS.
Example. Another student used weights 20, 30 wnd 32 grm.,
the last giving the vertical pvll. What were the angles between th
threads attached to the ring ?
Set ofF AB (Fig. 121) vertically downwards =3-2"; draw
circles with A and B as centres and of radii 2 and 3 inches to
intersect in C. Draw concurrent lines P-fi, P^O^ WO parallel to
the sides of this triangle.
A
J_
lbs. wt.
Force Scale
Fig. 121.
Evidently the angles between the threads are the supplements
of the angles of the triangle ; measure the angles of ABC by the
scale of chords. Approximately they are 37 -S", 65-8° and 76'7°.
Notice that since the construction may be done in two ways,
viz. the 2" circle from either A or B, it is impossible to say
which weight was put on the left-hand pulley.
In all solutions of statical problems by graphical methods it is
necessary to complete the solution either by drawing out the
force scale or by giving it in cms. or ins.
THREE CONCURRENT FORCES.
125
Example. A weight of 20 '6 tons is suspended by ropes of length
7 and 8 ft. from two hooks in a horizontal line distant apart 5 ft.
Find the pulls of the ropes on the weight {the tensions in the ropes).
First draw a figure ABG (Fig. 122) representing the position
of the ropes to scale (1" to 1').
A B
O
Fio. 122.
Then set off OP downwards to represent 20-6 tons weight
(1 cm. to 1 ton). From the ends of OP draw OQ and PQ,
parallel to AC and BC, and njeasure PQ and QO on the- ton scale,
126
GRAPHICS.
Example. From a telegraph pole radiate five lines {in the same
horizontal plane). The pulls of fowr of them on the pole are known,
find the pull of the fifth; given a pull due E. of 30 lbs. weight, one
due S. of 40, one S. W. of 2.5, one N. W. of U -i.
Force Scale
lbs. wt.
Pia. 123.
Set off OP (Fig. 123) horizontally to the left = 3", PQ vertically
downwards = 4", QR = 2-5" so that PQB= 135°, QB perpendicular
to .B5=4-14", then SO gives (3-2" in length) the fifth pull in
magnitude, direction and sense.
(2) In a tug-of-war A, B, G, D are opposed to ^i, B-^, C^, D,, D and
Z)i being the end men. The pulls oi A, B, G, D are given by lines of lengths
2'8, 3 "4, 3 '5 and 3 '9 ems. respectively, those ot A-^, B-^ and G^ by 2 '45, 3 '25
and 3"75 ; scale 1" to 100 lbs. wt. What is the least pull that Z), must
exert in order that his side may not be beaten ? If D^ exerts this force,
give the tensions of the rope at points intermediate between the men.
(3) Three strings are fastened to a ring as in Expt. IV. , two pass over
smooth pulleys and bear weights P and Q, the third string hangs vertically
and supports a weight R. If 6 and (p denote the angles between the
vertical and the threads attached to P and Q, find
(i) e and 4> when P=Q = 5 lbs. wt. and iJ=8 Its. wt. ;
(ii) B and <j> when P^Q=5 and 5=30°;
(iii) and f when P- 4, Q= 5 andi?=7;
FOUR CONCURRENT FORCES. 127
(iv) Q and when P= 7, R= 9 and 9 = 60°;
(v) P and <t> when Q = 12, R=\\ and 9 = 55°;
(vi) P and Q when i?= 7, 9 = 35° and = 50° ;
(vii) e and Q when 7?=13, P= 8 and (^=35°;
(viii) 9 and P when R=\\, Q= 6 and ^ = 30°;
(ix) P and Q when i?=14, 9=40° and Q is perp. to P ;
(x) R and Q when 9 = 35, = 70° and P=5-5. (Notice that
the angle between P and Q in the vector triangle is 75°. )
(4) Three threads fastened to a ring bear weights of 35, 27 and 25 grammes
as in Expt. IV. ; draw the vector triangle of the forces, and by measure-
ment determine the angles between the threads.
(5) Three threads are fastened to a ring, as in Expt. IV., the vertical
load is 135 grammes weight, the acute angles the sloping strings make with
the vertical are 25° and 50° ; find the other two weights.
(6) A load of 27 lbs. is supported by two strings attached to hooks ;
if the strings make angles of
(i) 27° and 48° ; (ii) 40° and 70° ; (iii) 50° and 50° ;
with the vertical, determine the pulls on the hooks.
(7) To two hooks A and B are fastened ropes which support a load of
3 cwte. The distance apart of A and 5 is 7 ft. and ^ is 3 ft. higher than
B ; if the lengths of the ropes attached to A and B are 5 ft. and 4 ft. , what
are the tensions in the ropes, i.e. what are the pulls of the ropes on the
load and on the hooks ?
(8) A weight of 12,000 grammes is supported by strings from two hooks
A and B in the same horizontal line. The distance apart of A and B is
15 ft. and the string attached to B is 12"2 ft. Find the pulls on the hooks
when the length of the string attached to A is
(i) 15 ; (ii) 14-4 ; (iii) 10 ; (iv) 5 ; (v) 4 ; (vi) 3 ft.
(9) Draw a graph shewing the relation between the pull on the hook in
Ex. 8, and the length of the string attached to it, as the string varies
in length from 3 to 15 ft.
Take two axes on squared paper and an origin. From the diagrams of
position and of vectors, the pull corresponding to the string length 3, 4, 5,
10 and 14 '4 ft. can be found. Plot points having as abscissae the lengths of
the string and as ordinates the corresponding pulls. Join the points by a
smooth curve. From the graph read off the pulls corresponding to string
lengths of 12 ft. and 7 ft. , and the lengths corresponding to pulls of 5000
and 17,000 grammes weight.
If the string could stand a pull of 12000 grammes only, what would be
the least length of string that could be used ?
(10) Draw a graph shewing the relation between the length of the
variable string and the pull on the other hook.
(11) Draw a graph shewing the relation between the two pulls on the hooks,
128
GRAPHICS.
(12) In repeating Expt.
V. a student used five
weights. The directions
and magnitudes of four of
the pulls being as given in
Fig. 124, what was the
magnitude of the fifth
weight, and in what direc-
tion and sense was the pull
exerted by it ?
(13) Four concurrent forces are in
equilibrium and act in the lines indi-
cated (Fig. 125). If P= 18 and Q =25
lbs. weight, find E and /S in magnitude
and sense. (Find the vector giving the
sum of the two known vectors, from its
end points draw parallels to E and 8.
This can be done in two ways ; but the
vectors parallel to E and 8 are the same
in each case. Since the forces are con-
current and the vector polygon is closed,
these vectors must give the forces in
magnitude, direction and sense.)
(14) A wheel has six central equi-spaced spokes, in four consecutive spokes
the pushes on the axis are 0-32, 0-72, 1-15 and 0-84 lbs. wt. ; what are the
actions of the remaining two spokes on the axis ?
(15) A weight of 10 lbs. hangs vertically by a string from a hook. The
weight is pulled horizontally so that the string makes an angle of 37° with
the vertical. What is the magnitude of the horizontal pull and what is
the pull of the string on the weight ?
(16) Draw a graph shewing the relation between the pull 8 on the hook
and the horizontal pull H in Ex. (15) as H increases gradually from
to 10 W.
For any given value of H the vectors form a triangle OAB (say), OA
representing the .weight W, AB the pull P and BO the pull of the string
on W. At B draw BP perpendicular to ^j8 and of length BO. Go
through this construction when AB represents 1, 2, ... 10 lbs. weight, and
join the points P so obtained by a smooth curve. This curve is the one
required, for A being the origin of coordinates and AB and AO the
axes, the coordinates of P are the values Qi'H and S necessary to give
e<juilibriura,
EXERCISES.
129
(17) In Fig. 126 A£ is a light rod with
a weight of 11 lbs. at B ; the rod can turn
freely round A. £ is pushed perpendicularly
to AB with a force of 4 lbs. weight. Find
the position of AB and the pull on A.
[Since the angle at £ is a right angle,
describe a semicircle on the vector repre-
senting 11 lbs. weight, and set off in this
a line representing a force of 4 lbs. weight ;
the closing line of the triangle gives the
direction ot AB and the force it exerts
on .8].
Vertical
Pig. 126.
(18) Draw a graph shewing the relation between the push at B (P) and
the pull SonAia Ex. (17) as P increases from zero to 11 lbs. weight.
(19) In Fig. 127 ^ is a
fixed hook and C a smooth
pulley, B a smooth ring to
which the threads AB, BG
and BW are attached. If
SOD =30°, ABC =85°, find
the pull on A and the weight
The angles remaining constant, draw a
between Q and W.
Q=27lbs.
W
Pio. 127.
shewing the relation
(20) Two cords are fastened to a ring at G, and, hanging over pulleys at
A and B, bear weights of 12 and 17 lbs. Find the force in magnitude,
direction and sense, with which C must be pulled in order that, with AC
and BG making angles of 60° and 80° with the vertical, there may be
equilibrium.
(21) A load of 5 ewts. is suspended from a crane by a chain of length
20 f t. ; a doorway is opposite the load and 5 ft. distant ; with what force
must the load be pulled horizontally to cause it just to enter the doorway ?
(22) Strings of length 5 and 3 '2 ft. respectively are fastened to a floor
at points distant 4 '3 ft. apart ; the other ends are attached to a smooth
ring which is pulled by means of a string making 30° with the vertical
with a force of 50 lbs. weight. The three strings being in one plane, find
the other pulls on the ring.
130
GRAPHICS.
(23) OA and OB (Pig. 128) are the axes of two forces, a is the vector of
the force in OA, c is the magnitude of a third force which, acting through
O, is in equilibrium with the other two forces. Find the vectors of the
Pig. 128.
other forces. How many solutions has the problem ? Can you choose a
magnitude for c, so that there shall be only one solution ? Can you choose
a magnitude for c so that equilibrium is impossible ? What is the least
magnitude of c consistent with equilibrium ?
(24) A weight of 50 lbs. is supported from A and B as in Mg. 129.
A\
(i) Find the pulls on A
and B. ■
(ii) If a man pulls in the
direction and sense
OG with a force of
10 lbs. weight, find
the alteration in the
pulls on A and B.
(iii) If he pulls in the
opposite sense, find
the alteration in the
pulls on A and B.
(iv) Find the pull so that
there may be no
tension in OB.
i
\ SO Ilia.
FlQ. 129.
DEDUCTIONS FROM EXPERIMENTS. 131
The Foundation of Statics. Such experiments as have
been performed cannot be considered in themselves as the best
foundation for the science of mechanics, the true basis for which
must be sought in far more generalised experience. Such
generalised experience was summed up by Newton (p. 135).
Deductions made from such experiments as those detailed must
consequently be regarded as tentative only. The experiments,
however, have the great advantage of giving a reality to
notions concerning the action of forces which descriptive matter
fails to impart.
Deductions from Experiments. The experiments now
performed all relate to the action of forces on rigid bodies.
Force without some body (mass) acted on is a meaningless term ;
forces do not act on points but on masses, and such an expression
as " forces acting at a point " means only that the lines of action
of the forces are concurrent.
Expt. I. shewed that a force is determined only when we
know some point in its line of action in addition to its magni-
tude, direction and sense
Expts. II. and III. shewed (i) that a body under the action of
two forces is in equilibrium when, and only when, the forces
differ in sense alone ; (ii) a force acting on a rigid body may be
supposed to act anywhere in its axis.
By a rigid body is meant one which retains the same relative
position of its parts under the action of all forces. Any body
; I which maintains its shape unaltered, or for which the change is
\ '• I too small to be measurable under the action of certain forces, may
! be considered as rigid for those forces. The paper in Expt. I.
■ -^was practically rigid for the forces acting on it; it is, however,
Iquite easy to apply forces to it that would change its shape. If
a set of forces deform a body, but after a time the body takes
up a new shape which does not alter while the forces are
unchanged, such a body after deformation may be treated as
rigid for those forces. For non-rigid bodies we must know not
only the axis of the force, but also its point of application.
132 GRAPHICS.
Expts. IV. to VI. shewed that if a rigid body, acted on by
ooncurrent forces, is in equilibrium, the sum of the vectors of
the forces is zero ; and conversely, when the sum of the vectors
of the forces is zero and the axes concurrent, the body is in
equilibrium.
Expt. VII. shewed that when a body is in equilibrium under
the action of three forces, their axes are concurrent ; but that
in general for four or more forces the axes are not concurrent
when there is equilibrium.
Rotors. Any quantity which, like a force, requires for its
specification the magnitude, direction, sense and a point on its
axis, is called a rotor quantity (Clifford).
Such quantities may be represented geometrically by rotors,
i.e. vectors localised in definite straight lines. The rotor may
be specified by giving its vector and a point on its line of
action. It is, however, usual and convenient to give
(i) the axis, (ii) the vector,
so that the direction is given twice over.
To avoid confusion in graphical work, the axes of the forces
(rotors) should be drawn on a different part of the paper from
the vectors giving the magnitudes, direction and senses of the
forces.
Equilibrant. When a body is in equilibrium under the
action of a number of forces, the forces themselves are, for
shortness, often spoken of as being in equilibrium. For such
a system of forces any one may be said to be in equilibrium
with the rest, and from this point of view is called the eq[iiilibrant
of the others.
Resultant. The equilibrant of such a system of forces would
be in equilibrium with a certain single force differing from it
only in sense (Expt. II.), and this reversed equilibrant would
have the same effect, so far as motion is concerned, as all the
rest of the forces together. The equilibrant reversed in sense
is called the resultant of the forces.
RESULTANT OF CONCURRENT FORCES. 133
It should be noticed that it has not been shewn that any
system of forces has a resultant, but simply that if a system of
forces is in equilibrium any one of them reversed in sense is the
resultant of the rest, and would produce the same effect as
regards motion as all the rest together.
Resultant of Concurrent Forces. To find the resultant
of a number of concurrent forces acting on a body, add their
vectors to a resultant vector and through the point of con-
currence draw the axis of the resultant force parallel to its
vector.
Force Scale
' P ? 8 ? 6 5 4 3
lbs. wt. I
Fm. 130.
Example 1. Find the resultant of two forces of magnitude 9'2
avd 12'1 lbs. weight acting towards the E. and towards a point 66-5°
N.ofE.
Draw the axes a and i (Fig. 130) and add the vectors a and j8
of the forces (scale 1" to 5 lb.), a + p = y, then y is the vector of
the resultant force. Through the point of intersection of a and b
draw the axis c of the resultant.
134 GRAPHICS.
Notice that if y be set off along c and a. and ^ along a and h,
they form two adjacent sides and the concurrent diagonal of
a parallelogram. That the magnitude, direction and sense of
the resultant of two intersecting forces can be found, by adding
the forces as vectors, is often, but badly, expressed by saying
that forces are combined by the parallelogram law.
Example 2. Fvnd the resultant offow forces of Tnagmtvdes 13, 11,
9, 7 kilogrammes iveight whose axes are the lines joining a point to
points A, B, C, D, the five points being the vertices in order of a
regular pentagon, and the senses being from to A, to B, to C
and D to 0.
Draw a circle of radius 2", divide the circumference into five
equal parts with dividers by the method of trial. Mark the five
points in order 0, A, B, C, D, then draw the vector polygon,
a. of length 13 cms. parallel to OA, /3 of length 11 cms. parallel
to OB, y of length 9 cms. parallel to 00, and, finally, S parallel
to OD, but having a sense from D to 0. The vector o- joining
the beginning of a to the end of S is the resultant force in magni-
tude, direction and sense. Finally, draw a line through parallel
to 0-, this is the axis of the resultant force.
(2.5) Find the resultant of two forces of magnitudes 16 and 18 kilogrammes
weight, if they are directed N. and 75° E. of N.
(26) Three concurrent forces have magnitudes 23, 18, 15 lbs. weight, find
their resultant in magnitude, direction and sense when the angles between
them are 120° and 100°, and the forces all act outwards.
(27) If two forces are equal, shew that the resultant must bisect the angle
between them.
(28) If the magnitudes only of
two forces are given, in what
relative directions should they
act so that the resultant is (i) as
big, (ii) as small as possible.
(29) a (Fig. 1.31) acts in the axis
Ox, another force in Oy ; find
graphically the magnitude and
sense of this force so that the,0
resultant may be as small as ■
(30) Concurrent forces of magnitudes 12, 17, 10 and 8 lbs. weight are
directed towards N., N.E., S.E. and 30° W. of S. respectively; fcid the
resultant in magnitude, direction and sense.
NEWTON'S LAWS OF MOTION. 135
(31 ) A wheel has six equi-spaoed radial spokes ; four consecutive spokes
are in tension and pull on the hub with forces of 10, 15, 12 and 7 lbs.
weight ; find the resultant pull on the hub due to these spokes.
(32) A string ABO is fastened to a hook at A, passes round a free
running pulley at B (AB horizontal), and is pulled in the direction BG
where ABO= 105° with a force equal to the weight of 17 lbs. Find the
resultant force on the pulley at B.
(33) ABGD is a square of side 2", a force of 11 lbs. weight acts from A
to B, one of 7 lbs. from D to A and one of 3 lbs. from G to A ; find the
resultant force.
FORCE, MASS AND ACCELERATION.
Newton's Laws of Motion. The laws for the combina-
tion of concurrent forces deduced from Expts. I. to VII. are
immediate deductions from Newton's famous Second Law of
Motion. Stated shortly in modern language the law is — a force
acting on a particle (or body, if the axis passes through the
M.c), is measured by the product of the mass of the body and
the acceleration produced.
Acceleration being a vector quantity, force is a vector quantity,
and since the force must act on the mass moved, it is a localised
vector quantity or rotor.
The effect of two or more concurrent forces is found, therefore,
by adding the corresponding accelerations as vectors. The single
force, which would produce this resultant acceleration, is called
the resultant force, and is measured by the product of the mass
and this acceleration. To find, then, the resultant of a number of
concurrent forces, add the forces as vectors ; the sum gives the
vector of the resultant force, and the axis of the force passes
through the given point of concurrence.
A mass being in equilibrium when it has no acceleration, we
see this will be the case, when, the axes being concurrent, the
vector sum of the forces is zero, and conversely.
The equation connecting the three quantities, mass, force and
acceleration is
Force = Mass x Acceleration.
136 GRAPHICS.
^Mass and Weight. If a foot and a second are units of
length and time, a foot per second is the unit of speed, and
a speed of a foot per second added per second (or a ft. per sec. per
sec.) is the unit of speed acceleration. Further, if the unit of
mass be a lb. mass, the unit of force must be that force which
would give a lb. mass a speed acceleration of a ft. per sec. per
sec. ; or which would increase its speed every second by a ft.
per sec. This follows at once from the equation : if the mass = 1
and the acceleration = 1, then the force must = 1.
We know that a body falling freely has a speed acceleration
of 32'2 ft. per sec. per sec ; hence if the mass be a lb. mass, the
force acting on it is the lb. weight and is given by the equation,
lb.-wt.= force = 1 x 32-2.
In this system, then, the force on a falling lb. mass would
be 1 X 32'2 units of force ; this is the weight of a lb. mass
in these units. For statical purposes it is better, however, not
to use this system, but to take the weight of the lb. mass as the
unit of force.
The expressions lb. weight, force of a lb. weight, and lb. mass
will often be met with ; the first denotes the force with which
the earth attracts the lb. mass ; the second a force equal in
magnitude to the weight of a lb. mass, but usually having a
different direction.
In the c.G.S. system, similar double terms occur. The unit of
mass is here a gramme, and the unit of length and time a
centimetre and a second.
Unit force is then = gramme x an acceleration of a centimetre
per sec. per sec, and is called a dyne.
The acceleration due to gravity in centimetres per sec. per sec.
is 981, and, therefore, the weight of a gramme mass is 981
dynes. In statics, however, it is usual to consider the gramme
weight as the unit of force, and thus we meet with the terms,
gramme weight, force of a gramme weight, and gramme (or
gramme mass).
ACTION AND REACTION.
137
W
Action and Reaction. In Expt. II. (p. 119) the ring was
found to be in equilibrium under the opposite pulls of the
threads BA and CD. Consider the bit of thread AB, it is in
equilibrium under the pull ( = W-^) from A upwards and the pull
( = ;Fj) from B downwards. At B the action of the thread on
BO is equal and opposite to that on BA. At every point of the
thread a similar argument holds, i.e. there are two equal and
opposite forces pulling away from each other. This double set
of forces is called a stress, tensile stress in this particular case.
If a column (Fig. 132) supports a load JV, then the action
of the upper portion on AP is (neglecting the weight of the
column itself) a downward push = JV, and the
upper part is in equilibrium under the load W
and the reaction of AP. The action at P,
therefore, on the upper part must consist of
an upward push = W. Whatever part of the
column be considered, the result is the same, at
every point there are two equal and opposite
pushing forces. This double set of forces is
called a compressive stress.
No force can be exerted without the presence
of an equal and opposite one. If a body be
pushed, the body will push back with a force (called the resist-
ance) equal in magnitude and opposite to it in sense.
If a spiral spring be pulled out beyond its natural length it
tends to shorten and pulls back with a force of equal magnitude.
Again, the wind only exerts force in so far as its motion is
resisted, and the resisting obstacle reacts on the moving air with
a force of equal magnitude.
Put shortly as in Newton's Third Law of Motion : the action
of one body on another (or of one part of a body on another
part) is equal in magnitude and opposite in sense to that of the
second body on the first, or still more shortly : action and re-
action are equal in magnitude, Iiave the same axis, but are of
opposite sense.
A
Fia. 132.
138
GRAPHICS.
Example. A hook is in equilibrium on a horizontal table, not
because there is no force acting on it, but because the pressure of the
book on the table, due to its weight, is exactly equal in magnitude and
opposite in sense to the reaction of the table on the book.
■fR
R
(
(
W
Flo. 133.
Suppose the table to be tilted ; then if the book still remains in
equilibrium it must be because the reaction of the table is still
vertical, and of the same magnitude as before. * The reaction is
therefore no longer normal to the table, and hence there must
be some force along the common surface of table and book; in
fact, there is friction.
Ideal surfaces between which normal action alone is possible
are called frictionless or smooth. Smooth as applied to one body
only is, strictly speaking, meaningless ; it is a term relating to
the action and reaction of two bodies. If in any problem one
surface is spoken of as being smooth, it is meant that the action
between that surface and any other body considered in the
problem is wholly normal.
Since the action between any two bodies is never wholly
normal, problems involving the supposition that certain surfaces
are smooth are to a great extent academic, and the results
obtained must be regarded as only first approximations to the
real state of things.
Note. In this chapter the weight of a body will be supposed
to act through its mass-centre.
*If a body is in equilibrium under two forces— the weight and the table
reaction— these forces can only differ in sense.
ACTION AND REACTION.
139
Example. A body of weight W (2-5 Mlogrms.) is kept in position
on a smooth plane of inclination 30° by a hmizontal force a. What
must be the magnitude and sense of a, ami what is the reaction of the
plane?
The body is in equilibrium under the action of three forces,*
viz. the weight, the force a, and the reaction y of the plane.
The latter is perpendicular to the plane, since the plane is
smooth.
Set oS JB= 2-5" (Fig.
134) vertically downwards,
draw through A, AC mak-
ing 30° with the vertical,
and through B, BC hori-
zontal.
Then
BC measures the pull
= 1 -44 kilogrms.,
and
CA measures the reaction
= 2-89 kilogrms.
Force Scale
Kilogrammes wt.
FlQ. 134.
* The axes must bo concurrent, p. 132.
140 GRAPHICS.
(34) If the plane is inclined at 75°, find a and 7.
(35) If the plane is inclined at 30° and the direction of a makes 15° above
the horizontal, find a and 7.
(36) If the plane is inclined at 30°, and a's direction is 15° below the
horizontal, find a and 7,
(37) Shew from the vector polygon that, whatever the inclination of the
plane, the pull will be a minimum if it be applied parallel to the plane.
(38) A garden roller of weight 2 cwts. is hauled up a slope inclined 1 in 5
(1 vertically to 5 horizontally) and. held with the handle horizontal. What
is the horizontal pull on the handle ?
(39) A body weighing 7 cwts. is kept in position on a smooth inclined
plane by a force of 2 cwts. parallel to and up the plane and another force
inclined at 15° below the horizontal. The ratio of the height and the
base of the plane being 0'7, find the force inclined at 30° and the reaction
of the plane.
Example. A mass of 5 lbs. weight is attached to a string of length
1 ft. The string is fastened to a point on the circumference of a smooth,
fixed horizontal cylinder of radius 2 ft. The point of attachment being
r2 ft. from the top of the cylinder ; find the tension in the string and
the reaction of the cylinder.
The direction of the string at C is along the tangent to the
circle, the string being supposed quite flexible.
The tension in the string being the same at all points of BC
(see formal proof on p. 162) it is immaterial at what point we
suppose it fastened to the cylinder; in fact the length of the
string may be anything, provided one end is at C and it is
wound on the cylinder from the fastened end towards C in a
clockwise sense.
Draw a circle of radius 2" to represent the vertical section of
the cylinder containing the weight and string. Step off, from
the highest point A (Fig. 135), the are ^C=2-2". Join to
the centre of the circle. Then draw the vector polygon of the
forces ; a vertically downwards of length 5 cms., y parallel and jS
perpendicular to 00. Then measure y and /? in cms to obtain
the reaction of the cylinder and the tension of the string.
ACTION AND REACTION.
141
Position Scale
3' 4'
I L
I 1 I 1 1 1_
Force Scale
Pio. 135.
142
GBAPHICS.
Example. A body of 15 lbs. weight is sustained on a smooth
vndined plwne by a horizontal farce of 7-2 lbs. weight and a force
■parallel to the plane of 3-7 lbs. weight. What is the inclination of tJw
plane and its reaetim,? „^ ,^. ^„„,
^ Draw OA (Fig. 136)
Ok downwards of length 15
cms., then AB horizontally
of length 7-2 cms. With B
as centre, describe a circle
of radius 3-7 cms., and,
by the aid of set squares,
draw a tangent to it from
; let BD and OD be the
radius and tangent.
Fig. 136.
Measure OD on the cm. scale, this gives the reaction ; measure
the slope of BB by finding how many inches it rises for 1"
horizontally, or use a protractor and obtain the angle DBA.
(Eeaction= 16-25 lbs. and angle of plane 38'2° approximately.)
(40) A mass of 7 lbs. weight is to be attached to the highest point of a
smooth horizontal cylinder (radius 2') by a string which can only bear a
tension of 4 lbs. ; what is the greatest length of string that may be used ?
(The reaction being perpendicular to the tension, the vector triangle is right-
angled, and since a is known and the magnitude of /3, y is determined. )
(41) Find the inclination of a smooth plane so that a body of 5 kilogrms.
weight may be supported on it by a horizontal push of 2 kilogrms. weight.
EXERCISES.
143
(42) Knd the inclination of a smooth plane bo
that a body of 17 lbs. weight may be supported
on it by a force of 7 lbs. weight applied parallel
to the plane. (In this case we know the reaction
of the plane is perpendicular to the applied force
of 7 Ids., so set off 17 cms. vertically down-
wards for the weight. From the lower end of
this line describe an arc of radius 7 cms. as in
Fig. 137, and draw a tangent to it from the
upper end. The length of the tangent gives
the reaction R. The reaction, and therefore
the normal to the plane, is now known. )
Fig. 137.
(43) A body of weight 15 lbs. is supported on a smooth inclined plane by
a horizontal force of 7 lbs. weight together with a force of 4 lbs. weight
acting parallel to and up the plane ; find the inclination of the plane
and the reaction.
(44) A truck weighing 15 cwts. is kept at rest on an incline of 1 in 5 (one
vertical to five horizontal) by a rope 6 ft. long attached to the truck 3 ft.
above the level of the rails and fastened to a hook midway between them.
Find the pull on the rope.
(45) A smooth ring weighing 3 kilogrms. can slide on a vertical circular
hoop of radius 2 ft. It is attached to the highest point of the hoop by a
string 3 ft. long. Find the tension in the string and the reaction of the
hoop on the ring. (The reaction is along a radius of the circle since the
ring is smooth. )
(46) A string with equal weights of 11 lbs. attached to its ends is hung
over two parallel smooth pegs A and B in the same horizontal line ; find
the pressures on the pegs. (The tension of the string is the same through-
out ; the concurrent forces at each peg which are in equilibrium are the
reaction of the peg and the two pulls of the string, one on each side of
the peg.)
(47) If in Ex. 46 the line AB makes an angle of 40° with the horizontal ;
find the pressures on the pegs.
(48) A string with equal weights of 750 grms. attached to its ends
passes round three pegs in a vertical plane at the vertices of an equilateral
triangle. Find the pressures on the pegs when one side of the triangle is
horizontal and (i) the third vertex above, (ii) the third vertex below the
horizontal side, the string passing under this vertex and over the other two.
144 GRAPHICS.
Simple Bar Frameworks. In problems on the equilibrium
of very simple frameworks of rods we suppose at first that
(i) the weight of the rods may be neglected ;
(ii) the joint connecting two rods is made by a perfectly
smooth circular pin ;
(iii) the loads are applied only at the joints.
The action of the pin on the rod must then pass through the
centre of the pin (why T) ; hence, any rod is under the action of
two forces passing through the centres of the end pins, and for
equilibrium these forces must be equal and opposite, i.e. in the
line joining the centres of the pins. The bars may therefore
be represented by the lines joining the pin centres.
Example. A wall crane consists of two bars AC and BC pin-
jointed together at C and to the wall at A and B. (BC is called the
beam, AC the tie rod.) A load of 4-02 tons is suspended from C.
Find the stresses in BC and AC and whether they are tensile or
compressive, given that BC= 10-1 ft., AC= 15/if.
Since the forces on the pin at C are 4-02 tons downwards and
pulls or pushes along BO and OA, we have simply to find the
forces in the directions OA and OB which will be in equilibrium
with 4'02 tons downwards.
Draw first the crane to scale and then set off" Of = 4 '02 cms.
(Fig. 138) vertically downwards and draw PQ horizontally and QO
parallel to AO. The forces at are given by OP, PQ and QO in
magnitude, direction and sense. Scale these vectors ; PQ gives
3-64 tons. Notice that OPQ is similar to ABO, hence, if ABC
be supposed the vector polygon for the forces at 0, then AB
represents 4-02 tons. Measure the length of AB, and from this
determine the forces represented by BO and OA.
At G the beam BO pushes from left to right, and, therefore,
since C is in equilibrium, the pin must push the beam from
right to left and exerts a compressive force on it. Again, the
beam is in equilibrium and hence the pin at B must also exert a
force on the beam from left to right. Hence, BO is in a state of
SIMPLE WALL CRANE.
145
I
llil
o
l' J
1 1
f
i
Crane Scale
f ?' 7' ?'
?' 1°'
II' 12*
1 1
-p'-f*
1
,M?
X
1
Force Scale
2
t
f
4
1
5
1
Tons
FlQ. 138.
compression and the compressive stress is measured simply by
the force at either end.
T.6, K
146 GBAPHICS.
Again, QO measures the action of the bar AC on C, and since
it is upwards, the bar evidently pulls at C; and further, since the
bar is in equilibrium it must also pull the pin at A, and hence
the bar AC is pulled at C and A with forces tending to lengthen
it and must therefore be in a state of tensile stress or (shortly) in
tension.
(49) If BG=9 ft. and AG =12 ft., and the load suspended from O is
3 '78 tons, find the stresses in AC, and BO.
(50) If ABO=m', BaA=4:5°, and the load is 6-3 tons, find the stresses.
(51) If BC= 10 ft. and AB= 12 ft., and BO slopes downwards at an
angle of 30°, find the stresses in AB and BO due to a load of 2 '8 tons wt.
(52) If BO =10 ft. and is horizontal, find the stresses in BO and AO
when AB has the following lengths 10, 8, 6, 4 and 3 ft. ; the load is 1'7
tons wt. Draw a graph shewing the relation between the length oi AB
and the stress in BO.
Example. In a wall crane ACB (Fig. 139) the diain hearing the
load W passes over a smooth pulley at C and is fixed to the wall at E ;
f/nd the stresses in AC and CB given that AC is horizontal and
of length 9 ft., BG= 12 ft, AE = 4-4 ft., and W = 3-7 tons.
Draw the frame to scale, say 1 cm. to a foot. Since the pulley
is smooth, the pull on E, and therefore the tension in CE, is
measured by 3'7 tons weight.
Hence, set oif OF = 3-7" vertically downwards to represent the
load; then PQ = 3-7" parallel to CE. Through Q draw QR
parallel to AC, and through draw OR parallel to BC ; then
OPQR is the vector polygon of the forces keeping the pin in
equilibrium at C.
Measure the lines to scale. The senses in which the vectors
must be taken at are decided by OP and PQ. QE acts from A
to C, and the bar pushes at C and is therefore in compression.
RO acts from C to B, and the bar pulls at C and is therefore in
tension.
(53) Mnd the stresses when AE=3 ft.
(54) Find the stresses if A0=AB = 9 ft., AE=i-5 ft., and AO slopes
upwards at an angle of 20° with the horizontal.
WALL CRANE.
147
148
GRAPHICS.
10-
i
*-
i
an-
I?
•a-
w
w
1
8
11
[gsi
fan.
0.
i
10
;
■
0-
qJ
SIMPLE TRUSS.
149
Example. Two equal rods AB and AC are pin-jointed together
at A and their other ends connected by a cord BC ; the whole rests on a
smooth table in a vertical plane with a weight W= 29-4 lbs. suspended
from A. Given AB = AC = 3-2 ft. and BC = 5-9 ft., find the stresses
in AB, AC and BC.
First draw the frame ABC (Fig. 140) to scale (1" to 1'), and
then the vector polygon: OP =5-88 inches, PQ parallel to AB,
QO parallel to AC. Then these vectors measured on ^" scale
give the stresses in the bars in lbs. weight. Are the rods AB
and AC in compression or tension?
For the joint B the force QP pushes. Draw PiZ parallel to
BC and QR vertical, then the sense of the vector triangle for
B is QPJR, and PR measures the tensile (why tensile V) stress in
BC. Why was RQ drawn vertically upwards, and what does it
measure 1
(55) It AB = 3 ft., A0=2 ft., BG=S-5 ft. and W=2-3 kilogrms., find all
the stresses and the reactions of the table at B and C.
(56) The Derrick Crane. BC (Fig. 141) is the
post (kept vertical by some means not shewn),
AC the jib, AB the tie rod. Given AB = 6 ft.,
A0=13 ft. and BO=10 ft. A load W=7-i tons
is suspended from A, find the stresses in AB and
AC.
(57) If the supporting chain passes over a smooth
pulley at A and is fixed at D, where CD =3-5 ft.,
find the stresses in AB and AC.
(58) Given AB=n It, AC=25 ft., 5C=16 ft.,
GD=5 ft. and H''=14'5 tons", find the stresses in
^Sand^a
(59) A picture weighing 6 '5 lbs. is hung by a
wire over a smooth nail. If the distance apart of Fia. 141.
the points AB s.t which the wire is fastened be
1 ft. 7 in. and the length of the string 2 ft. 3 in. , find the pressure on the
nail and the tension in the string.
(60) If the length of the string in Ex. 59 vary, draw a graph shewing
the relation between the tension of the string and its length.
(61) Light rods AC, CB, of lengths 7'2 ft. and 5'7 ft. are pin-jointed
together and to two fixed points A and S distant 6'3 ft. apart. AB is
incUned to the horizontal at an angle of 25" [A being the higher) a load
of 4-7 owta. is suspended from the pin joining the two rods ; find the
stresses in the rods.
150
GRAPHICS
(62) ABC (Fig. 142) is a wall crane pin-
jointed a,t A, B and O ; a load W oi 5 tons
is suspended from a pulley D, which is
attached to the crane at B and Chy a, chain
BDG. ABD = T2°, AC=1-5 ft., ^.8= 12-9
ft. ; find the stresses in ^ B and BO. (Notice
that BD and OB must be equally inclined
to the vertical, since the tension throughout
the chain is constant. Hence first draw the
vector triangle for the forces at D, and
determine this tension. Knowing the pull
of the chain at B, the stresses in AB and
BO can be found. )
(63) A picture, of weight 11 lbs., is suspended from a smooth nail by a
continuous string passing through two smooth rings on the picture frame
distant apart 1 ft. 7 in. If the height of the nail above the two rings
be 3 ft., find the tension in the string and the pressures on the nail and
rings.
(64) AB and BO (Fig. 143) are light rods pin-jointed together at B, and
to fixed points at A and O. A load W
(7 cwts.) is suspended by a chain which ^B
passes over a smooth pulley at B and is
attached to M, the mid-point of AO. The
load is pulled by a horizontal rope until
the chain makes an angle of 30° with the
vertical. Find the stresses in the rods and ^-^ / / iiW
chain, given that AB= 10 ft.,
BO =5-i ft. and ^C=6-22 ft.
M
Horizontal
Fig. 143.
Components of a Force. In relation to their resultant the
forces of a given system are called the components. Finding
COMPONENTS OF A FORCE.
151
the resultant of a set of concurrent forces is a unique process ;
the converse problem of finding the components when the
resultant is known is not in general unique.
A force may be decomposed into two components having
given directions and passing through any point on the axis, in
one and one way only.
The proof is exactly the same as that for the decomposition
of vectors, on p. 84.
When the component of a force in a given direction is spoken
of without reference to the other component, it is always implied
that the two components are perpendicular.
Scalar Conditions of Equilibrium for Concurrent Forces.
For equilibrium under concurrent forces, it is a sufficient and
necessary condition that the vector polygon of the forces should
be a closed figure.
/y
Fia. 144.
Let the axes of the forces be supposed concurrent, and let
a, j3, ... cr he their vectors whose sum is zero {i.e. the vector
polygon is closed). Draw any line XX ; project the vectors on
to this line by drawing parallels through the end points of
the vectors. If aj, ;8j, ... be the projections, Fig. 144 shews that
152
GRAPHICS.
Similarly project on Yy a line parallel to the former direction
of projection, and establish a similar theorem for the projection
on It, VIZ.
"■2 +1^2 + 7-2 + ^2 + "'2 = 0.
Then a^ and a^ axe the components of a in the directions XX
and VY, and so for the other components, and the sum of the
components in any two directions is zero.
Conversely, if the sum is zero in any two directions the vector
polygon is closed, and the forces (if concurrent) are in equi-
librium. One direction is not sufficient, for it might happen,
as in Fig. 145, that though the polygon is not closed, the first
and last points of the projections are coincident
Fio. US.
The two directions being at right angles we have the theorem :
The sum of tW components in any direction of all the forces
acting on a body in equilibrium is zero.
Again, - o- is the resultant of a + ;8 + y + 8, and hence we get
the theorem : The sum of the components in any direction of
any number of concurrent forces is equal to the component of
the resultant in that direction.
(65) Five oonourrent forces in a horizontal plane have components 3 '7,
2'1, rS, r7 and 2-9 towards the N., and components 1'2, 3'7, 2'4, 3 and
3 '2 towards the E. Find the resultant in magnitude, direction, sense and
position.
COMPONENTS OF A FORCE.
158
(66) Mark on squared paper the four points whose ooordinates are (1, 0),
(17, 2), (2'3, 1), (3'2, 4) inches, and let the lines joining the origin to these
points represent concurrent forces to the scale of 1 cm. toakilogrm. Find
the resultant by (i) the vector polygon method, (ii) the component method.
(67) Forces of magnitude 5, 2'8, 3'1, 4'7 are concurrent and make angles
of 15°, 30°, 60° and 75° with a line through the point of concurrence. Find
the forces in this line and a line perpendicular to it which would be in
equilibrium with the given forces.
Example 1. A horse begins to pull a small tramcar with a
force P = 500 lbs. weight. The traces make an angle of 25° with
the horizontal, find the component of P in the direction of motion.
If the weight of the car be ^ a ton, what is the reaction of the
I i {Suppose no friction.)
Force Scale
lbs. wt.
Flo. 146.
Draw OA = 5" (Fig. 146) making 25° with Ox, and draw AB
perpendicular to Ox; then, since as vectors OA = OB + BA, OB
represents the forward pull on the car.
From B along BA set up 5(7=11-2", then AC gives the
reaction of the ground, for it represents the weight of the car,
less the vertically upwards component of the pull of the traces.
154 GRAPHICS.
Example 2. The points A and B are 5" apart and distant 1 and
3-7" respectively frim the line CD, and both on the same side of it.
Find the components through A and B of a force of 8 lbs. in CD, when
(i) the component through A is perpendicular to CD ; (ii) the com-
ponents through A ami B are mutually perpendicular; (iii) the
components are equal in magnitude.
(i) Draw, through A, AO perpendicular to CD. Join BO, and
find the components of the 8 lbs. weight along OA and OB.
(ii) Draw a semicircle on ^5, cutting CD in and 0^ ; then
0^, 05, O^A and OjiJ are possible directions for the components.
There are thus in this particular case two sets of components
which will satisfy the conditions of the problem. Find these
components.
(iii) Draw AM perpendicular to CD and produce it to A^,
where AM=MA-^. Join A^B, cutting CD in N; then AN
and NB are the required directions. Find the components
in these directions.
(68) In the above example, if B is on the opposite side of CD to A,
determine the components in the three cases (i), (ii) and (iii).
(69) The pressure of wind on a sail when the sail is perpendicular to the
wind is 500 lbs. weight ; find the normal pressure on the sail when the wind
makes angles of 15°, 40°, 65° and 75° respectively with the sail.
(70) Find the components of a force of 11 lbs. weight making angles of
30° and 75° with it.
(71) A force of 17 lbs. weight is directed due N. ; find the components in
the directions (i) N.E. and N.W., (ii) E. and N.W., (iii) S.E. and 30°
W. of N.
(72) Two ropes are attached to the coupling of a railway van and are
pulled horizontally with forces of 200 lbs. and 270 lbs. weight. The
lengths of the taut ropes are 18 ft. and 21 ft. and their ends remote from
the truck are at distances of 10 ft. and 7 ft. respectively from the centre
Une of the rails. Find the forward puU of the van and the side thrusts
on the rails when (i) both ropes are on the same side, (ii) on opposite sides
of the rails.
(73) On squared paper mark the positions of two points whose coordinates
are (1, 2) and (2'4, 1'2) inches; find the components through the origin
and these points of a force of 7 lbs. weight acting (i) along the axis of x,
(ii) along the axis of y, (iii) along the line bisecting the angle xOy.
(74) On squared paper mark the point whose coordinates are (2'7, 1"1),
and draw a Une parallel to Oy and distant 1" from it on the negative side.
Find the components of a force of 5 lbs. weight, one of which is along this
parallel and the other passes through the given point when the axis of the
EXERCISES.
155
force is (i) along Ox, (it) along Oy, (iii) a line making 30° with Ox and outs
Ox at 1 '5" from the origin on the positive side.
(75) A smooth inclined plane (Fig. 147) rising 1 in 3 has a smoothly
running pulley at the top.
A body of weight W (11-6
lbs.) is kept in equilibrium
by the pull of a string
parallel to the plane. The
pull being produced by a
freely hanging weight P,
find P and the reaction of
the plane.
What is the component of
the weight parallel to the
plane ? What is the vertical
component of the reaction of
the plane ?
(76) A man distant 13"5 ft. from a tree pulls at the upper part of the trunk
by a rope of length 40 ft. His pull is equal to a weight of 80 lbs. What is
the horizontal pull on the tree, and what is the force producing compressive
stress in the trunk ?
(77) A barge is towed by a horse with a pull P of 152 lbs. weight making
an angle of 20° with the direction of the bank. What is the force pro-
ducing forward motion, and what would be the side thrust of the water on
the barge if there were no side motion ?
(78) A block is partly supported by a smooth right-angled wedge of
weight 18 lbs. (as in Fig. 148) the height and base of the wedge being
Fia. 147.
Fio. 148.
3 '2 ft. and 6 ft. respectively. If, to maintain equilibrium, the wedge has
to be pushed with a horizontal force of 28 lbs. weight, what are the
reactions of the wedge on the body and on the horizontal table ?
(79) A uniform cylinder of
weight 57 lbs. rests on two
inclined planes as indicated
in Fig. 149. The planes are
hinged together at A ; what
is the tension at A , and what
is the pressure of each plane
on the ground, given that the
wedges are equal in all re-
spects, each weighing 15 lbs.,
and that
BO_S,
A£~T
156 GRAPHICS.
Example. OF (Fig. 150) represents the crank of an engine, P
moving in the circle DFE omd being fixed. OF is the connecting rod,
being the cross head of the piston rod which moves to and fro along AB
(AB = DE). C is kept in the line AB by guides. The forward
thrust on C being 5000 lbs. weight, find the force transmitted along
the connecting rod OF and the side presswe on the guides at C
(assuming no friction) given that CF = 6-5 ft., DE == 3 ft. and AC = 6"
the direction of motion F being as indicated.
Find also the components of the force transmitted, along CF in the
direction of the forwrnd motion of F and perpendicular to it, (i.e., along
the tangent and radius at F j.
First draw the position diagram to scale, say 2 cms. to 1 ft.
Next construct the vector diagram PQ = 5" to represent 5000 lbs.;
then QR and PB are perpendicular to AB and parallel to OF
respectively. QR is the thrust on the guides (and RQ is the
reaction of the guide on G keeping it in the path ACB) and PR
is the force transmitted along the connecting rod.
Draw P8 perpendicular and RS parallel to OF, then PR acting
along OF is equivalent to PS acting along the tangent at F and
SR acting from F to 0. PS then gives the forward thrust of F.
(80) Find the force on F urging it round the circle when AC=(i'2, 0'4,
0-6 and 0-8 times AB.
Example. Draw a graph shewing the connection, between the
position of C (Fig. 150) and the thrust on F wrging it round the circle.
Divide AB into ten equal parts and draw ordinates at A and B
and the points of division. Produce Ci^ to cut the ordinate at
in G. Project Q horizontally on to the ordinate at 0. Do this
for the eleven marked positions of C and join the points so deter-
mined by a smooth curve. The force scale for this representation
is OF to 5000 lbs.
Compare the results with those obtained by the vector polygon.
* Proof.
From the construction of Fig. 150 we have the following
relations between the angles :
BPQ= GCO =90° -OGF, PES = GFO,
FORCE ON ENGINE CRANK.
157
and hence ^ = cos RPQ = cos (90° - OGF) = sin OGF,
DO
^ = sin PRS= sin GFO;
. PS sin GFO OG
•• PQ sin OGF~ OF'
(S8.
Via. 150.
Hence, if OF be taken to represent PQ or 5000 lbs. wt., OG
will represent PS or the forward thrust on the piston,
158 GRAPHICS.
* (81) If the force on the piston decreases viniformly from 5000 lbs. at A
to zero at B, find by a graphical construction the forward thrust on
^when the cross head is at Oand AO=<i'^AB.
*(82) Construct the curve giving the relation between the forward
thrust on F and the displacement of G for the variable force given in
the last example. Set up, perpendicular ia AB, .4 i4i= radius of crank
circle and project from G to Gj on AA^ and join O-^B; the point of inter-
section of &iB and GCj gives the force required for displacement AG.
Example. AB (Fig. 151) represents a sail of a ship whose keel
line is as shewn. The thrust a of the wind on AB if perpendicular to
the mnd would be 500 lbs. weight. If AB makes an angle of 30° mth
the keel line and the relative velocity of the wind to the ship be in
direction CM, making 45° with keel line, find the thrust v/rging the ship
forwa/rd.
Resolve a into /? and y perpendicular and parallel to the sail
AB. y has no eifect on the ship's motion. Find the components
of /8 along and perpendicular to the keel line; the former, S
(approximately 64-7 lbs. weight) is the thrust urging the ship
forward, the latter, t, tends to produce lee-way and in good sailers
is nearly balanced by the resistance of the water to side motion
and the force of the current on the rudder.
In the vector diagram, since RPQ is a right angle, a circle
described on RQ as diameter will pass through P. Draw this
circle. As the direction of the sail line AB is changed the
point P will move on this circle. Evidently as P changes, the
length QT will alter and it will be greatest when PT is a tangent
to the circle.
Kow, the radius being perpendicular to the tangent at any
point, the line joining P (when PT is a tangent) to the mid-
point S of RQ must be perpendicular to PT and therefore parallel
to the keel line. Hence, to find the best position for the sail,
bisect RQ at S and describe a circle of radius SQ ; then draw SP
parallel to the keel line cutting the circle at P. RP gives the
direction in which the sail should be set and the greatest
possible forward thrust is given by QT. Since SRP = \QSP we
may give this direction as the one bisecting the angle between
the keel line and the direction of the relative wind.
SAILING AGAINST THE WIND.
159
No matter how small the angle between the keel line and the
wind direction, there will always be a force urging the vessel
on. If there be much lee-way, sailing close to the wind is impossible.
C,
<^
Fig. 161.
(83) Draw a figure for the case when the sail is set on the other side of
the keel line, and shew that this ease is an impossible one.
(84) The keel line being from W. to E. and the relative wind from the
N.W. ; find the forward thrust on the ship when the sail is set 25° S. of W.
(85) Shew from the vector diagram for given directions of the keel line
and stern wind that the greatest forward thrust would not be obtained
by putting the sail as nearly perpendicular as possible to the keel line.
(86) The keel line being from N. to S. and the relative wind from E. to
W., find the forward thrust when the sail makes an angle of 20° with the
keel line. Find the angle at which it should be set to give maximum
forward thrust on the ship.
160 GRAPHICS.
(87) The force of a current on a rudder when placed perpendicular to the
stream is 50 lbs. ; find the retarding force on the ship when the rudder
makes angles of 20°, 30° and 60° with the keel line. Shew that, in a race,
the rudder should be used as little as possible.
(88) Explain how it is that a kite, though fairly heavy, is enabled to rise
in the air.
(89) The force of the wind on a kite if placed perpendicular to it would
be 5 lbs. When the kite makes an angle of 35° with the horizontal, find
the force due to the wind iirging it upwards.
(90) In Ex. 89 if the kite be stationary and its weight 10 ozs., what is
the puU of the string on the kite in magnitude, direction, and sense.
Body in Equilibrium under Three Non-Parallel Forces.
Experiment VII. on p. 122 shewed that when a body is in equili-
brium under three non-parallel forces, the axes of the forces are
concurrent. The same result follows from the combination of
concurrent forces, deduced from Newton's Second Law of Motion,
since equilibrium is only possible under three forces when the
resultant of any two differs only in sense from the third.
This consideration enables us to draw the axes of those forces
in equilibrium when one force is unknown in direction.
Example. A uniform beam rests vnih rnie end against a smooth
vertical wall and the other on rough ground. Determine the readions
of the ground and wall.
AB (Fig. 152) is the beam of length 25 ft., ABC=&Q° and
the weight is 0'505 cwt.
Draw the beam in position (scale 1 in. to 5 ft.), then draw a
vertical through G the M.C. of beam, and a horizontal through
A, intersecting in 0. Join OB, then BO is the direction of the
ground's reaction.
Construct the vector polygon (scale 10 cms. to \ cwt.).
Draw P0=10 cms. downwards; then QB horizontal and RP
parallel to BO. Scale the lengths, QR and PR giving the re-
actions (Q/i = 0-145 cwt., i2P = 0-525 cwt. approximately). Why
is AO drawn perpendicular to A CI
(91) Determine the reactions of the wall and ground if .45 is inclined at
45° to the horizontal.
(92) Determine the reaction of the wall and ground ii AB is inclined at
40° to the horizontal and the mass centre G of the beam is at 9 ft. from
the ground, reckoned along the beam.
EQUILIBRIUM UNDER THREE FORCES.
161
(93) A uniform beam AB hinged 3.t A, is supported in an inclined
position to a vertical wall AC hy a, string GB fixed to the wall at O.
The weight of the beam is 17 kilogrms., AB=iit., AG=2it., BO =5-2 it.,
find the tension in BO and the reaction at A on the beam.
(94) With dimensions as in Ex. 93, find the stress in BO if the mass-
centre of the beam he i AB from A.
(95) Draw a graph shewing the connection between the distance of
G from B and the tension of the string BO.
(96) In Ex. 94, if ^£=30 ft. and AB0='16% find the reactions.
A lO
Force Scale
j_
_±.
ji
Fia. 162.
(97) A uniform beam AB of length 25 ft. and weight 70 lbs. is hinged to
a wall &t A (19 ft. above the ground at 0), the other end B rests on a
smooth inclined plane OB. Find the reactions at A and B when the
inclination of the plane is 30°, 15° and 60° respectively.
(98) Draw the two sides and base of a rectangle, the sides being 3" and
base 2" ; draw a diagonal and produce it 4'5". Let the two sides of the
rectangle represent vertical boards securely fixed in the ground (base), and
the diagonal produced a uniform beam. The beam being smooth, find the
reactions at its points of contact : weight of beam 2 cwts,
T.o. L
162
GRAPHICS.
(99) A uniform beam of length 3 ft. is hinged at one end to the lowest
point of a horizontal hollow circular cylinder of inner radius 2-5 ft. The
other end of the beam rests against the inner smooth surface of the
cylinder in a plane perpendicular to the axis of the cylinder. Find the
reactions at the two ends of the beam, the weight of the beam being
530 lbs.
/
A
sV
_^B
Fio. 163.
(100) AB (Fig. 153) is a weightless rod 5 ft. long, which can turn
about (7 as a fulcrum ; ^C=3-2 f t. ; it is acted on by two forces P
and y as shewn. P= ICK) lbs. ; find Q and the reaction at C.
Fig. 154.
(101) A uniform beam AB (Fig. 154), 13 ft. long and of weight 80 lbs.,
rests against a smooth inclined plane BC (rising 4 ft. vertically to 7 ft.
horizontally) and is prevented from
sliding by a peg at .4, AC =9 ft. Find
the reaction of the plane and the total
reaction at A.
The Smooth Pulley. A flex-
ible string of negligible weight is
fastened at 5 to a smooth pulley
(Fig. 155) and passing over it bears
a load W. Consider the equilibrium
of any part QP of the string. The
forces acting are the pulls (tensions)
at Q and P and the reactions of the
surface QP. The former are tan-
gential and their axes intersect at
G, the latter are normal and have therefore a resultant passing
Fio. 156.
tw
PROBLEMS ON THREE FORCES. 163
through 0. Since three forces in equilibrium must be concurrent,
the resultant reaction must pass through C as well as 0. But
CO bisects the angle QCP, and hence, from a trial stress diagram,
we see that the tensions at Q and P must be equal in magnitude.
In some problems the axes of one, two, or more of the forces
are unknown in direction, but other geometrical conditions are
given which, with the aid of a trial diagram, will enable the
solution to be found.
Example 1. A uniform heavy rod, of weight 9 lbs. and length 3ft.,
is suspended from a point hy two strings of length 2 -5 and 2 ft.
respectively attached to its ends. Find the equilibrium position, and
the stresses in the strings.
If AB, AC, BO (Fig. 156) be the two strings and the rod,
then on the system of strings and rod act two forces, the
weight of the rod at M, its mid-point, and the reaction at A.
These must be in a line, since there is equilibrium, and hence
AM must be vertical. Draw the triangle ABO, in any position,
and its median AM. Then, if AM be vertical we have the
required position, and the vector polygon can be drawn. (The
usual convention in books is to represent the vertical in space
by a line parallel to the bound edge of the paper; if this
convention be adhered to, another triangle A-^B-fi^ must be
drawn with AM vertical, and this can easily be done by con-
structing a parallelogram, whose diagonal is vertical and equal to
164
GRAPHICS.
2AM, and whose adjacent sides are equal to AB and AC.)
Complete the solution.
*ExAMPLE 2. A heavy unifmin smooth ring weighing 17 lbs. slides
on a shing of length 4 '9 2 ft./ the ends of the string are fastened to
two hooks A and B, whose distance apart is 3-95 //., A being 0'98 ft.
above B ; find the position of equilibrium and the tension of the string.
Since the ring is smooth the tension of the string must be the
same on both sides, and hence from a trial stress diagram we
see that the two parts of the string must be equally inclined
to the vertical. Draw ADB (Fig. 157), where AD = 0-98" and
is vertical, AB= 3-95" and DB is horizontal. With B as centre,
describe a circular are of radius 4-92" cutting AD produced in E,
or set an inch scale so that BE = 4:-92". Bisect AE at M and
draw MN parallel to DB, then ANB is the form assumed by
the string.
A
Fio. 157.
For AN^EN, and therefore AN + NB = ^-92" , and AN and
NB make equal angles with the horizontal. To find the tension,
PROBLEMS ON THREE FORCES. 165
draw, in the vector diagram, lines parallel to AN and NB from
the extremities of the vector, giving the load of 17 lbs.; complete
the solution.
* Example 3. A uniform beam weighing 105 lbs. rests with one
end A in contact with a smooth plane of inclination 35°, the other
end B rests on a smooth plane of inclination 50°. Determine the
reactions of the supporting planes and the position of the beam.
Draw first the vector polygon of the forces, PQ representing
105 lbs. weight to scale, then draw QE and PB making angles of
31° and 50° with PQ. These lines give the reactions at A and B.
Draw two lines AC and BC for the planes and at any two points
A and B, their normals intersecting in 0. Join to the mid-
point M of AB. Complete the parallelogram OATB of which
OA and OB are adjacent sides. Then 0TB and OAT should
be similar to DRQ ; hence if PQ be bisected at S, GB should be
parallel to SR, hence SB gives the inclination of the beam.
(102) A rod of length 7" lies in a smooth hollow horizontal cylinder,
perpendicular to its axis, of radius 9" The maas-oentre of the rod is at a
point distant 2 '5" from one end ; draw the position of £he rod in the
bowl when in equilibrium, and measure its slope.
Note. The m.c. of the rod must be vertically under the axis of the
cylinder.
* (103) A uniform rod, of weight 7 kilogrms., can turn freely about one
end in a vertical plane ; it is pulled by a horizontal force of 4"3 kilogrms.
weight at its free end. Draw the rod in its position of equilibrium, and
measure its slope.
Note. Three forces act on the rod and must pass through a point ;
knowing the vertical and horizontal forces, the reaction at the hinge can
be found. From any point draw two lines: (i) OB parallel to the
reaction, and (ii) OA horizontally. Bisect OA at M, and draw verticals
from M 3jaA A. Where the former cuts OB (at B say) draw BT, cutting
the vertical through A in 7', then OTis the direction of the beam. Measure
OT in cms. or inches, and determine the scale to which the figure is drawn.
* (104) Solve the previous exercise if the rod is not uniform and the
M.c. is at a distance of J of the length from the lower end.
* (105) A rod of length 6 ft. has its m.o. at a distance of 2 ft. from the
end which rests on a smooth plane of inclination 30°, the other end rests
on another smooth plane whose inclination is 45°. Draw the rod in its
position of equilibrium ; its weight being 6 cwts. , find the reactions of the
planes.
166
GRAPHICS.
*(106) A uniform beam AB (Kg. 158), of length 7 ft. and weight 27
kilogrms., can turn freely, in a vertical plane, about .4 ; to its upper
extremity B is fastened a cord which runs over a smooth pulley at O,
9 ft. vertically above A, and carries a, weight of 11 kilogrms. Find the
position of the beam and the reaction at the hinge.
Fig. 168.
Draw a trial figure A BC and a stress diagram PQR, where PQ repre-
sents the weight of the beam, RP the tension of the cord ( = 11 kilogrms.)
and QR the reaction at 4. Then PQR should evidently be similar to AOC
(O being point of concurrence of the axes of the forces). Hence
CO_PR_U
GA~PQ~Z7'
and hence 00 and therefore OB {=200) is known, and hence the triangle
AGB can be conatruoted to scale. Do this construction and determine
QR.
MISCELLANEOUS EXAMPLES. IV.
1. With the aid of your instruments find the resultant of the two forces
represented in magnitude and direction by the straight lines shewn in the
■<
Pig. 159.
MISCELLANEOUS EXAMPLES. 167
diagram (Fig. 159). Assuming that one inch represents 10 lbs. weight,
write down the magnitude of the resultant. Also, express in degrees the
angle the resultant makes with the greater of the two forces.
(Engineer Students, 1903.)
2. Draw diagrams to shew the directions in which each of the following
sets of forces must act so as to maintain equilibriiim, if they can do so.
Set A. 3, 4, 5. Set B. 1, 1, 3. Set 0. 4, 1, 3.
(Naval Cadets, 1904.)
3. Two men, who are lifting by ropes a block of wood, exert pulls of
45 lbs. and 65 lbs. respectively. The ropes are in the same^ vertical plane ;
the rope to which the smaller pull is applied makes an angle of 25° with the
vertical, and the rope to which the other pull is applied makes an angle of
33° with the vertical on the opposite side. Determine graphically, or in
any other way, the actual weight of the block of wood if it is just lifted by
these two men. (Naval Cadets, 1904. )
4. One of two forces, which act at a point, is represented numerically
by 7 ; the resultant is 14 and makes an angle of 30° with the force of 7 ;
find graphically the magnitude and line of action of the second force. Also
calculate the magnitude to two places of decimals, and measure the angle
between the two forces as accurately as you can.
5. Three forces, acting in given directions, are in equilibrium at a
point ; shew how to find the relative magnitudes of the forces. What
additional information suffices for the determination of the absolute
magnitudes ?
Two small equal brass balls, each weighing yij, oz. , are suspended by equal
silk threads, 12 inches long, from a single point ; the balls being electrified,
there is a force of repulsion between them so that they separate and remain
in equilibrium 4 inches apart ; find the force of repulsion and the tension
of each thread. (B. of E., I., 1904.)
6. A weight of 1 ton is hung from two hooks 20 ft. apart in a horizontal
platform by two chains 15 and 10 ft. long ; find by construction and
measurement the tension in each chain. (B. of E., II., 1903.)
7. A chain weighing 800 lbs. is hung from its two ends, which are
inclined to the horizontal at 40° and 60° respectively. What are the forces
in the chain at the points of suspension? (B. of E., A.M. I., 1903.)
Fig. 160.
8. The figure (Fig. 160) shows a bent lever AOB with a frictionless
fulcrum 0. AO is 12", BO is 24". The force Q of 1000 lbs. acts at A ;
what force P acting at B will produce balance ? What is the amount
and direction of the force acting at ? (B. of E., A.M. II., 1904.)
168
GRAPHICS.
9. A thread is fastened by one end to a fixed point A, and carries at its
other end a weight F of 20 lbs. To a point B of the thread a second thread
BG is fastened and this second thread is pulled at the end C by a force
equal to the weight of 8 lbs. ; when the system comes to rest it is found that
BC is horizontal. Shew the system when at rest in a. diagram drawn to
scale, find the angle which AB makes with the horizon and the tension set
up in AB. (B. of E., I., 1904.)
10. Draw two lines OA and OB and let AOB be an angle of 37° ; suppose
that -R, the resultant of two forces P and Q, is a force of 15 units acting
from O to B; suppose also that P is a force of 8 units acting from O to .4.
Find, by a construction drawn to scale, the line 00 along which Q acts,
and the number of units of force in Q. (B. of E., I., 1904.)
11. In a common swing gate the weight is borne by the upper hinge.
The distance between the upper and lower hinges of such a gate is 3 '5 ft.
If a boy weighing 119 lbs. gets on the gate at a distance of 8 ft. from the
post, find the magnitude and direction of the pressure he exerts on the
upper hinge. (B. of E., II., 1905.)
12. A machine of 5 tons in weight is supported by two chains ; one of
these goes up to an eyebolt in a wall and is inclined 20° to the horizontal ;
the other goes up to a roof principal and is inclined 73° to the horizontal ;
find the pulling forces in the chains. (B. of B., A.M. I., 1907.)
13. Fig. 161 shows a weight
of 500 lbs. supported by two
equally inclined poles. Find
the thrust on each pole.
(Naval Cadets, 1903.)
14. The bracket shewn in the
sketch (Fig. 162) carries a load of 100
kilogrammes at O. Find whether the
stresses in AC and BO are thrusts or
pulls and the amount.
(Military Entrance, 1905.)
Fig 162.
MISCELLANEOUS EXAMPLES.
169
15. Draw a triangle ABC with AB vertical, and let ^Cand BO repre-
sent two weightless rods, joined together by a smooth hinge at O and
fastened by smooth hinges to fixed points at A and B ; a weight W is hung
from C ; shew that one of the bars is in a state of tension, and the other of
compression ; also shew how to calculate the stresses.
Obtain numerical results in the following case : AB=4, BO =3, CA=2,
and ir= 18 tons. (B of E. , 11. , 1905. )
16. In Pig. 163 If is a weight of 170 lbs.
hanging from a joint at .4 by a chain that
weighs 20 lbs. The joint is supported by
rods AB and AG fixed at B and O. Find
the stress in each rod, and say whether it
is a thrust or a pull. (Naval Cadets, 1904. )
□ W
Pig. 16
17. The two bars AG
and BC (Eig. 164), hinged
at A and B, and hinged to-
gether at O, carry a load
of 170 lbs. at 0. Find the
stress in each bar.
(Inspector of Ordnance
Machinery, 1904.)
770 lbs.
Fig. 164.
18. A body whose mass is 2 cwts. rests on a smooth inclined plane ; it is
maintained in position by a force of 40 lbs. acting parallel to the surface
of the plane, and by a horizontal force of 110 lbs. Determine in any way
the angle of inclination of this plane. (Military Entrance, 1905.)
170
GRAPHICS.
19. A weight slides freely on a cord 2-6 metres long, the ends of which
are attached to fixed pegs P and Q; P is 1-4 metres from the vertical
through Q and 30 centimetres below the horizontal through Q. Draw to
a scale of rijth a diagram shewing the position of equilibrium. Determine
the tension in the cord and the proportion of the weight borne by each
peg.
Denoting the span or horizontal distance between the pegs by S, the
height of one peg above the other by H, and the length of rope by L, find
expressions for the horizontal and %'ertical distances of the weight from
the lower peg. (Military Entrance, 1905.)
20. A weight is supported by a tie and a horizontal strut ; find how the
pull in the tie varies as the inclination changes, and plot a curve giving
the puU as a function of the angle of inclination of the tie.
(Military Entrance, 1905.)
21. Mg. 165 represents a vertical section (drawn to the scale of 1 inch
to a foot) of the roof of a building, A CB being a window which can turn
about a hinge at A and which is opened by means of a rope tied to the end
F oi a, light iron bar CF, which is firmly fixed to the window at G. The
Fia. 165.
rope from F passes over a smooth pulley at K and is fastened to a hook E
in the roof. Find the tension in the rope when the position of the window
is that indicated in the figure. The weight of the window is 30 lbs. and
may be taken as acting at O. (Military Entrance, 1905.)
22. A uniform bar AB oi weight W is freely movable round a smooth
horizontal axis fixed at .4. It is kept at a fixed inclination i to the horizon
by resting against a peg P whose position along the under surface oi AB
is varied. Represent in a diagram the various magnitudes and directions
of the pressures on the peg and the axis ^ as P is moved along the bar.
(Inter. B.Se. (Eng.), 1906.)
23. Enunciate the triangle of forces. Shew how to find, by a graphical
(Jonstruotion, the angle at which two forces, each equal to 50 lbs. weight,
must act on a point that they may have a resultant equal to 75 lbs.
weight. (Inter. Soi., 1900.)
24. Draw a triangle ABC with AB vertical and A above B to represent
two bars AC and BC freely jointed at G and attached at A and B to
MISCELLANEOUS EXAMPLES. 171
points of a wall in the same vertical line. A given weight being suspended
from G, determine the natures and magnitudes of the stresses in ^0
and BG. (Inter. Soi., 1900.)
25. Draw a triangle ABC having the vertical angle A large and the
base BG horizontal ; produce BC to D. Let AB denote a rod connected
by smooth hinges to a fixed point B and to the end ^ of a rod A G, whose
other end G can move in a smooth groove BGD ; the weight of the rods
being negligible, a force F is applied in the plane A BG at right angles to
AB at A ; find the force transmitted by the rod .4 C along the groove.
(Inter. Sci., 1900.)
26. A light cord attached to a fixed point 0, passes over a fixed
pulley Q at the same level as and at a distance c from it, and supports a
weight W attached to its end ; another weight w smaller than 2W is
slung freely over the cord between and Q ; determine the depth below
OQ at which. this weight will rest in equilibrium. (Inter. Soi., 1900.)
27. Explain a graphical method of finding the resultant of a number of
fiven forces acting on a particle. A light string of length I has its ends
xed at A and .B at a horizontal distance a apart, and a heavy ring of
weight W can slide along the string. Prove that the ring can rest
vertically beneath S if a force W^ be applied parallel to AB.
» (Inter. Soi., 1904.)
28. A uniform bar A B, 10 ft. long, of weight W is freely movable in a
vertical plane, about a smooth axis fixed a,t A ; it is sustained at an angle
tan"' I to the horizon by resting against a fixed (smooth) peg at G, where
A 0=6 ft. Eind the magnitude and exhibit the lines of action of the
pressures at A and O. _ (Inter. Sci. , 1904. )
29. A man stands on a ladder which leans against a vertical wall.
Assuming the pressure on the waU to be horizontal, find geometrically the
horizontal thrust of the foot of the ladder on the ground. Length of
ladder 15 f t. , foot of ladder 5 ft. from wall, total weight of man and ladder
3 owts. acting 5 ft. from the ground (reckoned along the ladder).
(Inter. Soi., 1904.)
30. A drawbridge AB, hinged at A (the axis of the hinge being hori-
zontal and perpendicular to AB), is to be raised by a chain attached at B
and carried over a pulley C fixed vertically over A a,t a, height AG=AB.
The resultant weight of the bridge acts through the mid-point of AB.
Shew in a diagram how to find the varying tension in the chain due to the
weight of the bridge as it is slowly lifted, neglecting the weight of the
chain and all friction.
If the bridge weighs 2 owts., find the tension of the chain and the
direction and magnitude of the reaction of the hinge when the bridge is
half -open, that is, when .45 is at 45° with the horizontal.
(Home Civil, I., 1905.)
31. A light bar AB can move freely about the end A, which is fixed,
and is supported in a horizontal position by a string GB, G being a fixed
point vertically above A. If a weight W be suspended from any point P
of the bar, find geometrically the direction and magnitude of the reaction
at A and the tension in the stay. W=V>, AB=1S,", AP=12!', AC=9".
(B.So., 1904.)
CHAPTER V.
THE LINK POLYGON.
Resultant of three Coplanar Forces (non-parallel).
Example. Draw a triangle ABC (Fig. 166) whose sides are 3, 4,
and 6 inches long. Take these as the axes of f wees whose magnitudes
are 7, 2, .5 lbs. weight, and whose senses are given by AB, BC, and
CA. To find the resultant of these forces in magnitvde, direction,
sense i
Draw the vector polygon for these forces
a, /3, y, finding
(r = o-j + y = a + ^ + 7.
Through B draw BD parallel
to o-j cutting AG at X>.
Through D draw DE parallel
to o- cutting BG in E.
RESULTANT OF FORCES. 173
Then DE is the axis and (t the vector of a force called the
resultant of the given forces. Measure the magnitude of n- and
the angle it makes with BC, and. the position of E with reference
to B and G.
Note that o- is independent of the order of addition of a, /8, y ;
it is not evident that E is independent of the order in which we
suppose the forces combined.
(1) Combine the forces in two difFerent ord<Jts, viz. (i) a and 7 to a
resultant through A and combine this resultant with ;S ; (ii) /3 and 7 to
a resultant through G and then combine this with a. Shew in each ease
that the resultant always outs BG at E.
Resultant of any number of Coplanar Forces (non-
parallel). When there are more than three forces the process
for finding the resultant, if there is one, is simply a continuation
of the process explained for three forces, and consists in finding
the resultant of two intersecting forces, then the resultant of this
and a third force intersecting it, and so on. The whole con-
struction is a repetition of that for two concurrent forces ; its
validity depends on the truth of the assumption that the order,
in which we suppose the forces combined, is immaterial.
(2) Draw an equilateral triangle ABG of side 4" ; forces of magnitudes
2, 5, 1 lbs. weight act in these sides with senses AB, BG, GA. Find the
resultant in magnitude, direction, sense and position.
(3) On squared paper take axes Ox a,nd Oy. Mark the points whose
coordinates are (3, 2), (1, - 1), (2, 3) and ( -4, 2). Forces of 3, 5, 1'54 and
2 lbs. weight act through these points, their directions and senses being
in order.
(i) parallel to Ox and in the positive sense.
(ii) making 45°, (iii) making 75°, (iv) making 120° with Ox and with
senses upwards.
Find the magnitude, direction and sense of the resultant and where it
outs the axis Ox.
Resultant by Link Polygon. The construction in the
previous examples fails altogether for parallel forces and in many
oases involves finding the point of intersection of lines which
are nearly parallel.
These difficulties can be overcome by the introduction of two
new forces which differ only in sense.
174 GRAPHICS.
The construction now to be explained depends for its validity
on the truth of the suppositions, (i) that such a pair of forces
will not affect the equilibrium, (ii) the order of the combination
is immaterial.
Example. On any straight line mark fou/r points an inch apaii,
and draw lines a, b, c, d through these as indicated, a, ^, y, S are the
vectors of the forces acting in these lines of magnitudes r98, 3'3, 4"15,
and 2 '05 lbs. weight. Find the resultant.
Draw the vectors, to the scale 2 cms. to 1 lb. weight, and add
them to a resultant vector o- (Fig. 167).
a- = a + ^ + y + S.
Mark a point on the concave side of the vector polygon
(called the pole). Join this point to the vertices of the vector
polygon P-^^P^P^P^Pi. The point should be chosen so that
these joining lines are not nearly parallel to any of the vectors.
For this reason the concave side is the better position for 0.
Mark any point A on a and through it draw a line, e, parallel
to OPj. The latter line is a vector, call it e.
Through A draw AB (cutting h at B) parallel to
OP2( = 6 + a).
Through B draw BO (cutting c at C) parallel to
OP3( = e + a + y8).
Through C draw CD (cutting d at D) parallel to
OP^( = € + a + ^ + y).
Through D draw BE (cutting e at E) parallel to
OP5( = £ + a + y8 + y + S).
Through E draw r parallel to
PjPj ( = o- = e + a + /) + y + S-€ = o + p + y + S).
Then r is the axis and o- the vector of the resultant of the
given forces.
Proof. Let two forces differing only in sense act in e, and
suppose their vectors to be e and -e. A.t A there are two
concurrent forces € and a ; these are combined to e + a acting in
AB. At B there are two concurrent forces ;8 and e + a ; these
THE LINK POLYGON.
175
FIO. 167.
are combined to e'+ « + /3 acting in BC. At G there are two con-
current forces y and e + a + jS; these are combined to e + a + ^S + y
176 GRAPHICS.
acting in GD. At D there are two concurrent forces S and
« + a + ^ + y ; these are combined to £ + a + ;8 + y + S acting in
DE. Finally, at E there are two concurrent forces - e and
6 + a + ;8 + 7 + 6; these are combined to a force a + |8 + y + 8
acting in r. Hence r is the axis of the resultant and o- is its
vector.
Note that the construction is always possible and never
awkward if the pole is chosen properly, for this means that
AB, BC, ... always intersect b, c, ... at angles never very acute.
(4) Repeat the construction, using a different pole. Is the same axis r
obtained ?
(5) Repeat the construction, choosing a different point A on a.
(6) Repeat the construction, adding the vectors in the order
a+7 + S + ^.
The figure A, B, C, . . . , constructed on the axes a, b, c, . . . , is called
the link polygon {sometimes the funicular polygon) ; the vector
polygon is often (but vrrongly) called tlie force polygon.
(7) Find the resultants in Exx. 2 and 3 by the link polygon method.
(8) A wheel has eight tangent spokes placed at equal distances round
the hub. The tensions in five consecutive spokes are 31, 2'7, 3'3, 1'8 and
2 '4 lbs. weight. Find the magnitude, direction, sense and axis of the
resultant pull on the hub due to these Ave spokes, the spokes being
tangents to a circle of radius 2'5".
Equivalent Forces. Any set of forces wMcli would produce
the same effect, so far as motion is concerned, as a given system
of forces is called ecLuivalent to the latter.
Resultant Force. If a single force would produce the same
effect as a given system of forces this ectuivalent force is called
the resultant of the given system.
Not more than one single force can be equivalent to any given
set of forces, otherwise forces differing in magnitude or direction
or sense or position, or in all together, could produce the same
motion in a body. The latter supposition is inadmissible (see
Expt. 1, p. 119, and Newton's Second Law of Motion, p. 135).
THE LINK POLYGON.
177
Equilibrant. Look at the matter a little differently. Sup-
pose a set of forces to be in equilibrium, then any one of the
set may be considered as the ectuilihrant of the rest. A force
differing only in sense from the equilibrant is the only force that
could produce equilibrium with it, and hence is the only single
force equivalent to all the rest (Expt. II., p. 119).
Unique Resultant. If, then, a set of forces has a resultant,
it can have one only.
That any set of forces has a resultant has not been proved;
as a matter of fact, two forces, which differ only in sense and
position have no resultant.
The construction given for finding the resultant of any number
of copknar forces consists in finding one after another the single
forces equivalent to 2, 3, 4, ... up to the last of the given set,
and including in this set two forces differing only in sense. At
each step of the process we find the resultant in conformity with
experimental results and with Newton's Second Law of Motion.
Since there can be one resultant only, the order in which we
suppose the forces combined is immaterial.
Expt. VIIL Punch four holes in a piece of cardboard,
and suspend it as in Expt. "VI. , Chap. IV. Mark the lines
of the forces and the corresponding magnitude and sense of
each pull.
By vector and link polygons construct the resultant of
three of these, and hence shew graphically that this
resultant differs only in sense from the fourth force.
Notation (Bow or Henrici). For graphical
work it is often (but not always) convenient to
have a different notation from that used hitherto.
The axis of the force is indicated by two letters
(or numbers), one on each side of the force, whilst
the vector of the force is indicated by the same
two letters in capitals placed at its ends. Thus ab (Fig. 168) i§
the axis, and AB the vector, of a force,
FlQ. 168.
178 GRAPHICS.
Example. Mark five points P, Q, E, S, T 3 cms. apart on a
straight line, and draw lines through these points at angles 65°, 90°,
70°, 90° and 65°, as indicated in Fig. 169. Letter the spaces between
the lin^s a, b, c, d, e, f, as indicated. The forces in these lines are
given by the vectors AB, BC, CD, DE, EF, and represent 3, 2-15,
2-08, 2-7, 2-85 lbs. weight respectively. Find the resultant force.
Choose a convenient pole (Fig. 169). Through the space a
draw B-^B parallel to OA ; through the space 6 draw B.fi^ parallel
to OB ; through c draw B^B^ parallel to OC ; through d draw iJgi?^
parallel to OB ; through e draw B^B^ parallel to OE ; and finally
through / draw B^^B parallel to OF. The lines through the first
and last spaces, i.e. B^R and B^B, intersect at B, a point on the
resultant.
Draw, then, through B a line parallel to AF ; it is the axis,
and AF is the vector of the resultant.
Note that it is not necessary to draw the radial lines
OA, OB, 00, ... ; in fact it is better not to do so, as the crossing
of the lines at tends to make the exact position of the pole
doubtful.
The advantage of the space notation consists in its rendering
mechanical the order of drawing the lines. A corresponds to a,
B to 6, and so on. The more important advantage of uniqueness
of construction will be better seen when stress diagrams are
under consideration.
In some cases where the axes of the forces cross, some little
care is necessary in choosing a good order in which to take the
lines and a convenient pole, so that the construction lines may
not intersect ofi' the paper.
Example. Draw a pa/rallelogram having adjacent sides of 2-76
and 2'53 inches, the included angles being 75° and 105°. Letter the
spaces as indicated (i.e. talcing the parallel sides in order and not the
adjacent sides). The forces acting in the sides are given by the vectors
AB, BC, CD, DE, and are of magnitudes 10-3, 4-2, 3-1 and 5-6
kilogrms. weight. Find the resultant.
Choose a pole somewhere near the position indicated (Fig. 170)
EXAMPLES ON LINK POLYGONS.
179
U.^
lbs. wt.
FlQ. 169.
180
GRAPHICS.
and construct the link
polygon and the axis
of the resultant as
shewn.
In Fig. 170 the lines
from to ^ and A to
are the arbitrary
vectors. The first line
in the link polygon is
drawn parallel to OA,
and the second is drawn
through the space h
parallel to OB. AE,
shewn dotted, is the
vector of the resultant
force ; its axis is the
dotted line in the link
polygon.
s \^l
^o
Fig. 170.
(9) Take the order in which the forces are combined differently; e.g.
take the adjacent sides in order contraolookwise.
(10) Combine the forces directly without using a pole.
Parallel Forces (Like). Parallel forces are but a particular
case, and the construction for their resultant does not differ in
any respect from that given for the general case.
Parallel forces having the same sense are said to be like;
if they have opposite senses they are unlike.
Example. Fig. VI \ is a diagram of the four pairs of driving
wheels and the trailers of a modern locomotive. The weights borne by
the wheels, taken in order from left to right, are 15, 17, 17, 17 and 13
tons weight, their distances a'pairt are 6', 5' 8", 6' and 7' 6". Find
the axis of the resultcmt thrust on the rail.
Letter the spaces as indicated, and draw the vector and link
polygons. Fig. 171 shews B, a point on the axis of the resultant,
distant 1" to the right of the centre of the third pair of wheels.
PARALLEL FORCES (ENGINE PROBLEM).
181
182 GRAPHICS.
(11) Find the resultant of two weights of 5 and 7 lbs., distant apart 11",
hung from a horizontal rod.
(12) Find the resultant of six equal weights (1'23 lbs. each) hung from a
horizontal rod at distances, from left to right, of 1'7, 1'04, 1'83, 2'02, 0'97
inches apart.
(13) Three men pull at parallel ropes attached to a block in a horizontal
plane with forces 51 '7, 65 2 and 55'4 lbs. weight ; the first two ropes are
2 ft. 8 in. apart; where should the third rope be so that the resultant
pull should be in a line midway between the first two ropes.
(14) Three weights Wi, Wg, Wg are placed in a line on a table, the
distance apart of Wi and W2 is 2 ft., and of W2 and W^ 3 '2 ft. If W'-i=7
lbs. and 1^2=* lbs., find W3 if the resultant push is to be midway between
W2 and Ws.
Parallel Forces (Unlike). Should some of the parallel
forces be of opposite sense to the rest, the corresponding vectors
in the vector polygon must be drawn in their proper sense, the
construction is otherwise exactly the same.
Example. Five men pull on a yacht which is stuck on a mud bank
by parallel ropes {in the same plane). Find the resultant pull on the
yacht, the distances apart of the ropes in ft. and the magnitudes in lbs.
weight and senses of the pulls being as given in Fig. 172.
Set oflf AB = 5-8, 5C=7-1, CD = 7-9 cms. downwards; and
DE = &-3 and EF= 5-5 cms. upwards. The vector sum is AF
and represents the resultant in magnitude, direction and sense.
Draw the link polygon as before, keeping to the order of the
letters ; finally, P is found as the point of intersection of the first
and last lines of the link polygon and therefore is a point on
the axis (r) of the resultant.
(15) Find the resultant of two parallel forces 10 and - 15 lbs. weight, the
axes being 3 ft. apart.
(16) Six parallel forces act on a rod ; the magnitudes are 10, - 15, 8, - 12,
7 and -20 lbs. weight at distances 2, 8, 9, 11, 12, 15 inches from one raid;
find the resultant force and where its axis outs the rod.
(17) PQSS is a square of side 3" ; a force of 15-3 lbs. weight acts along
PQ, one of 8-2 lbs. weight along QR, one of 9-8 lbs. weight along SR and
one of 18-4 lbs. weight along SP. Find the resultant in magnitude, direction
and sense, and the point where its axis outs QR.
PARALLEL TORCES (UNLIKE).
183
(18) Find the equilibrant of three parallel forces of magnitudes 8,-7 and
-2 cwts., the distances apart of their axes being 1 and 1'6 yards.
(19) In a certain locomotive there are four pairs of driving wheels whose
distances apart are all 5' 8" ; the distance between the last wheel and the
first wheel of a coupled goods truck is 9' 10". The truck has three pairs
of wheels whose distances apart, from front to rear, are 6' 3" and 6'.
■-8
B
■-F
-C
■-E
Fig. 172.
:+o
The thrusts of the wheels taken in order from the leading driving wheel
are 12 tons 10 cwts., 14 tons 8 cwts., 12 tons 14 cwts., 9 tons 13 cwts., 9 tons
12 cwts., 9 tons 15 cwts. and 7 tons 5 cwts. Find the axis of the resultant
thrust on the rails.
lU
Graphics.
Vector Polygon Closed. Two Forces. When the vector
polygon is closed there is evidently no resultant force, but it does
not follow that the forces are in equilibrium.
Example. Draw two parallel lines ab and be 3 inches apart,
and suppose forces of 10 and - 10 lbs. weight to act in these lines ; go
through the construction for finding the axis of the resultant.
Draw the vector polygon (Pig. 173), AB= 10 cms. downwards,
BC= 10 cms. upwards ; it is of course closed, since the starting
and ending points are the same. Choose a pole and draw
OA, OB, 00. Through space a draw K-^P^ parallel to OA;
through h draw P-^P^ parallel to OB ; and through c draw P^K^
paraHfel to 00.
Pio 178.
The theory of the construction is : in K^P-^ we suppose two
forces OA, CO differing only in sense ; OA is combined with. AB
to the resultant OB acting along P^Pj ; then OB is combined with
Be to the resultant 00 in P^K^; and we have, finally, CO in K^P.^
and -CO in PJC^.
Definition. Two forces which differ only in position and sense are
called a couple of forces or shortly a couple.
VECTOR POLYGON CLOSED. 185
The construction on p. 184 shews that the given couple is equi-
valent to the final couple, and, since the pole may be anywhere,
there is an infinite number of couples equivalent to any one couple.
To see the connection between the couples, measure the per-
pendicular distance between the forces. Shew that the product,
force X the perpendicular distance between the couple, is the
same, i.e. shew that AB xp = OC xq where p and q, the distances
between the axes, are called the arms of the couples.
AB .p measures the area of a parallelogram whose opposite
sides are AB and -AB, and is called the momental area of
the couple, if account be taken of the sense of the area.
An area is considered positive if its boundary is given a
contraclockwise sense, and negative if the boundary is clockwise.
Taking the sense of the momental area as given by the sense
of one of the forces we see that the momental area has the same
sense in the two oases.
(20) Use in turn four other poles for the vector polygon, taking at least
one pole on the opposite side of .45 to that in Fig. 173. Calculate in
each case the momental area of the equivalent couple, and see that it is
equal to AB .p, and of the same sense.
Proof that the construction does give couples of equal
momental areas. Produce K-^^P^ to cut he in Pg, and K^P^ to
cut ah in P^. Then P-^P^P^P^ is a parallelogram, and P-^P^P^ and
PjP^Pj ^^^ similar to the vector triangle OAB.
The area of P,P,P^P^ = P,P, . q = P.,Ps ■ P ;
:. pAB = q.OA;
or, denoting by F and P the magnitudes of the forces represented
by AB and AO pp^ qp^
i.e. the momental areas of the two couples are equal in magni-
tude. A simple inspection of Fig. 173 shews that the senses are
the same.
Unit of Momental Area. This unit has no special name;
if the force be measured in lbs. weight, and the distance in ft.,
186 GRAPHICS.
the momental area would be in lbs. ft. (Not ft. lbs. — a term
which has a totally different meaning.) The units employed
must always be distinctly stated.
Vector Polygon Closed (General Case). The vector
polygon being closed, the first and the last lines of the link
polygon are of necessity parallel, and the simplest equivalent
set of forces is a couple (except in the special case when the
first and last links are coincident).
Example. Draw a closed vector polygon ABCDEA swch that
AB = 2-2", BC=l-82", CD = 2-8", DE=1" ard EA = 4" and
BE =.4-1" arid CE = 3-2".
Draw any line cd (Fig. 17 i) parallel to CD, and (on the left-hand
side of the paper) ab parallel to AB cutting cd at P. On cd mark
points Q, R and S where PQ = 2-64", PR = 4-61" and PS = 5-93".
Through Q, R and S draw be, de and ef parallel to BC, DE and
EF respectively.
Let the vectors represent in magnitude, direction and sense farces
to the scale of 1 cm. to 1 lb., and let the lines drawn through
P, Q, R, S be their axes.
Find the equivalent couple to the forces whose vectors are AB, BC,
CD, DE and EA, and whose axes are given.
Choose some convenient pole within the vector polygon.
Through any point B^ in a6 draw B^R parallel to OA, and
B^B^ parallel to OB, and proceed as usual with the link polygon
construction until B^ on ef is reached, and B^B^ is parallel to OE.
Finally, draw B^R parallel to OJ.
The theory of the construction is just as before : in B^B we
suppose two equal and opposite forces OA and AO ; the former
we combine with AB to a resultant OB in B^B^; OB is combined
with BC to a resultant OC in B^B^; 00 is combined with CD to
a resultant OD in B^B^; OD is combined with DE to a resultant
OE in B^B^; and, finally, OE is combined with EA to a resultant
OA in B^B.
The given set of forces has thus been replaced by a force AO
or - OA in Bj^B and AO in B^B, i.e. by a couple.
VECTOR POLYGON CLOSED.
187
Measure the perpendicular distance between R^R and R^fl in
inches, and multiply the result by the number of lbs. represented
by OA (either graphically or by actual multiplication of numbers).
Pio. 174.
The product is the momental area of the couple in lbs. and
inches. Notice that the sense of the couple is contraclockwise
and therefore the sign of the momental area is positive.
188 GRAPHICS.
(21) With the same vector polygon and axes, take a new pole Oj outside
the vector polygon and shew that the momental area of the couple obtained
is the same in magnitude and sign to that obtained previously.
(22) Draw four lines at distances apart of 0'5, 1 and 1 -5 inches, and
suppose parallel forces of 2-3, 3'7, -1'8 and -4'2 lbs. weight to act in
them. Go through the process of finding the resultant and shew that the
given set of forces is equivalent to a couple, and find its momental area.
Closed Vector Polygon and Couples. Since the pole
may be taken in any position, OA may have any magnitude and
direction, and i?j may be any point on ab ; hence the couple
equivalent to the given set of forces may have any position in
the plane and the forces constituting it may have any magnitude
and direction. All the couples found by changing and ^j are
therefore equivalent, and the connection between the couples is
that they all have the same momental area.
Momental Areas are Vector Quantities. Since a
momental area has no definite position in space, but is fixed
when its magnitude, direction (or aspect of its plane) and sense
are given, momental areas are vector quantities.
For coplanar forces the momental areas are all in one plane,
and hence they are added by adding their magnitudes alge-
braically.
Addition of Momental Areas. Example. To find a couple
equivalent to three couples having the same sense.
The magnitudes of the six forces of the three couple aire given hy
lines of length 10'7, 8'6.5 wnd 12'8 cms. {to a scale of 1 kilogrm. weight
to an inch) the perpendicular distance between the farces constituting the
couples are 12'5, 10'8 and 6'25 cms. The senses of the couples are all
clockwise.
Mark off along any line on a sheet of squared paper (Fig. 175)
0^ = 12-5, OB =10-8, 0C= 6-25 cms.,
and on a perpendicular line through
0^1 = 10-7, 0^1 = 8-65 and 0^1 = 12-8 cms.
On the former mark off 0?7= 10 cms.
Draw AA2 parallel to A^U, BB^ parallel to B-^U, and CC^
parallel to G^U, cutting the force axis in A^, B^ and C^. Add
by the strip method OG^ + OB^ + OA^ and scale this with the
MOMENTAL AREAS.
189
tenth of an inch scale. It is the momental area of the resultant
couple in kilogrms.-cms. (viz. - 121 approximately).
Proof. A couple may be supposed to occupy any position in
the plane, hence all the couples may be supposed placed so that
one force of each lies along OA-^, the other forces will then be
parallel to OA^ and pass through A, B and G respectively.
A=
s.
c,
\
s
\
A,
\
\
b:
S
\,
\
>
s
\,
\
<-=
^
s
s
\
\
\,
\
\
s
\.
\
s
\
\
\,
\
\
\,
\
\
\,
\
\
s^
c
Pro. 175.
U B
Further, a couple may be replaced by any other of equal
momental area, hence the couple of force OAj and arm OA
may be replaced by one of force OAg and arm OU. Similarly,
the others may be replaced by forces OB^ and OCg and arm U.
The construction is simply our old construction (p. 46) for
reducing an area to unit base (in this case 10 unit base).
Finally, we have forces given by OA^, OB, and OCj along OA-^,
and parallel forces of opposite sense through U, and the three
couples have been replaced by one of force given by
OA2+OB2+OC2 and arm OU (10 cms.).
190
GRAPHICS.
Force.
Arm.
23-6
2-84
7-9
4-65
15-4
2-26
10-8
1-92
Should all the couples not have the same sense, the distances
OA-^, OB^, ... must be set off from with their proper senses and
the corresponding subtraction made by the strip method.
(23) Couples having positive momental areas are
given by the annexed table ; find the resultant couple
(i) geometrically by reducing each couple to
forces distant apart 1",
(ii) algebraically by adding the momental areas.
(24) The couple given in the first and last lines of the above table are
negative. Knd the resultant couple and its momental area in lb. inches.
Vector and Link Polygons Closed. Refer back to Fig.
174 on p. 187. Imagine B^Ii^ produced to cut B^B in B^, then
if ef be supposed moved parallel to itself to cut B^B in B^, B^R
would be the same line as B^B, and hence the forces OA and AO
would cancel and there would be equilibrium.
Example. Parallel forces act in the lines and have magnitudes and
senses as indicated in Fig. 176. To show graphically that the forces
are in equilibrium {approximately).*
Draw the vector polygon starting with the downward forces
BC, CD, DE, then EF and FB upward-s (Fig. 176). The upward
force AB is the same as FB. Hence A and F are coincident,
and a and / must be considered the same space. The vector
polygon is closed. Choose a convenient pole.
Draw through the space a P^P^ parallel to OA.
c P,P,
d P^P,
OB.
00.
OD.
OE.
OA.
e P^t^i „
f a line „
The construction (if properly done) gives the first and last
lines identical, viz. P^P^ and P^P^. But in P^P^ acts the force
* Graphical work will not as a rule give results correct to more than 3 figures,
the numbers given in the example are correct to 1 in 1200.
LINK POLYGON CLOSED.
191
whose vector is AO, and in P^P^ the force whose vector is - AO.
Two such forces in the same axis must be in equilibrium, and
therefore the whole set of forces is in equilibrium. The
cancelling of the forces in P5P4 is due to the fact that the first
and last lines of the link polygon are coincident, i.e. the link
polygon forms, like the vector polygon, a closed figure.
A--
E-l-
+
Fig. 176.
If both the vector and the link polygons for a set of forces
are closed, the forces are in equilibrium.
Proof. The general proof of this theorem is seen easily from
the construction when the vector polygon is closed.
192
GRAPHICS.
Let JBCDEA (Fig. 177) be the closed vector polygon for certain
forces. Then if be the pole, the first link for the link polygon
is parallel to OA, and has,
finally, a force whose vector is
AO acting in it. The last link
is also parallel to OA, and has
a force whose vector is -AO
acting in it. These form a
couple (in general), but if the
first and last links coincide,
the forces whose vectors are
AO and OA cancel, and the
whole set must be in equili-
brium.
Via. m.
ExPT. IX. Punch four holes in an irregular shaped piece of oardhoard,
and suspend it in front of a drawing board, as in Expt. VI., p. 122.
Mark on the card the lines of actions of the forces, and their magnitudes
and senses. Remove the card, and draw the vector and link polygons.
Both will be found closed; or as nearly closed as one can expect from
the errors incidental to the experiment.
Perform an experiment similar to IX., with five forces.
Expt. X. Draw a closed four-sided vector polygon on cardboard, the
sides being of such lengths that they represent to scale obtainable
weights. Draw on the card four non- concurrent lines parallel to these.
Fix the card to the drawing board by two pins. Adjust the position of
the pulleys so that the corresponding weights may pull on the card along
these hnes. Remove the pins and see if the card moves.
Expt. XI. Draw on stiff cardboard a closed four-sided vector polygon
ABGDA (Fig. 178), the sides being of convenient lengths to represent to
scale obtainable weights.
Draw three non-concurrent lines ah, he, cd parallel to the corresponding
vectors. Take a pole O inside the vector polygon (say the point of inter-
section of the diagonals).
Mark any point S^ on ctb, and through it draw R^R^ parallel to OA ;
draw SiR^ parallel to OB, and cutting be in B^ ; draw R^R^ parallel
to 00, cutting cd in R^; then draw R^R^ parallel to OD, cutting RiRi in
ifj. Through R^ draw the axis ad parallel to AD.
Then R^R^ cuts ad at R^, and on drawing through Ri a line parallel to
OA we come to RiR^ again. Hence the first link R-iRi, in which lies the
force given by AO, coincides with the last link RiRi, in which lies the
force given by OA. Hence both the vector and the link polygons are
closed. Fix the card (with the axes of the forces marked on it) on the
EXPERIMENTS.
193
drawing board by two stout drawing pins, and adjust the pulleys so that
the threads (with their proper weights attached) lie over the axes. Remove
the pins, and see that the card doesjiot move.
Devise an experiment for shewing that couples of equal momental areas
are equivalent.
R4
Fig. ire.
Expt. X. shews that in general there is not equilibrium when
the vector polygon only is closed.
Expt. XI. shews that there is equilibrium when the link and
vector polygons are both closed, and Expt. IX. shews the eon-
verse.
194
GRAPHICS.
Determination of Reactions.
Example. A locomotive has three pairs of driving, one pair of
leading, and one pair of trailing wheels, ami, is stopping on a short
bridge of 40 ft. span. The centre of the leading wheels is 6' 8" from
one end of the bridge (the left in Fig. 179) and the distance between
the centres of the wheels are, from the leading to the trailing wheels,
8' 9", T, 7' 9" and 8' 3". The load each pair of wheels carries is, in
the same order, 9 tons, 17 tons 13 cwts., 18 tans 4 ewts., 18 tons
4 cwts. and 1 1 tons 9 cwts. Determine the reactions of the supports
(supposed vertical).
Draw the position diagram (Fig. 179) to scale, say 1 cm. to 20
inches, with the reaction and load lines, and letter the spaces
0, a, b, c, d, e, f, o referring to the spaces outside the reaction
lines. Draw next the load vectors, say to the scale 0-1 inch
to 1 ton, so that AB is of length 0-9", BO of length 1 -765 inches
(i.e. nearly 1-77 inches), etc. Choose some convenient pole P and
draw through the space a a line parallel to PA cutting the
reaction line oam R^; through the space b draw a link parallel to
PB, and so on to the link through the space / parallel to PF cut-
ting the reaction line fo in R^. If the reactions in oa and of to
maintain equilibrium had been known, the first line of the link
polygon would have been through R^ and the last through R^,
REACTIONS BY LINK POLYGON. 195
and these would have been coincident and therefore would have
been the line R^R^ itself.
Hence join R-Jl2y *•*• close the link polygon, and through P
draw a line parallel to this closing line cutting the load vectors
in 0; then FO is the reaction in fo and OA that in oa, and these
are the forces necessary to maintain equilibrium.
Three cases have now been considered :
(i) If a set of forces acts on a body, and the vector polygon
is not closed, there is a resultant force whose vector is the sum
of the given vectors and whose line of' action is determined by
the link polygon.
(ii) If the forces have a closed vector polygon they are (in
general) equivalent to a couple whose momental area can be
found from the link polygon.
(iii) If the forces have both the vector and link polygons
closed, the body is in equilibrium.
Incidentally (ii) shewed that all couples which have the same
momental area are equivalent.
The third case enables an unknown reaction (or reactions)
which keeps a body in equilibrium when under the action of
known forces to be found.
(25) A horizontal beam 17 ft. long is loaded with weights distributed as
in the Table. The beam being supported on knife edges* at its ends, to
find the reaction of these knife edges (neglecting the weight of the beam
itself).
Weight in tons,
3-2
3-5
21
1-9
Distance from left end in feet.
4-2
7-3
8-6
12-3
(26) A horizontal beam is supported on knife edges at its ends ; the
length of the beam is 60 ft. and at distances 7, 20, 25, 32, 40 and 49 ft.
from the left-hand end are hung weights of 3, 10, 8, 7, 12 and 6 cwts. ;
find the reactions of the supports.
(27) A beam loaded as in Ex. 26 is supported at the left end and at a
point distant 36 ft. from it ; find the reactions due to the loads.
* The knife edge, shaped like A, is simply to ensure that the end is f eely
supported at one point only or (taking in,to. account the breadth of the beam)
on a line perpendicular to the length of the beam.
196 GRAPHICS.
(23) The beam, loaded as before, is freely supported at a distance of
30 ft. from the left end ; can it be supported in equilibrium at the left
end, and what is the reaction (given by the polygons) there ?
(29) The centre lines of the wheels of a locomotive and tender are from
the leading wheel backwards 12', 9', 10' 3-25", 6' 10-5" and 6' 10-5" apart ;
the loads on the wheels are 20 tons 14 owts., 19 tons 11 owts., 19 tons
11 owts., 14 tons 5 owts., 14 tons 5 cwts. and 14 tons 8-5 cwts. The engine
and tender are stopping on a bridge of 60 ft. span and the leading wheel
is 7 ft. from one support of the bridge ; find the reactions (supposed
vertical) of the supports of the bridge.
Non-Parallel Reactions.
Example. A beam is pin-jointed to a suppm'ting wall. Its
length is 25 feet and it is swpp&rted at the other end by a chain of
length 37 ft. attached to a wall hook 21 ft. vertically above the joint.
Weights of 54, 58-5, amd 45 lbs. are hung from it at points distant
{along the beam) 7, 12, and 20 ft. from the pin. Find the tension of
the chain and the reaction of the pin.
Draw first to scale the position of hook, pin-joint and chain
Z, X and YZ (Fig. 180).
Then draw the axes ab, be and cd of the forces; then the vectors
of the forces AB, BC, CD acting in ab, be, cd, and, finally, the
link polygon corresponding to any pole 0. The resultant of the
loads is thus obtained and its axis passes through E.
In Fig. 180 the line ZX gives the vertical. To obtain the conventional
position, the book must be turned round.
Find the point P where this axis cuts the chain ZY. Join
XP, and in the vector polygon draw AE and DE parallel to
XP and PY respectively.
Then DE gives the tension in ZY (why tension and not
compression ?) and EA the reaction of the hinge.
Since the force whose vector is AD and axis PB is equivalent
to the given three loads, the beam may be supposed to be in
equilibrium under the action of this force, the tension in the
chain and the reaction at the pin.
These forces must pass through a point, viz. P, and then the
necessary and sufiicient condition for their equilibrium is that
their vector polygon should be closed. Hence DE must give the
force along YZ, and EA that along XP.
REACTIONS BY LmK tOLYGON.
197
Find the vertical reactions through X and F that would be in
equilibrium with the given loads in ab, be and cd. Project E
horizontally to £j on AB and see that DE, and EjA give these
vertical reactions.
Via. isa
(30) Produce the first and last links to out XP and PY, and hence shew
that the closing line of the link polygon is parallel to OB. Why is this
198
GRAPHICS.
(31) A horizontal beam 45 ft. long is pin-jointed to a supporting pier at
one end (the left), and rests on a smooth horizontal roUer at the other.
Forces of 12, 8, 20 and 15 owts. act downwards at points distant 6, 10, 28,
and 32 ft. from the pinned end and make angles of 30°, 45°, 90° and 60°
with the beam line from left to right. Find the reactions at the ends.
Note. The object of the roller is to make the direction of one reaction
known, viz. vertical ; if neither reaction be known in direction the
problem is indeterminate.
First draw the line of the beam and the axes of the known forces, add
the vectors of the forces, draw the link polygon and obtain the position
of the resultant force of the given set. Mark the point of intersection of
the vertical reaction and this resultant, and join this point to the pin.
The last line gives the direction of the reaction at the pin. The unknown
reaction may then be found from the vector polygon.
(32) Solve the problem in Ex. 31 by projecting the vectors on to a
vertical line and drawing the link polygon for these vertical components.
Determine thus the reaction of the roller and the vertical component of
the pin reaction. The latter combined with the reversed horizontal com-
ponent of the vectors gives the pin reaction.
(33) A swing gate is hinged at A (Fig. 181) to a post and rests against
a smooth iron plate &t B. AB = S'5 ft., GD = Q ft. and the gate weighs
200 lbs. Supposing the weight of the gate to act at the centre of the
rectangle CD, find the reactions at the hinge and plate. The distance
between AB and the gate is 3".
Fio. 181.
(34) A boy of weight 100 lbs. hangs on the gate at D. Find the total
reactions at A and B.
(35) The post AB is not vertical, but inclined at an angle of 15° to the
vertical, so that CD slopes (i) downwards, (ii) upwards. Find the reaction
when the boy is on the gate in the two cases, if AC=BE=9'.
(36) In a wall crane ABG the beam BO is loaded at equal distances as
in Fig. 182. Find the tension in the tie rod A C and the reaction at A and
B. (The beam BO and the tie rod ^G are pin-jointed at A, B and C.)
(37) AB is a uniform beam hinged at A (Fig. 183) and weighing 1 -6 cwts.
It rests on a smooth fixed cylinder D, and a load of 0-7 cwt. is suspended
from B. liAE is horizontal and if AB=W, AO=T and DE=2', &id the
reactions at A and O.
DECOMPOSITION OF FORCES.
199
(38) A uniform ladder rests against a smooth vertical wall at an angle
of 27° with the vertical. The weight of the ladder is 69 lbs. and may be
supposed to act at its mid-point. Mnd the reaction of the ground when a
man weighing 12 stones and a boy weighing 7 stones are § and J up the
ladder respectively.
Tft-
0-1 ton
0-21 ton 0-363 ton O-t-TSton OSS ton
Fig. 182.
Decomposition of Forces. Any force may be decomposed
along any two given axes if they intersect on the axis of the
force (see p. 151). Any force may be decomposed into two
forces parallel to it, having axes in assigned positions.
Draw any line be (Fig. 184) as the axis of the force, and BC
its vector. Draw two lines parallel to be, viz. ab and ca, on
200
GRAPHICS.
opposite sides of he. Choose any pole 0, and draw through any
point P of Ic, PP^ and PP^ parallel to OC and OB respectively.
Join P-^Pi, and in the vector polygon draw OA parallel to P^P^.
Then BA and AO are the vectors of the required components.
FlQ. 184.
Proof. At P, BC may be decomposed into two, BO and 00,
acting in PP^ and PP^. At P^, 00 may be decomposed into two,
OA in PjPg ^^^ -A-O in o£- At Pj, BO may be decomposed into
two, BA in ah and AO in P^P^- OA and AO in P-^P^ cancel, and
we are left with BA in ah and AO in ca.
Shortly put, the construction is that for finding the reaction
in ah and ca which will be in equilibrium with the given force
in he. These reactions will be the same in magnitude but of
opposite sense to the components.
(39) Choose two other poles (one on the side of BO opposite to O) and
see that the construction gives the same components.
(40) D .compose a force of given axis and vector into two parallel axes,
both axes being on the same side of the force.
(41) Find graphically the components of a force which pass through
given points, one direction being fixed.
PARALLEL COMPOiSTENTS.
201
Example. Find (he components, passing through two given points
of a force when one of the components has the least possible i
Let xy (Fig. 185) be the axis and XY the vector of the force,
and suppose A and 5 to be the given points. Join any point
P on a^y to A and B.
Through X and Y draw XO and YO parallel to PA and PB
respectively. Through 0, the point of intersection, draw OZ
parallel to AB, cutting XY in Z.
If the component through B is to be a minimum, draw YT
perpendicular to 02 and join XT; then XT and TY are the
required components.
Proof. To find components through A and B parallel to xy,
we may, instead of taking any pole for the vector polygon,
first draw from any point P in xy, PA and PB, and then find
the corresponding pole 0. The closing line of the link polygon
is AB ; and hence, on drawing OZ parallel to AB, we get the
reactions YZ and ZX at B and A in equilibrium with XY in xy.
These reactions must evidently be independent of the pole
used to find them, i.e. Z is a. fixed point on XY. If any other
202 GRAtHICS.
point P on xy be chosen, then the new pole must be on ZO, for
Z is fixed, and ZO is a fixed direction parallel to AB.
Further, XO and YO are two components of XY through
A and B, hence the smallest component through B will be such
that it is perpendicular to ZO.
Note that since ZO is a fixed line, a simple construction will
give the components through A and B which have any desired
relation, say that of equality, or the B component twice the
A component, etc.
(42) Solve graphically the example on p. 201 when A and B are on the
same side of xy.
(43) Find components of a given force such that one has an assigned
direction and the other is to be aa small as possible.
(44) Decompose a given force into two forces equal in magnitude passing
forces through points A and B when A and B are (i) on the same,
(ii) on opposite sides of the given force. (See also Chap. IV., p. 154.)
When does the construction fail ?
(45) Decompose a given force into two passing through two given points,
the magnitudes of the forces having the ratio of a to h.
(46) A man carries a pole across his shoulder at an angle of 25° with the
horizontal. The pole is of length 15 f t. , and the distance of the mid-point
of the pole from his shoulder is 5 ft. He keeps the pole in position by
hard pressure on the front end. In what direction should this pressure be
applied so that it may be as small as possible? What direction would
make the pressure on his shoulder as small as possible? (Assume that
there is sulHcient friction at the shoulder to prevent the pole sliding.)
Find the pressures in the two oases.
Any force may be decomposed into three forces lying in non-
concurrent and non-parallel axes.
Draw any straight line ah (Fig. 186) for the axis and a parallel
line AB for the vector of a force. Draw any three non-con-
current lines he, cd and da forming a triangle XYZ. •
Suppose ab cuts XY in P. Then, at P, AB may be decomposed
into two, AC and CB, acting along PZ and PF. At Z, AC may
be decomposed into two,- AD and DC, acting along XZ and YZ.
Hence, AB has components AD, DC and CB having ad, dc and
cb as axes.
Evidently unless ab is parallel to one of the sides of the triangle
XYZ, we may take as the starting point for the decomposition
THREE NON-CONCURHENT COMPONENTS.
203
any of the three points in which ah intersects the sides. If the
decomposition is unique, the components determined in the three
ways should be the same.
The proof that the decomposition is unique will be found in
the Chapter on Moments (p. 297).
Pio. 186.
(47) Start the decomposition at (i) Q, (ii) R, the points of intersection of
ah with TZ and XZ, and shew that the same components are obtained.
(48) Draw any four non-ooneurrent lines. Assign any value to the force
in one, and find the forces in the other three so that the four forces may be
in equilibrium.
204
GRAPttlOS.
(49) A weight of IQ
tons is suspended from
a crane ABGD (Fig. 187)
at A. Find the com-
ponents along BG, CD
and DB ; find also the
vertical components of
W through B and G,
and shew that the com-
ponent along CD is the
same as that found by
the first method.
Also, resolve the B
component along BG
and BD and compare
with previous results.
(50) AB, BG, CA (Fig. 188) are three light rods pin-jointed together and
supported at B and O in a horizontal line. Find graphically the com-
ponents along the rods due to a load of 1 owt. at D. Find also the
loads at A and G equivalent to
that at D, and hence find the ^
components along the rods. Com-
pare the two sets of results.
(51) ABG (Fig. 189) is a waU
crane, find the components in A B,
BG and GA due to a load of 1 ton
applied at D ; (i) by resolving W
along AD arid GD and (ii) by
finding the parallel components
of W through A and B. ,
Fio. 189.
THE TOGGLE JOINT.
205
(52) In Exercise 51 decompose the load at D into equivalent loads
at A and B and find the components of the latter along BG and BA,
Compare with the previous results.
A
*(53) ^fi and 40 (Fig. 190) are rafters of
a roof ; find the total thrust on the ytaWs
due to loads of 10 tons at the mid-point of
each rafter. (Eesolve the load at M in
directions MB and MC, and at O resolve
the latter along GB and GA. The com-
ponent along GB gives the outward thrust.
B/
Pig. 190.
Example. {The Toggle Joint.) AB isa beam hinged at A to
fixed masonry. CD is a bar pinrjointed to C {in AB) amd to D.
D is constrained to move along AD ly means of smooth guides. A
force P is applied at B perpendicular to AB, and a force Q ai D along
AD so that there is equilibrium. AB = 3-58", BC = 1-64", CD = 2-7"
and DA = 3-9". P = 1 lbs. weight. Find the components of P along
AC, CD and DA.
Fig. 191.
The beam AB (Fig. 191) is in equilibrium Uiider P, a force
along CD, and the reaction at A. The^ must be concurrent;
find the point of concurrence, and hence, in the vector polygon,
find the reaction at A and shew that this reaction is the equi-
librant of the forces at A found by resolving P along AC and
AD and CD.
Find the value of Q and the reaction of the guides on the
slide at D.
206
GRAPHICS.
(54) Deoompoae P into parallel forces at G and A. Find the components
of the former along CD and CA, and the resultant of the latter and the
force along GA. Compare with the former result.
(55) AB (Fig. 192) represents an open French window (plan or trace of
on a horizontal plane). It is kept in position by a bar CZ5, freely jointed
at G, The bar has a number of holes in it, any one of which can be fitted
over a peg at D so that the angle BAD may have any value from to 120°.
The wind is blowing parallel to
AD and would exert a force of
30 lbs. weight ii AB were perpen-
dicular to AD. Suppose the re-
sultant force of the wind on the
door to act at the mid-point M oi M,,
AB.
Find the pull in GD, given that
AB='2.'r, AD=rS-5"=AM, and
BG= 1 ft. when
.(a) GAD
(6) GAD
(c) GAD= 90'
{d) CAD = im
20°,
45°,
GENERAL ANALYTICAL THEORY OF THE COMPOSITION
OF COPLANAR FORCES.
^Theorem. Any force is
eoLuivaleut to a force through
any assigned point together
with a couple, if the forces
have the same vector.
If a (Fig. 193) is the axis
and a the vector of the given
force, and any assigned
point, draw through the
axis aj parallel to a.
Then we may suppose at
in dj forces u, and -a,
i.e. at we have a force a
which together with the couple a in a and -ain a^ are equivalent
to a in a. The cmple is called the couple of transference.
Pro. 193.
COMPOSITION OP rORCES.
207
"^ Theorem. Any set of coplanar forces is eoLuivalent to a
resultant force through some assigned point and a couple.
By the previous theorem each force of the set may be replaced
by an equal vectored force through and a couple. The forces,
being now concurrent, have in general a resultant found by the
vector polygon. The momental areas of the couples may be
added to the momental area of a resultant couple, i.e. the couples
are equivalent to a resultant couple.
"^Theorem. Any set of forces reduces to
(i) a single resultant, or
(ii) a couple, or
(iii) is in eciuilibrium.
Since a couple may have any position in its plane and may
have its forces of any magnitude, provided the momental area is
constant, we may replace the
resultant couple of the last
theorem by an equivalent couple
having its forces "■ (Fig. 194)
and - (T, where <r is the resultant
of the concurrent forces at 0.
If the arm of the couple is p,
and S is the magnitude o-, then
M being the known momental
area.
If, now, the couple be sup-
posed placed so that its force
- 0- passes through and is in a line with the resultant force a-
there, then a- and - cr cancel, and we have a single resultant
cr at a distance p from 0.
Should the resultant of the concurrent forces at be zero, the
set of forces reduces to the couple of momental area M.
If M is also zero, there is equilibrium.
Via. 194.
208 GRAPHICS.
These theorems are only what we had before as direct
deductions from the geometrical constrtictions. The actual
determination of the resultant, or the resultant couple, should
be effected by the link polygon construction.
MISCELLANEOUS EXAMPLES. V.
1. Draw a triangle ABG, having an angle of 45° at B and one of 30° at
C. Let forces of 9, 7 and 4 units act from A to B, B to G and G to ^
respectively. By construction or otherwise, find their resultant com-
pletely, and shew it in the same diagram as the triangle.
(B. of E., Stage II.)
2. Draw a square A BCD and a diagonal AG ; forces of 1, 2, 3, 4 units
act from A to B, from B to G, from G to D and i> to ^ respectively ;
find the sum of their components along AC, and also the sum of their
components at right angles to AG. (B. of B., Stage II.)
3. Draw a triangle ABG, and take D and E the middle points of BC
and GA respectively ; if forces P, Q, R act from A to B, A to C and
G to B respectively, and ate proportional to the lengths of the sides along
which they act, shew that their resultant acts from E to D, and is equsS
to2P. (B. of E., Stage II.)
4. Define a coilple. Explain how to find the resultant of two forces
which form a couple and a third force.
Draw a square A BCD; a force of eight units acts from A to B and
G to D respectively ; find the resultant. Also find what the resultant
would be if the first force acted from D tx> A. (B. of E. , Stage II. )
5. A horizontal beam 20 ft. long is supported at its ends, loads of
3, 2, 5 and 4 cwts. act at distances 3, 7, 12, 15 ft. from one end. Find,
by means of a funicular (link) polygon, the pressures on the two ends.
(Inter. Sci. (Eng.), 1904.)
6. Find the resultant of two parallel forces by a graphical construction.
Extend this to find the resultant of three or four parallel forces.
(Inter. Sci., 1902.)
7. Find the resultant of three parallel like forces of 2, 4 and 3 lbs.
weight acting through points in a straight line, distant 1, 3 and 7 ft.
from an origin in that line. (Inter. Sci. (Eng.), 1905.)
8. Draw a triangle ^BO, such that ^5= 10 cms., 50= 14 and 0.11=12;
take B' in AG, such that AB' = Z; and A' in BG, such that BA'=%. A
force of 20 lbs. weight acts in B'A'; shew how to replace this force by three
forces acting along the sides of a triangle by simple drawing, without using
any of the numerical data concerning lengths. (Inter. Sci. (Eng.), 1906.)
9. ABG is a right-angled triangle, AB=12 and BG=5. Forces of
52, 24 and 27 lbs. weight act from A to G, B to A and G to B. Find the
resultant of the forces and exhibit its line of action. (Inter. Sci., 1900.)
MISCELLANEOUS EXAMPLES. 209
10. ABO is an equilateral triangle, P is the foot of the perpendicular
from O on AB. Find in magnitude and line of action the resultant of
forces : 10 from ^ to fi, 8 from B to G, 12 from A to C and 6 from G to P.
(B.So., 1905.)
11. A locomotive on a bridge of 40 ft. span has the centre line of its
leading wheels at a distance of 11 ft. from one abutment, the distance
between the centre lines of the wheels are, from the leading wheel back-
wards towards the far abutment, 9' 10", 6' 8" and 6' 8". Find the pres-
sures on the abutments if the loads on the wheels be 15 tons 10 cwts.,
17 tons 10 cwts., 17 tons 10 cwts. and 16 tons 10 cwts.
12. State and prove the rule for finding the resultant of two unlike
parallel forces.
Given a force of six units, shew how to resolve it into two unlike parallel
forces, of which the greater is ten units ; and explain whether the resolu-
tion can be made in more ways than one. (B. of E. , II. , 1904. )
13. Let a horizontal line AC represent a rod 12 ft. long, resting on two
fixed points A and B, 10 ft. apart. Each foot of the length of the rod
weighs 12 ozs. ; a weight of 16 lbs. is hung from C. Shew that the rod will
stay at rest, and find the pressure at each of the points of support.
(B. of E., L, 1904.)
14. ABG is an equilateral triangle, and forces P, Pj and 2P act from
B to 0, C to A and A to B respectively. Find their resultant, and shew,
in a carefully-drawn diagram, its direction and line of action.
(B. ofE.,L, 1903.)
15. Forces P, Q and 3R act in order along the sides BO, OA, AB oi a
given triangle. If P, Q, R are proportional to the sides respectively, find
completely the force which would balance the three given forces, and shew
your result in a carefully-drawn diagram. (B. of E., II., 1907.)
T.G.
CHAPTER VI.
STRESS DIAGRAMS.
In Chapter IV. the stresses in simple frames, due to loads applied
at the joints, were considered. The present chapter is a con-
tinuation of the subject. It is shewn that the vector polygon
for the forces acting at a point is the stress diagram for the
bars meeting there, and that the space notation enables us to
draw in one figure — the stress diagram — lines giving the stresses
in all the bars of more complicated frames.
Three Bar Equilateral Frame.
Example. Three bars, each 3 ft. long, are pin-jointed together to
form an equilateral triangle. The frarhe is suspended ly one vertex,
and 5-5 lb. weights are hrnig from the others ; to determine the stresses in
the bars due to the weights.
Draw the frame PQB (Fig. 195) to scale (say 1" to 1') and letter
the spaces as indicated.
Draw the vectors of the two external forces at B and Q (scale
say 2 cms. to 1 lb. weight); then, without drawing the link
polygon, notice that the upward reaction CA at P (=11 lbs.
weight) will keep the frame in equilibrium.
Draw AD and BD parallel to ad, bd, then CD is parallel to cd ;
and AD, DB, and DC measure the magnitudes of the stresses in
the corresponding bars. Scale these lines and tabulate the
Bar,
ad
bd
cd
Stress in lbs. wt. ,
6-35
318
6-35
At the point P of the frame act three forces, viz. the reaction
CA vertically upwards, and the pushes or pulls of the bars ad and
cd. Hence CADG is the vector polygon for P. The sense of
the force at P due to cd is given by DC, and hence the bar must
THREE BAR EQUILATERAL FRAME.
211
Fig, 195,
pull at P. Similarly,
the sense of the force
in ad is given by AD,
and hence this bar
also pulls at P.
Now consider the
equilibrium of B.
The forces there are
the weight of 5'5 lbs.
(AB) downwards and
the forces due to ad
and hd. The vector
polygon is ABDA,
and, sintie AB is
downwards, the force
in bd is given by BD
and pushes at B;
similarly, DA pulls at B.
Thus, in the vector polygon the
line AD gives the pull at P or at
B according to the sense it is taken
in. This is as it should be, for the
bar ad is in equilibrium and must
be pulled with equal and opposite
forces at its ends. The line AD
thus gives the stress in ad, and the
figure ABGD is called now, not
the vector polygon, but the stress
diagram of the frame PQB and the
forces acting on it.
Finally, consider the point Q. It
is in equilibrium under BO, and the
forces along cd and dl ; the corre-
sponding lines in the stress diagram,
have been already drawn, and it
212 GRAPHICS.
only remains to determine the senses of the forces at Q. The
vector polygon for Q is BCDB, hence a force in cd (CD) pulls at
Q whilst DB pushes. Hence Id pushes at its two ends, Q and R,
and is therefore in compression, whilst PQ and RP pull at both
ends, and are therefore in tension. As vectors, therefore, the
lines AD, DC and DB should have double arrow heads; this
may easily lead to confusion, and it is, therefore, better to avoid
them altogether and to indicate in the frame those bars which
are in compression by drawing iine lines parallel to them. Mark
the bar hd as being in compression.
Change of Shape in Frames under Forces. The simple
triangular frame, just considered, was treated as a rigid body;
this was justifiable, since, although some bars may elongate a
little and others contract, yet the bars will always adjust their
positions to form a closed triangle, and when once the deformation
has taken place, the parts retain their relative positions unaltered.
That is, for the given forces the frame— after the elongations, etc.,
have taken place — is like a rigid body. In nearly all prac-
tical cases the elongations, etc., are so small that the frame may,
so far as change of form is concerned, be regarded as unaltered.
Only frames which have just a sufficient number of bars or
strings to keep them rigid under the given applied forces will
be considered, and of these only simple cases which can be solved
diredtly by vector polygons will be taken.
The weight of the bars will be neglected unless expressly
included in the problem.
Example. Three bars, PQ, QE. and EP, of lengths 4, 7, and
6 ft, are freely pinpointed together to form a triangular frame and
the frame is suspended by P. Weights of 4 and 6 lbs. are
suspended from Q and E. Draw the frame in its position of equi-
librium and find the stresses in the bars.
The resultant of the weights 4 and 6 at Q and R (Fig. 196) must
OS
act through a point .S' such that |p = | ; hence, by construction,
find the position of S in QB,
FRAMES AS RIGID BODIES.
21S
Then, if FS be regarded as vertical, the angles the sides make
with the vertical or horizontal can be measured. (If the con-
ventional position be desired, the length PS must be set off
vertically and then the triangle drawn in.)
214 GRAPHICS.
Letter the spaces as in Fig. 196 and draw the vector polygon
ABD for Q.
Draw next the vectors for the forces at B ; the vector triangle
is DBG where DG is parallel to dc. ADC is then the vector
triangle for the point P. See that GA gives 10 lbs., the reaction
at P, and that the senses of the forces at the joints are consistent
with one another.
Measure the stresses and see which bars are in compression
and which in tension, and tabulate the results. Measure the
angle QR makes with the vertical.
(1) Determine the position of QR (Fig. 196) and the stresses in the
bars if the weights at Q and B are 7 and 4 lbs.
(2) An equilateral framework PQE of three bars is suspended by means
of two vertical strings attached to P and Q so that PQ is horizontal. A
load of 17 lbs. is suspended from R. Find the stresses in the bars of the
frame, stating which are in compression and which in tension.
(3) The equilateral frame of Ex. 2 being suspended from a hook at P
and a vertical string at Q, so that PQ makes 15° with the horizontal,
determine, by the link polygon, the reaction at P, the tension in the
string, and find the stresses in the bars. Determine also the reactions at
P and Q from the stress diagram.
Braced Quadrilateral Frame.
Example. Draw to scale the frame PQRS (Fig. 197), mppmted
at Q and E in the same horizontal line, given that PQ = 1 '5, PS = 2,
A
QE = 3-7 metres, and PS is parallel to QR and PQR = 60°. A load of
121 lbs. is placed at P ; determine the stresses in the bars due to this load.
Letter the spaces as indicated and draw in the stress diagram
AB=\'i\ cms.; then BD and AD are parallel to hd and ad.
ABDA is the vector polygon for P and gives the stresses in
the bars hd and ad. BD evidently pushes at P, so that, since
bd is in equilibrium, it must push at both ends and be in com-
pression. Similarly, DA pushes at P, and hence ad must also
be in compression. At iS^ we know the force AD (pushing at S),
hence we can find the stresses in dc and ca which will give
equilibrium. Draw DC and AG, parallel to dc and ac, intersecting
at G; then ADGA is the vector polygon for S, in which we know
the sense AD of the force in ad at 8.
BRACED QUADRILATERAL FRAME.
215
Hence, the force DO
pulls, and the force CA
pushes at S, and the bars
dc and ca are consequently
in tension and compression
respectively.
At E, AO pushes and
the forces in oc and ao
must, therefore, be given
by AGOA where 00 is
parallel to co.
Hence, CO pulls at R,
and OA pushes upwards,
so that CO must be in
tension, and OA must be
the reaction at R.
Finally, at Q we have
CO pulling, CD pulling,
DB pushing and the reac-
tion, which must therefore,
be BO.
Pig. 197.
Notice that given the reaction BO, the senses of all the other
216
GRAPHICS.
forces necessary to produce equilibrium are consistent with the
senses previously obtained.
Scale all the lines in the stress diagram and tabulate the
stresses.
aa
ad
bd
dc
CO
Bar.
Stress in lbs. wt.
In the frame diagram mark those bars which are in compression.
The reactions BO and OA have been obtained, without drawing
the link polygon, on the supposition that there is equilibrium.
(4) Find the reactions in ob and oa due to the load in cib by the link
polygon, i.e. find the components of AB in two parallel lines through
Q and B, and compare with the previous results.
Cantilever.
Example. PQEST (Fig. 198) represents a cantilever pin-jomted
to a vertical wall at E and S. ES = 5, ST = 2-25, SP = 8-25,
PQ = 5-88, QE = 4-4/i!. The loads at T and P are 2300 and 3400
lbs. weight respectively. Find the stresses m the hwrs.
Draw the load vectors AB and BC. Then, since there is
equilibrium at P, draw BD and GD parallel to bd and cd. BGDB
gives cd in tension and db in compression.
For Q, draw GE and DE parallel to ce and de. Then, from
GDEC, we know that DO gives the sense of the force in dc on Q,
and hence DGED is the correct sense of the diagram, and ce is
in tension and ed is in compression.
Finally, for T draw AF and EF parallel to af and ef. We
may determine the senses of the forces at T either from knowing
that in ab or that in ed. The vector polygon is ABDEFA in the
sense given by the letters, and hence ef is in tension and fa in
compression.
Indicate on the frame figure the bars in compression, measure
the stresses from the stress diagram and tabulate the results.
Bar,
cd
bd
ce
ef
/«
Stress in lbs. wt..
Cantilever.
217
(5) Four equal bars are pin-jointed together to form a square and a
fifth bar is introduced diagonally. The frame is suspended by a vertex so
that the diagonal bar is horizontal, and the three remaining vertices are
loaded with 7 lbs. weight each. Find the stresses in all the bars.
(6) As in previous exercise only the diagonal bar is vertical.
3000 4oqo
*(7) If the weights at the two vertices, where three bars njeet, be 4 and
8 lbs., and a third vertex sustain 7 lbs., find the position of equilibrium
when the frame is suspended by the remaining vertex. Find also the
stresses in the bars. (Draw the frame with the diagonal bar, find the
M.c. of the three masses. Join the M.o. to the fourwi vertex, this line
relatively to the frame is the vertical line. The formal proof for the
Centre of Parallel Forces is on p. 298.)
*(8) Five bars of lengths r2, 1 2, I'B, r6 and 2 ft. respectively when
jointed together form a rectangle PQRS and one diagonal QS. The frame
is suspended by P and loaded at Q, B, S with weights of 7, 5 and 11 lbs.
Find the position of the frame and the stresses in the bars.
218 GRAPHICS.
(9) Three rods AB, BO, CA of lengths 7, 6-2 and 5-8 ft. respectively are
pin-jointed together. A is fixed and B rests on a smooth horizontal
plane so that ^ is 2 ft. above B, and a load of 70 lbs. weight is hung
from O. Determine the stresses in the bars and the reactions of the
supports on A and B.
S
(10) The frame PQRS (Fig. 199) is
loaded at Q and R with 100 lbs. weights
and supported at P and S. If PS= 15,
QR = 9, and PQ=8R=Q-Q, find the
stresses in the bars.
Bridge Girder.
Example. PQRSTUV (Fig. 200) represents a short N girder,
the bars forming right-angled isosceles triangles. It is freely supported
at P and T and loaded at each of the joints Q, E, and S with 1-5
tons. Determine the reactions at VMnd T and the stresses in the bars.
The loading being symmetrical, the reactions at P and T must
each be equal to 2'25 tons and there is no occasion to draw the
link polygon.
Letter the spaces as indicated and draw the vectors AB, BC
and CD of the loads; then, bisecting AD at 0, DO and OA
must be the reactions at T and P. At P the known force OA
acts and the forces in aj and oj ; these being in equilibrium
draw A J parallel to aj and OJ parallel to y ; then OAJ is the
vector triangle for P. The force AJ pushes P, and JO pulls,
hence aj is in compression and o? in tension.
At Q there are three bars with unknown forces, and as
resolution into three concurrent straight lines is not unique, we
must try some other point U. At U one known force OJ acts
and two unknowns ; draw, then, in the stress diagram 01 parallel
to oi and J I parallel to ji ; then OJI gives the stresses in the bars
meeting at U. Also OJ gives the sense of the force at U along
oj ; JI pushes at U and 10 pulls, hence ji is in compression and
io is in tension.
Now return to Q, where there are only two unknowns remain-
ing, and draw the vector polygon ABHIJA (most of it is already
BRIDGE GIRDER.
219
drawn, IB and BH being the only two lines necessary). See
from the sense of this polygon that hh and ji must be in
compression and ih in tension.
a
6
r \
c
d
\ J
N.
e /
S /'
/
J
\
IK
B
\
\
^
F }
c
D
E
/
Pig. 200.
Draw the rest of the stress diagram,, noticing, from the
symmetry of the loads and the frame, that the stress diagram
must also be symmetrical. In consequence of this symmetry the
points I and F of the stress diagram are coincident.
220 CtRAtHlCS.
However many bars the frame contains, the method of solution
always follows the same lines. The stress diagram is started by
drawing the vector triangle for three concurrent forces, of which
one is known and the directions of the other two are given by
bars in the frame. Sometimes this start can be made at once,
but more generally the reactions at the points of support have
first to be determined, either by drawing the link polygon for the
external forces or by taking moments. For the latter method
see Chap. VIII.
Roof Truss.
Example. The frame diagram shevm (Fig. 201) represents a
bowstring roof truss supported freely at P arid T in a horizontal line ;
PQ = 27', QR=19', PU = 19-4', EV = ll-4' and PT = 63', and the
ungle UPT = 44°. The loads at Q, R and S are 1-5, 2 and 2-5
tons; determine the reactions at the supports and the stresses in
Draw the frame to scale. Set off the vectors of the loads
AB, BC, CD, and number the spaces of the frame as indicated.
Choose any pole Oj and through Q^ any point on the vertical
QQ^, draw Q-^P-^ parallel to O^J, cutting the reaction line in
Pj. Then draw Q-^R^ parallel to O^^B, B^S^ parallel to OjC, and
S^T^ parallel to O^^D cutting the reaction line at T-^. Join Pil\,
thus closing the link polygon, and draw 0^0 in the vector poly-
gon parallel to P-^T-^. The reactions at P and T are, therefore,
OA and DO.
Now draw the stress diagram, starting with the vector
triangle for the forces at P. The vectors are OA, AJ and JO,
shewing that the bar aj is in compression and jo in tension.
At the point Q are three unknown forces ; and since the known
forces in aj and ab cannot be decomposed in three directions,
nothing can be done there at present. But at U we have only
two unknowns ; hence draw 01 parallel to oi and JI parallel to
ji to determine the point /; and the stresses in oi and ij are
found. The point Q may now be attacked, for there are now
only two unknown forces there.
ROOF TRUSS.
221
Fia. 201.
Complete the stress diagram, mark the bars which are in
compression, scale the lines in the stress diagrams and make a
table of the stresses as before.
222
GRAPHICS.
(11) The frame PQRST (Fig.
202) is loaded at T and S with
1 ton weights and supported at
P and B. Find the stresses in
the bars if
PT=TS=SR=9it.
and RPQ=3Q°.
(12) PQRST (Fig. 203) is a short Warren girder consisting of three
equilateral triangles. A load of 4 tons is suspended from R and the
girder is supported at Q and 8. Find all the stresses.
Fra. 203.
Fio. 204.
(13) PQRST (Fig. 204) is a short Warren girder of three equilateral
triangles supported at P and S, and with loads of 2 tons at Q and R.
Find the reactions of P and S, and the stresses in the bars.
Pio. 205.
(14) The figure PQRST (Fig. 205) represents a king post (roof) truss
supposed freely jointed at all the points and supported at P and Q. K
EXERCISES.
223
PQ =25 ft. and SPQ = 30°, find the stresses in all the bars when the loads
at T, S, R, are 3, 4 and 3 tons respectively.
(15) Find the stresses in the bars of the roof truss shewn in Fig. 206
when loaded at R, S, and T with 3, 4 and 2 tons respectively.
PT=T8=8-5.
PQ=29-5.
jPi7=10-8.
SU=7-8.
(16) Fig. 207 represents a queen post (roof) truss, supposed freely
jointed at all points, supported at P and /S', and loaded at T, U, V, W and
X, with 2, 2, 3, 2 and 2 tons. Find the stresses in all the bars if
PX = XW=WV and VPS=ZQI'.
Fig. 207.
(The figure shewn is not rigid for anything but symmetrical loads ; in
practice the stiffness of the joints prevents distortion when the loading is
not symmetrical. )
224
GRAPHICS.
(17) PQR8TU (Fig. 208)
represents a non-syrametrieal
king post truss loaded at Q,
8, T, U with 1, 2, 2, 1 tons
weight. Find, by the link
polygon, the reactions at the
points of support P and R,
and then determine the stresses
in the bars ; given that PR =19,
iJ7'=16-6, PTR=^\ and QU
parallel to RT, QS parallel to
PT.
Pig. 208
Q
R
S
\
X
/\
Fio, 209.
(18) Fig. 209 represents a short N girder consisting of right-angled
isosceles triangles. Find the stresses in the bars when supported at
P and T, and loaded at Q, R, S with 2, 3 and 4 tons weight.
Fra. 210.
(19) PQRSTUV (Fig. 210) represents a short Warren girder of five
equilateral triangles, loaded at Q and R with 7 and 10 tons. Find the
reactions at the points of support P and S, and the stresses in all the bars.
WEIGHT OF BARS.
225
(20) Find the stresses in the bars
of the cantilever in Fig. 211, loaded
at Q and R with 2 and 3 tons, and
supported by a vertical wall at P
and U; given
PU=8,
PT=P8=S%
PQ = 5,
Q8=4:%
PE= 8-8 ft.
(21) The queen post truss of Ex.
16, with a diagonal bar WB, is
loaded at T, U, V, rand X with
2, 3, 3, 1-5 and 1-5 tons; find all
the stresses.
Fib. 211.
Weight of Bars in a Framework. In many engineering
structures, the weight of the framework is very small compared
with the loads it has to carry, and in such cases no appreciable
error is made by neglecting the weights of the various parts.
If the weights of the bars are not small in comparison with
the loads, we shall suppose half the total weight of each bar
concentrated at its ends.
Every particle of a body is acted on by a vertical downward
force ; the resultant of all these parallel forces is a parallel force
through the M.C. of the body equal to the whole weight (see
Chap. VIII., p. 307). This resultant is merely the single force
which would produce the same motion as the actual forces on
the particles, and it by no means follows that the other effects
produced by it would be the same, e.g. as in bending the body
or in producing internal stresses. Consider a vertical uniform
column ; the resultant force is a force through the mid-point of
the column equal to the total weight. This force, if it acted at
the M.C., would not produce any stress at all in the upper half,
and a uniform stress for all sections in the lower half, whereas
226 GRAPHICS.
the actual stress must vary gradually from zero at the top to
the total weight at the bottom. The average value of the
compressive stress for the whole column would be the same in
the two cases, viz. half the total weight.
The simplest way to suppose this average but constant stress
produced would be to consider half the total weight concentrated
at the top and half at the bottom, giving a constant stress
measured by half the total weight and a pressure on the
supporting ground equal to the total weight.
This is the approximation we shall adopt : by the stress in a
bar of a framework, due to its own weight, we shall mean the
average stress, produced by a load of half the total weight of
the bar concentrated at each end.
When the bar is a sloping one, the weights of the various
particles tend to bend the bar, and thus set up additional stresses.
Except in Ch. IX. on Bending Moments, the bars of the frame
will be supposed straight, and these additional stresses neglected.
A similar supposition will be made as to the effects produced
by other forces acting on the bars at points other than the ends,
viz. they will be supposed replaced by parallel and equivalent
forces at the ends. Such suppositions have the additional
advantage of making all the forces act at the joints of the
framework, where they may be combined directly by the vector
law of addition.
(22) Find the average stresses due to the weights of the bars set up in
the Warren girder of Ex. 13, each bar weighing 530 lbs.
Reactions at Joints (two bars). Consider the simple
cantilever PQB (Fig. 212), PQ and PR being uniform equally heavy
rods of equal length. Suppose the weight of each bar to be W
(15 lbs. weight). Replace the bars by weightless ones, having
7 '5 lbs. weight concentrated at each end; then at P a verti-
cally downward force of W {\f> lbs. weight) acts. The average
stresses BG and CA are obtained in the usual way, and are
exactly the same as if the bars were weightless and a load BA
were suspended from P.
REACTIONS AT JOINTS.
227
To find the reactions of one body on another we may suppose
the second body removed ; then the force which has to be applied
at the old point of contact, to maintain equilibrium, is the reaction
of the second, body on the first.
Hence the reaction of cb at F is found by supposing the bar
cb removed, and seeing what force must be applied at P to keep
the rest of the frame in equilibrium.
Pig. 212.
But if we remove cb we must suppose the load ^Walso removed
from P, hence we bisect A'B at M, and CB + BM or CM is the
reaction of cb on the pin at P ; similarly, the reaction of the pin
on be is MC.
If the bars had been supposed weightless and a load W lb.
suspended from P, then the stresses found in be and ca would
have been the same as before, but the reactions of the pin on
be and ca would have been quite different. Por, on removing
be, no weight is taken away from P, and the reaction of cb on
P would be simply CB. Similarly, on removing ae, the reaction
on ea is seen to be CA.
228
GRAPHICS.
(23) If PQ=--'7-2 ft,, P2?=5-6 ft. and QB=6-3 ft., find the reaction at P
on PQ if each bar weighs 37 '24 lbs.
(24) As in the previous question if, in addition, a load of 40 lbs. hangs
from P.
Reaction at a Joint (two unequal bars). PQR (Fig. 213)
is a small wall crane; the bars PQ and QR are uniform and the
weights are 2 lbs. per foot. Find tlie pin reactions at Q given
PR = 9', RQ = 3-9' and QP = 7-1'.
Draw the total load vector AB of length proportional to
half the weight of PQ and QR, and draw the stress diagram
ABC as usual. Divide AB at D, so that ^ = ^- Join ^D,
and measure it by the force scale; it gives the reaction of the pin.
Suppose the bar PQ removed, then the force that must be
applied at Q to maintain equilibrium is given by the resultant of
the force OA and the weight AD (half the weight of PQ), i.e. by
CD.
At
Fill. 213.
This vector CD gives, therefore, the reaction of the bar PQ on
the pin at Q, and DC gives the action of the pin at Q on the
bar PQ.
Similarly, if we suppose QB. removed, we must remove the
forces BC and DB and replace them by their resultant DC.
REACTIONS AT JOINTS.
229
Hence DC is the reaction of the bar BQ on Q, and CD is the
action of tjie pin at Q on the bar.
Evidently these results are consistent, for, the pin at Q i)eing
supposed weightless (or so small that its weight may be neglected),
the total forces acting on it, viz. DC and CD, must be in equi-
librium.
(25) Find the reaction of the pin on be if a load equal to half the total
weight of the bars be suspended from it. v
Reactions at a Joint (three bars). If the bars, have no
weight, the reactions are simply along the bars; if they have,
then the reactions of the pin on the three bars may be found
as above by supposing the bars removed one by one, the reaction
being the sum of the remaining forces, or the sum of the half
weight of the bar and its force on the joint reversed in sense.
■w,+w^+w^
Fig. 214.
Thus, suppose be (Fig. 214), cd,-da are the bars, and Wj, w^, Wg
their half weights, and that the vector polygon for P is as
shewn, where AM=w^, MN=w^, NB = w.^^.
To find the reaction of the pin at P on le, suppose ho
removed, i.e. remove NB and BC in the vector polygon ; then NO
is their resultant and ON the resultant of the remaining forpes
and is the reaction of the pin on he.
Similarly, remove DA and AM (the force in da and half its
weight) ; the sum of the remaining forces is MD and gives the
reaction of the pin at P on ad.
230 GRAPHICS.
Finally, remove MN and CD and the sum of the remaining
forces is given by the sum of DM and NC. ,
Hence the sum of the three reactions of the pin is zero as
it should be.
(26) Find the reactions of the pin at V, of the girder of Ex. 18, on the
three bars meeting there, if the bars weigh 25 lbs. each.
Example. AB and AC (Fig. 215) represent hemy unifm-m beams,
each 7 ft. long, pin-jointed at A and resting on rough walls at B and
C in the same horizontal line, i/ BC = 9 ft. and the beams weigh 500
and 700 lbs., determine the reactions at A, B, and C, and the average
stresses in AB and AC.
It is assumed that the walls are sufficiently rough to prevent
the beams sliding down.
Suppose the beams replaced by light rods loaded at their ends
with 250 and 350 lbs. weight respectively. Then at A there is
a load of 600 lbs. weight, at B 250 and at C 350 lbs. weight.
Draw the beam diagram to scale and letter the spaces as
indicated; then draw the vector polygon XYZU for the loads.
Then for the point A, YVZ is the stress diagram, and VY and
VZ give the average values of the stresses in AB and AG. For
B, XYV is the stress diagram. Since XY is the load at B and
YV the push of the beam there on the wall, therefore VX gives
the reaction of the wall at B on the beam AB. Similarly, UV
gives the reaction at C. To find the reaction at A on AB suppose
AC removed, then we must also suppose a weight of 350 lbs.
removed from A ; set off then ZM=ZU and the resultant of the
weight MZ and of the force ZV is the force MV, which is the
reaction at A on the bar AB.
Measure the magnitudes of the reactions and the angles they
make with the horizontal ; measure also the average stresses in
the beams.
Find the resultant of the weights of the beams (supposed to act
through their mid-points) and shew by construction that the re-
actions at B and C, previously determined, are concurrent with it.
Of what general theorem is this concurrency an example ?
REACTIONS AT JOINTS.
2ai
If the walls were smooth, and B and G were connected by a
light rod, what would be the stress in that rod 1
(27) Find the reaction
on the bar AC at A.
(28) Shew (by drawing)
that the reactions at JB
and A on AB are eon-
current with the vertical
through the M.c. of AB.
232 GRAPHICS.
Quadrilateral Frame and Reactions at Joints.
Example. Fmr equal bars, each of weight 0-5 lb. and length 6",
are pin-juinted together to form a square. It is suspended by one
vertex and its form maintained by a light string connecting the upper
amd lower vertices. Determme the stresses in the string and rods and
the reactions of the pins on the bars.
Each bar may be supposed replaced by a weightless rod if
half the weight be supposed concentrated at its end. This
reduces the case to a frame loaded at the joints only. Further,
if the string be supposed cut, the geometrical form will be
maintained if we suppose a force equal to the tension in the string
to be applied upwards at the lowest joint and downwards at the
highest one. Letter the spaces as in the diagram (Fig. 216) and
draw the vector polygon for the loads at Q, B, S, viz. 0-5 lb. at
each ; then the reaction, due to these, of the support at P is AB
or 1'5 lbs. upwards.
At Q we have a force 0-5 lb. downwards (BO) and the forces
due to the rods bf and cf. Draw, then, £F and CF (parallel to
these rods) to intersect in F, the sense of the vectors being
BGFB. Hence c/is in compression and bf in tension. Similarly,
at S we get DAED for the vector polygon.
At a we have then FO, CD, DE and the tension in ef. Join,
therefore, E and F and CDEFG is the vector polygon for the
point. The tension in the string is twice the weight of a rod.
Tabulate the stresses- and mark the bars which are in com-
pression.
(29) Shew that the stress diagram for the point P gives results consistent
with those already obtained. Find the stresses if PS=2Q8.
For the reactions of the joints on the bars all we have to do is
to suppose the corresponding bar removed and find the resultant
of the forces acting at the joint, or find the force which would
produce the same action on the joint as the bar, and then change
its sense. It must be remembered that when a bar is supposed
removed, we must take away the half load concentrated at the
end under consideration.
REACTIONS AT JOINTS.
233
At Q, suppose the
bar hf removed, i.e.
take away the pull
FB and one half of
BO ; the resultant of
the remaining forces
GO and CF is GF, the
reaction of the joint
on the bar If. The
reaction on cf is simi-
larly seen to be FG.
At R, suppose the
bar de removed ; then,
in the stress diagram,
take away DE and
|CD and there re-
mains EFCH, the re-
sultant vector being
EH. Similarly, the
reaction one/is found
by removing FC and
I CD, and is therefore
Fig. 216.
■234 GRAPHICS.
These results contrast with the case in which the joints
are loaded and the weights of the bars are negligible, for in this
latter case the action on any bar must be equal and opposite to
the force of the bar on the joint, and is therefore given by the
stress in the bar itself.
* Pentagonal Frame and the Reactions at the Joints.
Example. A regular pentagonal framework is suspended by one
vertex, the regular form, being maintained by a light string joining the
top vertex to the middle point of the opposite side. Each bar weighs
1 lb. ; find the stresses in the bars and string and the reactions of the
pins on the bars.
Proceed as in previous example, but now the pull due to the
string at the raid-point of the lowest side must be replaced by
equal upward forces at R and S (see Fig. 217) of magnitude to be
determined. Letter the spaces as indicated, and draw the vector
polygon BGBEAB for the external forces. Then draw BOH for
the point Q and EAF for T, then HGDGH for R. The last
gives G-H as half the tension in the string. Complete the stress
diagram and see that the part for the point P gives results con-
sistent with those obtained before.
The reaction at Q on ch is HL, where L is the mid-point of BO.
For the reaction at R on ch we must not only remove half the
weight of the bar RS and its stress, but also half the tension of
the string (since the string does not really pull at R) ; the
reaction is therefore KH on ch, where K is the mid-point of CD.
Tabulate the stresses as usual.
(30) Two heavy uniform bars AB and BC are jointed together at B and
to supports at A and Cin the same horizontal line. If AB=i, BC=i'5
and AG =T ft., and the bars weigh 7 lbs. per ft. , find the average stresses
in AB and BC, and the reaction at B, A and C.
(31) A rectangular framework of four heavy bars is hung by one
vertex, the rectangular state is maintained by a light non-vertical rod
joining two vertices. The sides of the frame being 7 and 4'5 ft., and the
bars weighing 25 and 16 lbs. respectively, find the average stresses in the
bars and the compressive stress in the light rod.
(32) AB, BC and CD (Fig. 218) are three uniform heavy iron bars of
weights 15, 10 and 15 lbs., hinged a,t A, B, G and D, and hung from A
and D in the same horizontal line. Find the average stresses in the bars
aiid the reactions at A, B, C and D if the bars weigh 1 lb. per ft.
REACTIONS AT JOINTS.
235
^B^
'c;
KC
■D)
•^A^
Fig. 217.
B C
Fig. 218.
236 GRAPHICS.
(33) AB = 6, BG = 4-5, GD = 6,
BD = 9-5 and DAB= 45°. The shape
in Fig. 219 is maintained by a light
string BD ; find the average stresses
and the 'reactions if the bars weigh
3 lbs. per ft.
(34) JFive equal bars are freely jointed to form a pentagon, which is
suspended by one vertex. The frame is maintained in the regular penta-
fonal form by a light horizontal bar connecting two vertices. If the five
ars weigh 7 lbs. each, find the stresses and the reactions at the joints.
(35) As in the previous example, only the bars are light and weights of
7 lbs. are suspended from the five vertices.
(36) A heavy triangular framework is suspended by one vertex, the sides
are 3, 4 and 3 '5 ft. long. The bars are uniform and weigh 0"2 lb. per foot.
Find the position of equilibrium, and the stresses in the bars and the
reactions at the joints.
(37) Three equal rods each weighing 2'5 lbs. are freely jointed together.
The frame is supported at the mid-point of a horizontal side. Find the
stresses in the bars and the reactions of the hinges on the bars.
(38) A regular hexagon of uniform bars, each of weight 3 lbs. , is suspended
from two vertices in the same horizontal line, the form is maintained by a
light string connecting the mid-points of the top and bottom bars. Find
the tension in the string and the reactions at the vertices.
(39) Find the average stresses in the bars of the king post truss of
Ex. 13 due to the weight of the bars alone if they weigh 10 lbs. per ft.
The Funicular Polygon. The link polygon for like parallel
forces can easily be constructed by means of weights and strings.
The actual string polygon is called a funicular polygon (funi-
cula = a little rope), and sometimes the meaning is extended to
cover all the geometrical figures we have called link polygons.
Example. BC and CD (Fig. 220) are the vectors of forces whose
axes are be and cd. Choose omy pole between the perpendiculars at
B and D to BD. Through the space b draw a line parallel to OB,
through c a line parallel to OC and through d a line parallel to OD.
In Fig. 220, PB^RJ^ is the link polygon, P and Q being any
points on the first and last lines drawn.
If the paper, on which the drawing has been made, be fixed to a
vertical drawing board, so that he is vertical, and a string of length
Pi?i + R1B2 + R2Q be fixed to the board by stout pins at P and Q,
FUNICULAR POLYGON.
237
then if weights proportional to BC a.nd CD be fixed, hung by threads
knotted to the string at .Bi and R^ the forces acting on the string
will cause it to be in equilibrium in the given position PB^R^Q.
To prove this requires merely a statement concerning the
equilibrium of concurrent forces, ^j is in equilibrium if three
forces BO, CO and OB act there ; the last two will therefore give
the pulls which must be exerted on B^ to balance BC and can be
supplied by the tensions of the strings attached to R-y in the
positions given.
On joining PQ the link polygon is closed, and by drawing OA
parallel to it in the vector polygon we get AB, the vertical
Plo. 220.
reaction at P, and DA the vertical reaction at Q ; moreover,
AO gives the components of the reaction at P and Q along the
line PQ. Hence, if PQ be a light rod suspended by vertical
strings at P and Q, then AO would give the stress in PQ.
Notice carefully that we are able to draw the correct position of
the funicular polygon corresponding to a certain arbitrary vector
polygon, we can also, with certain exceptions, do the converse
construction ; i.e. given an arbitrary form PB^R^Q for the string,
238 GRAPHICS.
I
we can determine the weights BC and CD which will produce
equilibrium and find the corresponding tensions in the string.
It is essential that the form of the string polygon assumed be
such that equilibrium can be obtained by tensile stresses only ;
hence in the arbitrary vector polygon the angles at B and D must
be acute. Hence, the limitation as to the position of 0. It is,
of course, true that an infinite number of weights can be found,
but the sets will all be proportional; i.e. taking the first, BC,
arbitrarily, the other, CD, is fixed by the construction.
We cannot assume both the form of the string and the
magnitudes of both the weights.
Example. PQES (Fig. 221) represents a string attached to the
points P and S, P being 3 ft. above S, and PS = 15 ft., PQ = 5,
QiEl = 8-5 and, ES = 7 //. At Q is tied a weight of 4-7 lbs.; find
the weight which must be attached to E so that the angle QES may
be 120°.
Draw first the polygon PQES to scale ; to do this construct
the triangle Q^B^Sj^, where
Q^R^=QR, Bj^Sj=RS and 0ASi=120°;
this gives the length of QS, and hence the point Q can be
constructed.
Construct next the vector polygon for Q, viz. BCAB. At R
we know the pull AC in ac, and can therefore complete the
diagram for that point, viz. ACDA. The weight at R is
therefore given by CD, the other lines give the tensions in the
three parts of the string.
Notice that the horizontal components of all- the tensions are
equal.
(40) PQEST (Fig. 222) is a funicular polygon, PT is horizontal ; a load
ol 5 lbs. is suspended at Q ; find the loads at E and S necessary to maintain
the given shape. PQ = ^-5", QR=i-5", RS=&', PT=\r'-
(41) If the strings PQ and ST be suspended from a light rod PT hung
by vertical strings, find the tension in the strings and the compression in
the rod PT.
(42) The tension in PQ being 18 lbs. , what must be the weights suspended
from<2, ifandiS?
FUNICULAR POLYGON.
239
Fig. 221.
240 GRAPHICS.
Funicular Polygon for Equal Loads, the axes being at
equal distances.
Example. P and Q (Fig. 223) are two fixed points in a
horizontal line, distant apart 2'5 ft. ; a string is to be fixed to P and
Q, and loaded vnth weights 2, 2 "5, 1'5, 3 lbs., whose axes are
equidistant {viz. 6"). The pu/i-t of the string attached to P is to
make 30° with the vertical. Find the length of string necessary,
the form of the funicular polygon and the points on it where the
weights have to be fastened.
Draw first the vector polygon of the weights BGDEF, choose
a pole A for the vector polygon, and find the position of the
resultant by the link polygon. Find the point R of intersection
of the known direction of the string at P and the axis of
the resultant. Join R to Q; then RQ is the direction of the
string at Q. Draw, then, BO and FO parallel to PR and QR to
intersect at 0. is then the pole of the vector polygon, and by
joining to ODE we get the directions of the remaining segments
of the string. Draw these directions on the link polygon, and
measure the segments between the given vertical lines.
It may appear at first sight that, although we can determine
the point in the manner given, yet when we come to construct
the corresponding link polygon — PR^ parallel to OB, R^R^ parallel
to OC, R^R.^ parallel t« DO, R^R^ parallel to EO — the last link
through R^ parallel to OF would not necessarily go through Q.
A little consideration of the properties of the link and vector
polygons will shew the necessity of this.
We have a number of parallel forces in be, cd, de and ef, and
BF is their vector sum; through b draw PR^ parallel to BO,
and suppose in PR-^ two forces OB and BO, differing only in
sense, to act. The resultant of OB and BC is 00 along R-^R^,
and so on to the resultant of OB, BC, CD, DE and EF is OF
acting along R^Q■^, where Q^ is the point on the vertical through
Q, where R^Q^, parallel to OF, cuts it. Combine, then, BO in PR-^
with OF to a resultant BF acting through the point of intersection
of PiJj and Q^R^. But the point in PR^^, where this resultant
FUNICULAR POLYGON.
Sll
cuts it, has been determined already, viz. the point B, and EQ
is parallel to OF, hence BQ and BQ^ must be both parallel to
OF, i.e. Q-^ is at Q.
(43) The distances apart of
the verticals are 1', 6", 4", 4"
and 8", PMi is to be inclined
at 45° to the horizontal, and
the weights are all equal ;
find the form of the string
polygon, the total length of
string and the tensions.
(44) The loads being 2'1, 2'2, TS and 2'9 lbs., and the vertical axes
equidistant (6"), find the form of the link polygon and the total length of
string necessary if ad is to be horizontal ; Pi?, makes an angle of 45° with
vertical and Q is fixed only inasmuch as it must lie in a certain vertical.
(45) As in Ex. 43, but now ac and ae are to be equally inclined at
45° to the horizontal.
T.O, Q
242 GRAPHICS.
The theory of the funicular polygon is of great practical
importance, since it is immediately applicable to the deter-
mination of the form and stresses of suspension bridges.
Funicular Polygon satisfying prescribed conditions.
A funicular polygon is to be constructed between two points,
P and Q, distant 3 ft. apart in a horizontal line. It is to be
loaded with seven equal weights of 5 lbs., the axes are to be
equidistant and the lowest vertex, V, \\\ inches below M, the
mid-point of PQ.
Since the axes and loads are symmetrical with regard to MV,
only half the funicular and stress diagrams need be drawn.
Set offPilf and MV (Fig. 224) to scale (say half full size) and
letter the spaces.
Set off to scale AB, BG, CD for the loads in ab, he, cd, and
DK=^AB ; draw KO perpendicular to AK; then the pole of
the vector polygon must be in KO. This follows from the
known symmetry of the funicular polygon. Take any point
in KO as pole, and proceed to draw the link polygon, starting
at F, viz. TFi parallel to OB, V^V^ parallel to OC, F^Fg
parallel to OB and F^F^ parallel to OA.
If F^ is at P we have solved the problem.
Evidently by taking different poles along KO, and constructing
the corresponding link polygons, we should get a series of points
like F^, and by trial we could find the position of the pole so
that F^ should be at P.
Through F draw a line Fmj parallel to PM and produce the
links FgFj, F^F^, F^F^, to cut it at u^, u^, u^.
Join P to Mg cutting ah in W^ ; join W^ to u^ cutting he in
W^ ; join W^ to Mj cutting cd in JV^; and, finally, join JF-^ to F.
Then PW^W^PF^F is the half funicular polygon required.
Through A, B, C and D draw lines parallel to PfF^, W^TF^,
^2^i> ^1^; these should all be concurrent to a point Oj
in^O.
Measure AO^^, BO^, 00-^ and DO^ on the force scale; these are
the tensions in the various parts of the string polygon.
FUNICULAR POLYGON.
243
^ M
244 GRAPHICS.
* Proof. Suppose that corresponding to the two poles and 0^
there are two link polygons VJV-^ . . . and FF^ Through F
draw a line Fu^ parallel to OOj, cutting W^W-^, ... in Mj, Mg and Wg.
Let OOj be the vector of a force in Fu^, then OOj, OJ) and
DO in Fu^, FW.^ and VF^ are in equilibrium. To OjD add DC
and to DO add CD, both forces acting in cd, then OOj , OjC and
CO are in Fmj, W^u^ and F^F-^ are in equilibrium. But three
forces in equilibrium must be concurrent, hence F^Ft^ must pass
through u-^. A similar argument shews that F^F^ and FJ''^
pass through u^ and Wg respectively.
Generally, then, we see that wherever the pole of the vector
polygon be taken in the line KO, the corresponding lines of
the link polygon must intersect on the line through F parallel
to 00-^. This is really a special case of a more general theorem :
Given the axes and vectors of a set of forces, if two link polygons be
drawn for poles and Oj, then the corresponding sides of the link
polygons intersect on a line parallel to OOj .
The Funicular Polygon and Parabola. The vertices V,
Wj, Wg, Wg, ... , of p. 243, lie on a parabola having V as its
vertex and VM as its axis of symmetry. Constrwi the links of the
fwmmilar as follows :
Fig. 225 shews the right-hand half of the funicular of which
the left half has already been constructed. PiM^ or ef, P^M^ or
fg are the load lines on which the vertices are to be found.
Join FQ and construct the vertices on P^M-,, P,^^, ... of the
parabola going through Q (see pp. 29-31).
Curves whose equations are like y = x^, y = 2x^ or y = Ax^ are
called 'parabolas.'
Go vertically from M-^ to Q, , horizontally to R^ and mark Pj
where FR^ cuts M-^P-^ . Similarly, determine the points P^ and Pg.
Set off the load vectors vertically downwards DE, EF, FG and
GH, and mark K, the mid-point of DE. Then draw, through E,
F, G and H, lines parallel to the links of the funicular and see
that they all intersect (or nearly so) on the line KO perpendicular
ioDE,
FUNICULAR POLYGON AND PARABOLA.
245
(46) Draw to scale the funicular polygqn joining two points P and Q in
the same horizontal line, PQ = 2-5 ft. The string is to be loaded with six
eqiial weights (1-5 lbs. each), and the axes are to be equidistant and the
lowest link is to be 2 ft. below PQ. Find the tensions in the various string
ts.
(In this case the lowest link is horizontal, and if the pole of the vector
polygon be chosen on the perpendicular through the mid-point of the
resultant load vector, the corresponding sides of the two link polygons
must intersect on the lowest link.)
D
K
M
Q
:: ::: ::::::;;;; = --;|r¥:: :: :
=:S:S;;=^f;i;::
f""-- I»??*^"" "
i:::::^:-.:;t::±:R
J ' n ' . if -^ '
vr \
" ._! J3
Fig. 225.
*(47) Draw to scale the funicular polygon joining two points P and Q
where PQ = 4 f t. , and makes an angle of 30° with the horizontal through' P
and below it. There are to be six equal loads in equidistant axes, and the
first link is to make an angle of 40° with the vertical. Find also the
tensions.
(In this case, through the beginning of the total load vector draw a
lino making 40° with the vertical and take the pole on this. Draw the
first link of the funicular through P, then wherever the pole of the vector
polygon be taken on the line drawn, the corresponding sides of all funiculars
starting at P must intersect on the first link. Hence it is easy to draw
that particular funicular which will pass through Q. )
*(48) Funicular as in Exercise 47, but the vertex bearing the middle load
is to be 2 ft. below the mid-point of PQ.
*The Funicular Polygon and Parabola. The vertices
of a funicular polygon for wMcli the loads are all ec[ual and
at equal distances apart lie on a parabola.
(a) Suppose the number of loads odd.
Let w be the load at each vertex, h the distance apart of the
loads, F the lowest vertex and Q any other, say the m"". Take
the axes of coordinates as the horizontal and vertical through F
246
GRAPHICS.
(Fig. 226) and suppose the coordinates of Q, x and y. Between Q
and V are «.- 1 loads and n spaces, hence the resultant load is
n-lw whose axis is mid-way between f^and Q.
The part of the chain between Fand Q is, in equilibrium under
the load n- lie and the tensions in the lowest and highest links.
Hence, since three forces in equilibrium must pass through a
^(n-i)w
w(n-i)
point, these links must intersect on the axis of the resultant load,
viz. at M in Fig. 226 whose abscissa is -.
In the stress diagram T and T-^ are the supposed tensions in
the links, then T, Tj and w{n-l) form the vector triangle for M.
For the point V, 7^, w and by symmetry a force of magnitude
1\ form the vector triangle. Hence T^ denoting the horizontal
component of all the tensions, we have T-^ equivalent to T^
and |mi.
Comparing the similar triangles of the funicular and vector
polygons we have at once
FUNICULAR POLYGON AND PARABOLA. 247
X - To '
2
w
M.N 2"
X -T,
2
Therefore, by the properties of fractions (adding numerators
and denominators),
y icn
But nh = x;
w „
This is the equation to a parabola, the axis of y being the
axis «of symmetry. Hence all the vertices lie on the parabola,
for the equation remains exactly the same whatever particular
value n may be supposed to have.
When the dip and span of the funicular and the number of
loads are known, Tg can be calculated.
Let 2s = the span,
d = the dip,
N= number of loads
{^'M
Then d- "" =2_<?y±i>
or Jo- 4 ^•
(b) Suppose the number of loads even, then the middle link is
horizontal. First take the origin at the middle point of this
link. Then for the equilibrium of the piece between V and Q
the link through Q must, when produced, pass through M the
mid-point of the space between the P' and n - 1'" verticals.
^8
GRAPHICS.
In Fig. 227,
a; = («.-l)A + H!
2'
and
y w{n-\) ,
vmh,
h
and
But
h
i = x +
2'
y+^
8T„ 2hT,
x\
If the origin be taken at a distance ^=- below V, the
w
8r„
equation becomes y=-^rrrp-x'^, a parabola.
* Method of Sections. If any closed curve be drawn
cutting some of the bars of a frame in equilibrium, then the
part inside the curve may be looked upon as a rigid body
in equilibrium, under the action of the known external forces
on that part of the frame, and certain forces acting along the
bars at the points where the curve cuts them. If the forces
necessary to produce equilibrium acting along these bars can be
found, they will give the magnitude of the stresses in those
bars, and whether the stresses are tensile or compressive can be
determined by the senses of the forces.
Fig. 228 represents an ordinary queen post truss. If we
consider the part enclosed by the dotted curve on the right, it
will be in equilibrium under the action of the external forces in
STRESS DIAGRAMS AND METHOD OF SECTIONS. 249
oa, ah and Ic, and certain forces in ch, hi and io. Suppose these
bars cut; then, to maintain this right-hand portion in equilibrium,
we must apply to the cut surfaces of the bars forces equal in
magnitude to the stresses in them before cutting, and of the
proper sense. For instance, if ch be in compression we must
take the sense of the force at the cut surface from left to right ;
Fig. 228.
if in tension, from right to left. Conversely, if we can find the
forces in these cut bars which will keep the right-hand portion
in equilibrium, these forces with their proper senses will give the
stresses in the bars.
If the curve can be drawn so as to cut only three non-
concurrent and non-parallel forces, we can always find by this
means the stresses in the three bars. (In special cases we may
be able to find the stresses in one or more bars when more
than three are supposed cut).
The method of sections enables us (i) to test the accuracy of
the stresses already found by independent means ; and (ii) to
draw stress diagrams for various frames in which the ordinary
method of procedure fails.
To determine the stresses, we either use the construction for
resolving a known force into three components acting in three
non-concurrent and non-parallel lines, or the construction for
finding the sum of the moments of a number of forces. The
latter method will be explained in Chap. X.
250 GRAPHICS.
* Example. Determine the stresses in the bars ch, hi and io of the
queen post truss {Fig. 229) »/ VZ = 40, VW=12, UW=12/i!., the
loads in ed, dc, cb and ba being 0-5, 0'6, 1"5 and 1 tons.
Draw the truss to scale and the load and reaction lines.
Draw the vectors ED, DC, CB and BA of the loads in ed, dc,
cb and ba. Take any pole P of the vector polygon and construct
the link polygon in the usual manner. Close the link polygon
and draw PO in the vector polygon parallel to the closing line.
Draw any closed curve cutting the bars ch, hi and io. Then
the forces acting on the enclosed body are OB, ED and DO and
the forces in the bars supposed cut. For these forces the first
line of the link polygon drawn is parallel to PO, and the last is
parallel to PC; these lines intersect (when produced) at R,
consequently the resultant of OE, ED and DC, viz. DC, acts
through R parallel to OC.
Produce io to cut the axis of this resultant in Q. Join Q to U,
the point of intersection of ih and ch. From C and draw lines
parallel to io and QU, viz. GX and OX. Then CX and XO are
in equilibrium with 00, their axes being h, Q U and RQ.
From and X draw OY parallel to ch and XY parallel to ih ;
then 00 acting along RQ is in equilibrium with OX, XY and YD
acting along io, ih and ch respectively.
Since CX pulls at the cut end of the bar io (enclosed by the
curve in Fig. 229) io must be in tension. Similarly ih is in
tension and ch is in compression.
When the angle is small between the sides of the link
polygon, for which the point of intersection is required, the
method becomes untrustworthy. In that case the moment
method explained in Chap. X. is used.
Cases in which the method of sections enables us to draw
stress diagrams for which the ordinary method fails will be
considered also in Chap. X.
(49) Draw the frame in Fig. 230 and determine the stresses in hg, 3/ and
fd by the method of sections. The load at P=\ ton and at Q = 0'5 ton.
RS=^, RP=2i, RQ = QP = nit. •
METHOD OF SECTIONS.
251
Fio. 229.
(50) rind the stresses in bj, jk and ko in the queen post truss of
Fig. 229.
252
GRAPHICS.
MISCELLANEOUS EXAMPLES. VI.
1. ABC is a triangle having a right angle at B. The sides AB= 12",
BG= 5". A force of 52 lbs. weight acts from A to C, one of 24 acts from
B to A and one of 27 from O to B. Find the magnitude and line of action
of the resultant. Draw the figure to scale and exhibit the resultant.
(Inter. Soi., 1904.)
2. ABG is a light frame in the form of a right-angled triangle, A being
the right angle. It can turn in a vertical plane about a fixed horizontsu
axis at A ; and, when a. weight W is suspended from C, the corner B
(which is vertically below A ) presses against a fixed vertical plate. Find
graphically the stress in each rod and the reactions at A and B.
(Inter. Sei., 1906.)
3. The pair of rafters shewn in Fig. 231 carry a load equivalent to
150 lbs. at the middle of each. Find the direction and magnitude of, the
thrust on each wall.
Pig. 231.
If the walls are relieved of horizontal thrust by a rod joining the lower
ends of the rafters, what is the pull on the rod ? (War Office, 1904.)
4. A triangular frame ABC can turn freely in a vertical plane about A
(the right angle). The side AB is horizontal, and the corner G rests
against a smooth vertical stop below vl. Find, graphically or otherwise,
the stresses in the various bars due to a weight W suspended from B.
AB=3 ft., AG=1 ft., r=50 lbs. weight. (Inter. Sci., 1903.)
5. The beams AB and AG (Fig. 232) are part of a roof and carry a load
of 160 kilogrms. at the ridge. The span of the roof is 10 metres, and the
beams make 28° with the horizontal. Find the thrust this load causes
on each wall.
Pig. 232.
If the walls are relieved of the outward thrust by a rod joining the ends
B and G of the beams, what is the thrust in this rod, and what thrust do
the walls still bear ? (Military Entrance, 1906.)
MISCELLANEOUS EXAMPLES.
253
6. A heavy straight rod AB, 12 ft. long, turns on a pivot at A, and is
supported in a horizontal position by a vertical force of 15 lbs. weight
applied at B. If the weight of the rod is known to be 90 lbs. , find the
pressure on the pivot ajid the position of the centre of gravity.
(Naval Cadets, 1904.)
7. A heavy uniform beam AB rests in a vertical plane against a
smooth horizontal plane CA and a smooth vertical wall GB, the lower
extremity A being attached to a cord which passes over a smooth pulley
at O and sustains a given weight P. Find the position of equilibrium.
(B. of E.,IL)
8. Find the stresses in the bars of the short Warren girder PQEST
figured (Ex. 12, p. 222). Loads of 2 and 1 tons act at P and T, and the
frame is freely supported at Q and jS in same horizontal line.
(Inter. B.Sc. (Eng.), 1905.)
9. ABGD is a thread suspended from points A and D, and carrying a
weight of 10 lbs. at B and a weight W at C ; the inclination to the vertical
of 2l-B and CD are 45° and 30° respectively, and ABC is an angle of
165°. Find, by construction or calculation, W and the tension of BC.
(B. of E., IL)
10. A uniform bar is bent into the shape of a V with equal arms, and
hangs freely from one end. Prove that a plumb line suspended from this
end will cut the lower arm at J of its length from the angle. (B.Sc, 1904.)
11. Fig. 233 shews (drawn to scale) a rectangular framework of four bars
freely hinged. It is supposed to lie on a smooth table with AB fixed.
A piece of elastic cord is stretched, and its ends fastened at E and F.
For what positions of E and F will the frame remain rectangular, for
what positions will the frame move to the right and for what positions
to the left ? Justify your statements.
Pia. 233.
If the frame is held rectangular by a pin driven through the comer C,
and the elastic cord stretched till it exerts a pull of 10 lbs. , and fastened
to the frame at the corners A and C, what force does the pin at C exert
on the frame? (Military Entrance, 1905.)
254
GRAPHICS.
12. Fig. 234 is a rough
sketch of a crane. If the
weight hanging at the point
A is 1000 lbs. , find, graphic-
ally or otherwise, the forces
actipg along the bars AC
and AB; and if the post
BC is free to move in the
vertical plane, find the pull
in the tie CD which will
prevent the crane from top-
pling over.
(Military Entrance, 1905.)
Fig. 234.
13. AB and BO are two uniform rods fastened by smooth joints to each
other at B, and to fixed points A and C; the point O being vertically
above A and GA =AB ; given that the weight of BG is twice that of AB,
find the reaction at G and A by which the rods are supported.
(B. of E., III., 1904.)
14. Draw a circle and from a point A outside the circle draw two
tangents and produce them to B and G. Suppose that AB and BG are
two equal uniform rods connected by a smooth hinge at A at rest on a
smooth vertical circle ; find the position of the rods when .4^ is 10 times
as long as the diameter of the circlfe. (B. of E., III.)
15. A, B, G are fixed smooth points, such that ABG is an equilateral
triangle, with BG horizontal and above A. A fine thread is fastened to
A, passes over B and G and carries a weight of 10 lbs. ; find the pressures
on B and G produced by the weight. (B. of E. , I. , 1903. )
16. A uniform rod AB can turn freely, in a vertical plane, about a
hinge at A ; the end B is supported by a thread BG fastened to a fixed
point G; AG is horizontal and equal to AB. Draw a triangle for the
forces which keep the rod at rest, and shew that in any inclined position
of ^^ the reaction of the hinge is greater than the tension of the thread.
(B. of E.,II., 1903.)
17. A, B, (7 and D are four points in a horizontal line. Two weights,
P and Q, are to be supported by three strings, AP, PQ and QD, in such
a way that P is vertically below B and Q below G. Give a construction
for finding suitable lengths of string for the purpose.
Also shew that in any configuration in which the stated condition is
satisfied, the string PQ will intersect AD produced in a fixed point.
(Patent Office, 1905.)
18. The outline of a crane is as follows: ABG is a right-angled triangle,
AB being horizontal and AG vertical, and AB is equal to AG ; on the
other side of. AG is a, triangle ADG, the angle A being 45° and the angle G
being 120°. A load of 10 tons hangs from D, and the crane is supported
MISCELLANEOUS - EXAMPLES.
2C5
at A and anchored at B. Find the reactions at the supports, and the
stresses in AC, CD and BO. (Patent Office, 1905.)
19, Draw five parallel lines at distances apart 4, 5, .3 and 6 inches from
left to right. Construct a funicular polygon for loads 2W, W, ZW, 2W
and W suspended from vertices on these lines, the sides of the polygon
which stretch across from the middle line to its neighbours being eaoli
inclined at 60° to the vertical. . (Inter. Sci. B.Sc. (Eng.), 1904)
20. Draw the stress diagram for the following truss, Fig. 235, indicating
which members are in tension and which in compression. Assume equal
vertical loads. (Admiralty Exam., Assistant Engineer, 1904.)
Pia. 235.
21. A king post roof truss is loaded with 1 ton at each of the five joints ;
find graphically the amount and nature of the force acting in each member.
Span=30ft., pitch=7' 6".
22. Draw the stress diagram for the frame shown in Fig. 236, assuming
the joints to be loaded with equal weights in the directi(m of the arrows.
The arrows A and B indicate the direction of the reactions.
(Admiralty, 1905.)
CHAPTER VII.
FEICTION.
However carefully the Experiments I.-VI. are performed, there
are sure to be some slight divergencies from the results stated.
These discrepancies are chiefly due to the friction at the pulley,
and but to a very small extent to the weight of the ring or card.
Resistance to Motion. In all cases where one body slides
or tends to slide on another, a resisting force called friction is
called into play. This force has the same direction as that of
the motion, or attempted motion, but is of the opposite sense.
ExPT. XII. Apparctlua : A board with a smoothly running pulley
screwed or clamped on to one end ; an open box with a hook for fastening
a string a set of weights and a scale pan. A board of varying width is
best, as this will give different areas of contact between the box and the
board.
Arrange the board horizontally and the box, etc., as in Fig. 237. See
that the hook is screwed in ". position so that the pull of the string is
Pig. 237.
horizontal. Put a 500 gramme weight in the box and load the scale
pan 20 grammes at a time until motion ensues. Find the greatest weight
that can be put into the pan without moving the box.
Starting with no weights in the box, increase the load by 100 grammes
at a time up to 800, and find in each case the maximum weight that can
be placed in the scale pan without moving the box.
Normal
Pressure
Pull of
Pan
EXPERIMENT ON FRICTION. 257
Tabulation of Remits. Add to each load put in the box the
weight of the box itself, and similarly the
weight of the scale pan to its load in each
case, and tabulate the results as indicated.
Graph. On squared paper take two axes
Ox and Oy and represent on these, to scale,
the loads.
Plot points like P (Fig. 239) where OM
represents the weight of the box and its load, ^^ ^^
and PM the pan weight and its load to scale.
All the points like P will be found to lie approximately on a
straight line.
By aid of a stretched black thread, determine the straight line
lying most evenly amongst the plotted points, and draw this
line. From this line find the greatest load that can be placed
in the pan so that the box will not move when it contains
270 grammes, and test the result by experiment.
Deductions. In the first experiment with 500 grammes load
in the box, at every instant before motion there was equilibrium.
For equilibrium the vector polygon must be closed, and there-
fore the sum of the components of the forces in any direction
must be zero. Hence, the friction resisting the motion of the
box must have been always equal to the pull of the string on
the box. Hence, for a given pressure between the box and
plank, the friction may have any value from zero up to a certain
maximum value, called the limiting friction. If there was no
friction at the pulley, the load due to the scale pan and its
contents would measure the friction; if there was friction at
the pulley, the pull of the string on the box — which measures the
friction between the box and the board — would be slightly less
than the load. If the straight line goes through the origin, it
shews that the pull necessary to turn the pulley is too small
to be measurable ; if it does not, then the intercept on the
y axis gives the friction due to the pulley {OA in Fig. 239).
Draw through A a line parallel to Ox.
258
GRAPHICS.
Distances measured upwards from this line to the sloping one
give to scale the pull on the box. In Fig. 239 PM-^ is the pull
on the box for a normal pres-
sure AM-^, and Pi/j therefore
measures the limiting friction
in this case.
Coefficient of Friction.
Since AP is a, straight line, the
PM, limiting friction
ratio .-' or ^
AM^ normal pressure
is always the same, no matter
what the pressure. This ratio
is called the coefficient of fric-
tion for the two surfaces, and is
always denoted by the letter /«..
ExPT. XIII. Place the box at the narrow end of the board so that it
overlaps and the area of contact is leas than before ; shew that ix is
unaltered.
ExPT. XIV. Pin a sheet of paper to the board and shew that /i has a
different value from that formerly obtained.
Laws of Statical Friction. The laws thus roughly estab-
lished are :
(i) Friction is a passive force, only called into play by the action
of other forces ; it tends to prevent motion and may have any vahie from
zero up to a certain maxinmm, depending on the normal pressure and
the nature of the swrfaces.
(ii) Limiting friction is independent of the area of contact.
(iii) Limiting friction is dependent on the nature of the surfaces.
(iv) Limiting friction is proportional to the normal pressure,
F = fiW.
Friction and Stress. If F is the force of friction on the
box, - i^ is the force on the plank, and friction is therefore of
the nature of a stress.
Angle of Friction. Draw PQ (Fig. 240) vertically down-
wards to represent to scale one of the normal pressures, and QE
horizontally for the pull on the box (-the limiting friction);
LAWS OF STATICAL FRICTION.
259
then, since there is equilibrium, the closing line RP must
represent in magnitude, direction, and sense, the total reaction
of the plank on the box.
The angle e, between RP and
the normal to the plank, is called
the angle of friction, and /i = tan€.
Measure e and /*. £ is always the
same, no matter what the pressure,
if the box is on the point of
W
moving, for ■=. is constant.
When there is no pull on the
box, RP is vertically upwards {QP);
as the pull increases, RP slopes
more and more away from the
normal and away from the sense
of the attempted motion, until
the maximum angle e is reached.
Expressed in slightly different
words, the total reaction of the surface may be inclined to the
normal at any angle between zero and «.
(1) If the box weighs 6 ozs., and it is loaded with 1'5 lbs., and the
horizontal pull on it when on the point of motion is 15 ozs., find e and /i.
(2) If the box weighs 10 ozs. and is loaded with 1 lb. , and the ooefKoient
of friction is J, find the pull on the box and the total reaction of the
surface of the plank when motion is about to take place.
(3) If the box weighs 15 lbs. and 11='^, would equilibrium be possible
with a horizontal pull of (i) 3 lbs. weight, (ii) 6 lbs. weight ?
(4) A block of stone weighs half a ton and rests on a fixed stone with a
horizontal top ; a horizontal push of 500 lbs. weight just causes the block
to be on the point of motion. What is the angle of friction and what is fit
(6) A 1000 gramme weight rests on a table ; the angle of friction is 25°.
What is the least horizontal force that will produce motion ?
(6) A cube of side 2" rests on a, horizontal plane. The weight of the
cube is 7 lbs., and it is pushed with a horizontal force of 3 lbs. weight
without producing motion. The line of action of the force passes through
the centre of the cube and is perpendicular to one face. Draw a diagram
of the axes of the forces acting on the block (they are concurrent), and find
the reaction of the surface.
(7) Draw a diagram of the forces in Ex. 6 when the applied force is
J of the way up, and find the surface reaction.
Fig. 240.
260
GRAPHICS.
Force not Parallel to the Surface. If the string in the
friction experiment is not quite horizontal, the normal pressure
between the surfaces in contact is no longer the weighted box.
Knowing, however, the slope of the string, the normal pressure
and the friction can be found.
Example. The string slopes up-
wards at an angle of 15° with the
horizontal, the weight of the box, etc.,
is 17 lbs., and a pull of 5'5 lbs.
weight causes the box to be on the
point of motion. Find the angle of
friction e, and the frictional force
called into play.
Draw AB (Fig. 241) vertically
downwards of length 17 cms.,
BG=5-5 cms. sloping upwards at
an angle of 15°. Then CA is the
total reaction of the plank on the
A
box, and GAB = e.
A
Scale GA, and measure GAB
by a scale of chords. Set off
AE =10 cms. along AB, and draw
EF perpendicular to AE ; then
EF on the 10 cm. scale measures
ju ( = tan e). Look up a table of
tangents, and compare e thus
determined with the value ob-
tained by the scale of chords.
Draw GD perpendicular to
AB; then BD and DC are the
vertical and horizontal compon-
ents of BC ; hence the friction
on the box is given by CD. ^''' ^""
BD is the upward pull on the box, due to the string tending
MINIMUM fORCE fOU MOTION. 261
to lift it off the plank ; hence the normal reaction of the plank
is not BA but DA.
Draw BG perpendicular to AC. If the inclination of the
string had been that of BQ, the force given by BG- would have
been the pull that would have caused the box to be on the point
of motion.
Evidently, this is the least pull possible when the box is in
limiting equilibrium. Any force greater than BG, if in the same
direction, will cause motion ; any force less than BG will not
cause the box to be on the point of motion. Scale BG and
measure the angle it makes with the horizontal.
(8) The box, etc., weighs 11 lbs. ; the coefficient of friotion is 0'4. Find
the least pull on the box that will cause it to be just on the point of
motion, and the amount of friction called into play.
(9) The box, etc. , weighs 21 '5 lbs. ; a pull of 12 lbs. applied at an angle
of 20° with the horizontal just causes the box to be on the point of motion.
Find the least force that will move the box, and the corresponding normal
pressure and friction.
(10) The box weighs 9 lbs. and /it=0'3. A horizontal force of 27 lbs. is
applied to the box. Is there equilibrium ; and if so, what is the angle the
total reaction of the plank makes with the vertical ?
(11) A harrow weighs 6 cwts., the chains by which a horse can pull it
along make 20° with the horizontal. If the horse exerts a pull of 2 cwts.
along the chains, find the total reaction of the ground in lbs. wt., the
angle it makes with the vertical, and the resistance corresponding to
■the friotion between harrow and ground.
(12) A block weighing 37 lbs. is to be pulled along a horizontal plane by
a rope ; find the least possible pull and its direction if the coefficient of
friction is 0-37.
(13) The. least force that will move a chair weighing 10 lbs. along a
rough floor is 6 lbs. ; find the angle and coefficient of friction and the
reaction of the ground.
(14) Two men push a side-board along, the side-board weighs 7'5 cwts.
and the angle of friotion is 25°. Find the force the men must exert if
(a) they push horizontally, (6) downwards at an angle of 20° with the hori-
zontal, (c) upwards at 20° with the horizontal, (d) find the direction in
' which they must push in order that the force they exert may be as small
as possible.
(15) A block weighing 17 kilogrammes rests on a rough horizontal table,
the angle of friction being 37°. If a horizontal force of 5 kilogrammes
weight acts on the block, find the least additional force that will cause
motion. What is the greatest horizontal force that can be applied
without motion taking place ?
262
GRAPHICS.
The Inclined Plane with Friction.
Example 1. A block rests on a hoard ; the latter is tilted about
a, horizontal axis through its end. The coefficient of friction being 0%
find the angle at which the block begins to slide and the greatest friction
called into play.
Whilst there is equilibrium, the reaction of the board on the
block must be equal to the weight of the block and in the same
line (since these are the only two forces acting). The angle
between the vertical and the normal to the plane is the same as
the angle between the plane and the horizontal, and as the
former is equal to € when the block is on the point of sliding
the angle of the plane must also be «, and tan£ = 0'19. Draw
AB (Fig. 242) horizontally of length 10 cms., and BC vertically
of length 1-9 cms.; then CAB is the angle of the plane.
Draw any length
LM vertically
downwards to re-
present W, LN at
an angle £ to LM,
and MN perpen-
dicular to LN, as
indicated in Fig.
242. MN is the
friction and NL the
normal reaction.
Measure the fric-
tion as a decimal
of W.
To do this, set
off as indicated
MW=ML, MI =5
inches, and draw
through N a line
parallel to WL
scale.
W
Fig. 242.
Measure the intercept on ML on the |"
INCLINED PLANE AND FRICTION.
263
Example 2. A block of weight 10 lbs. is s'wpported on an inclined
plane by a horizontai force. If 1^ = 0-3, and the plane rises 1 vertically
in 2 horizontally, find the value of the horizontal force that will just
cause the block to be on the point of motion, (i) upwards, (ii) downwa/rds.
First draw the plane
AG (Fig. 243) by drawing
AB = i" horizontally, and
BO =2" vertically, then a
normal MJV to the plane.
Set off along the normal
MN= 2", and (i) NK= 0-6"
down, and (ii) iVZi = 0-6"
up the plane.
When the body is about
to move up the plane, the
total reaction of the surface
has the direction MK, when
dovm the direction is MKj^.
Next draw the vector
polygon, a line FQ verti-
cally downwards 10 cms.
long, to represent the
weight /F of the block,
then from the two ends of
PQ, a horizontal line QR
and one PR parallel to
KM. Then QR gives the
horizontal force which will
just cause the block to be ~ -_ - P
on the point of motion up fig. 243.
the plane, and RP is the total reaction of the surface.
Draw PB^ parallel to MK^. Then QRj measures the horizontal
force which will just cause the block to be on the point of motion
down the plane, and R^P the corresponding reaction. Would the
block rest on the plane if the horizontal force were zero ?
264
GRAPHICS.
Example 3. The problem as before, but the slope of the phme
is now 1 in 8.
The graphical work is as in the previous example, but notice
now that iJj comes to the left of /Fand QEj is from right to left,
shewing that the pull is to be replaced by a push down.
Notice that whether the force pulls up or pushes down depends
on the relative magnitudes of e and o, where e is the angle of
friction and a the inclination of the plane.
Minimum Force and Inclined Plane. If a>E then the
body will not rest on the plane without a supporting force.
(Why ?) This condition being fulfilled, the total reaction QE
(Fig. 244) of the surface makes an angle of a-e with the
vertical when sliding down is about to take place; hence the
least force EP, which will prevent motion down the plane,
must be perpendicular to BQ or make an angle a-e with the
horizontal, or e with the plane and below it
Pig. 244.
When motion is about to take place up the plane, the total
reaction makes an angle a + e with the vertical, and the least
force is perpendicular to this reaction and makes an angle a + 1
with the horizontal, or e with the plane and above it.
INCLINED PLANE AND FRICTION.
265
If a<e (Fig. 245), then, when the body is on the point of
motion down the plane, the total reaction makes e - a with the
vertical, and the least force makes e- a with the horizontal, or
E with the plane downwards but above the plane.
Fig. 245.
When motion is about to take place up the plane the least
force makes e with the plane upwards.
(16) A gun has to be dragged up a steep hill (slope 1 in 5) ; the surface
resistance is equivalent to an angle of friction 40°. Find the best angle to
which the ropes should be adjusted. If the gun weighs 1 ton find the value
of the least force. What force would have to be applied if the ropes were
pulled (et) parallel to the ground, (ft) at an angle of 20° with the ground ?
(17) A weight of 3 kilogrammes is supported on an inclined plane (rising
I in 4) by a force parallel to plane. Find the greatest and least values this
force can have so that the weight may not move if /i=0'2.
(18) A weight of 7 owts. is supported on an inclined plane, risjng 1 in 1 ;
if /tt=0'3 find the least force that can support the weight.
(19) In the last example find the least and greatest forces parallel to the
plane so that the weight may be (i) on the point of moving up, (ii) on
the point of moving down.
(20) Find the least horizontal force that can move a block weighing
II lbs. up an inclined plane of inclination 30°, the angle of friction being 15°.
Find also the least force that will prevent motion downwards and cause
motion upwards.
What are the values of the friction called into play in the three cases ?
266 GRAPHICS.
(21) Find the force parallel to a plane of inclination 60° that will
support a block of weight 15 lbs. on the plane, the coefficient of friction
being 1/3. Find also the least force that can move the block up the plane.'
(22) A force of 17 lbs. weight will just support a block of weight 40 lbs.
on a plane inclined at an angle 55° if applied parallel to the plane. What
is the greatest force that can be applied parallel to the plane without
causing motion ?
(23) What is the inclination of a plane, coefficient of friction J, if the
minimum force necessary to move a block weighing 25 lbs. up it is 15 lbs. ?
(Remember the direction of the minimum pull is perpendicular to the
total reaction of the surface and this makes an angle e with the normal.)
For this plane what is the force that will just support the block ?
(24) A block of weight 5 cwts. is kept at rest on a rough inclined plane
by a rope AB fastened to - point A on the block and to a point on the
plane. The plane rises 3 vertically to 5 horizontally and /4=0'38. Find
the length of^the rope AB that will give the least tension if .4 be 1 foot
distant from the plane.
(25) Two light rods are pin-jointed together and rest in a vertical plane
on a rough board (/i=0'4). A weight W=9 lbs. is suspended from the joint ;
find the greatest angle between the rods consistent with equilibrium.
(26) In the example on p. 156, if the coefficient of sliding friction for the
piston be J, find the total reaction of the guides and the force transmitted
along the connecting rod. Find also the tangential force urging the crank
forward for the various positions given.
Harder Problems on Friction. In some cases a little
ingenuity is necessary to eflfect the graphical construction.
Example. A uniform beam, of length 11 ft. and weight 3 cwts., rests
against a smooth vertical wall and a rough floor for which the coefficient
of friction is 0'4. Find the position of the beam when it is just on the
point of slipping down, and the friction which prevents the motion.
Notice first that the reaction at A is horizontal, and that at
B inclined at « to the vertical. First draw the vector polygon,
XY (Fig. 247) vertically down, of length 15 cms. ; YZ hori-
zontally, of length 6 cms.; and join XZ. Join Z to the mid-
point M of XY.
From any point A draw AB parallel to MZ (Fig. 246) and of
length 1-7", AG vertical and GB horizontal. Then AB gives the
position of the beam relative to the wall AG and the ground CB.
*Proof. Since -j=p. = yg = 0-4, and YZ is horizontal, .^Z gives
the direction of the reaction of the ground at B, and XYZ
must be the vector triangle for the forces on the beam.
BEAM AND ROUGH FLOOR. 267
The three forces on the beam, viz. the weight through G (the
mid-point) and the reaction at A and B, must pass through a
point 0. Produce OG^ to .K" (as in Fig. 246), then 0KB and XYZ
are similar;
OK XY
^k=yE' ^^' *^^°' "^^ ^^ bisected at M,
MY GK
'Yy~K^ *°*^ therefore MZ is parallel to AB.
Pig. 247.
A Simpler Proof. Replace the uniform heavy beam by a
light rod having 1-5 cwts. concentrated at its ends; then for the
equilibrium at A we have the load given by MY, the reaction
at ^ given by YZ, and therefore the push of the beam along
BA must be parallel to ZM the closing line of the vector
polygon for A. YZ is known because it equals 0-4 XY.
Similarly, at B we have the load XM, the ground reaction ZZ
and the push MZ of the beam along AB.
268
GRAtHlCS.
Example. A uniform ladder rests against a wall at an angle of
30°. If it he just on the point of slipping down, and the angle
of friction is the same for wall and ground, find the coefficient of
friction.
B—
Fig. 248.
JB (Fig. 248) represents the ladder and G its M.c. through
which its weight is supposed to act. Then, since B is on the
point of moving to the right, the friction acts from right to left
and the total reaction at B makes some (unknown) angle with
the normal, and slopes towards A. At A, the total reaction
slopes upwards, for the friction acts upwards. Since the angle
of friction is the same at A and at B, these reactions must
intersect at right angles ; the point D of intersection, therefore,
lies on a semicircle having AB as base. For equilibrium, the
vertical through G must be concurrent with the reactions, hence :
Describe a semicircle on AB, draw the vertical through G to
intersect it at D, draw AD and BD the reaction lines, and
measure ;" ( = tan e).
(27) As in previous example, only the M.c. of the ladder is J of the way
up from the bottom.
LADDER AND FRICTION.
269
(28) Find where the M.c. of the ladder must be if the ooefRoients of
friction for the wall and ground are 0'3 and 0'5 respectively.
(29) A uniform bar AB of weight 27 lbs. rests on rough ground at A,
asid against a smooth bar at O. The inclination of the oar is 30° to the
horizontal ; AB=i ft., A0=5 ft. ; find the reaction of the ground and the
coefficient of friction if the bar is about to slip.
Example. The coefficients of friction for a ladder resting against a
wall and on the ground are 0-3 and 0'6. Find the limiting "position of
the ladder supposed imiform, and the friction on the wall, if the ladder
weighs 120 lbs.
Draw a vertical line PQ
(Fig. 249) of length 12 cms.
to represent the weight, and
from the ends draw PR and
QR parallel to the reactions of
the wall and ground. Bisect
PQ at S and join RS ; then
RS is the direction of the
ladder.
Proof. Replace the beam
by a light rod with equal
loads at the ends.
In Fig. 249 SQ is the load
at the bottom, QR the reaction
of the ground, and, therefore,
RS must give the push of the
beam on the ground. Hence
RS is parallel to the beam.
Fio. 249.
(30) Shew, by drawing ST parallel to EQ, and TG parallel to PQ, that
BS will represent the beam position. Notice that the reactions ET and
TS are in the right directions, and are concurrent with the vertical through
G the raid-point of BS.
270
GRAPHICS.
* Beam on Two Rough Inclined Planes.
Example. A beam rests on two planes of inclinations 30° and 45°
for which the coefficients of friction are 0-12 and 0-2. Find the two
positions of the beam when in limiting egwilibriv/m, the mass-centre
of the beam being ^ of its length from, the 30° plane. If the beam
weigh 700 kilogrammes, find thefrietion on the planes in the two cases.
(The vertical plane of the beam is supposed to intersect the
planes in their lines of greatest slope.)
Suppose the 30° plane to be on the left.
Set off PQ (Fig. 250) downwards of length 7" and draw PS and
QS parallel to the normals to the planes, i.e. PS making 30° and QS
making 45° with the vertical. Mark the point R where PB = 4:"
and QR = 3". Then set off the friction angle e, where tan e = 0"1 2,
on both sides of PS, and, similarly, set off tj on both sides of QS.
When the beam is about to slide down the 45° plane the
reaction of the plane tends to prevent the sliding and is, there-
fore, to the right of the normal ; at the same time the reaction of
BEAM ON TWO ROUGH PLANES. 271
the 30° plane tends to prevent sliding up and is, therefore, to the
right of its normal. For this case then QO^ and POj (Fig. 250)
are the directions of the reaction lines.
When the beam is about to slide down the 30° plane the
reaction lines will be parallel to PO and QO for the 30° and
the 45° planes respectively.
Now suppose the beam replaced by a light rod having 400 kilo-
grammes concentrated at its end A in contact with the 30°
plane, and 300 kilogrammes at the other end B. Then for
the equilibrium at A we have PR ( = 400 lbs.), the reaction of
the 30° plane and the push of the rod.
When A is about to slide up, the reaction of the plane is OjP,
and hence ROj gives the push of the rod at A ; therefore the rod
must be parallel to 0-^R.
Similarly, when A is about to slide up, PO is the reaction and
RO must give the push of the beam at A ; hence the beam must
now be parallel to OR.
Draw the planes XA of 30° inclination to the left, and XB of
45° inclination to the right. At any point A on the former,
draw ^F parallel to PO^, and AB parallel to O^R. From B on
the plane XB draw .BF parallel to QO-^. Then through Y, the
point of intersection oi AY and BY, draw a vertical cutting AB
inZ. Seetha.tAZ/ZB = ^.
In a similar manner, draw the other limiting position of the
beam A^B^, and verify the accuracy of the vector polygon
construction again.
To determine the friction in the first case, draw Oji'' perpendicular
to PS; then FP is the normal reaction and OjF the friction at A.
It is evident that there are not always two positions of
limiting equilibrium, e.g. if RO is steeper than 45°, or RO^ steeper
than 30°, there will only be one position ; if both happen together
there is no position of limiting equilibrium.
(31) A ladder rests against a vertical wall. The angles of friction for
the wall and ground with the ladder are 20° and 40°. Find the position of
the ladder when just on the point of slipping down, if the position of the
M.c. is J up the ladder from the ground.
272 GRAPHICS.
(32) A heavy beam weighing 1000 lbs. rests in limiting equilibrium with
one end on the ground and the other on a plane of inclination 30°. If the
ooeifioient of friction =0-4 for both ends, tod the beam be about to slip
when inclined at 20° to the horizontal, find the position of the M.c.
(33) As in previous exercise, only li is 0'4 for the ground and 0'3 for the
plane.
(.34) A heavy uniform beam AB weighing 700 lbs. rests with its end A
on a plane of inclination 30° and coefficient of friction 0'3. The other end
JS is on a plane of inclination 40° and coefficient of friction 0'4. If the end
B is about to slide down when the beam is horizontal, find the position of
its M.c.
(.35) A uniform rod of length 7" rests inside a vertical rough hoop of
radius 5". It is found that the greatest inclination that the rod can have
to the horizontal is 30'. Find the coefficient of friction.
(If O is the centre of the hoop and AB the rod inclined at 30° to
horizon, draw the circle circumscribing OAB ; this cuts the vertical
through M (the mid-point oi AB)m O; then GA and GB are the reaction
lines at A and B. Measure the tangent of the angle between each of these
lines and the corresponding radius. )
(36) In the previous exercise if the coefficient of friction at the ends are
0'3 and 0'2 (lower and upper ends), find the position of the M.c.
(37) The angle of friction being 20° at each end, and the rod uniform,
find the position of the rod in the loop when on the point of slipping down.
Draw AB in any position in the circle. Join OA and OB and draw the
reaction lines at A and B. Join the point of intersection of these lines
with M the mid-point of AB ; this last line represents the vertical.
Measure the angle between it and AB ; this gives the position of the beam,
(38) The angle of friction being 25° at each end, find the limiting position
of the rod when the mass-centre is distant 2" from the upper end of the
rod.
(39) A heavy beam weighing 1050 lbs. rests in limiting equilibrium with
one end on the ground and the other on a plane of inclination 60°. If the
coefficient of friction is 0'4 for both ends, and the M.c. is -^ up from the
ground, determine the position of equilibrium and the frictions at the ends.
(40) As in previous case, but the coefficient of friction for the end in
contact with the inclined plane is 0'25.
(41) A heavy uniform beam weighing 570 lbs. rests with one end A on
a plane of inclination 30° and the coefficient of friction 0"2. The other end
fi is on a plane of inclination 50° and the coefficient of friction is 0*4. If
the end B is about to slip down, find the position of the beam a'nd the
friction on the two planes.
Find if another position of limiting equilibrium is possible.
Example. Fig. 251 represents part of an ordinary bicycle screw-
spanner. By means of the screw thread S amd the rack BOj, an
upward force a is given to the movable piece. If a short rod DE is
placed between the jaws, required to find the force /? which is exerted on
the rod when the magnitude of a is 3 lbs. weight, and the coefficient of
FRICTION AND BICYCLE SPANNER.
273
friction between the movable and fixed parts is 0-4. The distance
between the axes of a and /3 is 1", between A and G 1-5", between B
and G 0-6".
The effect of - ^ downwards on the movable piece, is to press
the latter against the fixed part at A and B, and since the lower
jaw is tending to move upwards the friction acts downwards, and
hence the total reactions at A and B slope as in Fig. 251.
Draw the axes of the four forces, a, ^ and the two reactions,
the distances being taken double the actual ones. Find the points
FlQ. 252.
Fia. 251.
of intersection and Oj of /3 and the reaction at B, and a and the
reaction at A. For equilibrium, the resultant of a and the reaction
at A must balance /8 and the reaction at B, hence this resultant
must have the direction 00^ .
Draw a vertically upwards of length 3", and through its end
points draw <r, and o-^ parallel to AO-^ and OOj respectively. This
274
GRAPHICS.
lo:
A
gives a■■^ the reaction at A. From the ends of a-^ draw o- and - ^
parallel to OB and P respectively. Then a + a-- P + a■-^ = and
P is the force of the moving piece on the rod.
Another, and perhaps slightly simpler, way of solving the
problem is as follows. Find the point F of intersection of the
reactions at A and B. Eesolve a acting along BO-^ into two
parallel forces along ED and the parallel through F. The former
component is the reaction of the movable piece on the rod.
Compare the results obtained by the two methods.
Example. Fig. 253 represents the load stage and part of the
rack of a screw jack for raising loads eccentrically. AB is the pitch
line of the rack along which the lifting force
a acts, the load /i is carried by C. D and
E are pa/rts of the casing against which the
rack presses when a load is being raised.
AD = 0-5", DE = 5" and the distance of
the load line from AB is 3-5". When the
load is 2 '6 cwts., fimd the smallest magni-
tude of a necessary to raise the load if the
coefficient of friction between the rack and
casing is OS.
Draw the axes of the forces u and P
and the reaction lines of D and E.
Find the point F of intersection of the
two latter, and resolve yS through C into
two parallel forces, one through the
point F and the other along AB. The
compoaent along AB is a.
(42) Find u, also by the method first given for the previous example and
compare the results.
(43) Find the least value of a necessary to prevent the load descending.
Example. ABCD (Fig. 254) represents a horizontal d/rawer. It
is attempted to pull the drawer out of its case by a non-central handle
P. Neglecting the friction at the bottom of the drawer, how far may
the drawer be pulled out before jamming, «/ ju = 0"6 ?
/VY
if
Fig. 258.
FRICTION ON A DRAWER.
275
Suppose the drawer is pulled out a distance AA-^ ; then, owing
to the pull being non-central, the drawer is pressed at C against
the right slide, and at A■^ to the left. The reaction at C makes
angle € with CD (tan e = 0'6 : actually « for motion is a little less
than £ for rest) ; the axis of the pull meets this reaction line at
0. At ^j, the reaction is along A-J{, making e with A-^B-^^.
Neglecting the weight of the drawer there are three forces, and
Fia. 264.
three only, acting on it, viz. the pull and the two reactions.
For equilibrium these must pass through at point.
Now is fixed relatively to the drawer, hence A^ moves until
A-^K passes through 0, i.e. the drawer can be pulled out a
distance AA^, where QA^ makes an angle e with CD.
Hence to find point A^ : Divide CL into 10 equal parts, set
ofiF iO=6 of these. Mark N on CD, where LN=LC; join ON,
cutting AD a,t A^; then the drawer can be pulled out a distance
AA,.
276
GRAPHICS.
(44) Find the farthest distance that the handle may be from the centre
line in order that the drawer may be pulled out to within a quarter of its
length, /i=0'6.
(45) The handle being midway between the centre line and a, side, find
/i if the drawer jama when pulled out half its length.
Example. ABC (Fig. 255) represents a wedge, DE a wniform
iron rod of weight 7-2 lbs., hinged at D and resting on the wedge at E.
The coefficient affliction between the rod <md wedge is 0"3, and between
the wedge and ground it is 0'4. The wedge is pushed by a horizontal
force a so that it is just on the point of motion ; determine a if the
weight of the wedge may be neglected ; given
DE = 4-6'/i!., DB = l-9/i(., AC = 3-2ft., BC = 4//.
»
A
A
/ *
f ,'
^^^^
K -a
/
G y/^
Pio. 266.
The line of the reaction at E is known, also the weight of the
rod acts through its mid-point, and since the rod is in equilibrium
the direction of the reaction at B is known; hence find the
forces by the vector polygon. The force at E on the wedge is
now known, the axis of a. is known, and 'also the angle the
reaction of the plane makes with the vertical; hence draw the
vector polygon for the wedge, and determine a. from it.
(46) The weight of the wedge being 3 lbs. , acting through the m.c. of the
triangle ABC, determine a.
(47) Determine a when AC=BO= 1 ft. ; |U is the same for both s^rfaoes,
and the weight of the wedge is 4 lbs,
FRICTION ON A CUBE.
27?
In the problems on friction so far discussed it has been
supposed that the body considered remains in equilibrium until
the total reaction of the surface makes the angle of friction with
the normal. This is not always possible, as the body may begin
to turn about some point or edge before sliding commences.
Example. A cube of 2" side' rests on a rough horizontal plane,
for which jli = 0'6 ; it is acted on ly a horizontal force perpendicular to
a face and passing through the mid-point of the top face. Shew that,
however large this force, the cube will not slide, but that equilibrium
mil be broken by the cube twrning about an edge.
Draw a square of 2" side to represent the central section of
the cube containing the axis of the horizontal force.
While the cube is in equilibrium the three forces — horizontal
push, weight, and reaction of the plane — pass through a point.
This point 0, Fig. 256, is determined by the intersection of the
horizontal axis and the vertical through the M.o. of the cube.
If about to slide, the reaction of the surface makes an angle with
the vertical whose tangent is 0'6 ; draw this line OA-^ through
0; it cuts the plane outside the base of the cube. But
the plane reaction must act within the base, and hence it is
impossible for the cube to be on the point of sliding.
As the push u. increases from zero, the total reaction of the
278
GRAPHICS.
surface makes a larger and larger angle with the normal until
it comes to the position OA. Evidently when the reaction is
at A, AO must barely touch the ground except at A, and the
cube must be on the point of rotating about the edge through A.
Draw the vector polygon of the forces when OA is the line
of reaction of the ground, and determine the greatest value of a
consistent with equilibrium.
Example 2. A cable reel has an outside diameter of i ft. ; the
radius, from which the cable is un-coiled, is at a certain moment 1-36
ft. The reel is placed on the ground {coefficient of friction = 0-28) ;
find the point at which the cable must be taken off, and the direction
of the cable at that point, so that the reel may be just on the point of
slipping, and find the force necessary to effect this if the reel and
0'32
tons.
Draw (Fig. 257)
to scale circles
representing the
reel AB which
rests on the
ground, and the
layer from which
the cable is being
uncoiled. Then,
since both the
weight and the
reaction of the
ground must act
through A, so
must the pull of
the cable. Draw
AB^& tangent to
inner circle, which gives the direction in which the cable is taken
off, and the point B^ (there are two such points) at which the
cable leaves. The vector polygon must now be drawn, the three
FRICTION AND REEL.
279
sides being parallel to a, (3 and y respectively ; € is the angle of
friction.
Measure the pull of the cable and the reaction of the ground.
Since for equilibrium the three forces must pass through A if
the cable be taken off at any other point than B-^ the reel will roll.
Try an experiment, illustrating this, with a reel of cotton.
Example. The ground slopes at an angle of 30° and the reel
is just on the point of sliding doion, find the point at which the
cable leaves the reel and
the direction and magni-
tude of the pull on it.
Draw (Fig. 258) the
plane and reel in posi-
tion as shewn, also
the vertical through
the centre G of the
reel. Since the reel is
by supposition about
to slide down the fric-
tion must act upwards ;
draw then, at A,& line
making the angle of
friction with the nor-
mal. This line cuts
the vertical through G
at a point ; from
draw tangents to the
inner circle touching
it at points B and B^ ;
these points of contact
are the points at which
the cable may leave the reel, and the tangents themselves are the
directions of the cables.
Now draw the vector polygons for the two cases and determine
the senses of the pulls and their magnitudes.
280 fiRAtHlCS.
(48) One end of the cotton on a large reel is fixed to a vertical rough
wall ; if the reel rests in equilibrium against the wall and is just on the
point of slipping down, find the point at which the cotton must leave
the reel; given /t=0'6, the outer radius of reel=l", the inner radius at
which the cotton unrolls =0'45".
If the reel weigh 7 ozs. find the tension in the cotton.
(49) One end of the cotton is fastened to a rough plane, /i = 0'3, of
inclination 45°. Find the inclination of the cotton if the reel is about to
slide down the plane.
(50) Why is it not possible to cause the reel to be about to slide up the
plane by pulling at the cotton.
(51) Stand a thick book upright on a table and push it, perpendicular
to the cover, with a pencil, first near the table and gradually increasing
the distance until the book topples over. Measure the thickness of the
book and the height of the push and calculate ju.
(52) Draw a rectangle of height 4" and base 2" to represent the right
section of a cuboid through its centre, at points distant i, 4, |, ... inches
from the base ; suppose horizontal forces applied in turn until equilibrium
is broken. Mark the corresponding pbints on the base where the total
reaction of the surface cuts it, the coefficient of friction being 0'3. Find
the highest point at which the force may be applied.
(53) Draw a square of aide 3" to represent the section of a cube through
its centre, and a line through a top corner inclined at an angle of 30° with
the horizontal, to represent the line of action of a pull on the cube. The
coeificient of friction being 0'4, find the greatest possible pull, if the
equilibrium remains unbroken. Draw the axis of the total reaction of
the surface. Weight=10'7 lbs.
(54) In the last example, if the pull be exerted at an angle below the
horizontal, find the greatest angle for which sliding is possible and the
greatest pull possible if the cube does not move.
(55) ABO (Fig. 259) is the right central section of a triangular prism,
ABO being a right angle. BO rests on rough horizontal ground. A rope
is fastened to A and pulled horizontally as indicated parallel to BO. If
AB=i'l', BO=3'5', and /i4=0"3, determine if sliding is possible. If so,
find the least value of a that will cause sliding.
(56) If o be reversed in sense and /t=0-4 will sliding be possible?
Give reasons. What value of a will cause the equilibrium to be broken
in this case, and what value of n would cause it to be broken by sliding
with this value of o ?
(57) BO (Fig. 260) is part of an inclined plane of inclination 20°. ABGD
is a cube kept in position by a string parallel to BO and fastened to D.
What is the greatest value of /u. consistent with equilibrium? If it. has
this value find the greatest value of the pull in the string consistent with
equilibrium if the cube weighs 11 lbs.
(58) As in last example, but let string slope upward at an angle of 45°
with, horizontal.
(59) In Fig. 260 make AB=2B0. If fi.=0-3, find the direction of the
string attached to Z) so that the prism may be on the point of turning
about the edge at O.
MISCELLANEOUS EXAMPLES.
281
Fia. 260.
MISCELLAITEOUS EXAMPLES. VII.
1. An experiment was performed in which a loaded slider was, by a
suitable horizontal force P, caused to be just on the point of motion.
Plot the values of P and the weight of the slider W given in the table on
squared paper, and determine approximately the value of the coefficient
of friction.
P in lbs. wt.
0-45
0-65
0-84
11
1-3
1-4
1-7
W in lbs. wt.
2-3
3-3
4-3
5-3
6-3
7-3
8-3
282 GRAPHICS.
Direction of Motion
2. In Fig. 261 the circles represent "^
the coupled driving wheels of a railway
engine. If the engine is starting, show
roughly the direction of the pressure
between the wheels and the rail. Give
a reason for your answer.
(Naval Cadets. )
FlQ. 261.
3. A particle whose weight is 10 lbs. is placed on a rough plane inclined
at an angle of 30° to the horizon ; it is acted on by a force up the plane
equal to the weight of 6 lbs. , acting along the plane ; the particle does not
move ; find the friction between the particle and the plane.
If the particle is just on the point of sliding, find the coefficient of
friction. (B. of E., II.)
4. A uniform rod rests with one end against a smooth vertical wall,
and the other end on a rough horizontal plane ; it can just stand without
sliding when its inclination to the horizon is 45° ; find the coefficient of
friction ; also find the inclination when the friction called into play is one-
half of the limiting friction. (B. of E. , II. )
5. If the angle of friction for an inclined plane be 45°, determine
completely the least force that will drag a weight of 100 lbs. down a plane
inclined at 30° to the horizontal. (B. of E., II., 1906.)
6. Find graphioalliMihe magnitude of the least horizontal force which
a < tan"' /It.
(B.So., 1904.)
7. Define the coefficient and the angle of friction. A body weighing
600 lbs. is sustained on a rough inclined plane (base twice the height) by a
rope pulled in a horizontal direction. Prove that the greatest and least
tensions of this rope consistent with equilibrium are about 389 and 134
lbs. wt. (Inter. Soi., 1904.)
8. Find, graphically by .preference, the direction in which a force of
given magnitude must act if it is just able to move a body of given weight
up a rough inclined plane, the coefficient of friction being known.
Shew that when motion is possible there are in general two such
directions. Hunter. Soi., 1900.)
9. A beam rests against a smooth vertical wall and a rough inclined
plane of inclination a passing through the foot of the wall. Determine the
greatest angle the beam can make with the vertical.
(Inter. B.So. (Eng.), 1905.)
10. Define friction and limiting friction. Explain briefly what is
meant when friction is said to be a passive force.
.4 B is a uniform rod of weight 10 lbs. ; it lies on a rough horizontal table,
and is pulled at the end B in the direction of its length by a force of 2 lbs.
If ^5 stays at rest, how much friction is called into play?
Everything being as it was, a thread is tied to the end B and is pulled
vertically upwards by a gradually increasing force P ; find the least
coefficient of friction for which P will begin to lift the point B. How will
the rod begin to move if the coefficient of friction equals 0-25 ? (B. of E., II.)
MISCELLANEOUS EXAMPLES. 283
11. A body is placed on an inclined plane and the coefficient of friction
is J ; it is acted on by a force along a line of greatest slope ; find the force
when it is on the point of making the body slide up the plane.
(B. of E., IL, 1903.)
12. A ladder A B rests on the ground at A and against a vertical wall at
B. If AB is inclined to the vertical at an angle less than the angle of
friction between ladder and ground, shew geometrically that no load, how-
ever great, suspended from any point in the ladder will cause it to slip.
(B.Sc., 1905.)
13. A weight rests on a rough inclined plane, whose inclination (a)
exceeds the angle (\) of friction, being prevented from sliding by a force P.
Find (geometrically or otherwise) the direction and magnitude of the
least force which will suffice for this purpose. (Inter. Sci., 1906.)
14. A uniform circular hoop is weighted at a point of the circumference
with a mass equal to its own. Prove that the hoop can hang from a rough
peg with any point of its circumference in contact with the peg, provided
the angle of friction exceeds 30°.
(Relative to the point of support the M.c. of the hoop and particle lies
on a circle of radius half that of the hoop.) (Inter. Sci., 1905.)
15. Draw a horizontal line ABO, AB = l" and B0=3". Let ABO
denote a uniform beam of Weight w resting on a rough prop at B, 4nd
underneath a rough prop at A . Find the ifirection and magnitude of the
least force applied at the end which will just begin to draw out the beam
from between the props. (B. of E., II., 1906.)
(Draw the reaction lines at the points A and B to intersect in D,
resolve the weight of the beam acting at its M.c. into two, one passing
through D and the other through C, the latter component to be the
least possible. )
16. Prove that a sash window of height a, counter-balanced by weights,
cannot be raised or lowered by a vertical force, unless it is applied within
a middle distance a cot (0 the angle of friction).
If the cord of a counter-balance breaks, the window will fall unless the
width is greater than a cot 0. (B. So. , 1902. )
17. A square window sash weighing 30 lbs. slides vertically in grooves.
From the two upper corners sash cords are carried over pulleys and carry
two counterpoises each of 15 lbs. Shew in a diagram the forces acting
on the sash when one of the sash cords breaks, and find the least coefficient
of friction between sash and grooves that will keep the sash from sliding
down, if all other friction may be neglected. (C.S., Div. I., 1905.)
18. If a body having a flat base is placed on a rough inclined plane of
inclination i and angle of friction X, and the body is pulled by a horizontal
force P, prove that for equilibrium P must lie between the values
Wtaa(i + \) and HKtan(j-X) where X= weight of the body. If X > i,
explain the second case. (Inter. Sci. , 1906. )
19. Define the coefficient of friction and the angle of friction for two
rough bodies. A mass of 500 lbs. on a rough inclined plane for which the
coefficient of friction is i and whose inclination is tan"' ^ is sustained by a
rope which is pulled in a horizontal direction ; prove that the greatest and
least tensions of this rope are about 389 and 136 '4 lbs. wt. respectively.
(Inter. Sci., 1904.)
CHAPTER VIII.
MOMENTS.
To obtain a real grasp of the theory of moments the experiments
described in Appendix I. should be performed. It is only by
actually performing such experiments that the physical meaning
of turning moment or torque becomes realised.
Definition. The moment of a force about a point is the
product of the force and the perpendicular distance of its axis
from the point. It is positive if the direction and sense of the
force relative to the point is contraclockwise, negative if clock-
wise.
Geometrical Representation. From the point draw any
line to the axis, then the area of the parallelogram which has
this line and the vector of the force as adjacent sides, measures
the moment. If the sense of the boundary, as given by the
vector, is contraclockwise, the moment is positive.
Thus, if (Fig. 262) is the point, a the axis, and a the vector
of the force, the positive area OABC measures the moment.
Whatever the direction of 0^, the area and the sense of the
boundary remain the same.
For a given force, the moment in general changes when the
point is changed in position. In Fig. 262 the farther is to
the left of a the greater the positive moment. When is on a
the moment vanishes, and on crossing to the right of AB, the
moment is negative. If, however, moves on a line parallel to a,
the moment remains unaltered.
MOMENTAL AREAS. 285
It is important to notice that when the moment of a force is
zero about a point 0, we may have either OA or AB zero, i.e.,
the force itself may be zero, or it may pass through the point ;
to decide which alternative is correct further infornfiation is
necessary.
•a
Fig. 262.
Taking account of sense, OABC or an area equivalent to
it is called the momental area of the force u, in a about the
point 0.
Unit Moment. There is no special name in general use for
the unit moment. If we use a lb. wt. as the unit of force and a
ft. as the unit of length, then the unit moment may be called
one lb. ft. moment. This of course might mean a force of 1 lb. wt.
at a ft. distance, or 2 lbs. wt. at 6" distance, etc. ; later we shall
see that these are really equivalent. Whatever the units of force
and length used, it is necessary to specify them in giving a
number as the measure of a moment.
Graphical Measurement of a Moment.
Example, ab (Fig. 263) is the axis, AB the vector of a force.
Required the measure of the moment of the force about a point O
{distant p from ab). AB = 2'65", scale \ cm, to \ lb. wt., cmd
P = 4-36",
286
GRAPHICS.
Take a pole P of the vector polygon at a distance h (2") from
AB. Through draw a line A■^B^ parallel to ah, and through
any point Q in ah draw Q^^ parallel to PA, and O^i parallel
to PB, cutting A-yB-^ in ^i and B-^.
Measure A-^B^ on the \ cm. scale ; it is the moment of the
force AB about in lbs. inches.
lbs. inches (moments)
4 6 8 lo 12 14 i6 i8
34567
lbs. wt
Proof. PAB and QA-^B^ are similar triangles, and therefore
p _ h
A^~AB'
:. 4:-S6AB = 2AyBy.
From the ratios it is seen that p and h being in inches, A^B-^
must be measured on half the AB scale-
GRAPHICAL MEASUREMENT. 287
(This is really our old argument of p. 46 ; the moment is represented by
a rectangle p . AB, or by an equal rectangle AiB^h. If the base h is the
unit of length then the altitude -4]S] measures the area ; if A be twice the
unit of length, the altitude is only J what it was before, and to obtain
the old altitude we must multiply by 2, or use a scale with J the old unit. )
Sense of a Moment. If the vector polygon and moment
diagram be drawn according to rule, an inspection of the latter
will shew whether the moment is positive or negative.
The radial lines of the vector polygon are always supposed
drawn in the order of the vectors, and the link polygon lines in
the same order.
Hence the order in which the points A-^, B-^, ... are drawn gives
the sense of the intercept — downwards in the case considered.
The force being in ab and downwards, the moment about is
negative ; hence the rule : if the intercept has a downward sense the
moment is negative, if upwards, a positive sense.
(1) Verify this rule by taking on the other side of ab.
(2) Verify again by taking P on the other side of AB.
(3) A force is given by a length 3'48 inches (scale 10 lbs. to 1'25 cm.),
find graphically its moments about points distant 6 '72 ft. from it on
opposite sides of the force axis.
(4) Find the moments in Ex. 3 by drawing through O lines parallel to
PA and PB and measuring the intercept on ab.
Another Graphical Construction. With as centre — on
your drawing for Fig. 263 — describe a circle of 1" radius. From
any point Q on ab draw QO and a tangent QT to this circle.
Find the components of AB in db along QO and QT. Measure
the latter component on the cm. scale ; it is the moment of AB
about in lbs. ft. units.
Proof. The sum of the moments of the components of a force
is equal to the moment of the force itself (see p. 296). As one
component passes through and the other is at unit distance
from it, the second component must measure the moment.
Sum of Moments of Like Parallel Forces.
Example. Draw fow pa/rallel lines distant apart 1'22, 2 '38 and
1-94 inches, and let downward forces represented by lines of 3, 4, 2 -5
288 GRAPHICS.
and 3-7 cms. {scale 1" to 10 lbs. weight) act in these. Take a point
1-43 inches to the left of the first axis. Find the sum of the moments
of these forces about 0.
Add the vectors of the forces, and choose a pole P distant
3 inches from the vectors (Fig. 264).
Draw the link polygon R^B^R^R^R in the usual way and
produce the links to cut the line through parallel to ab in
Xj, Xj, ..., Z5.
Measure X-^X^ in inches and multiply by 30 ; the product is the
sum of tTie moments in lbs. inches.
Proof. Let x^^, x^, x^ and x^ be the distances of from ab, be,
ed, de. Since the A PAR is similar to R^X-^X^,
AB^X^ ._^_ AB.x^ = hX,Xo.
A Kj 112
Again, PBC, POD, PDE are similar to R^X^^, B^X^X^,
R^X^X^, respectively;
.'. BC .X2 = h.X^g,
GD.x^ = h.X^X^,
DE.x^ = h.X^X,.
Adding, we get
Sum of the moments of the forces about
= h{X,X, + X,X, + X,X, + X,X,)
= hX-^X^.
And since X-^Xr, is downwards, the sum is negative and the
moment cloekwise.
Obviously, for more than four forces we only need to extend
the construction ; the method will be exactly the same.
The sum of the moments is thus represented by a rectangle of
height X-^Xc, and base h. The moment of 1 lb. inch is represented
by a rectangle of height O'l" and base 1"; hence, since A. = 3",
to reduce h . X^Xr, to unit base we must treble the altitude,
i.e. 3 . X-^X^ measured on the tenth inch scale gives the sum of
the moments in lb. inches,
MOMENTS OF PARALLEL FORCES.
289
The moment of the resultant force AE acting through E is equal to
the sum of the moments of the components.
If X is the perpendicular distance from on the axis of JE,
then
rM6 = 4^ or JE.x = h.X,X,.
X h ^ ^
30 60 90 120
lbs. inches (moments)
Fla. 264.
If the point about which moments are to be taken is at Oj, in
the space c, then the moment of AB is h.SjS2 and is positive; the
moment of- BC is h . S^Sg and is positive ; the moment of CD is
h . S^^ and is negative ; and the moment of DE is h . SJS^ and
is negative : the algebraic sum of the moments is thus h . S^S^ or
the intercept between the first and last lines of the link polygon
multiplied by h.
For all positions of then, the intercept, between the first and
last lines of the link polygon multiplied by h, measures the sum
of the moments.
A simple inspection of the figures shews that this sum must
always be equal to the moment of the resultant.
290 GRAPHICS.
Fig. 264 has been drawn for parallel forces having the same
sense; the conclusion applies to all parallel forces which have'
a resultant.
The sum of the moments of a number of parallel coplanar
forces about any point in their plane is eqtual to the moment
of the resultant about that point. The algebraic sum of the
moments is given by the intercept, between the first and last
lines of the link polygon, on a line drawn through the point
parallel to the forces.
Notice that, the sum of the moments about any point in tk
resultant is zero.
(5) Find the sum of the moments about points in the spaces a, d and e.
Sum of Moments of Unlike Parallel Forces.
Example. Draw six pa/rallel lines ab, be, cd, de, ef (from left to
right), the spaces b, c, d, e and f being 0-82, 1-2, 1-46, 1-79 and
ri3 inches wide; forces of 4-6 (down), 1-5 (vp), 5-45 (down),
2-8 (down), 5 (up) and 3'2 (down) lbs. weight act in these lines.
Find the sum of the moments about a point distant 1-18 inches to
the left of ab.
Take a pole P (Fig. 265) at 2 inches distance from the sum AE
of the vectors, and proceed exactly as before, the only difiference
being that the link polygon is now re-entrant. Measure XjJj
on the force scale and double the number obtained ; the result
is the sum of the moments in lbs. inches.
(6) Parallel forces of 3-3, 4-1, 2-3, 1-5 and 2-8 tons weight aot on lines
distant 7, 8, 4 and 6 ft. apart from left to right, the first and last forces
being downwards and the rest upwards; find the sum of the moment
in ton ft. units about the points,
(i) distant 3 ft. to right of axis on extreme right,
(ii) distant 4-5 ft. from first and 2-5 from second axis,
(iii) distant 1-8 ft. from third and 2-2 ft. from fourth axis.
(7) Find separately in cases (ii) and (iii) the sum of the moments of all
the forces on the left and on the right of the point.
MOMENTS OF PARALLEL FORCES.
291
(8) The distances apart of the centres of the wheels of an express engine
and tender are 9' 8", 5' 3", 6' 0", 11' 2-5", 6' 6" and 6' 6" from the leading
wheels backwards. The loads carried by these wheels are 14 tons 10 ewts.,
17 tons 8 owts., 14 tons, 14 tons 10 ewts., 12 tons 5 ewts., 12 tons 10 ewts.
and 13 tons 5 ewts. The engine is partly on a bridge, one bridge support
being mid-way between the centres of the fourth and fifth wheels. Find
the sum of the moments of the loads about the support.
r— xp
6 8
lbs. inches (moments)
Pig. 265.
Sum of the Moments of Parallel Forces in Equilibrium.
The sum of the moments of such a set of forces is zero for all
points in their plane, and the sum of the moments of all the
forces on one side of a point is equal in magnitude — but opposite
in sense — to the sum of the moments of all the forces on the
other side.
Any one force of the given set reversed in sense is the resultant
of the rest. Its moment, for all points, is minus the moment of
the resultant and this is equal to the sum of the moments of the
rest. Hence the sum of the moments of the given set of forces is
zero. A proof from the link polygon-is given on p. 292.
292 GRAPHICS.
Example. A locomotive has the centres of the wheels from front
to rear at the follomng distances apart, 8' 9", 5' 5", 5' 5", 6' 0". The
loads borne by these wheels are 6 tons 8 cwts., 14 tons 6 cwts., 14 tons
8 cwts., 16 tores 7 cwts., 16 fo?is 7 cwjfe.; the engine is on a freely
supported bridge of length 40 ft., and the leading wheel is at a distance (
of 9' from the left-hand pi&r. Find the sum of the moments of all the
forces to the left of a point mid-way between the third and fowrth
wheels about that point.
Draw the reaction and load lines of the bridge to scale (say,
1 cm. to 1 ft.) ; then set out the load vectors AB, BO, CD, DE and
EF to a scale of (say) 1 inch to 10 tons (Fig. 266). Take a pole
P at a distance of 20 cms. from AF, and draw the link polygon
XR-iR2B^RiR^Y ; X and F being the points on the reaction lines.
Since there is equilibrium, XY must be the closing line, and
the reactions are determined by drawing PO in the vector
polygon parallel to XY.
Through Z, the mid-point of the space d, draw a line parallel to
the axes and cutting XY in M and R^R^ in N.
Measure MN on the force scale and multiply by 20 ; the pro-
duct is the sum of the moments, about Z, of all the forces to the
right or left (taking account of sense) of Z.
Proof. To find the sum of the moments of all the forces to
the left of Z, we find the intercept between the first and last
line of the link polygon. The first force (the reaction) has OA
as its vector, and therefore the first line of the link polygon
(drawn according to rule) is XY, the second is X^R-^, the third
i?ii?2, the fourth is R^R^ and the final (taking into account only
forces to the left of Z) is R^R^. Hence MN gives the sum of the
moments of all forces to the left of Z, its sense is downwards and
the moment therefore negative.
The unit moment of a ton ft. is represented by a rectangle of
height 0-1 inch and base 1 cm., the sum of the moments is given,
from the construction made, by a rectangle of height MN and
base 20 cms. If we measure MN, therefore, on the force scale and
multiply by 20, we have the sum of the moments in tons ft.
MOMENTS OF PAEALLEL FORCES.
•293
(20 cm. was taken as h instead of 10 to avoid the lines of the
link polygon being too steep.)
For the sum of the moments about Z of all the forces on the
right we have R^B^ as the first link, and since FO is the last
force, XY is the last link, and the intercept is now NM. The
sum of the moments is, therefore, of the same magnitude, but of
the opposite sense.
The total sum of the nioments is therefore zero.
Evidently, the deduction is- true for all parallel forces in equilibrium.
oX
^-- a
_
d
c
l'^
e
/
A
R,
^ivfr^
B
\
^
C
\
\
R3
R..
Ra
n;
--.^
k
Tons wt.
ID
10 20 30
40
200 100
200 400 600
Tons ft. (moments)
Scale of Feet
800
2
?.?.4 .6.8.1.0.13.14
Fin. 266.
(9) The distances apart of the centres of the wheels of an express engine
and tender are (from the leading wheel backwards) 12' 0", 10' 0", 8' 7'25",
6' 9" and 6' 9" The loads borne by these wheels are *21 tons 15 owts.,
* This load of 21 tons 15 cwts. is really that borne by the two pairs of bogie
wheels ; they are taken as one load to make the example simpler. The first
distance, namely 12' 0", is from the centre of the bogie truck to the front driving
wheel.
294
GRAPHICS.
Ri
19 tons, 19 tons, 12 tons, 12 tons 5 cwts., 12 tons 15 owts. The engine
stands on a bridge, the left-hand support being 11' 6" from the leading
wheel (centre), and the bridge is 75 ft. long. Find the sum of the moments
of all the forces to the left of the centre of the bridge about the centre.
Two Parallel Forces, (i) Of same sense. This is only a
special case of the general construction, but it is worthy of
separate consideration.
Draw (Fig. 267) any two
parallel lines ah and he, and
vectors AB, BC of the forces
supposed to act in those lines.
Construct the axis of the
resultant in the usual way
and mark R^, R^, R and R^
where the axes of the forces
cut the links. R^R x h then
gives the moment of the force
in he about any point in axis
of resultant, and RR^ x h gives the moment of the force in ab.
From the triangles R-^R^R and R^RR^, which are similar to
triangles in the vector diagram, we obtain
..^Trr^i^P
:5A
RoR
-^^ and ^3^2-^^
and therefore
R^R^
BG
'AB'
or, the resultant divides the distance between the axes inversely
as the magnitudes of the forces.
Notice that iJjiJg i® ^^Y ^^"^^ '> i^> then, the axes turn round any
points i?i and R^ and remain parallel, the axis of the resultant
turns round a fixed point in R-^R^, viz. R^.
(ii) Of opposite senses. Construct as in the previous case ;
now BG is upwards and the axis of the resultant is external
to the other axis and nearer the greater force.
See from the similar triangles that the axis of the resultant
divides externally the distance between the other axes inversely
as the magnitude of the forces.
MOMENTS OF A COUPLE OF FORCES.
295
(iii) Equal in magnitude but opposite in sense. This is the
case already considered in the chapter on the link polygon (p. 184).
(10) The axes of a couple are 3 '76" apart, each force is of magnitude
7 "21 lbs. -vreight. Take the pole at unit distance from the vector line and
find a line giving the momental area of the couple in lb. inches.
(11) With the same forces as in Ex. 10 find a line giving the momental
area of the couple in lb. centimetres.
(12) If a kilogramme =2 '204 lbs. find the momental area of the couple in
Ex. 10 in kil-centimetre and kil-inch units.
Moments of a Couple. — Direct Proof. The sum of the
moments of a couple of forces is the same for all points in the plane
and is equal to the momental area of the couple.
u, and - a being the forces and
(Fig. 268) any point, the moment
of a about is given by OABEO,
that of - a about is given by
OEGDO. The algebraic sum of
these is ABCD, which is the mo-
mental area of the couple, and this
result is quite independent of 0.
The moment and the couple are two distinct things. The
couple is simply the pair of forces, the momental area of the
couple measures the sum of the moments of the pair of forces
about every point in the plane.
Sum of Moments for Concurrent Forces. Draw any
parallelogram OAGB (Fig. 269) and let
OA and OB represent concurrent forces,
then 00 represents their resultant.
Take points P, P^ and P^ as indicated,
and measure the perpendiculars from
them on OA, OB and 00. Find the
algebraic sum of the moments of OA
and OB about P, P^ and P^, and com-
pare with the moments of the res'ultant 00 about those points.
(13) Draw any three non-concurrent lines and take an arbitrary vector
polygon for the forces in them. Find the axis of the resultant by the
link polygon. Measure the forces and resultant to any scale. Mark some
296
GRAPHICS.
point on the paper and measure the perpendiculars from the point on the
four forces and calculate the moments. What is the connection between
these moments ?
The algebraic sum of the moments of two concurrent forces
is equal to the moment of the resultant about all points in
their plane.
Proof. Let a, h and c (Fig. 270) be the axes of the two forces
and their resultant, the senses being as indicated. P is any point
in their plane.
Through P draw a
line parallel to c, cutting
a and 6 in ^ and B.
Then AB may be
taken to represent the
resultant in magnitude,
direction and sense, and
AOB is the vector poly-
gon for the forces.
Then the moment of
AO about P is twice
AOP and is positive, the moment of OB about P is twice OBP
and is negative; therefore their algebraic sum = twice AOB and
is positive.
(Notice that the sum is the same for all points m AB or AB
produced, and that, whatever the position of P, it is always twice
the area of AOB and in the sense of the letters.)
Draw BO I parallel to AO ; then twice AOB = AOO^B and
therefore measures the moment of AB in c about A, B or any
point P in AB.
It will be seen that the proof is perfectly general for all
positions of P.
For any system of coplanar forces the sum of the moments
of the components is, for all poles, equal to the moment of
the resultant (if there is one), or to the momental area of
the resultant couple (if there is one) or is zero.
SUM OF MOMENTS OF COPLANAR FORCES. 297
Proof. This result follows at once from the link polygon
construction, which consists in finding the resultant of concurrent
forces two at a time, and since the theory of moments holds
at each new composition it must hold at the final step when the
resultant or resultant couple is found.
As a particular case consider the decomposition on p. 203.
Taking moments about X, we have moment of AB = moment
of AD, and hence AD is uniquely determined.
Sum of the Moments of any Number of Forces about
a Point. If the forces have a resultant, take the pole of the
vector polygon at unit distance (or a simple multiple thereof)
from the resultant vector. Measure, on the force scale, the
intercept between the first and last lines of the link polygon
on a line drawn through the given point parallel to the resultant
vector.
If the forces are equivalent to a couple (the vector polygon
closed) take a force of unit magnitude (or a simple multiple
thereof) as the arbitrary force vector and measure, on the
distance scale, the perpendicular between the first and last lines
of the link polygon.
(14) The magnitudes of five forces are given by 3'7, 4:'8, 6'1; 2'8 and
5'3 cms., the scale being 17 inches to 10 lbs. weight. The coordinates of
points on their axes are (0, 0), (1, 2), {.S, 2), (1, 4), (2, 3) inches respectively,
and the forces are directed towards N. (the y axis), N.E., 15° N. of E.,
S.W. and W. Find tlie sum of their moments in lbs. inches about a point
whose coordinates are (1'5, 27), the coordinates being all measured in
inches.
(15) Choose the magnitudes of the first and last forces in Ex. 14, so
that the vector polygon is closed. Find the momental area of the
equivalent couple.
(16) Find the siim of the moments of three forces of magnitudes 3, 7 '2
and 5 lbs. weight acting along AB, BC, CA the sides of a triangle about
a point whose coordinates are (4, 2) inches. The coordinates of the vertices
A, Band Cof the triangle are (2, 2), (1, 3), (3, 4) inches.
(17) Draw a triangle ABO of sides AB=3, BC=3-5 and CA =4 inches ;
through the vertices draw three parallel lines, and suppose forces of
magnitudes 6"2, 7'4 and 9 lbs. to act in these lines through A, B and G
and to have the same sense. Construct the line of action of the resultant
in the usual way. Draw three more parallel lines through A, B, O, making
45° with the first set, and suppose forces of the same magnitude as before
298 GRAPHICS.
to act in them ; construct the new axis of the resultant. Repeat the
construction for parallel lines which are perpendicular to the first set.
See that the three axes of the resultants are concurrent. This point of
concurrence is called the centre of the parallel forces.
(18) The coordinates of four points are (1, 2), (0, 3), (4, 0) and (2-5, 3-6).
Parallel forces of like sense act through these points and are of magnitudes
2 '2, 3 '5, 1'8, 3 lbs. weight. Construct, as in previous example, three link
polygons, the sides making 45° with each other respectively, and shew that
the three axes of the respective resultant forces are concurrent.
(19) Three parallel forces equal in magnitude and of same sense act
through the vertices of a triangle. Shew by three constructions that the
resultant passes through the m.c. of the triangle.
Centre of Parallel Forces.
A number of parallel forces pass through points A, B, C, ...
respectively. If the axes turn about these points so as all
to remain parallel, then the axis of the resultant turns about
a fixed point in itself, the centre of the parallel forces.
Suppose masses to be at A, B, C, ..., whose magnitudes have
the same numerical values as the forces acting through the
points. The mass-centre of these mass-points must lie on the
axis of the resultant force, for the sum of the moments of
the forces and of the masses have exactly the same numerical
value. Supposing, then, the axes of the forces to have a different
direction, the M.C. of the mass-points must be at the intersection
of the axes of the resultants. But the masses can have only one
M.c, and, therefore, whatever direction the parallel forces may
have, the resultant must always pass through a fixed point, viz.
the mass centre of the masses.
Since the mass-centre theorem remains true if some of the
points be considered as having negative masses, the above theorem
remains true if some of the forces have a different sense from
the rest.
(20) Parallel forces of like sense and of magnitudes 1-27, 2-18, 3-24,
4-1 lbs. weight act through the comers ABGD of a square of side 8-35 cms. ;
find the centre of the parallel forces.
(21) Find the centre of the parallel forces in the last exercise if the 3-24
lbs. weight force be reversed in sense.
(22) Draw a triangle ABC having ^5=10-2, 50= 11 '8, C.4=6-48 cms.
Parallel forces act through A, B and C of magnitudes (in lbs. weight) given
by the opposite sides. Find graphically the centre of the parallel forces
and shew that it is the centre of the circle inscribed in ABC.
CENTRE OF PARALLEL FORCES.
299
Mass-Centres by Link Polygons. The mass-centres of a
number of mass-points can be accurately and expeditiously
found by the aid of the link polygon construction.
Example. Masses of 1-32, 1-66, 2-15 and 1-67 lbs. are
concentrated at points whose coordinates in inches are (ri4, 0),
(2-2, 1-13), (3-36, 2-87), and (4, 2), respectively. Find the position
of the mass-centre.
Plot the points on squared paper (Fig. 271), and through them
draw lines parallel to the axes of coordinates; label those parallel
to the y axis wij, m^, wig, m^ in order from left to right, then
those parallel to the x axis from top to bottom must be labelled
m^, mj, mj, m^.
?K, m^ Wj m^
- - - - - ^ ^ -
5 " T
5 It "
A \- ] ^ -- ^ - .
"h _ , i^:^ S. X ._..]. ....
*> , »
. 1 -- _._ .?•-■,- ^, .
^ ^ -- - - -^--^ 55 H
_s__^ _ = s„s,.^ „L_ L = „ = .:„.„
„ ::^-:r:::::::::::::S:::?5^::: :i:^:;^'^S
'"a ^ ^ ^ -5i£^^ S^ V^' Z
s L _r _ ^^v^^i-'^i- j_- -
i,^ S 5 v i'/t^ t
\^ - - \ ^s3,?f »
^V ^.- ^ ^^ ..Jg^j.-, .. ^
'^,5 % _ . _: :ff"_ A
\ \ ^ .'l%_ 1
% i+,?^j
^7 - 7
m^- -p- ' '_ ' ~_' '_ \' fit'. ;: _ ^ ^
\ .. f^ ^
,', - -5 . ^ _ __ _ _
'' ^ - _t I - '- -
-± ^ -__-__: . , :
J - _^_- z
7 -A-- t
: I _-Si- --
»» y \l
*"*_ ^'^.I ^-:--± Si = ---.= ±
- Z _ - - -- ^ ^
-7 ^ - _ - X
y'^ - - ±. _ _
7 __ ±
_ _ -__ , _-_ ±i __ _ _ -
Set off the masses m.
Fio. 271.
( = 1-32),
ntn
m„
( = 1-66),
( = 167) to scale along a line parallel to the
Too ( = 2-15),
axis, and
choose some convenient pole P for the vector polygon.
Using one side of a set square bounding a right angle, draw-
through Pj, any point in the vertical mj line, a link PP^
parallel to the first line of the vector polygon. With the other
300 GRAPHICS.
edge of the set square draw through any point Q, in the hori-
zontal line «!, a link QQ-^ perpendicular to the first line of the
vector polygon.
In this way it is quite easy to draw correctly and quickly
two link polygons P-J'^P^P^, Q^Q^Q^Qi whose vertices lie on
the vertical and horizontal m lines, and whose corresponding
links are perpendicular.
The intersection at P of the first and last links of the polygon
P-^P^PJi'^ gives a line PR, parallel to Oy, on which the M.C. of
the points must lie, and the intersection at Q of the first and
last links of the polygon Q^Q^Q^Qi gives a horizontal line QR,
on which the M.C. must also lie. The point R of the intersection
of these lines is therefore the M.c.
Proof. Since the M.c. of a number of masses is the same as
the centre of the parallel forces whose magnitudes are pro-
portional to the magnitude of the masses, all we have to do to
find the former point is to find the axis of the resultant force
in two cases. This is done most conveniently by supposing the
forces acting through the mass-points to be (i) parallel to the
axis of y, (ii) parallel to the axis of x.
To find the axis in the first case the vectors of the weights
are drawn parallel to the y axis, and a link polygon constructed,
and the resultant force has PR as its axis.
To find the axis in the second case, the vectors of the loads
may be drawn parallel to the x axis and a second link polygon
constructed. It is, however, more convenient, instead of drawing
a fresh vector polygon, to suppose the first one turned through
a right angle. To construct the second link polygon, therefore,
we have only to draw links perpendicular to those of the first
link polygon. There is less chance of error if the two link poly-
gons be constructed simultaneously, by aid of two perpendicular
edges of a set square, than if one be drawn completely first.
To find the m.c. it will, in general, be necessary to draw two
link polygons, preferably at right angles, each determining a line
on which the M.C must lie.
MASS-CENTRES BY THE LINK POLYGON. 301
(23) Masses 3, 2-5, 5, 4, 3-7 lbs. are concentrated at points whose co-
ordinates are (1, 0), (2, ,S), (1, 1), (,3, 2) inches. Draw two link polygons
at right angles and find the mass-centre.
Test the accuracy of your results by taking moments about the axes of
coordinates.
(24) Choose another pole for the vector polygon in the previous question
and see that the same point is obtained for M.c.
(25) Masses given by lines of length 2-7, 1-86, 3-1, 1-72, 094 inches are
concentrated at the vortices, taken in order, of a regular pentagon of side
3 inches. Find graphically the position of the m.c. and test roughly by
measurement and by calculating the moments.
(26) Draw as follows five straight line segments to form a broken or
zig-zag line: (i) horizontally a length of 4 '8 cms., (ii) sloping upwards at
45° a length of 3'75 cms., (iii) sloping downwards at 15° a length of 5'3 ems.,
(iv) sloping downwards at 60° a length of 2 '8 cms., (v) horizontally a length
of 4'25 cms. Find graphically the position of the M.c. of the zig-zag line.
(Suppose each line to be concentrated at its mid-point.)
(27) Find graphically the M.c. of six sides of a regular heptagon. (Notice
there is an axis of symmetry in which the M.c. must lie.)
M.C. of Areas by the Link Polygon. When the area can
be divided up into parts for which the M.C.'s are found easily we
may apply the link polygon to find the M.c. of these mass-points.
At each of these m.c.'s we must suppose a mass concentrated
proportional to the corresponding a,rea.
Example. Find the m.c. nf the area given hy Fig. 272.
Draw the figure to scale and mark its axis of symmetry.
o.s"[i
Fig. 272.
Divide the area up into top and bottom rectangles and
construct their M.c.'s.
Construct the M.C. of the central trapezium.
Draw horizontal lines through the M.c.'s of the two rectangles
and the trapezium. Reduce the areas of the three parts to unit
302
GRAPHICS.
base and draw a vector polygon for masses proportional to the
areas, and finally, by a link polygon, determine the horizontal line
on which the M.C. of the whole must lie.
(28) Find the M.C. of the double angled iron in Fig. 273. (Divide up into
four rectangles and use two link polygons. )
(29) Find the mass-centre of the bar shewn in Fig. 274 ; the ends are
semi-circles of radii 2-9 and 2-1 cms., the distance of the centres apart
being 12'1 cm.
(30) Find the M.C. of the area in Ex. 29 when a circular hole of radius
1 cm. is cut out as indicated by dotted circle.
k"
•6'
S"
3-3*
Fig. 273. Pio. 274.
Irregular Areas. When the area is irregular, or cannot be
divided into parts for which the M.C.'s are known, we may resort
to the method of strip division. Divide the area up into a
number of equally narrow strips, take the mid-point of the
middle line of each of these strips as the m.c. of the strip, and
draw two link polygons for these mass-points. The masses at
the points are approximately proportional to the lengths of the
mid-lines of the strips. If the area has an axis of symmetry
only one link polygon need be constructed.
(31) Draw a semi-circle of radius 4", divide up into ten equally wide
strips parallel to the base, and determine the M.c. of these strips by the
link polygon method.
(32) Find the M.c. of the irregular figure given on p. 58.
CENTRE OF GRAVITY.
303
Centre of Gravity, Centre of Parallel Forces, Mass-
Centre, Centroid, Centre of Figure, Centre of Mean
Position. Every particle of a body near the earth's surface is
attracted towards the centre of the earth. The body being
small compared with the earth, the axes of these forces are
parallel (or nearly so). The centre of these parallel forces is
called the centre of gravity of the body. The centre of gravity
as thus defined is the same as the mass-centre of the body.
Moreover, if the mass be uniformly distributed throughout the
volume, it is the same as the centroid or centre of figure, and as
the centre of mean position.
Centre of gravity is not, however, a good term to use, since it
denotes a point whose position depends not only on the body
but on the earth also, whereas the mass-centre depends on the
body alone and would remain unaltered if the body could be
taken right away from all external forces.
Since the mass-centre or centre of gravity of a body is the
point through which the resultant of the weights of the particles
always acts no matter what the position of the body (near the
earth's surface), the body, if supported at that point, would be in
equilibrium.
This consideration leads to an easy experi-
mental way of finding the M.C. of many
bodies. Suppose a triangular board ABC
(Fig. 275) suspended by a string attached
to any point D oi it ; then, since only two
forces act on the board, viz., the pull of the
string and the resultant weight, these must
be in a line, and therefore if a line is drawn
on the board in continuation of the string
it will pass through the M.C. By suspending
the board from another point D-^ and mark-
ing the point where the vertical through D-^ cuts the line already
on the board, the position of the M.C. is determined.
The M.C. of such bodies as cardboard or wooden triangles.
304 GRAPHICS.
quadrilaterals, circular sectors, ... should be determined experi-
mentally, and the results compared with the graphical deter-
minations.
* Moment and Couple. The moment of a force a in AB
about a point is the same as the momenta! area of the couple
u, in AB and - a in CO (where CO is parallel to AB), since both
are given in magnitude and sense by the parallelogram OABC.
* Resultant Couple and Moments. Any set of forces
(coplanar) can be reduced to a resultant force through any
chosen point and a number of couples whose momental
areas are added algebraically to the momental area of the
resultant couple. This couple has therefore a momental area
given by the sum of the moments of the forces about 0.
* Moments and Equilibrium. If there is equilibrium, the
resultant force and the resultant couple must vanish for any
point 0, hence the sum of the moments about any point must
be zero.
^Moments of Resultant and Components. If a system
of forces has a resultant, this resultant reversed in sense must
be in equilibrium with the components, and therefore the
sum of the moments of. the given forces minus the moment
of the resultant must be zero for every point. Hence,
2 moments of components = moment of resultant for all points.
* Theory of three Moments. Any system of coplanar
forces can be reduced to either
(i) a resultant force,
(ii) a resultant couple,
(iii) or there is equilibrium.
If, then, the sum of the moments of the forces about one
point be zero, there is either equilibrium, or there is a resultant
force passing through the point. (There cannot be a resultant
couple because the sum of the moments = the momental area
of the couple.)
MOMENTS OF EQUILIBRIUM.
305
If, then, the sum of the moments is zero for three non-
coUinear points, the forces are either in equilibrium, or there is
a resultant passing through three non-collinear points. The
latter alternative being impossible the forces must be in equili-
brium.
MISCELLANEOUS EXAMPLES. VIII.
1. A BCD is a rectangle, AB=12, BG=S. At A, B, C and D are
masaes 8, 10, 6 and 11 lbs. Find the M.c. by the funicular polygon.
(B.Sc., 1905.)
2. Prove that the sum of the moments of two forces in a plane about
any point in their plane is equal to the moment of their resultant about
that point. Can the conditions of equilibrium of a body acted on by a
system of forces in one plane be expressed solely by the principle of
moments? (Inter. Sei., 1906.)
3. What do you understand by "the moment of a force"? A down-
ward push of 40 lbs. acts on a 6" bicycle crank which is 50° below the
horizontal position. What is the magnitude of .the moment produced
about the axis ? If, by suitable ankle action, a push is produced at right
angles to the crank in the same position, how great must this be to
produce the same moment as the downward push of 40 lbs. ?
(Naval Cadets, 1904.)
4. Indicate the method of finding the resultant of two parallel, unequal,
unlike forces acting upon a rigid body.
A uniform bar 12 feet long, weighing 56 lbs., rests horizontally upon
two supports, one being under one end A and the other being 5 feet from
the other end B ; supposing a, weight of 10 lbs. to be hung from the end
B, find the pressures on the two supports. (B. of E., Stage I., 1904.)
5. Define the moment of a force about a point and state any theorem
concerning moments.
ABG is a triangle with a right angle a,t A, AB is 2 feet and AC ib
3 feet ; a force of 5 lbs. acts from A to B and
one of 4 lbs. from .4 to C; find the moment _
of the forces about the middle point of BG. M.*'
If the point in question were fixed, indicate
on a diagram the direction in which the tri-
angle (supposed to be a lamina) would revolve.
6. When a body capable of turning about
an axis is at rest under the action of two
forces perpendicular to the axis, what is the
relation between these forces ? (State the
relation, no proof is wanted. )
The disc in Fig. 276 weighs 2 lbs. and turns
about the point O. What force P, acting in
the position shown, is required to hold the
disc in the position shown !
(Naval Cadets, 1903.)
T,G, V
Fig. 276,
306
GRAPHICS.
7. Fig. 277 represents a form of wheel and
axle, drawn to the scale of one-tenth. A man
sits on a platform suspended from the end A and
raises himself by pulling on the end B. What
force must he exert to support himself, and how
much work must he do to raise himself a distance
of 2 metres? The weight of the man with the
platform is 100 kilogrammes, and the pull at
either end of the tackle to raise a given weight
at the other is twice as much as it would be
without friction. (Military Entrance, 1905.)
8. An hexagonal table, diameter 3 ft. and weighing 50 lbs., has weights
5, 10, 15, 20, 25 lbs. placed in order, etc. at five of the angles. Determine
the centre of the parallel forces, and its distance from the centre of the
table. (B. of E., II., 1904.)
9. Mark five points in a line PQSST, the distances apart representing
3-2, 4-7, 1-8, and 2-6 ft. from left to right. Through P, Q, JR, S and T
act forces of magnitude 1510, 2150, 750, 18.30 and 1980 lbs. weight. The
forces make angles of 15°, 50°, 80°, 140° and 250° with FT reckoned contra-
clockwise.
Find the sum of their moments in lbs. weight about a point U, distant
7 ft. to the left of P, (i) by decomposing each force into two components,
one of which is parallel to PT, and the other passes through the point U ;
(ii) by finding the resultant force.
10. On squared paper mark five points whose coordinates are (2'1, 3'3),
(0-3, 5-2), (3-7, 2-1), (3-9, 4-5), (57, 0-8) inches. Masses given by lines of
length 2-35, 1-82, 4-16, 3-05, 2-18 centimetres (scale 15 cms. to 10 lbs.) are
at these points. Find the coordinates of the mass-centre by construction.
CHAPTER IX.
BENDING MOMENT AND SHEARING FOECE.
Hooke's Law. If a bar be subjected to tensile or com-
pressive stress its length changes; the relation between the
stress and the elongation, or compression, was discovered by
Hooke, and is usually called Hooke's Law. The law is purely
an experimental one.
If I be the original length of the bar, T the force producing
extension, and e the elongation, then T is proportional to 1 or
If
T= X-, (Hooke's Law)
and A. is called the modulus of the bar.
If A is the cross sectional area of the bar, then
T ■
—f is the stress per unit area,
." .
A T X e
"^^ ' J = JT
li -j — E then E is called Young's Modulus for the material
of which the bar is made.
An exactly similar law holds for compression. Always, then,
if a bar is in a state of compressive or tensile stress it is shorter
or longer than its natural length, and this stress is proportional
to the compression or extension.
For very large forces the law ceases to hold and the elastic
limit of the material is said to have been passed.
308 GRAPHICS.
This law is of great importance in many ways; it has
important bearings on the stresses set up in many frames as
well as on the bending of beams.
Simple Cantilever. AB (Fig. 278) represents a horizontal
beam fixed in a wall at A and loaded at its free end B with a
weight W. (The weight is supposed so large that the weight of
the beam itself may be neglected in comparison with it.) Consider
the equilibrium of the part BP of the beam. It is evident that
tw
Fio. 278.
the force or forces which the part AP exerts on PB must be in
equilibrium with the load W at B. Suppose the beam cut
vertically through at P, then the forces which we have to apply
to the cut surface at P to keep PB in equilibrium must be
equivalent to the reactions of AP on PB.
No single force at P can be equivalent to ^ at i?; if we
suppose a force - /F to act at P, then PB is under the action of
a couple whose momental area is - W. PB (clockwise). Hence
for the equilibrium of PB, we must apply at P an upward force
of magnitude W and a couple of momental area }F. PB. The
reaction forces of AP on PB must therefore be equivalent to
-Wa,tP and a couple whose momental area is W. PB.
This is simply another way of looking at the theorem on
p. 206, viz. a force ^ at ^ is equivalent to a force W a,t P and a
couple of transference -W.PB\ it is this force and couple
which are equivalent to the reaction forces of PB on AP.
This theory can actually be demonstrated by connecting the
two parts of the beam by a rod EF, hinged at its ends, (see
STRESSES IN A LOADED BEAM.
309
Fig. 279) and a horizontal string DC passing over a pulley G and
bearing a load Q, whilst a vertical string at D passes over another
pulley and bears a load R.
When the vertical and horizontal pulls on D are adjusted so
that BD is horizontal and in a line with C, it is found that the
vertical pull R is of magnitude W, and the horizontal pull Q
is such that q ^ j)p^ w x PB.
o\
C D
m
13
w
E F
tw
Fig. 279.
Since the part PB, under the action of R, W, Q and the force
in EF, is in equilibrium, and since R and IV constitute a couple,
Q and the force in EF must also form a couple, and therefore
EF is in compression and the stress in it is measured by Q.
The upper fibres of the beam (Fig. 278) must therefore be in
tension and the lower ones in compression, and hence the upper
fibres are elongated and the lower ones shortened. The beam
itself must therefore be bent more or less, the loaded end being
lower than the fixed one. It is these tensile and compressive
forces in the fibres of the beam itself that prevent further
bending, and it is the moment of W about P which tends to
produce the bending. Hence JVxPB is called the Bending
Moment at P.
The upward force W at P, of AP on PB, prevents PB sliding
downwards relatively to AP, whilst the external load JF tends
to make it do so, hence W is called the Shearing Force at P.
For the portion AP on the left the shearing force and bending
moment at P have the same magnitude but are of opposite senses
to those on the right.
310
GRAPHICS.
Bending Moment and Shearing Force Diagram for a
Simple Cantilever.
Example. A horizontal beam is fixed in a wall, the length from
the wall A to the loaded end B is 19 -4 /if. If the load he 5 -IS tons
draw diagrams giving the bending moment and shea/riv,g force at every
point of the beam.
Draw AB (Fig. 280), of length 3-88", to represent the beam
(scale 1" to 5 ft.), and then PQ the load vector, of length
5'18 cms. (scale 1 om. to a ton).
Fro. 280.
Through P draw PO, of length 2", perpendicular to PQ.
Through B draw BG parallel to OQ, and above AB draw a
rectangle of height PQ (BS in Fig. 280) and base AB.
The triangle ABC is the bending moment diagram and the
rectangle RS is the shearing force diagram.
The bending moment at any point S of the beam is given in
tons ft. by ST measured on the mm. scale.
BENDING MOMENT AND SHEARING FORCE. 311
Draw the force and moment scales and measure the bending
moments at points 5-36 and 10'28 ft. from A.
Proof. Consider any point S of the beam ; the moment of the
load W &t B about S is given by (Chap. VIII., p. 288) ST . PO
where ST is parallel to AG.
Since PO represents 10 ft., if ST be measured on the force scale,
i.e. in centimetres, and multiplied by 10 the result will be the
bending moment at S in tons ft.
Hence, wherever S may be in AB, the vertical intercept ST
of ABG gives the bending moment (b.m.) in tons ft.
Again, since the load to the right of S is always PQ, the
shearing force (s.F.) is constant and, therefore, the diagram for
all points is the rectangle BS.
At S the part on the left tends to slide upwards relatively
to the part on the right and we may regard SB as being drawn
upwards to indicate this. If we wished to indicate that the
part on the right tends to slide downwards relatively to the
left part then SB would have been drawn downwards.
Authorities differ in this matter. In Cotterill's Applied
Mechanics the s.F. ordinate, at any point, is set upwards when
the part on the left tends to slide upwards relatively to the
right hand part and is set downwards in the other case. In
the article " Bridges '' in the Encyclopmdia Britannica, on the other
hand, the ordinate is set upwards or downwards, at any point,
according as the right hand portion tends to move upwards or
downwards relatively to the left.
Fig. 281 is an example of Cotterill's method of construction.
Fig. 282 an example of the Encyclopcedia method. With the
exception of Fig. 282 we shall adhere to the former way, i.e. to
Cotterill's method.
B.M. and S.F. diagrams for a beam freely supported
at the ends and loaded at any one point.
Example. A beam LM (Fig. 281), of length 21-5 ft., is sup-
pmied freely at its ends in a horizontal position. It is loaded with
312
GRAPHICS.
a weight W (3470 lbs.) at a point distant 13-2 from L. Draw the
bending moment and shearing force diagrams.
Draw LM to represent to scale the beam length, and mark the
point N on it where the load acts. Choose a pole P for the
vector polygon at 10 units of length from the load vector AB.
X
N,
N
M
V
a
a
b b
\
T
Vl
V. B
lbs. wt.
aooo
3000
10,000 20,000 30,000
lbs. ft. (moments)
Fig. 281.
40,000 50.000
Draw the link polygon and close it ; in the vector polygon draw
PO parallel to the closing line, so that OA is the reaction at L,
and BO that at M.
At iVj a point on LM where LN-^ = 7-72 ft., draw a vertical
cutting the link polygon in S and T.
FREELY SUPPORTED BEAM. 313
Measure ST on the force scale and multiply by 10, this product
is the bending moment at iV, in lbs. ft. Draw a scale of bending
moments and measure the moment at a point distant 15-7 ft.
from L.
Draw a horizontal line UF between the reaction lines at L and
M, V being vertically below M. Set downwards VV-^ = OP, and
upwards UU-^ = OA.
Complete the rectangle VV-JFXU-^U as indicated; this is the
shearing force diagram. The shearing force at any section is
the ordinate of this diagram reckoned from UV.
Proof. Suppose the beam cut through at iV^j, then the external
vertical force acting on LN-^ is OA, and on iVjilf it is AO, hence
the part LK-^ tends to slide upwards relatively to N-^M.
To keep N-^M in equilibrium, we must replace OA at L by
OA at JVj (the shearing force at iVj) and a couple whose momental
area is OA . LN-^ . Since this momental area is measured by the
moment of OA about iV, it is given by ST and 10 . ST measures
the moment.
Similarly, to keep iiV, in equilibrium, we must replace AB at
N and BO at" Af by AB + BO ( = A0) at iVj, and a couple of
momental area AB . N^N+ BO . N-iB, i.e. by a couple whose
momental area is the sum of the moments of AB and BO about
i\^j, and this moment is measured by 10 . ST.
Hence, before cutting, the material of the beam at JVj must
exert shearing stress given hy AO or OA, and stress couples
whose momental area is measured by 10 . Sl\
The shearing stress prevents the shearing of the beam at iVj,
L2Vj upwards N^M downwards; the stress couples prevent the
part Nj^M rotating con traelock wise, i.e. prevent N■^ sagging.
Hence the measure of the momental areas of the stress couples
at iVj — which prevent the beam bending — is 10 . ST.
Similarly, the shear stress at iVj is measured by OA.
The maximum bending moment is at N, where the load is, the
shearing force is constant from L to N, changes suddenly at N
and is constant again from N to M.
314 GRAPHICS.
However complicated the loading on a beam or girder, the
process for finding the shearing force and bending moment at
any section is similar to the above.
Definition. The shearing force (S.F.) at any section is
defined as the sum of all the external forces perpendicular to
the beam on one side of sectionj and is considered positive when
the right-hand part tends to move upwards relatively to the
left-hand part.
Definition. The bending moment (B.M.) is defined as the
sum of the moments of all the external forces perpendicular to
the beam on one side of the section, and is considered positive
when the right-hand part tends to rotate or bend contra-
clockwise.
B.M. and S.F. Diagrams when there is more than one
Load.
Example. A bridge 80 ft. long is supported freely at its ends.
The leading pair of wheels {centre line) of a locomotive and tender is
15 ft. from one support (abutment) of the bridge, the distances apart of
the centre lines of the wheels are 10' 5", 8' 9", 10' 8", 6' 6" and 6' 6",
reckoned from the leading wheels. The loads borne by the wheels are
16 tons, 17 tons, 16 tons, 10 tons 7 cwts., 9 tons, 9 tons. The engine
a/nd tender being wholly on tlie h'idge, draw the B.M. and s.F. diagrams.
Set off the load vectors AB, (Fig. 282), BC, CD, DE, BF, FG
to scale. Choose a convenient pole P and draw the link polygon
as usual. Close the link polygon and draw FO in the vector
polygon parallel to the closing line. The intercept on any vertical
line between the first and last lines of the link polygon gives the
sum of the moments of all the external forces, including the
reaction, on either side of the vertical line.
Hence the link polygon gives the B.M. at any point of the
bridge. In this connection the closed link polygon is called the
bending moment diagram.
In Fig. 282 the pole P is taken as four units of length from
AG, and hence the B.M. diagram must be measured on the force
BENDING MOMENT DUE TO SEVERAL LOADS. 315
scale and the measurement multiplied by 4 ; this gives the b.m.
in ton feet.
Draw a horizontal line XYiov the datum line of the shearing
force diagram. At Y set YZ upwards (see p. 311), equal to GO ;
at Y^ on fg set upwards Y^Z^ = F0; at Y^ on ef set upwards
Y^Z^ = E0; at Tg on dc set upwards Y^Z.^ = D0 ; at Zg on cd set
downwards X^W^=OC; at Zj on he set downwards X^W^^BO;
and at X^ on ah set downwards X-^W^ = A0.
Ar
O I00 200 400 600 SOO ICXX3 I200 1400
Fig. 282.
Complete the zig-zag ZZ-^Z^Z^X^W^W^W-^JV ; it is the shearing
force diagram.
Evidently, for the space g, the shearing force is GO = YZ, for
the space / the sum of the forces to right is FG + GO, and the
shearing force is FO, and so on all along the bridge.
It will be noticed that the maximum bending moment is along
cd, and the maximum shearing force is through the space a.
(1) A horizontal beam fixed in a wall projects 12 ft., and it is loaded at
its far end witli 500 lbs. Draw the b.m. and s.r. diagrams and measure
the B.M. and s.r. at a point distant 4 ft. from the wall.
(2) A cantilever, whose horizontal distance between the free end and
the point of support is 25 ft. , is loaded at distances of 5, 10, 15 and 25 ft.
from its fixed end with 500, 300, 700 and 1000 lbs. weights. Draw the
diagrams of b.m. and S.F., and measure these quantities at distances of 8,
15 and 20 ft. from the fixed end.
316 GRAPHICS.
(3) A beam of length 30 ft. is supported freely at its ends in a horizontal
position. Loads of 1, 2'5, 3, 2 tons weight are applied at distances of 6,
10, 20, and 25 ft. from the left-hand end ; the beam is propped at the
centre, the upward thrust there being equal to a force of 1 '8 tons.
Draw the b.m. and s.r. diagrams and measure the b.m. and s.r. at
distances of 8 and 20 ft. from the left-hand end.
Bending Moment for non-parallel Forces. In such cases
the forces must be resolved into components along and per-
pendicular to the beam. The former tend to slide the beam
off the supports, consequently the beam must be fixed at one
end (say by a pin-joint) and supported at the other. The com-
ponents perpendicular to the beam are alone considered as
producing bending moment.
It is not necessary, before attempting to draw the b.m. diagram,
to find the reactions at the supports ; the b.m. diagram itself deter-
mines the components perpendicular to the beam.
(4) A horizontal beam PQ, of length 25 ft. , is pin-jointed to a support at
P, and rests freely on the support at Q. Forces of 2500, 2000, 1500 and
3000 lbs. weight act at points distant 5, 13, 18 and 20 ft. from P, and are
inclined to the vertical at angles of 15°, 30°, 60° and 45° towards Q. Draw
the B.M. and s.J. diagrams, and measure their amounts at points distant
7 and 19 ft. from Q.
(5) A beam PQ, of length 18 ft., is pin-jointed to a wall at P and
supported at Q by a chain of length 27 ft. which is fastened to the wall
at a point E, vertically P, and distant 12 ft. from it. Loads of 1, 1-2, 2-3
and 1'8 tons are hung at equal intervals along PQ. Draw the b.m. and
s.F. diagrams.
Reactions non-terminal. Occasionally it happens that the
order in which we have to draw the vectors in the vector
polygon, to determine the reactions of the supports on the
beam, is different from the order of the points on the beam
at which the forces are applied. In such problems it is necessary
to take care that the links of the link polygon, or b.m. diagram,
are drawn between the proper lines. If the diagram is too com-
plicated to be read easily the vector polygon must be re-drawn,
so that the vectors follow in the order of the points.
Example. PQ (Fig. 283) is a horizontal beam of length 25'3 ft.,
it is pin-jointed at P, and rests on a knife edge at E, and w partly
suppo^-ted by a rope fastened to it at Q. The rope QU passes over a
NON-PARALLEL FORCES.
317
smooth pidley at U, vertically above P, avd a weight W is attached to
the end. The beam being loaded at S, T and V, required to find the
bending moment and shearing force at any point.
Pe=25-3 ft., PU=U-1 ft., ^5=4-36 ft., PT=lO-5 ft.,
PB = 197 ft., PF=2-2-2 ft.; the loads at S, T and V are 1650,
1890 and 1340 lbs. weight, and the weight suspended at the
end of the rope is 3740 lbs.
u
^
^"^'^V^ H -^.
/-
S T
r">^;;-,
a
b
d
J
e
R
___
d.
'',
/
\
~~ — —— 5:
—7
Q,
^
\
^
RT
in
Q,
Ibs.wt.
lOOO 2000 3000
4000 5000
10,000 o 10,000 20,000 30,000 40,000 50,000
lbs. ft. (moments)
Fio. 283,
Draw the vectors of the loads at S, T and F, viz. AB, BC and
CD; then DE^ (E^ is not shewn in Fig.) parallel to QU for
the tension in the rope. Project horizontally E2 to E on AD.
Notice that the space c in the beam diagram must go from T
to v. Take the pole of the vector polygon 10 units of length
from JB. Draw the link polygon P,/S'j parallel to OA, S-^T^
parallel to OB, T-^F^ (through whole space c) parallel to OC,
F■^Q■^ parallel to OD, Qj^R^ parallel to OE (so that the space e
must be considered as going from QQ^ round the top of the
beam to the vertical through B). Close the link polygon by
Pj,Rj and draw OF in vector polygon parallel to it. Then EF
318 GRAPHICS.
is the reaction at R, and FA the vertical component of the
reaction at P. (The space / must be considered as going round
from PP■^ over the beam to the vertical through B.)
The vertical intercepts of P^S^T^V^Q^Ji-^ give the bending
moments in lbs. ft. when measured on the force scale and
multiplied by 10.
For the S.F. diagram set downwards from PQ a,t Q a.
distance = Z)^, at F a distance = C.S, at iJ a distance = i^'C i at
T set upwards a distance = 5F and at S, at distance AF,
complete the rectangles as in the figure. The vertical intercept
at any point between PQ and the thick horizontal lines gives
the S.F. at that point.
As regards the b.m. ; the fact that D^^Rj^ is (if we start with
OF in the vector polygon) the first and the last line of the link
polygon, and yet for the space RQ we measure intercept from
i?i4 is perhaps a little diificulty. The difficulty is due to the
links not being drawn in the order of the points. When we
come to the right of B the intercept between P^B^ and T-^F-^
does not take account of the b.m. due to the reaction at B, but
the intercept between B^Q^ and PiBj^ (produced) does so, and,
since, coming backwards from the right, Q^B^^ is before PiBj^, the
intercepts have to be added.
It is, however, clearer to redraw the latter part of the Imk
polygon by taking the forces in order. Letter the spaces
between B and V, F'and Q, d^ and e.^, respectively, and the space
c will now end at B.
In the vector polygon, from C draw GD^ upwards, equal to
the reaction at B, viz. EF ; from J>j set D-^E^ downwards, equal
to the load at F ( = CD) ; then E^F is the vertical reaction at Q.
The vector polygon is now FJ, AB, EG, GV-^, D^E^ and EJ'.
The link polygon is as before up to the space d-^; T-^V-^ stops at
B^ ; the next link through the space d^ is .Rg^s parallel to OD^ ;
and then V^Q^ through ej is parallel to OE-^. P-^Q^ should be the
same line as Pj^Bi if the construction is accurate. The B.M.
diagram is now P-^S^T-Ji^V^Q^P-^.
CONTINUOUS LOADING.
319
What are the B.M. and S.F. at distances of 12 and 21 ft.
from PI
(6) The span of a roof truss is 40 ft. : three equal loads are placed at
equal intervals of 10 ft., each load being 1-7 tons weight. The resultant
wind pressure on the roof ia equal to a force of 1 ton, and makes an angle
of 45° with the horizontal, and its line of action passes through tiie
mid-point of the line joining the points of support. The roof being
supposed pinned at the end facing the wind and freely supported at
the other, draw the diagram of the b.m. and s.p. for the forces perpen-
dicular to the line joining the points of support.
rw
(7) Find the b.m. and s.P. diagrams for the
vertical post of the derrick orane in Fig. 284,
due to a load W (3 tons) suspended at A.
Length of jib AG =IG'8 ft., length of tie rod
AB^U ft., BO=10 ft., BE=^ ft. The post
is kept vertical by a smooth collar at D and a
cup-shaped socket at E, and DE=4: ft.
Pig. 284.
(8) Find the b.m. and s.r. diagrams if the chain supporting Wis carried
over a smooth pulley at A, and is fastened to the post at F the mid-point
of BO, and the collar at D is replaced by a tie rod at Di, sloping down-
wards at an incline of 30°, ED^ = 12 f t.
(9) Find the b.m. diagram in Ex. (8) if the load is suspended from a
point A^ in BA produced, where fi^i = 17 ft.
Beam uniformly Loaded.
Example. A beam 20 ft. long is uniformly loaded with 50 lbs.
per foot run ; draw the shearing force and bending moment diagrams,
the beam being supported at its ends.
Draw a line PQ, 20 cms. long, to represent the beam ; draw a
vertical upwards from the beam 1" long to represent the load per
ft. run. Complete the rectangle of base 20 cms. and height 1".
The area of this represents a load of 20 x 50 lbs. weight.
Divide the rectangle into ten equal parts. Suppose the load on
each of these parts concentrated at its M.c.
The reaction at each end is 500 lbs. weight.
320 GRAPHICS.
Draw the link polygon for loads each of 100 lbs. weight
concentrated at the M.C.'s of the rectangles.
Then draw the b.m. diagram. This diagram gives only
approximately the b.m. at the various points, because the real
loading is uniform and not ten equal detached loads. But, the
vertices of the diagram are points on the true b.m. diagram.
For consider any point X on one of the M.C. lines on the beam,
the bending moment there = moment of reaction at P - 2 moments
of all the weights along PX. This last quantity is equal to the
weight of PX multiplied by the distance of its M.C. from X,
which is the sum of the moments of the partial system ; and
this difference is exactly what the b.m. diagram does give. On
the other hand, for points between two of the m.c. lines, the
diagram is wrong, since it neglects the load between them. The
true B.M. diagram is a curve passing through all the vertices of
the constructed diagram. Draw a smooth curve through the
vertices and measure to scale in lbs. ft. the B.M. at points distant
3, 11 and 15 ft. from one end of the beam.
To draw the shearing force diagram. At Q, the right-hand
end point of the beam, set downwards QQ^ representing 500 lbs.
to the proper scale. Join Q^ to the mid-point of the beam and
produce it to cut the vertical through P. This line, with the
datum line PQ, forms the shearing force diagram ; for the S.F.
must decrease uniformly from 500 lbs. weight at Q to zero at the
centre.
Beam continuously but not uniformly loaded. The
method of the previous section applies to this case also.
Example. A horizontal beam PQ (Fig. 285) supported at the
ends is continuously loaded, the load per foot run at any point being
given by the m-dinates of the triangle PQC. Find the s.F. and b.m.
diagrams — the scale of the figure being horizontally 11" to \0Q ft., and
vertically 1 cm. to 0-5 tons per ft. run.
Divide the load curve into eight equal parts; find the
vertical M.C. line of each part, and the load represented by the
CONTINUOUS LOADING. 321
area of each part. Set these off to scale and draw the vector
and link polygons as usual. Draw a curve through the vertices
of the link polygon — this curve will be approximately the B.M.
diagram.
Fio. 285
Draw ordinates for the shearing forces at the end points of
the sections from right to left, and draw a smooth curve through
their end points.
(Eemember that the s.F. diagram is such that the ordinate at
any point gives the sum of the forces on one side of the beam.)
(10) Draw the b.m. and s.F. diagrams for a horizontal beam fixed in a
vertical wall and projecting 25 ft., the load being uniform and 500 H)s.
per ft. run.
(11) Draw the b.m. and s.P. diagrams for a, cantilever due to its own
weight and a load of 3 tons at its end, the length of the cantilever being
30 ft. and its weight 100 lbs. per ft. run. (Draw the diagrams separately
and add the ordinate. )
(12) The length of a beam is given by PQ (7") (scale 1" to 8') the load
per J foot run is given by the ordinates from PQ to a circular arc on PQ,
the maximum ordinate being 4 eras, (scale 1 cm. to 0'5 ton per ft. run).
Draw the b.m. and S.F. diagrams and measure the b.m. and s.F. at points
distant 21 and 15 '3 ft. from the centre.
(13) Draw a right-angled triangle ABG having BC horizontal and
of length 6 •72", and BA vertical ot length 4-3". Let BO represent the
length of a cantilever from the fixed end B to the free end C, scale 1" to
10 ft. Let the ordinate of the triangle at any point M of BC represent,
to the scale of 1 cm. to 100 lbs. weight, the load per foot run there. Draw
the diagrams of the bending moment and shearing force.
Travelling Loads. We have now to consider how the b.m.
and S.F. at any point of a beam, bridge, or cantilever changes as
one or more loads travel along it. Many structures of large span
are now made with cantilevers connected by comparatively short
girders; perhaps the best example of this kind of bridge is seen
in the Forth railway. The whole bridge consists of two spans of
about 1700 ft. each, two of 67&ft. each, fifteen of 168 each and
322 GRAPHICS
five of 25 each. For the main spans there are three double
cantilevers, like scale beams, on supporting piers, and these
cantilevers are connected by girders each 350 ft. long ; the
length of the double cantilever is 1360. For such massive
cantilevers as those of the Forth bridge, the b.m and S.F.
due to the travelling train are small compared with those due
to the weight of the structure itself.
As the B.M. diagram is more easily constructed for a canti-
lever than for a girder, it is advisable to commence with the
consideration of the former.
B.M. and S.F. Diagrams for a Travelling Load on a
Cantilever.
Example. A ca/rdilever is of length 250 ft. from pier to free end.
Requi/red diagrams giving the b.m. and S.F. at any point as a load of
lt"3 tons travels from the free end towards the supported pier.
In Fig. 286 PQ represents the length of the cantilever, Q
being the free end, L the given point at a distance of 87 ft.
from P.
AB is the load vector. Suppose the load at Q. Choose a pole
at a convenient distance from AB. Through B, a point on
the vertical through L, draw BQ^ and BQ2 parallel to OA and
OB. Then Q-^Q^ gives the moment at L of the load at Q. When
the load is at S, the intercept S-^S^ on the vertical through S
gives the b.m. at L; hence, BQ-^Q^ gives the b.m. at L for all
positions of the load between L and Q.
When the load passes L the b.m. vanishes.
The shearing force at L is constant for all positions of the load
between L and Q; when the load passes L, the s.F. vanishes.
At all points, therefore, the maximum s.F. is the same and
is equal tr the load.
To find the B.M.'s for points other than L we have only to
notice that moving L to the left is equivalent to moving Q an
equal distance to the right. Through any point E on the
vertical through P, draw RP-^ and BP^, parallel to OA and OB
TRAVELLING LOAD ON A CANTILEVER.
323
and cutting the vertical through Q in Pj and Pj- The triangle
P-fiP^ gives the b.m.'s at P as the load travels from Q up
to P. For a point L, distant x to the right of P, mark a point
L', X to the left of Q, and draw the vertical L-^^L'L^ ; then L^RL^
is the B.M. diagram for L as the load moves from Q up to X {i.e.
in new figure from L' to P).
Pio. 286.
(14) What are the b.m.'s at points distant 30, 40, 100 and 150 ft. from
P when the load is 30, 50, 80 and 100 ft. from Q.
(15) Find the maximum bending moment at ten points between Q
and P. Set up, at these points, ordinates giving the maximum bending
moments to seale, and thus find the curve of maximum b.m.'s.
324 GRAPHICS.
*B.M. Diagram for Several Travelling Loads on a
Cantilever.
Example. Loads of 3, 5-2, 4-7 and 3-3 tms weight travel along
a cantilever of length 57 ft.. The distances apart of the loads being
5-3, 7-7 and 5-3 ft. from the foremost load of 3 tons backwardSy
determine the b.m. and S.¥. at amy point as the loads travel from the
free end up to the supporting pier or wall.
Draw a line PQ (Fig. 287) to represent the length of the
cantilever having Q for the free end, and mark the point L on
it where the b.m. is required. PL represents 15 ft. Choose a
pole at, say, 10 units of distance from the load vectors AB . . . E,
Draw the axis ab of the load AB in its farthest position from L,
and then on the other side of L draw axes 6jCj, Cjt^j and d-^e^ at
the proper distance of the loads apart from the axis through L
(i.e. draw the axes as if the leading load was at L, and the loads
were travelling towards Q).
Through any point ^j on the axis aJj draw A^Xj^ parallel to
AO, and ^1^2 parallel to £0. Produce the latter line back-
wards to cut ftjCj in JSj ; from ifj draw i^jXg parallel to CO ;
produce this line backwards to cut CjCij in Cj and draw ClX^
parallel to OD ; produce this line backwards to cut <?j«j in Dj,
and draw D^X^ parallel to OE.
Then X-^XJi^^G^yB-^A-^ is the b.m. diagram for L as the leading
load travels from ilfj(fl6) up to and past L, and the last load
comes up to L (the first load being then at M^.
When the leading load is at Afj , the b.m. is given by X-^Xe, ;
when at M^ it is given by Y^^Yr^ ; when at M^ by Z^Z,^ ; and when
at 71^4 it is zero.
Proof. Since A^X^^X^ is similar to OAB, the vertical intercepts
of the former give (p. 322) the b.m. at L due to the first load as
it travels from M-^ up to L.
Again, B^^X^^ is similar to OBC ; and since the distance from
6jCj to ah is equal to that between L and the second load when
the first load is at ilfj, the vertical intercepts of B^X^X^ must
give the b.m. at L due to the second load, as it travels from be
TRAVELLING LOADS ON A eANWLEVEtl.
3S5
up to X. A similar argument shews that C-^X^X^ and D^X^X^
give the b.m. due to the third and fourth loads, as these loads
travel from their initial positions up to L.
M« P Ma
«.
M, M,
Q
*,
^
Y,
XO >
a
b
X,
X3
x»
Xs
c
d
^
\
Y>
-D
Tons wl.
? t ^ ^ 7
100 50 O lOO 300 300 400 500 550
Tons ft. (moments)
Fig. 287.
Since on passing L the load ceases to have any B.M. at L, the
sum of the vertical intercepts of these triangles must give the
total B.M. at L as the loads travel.
(16) Find the bending moment in tons ft. when the leading load is at
(1)21-5, (2) 11-8 ft. ixoxa.L.
(17) Shew how to iind the b.m.'s at L aa the leading load travels from Q
to ilfj (the loads may be supposed travelling from a second cantilever over
a connecting girder to the one under consideration). Find the e.m. in
tons ft. when the leading load is 7'2, 14'1 and 16 ft. from Q respectively.
To find the b.m.'s at points other than L it is not necessary to
draw a fresh b.m. curve, because the lines A-J^-^, -^1-^2 ••■ ^"^^ ^
fixed relatively to one another ; and, therefore, instead of moving
326 • GRAPHICS.
L, it is sufficient to move the line X^Zj. Thus, supposing the
B.M. diagram for the travelling loads are required for a point
10 ft. to the right of L, then X-^X^ must be drawn 10 ft. to the
left of Ml ; if for a point 10 ft. to the left of L, X^X^ must be
drawn 10 ft. to the right of Tlfj, and the lines A■^X^ ... produced
to cut it.
The matter is thus mainly one of relettering the diagram
originally drawn. Change L to P (the fixed end of the canti-
lever) and move M-^ to the right a distance LP, and change the
letter to P- Produce all the lines A^X^, B-^X^ ... D^Xg to cut
the vertical through P in P-^P^ ...P^. Then P-^A.^B^ ...P^is the
B.M. diagram for P as the last load moves from the free end up
to the fixed end. For any point L, distant x ft. to the right of
P, draw a vertical through L, x ft. to left of P' and cutting
Aj^X^, B^Y^ ... in L.^L^ ... L^ ; then L^A^B^C^D^L^ is the B.M. for
L (at P) as the last load travels from the free end up to i.
(18) Find the maximum b.m.'s at points distant 5, 10, 15, 20, 25, 30, 35,
40 and 45 ft. from the pier, and draw the maximum b.m. curve.
(19) Draw the shearing force diagram for the point L as the loads travel
up to and past the point.
(Up to the leading load being at L the s.f. is AE ; immediately it passes
L, the S.F. drops to BE and so on.)
(20) Loads of 10 tons 17'5 owts., 10 tons 17"5 ewts., 19 tons, 19 tons,
12 tons, 12 tons 5 cwts. and 12 tons 15 owts. , due to an engine and tender,
travel from a girder over a cantilever of length 150 ft. (from free end to
pier). The distance apart of the loads from the leading one backwards
are 6' 6", 8' 9", 10', 8' 7-25", 6' 9" and 6' 9" respectively
Draw the diagram of the b.m. for a point distant 25 feet from the pier as
the engine and tender travel over the cantilever.
(21) Alter the lettering so that the diagram will give the b.m's. at any
point of the cantilever, and determine the b.m. at points distant 50 and
75 ft. from the pier when the leading wheels of the engine are at 110, 100,
90, 75, 50, 10 and 5 ft. from the pier.
Travelling Loads on a freely supported Bridge.
Example. AB (Fig. 288) is a beam freely supported at its ends.
A load W travels from A to B ; required the B.M. at any point Q for
every position P of the load W.
The reactions Pj and Pg ** ^ ^.nd B, due to W at P, are given
^y Pi . AB= W. PB and Ii^.AB= W. AP,
TRAVELLING LOAD ON A BRIBGE.
327
and the bending moment at Q is
R^.QB = ^.AP.QB.
A
Q
> ■
W
Pig. 288.
Now suppose W at Q, then the reactions R{ and R^ are given
%
R-l.AB^W. AB and R^.AB=W. AQ,
and the bending moment at P is
R^.AP-
Hence the b.m. at Q due to the load W at. P is the same as the b.m.
at P due to the load W a< Q.
B.M. and S.F. Diagrams for a Travelling Load on a
freely supported Bridge.
Example. A horizontal beam AB (Fig. 289), of length 50 ft., is
freely supported at its ends; a load of 3 '38 tons travels from, A to B;
required diagrams giving the b.m. arid S.F. at apoint Q (QB = l&'ift.)
for all positions of the had.
Draw the B.M. diagram for a load of 3 '38 tons at Q, and
through any point P in AB draw a vertical PP2P3 cutting the
B.M. curve at Pj ^"d P3. Then the intercept Pj^s gi^^s, to
scale, the b.m. at Q due to the load 3-38 tons at P.
Set upwards from A, to scale, AG = the load, join BC cutting
the vertical at Q in Q-^. From A draw AQ2 parallel to BQ^
cutting the vertical at Q in Q^. Then AQ^Q^B is the S.F. diagram
with AB as base line.
The vertical at P cuts the diagram at Pj , and PPj measures the
S.F. at Q due to the load at P.
Proof. That for the b.m. has already been given.
328
GRAPHICS.
For the S.F. we notice that when the load at P is between
A and Q the S.F. at Q is the reaction at B, and when P is between
, Q and B the s.F. at Q is the reaeMon at A.
If R■^ is the reaction at A when the load is at some point P'
between Q and B, then B^.AB^W.PB
S.F. at Q = i?i =
^ AB "^ AC
hence P'P/ measures to scale the S.F. at Q, and therefore the S.F.
at Q, when the load is between Q and B, is given by the ordinate
of the diagram BQQ^.
20 40 60 So
Tons ft, (mjoments)
Fio. 289.
Similarly, if the load is at P between A and Q, and E^ is the
reaction at B, R^- AB= W . AP;
AP PP
:. S.F. at Q = .B2 = ^jjg = ^ -jA, (since ^ Q^ is parallel to BC) ;
therefore the S.F. at Q, when the load is between A and Q, is
given by the ordinate of the diagram AQ^Q.
MAiCIMUM BENDING MOMENT.
S2d
Immediately before P comes up to Q the S.F. is QQ^, and
immediately after it is QQ^.
The maximum s.F. is therefore immediately before the load,
travelling from A, comes up to Q.
The maximum b.m. is when the load is at Q.
<22) What are the b.m. and s.P. at Q when the load is at P and
(1)4^=25-7 ft., (ii) .4P = 41-6ft.
Curve of Maximum Bending Moment for a Travelling
Load. To find the maximum bending moment for other
positions of Q, it is not necessary to redraw the whole bending
moment diagram. A new closing line only is wanted.
Keeping the pole fixed, the lines R-^R.^ and .^2^3 (^ig- 2^*^)'
being parallel to OX and OY, must always have the same
directions, and hence, instead of supposing Q moved relatively
to A and B, we may suppose A and B moved relatively to Q.
A, A g B, B
Fig. 290.
Suppose the maximum bending moment were required at a
distance x from A. ^e,t o^ QA.^ = x from Q towards A and
make A^B^ = AB. Mark the points S^ andiSg, where R^R^ and
R^^ cut the verticals through A-^ and ^j, join S-^S^ cutting
the vertical through QinS. Then SR^ gives the maximum B.M.
at a distance x from A.
(23) Find the maximum bending moment due to the load at points
distant 5, 10, 15, 20, 25, ... , 45 ft. from A (Fig. 289). Set up at these
points ordinates giving the maximum b.m.'s to scale and join them by a
smooth curve, which must evidently pass through A_ and B This curve
is the maximum b.m. curve for a single travelling load.
330 GRAPHICS.
*B.M. due to several Travelling Loads.
Example. To find the b.m. at any point of a bridge, freely
supported at its ends, as an engine travels from one end (abutment)
to the other.
A bridge is 50 ft. long amd an engine travels over it; draw a
diagram giving the B.M. at the mid-point oj the beam for all positions
of the engine whilst wholly on the bridge. The load on the leading
wheels is 17 tons 18 cwts., and on the following ones 19 tons 16 cwts.,
19 tons 16 cwts. amd 17 tons respectively. The distances between the
centre lines of the wheels are, from rear to front, 8' 3", 7' 0" and
9' 0" respectively.
Draw the bridge length, XY (Fig. 291), to scale and mark the
load lines when the engine is in one definite position, say with
the centre of the trailing wheels 2-7' from the left abutment.
Draw the load vectors ABODE and take a pole P at a convenient
distance. Construct the link polygon Bf^R-^R^R^R^R^ in the usual
way, close it by the link R^^, and draw in the vector polygon
PO parallel to R^R^. Mark the point R where the first and last
lines of the link polygon intersect. Bisect R^^Rf, at M; mark the
point M^ on BB^ so that the horizontal distance between R and il/j
is half the span (25 ft.) ; join M and M^ and produce both ways.
From the line aJ mark off to the right horizontally a distance = J
span and determine a point L; similarly, from de mark oif
horizontally to the left the same distance and determine a point
K; draw verticals through K and L; then the figure between
these verticals, MM^ and R^B^...B^, is the bending moment
diagram for the mid-point of the beam, from the trailing wheel
leaving X to the leading wheel reaching Y.
Maximum Bending Moment at the Centre. An in-
spection of Fig. 291 shews that M^B^. is the greatest value
of the B.M. and this occurs when the third wheel of the
engine is just over the mid-point of the bridge.
When the leading wheel is over the mid-point, the b.m. is
given by MJS,^■, when the trailing wheel is over the mid-point
the B.M. is given by M-^R-^ ; and so for all positions of the engine.
TRAVELLING LOADS ON A BRIDGE.
,0. O,. O3 0.0„
331
Fig. 291.
S32 Graphics.
(24) Find the b.m. diagram for a point distant 15 ft. from the left
abutment.
Mark the point on XT, project vertically to -B0-S5 cutting it at A, trom
R go horizontally 35' and then vertically to 1S5 on RBs- Join SS^ ; then
the vertical distances between SSp and i?„2?j...i?5 give the b.m. at the
point required. The horizontal distances between which this will hold
are 15 ft. to the right of ab and 35 ft. to the left of de.
(25) What is the maximum b.m. at this point?
If the pole P be kept fixed, then, whatever the position of the
engine on the bridge, the link polygon R^S^B^B^ and the lines
BB-^ and RB^ will be in exactly the same relative positions. The
only things. that alter as the engine moves are the reactions at
the ends, which alter the direction of the closing line R^B^-
Instead, therefore, of supposing the loads to move, we may
suppose the supports moved. Thus, to get the second pair of
wheels cd over a point Q (distant x from Z) of the bridge, we
have only to set off x to the left and 50 - a; to the right of cd, and
the points so determined are the new position X^Fj of the
supports. The verticals through Xj and Y-^ cut R^R and RR^
in, say, G^ and Gj, and GqG^ is the closing line. Gfi^ cuts cd in
(?3, say, then GJR^ gives the B.M. at cd (which is now the
vertical through Q).
Now <?3 divides G^^G^ in the ratio a; : 50 - a;. If, therefore, we
could find the locus of the points dividing the closing chords in
this ratio, we could read off at once the vertical distances between
the locus and R^R^ ■■■ R^, giving the b.m. at the point required
as the pngine travels along the bridge. This locus is a straight line.
(26) Find the closing lines of the b.m. diagrams for points distant 10,
20, 30 and 40 ft from X, and measure the maximum b.m. at those points.
(27) Set up ordinates at points along XT corresponding to the maximum
B.M.'s determined, and join the end points by a straight line. It is the
maximum b.m. diagram for all points on the bridge, as the engine, being
wholly on the bridge, travels from left to right.
*A straight line of variable length moves so that its end
points describe straight lines, the ratio of the distances moved
through by these end points being constant, the locus of the
point dividing the moving Une in a constant ratio is a straight
line.
MAXIMUM BENDING MOMENT ON A BRIDGE. 333
^fliJj and G^G^ (Fig. 292) are any two positions of the moving
line.
As G^ and G^ move y^ is always the same.
S divides R^B^ in a given ratio.
Join (?o^5, and draw SS^ parallel to G^R^ cutting G^R^ in S^,
then draw S^S^ parallel to R^G^^ cutting GJ}^ in S^.
Then,™ ^=11=11
e„ s, R>s
ElO. 292.
iSj must divide G^G^ in the given ratio.
But S/Sj bears a fixed ratio to G^^,
and S^S^ „ „ ^5-^5 ;
• • "oV ^^ constant if ^^^ is constant.
"1*2 ""s-^s
Also the angle 8,^-^8 is equal to the angle at R, and is there-
fore constant. Hence, whatever the position of S-^ and S^, the
triangle SS-^S^ retains the same shape ; the angle S^S-y is
therefore constant; and since iSS'j is always parallel to RiyR; S^
lies in a fixed direction from S, i.e. the locus of S^ is a straight
line through S.
Referring back to the b.m. diagram, we see that this locus
may be drawn by finding out where it cuts R^ or R^E, and
joining the point so determined to the given point in Rf^R^.
When (?Q comes to R, then G^ must be 50 ft. horizontally
to the light of R, and hence where the locus cuts RR^ is deter-
334 GRAPHICS.
mined by dividing RQ^ in the given ratio. Since, for the special
case drawn, the ratio is unity, S is aX, M and S^ is at M^^, where
the horizontal distance between B and M^ is 25 ft.
Similarly, if the point for which we want the b.m. is at a
distance x from the left-hand abutment, then, when G^ comes
to R, G^ must be 50 ft. horizontally from R, and the point
required must be x ft. from R (horizontally) ; hence set off x
ft. horizontally from R, and project vertically down to RR^.
As regards the limits within which the straight line locus
is available, we must remember that unless all the loads are
on the bridge, the link polygon will not be the same. When
the leading wheels come to the right-hand abutment, i.e. when
the right-hand abutment is at de, the point distant x from the
left-hand abutment will be 50 - a; from de, and when the last
pair of wheels is just on the left-hand abutment {i.e. when the
left-hand abutment is at ah) the point required is at x ft. to
the right of ah.
When only part of the train is on the bridge, only the part
of the link polygon corresponding to the load actually on the
bridge must be taken. Thus, if the leading wheels have passed
the right-hand abutment, R^R^ is the last line of the vector
polygon, and the line joining i2g to the intersection of R^Ri
and eo is the closing line. Similarly, if the first two pairs of
wheels have passed the right-hand abutment, RJi^ is the last
link, and the line joining ^„ and the point of intersection of
R^Ri and eo is the closing line.
By dividing these closing lines in the ratio a; to 50 - a;, points
on the locus from which the B.M.'s are measured may be found
for the various cases when all the engine is not on the bridge.
*S.F. Diagram for more than one Travelling Load.
The S.F. is determined at any point when we know the closing
line of the link polygon.
Take the case of the S.F. at the mid-point.
When the leading wheel is just coming up to the mid-point, M^
is a point on the closing line (Pig. 291) ; draw then in the vector
MISCELLANEOUS EXAMPLES. 335
polygon PO^ parallel to the closing line through M^ (this closing
line cuts RR^ at a point distant 25' horizontally from M^.
The S.F. is then OJi ; immediately the leading wheel passes the
mid-point, the S.F. drops to 0^D.
When the second wheel is over the mid-point, M^ is on the
closing line ; draw theif PO^ parallel to this line. Just before
the second wheel reaches the mid-point, the s.F. is O^D ; and
just after passing it the s.F. is OgC and has changed sign.
Similarly, when the third wheel is just coming to and just
going from the mid-point, the S.F.'s are O^C and O^B, and when
the fourth wheel is just going to and from the mid-points, the
S.F.'s are O^B and O-^A respectively.
Draw a horizontal datum line A-^E^ for the S.F. perpendicular
to the load lines «6, be, cd and de. At E-^ (on de) set down-
wards E-^D-i and E^D^ equal to O^E and O^D respectively, at Cj
(on cd) set downwards O^D^ = O^D, and upwards G-fi^ = O^C. At
B^ (on be) set downwards OgC, and upwards O^B. At A-^ (on ab)
set up A^Bg = Oj^B, and A^A2 = O^A.
Then E^D-fi^B^C^C^B^B^A^^ is the shearing force diagram
for the midpoint of the bridge as the engine travels over the
bridge from left to right, from the moment when the trailing
wheel is on the left-hand abutment to the moment when the
leading wheel is on the right-hand abutment.
The maximum value of the S.F. at the mid-point is seen
from Fig. 291 to be when the trailing wheel has just passed
the mid-point.
MISCELLANEOUS EXAMPLES. IX.
1. A beam 30 ft. long, supported at the ends and weighing 1000 lbs.,
carries a load of 1500 lbs. 10 ft. from one end. Shew how "to find the
moment of the force tending to bend the beam at any point ; shew in a
graph this moment for all points of the beam and find where the beam
is likeliest to break. (Home Civil, L, 1905.)
2. A beam 40 ft. long is loaded with three weights of 5, 15 and 10 tons
placed 10 ft. apart, the 15 ton weight being at the centre of the span.
Draw the diagram of the bending moments and the shearing stresses.
(Admiralty, 1904.)
336 GRAPHICS.
3. A beam is in equilibrium under any system of parallel forces, acting
in a plane which passes through the axis of the beam. Explain what
is meant by the shearing force and the bending moment at any seotion,
and shew how to determine their values.
The beam ^5 is 20 ft. long, and rests horizontally on two supports,
one at A, the other 5 ft. from B. There is a load of 2 tons midway
between the supports, and a load of 1 ton at B. Draw a diagram for the
bending moments along the beam. , (Patent Office, 1905.)
4. Find the stress diagram of a triangiilar crane A BO, of which AB ia
the vertical post, AG a, horizontal beam supported by a tie rod BG, due tq
a load of ff tons carried at a point D of AC. Prove that the bending
moment at D is W "^^'!"^ - ft. -tons. (Inter. Sci., 1902. )
AG
5. Draw a rectangle ABGD and its diagonals AG, BD intersecting at
E, the lengths ot AB and AG being 6 ft., and let its plane be vertical and
AB horizontal. Let AG and BD represent two weightless rods, turning
freely .round a pin at E, with their lower ends A, B connected by a thread,
and standing on a horizontal plane. If a weight is hung at G, find the
pressure on the ground, the tension of the thread, the stress on the pin at
E, and the stresses in the rods themselves.
How would the results be affected if G and D were connected by a
thread instead of ^ and .B? (,B. of E.,. 11., 1902.)
6. BG, GA, AB are three weightless rods formed into a triangular
frame ; their lengths are respectively 10, 8, 6 ; the frame is hung up by,
the angular point A ; a weight of 100 lb. is hung from the middle point of
BG. Eind the stresses in BG.
Find also what diflferenoe it would make in the stresses if 50 lb. were
hung at B and 50 lb. at G, instead of 100 lb. at the middle of BO.
(B. of E., 11., 1903.)
7. Draw bending moment and shearing force diagrams for a beam
loaded as follows :
A uniformly distributed load of 3 cwt. per foot run covers ^ of the span
from one abutment, and the span is 60 ft. Mark on your drawing the
position and amount of the maximum bending moment.
(B. of E., III., Applied Mechanics, 1904.),
8. A bridge has a span of 72 ft. Draw the bending moment and
shearing force diagrams for a, . point distant 9 ft. from the right-hand
abutment as an engine travels from the left to the right-hand abutment.
The distances apart of the centre lines of the wheels are 7', 5' 6", 7' and
8' .3" from the leading wheels backwards, whilst the loads on the wheels
aire 8 tons 19 cwts., 8 tons 19 cwts., 19 tons 16 owts., 19 tons 16 cwts. and
17 tons cwt. in the same order.
CHAPTER X.
STRESS DIAGRAMS {Continued).
In designing roof trusses, etc., the engineer has to take into
account not only the permanent loads but also the pressures due
to wind and snow.
Indeterminate Reactions. The wind pressure, being normal
to the roof, has a horizontal component tending to slide the
roof off its supports. This is, of course, resisted by the walls,
but how much is borne by each wall it is generally impossible to
say, since a given force may be resolved into two passing through
fixed points (the points of support) in an infinite number of. ways.
This does not mean that the reactions of the supporting walls are
indefinite, but simply that further information is necessary to
determine them.
A similar difiiculty occurs in the attempt to determine how
much of the weight of a door is borne by each of the two hinges.
Here the indeterminateness is due to not knowing the exact
relation between the parts of the hinges screwed to the door and
the parts screwed to the door post.
When the door is put into position, it may happen that the
upper hinge parts only come into contact, and then the whole
weight of the door is borne by the upper hinge ; similarly for the
lower hinge. Again, if the upper hinge parts come into contact
first, the wood may give slightly, so that finally the lower hinge
parts come into contact ; in this case we must know a good deal
about the elastic properties of the wood and hinge before the
hinge loads can be determined.
T.G. T
338 GRAPHICS.
The hinge parts on the post being at slightly different distances
apart from the corresponding parts of the hinges on the door, the
latter may have to be forced into position and any amount of
compressive or tensile stress may thus be brought into play at
the hinges. Finally, change of temperature, or accident, may
alter the initial positions of the various parts.
Reactions made determinate. In heavy gates it is not
uncommon to have one hinge only near the top, the lower
being replaced by a vertical iron plate against which the lower
part of the gate (or rather a rounded iron fixed to the gate)
presses. The forces in this case are determinate, since the
reaction of the plate must be horizontal.
A similar device is sometimes used for large roof trusses. The
truss is hinged (pin-jointed) to one wall, the other end of the
truss forms an iron shoe jointed to the axle of an iron roller,
which rests on a horizontal iron plate on the top of the wall.
The reaction of the plate on the truss is therefore vertical and
the resultant of the external forces being known, the reaction at
the hinge can be determined, and therefore the stresses in all
the bars.
Reactions and Stresses due to Loads and Wind
Pressure.
Example. PQEST (Fig. 293) represents a roof truss, pin^'mnted
at P, with an iron roller shoe at Q. The loads at T, S, and B due
to the roofing and snow are 1, 1-3 and 1-75 tons. The wind pressure
is equivalent to forces of 0-25, 0-5' and 0-2 tms at S, E «!«<? Q
perpendicular to the roof. Determim the reactions at P and Q, and
the stresses in the bars.
Given PQ = 30 ft., QB==RS=19 ft., and that the base and
altitude of the central triangle are 11-3 ft. and 8 ft.
Draw the truss to scale and letter the spaces ; then draw the
vector polygon AB, BC, CD, DE of the forces to scale.
Project on to the vertical and obtain ABC-^D^E^ ; then E,E is
the horizontal force borne by P.
STRESSES DUB TO WIND PRESSURE.
339
Take any pole Pj of the vector polygon and draw the link
polygon for the vertical forces AB, BCj, OjDj. Determine the
vertical reactions at P and Q due to' these by closing the link
Fig. 293.
polygon. In Fig. 293 these are DjOj and O^A. The total
vertical reaction at Q is EiOj. Draw 00-^ equal and parallel to
EE,, then OA is the total reaction at P, and EO that at Q.
340 GRAPHICS.
The vector polygon for the external forces is now OABCDEO.
Another way would be to letter the spaces between the
vertical loads and the wind pressures as indicated, Cg and d^, and
draw the link polygon for all the forces AB, BOj, O2O, ODj, DjD,
DE and determine the axis of the resultant. Then find the
point of intersection of this axis and the known line of reaction
at Q and join P to this point. The reaction at P is thus
determined in direction, and the forces at P and Q will be given
in the vector polygon by drawing lines from A and E parallel to
the reaction lines.
Draw now the stress diagrams for the bars in the usual way,
and tabulate the results. Shew in the frame diagram those bars
which are in compression.
(1) Find the stresses in the bars if Q be pin-jointed and P be on a
roller. Tabulate the results.
(2) Find the stresses in the bars on the supposition that P and Q each
bear half the horizontal thrust of the wind.
(3) Find the stresses on the stipposition that P and Q are fixed and the
wind pressures at 8 and R can be replaced by forces, through P and Q
parallel to them.
(Fixing Q, in addition to P, renders the frame over rigid, and merely
putting the frame in position may set up large stresses. Again, suppose
the temperature rises, then the bars tend to elongate whilst the fixed
points P and Q resist these elongations. Hence, stresses will be set up
quite independently of the loads. We are, therefore, driven to further
assumption that the truss can be fixed in position without causing stress,
and that the temperature does not change. These suppositions are some-
times made in books dealing with actual engineering structures, but there
is no real justification for them, see also pp. 337 and 338.)
(4) Find the stresses in the bars of the queen post truss of Ex. 21,
Chap. VI. (p. 225), if Q be pin-jointed, and P on rollers, and if the normal
pressures due to the wind at T, 8, R, Q be 0'25, 0"5, 0'5 and 0'25 tons
weight.
Stresses found by Moments. On p. 248 was given a
method of sections for finding the stresses in one or more
particular bars by the resolution of forces into three components
lying along three non-concurrent lines. A similar method of
sections combined with the moment construction will also often
give the stresses in particulars bars.
For this method to be effectual we must be able as before to
THE SUSPENSION BRIDGE. 341
make an ideal section of the frame cutting the particular bar
for which stress is wanted, whilst the rest of the cut bars are
concurrent.
A good example of this moment method is afforded by the
suspension bridge problem already considered from another point
of view in Chap. VI., pp. 242-248.
Suspension Bridges. In suspension bridges, the roadway is
supported by two sets of equidistant vertical rods (tie-i-ods) which
are attached, at their upper ends, to the pins of long linked
chains supported on pillars at the ends of the bridge. The
pillars are kept vertical either by passing the chains over their
tops and iixing them to blocks in the ground, or by means of
separate tie-rods (backstays).
In Fig. 284, PQ represents the roadway supported by eight
tie-rods (only a few are taken for the sake of simplicity). PB and
SQ are the supporting pillars, BT and SU the tie-rods keeping
the pillars in a vertical position.
The problem is : Given the span PQ (60 ft.) of the bridge and the
dip of the chain, i.e. the vertical distance of the lowest point 'L from
the highest S (15 ft.), tofmd the lengths of the various links, their slopes
amd the stresses in them.
If the roadway be uniform, each vertical tie-bar bears an equal
fraction of the total weight. Let this be w (3'5 tons).
Since the position of the chain cannot depend on the slope of
the tie -rods, we may suppose these rods replaced by a light rigid
strut joining BS. In this case PB and QS must react on B and S
with equal forces of iw (14 tons).
Number the spaces as indicated and set out the load vectors
AB, BC, CD, DB, EF, .... From E draw EO perpendicular to
JE, and make EO represent 10 ft. to the scale to which the span
and dip were drawn.
Draw the link polygon for the pole 0, starting with the first
link, parallel to OA, through B. The middle link eo will
evidently from symmetry be horizontal, but in all probability
will not go through L.
342
GRAPHICS.
Through any point Z in the lowest link draw the vertical
XYZ cutting RS in X and the desired position of the lowest
link in Y.
In the vector polygon draw OB, perpendicular to YS, and BO-^
perpendicular to ZS. Then 0-^ is the correct position of the pole
and EO, is the stress in the lowest link.
¥0 XZ
This last construction is simply to find EO-^ so that -^^ = -Yy-
and this is done by making the sides of REO and REO^
perpendicular to the sides of XYS and XZS.
B
R
C-
D-
E
F
^^ . Tons wt.
\ A o 1 2 4 6 8 10 T2 14
--o o,
Fia. 294.
Draw the link polygon with Oj as the pole of the vector
polygon, and see that if the first link be drawn through R the
middle one will pass through L.
Now draw to scale the supporting pillars RP and SQ, length,
say, 18 ft., and the tie-rods RT and SU making 40° with the
vertical. In the vector polygon AO^ gives the tension in the link
oa ; hence draw through 0^ a line parallel to RT (link polygon)
cutting AE in N; then NA is the reaction along PR and O^N is
the tension in RT.
Proof. If be the pole giving Z for the lowest link, and
Oj the pole for Y, then XY. EO-^ measures the sum of the
THE SUSPENSION BRIDGE. 343
moments of all the forces to the right of XY about any point
in XFZ (Chap. VII., p. 291). Similarly, XZ . EO measures
the sum of these moments, and, therefore, XY. EO^ = XZ . EO.
Oj was determined from this equation; therefore EO-^ gives
the stress in the link LY and the other links must be parallel
to the corresponding lines of the vector polygon.
The position of 0^ may also be determined by calculation.
The distances of the load lines from the middle point of the
lowest link are all known; the reaction at S is also known,
viz. half the sum of the loads. Hence, taking moments about
the middle point of RS, we have the sum of the moments of the
external forces = stress in the middle link multiplied by XY.
This stress must be set off from E along EO on the force
scale ; it should come to Oj .
Notice also that the horizontal component of all the tensions
is the same in . magnitude, hence the forces in the cut bars at
X and Y must form a couple of momental area equal in magni-
tude and opposite in sense to the couples formed by the reaction
at S and the resultant of the loads ef, fg, gh and'Ai.
If the number of the rods be odd no link will be horizontal.
The construction for finding, the pole of the vector polygon
is still the same. Z must be taken on the middle load line,
and on the perpendicular through the mid-point of the
resultant load vector.
(5) Draw the stress diagram for the same span, dip and load per
vertical tie-rod if there be 9 vertical tie-rods and 10 spaces.
' (6) The span is 100 metres, the dip 8 metres, and there are 12 vertical
tie-rods each bearing a load of 20,000 kilogrammes. Draw the chain to scale.
The Suspension Bridge and Parabola. The connection
between these was given in Chap. VI., p. 247. A much simpler
proof by moments can now be given.
Let V (Fig. 295) denote the lowest vertex and Q the w"" one
from V, so that between V and Q there are m - 1 equal loads each
of magnitude w. Since the loads are equidistant, the resultant
load must be vertical and midway between V and Q. Take the
344
GRAPHICS.
horizontal and vertical through F as the axes of coordinates.
Then if x and y are coordinates of Q, the resultant load is at
a distance - from V.
Let QM and VM be the
direction of the links at Q
and V; then since the chain
between V and Q is in equili-
brium under three forces, the
total load and the tensions
along VM and MQ, these
must meet at a point on the
resultant load, i.e. at M, whose
abscissa is \x.
Take moments about Q, then the moment of the total load
acting through M is equal to the moment of T-^ in VM ; but
T^ is equivalent to T^ horizontally and - vertically, hence,
taking account of sense,
and \ih = distance apart of tie rods
rih = x.
h'
(i)
-^=T,yory = ^^
a parabola.
To is determined when the span and dip and number of
spaces are known. Say span= 100 ft., dip = 20, A = 5,
(50)2
"0^'
then
To = 12-510.
If the middle link be horizontal (Pig. 296),
then
MN='^,
TELEGRAPH WIRE.
345
and, taking moments about Q,
and since re - ^ = (n - 1)^,
x + ^ = nh;
n-i ^
'(h-i)w
Fig. 296.
as on p. 248.
Links forming a continuous Curve. However large n
may be (i) will be true always, but both h and w become smaller
and smaller as the number of links is increased. When the
loading is continuous, -j- is the load per horizontal unit of
length (or foot run), and the chain assumes a continuously
curved form.
Such a loading and chain cannot be obtained easily but very
near approximations are possible. Telegraph and telephone
wires, for which the sag in the middle is small, are cases in
point. The sag being small, the distance between the points
of support is approximately the same as the length of the wire,
and hence the load (which is continuously applied) may be
taken as constant per ft. run.
If the wire or cable were of variable section, so that the
weight per horizontal ft. run was constant, the above supposition
would hold whatever the sag.
w W
It j-=W then y =-^^^-
The equation to the curve assumed by the chain may also be
determined directly, as in the following article.
346
GRAPHICS.
Uniformly loaded Chain. In this case the number of
vertical tie-rods is supposed so great that the roadway may
be regarded as being continuously supported. Let ^= weight
of roadway per unit length ; then the weight of any length x
is Wx, and the axis of the resultant weight acts through the
mid-point of the length x.
;.
//^
I
^
y
/
\
' T„
M
N
irvix
Let V (Fig. 297) be the lowest point of the chain, then the
chain is horizontal at that point. Let Q be any other point on
the chain. Take the horizontal and vertical through V as axes
of coordinates, x 4nd y being the coordinates of Q. Then, for
the equilibrium of the bit of chain VQ, the tensions at Q and V
must intersect on the axis of x mid- way between V and N, i.e.
at 2"
UNIFORMLY LOADED CHAIN. 347
Draw the vector polygon for M; then, from the similar triangle,
y Wx ^
X
' T,
2
w
y=
-%T,
We might have established this by taking moments about Q
whenro.2/=^x.|.l
The above construction shews us how to draw a tangent to
a parabola (for the tension J" at Q is in the direction of the
tangent at Q), viz. to draw the tangent at Q, first draw the
ordinate QN, bisect VN at M, and join QtoM; then QM is the
tangent at Q.
(7) A cable is to be made so that when erected its span will be 30 ft.
and dip 15 ft. It is to be of variable section so that the weight per
horizontal ft. run is constant and equal to 100 lbs. Draw the tangent at
the highest point, and determine the tension at the highest and at the
lowest point of the cable.
(8) The span of a suspension bridge is 100 ft., the dip 20 ft., the
number of vertical tie-rods for each chain being 8, and the load on each
5 tons ; determine the stresses in the links and uieir lengths.
(9) In Ex. 8 if the tie-rods RU a,nA SV be inclined at 45° to the vertical,
determine the stresses in them, and in the supporting pillars PR and QS.
(10) The span being 100, the dip 30, and the number of vertical tie-rods
13 each bearing a load of 4 tons, determine the stresses in them, and the
lengths of the links.
(11) Construct the parabola of Ex. 7 and find approximately the ratio of
the cross sectional areas at the highest and lowest points. (If O is the
lowest and H the highest point, join OH and take any point Pj on it.
From Pj go horizontally to P^ on the ordinate at H ; mark P whe«e OPo
cuts the ordinate at Pi . Then P is a point on the parabola. The ratio of
a small length of the curve at -ff to its horizontal projection is approximately
the ratio of the cross sections at O and H).
(12) The dip being 5 ft. and the span 100, draw the parabola. The
load being 10 lbs. per horizontal foot, find graphically the stresses at the
lowest and highest points of the chain. This example is ctpproximately
the telegraph line problem.
(13) The span being 50 ft. and the load per horizontal foot run being
15 lbs. , and the greatest tension allowable being a force of 500 lbs. weight,
find the dip, tension and lowest point, and draw the curve assumed by
the chain.
34S
GRAPHICS.
Stresses in Frames by Moments.
Example. The frame QKSTUVW (Fig. 298) is freely supported
at Q and T. The bars QV and UT simply cross at W and are iwt
pinned there. The loads at R and S are 4-8 and 6-2 tons respectively.
If qT=^30 ft., QV = 22-5//!., VT = ll-5/i!. and SV = ift.; find the
stresses in the bars.
Draw the frame to scale and letter the spaces. Choose a pole
P and draw the vector and link polygons. Close the link polygon,
and in the vector polygon draw PO parallel to the closing line.
In Fig. 298 the reactions at T and Q are CO and OA.
It will be noticed that at all the points QBSTU and F there
are three bars meeting (double joints), and that therefore the
usual method of resolution is not applicable.
•The construction for determining the stress in bg will first be
given and then the proof of its correctness.
Flo. 298.
Draw a vertical through JV cutting the link polygon and
US {bg).
Measure^, the length intercepted on this vertical between W
and BS, and k, the intercept cut off by the link polygon.
In the vector polygon draw BG horizontal, and construct (as
BG h
in Fig. 298) BG, such that
k p'
BG must be drawn from right to left, not from left to right
as BG,
METHOD OF SECTIONS AND MOMENTS. 349
Then BG- is the force exerted by hg on R, and GB is the force
on S. The stresses in the other bars meeting at iJ or iS may-
new be determined, viz. draw GF parallel to gf, and CF parallel
to cf, intersecting at F. The vector polygon for S is then BGFQB.
For the equilibrium at V, draw FH parallel to fh, and GH
parallel to gh, determining H, and the polygon for F is GFH.
For the equilibrium at T, we have CO, OH, EF, FG, and since
H and are already marked we must have HO parallel to ho.
If it is not so, some mistake must have been made in finding BG.
Since QV and UT have two pairs of letters to denote them,
eo and gh and ge and ho, the stresses in these bars will be given
twice over in the stress diagram. See that the results are
consistent.
Proof. Suppose the frame cut through as indicated by the
dotted curve (Fig. 298). Then any rigid body within may be
considered as in equilibrium under the action of the load at B,
the reaction at Q, and forces in the bars RS, UT and QF" applied
at the cut ends of the bars. The sum of the moments of these
forces about any point must be zero; hence, taking moments
about W:
Moment of force at Q + moment of force at ^ + moment of force
in RS must be zero.
The algebraic sum of the first two is given by
h .h,
and from Fig. 298 this moment must be negative or clockwise.
Hence if s denote the magnitude of the force in Ig acting on the
body hk=p.s,
and the force must push towards R, since its moment is positive
or contraclockwise.
The bar bg or RS must therefore be in compression. If, now,
we consider the equilibrium at iS*, the force at S must push, and
therefore, since BG is downward, GB must be from left to right
as indicated.
(14) Find the compression in hg by resolving the resultant of the external
forces at Q and M into three along BS, QF and UT.
350 GRAPHICS.
Example. The frams PQRST (Fig. 299) is freely supported at
P aThd Q, sknd loaded at T, S amd R with 1, 2 and 1-5 tons. Find
the stresses in the ha/rs.
PT=.TS=SR = RQ=9 and 7^^ = 45°
PZ7=10-6 and UPQ=n°,
5^=6-3 and VSW=iQ°.
Draw the frame to scale and then determine the reactions
at P and Q by the link polygon. It will be seen that the
usual process for determining the stresses stops at T and U,
since at those points are three bars with unknown stresses in them.
If we make an ideal section, as in Fig. 299, then any rigid body
within the dotted curve may be considered as in equilibrium
under the vertical forces at T and F and forces in the bars
bj, ji, io, and the sum of their moments about S must be zero.
The vertical through S cuts the link polygon in M and N;
and hence the sum of the moments of the vertical forces at F
and T is given by MN. F^X, where F-^X is the perpendicular from
F-^, the pole of the vector polygon, on to the load vector AD.
Also two of the bars hj and ji pass through S ; hence, if s is
the stress in io and SZ the perpendicular from S on the bar,
oi the sum of the moments of the stresses in the three bars
about S is given by s . SZ, and hence
s.SZ=MN.F.^X.
Construct, then, s = XY,
'°*^'* XPr^'
and the stress in io is determined.
Again, the sum of the moments of the external forces at T
and F is clockwise or negative, hence the moment of T must
be contraelockwise, or T must pull on the body enclosed by
the dotted line and the bar io must be in tension.
Hence set off 01 parallel to oi and of length XY.
The point L is determined by drawing AL and OL parallel to
al and ol. Hence K is determined by drawing LK and IK
parallel to Ik and ik. Since for the point L the action of the
STRESSES FOUND BY MOMENTS.
351
bar lo is given by LO {OAL being the vector triangle),, for the
point U we must have the sense OL, and since oi pulls at ZT,,
/ must be on the left of o, as in Fig. 299, so that the vector
polygon for U is lOLKI.
Fig. 299.
The point / is determined by KJ and I J parallel to hj and ij.
Hence B and / are known, and if the drawing has been done
correctly, BJ will be parallel to Ij.
Notice that, if a mistake bad been made in deciding, whether
352 GRAPHICS.
io is in tension or compression, the direction of BJ would have
been totally different from Ij ; this, then, gives a test as to the
correctness of the reasoning and of the drawing.
For the theory of reciprocal figures as applied to stress diagrams
the student is referred to Cremona's Graphical Statics,* edited hy
Professor Beare, and to a very elementary work by Professor
Henrici and Mr. Turner on Vectors and, Motors, j
(15) Fig. 300 gives a not unusual form of truss supporting the roof
shelter on a railway platform.
The loads at P, Q, S, S, T, U, V due to the roofing and snow are
0-8, 27, 3'8, 5-2, 5, 4-7, 1-5 owts. Find by moments the stress in WX and
compare ^ith that obtained by the usual graphical method. ET=5,
Pr=3-85, SW=2-3.
(16) Find the stress in SR, of the cantilever (Fig. 301), by the method of
moments. The load at Q is 3 tons. Test the result by resolving the load
at Q into three components lying along TV, TQ and SM. jSiJ=5'2,
EV=5-5, TV=5-25, TS=n, ,S'K=9-4, BQ = 13-5, TQ = ni. Determine
the stresses in all the other bars.
(17) As in the previous example, only the load is suspended from a chain
which passes over a smooth pulley at Q and is fastened to M, the mid-
point of TS.
MISCELLANEOUS EXAMPLES. X.
1. Prove that a, chain made of equal links will hang in equilibrium
in a vertical plane with the links parallel to lines drawn from a point
to equidistant points on a vertical line ; and determine graphically the
stress at a joint. (Inter Sci., 1902.)
2. A heavy chain is supported by its ends A and B, which are 12 ft.
above the lowest part of the chain. The horizonal distance between
A and B is 66 ft. and the weight of the chain is 20 lbs. per ft. of its
horizontal projection. Draw out to scale (10 ft. to an inch) the shape
of the chain and find the force on the chain at the lowest point. What
is the maximum force in the chain. (A.M. II., B. of E., 1903.)
3. A suspension bridge two hundred feet span between the centres of
the towers has cables having a dip of 30 feet ; the backstays are anchored
at a distance of 60 feet from the centres of the towers ; the load on each
cable is 4 tons per foot run. What is the stress on the cables at the centre
of the bridge, at the towers, and in the backstays ?
(Admiralty Examination, 1904.)
4. Find the stresses in the bars of the trusses shewn. Figs. 302 and 308
for equal loads.
■ 5. Find the stress in ZT of the French roof truss, Fig. 304, by the
method of moments and thence the stresses in all the bars. The loads
at E, 8, T, U, V, W, X being 1500, 2700, 1600, 3400, 1800, 3250, 1750
lbs. weight respectively, YZ=QZ=ZU=4:-^, QU=S-5, and QE=U, and
the loads are equidistant.
* Clarendon Press, t Published by Mr. Edward Arnold,
EXAMPLES.
CHAPTER XI.
WOEK.
A FORCE acting on a body is said to do work when the body is
displaced.
The work done by a constant force . acting on a body is
defined as the product of the displacement of any point on the
axis of the force, and the force component in the direction of
the displacement.''^
Thus, if in consequence of the motion of the body, the point
A (Fig. 305) on the axis of the force OF moves from A to A-^,
the work done by F is the product AA-^ . OF^, where OF^ is
the force component in the direction of ^j.
If the force component has the same sense as the displacement,
work is said to be done hy the force, and it is considered positive.
If the force component has a sense opposite to that of the
displacement, work is said to be done against the force, and this
is considered as negative work done by the force. The reason
for this sign convention is not difficult to see ; suppose two
forces differing only in sense act on the body, then, so far as
motion is concerned, these are equivalent to no force at all,
and therefore in any displacement no work is done on the whole.
But the components of the forces in the direction of any dis-
placement are equal in magnitude, and opposite in sense, hence
the work done by each force must be equal in magnitude, and
if one be considered as positive the other must be negative.
* The other component is supposed perpendicular to the first one.
UNIT OF WORK.
355
Unit of Work. In Statics the unit of work is usually taken
as the footpound, or the work done by a force of 1 lb. weight
when the body is moved 1 ft. in the direction of the force.
If the unit of force be a dyne, the unit of work is called an
erg, and is the work done by one dyne when the body is dis-
placed I centimetre in the direction of the force.
Fro. 305.
Graphical Representation. Work done is represented
graphically by the area of a rectangle of which one side represents
to scale the displacement, and the adjacent side the force com-
ponent in the direction of the displacement. To measure this area
the rectangle is reduced to unit base either (i) the unit of length,
when the altitude is measured on the force scale, or (ii) the unit
of force, when the altitude is measured on the length scale
(pp. 38-40).
Moment of a Force and Work done by a Force. These
are both represented graphically by an area, but have totally
different physical interpretations. The moment of a force is a
vector quantity, its plane is determined by the plane of the force
and the point, and its sense by the sense of the force. The area
representing the work done by a force has no special plane,
and may be supposed anywhere; it is a scalar area though it
may be positive or negative.
356
GRAPHICS.
Example. The shafts of a carriage are inclined at an angle of
1,5° to the hmizontal. If the pull traiismitted along the shafts from
the horse he 123 Ihs. weight, fmd the work done ly this force in
moving the carriage through 137 ft.
Draw vertically upwards a line OD (Fig. 306), of length 6-85",
and set up along it 011=5". Through draw (i) a horizontal
line OFj, and (ii) OF sloping at 15° and of length 12-3 cm.
Draw FFj^ perpendicular to Oi^,, and mark on OF^ the point
W, where DIV, parallel to UF^, cuts it.
Fig. 306.
Measure OW on the force scale, and multiply by 10 ; this gives
the number of foot-lbs. in the work done by the horse on the
carriage.
Proof. OFj is the component of OF in the direction of the dis-
placement, and hence the work done is measured by OF^.OD, i.e. by
the area of the rectangle having OF^ and OD as adjacent sides.
This rectangle is equal in area to OU. OFT, and OU represents
10 feet; hence measuring OfF on the force scale gives the work
done in 10 ft.-lbs., and therefore 10 x OW on the force scale gives
the work done in ft.-lbs.
FORCE AND DISPLACEMENT.
357
(1) The weight of a bucket of water is given by a line of length 47 cm.
(scale 1" to 10 lbs.). Find the work done in ft. -lbs. against gravity in
raising the bucket from the bottom to the top of a well 764 ft. deep.
(2) A horse pulls a canal boat with a, force of 151 lbs. weight, the tow
rope makes an angle of 25° with the bank. Find the work done in ft. -lbs.
on the boat in pulling it along 1 17 ft.
(3) If a hole is punched through a metal plate 078 inches thick, and
the average resistance to the force of the punch is of magnitude 23700 lbs.
weight, find the work done in ft. -lbs.
(4) A weight of 1720 lbs. by falling through 27 '8 ft. lifts, by means of
a machine, a weight of 970 lbs. through 47 "3 ft. Find the total work
done by gravity.
(5) The inclination of a plane is 25° ; find the work done against gravity
in pushing a body weighing 7 '3 owts. , 15 '7 ft. up the plane.
(6) If the body be pushed up the plane by a horizontal force of 8 '2 owts.,
find the work done by this force.
The work done by a force when a point in its axis is
displaced, is the product of the force and the component
displacement in the direction of the force.
This is really an alternative definition to that given on p. 354.
Let OB (Fig. 307) represent the displacement, and OF the
force under consideration ; then, according to the first definition,
the work done = OF^ . OD, where FF^ is perpendicular to OD.
Fia. 307.
According to the alternative definition the work done = OF. OD^,
where DD^ is perpendicular to ODj .
But OFj^F and OD-^D are similar triangles,
and hence ^"5 = ^ °^ ^^ " ^^i = ^^^ ' °^-
358 GRAPHICS.
More shortly, if 6 is the angle between the force and the
displacement, and /and d are their magnitudes, then the
work done=/. dcos 6 (according to second definition)
=/cos 6 .d (according to first definition).
Work and Motion. It should be observed that for a force
to do work or work to be done against a force, motion is essential.
Unless some point on the line of action of a force moves, and
the displacement has a component in the direction of the force,
no work is done by or against the force. Thus, however great
the force which a horse exerts on -a cart in trying to start
it, no work is done by this force on the curt unless the cart
moves. If by means of a second horse the cart be made to
move, then the first horse does work on the cart, the amount
being his pull multiplied by the component displacement. If
a force acts on a body at right angles to its displacement, no
work is done by the force; thus in the case of a body pushed
along the surface of a horizontal table no work is done by the
weight of the body because its line of action is perpendicular
to the displacement.
If, then, we know the work done by a force to be zero, we
may have either (i) no displacement, or (ii) a displacement
perpendicular to the force.
We are not here directly concerned with the force or forces
to which the motion as such may be due. For instance in
Exercise 1, the actual work done by the person raising the
bucket is not the same as the work done against gravity. To
find the former from the latter we must know the speed of
the bucket and the work spent in giving it kinetic energy as
well as the work done against the resistances.
It is true, of course, that if we knew the force exerted by the
man the work done would be this force x the displacement.
The difference between this work, and the work done against
gravity, gives the energy imparted to the bucket, and the work
done in overcoming resistances. Similarly, in the example on
p. 356, we are concerned with the work done by the force
DISPLACEMENT AND PATH. 359
applied to the carriage ; this may not be the same as the work
done against the resistances to the carriage motion, because the
carriage may be going faster at one time than at another.
Shortly put, we are concerned in Statics only incidentally with
the forces causing motion; our problem always is to find the
work done by certain given forces when the body is displaced,
the work being measured according to the definition on p. 354.
Displacement and Actual Path. The actual path of the
displaced point is immaterial; so long as the displacement is
the same, the work done will be the same.
Let F be the force and AAj
the displacement of A, then (Fig. .
308) the work done is F . AA^. B,^-^^;/^
If the displacement had been /\y^
first from A to B, and then from //^
-B to A^ the work done would /^ IB;
Ji^^^^een A ^^^^^^ F
F.AB^ + FB^A^ = F.AA^.
This decomposition of displacements may be supposed repeated
without limit, so that A may be supposed to move on any
curved path from A to A-^, and the work done by F will still
heF.AA^.
(7) Draw a circle of 3" radius, and suppose it to represent a vertical
wheel of radius 6'. Find the work done by gravity when a load of 0'34
ton is moved round the wheel from the lowest position through one, two,
three and four quadrants respectively.
Change of Direction of Force. If the force changes its
direction as the point on its axis moves, but the angle between
the force and the direction of the motion remains unaltered, the
work done, will be the product of the distance moved through
by the point and the force component in the direction of the
motion at any instant.
Thus, if the point move in a circle, and the force is always
a tangent to the circle, the work done in a complete revolution
will be the force x the length of the circumference.
360 GRAPHICS.
(8) A body is moved through a circular arc, of length 25 ft. and radius
19 ft., by a force of 34 lbs. weight, which always makes an angle of 70
with the radius. Find the work done on the body in ft. -lbs.
(9) A man pushes at a capstan bar with both hands. One hand, at a
distance of 9 ft. from the axis, pushes perpendicular to the bar with a force
of 30 lbs. weight, the other pushes with a force of 38 lbs. weight at a
distance of 7'8 ft. from the axis, and inclined to the bar at an angle of 72°.
Find the work done in ft. -lbs. during a complete revolution.
Work done against Friction. In the Chapter on Friction
it was explained that the coefficient of friction was the ratio
of the force tending to produce motion to the normal pressure
when the body was just on the point of motion. Such a
coefficient of friction is therefore not at once applicable to
bodies in motion without further experimental evidence.
Experimental Laws of Friction for Bodies in Motion.
It has been found for bodies actually sliding one on the other
that the friction between them is
(i) proportional to the normal pressure ;
(ii) independent of the relative speeds of the bodies ;
(iii) „ „ area in contact ;
(iv) dependent on the nature of the surfaces ;
so that for bodies in motion, if F denoted the friction and N
the normal pressure, F=uN,
where /j. is constant for any two particular surfaces, but varies
for different surfaces and is called the coefficient of dynamical
friction. This coefficient of dynamical friction is slightly less
than for limiting friction.
Example. A rough plane w of length 13 ft. and height 7 '8 ft.
Find- the work done hy the least possible equilibrating fmxe * when the
body of weight 24-8 lbs. is displaced from the bottom to the top of
the plane. The angle of friction for the plane and body is 18°.
Draw the plane ACB (Fig. 309) to scale (AC =13 cm. say),
and draw OiV perpendicular to AC.
* A body is not neoessarilj at rest when in equilibrium ; it may be moving
with constant velocity.
WORK AND FRICTION.
361
Set off OfF vertically to represent the load of 24 '8 lbs. weight,
and draw OP, making 18° with ON, on the side away from OfF,
and then ^P perpendicular to OP.
Project ?FP to W-^P^ on
the plane by lines parallel
toOiV. Along ^j^ set off
/FiQ=13cm.
and JF^U'=lOcm.
Draw QQj parallel to UP^ .
Measure JF^^Q^ on the
force scale, and multiply
by 10; it gives the num-
ber of ft. -lbs. of work
done in sliding the body
up the plane.
W,<'
(10) Find the work done against gravity and also that done against
friction. What connection is there between these and the work done
by the equilibrating force?
(11) Find the work done when the body is displaced up the plane by
the equilibrant when (i) parallel to the plane, (ii) horizontal.
Find also the work done against friction.
(12) A man pushes a roller up a hill rising 1 in 7, and keeps the handle
horizontal. The resistance is equivalent to a force of 24 lbs. acting down
the hill. Find the work done on the roller by the man in moving it 123'
up the hill, if the roller weighs 268 lbs.
(13) The axle of a fly-wheel has a radius of 2", the weight of the wheel
is 1780 lbs. and the coeiBoient of friction for the axle and bearing is 0"18.
Find the work done in ft. -lbs. against the friction per revolution of the
wheel.
If a given set of forces acting on a body would keep it in
equilibrium, then the total work done by all the forces is
zero during any displacement of the body, the forces being
supposed constant.
362 GRAPHICS.
This is a direct consequence of the fact that the vectors of
the forces form a closed polygon, and therefore the sum of
their components in any direction is zero.
In many cases the forces cannot be supposed constant for
finite displacements, and we have to consider infinitely small
displacements. In the case of a beam leaning against a
smooth wall and kept from sliding down by a peg at its
foot, the reactions depend on the slope of the beam, and
we cannot at one and the same time suppose the slope altered
and the reactions unaltered.
M.C. and Work Done. The work done against gravity
in raising a body of weight W is eciual to the work done in
raising a mass of weight W supposed concentrated at the
mass-centre of the body.
Let Wj, Wj, Wg, ... be the weight of the particles of the bodies,
2/j, y^, i/g, ... their initial vertical distances above some horizontal
plane, and Pj, Y^, ... their final distances above the same plane.
Then the work done against gravity is
«'l(i^l-2'l)+«'2(I'2-y2)+---=2«CiFi-St«jyi.
But SwjT/j = </Swj where y is the initial vertical distance of the
M.c. above the plane
and 2wjFj = 3''Swj where Fis the final vertical distance.
Therefore, total work done against gra,vity = (Y - y) W the
work done on a particle of weight W in lifting it through
the distances Y-y.
(On each particle of the body other forces than the weight act,
viz. the pushes and pulls of adjacent particles. These pushes
and pulls constitute the stress of the body, and since the stress
consists of equal and opposite pairs of forces, the total work
done by these is zero.)
(14) Find by calculation the work done in emptying a cylindrical well
shaft of diameter 3 ft., the depth of the well being 110 ft. and the top of
the water being 26 ft. below the surface ; the weight of a cubic foot of
water is 62*5 lbs. approximately.
VARIABLE FORCE.
363
(15) ^BC is a triangular prism weighing 78'3 lbs. It rests on the
ground with the face BC, of length 2-37 ft., in contact with it. BA is
vertical and of length 3 '16 ft. Find the work done against gravity in
turning the prism so that it is about to fall over (i) round the edge at B,
(ii) rovmd the edge at G.
(16) Find by calculation the work done against gravity in raising a
cage of weight 727 lbs. from a depth of 236 ft. by a wire rope of that
length, the rope weighs 5 '7 lbs. per yard.
Work done by a Variable Force.
Example. A force mooes a hody in its line of action; for suc-
cessive displacements of 1 ft. the magnitude of the force is 5r3, 72-4,
65-7, 42-6, 31-5, 27-1, 30'3, 39-2, 46-9 lbs. weight. Represent the
work done graphically and find its amount in ft. -lbs.
(
,-»
70
#
..%
^^
60
/
1
1
\
tl-,
_
..\
gbu
\
*
CO
:S40
.5
\
^'
■-
^
__N
,'
*•
^
...
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J
A
3
e
7
J
Displacement in ft
Fia. 310.
On squared paper take two axes of co-ordinates, the horizontal
one to represent displacements to the scale of 1" to a foot and
the vertical one to represent forces to the scale of 0-1" to a
lb. weight.
Plot the points corresponding to the numbers, and complete
the rectangles as indicated in Fig. 310. Evidently the whole
area of the figure represents the work done. Find the area
in sq. inches ; the number of sq. inches multiplied by 10 gives
the work done in ft. -lbs.
364
GRAPHICS.
Suppose we knew that the force changed for every displace-
ment of 6", being 60, 68, 52, 36, 29, 28, 35, 43 and 47 lbs. weight
at the intermediate points. Then, plotting these points, we see
that the total work done is represented by the area of the new
figure.
Further, if the force changed continuously instead of suddenly,
we should have, instead of a succession of points forming a
zig-zag line, a continuous smooth curve, and we see that the area
enclosed by this curve, the axes of coordinates and the ordinate
at 9 represents the work done.
(17) The resistance to the motion of a car for various displacements
is given in the accompanying table. Draw a curve giving the relation
between the displacement and the resistance. Divide the total displace-
ment into ten equal parts and erect ordinates at the mid-points. Add the
mid-ordinates by means of a straight strip and measure on the force scale ;
multiply by the width of the strips in feet. The product is the work done
against the resistance in ft. -lbs.
Displacement
in ft.
10
30
50
70
90
110
130
150
170
190
Resistance
in lbs. weight
160
184
231
289
394
540
641
709
751
776
(18) The force in lbs. weight acting on a body is always twice the
magnitude of the displacement in feet and acts in the direction of the
displacement. Find the work done by the force for a total displacement
of 17-4 feet.
Work done in Spring Extension or Compression.
Experiments shew that when a helical spring is extended beyond
its natural length, the extension is always proportional to the
force applied (up to the elastic limit of the spring), so that if
W
e denote the extension, and W the load, — is, for the same
spring, always constant (Hooke's Law).
If the results of an experiment be plotted, say extension in
cms. horizontally, and load in grms. weight vertically, the points
will be found to lie approximately in a straight line — as in
Fig. 311 — passing through the origin of coordinates. From the
graph we see that — is always constant and is measured by tan a.
SPRING EXTENSION.
365
The work done in extending the spring a distance OA is
therefore given by the area OAB to scale, and is measured
by ^OA . AB, i.e. \ the product of the maximum extension
and maximum load.
5s
II
1
p
^
eg
-<i
-^
.e
>*^
1
>
^
^
^iC
^
a
U Extension in cms. = e "
Fia. 311.
(19) A spring is found to extend a distance of 12 '7 cms. beyond its
natural length under a load of 46 '3 lbs. Find the work done in inoh-lbs. in
gradually extending the spring from its natural length to 8 '7 ems. beyond.
(Draw the straight line graph by setting off 0.4 = 12-7 ems. horizontally,
and AB vertically a distance of 4'63 inches, and join OB; OB is the
graph. Find the area enclosed by OB, OA, and the ordinate at a point
8-7 from O.)
(20) A spring is found to extend a distance of 15 cms. under a load of
17 '6 lbs. Find the work done in gradually extending it from an extension
of 7"3 cms. to one of 17"4 in ineh-lbs.
(21) A bar is fixed at one end, and twisted by means of an arm of length
1 ft. , fixed at right angles to the length, at the other end. To keep the
free end of the bar twisted through a radian requires a force of 27 lbs.
weight to be applied to the end of the arm. The force applied being
proportional to the angle of twist, find the work done in twisting the free
end from 0'56 to 1"32 radians.
Work done in compressing Gases. For many gases at
ordinary temperatures and pressures the relationship between
the volume they occupy and the pressure to which they are
subjected is given by the law
Pressure x volume = C (where is constant)
(known as Boyle's or Marriotte's Law).
Example. A gas obeying Boyle's Law is enclosed in a cylinder filled
with an air-tight piston ; to find the work done in compressing the gas.
Suppose when the compression starts at the atmospheric
pressure of 14-3 lbs. per square inch, the volume of the
366
GRAPHICS.
enclosed gas is 100 cubic inches, then the constant above is
14-3x100=1430. Suppose the volume is reduced to 20 cubic
inches.
On squared paper take two axes of coordinates, the vertical
one for pressures, and the horizontal one for volumes. Along
the latter set off OM to represent to scale 100 cubic inches, and
set up vertically MP to represent the pressure of the gas
(14'3 lbs. per sq. inch).
70
Q
\
60
\
\
1-
\
\
\
|40
s
\
s
s \
/
>
f
^
--
--
■-•
--
—
■-•
--
■-;
.g20
^^
,
,-
--
,-
'•
"
■~-
__
P
10
,
.-
V,
,.,
-■
feS.
.--
""
N
J
1
V
,„
M,
40 50 60
Cubic fhches
Fio. 312.
70
80 90
Draw a horizontal line through P (Fig. 312), and mark any
point Fj on it; put a straight edge along OF^, and mark the
point Tj where it cuts the ordinate PM. From F^ go horizontally
to Fg on the ordinate FFj through Fj.
Fg is a point on the curve whose equation is
P. F=U-3x 100 = 0M. MP.
Repeat the construction for a number of points like F-^, and
obtain, say, nine points between F=20, and r= 100. Join the
points by a smooth curve QF^P. This curve has for its equation
P. r= 14-3x100.
Find the area enclosed between the curve, the axis OM, and
the ordinates QN and PM, by the mid-ordinate method, i.e. add
GAS COMPRESSION. 367
the mid-ordinates of a number of equally wide strips between
JSf and M by the strip method, and multiply by the common
width of the strips, the former being measured on the pressure
scale, and the latter on the volume scale.
The product is the work done in compressing the gas in inch-
pounds.
Proof. Pressure meaning force per sq. inch, then if A is the
area of the piston in square inches, the total force acting when
the pressure is P is P . A =F say.
If L is the length of the cylinder in inches, then
V=AL,
and PV=^.A.L==F.L,
A
and we may take the ordinate to represent the force on the
piston, and the abscissae to represent displacement.
(22) Find the work done in compressing a gas from a volume of 7 '32 to
3 '64: ou.-ft. , if the initial pressure of the gas was 5200 lbs. weight per
sq. ft.
* (23) The resistance to ithe motion of a body in a liquid varies as the
square of the speed, find the work done in reducing the speed from 20 to
11 "5 miles per hour in a distance of 1'6 miles, if the resistance to the
motion is a force of 12 '3 tons weight, when the speed is 8 '6 miles per
hour.
* (24) If Wo is the weight of a body on the earth's surface, and W the
weight of the same body at a, distance R from the earth's centre, then
WR^= WqR^ where i?„ is the radius of the earth. Find the work done
on a meteorite by the earth's attraction in moving it from iJ=8000 miles,
to 7?=iJo=4000 miles, if the meteorite weighs J » ton on the earth'::
surface.
MISCELLANEOUS EXAMPLES. XI.
1. Explain the term work.
Find the work done in raising a lift full of people and weighing 2 tons
through a height of 80 feet. Explain the system of units you use.
(Engineer Students, 1903.)
2. A uniform rope hangs by one end, and carries a weight at the other ;
shew how to draw a diagram to represent the work done in winding up the
rope, and thereby lifting the weight. (B. of E., II.)
3. Draw a line Ax, and take in it five equidistant points, B, G, D, E,
F; suppose a force (P) to act along Ax, and that its value at the points
B, G, D, E, F are respectively 50, 35, 28, 25, 24 lbs., let the distances
368 GRAPHICS.
BC, CD, ... represent 3 ft. apiece ; draw a diagram of the work done by
the force, and calculate (by Simpson's rule, if you know it) in foot-pounds
the work done by the force while acting from B to If. (B. of E. , II. )
4. Draw a diagram of work in the following case : Six equal weights
( W) are fastened to a rope in such a way that one follows another at
distances of a foot. The rope hangs vertically with the lowest weight
3 ft. above the ground ; if the rope be gradually lowered draw a diagram
for the work done by gravity on the bodies, when all have come to the
ground. (B. of E., II., 1903.)
5. A load of 10 cwts. is raised from the bottom of a shaft 500 feet deep
by a wire rope weighing 2 lbs. per foot. The rope is wound up on a drum,
3 feet in diameter. Draw a curve, showing the moment exerted at the
drum throughout the motion, and find the whole work done during the
lift. (Patent Office, 1905.)
6. Shew how to represent in a diagram the work done by a force P o£
variable magnitude, which displaces its point of application in its own fixed
line uf action from A to B. Let P begin with the magnitude 50 lbs.
weight, and keeping its magnitude constant, displace its point of application
from .4 to C, a distance of 2 feet ; then from C to D, a distance of 8 ft. , let
P vary inversely (without discontinuity) as the distance of its point of
application from A. Draw the work diagram and find the total work done
from A to B. (Inter. B.Sc. (Engineering), 1906.)
7. Give the vector definition of the mass-centre of a system of particles.
Prove that the work done against gravity in moving a system from one
configuration to another is equal to the work done in lifting a, particle
equal to the total mass from the first position of the M.c. to the second.
(Inter. B.Sc. (Engineering), 1905.)
8. The table below gives the relation between pressure and volume for
1 lb. of saturated steam, between certain limits of pressure. Plot a graph
which will show this relation, and by counting squares on the sectional
paper, determine the area bounded by the curve, the horizontal axis or
line of zero pressure, and the limiting ordinates (parallel to the line of zero
volume).
If for any small change of volume of the steam the product of pressure
m lbs. per square foot and the change of volume in cubic feet represents
the work done in foot-lbs. , find how many foot-lbs. of work will be done in
compressing the steain from a volume of 4 '29 o. ft. to a volume of 1'53 c. ft.
Pressure in lbs. per square inch. Volume in cubic feet.
101-9 4-29
1151 3-82
129-8 3-42
145-8 3-07
163-3 2-76
182-4 2-48
203-3 2-24
225-9 2-03
250-3 1-84
276-9 1-68
305-5 1-53
(Military Entrance, 1905.)
APPENDIX.
EXPERIMENTS ON MOMENTS.
The Lever. A good simple lever can be made from a metre scale.
Leaving about 3 cms. untouched at the centre, cut away from one edge
down to the middle line and remove the end portions as in Fig. 313.
Bore a hole just above the middle line at the mid-point of the length, and
fix a steel cylindrical peg firmly in the hole so as to protrude about j"
on both sides.
FlQ. 313.
The lever should be supported between two wooden blocks of the same
height or on a special stand. Weights may be suspended from the lever
by looped threads, either directly or by scale pans.
W,
W,
FlQ. 314.
Stops should be provided to prevent the lever overbalancing when the
weights are not properly adjusted.
If the lever is not horizontal when placed on its supports, cut away a-
little more wood from the heavier side until it is.
ExPT. I. Suspend, by means of a silk loop, a 100 gramme weight from
the left-hand side of the lever at 20 cms. from the fulcrum. From the
right-hand side suspend a weight of 40 grammes and adjust its position
until the lever is horizontal.
Do the same with weights of 50, 60, 100, 150, 200, 250, 300, 400 and
500 grammes on the right-hand side, noting in each case the distance of
the supporting thread from the fulcrum.
Tabulate the results thus :
Weight in grammes = W,
40
50
60 100
etc.
Distance from fulcrum in
cms. —d,
2a
370
GRAPHICS.
On squared paper take a horizontal line to represent distances from the
fulcrum, scale 1 in. to 5 cms., and vertical distances to represent the weights,
scale 1 inch to 50 grammes, and mark the divisions as indicated in the
figure.
bUU
1
\
40U
1
\
W
\
\
\
\
N
^^
■--.
p
bo
d
M
10 15
20 25 30
Cms.
Fig. 316.
35 40 45
Plot the points whose coordinates are given in your table and draw
a smooth curve lying as evenly as possible amongst them.
The curve can be recognised as like the one constructed on p. 34.
Verify this by multiplying the weights by the distances (take the weights
from the curve).
Find that always ^^.^^^ ^ ^^^ distance from fulcrum
is the same.
See if this constant product is also the product of the left-hand weight
and its distance.
These products are called the momenta or turning moments of the weights
about the fulcrum, and we have TnomeiU ofweigfit on one side—moTnerd of
weight on the other.
From the graph determine
(i) At what distance from the fulcrum a weight of 120 grammes on
the right would balance the left-hand weight ?
(ii) What weight must be placed at 27 '5 cms. on the right to balance the
left-hand weight ?
(iii) Suspend a ■^ lb. weight from the right arm and determine by trial
the point at which it is in equilibrium. Read off from the curve its weight
in grammes.
EXPERIMENTS.
371
(iv) If all the weights used on the right in the experiment were suspended
at the same time from their old positions, what weight would have to be
suspended at 30 cms. from fulcrum on the left ?
(v) Where would a 500 gramme weight have to be suspended on the left
to balance all the weights used on the right ?
ExPT. II. Generalise Expt. I. by taking three weights on one side and
two on the other at the same time.
If Mr and Ml mean the moment of a weight on the right and left
respectively, then the result of this experiment may be symbolised by the
equation ZJfi- S71fjj=0.
Notice that the weights on the left tend to produce a oontraolockwise
rotation of the lever, and those on the right a clockwise, hence if the
former be reckoned positive and the latter negative, then, as an algebraic
sum, tilt mim of all the moments ahmit thefvlcrum is zero, or Silf =0.
Repeat this experiment twice, using different weights, and see if Sif =0
in these cases.
(1) Weights 20, 25 and 100 grammes are placed at distances 45, 30 and
20 cms. to the right of the fulcrum. A weight of 70 grammes is placed at
10 ems. on the left ; where must 20 grammes be placed to balance the others,
and what weight must be placed £,t 40 cms. to the left to balance the same
set?
(2) Would an upward push of 200 grammes at 15 cms. on the right be
likely to have the same balancing power as 200 grammes hanging at
15 cms. on the left ? (A 200 grammes upward push at any point would be
counter-balanced by 200 grammes weight hanging vertically below.)
(3) AB represents a lever 25 ems. long, the fulcrum being somewhere
between A and B. A weight of 170 grammes at A balances 60 at ^ ;
determine graphically the position of the fulcrum.
1
1
..p
c
-4—
w.
w.
Fia. 316.
ExPT. III. Generalise the results of Expt. II. by applying an upward
force on the right side by means of a spring balance {Pig. 316).
T,G.
2a2
372
GRAPHICS
Shew in several oases that
W^.AC=W2xGB-P.GD (where C is the fulcrum),
P being the upward pull of the spring as registered on the scale.
(4) Forces in lbs. weight,
10
-15
25
-12
20
Distances from it in inches.
2
3
-10
7
The above table gives the parallel forces acting on a lever, and the
distances of their points of application from fulcrum. Distances to the
right are given as positive, those to the left as negative ; a - force means
an upward push. Find the distance of the 20 lbs. weight from the fulcrum
for equilibrium. On which side must it be placed ?
(5) Expt. I. Chap. IV. shewed that a force acting on a rigid body
may be supposed applied at any point in its line of action. Suppose, then,
a weight on one side of a lever were suspended by a thread cemented on
to the lower edge (instead of strung from the upper edge), would this make
any difference in its balancing power ? If not, the moment would in this
case be expressed by weight x ?
Expt. IV. A farther generalisation of Expt. III. can be made by
changing the shape of the lever.
Fix a cylindrical peg through the centre of a rectangular piece of planed
wood (say 12" x 10" x J"). Fix two drawing pins firmly near opposite
edges of the board, leaving sufficient space between the heads of the pins
and the board to insert a loop of thread. Balance the board by means of a
peg on two blocks as in the case of the metre scale lever. Hang weights of
400 and 500 grammes on the pins, and let the lever take up a position of
equilibrium. Mark the lines of the strings on the board. Measure the
perpendicular distances of the strings from the fulcrum. See if the moment
law, S (weight X perpendicular distance from fulcrum) = 0, is true.
Remove the pins and mark their old positions ; set the pins in other
positions on the two lines drawn on the board, and suspend the weights
as before. Is the lever in equilibrium in the same position as before?
How can you tell that the position is the same ?
How does the experiment verify the deduction from Expt. III. Chap. IV.
Notice that when the line joining the two pins is above the fulcrum the
equilibrium is unstable.
Expt. V. A final generalisation is efifected by arranging an experiment
in which the forces have different directions.
Cut out a piece of irregular shaped cardboard. Punch five holes in it,
and fix it on the vertical drawing board (Expt. II. Chap. IV.) by means
of two stout pins. Attach hooks, threads and weights as in Expt. IV. of
Chap. IV., one weight hanging vertically. Remove one of the pins; tho
card will probably turn round the remaining pin, and take up some
position of equilibrium. The card is now a lever in equilibrium, the pin
being the fulcrum.
Draw lines on the card shewing the axes of the forces, and indicate on
the lines the magnitudes and senses of these forces,
DEDUCTIONS. 373
Dismount the card. Measure the perpendiculars from the fulcrum on
the axes of the forces, and calculate the sum of the moments.
Find the sum of the vectors of the forces.
Draw through the fulcrum a line parallel to the sum of the vectors.
Mark a point on this line, and mark also a point not near the line.
Calculate the sum of the moments about these two points as if they
were fulcra.
If a force equal in magnitude and direction, but opposite in sense, to the
resultant vector acted in the axis drawn through the fulcrum, what would
be the sum of the moments of the whole six forces about these three points,
viz. the original fulcrum, and the two marked points ?
What was the force acting at the fulcrum on the card, i.e. the reaction
of the pin ?
IJxPT. VI. Fix the same card to the drawing board (by two pins) in
such a position that the axis originally vertical is vertical again.
Arrange the pulleys so that the same forces may act on the card as in the
last experiment, and in addition a sixth force given by the reversed
resultant vector whose line of axis has been already drawn.
Remove the pins and see if the card is in equilibrium.
If in equilibrium any point on the card pinned to the board would do
for a fulcrum. Calculate the sum of the moments of the six forces about
several points.
Was the vector polygon closed for all the forces ?
Would there have been equilibrium if the sixth force had been applied
in any other line than the one through the fulcrum ?
Deductions.
(1) If a rigid body is free to turn about an axis (fulcrum) the rotative
tendency of any force acting on the body is measured by the product of
the force and the perpendicular on its axis from the fulcrum, and this
moment is positive or negative according as the rotative tendency is eontra-
clookwise or clockwise. Expts. I. -VI.
(2) If a number of ooplanar forces act on a body free to turn about an
axis, the body will turn until the sum of the moments of the forces about
the axis is zero. If the body, free to turn about an axis, does not, the sixm
of the moments of the forces about that axis is zero. Expt. V.
(3) If the sum of the moments of the forces is zero about three points
(non-coUinear) in the plane, the body will be in equilibrium (Expts. V.
and VI.), and if the body is in equilibrium, the sum of the moments
is zero for all points. Expt. VI.
374 GRAPHICS.
NOTE A.
CIRCULAR MEASURE OF AN ANGLE.
Draw a circle of radius 3" and mark by radial lines angles of 30°, 60° and
90°. Step oflf the arcs, corresponding to these, along a tangent and measure
their lengths in inches. Divide these lengths by the radius. The numbers
so obtained are approximately the circular measures of the angles. See
that the last number is 1 -57 and that the numbers are in the ratio 1:2:3,
or nearly so.
Whatever the radius of the circle the circular measure of these angles
will be the same.
If an arc be stepped off equal to the radius, the angle at the centre will
have unity as its circular measure. This angle is called a radian and is the
unit of circular measure. Construct a radian and see that it is nearly 57 '3°.
The measure of an angle in radians is always — — — , which being the
ratio of two lengths is a number. radius
Mathematicians formerly went to enormous trouble to calculate the
measure of two right angles in radians ; it has been worked out to 707
decimal places. The number of radians in two right angles is alwaj's
denoted by the letter tt. Correct to 5 figures its value is 3'1416 ; the
fraction -^^ gives tt correct to three figures. Instead of giving the radians
in an angle as a number, approximately correct, it is sometimes convenient
to express it as a fraction of t ; thus the number of radians in 15°, 30° and
45 are ^, ^ and j.
„. semicircular are ., . „
hinoe 7r = ,. , the circumference of a, circle of radius r
is 2irr.
NOTE B.
GREEK LETTERS USED IN THE TEXT, WITH THEIR
USUAL PRONUNCIATIONS.
o (Alpha). e (Theta).
j3 (Beta). i- (Pi).
y (Gamma). p (Rho).
'irf, :} (Sigma),
e (Epsilon). SJ *
(Phi).
ANSWERS.
CHAPTER I.
Exercises. Paqbs 1-41.
3. 0-62"; 38-4'. 5. Total' load, 15-1 ozs. 6. 11-3 ozs.
7. 123. 8. 27-4. 9. 12. 10. -9 7.
15. Complete the triangle on a, and measure the angle between u and the
third side. Through the ends of b draw lines making and 60°
with b, then c is the side bounding the 60° angle.
16. 8-88. 25. 1-92. 29. 1-36, 1-48, 0-57. 32. 1-47, 0-83. 34. 915.
35. 6-38. 36. 6-06. 38. 0-575. 39. 288. 40. 225.
41. 1-28, 1-64, 2-08, 2-67, 3-43, 4-39. 42. 0-73, 0-62, 0-53.
43. 1-13, 1-06, 0-92, 0-96.
46. 0-04, 0-13, 0-21, 0'82, 0-97, 1-74, 1-90.
47. 0-63, 0-87, 0-96, 1-17, 1-30, 1-43, 1-48.
50. y = ^=—, where M is 1"; y=x^; 9".
51. f =-t; ; 1'5" and 13'5" and are independent of a.
b a?
64. 3-8. 65. 8130. 66. 10-9". 67. 257 ; 694.
69. 31-2 million ergs. 70. 10-4 sees. 71. 2 94 ft. -lbs. 72. 17 -4 ft. -tons.
73. A line of length 8 '96 cms., scale 1 cm. to 1 lb. -ft. ; a line of length
8-96", scale 1" to 1 lb. -ft.
Miscellaneous Examples I. Pages 41, 42.
1. 0-6", 1-42", 0-7". 2. 5-14 cms. . 4. 1-88".
5. 7-85, 12-6, 15-7. 6. 45-3, 35-0, 17. 7. 107, 138, 236, 284.
8. J/ = j|^, where Mis an inch. 9. 2-47. 10. 21 ft. -lbs.
11. 2-46, 1'19, 0-76, 4-0. 13. 0-31 cub. ft.
aio
Ij r\i Ji JT XliOO.
CHAPTER II.
1.
6.
9.
13.
25.
Exercises. Pages 43-64.
5-12 sq. ins. 2. 20-4 sq. cms.
9-1 sq. ins. 7. 2420 sq. ft.
ABGD is - 1-03 sq. ins. 11. 9 sq. ins.
8-87. 16. 5-9 sq. ins.
21-6 cub. ins. 26. 9-75 cub. ins.
4.
8.
12.
24.
2'28 sq. ins.
1-67 sq. ins.
4950 sq. yds.
37 '6 cub. ins.
8.
5.
8.
Miscellaneous Examples II. Pages 65,
12sq. ins., l-2sq. ins. 3. 5'4 sq. ins. 4.
1-5 sq. ins. 6. 9-9 sq. ins. 7.
3560 cub. eras. 9. 22200 sq. ft. 10.
66.
4-1 sq. ins.
152 sq. ems.
40 '7 sq. ems.
12. Area is 5-93 sq. ins., angles are 71-8°, 61-8°, 46-4°.
CHAPTER III.
Exercises. Pages 69-114.
1. 113 ft. N.W. ; 160 ft. N. ; 113 ft. N.E. ; 0.
2. (i) 4-95', 0-6° W. of S. ; (ii) 6-97', 11-4° W. of S.
7. 14-3 eras., 23-2° S. of E.
9. 3-69m./hr. S.W.; 3-7m./hr.,25-9° S. of W.; 1-97 m./hr., 32-2°N. of W.
Speeds 3-69; 4; 3-86 m./hr.
11. 5-83 ft./see., 31° E. of N. 12. 36-9° with the up-stream line.
13. 24'3° with the up-stream line. 14. 258 m./hr., 22-8° N. of E.
15. 16'6 m./hr., 252° W. of S. 17. 14'3 m./hr. from 36-5° N. of E.
18. 17 m./hr. (nearly). 19. From 28-4° N. of E. and 15-6° W. of N.
20. 3-3 miles. 81. 13-5 m./hr. 47-5 E. of S.
88. 9-2 m./hr., 14-7° E. of S. ; 157 m./hr., UT E. of S.
83. The total accelerations are 17 ft. /sec, 48-3° S. of E., 26-1 ft./seo. E.,
16 ft. /see., 45° N. of E. and 0.
24. 41 '9 ft. per sec. per sec. at 22'5° with its own direction of motion.
26. 7-07 ft. each. 27. 10-6 and 5-5 ft.
28. 27 "9 and 16"1 ft. per see. per see.
29. 4 '93 and 0'83 ft. per sec. per sec.
30. 47 miles. 31. 29 '2 ft. per see. per sec.
32. 1"61 ft. per see. per see. down the incline.
33. 2'31 ft. per sec. per sec. making 43'4° with the vertical.
48. 0'85" from the centre.
43. I'l" and 0'8" from the sides containing the second and third, and the
third and fourth masses respectively.
48. 5 '93 cms. from origin. 49. 4" from origin.
53. 1'92" and 0-29". 57. 0-608 of radius from centre.
ANSWERS.
377
58. 0'84" from centre of circle.
66. 0'19" from the centre of the 3" circle away from the hole.
69. 0'33" from the centre of the rectangle away from the hole.
70. 0'44" from centre of rectangle.
71. 0-57" from centre of rectangle. '
Miscellaneous Examples III. Pases 115, 116.
3. 26 '8 ft. per see. at 52 -6° with the horizontal.
4. n-2m./hr., 49-r N. of W. 5. Coordinates are 2-42 and 3-54 cms.
6. Coordinates are 2-32 and 3 '88 cms. 8. 2-25" from centre.
9. 0'78 and 0'61 of the radius from AB and £C respectively.
11. 7 '17 ft. from base of block.
CHAPTER IV.
Exercises.
166 lbs. wt.
(i) e = = 36-9°.
(iii) = 34-05°, 9 = 44-4°.
(v) P=14-21bs. wt. = 76-3°.
(vii) 9 = .33 -8°, g = 7-76 lbs. wt.
(ix) P= 10-73, Q = 9.
9S-5°, 134-7°, 129-8°. 5.
(i) 12-7 and 20*8 lbs. wt. respectively.
2-98 cwts. at A and 1 -57 at B.
Pages 119-166.
(ii) iJ = 8-7 lbs. wt., = 30°.
(iv) Q = 8-18 lbs. wt., = 48°.
(vi) P = 5-38, Q = 4-121bs. wt.
(viii) P = 6-54.
(x) i? = 5-66, e = 3-35.
59 and 107 grms. wt. respectively.
A
5,340
5,880
8,980
12,600
14,800
30,600
B
8,790
8,720
7,560
9,080
11,400
29,000
12. Ill making 112-75° with the 30 force.
13. JJ=37-4, S=6-3. 14. 0-2 and 1-67 lbs. wt. respectively.
15. 7-54 and 12-52 lbs. wt. respectively.
17. AB makes 21-3° with the vertical, the pull on A is 10-25 lbs. wt.
19. ^r=32-8 lbs. wt. 20. 11 lbs. wt. at 35-3° with the vertical.
21. 1-29 cwts. 22. 20-5 and 36 lbs. wt.
23. Through one end of a draw a line parallel to OB, describe a circle
of radius -c with the other end of a as centre. The circle cuts the
line drawn parallel to OB in two points, henoe c has one of two
directions. If c has a direction perpendicular to OB the magnitude
of c will be the least possible.
24. (i) 13-4 and 44-8 lbs. wt. (ii) Diminished by 5-3 and 7-3 lbs. wt.
(iii) Increased by 5-3 and 7-3 lbs. wt. (iv) 18-3 lbs. wt.
378 GRAPHICS.
25. 20-7 kilogrms. wt., 40-1° E. of N.
26. 6-5 lbs. wt. at 67-2° with the 23 force.
30. 18'2 lbs. wt. towards a point .S3-7° N. of E.
31. 23-8 lbs. wt. at 79-1° with the 10 lb. wt. force.
32. 20-7 lbs. wt. bisecting the angle ABC.
33. 12-7 lbs. wt. at 45-05° with AB. 34. 9-33 and 9-66 kilogrms. wt.
35. 1-29 and 2-5 kilogrms. wt. 36. 1-77 and 3-42 kilogrms. wt.
38. 0-4cwts.
39. First draw the plane. Then set off ^ C vertically downwards for the
weight of the body and draw A B and BG parallel and perpendicular
to the plane. Set off ^^ along AB for the pull parallel to the plane,
and then draw MD at 15° with the horizontal cutting CB at D.
Scale DE and DO ; they give 10'9 and 14"3 owts. as the required
forces.
40. Braw parallel to 7 a radius of the circle and measure the length of the
arc to the highest point (1'22').
41. The slope is 041 the angle 21-8°.
42. 24-3°. 43. 39° ; 16-1 lbs. wt. 44. 3 39 cwts.
45. 4 '5 and 3 kilogrms. wt.
46. 15'6 lbs. wt. at 45° with the vertical.
47. 9 3 and 19 '9 lbs. wt. bisecting the angles.
48. (i) 388,1300 and 388 grms. wt. bisecting the angles,
(ii) 1449,1299 and 1449 grms. wt. bisecting the angles.
49. 5-71 tons wt. in AG, 4-28 in BG. 50. 8'61 tons wt. in BC.
51. 2 '33 tons. wt. in BC, 4-46 in AC.
53. 6-39 tons wt. in AC, 3-83 in BC.
54. 4-3 tons wt. in AG, 0-92 in BG.
55. 1-89 kilogrms. wt. in AC, MQ in AB, 0-98 in BC, 068 kilogrms. wt.
reaction at B, 1'62 at A.
56. 4-44 tons wt. in AB, 9-62 in ^C.
57. 2-8 tons wt. in AB, 15-9 in AC.
59. Reaction of nail=weight of picture ; tension=4-62 lbs. wt.
61. 1-72 owts. in BC, 3 5 in AG, G being below AB.
62. Stress in BA, 529 tons wt. 63. Tension is 15-5 lbs. wt.
64. First find the tension in the chain by drawing from the ends of the
vertical load vector a horizontal line and one making 30° with
the vertical. Find the total pull at B of the two parts of the chain
and then the forces in ^ S and CB which are in equilibrium with
this. Stress in AB is 6-6 owts., in CB 17-3 owts.
65. 18-18, 42-1° N. of E. 67. 10-02 and 9-92.
69. 129, 321, 453, 483 lbs. wt. respectively.
70. 11 and 5-69 lbs. wt. respectively.
72. (i) 421 lbs. wt. along the rail, side thrust 201.
(ii) 421 lbs. wt. along the rails, side thrust 21.
ANSWERS. 379
tS. (i) 5-3 and 10-55 lbs. wt. (ii) 10-55 and 5-3 lbs. wt.
(iii) 3-7 lbs. at each.
75. i'=3-67 lbs. wt., reaotion = ll lbs. wt. Component parallel to
plane = 3-67 lbs. wt., vertical component of reaotion = 10-4 lbs. wt.
76. 27 lbs. wt., 75-3 lbs. wt. 77. 132-7 lbs. wt., 55-8 lbs. wt.
78. The wedge is in equilibrium under the horizontal push of 28 lbs. wt. ,
the reaction of the block and a vertical force equal to the table
reaction less the weight of the wedge. Draw a horizontal line
representing to scale the push of 28 lbs. wt. From its end points
draw lines vertical and perpendicular to face of wedge supporting
the block, the latter gives 59-6 lbs. wt. as the reaction of the block
on the wedge. The former gives 52-5 as the vertical force on the
wedge, and this consists of 18 lbs. downwards and the reaction of
the table upwards. The table reaction is therefore 70-5 upwards.
79. 12-2 and 43-5 lbs. wt. respectively.
84. 199 lbs. wt. 87. 5-85, 12-5, 37-5 lbs. wt. 89. 2-35 lbs. wt.
90. 2-36 lbs. wt. at 43-7° with the vertical.
91. 0-252 owts. at wall, 0-564 at ground, making 63-45° with the horizontal.
98. 0-217 owts. at wall, reaction at ground makes 23-25° with the vertical.
93. Tension is 221 kilogrms. wt.
97. For the 60° plane the reaction is 29-9 lbs. wt., the reaction of the
hinge is 60-8 lbs. wt., making 25-2° with the vertical.
98. 1 25 cwts. at the top and 1 -66 cwts. at the bottom, the latter making
51-6° with the horizontal.
99. 521 lbs. wt. at hinge, making 24° with the beam, and 265 lbs. wt. at
the top.
100. Q = 212 lbs. wt., reaction at G 232 lbs. wt. passing through the
intersection of P and Q.
101. 42 lbs. wt. at plane and 48 at the peg, making 25-6° with the vertical.
102. 83-1° with the vertical. 103. 50-8° with the vertical.
104. 42-7° with the vertical.
105. 5-1° with the horizontal ; reactions of the planes are 3-66 and 2-59 cwts.
CHAPTER V.
Exercises. Pages 172-206.
2. Resultant of magnitude 3-62 lbs. wt. inclined at 13-8° with BC, the
axis outs A B produced at 1" from B.
3. Resultant of magnitude 8-99 lbs. wt., making 48-7° with Ox and
cutting it at - 0-31 units from origin.
8. Magnitude 6-62 lbs. wt. making 78-3° with the iirst spoke, axis cuts
first spoke at 4-6" from its point of contact with the hub.
11. 12 lbs. wt. at 6-42" from the smaller weight.
12. 3-86" from the left end weight.
13. First find the resultant of the first two forces and the given equilibrant.
The third rope is 3 -9" from the mid-point and nearer the smaller weight.
380 GRAPHICS.
14. Draw the link polygon for the weights Wj and W^, mark the points
where the first link cuts the known axis of the resultant, and
where the last link outs the axis of Wg. A line through the pole of
the vector polygon parallel to the line joining these two points
determines the magnitude of W^ (19 '75 lbs. wt.).
15. - 5 lbs. wt. at 9 ft. from first force and 6 ft. from second.
16. -22 lbs. wt. at 17-1" from the end.
17. Magnitude 27 '2 lbs. wt., the axis making 22-1° with PQ and cutting it
at 0-47" from P.
18. 1 cwt. at 0"72 yds. from first and 0'28 yds. from second force.
19. 197" from leading wheel. 25. 4-73 and 5-97 tons.
26. 23-65 and 22-35 owts. 27. 39-4 and 6-6 cwts.
28. No ; reaction 1 -3 owts. downwards.
31. 23-7 owts. at roller, reaction at pin makes 41-6° with beam and is of
magnitude 31 '5 cwts.
33. 185 -7 lbs. wt. at plate, 273 lbs. wt. at hinge, making 42-9° with
vertical.
34. 364 lbs. wt. at plate, 472 lbs. wt. at hinge, making 50 '5° with
vertical.
35. (i) ,327 lbs. wt. at plate, 499 lbs. wt. at hinge, making 54-5° with the
gate post.
37. Reaction at cylinder 1-82 owts., reaction at hinge 1-22 cwts., making
51 -9° with the horizontal.
38. 348 lbs. wt., making 15-2° with the vertical.
46. (i) Perpendicular to the pole with a force 1-81 times the weight of the
pole ; (ii) a force of 1 -86 times the weight of the pole at an angle of
76-9° with the pole.
53. 10-3 tons wt., making 16-1° with the vertical.
CHAPTER VI.
Exercises. Pages 210-251.
2. BQ and PJR in tension ; stress 9-82 lbs. wt. in each. PQ in com-
pression ; stress 4-91 lbs. wt.
3. Reaction at P is 12-45 lbs. wt., stress in PO = 3-72 lbs. wt. and at
QR=5-0S lbs. wt.
5. A tensile stress of 4-97 lbs. wt. in the lower bars ; a compressive stress
in the horizontal bar of 14 lbs. wt.
6. Lower bars have a tensile stress of 4-97 lbs. wt., the vertical bar one
of 14 lbs. wt.
7. The diagonal bar makes 4-75° with the horizontal ; the stresses in the
two lower bars are 4-43 and 5-42 lbs. wt.
9. The stresses in AB, BO and CA are 21-7, 56-3 and 27-3 lbs. wt.
11. Tensile stress in PQ of 2 tons wt., compressive stress in PT of 1'73
tons wt., compressive stress in TS of 1-15 tons wt.
ANSWERS. 381
12. Compressive stress in PQ of 2 'SI tons wt., in QS a tensile stress of
1 "15 tons wt.
13. Stress in PQ is 2-31 tons wt. 14. Stress in PU is 8-66 tons wt.
23. 33'1 lbs. wt. along the line joining P to the mid-point of QS.
30. Average compressive stress in .4 £ is 3 '93 lbs. wt.
CHAPTER VII.
Exercises. Pages 256-280.
1. /i = !s, e=26-6°.
2. Reaction 27"4 ozs., making 18'4° with the vertical.
3. (i) Yes ; (ii) no. 4. /i=0-447, 6 = 24-1°.
5. 467grms. wt. 8. 4-09 and 379 lbs. wt.
9. Least force 11 '7 lbs. wt. at 32 '4° with the horizontal, corresponding
friction is 0'98 lbs. wt.
10. Yes ; 15 '5° with the vertical.
11. 5-64 cwts. at 19-5° with the vertical, horizontal resistance TSS cwts.
12. Least pull is 12-8 lbs. wt. at 20'3° with the horizontal.
13. Angle of friction 36 "9°, reaction 8 lbs. wt.
14. {a) 3-5. (6) 4-14 cwts. (c) 2-98. [d) At 25° with the horizontal.
15. 6-23 kilogrms. wt. is the least force, 17'81 in the opposite sense to the
5 force.
16. 0-78 tons wt., making 38-7° with the vertical; (a) 1-02 tons wt. ;
(6) 0-83 tons wt.
17. Greatest force 1 "31 kilogrms. wt.
18. 3 32 cwts. 19. 6-43, 3-46 cwts.
20. 11, 2-85 and 7'78 lbs. wt. For the friction in the three cases resolve
the total reaction of the surface into two components along and
perpendicular to the plane, the former components give the friction.
21. 10-5 and 14-7 lbs. wt. 28. 48-5 lbs. wt.
23. 10'3°. 24. 2-85 ft., least tension 0'896 cwts.
25. 43-6°. 27- A=0-14.
28. The M.c. divides ladder in ratio 13/100.
89. 14-29 lbs. wt. ; ^=0 866.
31. 8-8° with the horizontal.
32. The M.c. divides the beam in ratio 342/100.
33. The M.c. divides the beam in ratio 211/100.
34. The M.c. divides the beam in ratio 323/100.
36. The M.c. is nearly at the midpoint.
38. 63° with the vertical.
39. 76-7° with the vertical.
41. 70° with the vertical, no other possible position for limiting equili-
brium.
382 GRAPHICS.
CHAPTER VIII.
Exercises. Pages 284-302.
3. 475 lbs. ft.
6. (i) -38-7. (ii) 13-7. (hi) -8-5 tons ft.
8. 5450 tons inches. 9. 13700 tons inches.
11. Take the pole 10 cms. from the force vector.
12. Measure the line giving the momental area on the kilogramme weight
scale.
14. 15-7 lbs. inches. 16. - 1 1 -4 lbs. inches.
80. 2 '67 cms. from CD, 4-2 from AD.
21. 6-66 cms. from CD, 2-05 from AD outside the square.
29. 5-4 cms. from lower centre. 30. 5-62 cms. from lower centre.
CHAPTER IX.
Exercises. Pages 315-332.
1. B.M. is 4000 lbs. ft., s.F. is 500 lbs. wt.
2. b.m's are 22,500, 10,000 and 5000 lbs. ft.
s.f's are 2000 lbs. wt. at 8 ft.
1700 lbs. wt. at just under 15 ft.
1000 lbs. wt. at just over 15 ft.
1000 lbs. wt. at 20 ft.
3. B.M. at 8 ft. is 21-2 tons ft.
s.F. at 8 ft is ] '9 tons wt.
4. At 7 ft. from Q the b.m. is 21,100 lbs. ft. and the s.P. is 1500 just
under, and 750 just over 7 ft. from Q.
I
CHAPTER XI.
Exercises. Pages 354-367.
1. 1420 ft. -lbs. 2. 16,000 ft. -lbs. 3. 18,500 inch-lbs.
4. 1930 ft. -lbs. 5. 48-4 ft.-owts. 6. 117 ft.-owts.
8. 799ft.-lbs. (independent of the radius). 9. 3470 ft. -lbs.
10. 193 ft. -lbs. against gravity, 57 against friction, 250 by the equilibrating
force. Total work is zero.
11. 278 and 367 ft. -lbs. 12. 765 ft. -lbs.
13. 335 ft. -lbs. 14. 2,520,000 ft. -lbs.
15. 207 ft. -lbs. round B and 66-2 round G.
16. 224,500 ft. -lbs. 18. 303 ft. -lbs. 19. 54-6 inch-lbs.
20. 57'9 inoh-lbs. 21. 19-3 ft. -lbs.
INDEX.
Abscissa, 13.
Acceleration, average, 82.
definition, 83.
due to gravity, 83, 116.
mass and force, 135.
total, 82.
Action and reaction, 137.
examples on, 138-142.
Addition, graphical, 4.
of accelerations, 83.
of displacements, 70-72.
of momental areas, 188-190.
of moments, 287-297.
of velocities, 78.
of work done, 361.
Areas, circular sector, 55.
circular segment, 56-57.
equivalent figures, 62, 113.
irregular figures, 58-61.
mass-centres of, 98, 114.
mid-ordinate rule for, 58.
negative, 48, 49, 106, 107.
parabolic segment, 60.
polygons, 50-52.
quadrilateral, 46-48:
cross, 49.
re-entrant, 47.
rectangle, 46.
Simpson's rule for, 59.
to scale, 36, 37.
triangle, 43-45.
Arm, of a couple, 185,
Average acceleration, 82.
speed, 77.
stress, 226.
velocity, 77.
Axes of coordinates, 13.
Axis of rightand skewsyrametry, 95.
Backstays, ,341.
Bending moment, definition, 314.
diagrams, 314.
experiment and explanations, 309.
for cantilever, 310.
for continuously loaded beam,
319-321.
for freely supported bridge, 312.
for non-parallel forces, 316.
for several loads, 314.
for travelling loads, 321-334.
maximum, 313, 322, 329, 330.
Bending of beams, 309.
Bicycle spanner (problem), 273.
Bow, notation due to, 177.
l^owstring roof truss, 221.
Boyle's law, 265.
Cantilever, 216, 225, 308-311, 322,326.
Centre of figure, 89, 303.
of gravity, 303.
of mean position, 85-90, 303.
of parallel forces, 298, 303.
Centrold, 89, 303.
Chain, uniformly loaded, .346.
of suspension bridge, 344.
Circular arc, length of, 53-54.
M.c. of, 98.
measure, 374.
sector, area, 55.
M.O., 105.
segment, area, 55.
M.c, 108.
The numbers refer to pages.
384
GRAPHICS.
Components, vectors, 84.
forces, 151-154, 199-205.
parallel forces, 201.
three non -concurrent forces, 202.
Composition of forces, 206-208.
Conjugate direction, 99.
Continuously loaded beams, 319-321.
chains, 345, 346.
Coordinates, rectangular, 13.
Couples, 184, 207, 304.
arms of, 185.
momental areas of, 185.
of transference, 207.
Crane, simple wall, 144-147.
derrick, 149.
Cremona (Professor), 352.
Cubes, 33.
Cube roots, 33, 34.
Curve of cubes, 33.
of gas compression, 366.
of reciprocals, 34.
of squares, 29.
of suspension bridge chain, 344.
Decomposition of forces, 151-160,
199-206.
Derrick crane, 149, 204, 319.
Dip (of a suspension bridge), 341.
Direction, 1, 69.
Displacement, 69, 70.
addition of, 71, 72.
minimum, 81.
relative, 73.
Division, 15, 16, 17.
and multiplication, 18, 19.
on squared paper, 17.
Dyne, 136.
Ela.stic limit, 305.
Engine crank, force on, 156.
Equation, to a straight line, 13.
to curve of suspension bridge, 248,
344, 345.
1 1
y = x', a;",
„ 31-36.
Equilibrant, 132, 177.
Equilibrium under concurrent forces,
132, 135.
under two forces, 118.
under three forces, 132, 160-165.
and friction, 257, 258.
Equilibrium and link 'polygon, 190,
195.
and vector polygon, 184, 186, 195.
broken by rotation, 277-279.
scalar conditions, 151.
Equivalent figures, 62, 113.
forces, 176.
Erg, 40, 355.
Experiments I. -VII. on concurrent
forces, 119-122.
deduction, 131, 132.
VIII. on link polygon construc-
tion, 177.
IX. , X. , XI. on general conditions
for equilibrium, 192.
deductions from, 193.
XII., XIII., XIV. on friction,
256, 258.
deductions from, 258.
with lever, 369-373.
deductions from, 373.
Palling bodies, 83.
Foot-pound, 38, 355.
Force, components, 151-156, 199-
205.
equilibrant, 132, 177.
equivalent, 176.
mass and acceleration, 135.
moment of, 41, 284, 370-373.
on crank, 156.
on sail, 159.
polygon, 176.
unit, magnitude of, 136.
Forces, concurrent composition of,
133.
decomposition of , 151.
examples on, 123-126.
experiments on, 119-123.
resultant, 132, 135.
coplanar composition of, by link
polygon, 174-176.
decomposition of, 199-203, 206.
like and unlike, 180.
parallel, 180-185.
reduction of any set, 207.
resultant, 176, 177.
Framework, simple bar, 144-149.
and weight of bars, 225-2.35.
French window problem, 206.
roof truss, 353.
The numbers refer to pages.
INDEX.
385
Friction and beams on two planes,
270.
and bicycle spanner, 273.
and cube, 277.
and drawer, 275.
and inclined plane, 262-265.
and motion, 360.
and ladder, 266-269.
and lifting jack, 274.
and reel, 278-280.
and wedge, 276.
angle of, 259.
coefficient of, 258.
laws of, 258, 360.
minimum force for motion under,
261.
Funicular polygon, 176, 236-248.
and parabola, 244-248.
for equal loads, 240.
Gas, work done in compression, 366.
Girder, Warren, 222, 224.
N, 219.
Graphical addition, 4.
subtraction, 5.
Graphical measure of angles, 374.
of areas, 36-38, 43-61.
of bending moments, 310.
of circular arc, 53.
of moments, 285.
of powers, 26.
of products, 8-13.
of quotients, 15-17.
of reciprocals, 35.
of shearing force, 310.
of volumes, 62.
of work done, 38, 355.
Graphical representation of a
moment, 285.
of momental area, 185, 285.
of work done, 358.
Graphs of y=mx, 3?, 3?, -, etc., 13,
31-36. *^
of force on engine crank, 156,
157.
of force on meteorite (problem),
367.
of friction and normal pressure,
258.
of gas eompres-sion, ,366.
Graphs of lengths of ropes support-
ing loads, 127-129.
of lever law, 369.
stresses in simple wall crane,
146.
of spring extension, 365.
of work done, 363.
Gravity, acceleration due to, 83, 136.
centre of, 303.
Heaviside, vector notation, 74.
Henrici (Prof.), notation for graphi-
cal statics, 177.
vector notation, 74.
vectors and rotors, 352.
Hinge, door, reactions, 335.
reactions on bars, 226-230.
Hooke's law, 307, 364.
Inclined plane and beam in equili-
brium, 270, 271.
and coefficient of friction, 262.
and work done, 361.
problems on, 139-142.
and reel, 279.
with friction problems, 262-265.
Independent vectors, 110.
Integral powers, 26.
Irregular figures, area of, 58, 61.
M.c. of, 302.
Joints, pin, 144.
two bar reactions, 226, 228.
three bar reactions, 229.
King post truss, 222, 224.
Kite problem, 160.
Knife edge, supports, 195.
Law, Boyle's or Marriotte's, 365.
Hooke's, 306, 364.
Newton's second, 135.
third, 137.
Length of circular arc, 53.
Lever, experiments on, 369-373.
law of, 373.
simple, 369.
Like forces, 180.
vectors, 110.
Line, straight, 13.
M.o. of, 89.
yAe jiiimbers refer to pages.
386
GRAPHICS.
Line in division and multiplication,
17-22.
Link, polygon oonstructions, 173- 176,
178-180.
closed, 190.
for parallel forces (like), 180-182.
(unlike), 182.
Localised vectors, 132.
Mass, force and acceleration, 135.
moments. 111.
negative, 106, 107.
points, 91.
and weight, 136.
Mass centres and centres of gravity,
303.
by link polygon, 299-302.
definition, 91.
formulae for, 93, 97, 98, 104, 105,
111.
graphical construction, 91-93, 111,
112.
of circular arc, 97, 98.
sector, 105.
segment, 108.
of irregular area, 113, 302.
of lines, 96.
of points in a line, 92-94.
of quadrilateral, 100-101.
of trapezium, 104.
of triangle, 99.
scalar equations for, HI.
and work done, 362.
Masses to scale, 2.
Maxwell vector notation, 70.
Mid-ordinate rule for areas, 58-59.
Modulus, of a bar, 305.
Young's, 305.
Moment, Bending, 309, 314.
geometrical representation of, 285.
graphical measure of, 285, 286.
mass. 111.
of a force, 284, 290, 370-373.
sense of, 287.
unit of, 285.
Momental areas, 185.
addition of, 185.
unit of, 185.
Moments and couple, 295, 304.
and equilibrium, 304, 373.
bending, 308.
Moments, sum of, 296, 297.
by link polygon, 287-294.
theory of, 304.
Multiplication of lengths, 37.
of numbers, 8-11, 20.
of vectors by soalars, 85.
on squared paper, 12, 22, 23.
Newton's second law of motion,
135.
third law of motion, 137.
Notation for displacements, 70.
for lines representing numbers,
8.
for numbers, 8.
for vectors, 74.
space for forces and vectors, 177.
Numbers to scale, 6.
Ordinates, 13.
Origin of coordinates, 13.
Parabola, equation to, 244.
and funicular polygon, 244-248.
and suspension bridge, 343.
and telegraph wire, 345.
Parallel forces (engine problem),
180-182.
like and unlike, 182.
vectors (like), 74, 1 10.
Parallelogram, area of, 46.
centre of figure of, 89.
law, 134.
Particles, 91.
Path and displacement, 69.
and work done, 359.
Pentagonal frame stresses, 234.
Plane, inclined, 139-142.
and friction, 262-265.
Polygon, area of 50
force, 176.
funicular, 176.
link, 173-176.
vector, 176.
Powers of numbers, 26, 27.
Product of force and length {see
work done and moment).
of two lengths, 36-38.
of numbers, 8.
of ratios, 25.
Pulley, smooth, 119, 162.
Tlie numbers refer to pages.
INDEX.
387
Quadrilateral, area of, 47.
cross, 48.
frame, 214, 230.
M.c. of, 100, 101.
re-entrant, 47.
Quantities, scales and vector, 1.
to scale, 1.
Queen post truss, 223, 250.
Radian, 374.
Rankine, 53.
Reaction, determination of, 138-142,
160.
normal, 138.
of a surface, 138.
Reaction and action (see Newton).
Reactions, at joints, 226, 234.
by link polygon, 194, 196.
indeterminate, 317.
made determinate, 338.
of walls, 205, 230, 337-339.
Reciprocal figures, 352.
Reciprocals, 35.
squared, 36.
Rectangle reduced to unit base, 46.
Reduction of a set of forces, 195,
206, 207.
Reel, problems on, 278.
Relative velocity and displacement,
73, 79-82.
Resistance, 137.
Resultant force, 132, 133, 176.
by link polygon, 173-176.
examples in concurrent forces,
133, 134.
three forces (non-parallel), 172.
uniqueness of, 177.
Rigid body, 131.
frame as a, 212.
Roots, square, 30.
cube, 33.
Rotors, 132.
Sag, of a telegraph wire, 345.
Sailing against the wind, 159.
Scalar quantities, 1.
Scale, areas to, 36.
masses to, 2.
numbers to, 6.
Scales, change of, 22, 24.
different, 15.
Screw jack and friction, 274.
Second law of motion, 135.
Sections, method of, 248-250, 340-
352.
Sense and sign, 70.
of an area, 49.
of a line, 68.
Shearing force (s.F.) definition, 314.
experiments and explanations,
309.
for continuously loaded beams,
320-321.
for freely supported bridge, 312.
for several loads, 314.
for simple cantilever, 310, 326-335.
for travelling loads, 321-326.
Similar triangles, 7.
Simpson's rule for areas, 59.
Skew symmetry, 95.
Smooth (bodies), 138.
Speed, 76, 77.
Square roots, 29, 30.
Squared paper and division, 17.
and multiplication, 12.
and powers, 29-35.
Statics, foundation of, 131.
Straight line, as a vector, 74.
equation to, 13.
locus in B.M. diagram, 332.
M.c. of, 89.
in division, 17-19.
in multiplication, 12, 21, 22.
Stress, compressive, 137.
diagrams, three bar frame, 210-
214.
braced quadrilateral, 214-216.
bridge girder, 218.
cantilever, 216, 225.
examples, 221-225.
pentagonal frame, 234.
roof truss, 220, 338.
and vector polygon, 210.
tensile, 137.
Stresses by moments, 340-343, 348-
352.
Strip division method for areas,
58, 62.
Subtraction, 5.
Suspension bridge, 341.
and parabola, 343.
Symmetry, right and skew, 95.
Tlie numbers refer to pages.
388
GRAPHICS.
Telegraph wire, 345.
Three moments, 304, 373.
Toggle joint, 206.
Torque, 41.
Triangle, area of, 43-46.
M.c. of, 99.
Triangles, similar, 7.
Truss and wind pressure, 338.
bowstring roof, 221.
French roof, 352, 353.
king post, 222, 224.
quadrilateral, 251.
queen post, 223, 250.
railway platform, 223.
simple three bar, 149.
taking account of weight of bars,
353.
various examples, 215, 222, 223,
251, 339, 348, 351, 353.
Turning moment, 41, 370.
Unit area, 36.
base and reduction of areas to, 43.
force (magnitude), 136.
length, 78.
moment (magnitude), 285.
speed acceleration, 1,36.
speed, 78, 1.36.
work, 38.
Unlike forces, 180.
Vector addition, 74, 75.
definition, 74.
notation, 74.
polygon, closed, 106, 186, 190.
quantities, 1.
Vectors, components, 83, 84.
independent, 110.
Uke, 110.
multiplication by scalars, 85.
and vector quantities, 73.
Velocities, addition of, 78.
Velocity, average, 75.
definition of, 78.
relative, 79-82.
and speed, 77.
units, 78.
Volumes, of revolution, 62.
to scale, 36.
Wall crane, 198, 204.
reaction of, 205, 230, 337-339.
Warren girder, 222, 224.
Weight and mass, 136.
of oars in frames, 225-235.
Wind pressure on roofs, 338.
on a kite, 160.
on a sail, 159.
Work done, 38-40.
and M.c, 362.
and friction, 360, 361.
and motion, 358.
by component forces, 361, 362.
by variable force, 363.
definition, 354, 357.
graphical representation, 355.
in compressing a gas, 366.
in spring extension, 364.
negative, 354.
unit of, 38, 354.
Young's modulus, 305.
The nwmhera refer to pages.
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