1 20 THE PHYSICS OF PARTICLES

object, there are other physical. properties which —

Forces Acting on an 20IbE —__ can be represented as vector quantities. Other

such’vector quantities which will be discussed in
detail later are displacement, velocity, acceler-
Object: Vectors 2 in = 20 lb
ation, weight, momentum, torque, and electric
field intensity. Once we have established proce-
dures for solving force problems vectorially, these
Figure 1-1
Representation of a vector quantity.
same procedures will enable us to solve problems
involving these other vector quantities.
In addition, there are other physical quantities
which can be described in detail by using just the
example, if a 20-lb force is applied to a block, magnitude of the quantity. These quantities
we can represent the force by a scaled line with which can be completely specified by a number
a sufficient length to represent the 20 Ib. If 1 in and a unit are called scalar quantities. Typical
of the line is set equal to 101b, then the line scalar quantities are mass, volume, temperature,
would be drawn 2 in long. We have a problem, energy, length, and speed, all of which can be
In the study of physics we are concerned with ence of an energy-bearing object on another ob- however, in that whereas we know the magnitude added by ordinary arithmetic: 3s + 5s = 8s.
the exchange of energy between objects which ject. Force is one of the most important physical of the force, we are unable to predict the direction
make up our universe. The effect of an energy concepts, but since it is based on evidence pro- in which the object will move unless we know
vided by our senses it is difficult to define satis- the direction, or “sense,” of the force. Therefore,
exchange may have great influence on our lives, 1-2
factorily. A force is arbitrarily defined as any we need to know whether the 20-Ib force is acting
but the fact that an energy exchange is taking VECTORS
influence capable of producing a change in the easterly or northerly or any other compass direc-
place may not always be obvious to an observer.
tion from a reference direction. For example, if
This is partly because energy may exist in a num- motion of an object. Forces can be expressed in If two forces of 20-Ib magnitude are applied to
the 20-Ib force acts easterly, we can represent the
ber of forms—as heat, light, and electric, me- the familiar unit pounds and also in the units of an object, one acting easterly and the other
force with, let us say, a 2-in line with an arrow-
chanical, and nuclear energy, for example. We the cgs system, dynes, and the mksa system units, northerly, and these are the only two forces acting
shall see later that mechanical energy is defined newtons. Both dynes and newtons will be defined head indicating the direction (easterly) of the
on the object, it is possible to find a third force
later in this book. In order to analyze and predict force (Fig. 1-1). This scaled line with arrowhead
as the ability to do work. We usually think of which has the same or equivalent effect as the
this form of energy when we use the term, for the change resulting from the forces acting on a is called a yector. The word “vector” comes from
the Latin vehere, meaning to carry, which indi- two forces acting on the object. This third force
it is easy to observe the effect of the transfer of particle of matter or as far as we are concerned is called the resultant force of the two forces. If
the energy possessed by a body in motion when on a group of particles organized into bodies, it cates that the force is being carried from one
place to another—20 Ib easterly in this case. If
one wishes, this third force could be substituted
it reacts with another body. The effect of a mov- is necessary to develop certain mathematical and for the two original forces.
this force were the only force acting on the block,
ing car striking a parked car or a falling object graphical tools.
Simon Stevin, a Flemish scientist who lived it would move in an easterly direction at a rate
landing on an object at rest on the ground is quite
obvious. Since a goal of this;book is to relate in the northern Netherlands from 1548 to 1620, that could be predicted by using the appropriate
when offering mathematical proof of the law of
equations of motion to be described later in the
physics to the real-life world as well as to assist
the inclined plane, introduced a technique which book.
the student to build a knowledge of physics, we
. It is sufficient for us to realize at this point
shall use this type of familiar event as a point we use today: the triangle of forces method
that we can represent forces by vectors, It
of departure in the book. We shall consider the (the closed polygon of forces method) which re- is now
necessary to see how we can employ vectors
energy possessed by a body due to its state of vealed the additive properties of force. Through , as 20N
Stevin did, to enable us to make predic
motion or its position and then explore what his work, forces were shown to be vector quanti- tions
about the effect of forces on a particle.
happens when this energy is transferred to an- ties. We shall use his discovery to study the effect
other object. of forces on a point or on a body.
Before considering the exchange of mechani- If we consider an object which has a force, 1-1
cal energy, however, we must consider how this often described as a push or pull, acting on it, VECTORS AND SCALARS oO 20 N east A
energy interacts between objects. The term we can predict something about its resultant mo- It is worth noting here that although
“force” has been invented to describe the influ- tion if we can represent the force graphically. For we ar con- Figure 1-2 i
emod for the moment with Parallelogram of forces diagram,
mg acting ai i e
19 ee e
\
 FORCES ACTING ON AN OBJECT: VECTORS 21 THE PHYSICS OF PARTICLES

According to the work of Stevin it is possible in the next few chapters of this book. When the marizing these rules, it can be shown that any
to add these two forces vectorially. By con- vector sum of the forces acting on an object is two nonparallel lines will intersect and form an
structing a parallelogram of forces, it is possible equal to zero, the object is in equilibrium insofar angle @. If the two lines are truncated by a line
to show that the sum of vector OA and vector as linear motion is concerned. This statement is drawn at 90° to one of the lines (line DE or AB
OC is equal to vector OB (Fig. 1-2). It should called the first condition for equilibrium. in Fig. 1-4), a right triangle will be formed. In
be noted that the vector OB is not equal to the It should be noted also that had the equili- solving physical problems, especially problems
sum of the absolute values of OA and OC, 20 Ib + brant in the above example (12 lb, 60°S of E) involving vector quantities, it is often helpful to
20 lb # 40 Ib in this case, because of the vector been an original force acting on the object, the know the ratio of the sides of a right triangle to
nature of the forces. (Note: In this book boldface vector diagram would have been a closed poly- each other. Direct measurement will show that
type is used to represent vector quantities and to gon. In fact, the purpose of the equilibrant is to the ratio of the side opposite the angle to the
portray their vector properties clearly.) That is, cancel out forces and to put the object in a state longest side in a right triangle, called the hypote-
we have to consider not only the magnitude but of equilibrium. nuse, will be the same in the example shown in
the effect of direction in determining the resultant Graphical procedures have-an advantage over Fig. 1-4 whether the ratio ED/OE or that of
OB. A vector combines with another vector by mathematical procedures in that they may be AB/OB is considered. For the given angle @, the
geometrical addition to form a resultant vector Figure 1-3 used easily for multivector problems and for ratio will always be the same regardless of the
which represents the combined effect of the A vector polygon diagram. cases in which two forces are not acting at right _ size of the triangle. For convenience, the ratio of
quantities represented by the original vectors. angles to each other. Both types of problems the side opposite the angle of reference, 0, to
require fairly lengthy mathematical procedures to the hypotenuse is called the sine of the angle.
If it were desirable to cancel out the effect of solve analytically. The disadvantage to using In a similar manner, the ratio of the side
1-3 this resultant, as is often the case, a force equal graphical procedures is that the accuracy of the adjacent to the angle to the hypotenuse can be
ADDITION OF VECTORS in magnitude but opposite in direction would be results depends on the care taken in drawing the shown to be constant for a given angle 8. The
applied. The force, called the equilibrant, would vector diagram. Carelessly drawn diagrams will
In solving all vector problems, either graphical in this case be OD, 28 Ib, 45°S of W. ratio of the side adjacent to the hypotenuse is
lead to unacceptable errors. For accurate solu- called the cosine of the angle.
(geometrical) or mathematical procedures may be Another graphical procedure for solving vec- tions to problems, it is wise to be familiar with While other ratios can be set up between sides
used. The graphical procedures involve carefully tor problems is the polygon-of-forces technique
accepted mathematical techniques. of a right triangle, the only other one of much
drawn diagrams showing the magnitude and di- which is used when more than two forces are The mathematical, or analytical, method for value in this study of physics is the ratio of the
rection of each vector. In solving the problem acting on an object. determining the resultant R of any two vector side opposite the angle to the side adjacent to the
mentioned above, a scale is established, for ex-
quantities A and B employs rules of trigonometry. angle. This, too, will be a constant for a given
ample Lin = 10 1b, and by constructing a paral- Example What is the resultant of the follow-
In the simplest case of two vectors acting at right angle. The ratio of the side opposite to the side
lelogram, the length and direction of the resultant ing forces acting on an object at 0? A = 6 lb W;
angles to each other, use is made of the ratios adjacent is called the rangent.
OB is then determined with a ruler and a pro- B = 41b NW, C = 81b N, and D = 3 1b E.
tractor. The magnitude of OB is thus obtained
of the three sides to each other. _ The length of any of the three ‘sides in a right
First construct the vector diagram to scale
The student should review the trigonometric triangle can be determined, provided that the
in terms of the established scale (Fig. 1-2). (Fig. 1-3). Then, using any one vector as a base
rules for determining the ratio of the sides in a length of two of the sides are known, by the
(D in this illustration), draw vector C’ at the end of
Let lin = 10 lb right triangle (a triangle with a 90° angle). Sum- pythagorean -theorem,
vector D, then. draw vector B’ at the arrowhead which states that the
of vector C’, and vector A’ at the end of vector square of the hypotenuse is equal to the sum of
Therefore, OA = 2in B’. Since this is the last vector, the resultant is the squares of the other two. sides:
OC =2in determined by a vector drawn from the origin O
to the last arrowhead, vector A’. The length and (OB)? = (AB)? + (OA)?
and OB = 2.8 in direction of the resultant can be found by the use
of a scale and a protractor. In this case the result- Example Find mathematically the resultant
Therefore, OB = 281b ant is a force of 12 lb, about 60°N of W. force when a 20-N force east and
a 20-N forge
Should the last vector drawn come back to the north act on an object (see Fig. 1-5). Since the
A protractor shows angle @ to be equal to 45°. origin, the figure is called a closed polygon and gie is a parallelogram, OC = AB = 20N
According to this graphical solution, the re- the resultant is zero because the forces canceled north.
sultant is a force of 28 lb acting at an angle of 45° each other out. This is an example of linear equi-
north of east. librium, a concept we shall encounter many times Therefore,
nee side Opposite 20N
gee me 1 L444 side adjacent ~ 20N = |°
 FORCES ACTING ON AN OBJECT: VECTORS 24 THE PHYSICS OF PARTICLES

follows the rules of signs for angles in quadrant be greater than, equal to, or less than either one
In the case in which AB in Fig. 1-4 is not at
Tight angles to OA, a new set of relationships exist
I; and since angle a is between 90 and 180°, it of them, depending on the angle between them,
follows the rules of signs for quadrant II. In Fig. 1-84, two vectors A and B of magnitude
(Fig. 1-6). The pythagorean theorem now ex-
Therefore the cosine, which is — in quadrant II, 2 and 3 units, respectively, are shown separately,
pands into a new form, called the Jaw of cosines,
which appears as is + in quadrant I; hence the sign change in the In b, where the vectors are in the same direction,
two forms of the law of cosines. the magnitude of the resultant is merely their
(OB)? = (AB)? + (OA)? — 2(AB\(OA) cos a The law of cosines will enable us to find the arithmetic sum. As the angle between the vectors
value of the side OB, which is the resultant; but increases, the resultant becomes less, as in c, d,
Since the angle ¢ is often given in problems since we need also to determine the sense (direc- and e. In f,the magnitude of the resultant is the
involving vectors, the law of cosines can appear tion) of this resultant, it is necessary to determine difference of the magnitudes of the two vector
as the angle 8. The angle can be found by using the quantities.
o 20 N east A
law of sines, which relates the sides of a triangle
Figure 1-5 (OB)? = (AB)? + (OA)? + 2(AB\(OA) cos $ to the angle opposite each side: Example Two vectors of 8.0 units and 5.0
Parallelogram of forces diagram. units make an angle of 60° with each other. What
The change in sign in the last term of the is their vector sum?
equation is due to the rules establishing the sign Using a convenient scale, draw vector A 8.0
and 0@= 45° values of the trigonometric functions as they ap- If three of the four variables in two of these units long (Fig. 1-9). Beginning at the end of A,
‘ pear in various quadrants (quarters of a 360° equivalencies are known, the fourth variable can draw B 5.0 units long so that there is a 60° angle
Then, reference circle) (see Fig. 1-7). be found. between the direction of A and that of B. There
Since angle ¢ in Fig. 1-6 is less than 90°, it In magnitude, the resultant of two vectors may is, of course, a 120° angle between the line seg-
sin 45 -AB
= Op ani OB==R=-
= Suse oo

l Rg
90°
= 20N_
_ 95
mO. 2

This is the same answer obtained when we solved 0

the problem graphically earlier: 28N 45°N of E. 180° 0
The trigonometric rules stated above hold only
for right triangles.
(a) Two vectors, A and B (b) Vectors (à Angle @ > 90°
270° acting in same
J B direction
Figure 1-7a
Quadrants in a circle. A
R
Quadrant Angle sin cos tan
R
I ` 0-90° + + + (d) Angle @ = 90° |
Il 90-180° + = Fe B :
Il 180-270° - = + (e) Angle 0 < 90°
IV 270-360° = Jk T
(f) Vectors opposite
each other
Figure 1-7b Figure 1-8
Figure 1-6 Table showing sign of trigonometric functions for The resultant of two vectors depen
A triangle not containing a right angle. various quadrants. ds on the angle between them
 FORCES ACTING ON AN OBJECT: VECTORS
26 THE PHYSICS
OF PARTICLES

is numerically equal to the same function of its

related angle. For example, the sin 120° = sin the vector A can be considered as the resultant
(180° — 120°) = sin 60°. Since sin 120° is for of the vectors A, and A,. The magnitudes of
an angle in the second quadrant, the sign is pdsi- the horizontal and vertical components are
tive. Also, the angle sin 258° = sin (180° + Acos 45° and A sin 45°, obtained by recalling that
78°) = sin 78°; but since the sin 258° is in the sin 45° = A,/A;. A,=A sin 45°; and that
third quadrant, the sine has a negative value cos 45° = A,/A; A, = A cos 45°. The directions
l
and sin 258° = —sin 78° of the arrowheads are important, for we are now !
considering that A, has been added to A, to give |
|
Figure 1-9 the resultant
A; therefore,
the arrows must follow |
!
Addition of two vectors in given example. head to tail along A, and A, so that A can prop- |
1-4 i erly be considered as a resultant drawn from the
|
COMPONENTS OF A VECTOR tail of the first arrow A, to the head of the last i1
There is frequently occasion to perform the oper- arrow A,. This resolution into components now pgES ee
ments. Complete the triangle by drawing the
ation inverse to geometric addition, namely, to allows us todiscard the vector A in our problem
resultant vector R from O to the head of B.
determine a set of vectors whose effect when and keep only the two components A, and Ay.
Measure the magnitude of R, in the scale se- These two taken together are in every way equiv-
lected, and the angle 0 between R and the refer- acting together is the same as that of a given
vector. This process is called the resolution of a alent to the single vector A.
ence direction, say, that of A. It is found that R What is the advantage of having two vectors
is 11+ units, at 20°+ to the direction of A, the vector into its components. A set of components
to deal with where there was only one before?
8.0-unit vector. of a vector is a set of vectors whose resultant is
The advantage lies inthe fact that several vectors
Applying the analytical method to obtain the original vector. Figure 1-11
making various odd angles with each other can
greater precision, we have A given vector may have any number of com- Addition of vectors by component method.
be replaced by a setofvectors along one direction
ponents. The most useful choice of components and a setina direction atright angles tothe first.
is usually that in which the x and y components
R? = A? + B? + 2ABcos¢ The magnitudes of each of these two groups of
are found, where x is the horizontal component vectors can then be summed up separately by A, = Asin
70°
= 8,0? + 5,02 + (2)(8.0)(5.0)(0.50) = 129 and y is the vertical component, as in Fig. 1-10. ordinary arithmetic, thus reducing the problem
Consider the vector A, which makes an angle = 10 (0.940) = —9.4N
R = 114 units Bee of twovectorssetingat rightanglesw each
of 45° with the horizontal, as is shown in Fig.
1-10. In order to obtain a set of rectangular 0 VectorB:
The angle @ may be found from the sine law ponents of A, one of which shall be horizontal,
draw a horizontal line through the tail of the B,=0 Bx=B=4+5N
Re) 5 vector A. Now from the head of A, draw a line 1-5
perpendicular to the horizontal line. We see that COMPONENT METHOD VectorC:
sin 120° ~ sin@
OF ADDING VECTORS
«
sinĝ
_ Bsin120°
R
_ 5.0 E
PETTY 9.87 = 0.38
C, = Coos 45° =15 (0.707) = +10.6 N
To add a number ofvectors A,B,C,and D (Fig,
C, = Csin 45° = 15 (0.707) = +10.6 N
1-11), where A = 10N, 70°S of E; B = 5N, E;
Oi 22° C = ISN 45°N of E; and D = 15N 60°N of W, Vector D:
we proceed as follows, Place the vectorsat the
Hence R is 11.4 units at 22° to the direction of A. origin on a setofrectangular coordinates (xand
Y). Next resolve each vector into x and y compo- D, = D cos60° = 15 (0.5) = —7.5 N
It is sometimes confusing to find the trigo-
nents (Fig. 1-12), D, = D sin60° = 15 (0.866) = +13.0N
nometric values of an angle greater than 90°. A VectorA:
table of rules for determining the trigonometric Figure 1-10
functions of obtuse angles is presented in the Vertical and horizontal A, = A cos 70°
Appendix. It might prove helpful to remember components of a vector. and then add the forces along the y
axis,
that any trigonometric function of a given angle —_—_——— = 10(0.342) = +3.4N the sum Zy. Some of the components calling
may be
 FORCES ACTING ON AN OBJECT: VECTORS 27 y 28 THE PHYSICS OF PARTICLES

duce no change in the motion of the body. The

state in which there is no change in the motion
of a body is called equilibrium. A body in equilib-
rium may be at rest but does not necessarily have
to be at rest; it may be moving with uniform
speed in a straight line or rotating uniformly
around a fixed axis. An example of this situation
occurs when a parachute reaches its terminal
velocity when falling (Fig. 1-14). It continues to
fall toward earth with a constant speed but, as
we shall see later, no longer has a net force acting
on it to cause it to accelerate.
In this chapter the discussion will be restricted
Figure 1-12 to the action of forces that are in equilibrium and
Resolution of vectors into x and y components. also that have lines of action which pass through
the same point. Forces whose lines of action
intersect at a common point are said to be con-
negative (for example, D, and A,). This is due Table 1 current.
to the sign convention which says that x to the x AND y COMPONENTS OF VECTORS IN We have seen earlier that a body in linear
right of they axis is + and to the left is —; y THE EXAMPLE motion is in equilibrium if the vector sum of the
above the x axis is + and y below the x axis forces acting upon it is zero. This statement is
is —, It should be noted that a vector has a zero Vector x Component, N y Component, N
known as the first condition for equilibrium.
rectangular component at right angles to itself. A +34 ' = 94 Several forces may be added by the use of any
For instance, vector B has a zero y component, of the methods previously described. When the
We have now resolved the four vectors into B + 5.0 0
vectors are added by the graphical method, the
two vectors acting 90° apart and can solve for c +10.6 +10.6 vector sum is zero if the length of the arrow
the magnitude of the resultant by using the py-
D -75 +13.0 representing the resultant is zero. It was shown
thagorean theorem or by the addition of vectors
-2x = + 115 -2y = + 142 that this can occur only if the head of the last
(see Fig. 1-13). Using the latter gives vector to be added comes back to touch the tail

aes
= 42. of the first vector, The vector sum is zero if the
vector diagram is a closed polygon. This method
is especially useful when there are three or more
@=5I°NofE and sin51° = 142 concurrent forces.
Mar TE peg R y Example An object weighing 100 Ib and sus-
R= ainsi = 0777= SAN Ly= +14.2N pended by a rope A (Fig. 1-15a) is pulled aside
The resultant is an 18.3-N force acting 51°N by the horizontal rope B and held so that rope
of E. A makes an angle of 30° with the vertical. Find
Figure 1-14 the tensions in ropes A and B.
After a parachute opens and falis a certain _ Consider the junction O as the body in equi-
1-6 Ex = +11.5N distance, it moves downward thereafter with librium. It is acted upon by the three forces w
EQUILIBRIUM uniform velocity. Such a system of balanced (known), F,, and F, (unknown). These forces afè
Figure 1-13 forces is in equilibrium. represented in Fig. 1-15b, with W scaled to repre-
Many important problems confronting the physi- Vector diagram using the sums of the x sent the known weight and only the directions
cist and engineer involve several forces acting on and y components. and not the magnitudes of F, and F, known. The
a body under circumstances in which they pro-
junction O is in equilibrium, since it is not mov-
ing due to the influence of these three forces;
 FORCES ACTING ON AN OBJECT: VECTORS 30 THE PHYSICS OF PARTICLES

That is, in order to hold the system in the point of concurrence, aswell asboth the magni-
position of Fig. 1-15a, one must pull on the hori- tudes and directions of the known forces, being
(b) Forces acting on O zontal rope with a force of 58 lb. The tension of sure that only forces acting on the body are in-
rope A is then 115 lb. The tension in the segment cluded.
of rope directly supporting the weight is, of 3 Draw a closed vector polygon, such as Fig.
course, just 100 Ib. 1-15c, scaled to show both the magnitude and
direction of each of the forces.
4 Finally, solve the vector problem by suitable
1-7 mathematical methods.
HINTS FOR SOLVING
PROBLEMS INVOLVING A BODY Example By the method of components find
IN EQUILIBRIUM the resultant and the equilibrant of a 7.0-1b hori-
zontal force and a 12.0-Ib force making an angle
One of the first challenges that students face in of 60° with the horizontal (Fig. 1-16).
a course in physics is that of learning to apply The horizontal and vertical components of the
the proper techniques to solve physical problems. 12.0-lb forces are
(a) Schematic showing forces (c) Closed polygon While there are many ways in which such prob-
acting on O lems can be approached, experience will show H = 12 cos 60° = 6.0 lb and
that certain techniques work best for particular
problem-solving situations. The challenge is to V = 12 sin 60° = 10.4 1b
Figure 1-15 :
Finding an unknown force by the vector method. “recognize or identify the type of problem and We thus replace the original two forces by three
then to apply the proper technique to solve it, The horizontal and vertical components of the forces, one vertical and two horizontal. Since the
The student must analyze each problem to deter- 7.0-lb force are two horizontal forces are in the same direction,
their resultant must therefore be zero. The vectors mine what data are given and what data are
they may be added as ordinary numbers, giving 4
representing W, F,, and F, can be redrawn as in sought. This-analysis coupled with a knowledge H = 701b and V=0
total horizontal force of 6.0 Ib + 7.0 1b = 13.0 1b.
Fig. 1-15c so that they combine to form a closed of the formulas, either defined or derived, which
triangle. Note’ the fact that each vector is drawn relate to the problem provide the tools needed The problem is now reduced to the simple one
to solve all physical problems.
of adding two forces at right angles (Fig. 1-16c),
parallel to the force that it represents. It is also 7.0 Ib
The authors of this book have made an at- nella giving the resultant
important to draw the vector diagram so that the
forces considered are those which act upon the tempt to assist the student in developing the tech- (a) Components of the 7-lb horizontal vector
niques of problem solving by providing in several R = V(04? + 13.02)Ib = 16.6 lb
body that is in equilibrium.
In order to solve the vector triangle of Fig. strategic positions in the book procedures which
have proved to be quite effective in the solution The angle @ which R makes with the horizon-
1-15c, it may be observed that
of certain kinds of problems. It is strongly recom- tal is given by tan @ = 10.4/13.0 = 0.80, so that
F, mended that the student read these sections care- 6 = 387°.
int
101b A= tan 30 o==03% The equilibrant is equal in magnitude but
fully.
This is the first of such sections, directed to- opposite in direction to R. Hence it is a force of
so that F, = (100 1b)(0.58) — 58 1b. To get F}, we ward the solution of problems involving a body 16.6 lb at an angle of 218.7°.
can put in equilibrium. In such problems, the following
steps are recommended: Example An object weighing 100 Ib and sus
->
pended by a rope A (Fig. 1-15) is pulled aside
LLE cos 30° = 0.866 1 Draw a sketch to picture the apparatus, as in
H = 6.0 lb 6.0 Ib 7.0 lb by the horizontal rope B and held so that róp?
2
Fig. 1-15a. Include on the sketch the known data. (b) Components of (à Combined horizontal A makes an angle of 30° with the vertical. Find
12.0 Ib vector and vertical components the tensions in ropes A and B.
Therefore, Indicate by a suitable symbol, sugh as O, the at 60° of the two forces
point of concurrence of the several forces. We have previously solved this problem by the
2 Make a figure, like Fig. 1-155, to show: the Figure 1-16 straightforward method of adding the vectors to
directions of all the forces acting through the Resultant forces by component
method. form a closed figure. The method is quite appro-
a
Priate tosimple cases, but forthe sake of illus
 31
FORCES ACTING ON AN OBJECT: VECTORS 2 THE PHYSICS OF PARTICLES

tration, let us now solve the problem again by From Eq. (a), F, = F, Substituting in Eq. (b), in Fig. 1-205. In the vector diagram, B = W sin
the more general method of components. In Fig. 8. Since @ is 30° and W = 1,000 lb,
1-17 are shown the same forces, separated for F, sin@+ F,sin8 — 100 1b = 0
greater convenience of resolution. The horizontal 2F, sind = 100 1b
B = (1,000 Ib) sin 30°
and vertical components of the 100-Ib force are, = (1,000 1b)(0.500) = 500 Ib
respectively, 0 and 100 1b down. The horizontal 2F,(0.26) = 100 Ib
and vertical components of F, are, respectively, _ 1001b _ The value of A, the perpendicular force ex-
F, (to the right) and O. Although we do not yet F,= 3626) = 90% erted by the plane, can be found by observing
know the numerical value of F,, whatever it is,
that
the horizontal and vertical components will cer- and F, = 190 1b
tainly be F, cos 60° to the left and F, sin 60° up.
We now have four forces, two vertical and two
Figure 1-19
- Two things should be noticed about the prob- A = W cos 30° = (1,000 Ib)(0.866) = 866 Ib
Horizontal and vertical components of the
horizontal, whose vector sum must be zero to forces in a stretched rope. lem just solved: (1) the valueof a function of
ensure equilibrium. In order that the resultant Ce ee ee eee ete ee ee Tas CTT an angle in the vector diagram was needed in It should be noticed that W can be resolved
may be zero, the sum of the horizontal compo- order to carry out the solution; and (2) the value into two components that are, respectively, paral-
nents and the sum of the vertical components lel and perpendicular to the incline. These com-
it by sections A and B of the rope and the: 100-Ib of that function was determined from the geome-
ponents are equal in magnitude and opposite in
must each. be equal to zero. Therefore, weight. A vector diagram of the forces appears try of the original problem.
direction to B and A, respectively.
in Fig. 1-19. The horizontal and vertical compo-
E Fy = F sin 60° - W=0 nents of the 100-Ib force downward are 0 and Example Calculate the force needed to hold
F,sin60° = W 0.866 F, = 100 100 Ib, respectively. The horizontal and vertical a 1,000-Ib car on an inclined plane that makes
components of F, are, respectively, F, cos 8 to the an angle of 30° with the horizontal, if the force
F, = 115 1b left and F, sin @ upward, Similarly, the horizontal is to be parallel to the incline.
1-8
and vertical components of F, are, respectively, The forces on the car include (see Fig. 1-20)
DIFFERENCE OF TWO VECTORS
=0
Z F = A — F,cos60°
and F, cos @ to the right and F, sin 0 upward. In order its weight W vertically downward, the force B
F, = F, cos 60° for the resultant to be zero, the sum of the horizon- parallel to the incline, and the force A exerted We shall see later in the study of relative velocity,
tal components and the sum of the vertical com- on the car by the inclined plane itself. The last acceleration, and certain other properties of mat-
F, = (115 1b)(0.5) = 58 Ib ore,
ponents must each be equal to zero. Theref force mentioned is perpendicular to the plane and terwhich are vector in nature that the occasion
_ ‘is called the normal. arises when it is necessary to find the difference
Example A load of 100 1b is hung from the = Fy = F,cos0 — F,cos8 =0 horizontal (a) Since the car is in equilibrium under the action of two vector quantities of the same kind. The
middle of a rope, which is stretched between two '
walls 30.0 ft apart (Fig. 1-18). Under the load the ZF = Fsin0 +F, sin9 — 1001b =0 l of the three forces, A, B, and W, a closed triangle difference of two vectors is obtained by adding
middle. Find the tensions vertica (b) can be formed with vectors representing them, as one vector to the negative of the other (the vector
rope sags 4.0 ft in the equal in magnitude and opposite in direction).
in sections A and B.
The midpoint of the rope is in equilibrium Since these two equations involve three un- As in arithmetic, 5—3 =2 may be written
three forces exerted on known quantities F,, Fp and 0, we cannot solve 5 + (—3) = 2, so in the case of vectors we may
under the: action of the
them completely without more information. understand the vector difference A — B = C
as
The value of. sin 0 can be determined from A+ (—B) = C. The vector —B is equal
in mag-
the dimensions shown in Fig, 1-18: nitude and opposite in direction to B.
Bh ee ne Ee the direction of the differ-
«9 _ 4.0ft A and com it with the vector
sin 9 = -7
which would be the Koba and B (make
your
own diagram). The vector difference C
A= V5.0? + 4.0%) ft? = V241 ft? = 15.5 ft frequently defined as the vector which
is also
(a) (b) must be
100 lb and pein to B to give A. Does C satisfy this
defini-
Figure 1-20
Figure 1-18 Finding
nalts: the forces acting on a body on an
Finding the tension in a stretched rope. Example Two cars start from
eS the same iint
vite et) hoa SS so ie Jo ee Tee Ė—Ů
but one travels north at 50 km/
h and the fir
a
 34 THE PHYSICS OF PARTICLES
FORCES ACTING ON AN OBJECT: VECTORS 33

— 1-23b, where the boat is actually headed at some

(a) Resistance to wind provided by sail angle into the wind.
Here, the resultant of the force of the wind
on the sail can be represented as a force acting
perpendicularly to the sail. It is then possible to
resolve this resultant force into a force at right
angles to the boat, a tipping force which has no
value in moving the boat forward other than
putting the boat on its side and cutting down the
Wind
Accelerating component friction between the boat and the water (this type
(a) (b) (c) of sailing is called planing) and which must be
Figure 1-21 compensated for by the keel or centerboard. It
Difference of two vectors. ‘Resultant of force is also possible to resolve this resultant force into
acting on sail == Wind a force acting parallel to the boat. This is the
Sail accelerating component of the force of the wind
east at 30 km/h. What is the vector difference of Tipping component and will cause the boat to move forward. It
these two velocities? What does it signify? (See of the wind should be noted here that the sail does not move
Fig. 1-22). away from the wind as in the case of running
before the wind. Therefore the force of the wind
USO
=22vey
tan9 =167 —
0=59 59° remains in contact with the sail and the boat
© keepsaccel thataccordinng.
Weknowerati g to New-
ae Ig ile. SA ton’s second law the boat should keep accelerating
sin59i= m i Sage OS Figure 1-23
even after attaining the velocity of the wind.
Forces acting on a sailboat.
C = 58.3 km/h Further, as the boat travels faster, the force of
impact between the sail and the wind is increased
Therefore, the vector difference is 58.3 km/h 59°S because the boat is actually heading somewhat
of E. This signifies the relative velocity of thé two weekend sailor than this. Could this be the con- into the wind. Hence the boat can go faster than
cars to each other. figuration of the wind and sail to achieve the the wind in this case.
greatest speed? To answer this, let us analyze the Can the boat keep accelerating indefinitely?
forces that are acting on the sail. There is a force As we have seen in several other situations earlier
1-9 exerted on the sail by the wind and another re- in this chapter, the boat attains a terminal velocity
THE PHYSICS OF SAILING sisting force provided by the sail due to the fric- for the same reason that the falling raindrop does.
Lovers of the sport of sailing may claim that tion of the boat in the water and also initially That is, at a certain velocity the forces opposing
physicists are taking the fun out of sailing by due to the inertia of the boat. If the force of the the motion, mainly water resistance on the hull
attempting to analyze what happens when a sail- wind is greater than the resisting force, the boat of the boat, cancel the accelerating force due to
boat is driven by the wind. Despite this objection, will accelerate according to Newton’s second law the wind. In the design of boats used for tacing,
a sailhoat in motion does provide an excellent C= A + (~B) of motion. This acceleration will continue until the hulls are carefully built to reduce this friction”
example of forces acting on an object. the boat starts moving as fast as the wind; then to a minimum. Anyone who has seen 4
A question that is often asked is whether a when the sail is moving at the same speed as the catamaran-style sailboat, with its twin hull de="
sailboat can go faster than the wind. Let us con- wind, the accelerating force disappears because signed to keep friction down, moving with such |
sider a sailboat initially at rest with its stern the sail no longer provides a surface for the wind great speed will recognize just how fasta sailboat
(back) to the wind, Fig. 1-23a. If the sail is ex- to push against. So, when running before the can move when resistance is reduced. A further
tended so that it makes a right angle with the ~ wind, the maximum velocity that can be attained
boat, the boat will start to move, in’a form of Figure 1-22 by the boat is the velocity of the wind itself.
example can be seen in iceboats, which have
practically no ‘hull resistance. These “boats” call
sailing known as running before the wind. Per- Finding the vector difference. Suppose that we now turn the boat so that the attain a velocity several times faster than the
haps no type of sailing is more enjoyable to the sail makes an angle with the wind, as.in Fig.
wind. In answer to the question posed at the start
 36 THE PHYSICS OF PARTICLES
FORCES ACTING ON AN OBJECT: VECTORS

9 What are the handicaps which one has in Problems

of this section, sailboats can attain speeds of two The component method of adding vectors is
to resolve each into its rectangular components,
solving complicated vector problems by the pate
to three times the speed of the wind. Further- allelogram method?
more, an analysis of forces that are acting will which are then added algebraically and the re- 1 Resolve a force of 100 N acting at an angle
sultant found. 10 Show by a series of vector polygons that it of 37° with the horizontal into horizontal and
show that the maximum velocity will be attained is immaterial which order is used in laying off
when the sail makes a 45° angle into the wind. vertical components.
R? =R} +R? the vectors end to end and that the resultant is 2 An object weighing 40 N rests on an inclined
always the same. plane making an angle of 30° with the horizontal,
A body is in equilibrium when there is no 11 Show how the parallelogram method of Resolve the force into components parallel to the
SUMMARY change in its motion. solving vector problems can be simplified by the plane and perpendicular (normal) to the plane,
A force is arbitrarily defined as any influence When a body is in equilibrium, the vector sum use of the polygon method. Ans. Force parallel = 20 N, force perpendic-
capable of producing a change in the motion of of all the forces acting on it is zero. This is known 12 An automobile is acted upon by the follow-
ular = 34,64 N,
an object. as the first condition for equilibrium. When a body ing forces: a Horizontal force due to air resist-
3 Two forces of 10N each are acting on a
Forces are usually expressed in pounds (in the is in equilibrium, the force diagram is a closed _ ance; the weight of the car; a force almost verti-
point. One force acts E and the other 70°N of
fps system), dynes (in the cgs system), and newtons polygon, or the sums of the rectangular compo- cally upward on the front wheels; the force of
E. Find the resultant by using the law of sines
(in the mks system). nents of all the forces must each equal zero: the ground on the rear wheels: Draw a vector
and cosines.
=F, = 0 and =F, = 0. polygon to show these forces in equilibrium.
Quantities whose measurement is specified by 4 Resolve a force of 100N into two compo-
The difference of two vectors is obtained by What does this imply with respect to the velocity
magnitude and direction are called vector quanti- nents which lie on opposite sides of the force and
ties. Those which have only magnitude are called adding one vector to the negative of the other of the car?
(the vector equal in magnitude and reversed in each of which makes an angle of 30° with the
scalar quantities. 13 When two vectors are drawn from a common
sense).
force. Ans. Each is equal to 57.7N.
A vector quantity may be represented graphi- origin in a vector parallelogram, what quanti
ty 5 Two airplanes start from the same point, one
cally by a directed line segment, called a vector. does each diagonal of the parallelogram repre-
traveling 250 km E and the other 180 km N. What
A vector is the line whose length indicates to scale sent?
is the vector difference of these two displace-
the magnitude of the vector quantity and whose Questions 14 A wire is stretched horizontally between
two ments? What does it signify?
direction indicates the direction of the quantity. Supports (Fig. 1-18), with a load applied
The resultant of two or more vector quantities 1 What items must be stated to specify a vector center. Assuming no stretching in the wire,
at the 6 Three cities X, Y, and Z are connected by
quantity completely? plot straight highways. X is6km from Y, Y is 4km
is that single quantity of the same physical a rough graph to show how the force in
2 Give several examples of scalar quantities; the wire from Z, and X is 5km from Z. Find the angle
makeup that would produce the same result, would vary as the angle @ is varied from
of vector quantities, 0 to 90° made by the highways XY and YZ.
The resultant R of two vectors M and N hav- by moving the supports. «
3 Show that the maximum value that the re-
ing angle 0 between their directions is conve- 15 Explain the difference between vector Ans. 55.7".
sultant of two vectors can have is the arithmetic and
niently found from the law of cosines and the algebraic sums. 7 What is the angle between a 2.00-unit vector
law of sines,
sum and that the minimum value is the arithmetic and a 3.00-unit vector so that their sum is 4.00
16 Three forces of 12, 15, and 20 N are
difference, in equi- units?
R? = M? + N? + 2MN cos 9 4 Show why the vector addition of a 5-N force librium. If the 12-N force is directed horizontally
to the right, what two Configurations in a 8 What effect do the following forces have on
E and a 5-N force N is not 10N NE. vertical
Mire R etl 2 5 The sum of two vectors is at right angles to plané may the other two forces have? a point: 100 N, 30°E of N; 200N, 80°S of E;
sinp. sin (180° — 8) 17 Give several examples of moving 150N, 45°S of W; 175 N, 25°W of N; 50N, due
their difference. Show that the vectors are equal bodies
Vectors are conveniently added graphically by. in magnitude. $ which are in equilibrium. N. Solve graphically. Ans. 95.5 N, 4.5°N ofW.
6 Itis found that the sum and the difference 18 Why does an airplane pilot prefer to take 9 Solve Prob. 8 using the component method.
placing them head to tail and drawing the result- off 10 A 20-m-long Tope attached at
ant from the origin to \he head of the last vector, of two vectors have equal magnitudes. Prove that and land into the wind? Explain by
the use of the top and
the two vectors are at a 90° angle to each other. a vector diagram. the bottom of a flagpole is pulled 2 m away from
closing the polygon. the pole by a 100-N force acting at right angles
The rectangular components of a vector are its 7 Are vectors necessary, or is the concept of 19 In moving a sled over the snow
is it bet-
projections on a set of right-angle axes, for exam- a vector merely a convenience in expressing terto pull the sled with a rope
or to to the pole at its midpoint. What is the tension
on
ple, the horizontal and vertical axes. physical quantities? the sled with a pole? Explain by the use of on the segments of rope on each side of the 100-N
8 Two forces of 50N and 80N act upon a vectors. force? $ Ans. 255 N.
R, = Rosé body. What are the maximum and minimum 20 When telephone wires are Cove
red with ice,
11 An automobile which weighs 3,200 Ib is on
R,= Rsin@ possible values of the resultant forces? is there more danger of their breaking when they & road which rises 10 ftfor each 100ftof road.
are taut than when they sag? Why? What force tends to move the car down the hill?
12 Four boxes each weighing 100N are
sus-
 FORCES ACTING ON AN OBJECT: VECTORS
37 38 THE PHYSICS OF PARTICLES

grasped at the middle, if possible, and pulled 5.0 km/h south. What is the actual speed of the
away from the mast, after which the slack so man relative to the earth? What is the direction
gained is taken up at the cleat. If the halyard is of his velocity? Ans. 16 km/h at 21°N of W.
20m from pulley to cleat andis pulled at the 25 A particle in a cyclotron travels in a circle
middle, so that it is 1.0 m out from the mast, with of 2,000-m radius. What is the difference in th
a force of 20N, what is the tension in the magnitude of its displacement (chord) and it
` Figure 1-24 halyard? distance of travel (arc) in one-eighth of a circle!
ee
eee 22 A person
who is 35ft east of you runs north 26 Three men pull on ropes attached to the top
at 16 ft/s. At what angle north of east would you of a heavy object which is level with the ground
pended from a beam (Fig. 1-24). What is the (Fig. 1-26). When a 20-N force is exerted on the throw a ball at 60 ft/s groundspeed in order to „ofE
Man A is’6 ft tall, stands 6 ft away, 45°N
tension in each of the wires? joint along the angle bisector, the opposite ends hit him? Ans. 15.4°. from the center of the object and exerts a force
Ans. 51.8 N; 100 N; 50N; 193 N. of the bars may exert very much larger forces on 23 A rope 10 m long is stretched between a tree Bis 5ft6 in tall, is 6ft away, 60°N)
72 Ib. Man
of
13 A boat which can travel at 10 m/s in still and a car. A man pulls with a force of 10N at of W and pulls with a force of 60 Ib. Man Ci
water attempts to reach a point directly across a right angles to and at the middle point of the of Eand pull
5 fttall, also stands 6 ft away, 30°S
river in which there is a current of 8 m/s. At what rope, and moves this point 0.5 m. Assuming no with a force of 80 lb, Assuming the ropes to be
20N
angle to the shore must the boat be steered stretching of the rope, what is the tension in the attached to their shoulders which are & of theif) )
reach that point? f rope at the final position? height from the ground, what is the horizont
14 Two similar cylindrical polished bars weigh- 24 A man walks westward on a boat with a resultant of these forces on the object? b
. ing 5.00 N each lie next to one another in contact. speed of 4.0 km/h; the ship’s propeller drives it Ans. 93.5 1b, 33°N of È
A third similar bar is placed on the other two 15 km/h northwest; tide and wind drive the ship

their restraints. If the angle at the joint is 170°,

find the two horizontal forces necessary to hold
the system in equilibrium. Ans. 714 N.
17 Add the following forces by the component
method: 12.0N at 30° to an x axis, 15.0N at
140°, and 8.0 N at 290°.
18 An iron sphere weighs 10N and rests in a
V-shaped trough whose sides form an angle of
Figure 1-25
60°. What is the normal force exerted by the
sphere on each side of the trough?
Ans. 10 N.
in the groove between them. Neglecting friction, 19 As a flag is hoisted up the mast at 15 ft/s,
what horizontal force on each lower bar is neces-
a ship goes south at 22 ft/s, and the tide moves
sary to keep them together (Fig. 1-25)? east at 4.0 ft/s. What is the speed of the flag
Ans. 2.5 N. relative to the earth?
15 A train has a velocity of 150 km/h due east.
20 A boat travels 10.0 km/h in still water. If it
A person on the train walks toward the front of
is headed 60°S of W in a current that moves at
the train with a velocity of 15 km/h relative
to 12.0 km/h due East, what is the resultant velocity
the train. What is this velocity relative to an of the boat? Ans. 11.1 km/h at 51°S of E.
observer on the ground? 21 When a sail is hoisted, the rope or halyard
16 Many presses and other tools make use of passes over a pulley and down the mast, where
the toggle, which is a pair of bars jointed together it is often wrapped around a cleat at the bottom.
So as to form a very obtuse angle between them To effect the final tightening, the halyard is
 Henri Antoine Becquerel, 1852-1908

Born in Paris. Professor at the Paris Polytechnic

School. Awarded, with the Curies, the 1903 Nobel
Prize for Physics for his discovery of spontaneous
radioactivity.

Plerre Curle, 1859-1906

Born in Paris. Professor at the Municipal School

of Industrial Physics and Chemistry. Pierre and
Marie Curie shared the 1903 prize with Henri Bec-
quere! for their work on the phenomena of the ra-
diation discovered by Becquerel.

Marie Sklodovska Curie, 1867-1934

BorninWarsaw. Director of applied physics in the

University of Paris. Marie Curie shared the 1903
Nobel Prize for Physics with Pierre Curie and
Henri Becquerel and was awarded the 1911 Nobel
Prize for Chemistry for her discovery of radium
and polonium.
 42 THE PHYSICS
OF PARTICLES

2 it moves per unit oftime. If the speed isuniform,

the object moves equal distances in each succes-
as the time rate of change of displacement. The
definition of average velocity is given by

Velocity and Acceleration sive unit of time. Whether or not the speed is
constant, the average speed is the distance the
body moves, its displacement in space (which is Se ny
~ displacement
= ine elapsed
defined as its change in position specified by a
length and a direction), divided by the time re- put
eer (1b)
quired for the motion.

Constant velocity is a particular case of con-

stant speed. Not only does the ‘distance traveled
per unit time remain the same, but the direction
d= a (la) as well does not change. An automobile that
travels for 1 h at a constant velocity of 20 mi/h N
reaches a place 20 mi N of its original position.
where s is the distance traversed, 0 the average If, on the other hand, it travels around a racetrack
speed, and ¢ the time. In this book, a bar placed at a constant speed of 20 mi/h, it may traverse
In Chap. 1 it was shown that forces acting on an tion and learn to control them. While it might over a symbol, as with 0, indicates that this is an the same distance without any final displacement.
object can cancel each other out and cause the be more exciting to begin our study by observing average. The mks unit of speed is the meter per At one instant its velocity may be 20 mi/h E; at
object to be in equilibrium. In this chapter we motions such as that of a satellite orbiting the
earth or a rocket traveling to the moon, this type
second (m/s); the fps unit is the foot per second another, 20 mi/h S.
will see what happens to an object if the forces (ft/s); and many of the other units are common, The statement “An automobile is moving with
do not cancel each other out, that is, if there is of motion will be considered only after we have
studied some simpler cases. It is surprising to
such asthe mile per hour (mi/h), centimeter per a velocity of 20 mi/h” is incorrect because it is
a net unbalanced force acting upon the object.
As an illustration of this, common sense will tell most students of pħysics to find that even the second (cm/s), knot, etc. Equation (la) may be incomplete, inasmuch as the direction of motion
most complicated motions can be analyzed and put in the following form: must be stated in order to specify a velocity. For
us that if two men push on opposite ends of an this reason one should always use the word speed
object with equal force, the object will not move- represented in terms of a few elementary types,
However, if one man stops pushing, the force still when these simple types of motion are thoroughly s=0t (1) when the direction of the motion is not specified
being applied by the other man will cause the understood. In fact, it will be shown that all or when the direction is changing.
object to move in the direction of the remaining complex motion is merely a combination of two
If the speed is constant, its value is, of course,
force. In the next chapter we will see just how or more simple motions. Example An automobile traveled by a cir-
identical with the average speed. If, for example,
this unbalanced force is related to the movement an automobile travels 200 km in 4h, its average
cuitous path for a time of 3.0h and covered a
of the object. For the present, however, we shall speed is 50km/h. In 6h it would travel 300 km.
total distance of 180 mi. (a) What was the average
be concerned with the description of the-resultant 2-1 As we saw earlier, speed is a scalar quantity, speed in feet per second? (b) If at the end of the
movement of objects due to the influence of SPEED AND VELOCITY and hence the concept of speed does not involve trip the car was exactly 60 mi N of its starting
unbalanced forces. $ the idea of direction. A body moving with con-
„point, what was the average velocity?
Just as the nature of the forces acting on an The simplest kind of motion that an object can (a) The average speed was
object can vary, the type of motion resulting from have is a uniform motion in a straight line. By stant speed may move in a straight line or in a
these forces can vary. While some motion in uniform motion is meant that in every second the circle or in any one of an infinite variety of paths
nature may be simple in form, most motion is body moves the same distance in the same direc- so long as the distance moved in any unit of time b= = lO mi L 60 mi/h = 88 ft/s
complex. In fact, we sometimes are endangered tion as it did in every other second. Every part is the same as that moved in another equal unit
by the motion of objects around us, especially if of the body moves in exactly the same way. An of time.
that motion is erratic and uncontrolled as we object moving in this manner is moving with The concept of velocity includes the idea of and
observe it in a flooded river, a hurricane, or a
constant velocity. It should be carefully noted that direction as well as magnitude and velocity and
runaway automobile. On the other hand, con-
the term constant velocity implies not only con- | is therefore a vector quantity. A car moving at
trolled motion can be of service to man. It is stant speed, that is, no change in the rate of move- constant speed along a winding road hasa chang-
necessary to study the motions of objects if we ment, but unchanging direction as well. ing velocity because the direction of motion is
are to understand the behavior of objects in mo-
The speed of a moving body is the distance changing. Velocity, a vector quantity, is defined
41

n
aaa
Mihei
aMŇ
mm
aa
 VELOCITY AND ACCELERATION 43 44 THE PHYSICS OF PARTICLES

in-
2-2 This statement means simply that the speed
INSTANTANEOUS VELOCITY gaa Qa creases 4.4 ft/s during each second, or 44 ft/s?.
We are frequently interested in knowing the
as Example A proton in a cyclotron changes in
speed and velocity of a body at a particular in- where Av is the vector difference of v, and Uo velocity from 30km/sN to 40km/sE in
t = 0,
stant and not merely the average value over a defined in Sec. 1-9, uo is the velocity when (1 ps = 107° s.) What is the
y at 20 microseconds (us).
considerable time interval. For this purpose it is t is the time elapsed, and v; is the velocit
convenient to consider the ratio As/At, where As average acceleration during this time?
time t.
The vector character of the velocities in this
is the change of displacement which the body has Since units of acceleration are obtained by the
case is important. The vector difference of
during the small time interval Az. If Ar is made dividing a unit of velocity by a unit of time, it
velocities is found as in Fig. 2-2.
smaller and smaller, approaching zero but never may be seen that the mks unit of acceleration is ies
The magnitude of the difference of velocit
reaching it, As becomes equally infinitestimal and Displacement
s
the meter per second per second. Recall from the by the
represented by |v, — Vol can be obtained
the instantaneous velocity v is the lowest possible law of multiplication of fractions that
value or limit of this ratio. From the calculus, (m/s)/(s/1) = m/s X 1/s = m/s?. Similarly, the
pythagorean theorem in this case.
this fact is written fps unit is the foot per second per second (ft/s),
Jv, — vol = V(40 km/s)? + G0 km/s)?
and the cgs unit is the centimeter per second per
second (cm/s*).
v== As = 50 km/s
re (Ie)
Although we use the word speed to describe 30 km/s ae
the magnitude of a velocity, there is no corre- 30km/s = sin 37
sponding word for the magnitude of an acceler-
Thus instantaneous velocity is the time rate of
change of displacement.
Figure 2-1
Displacement-time curve. ation. Hence the term acceleration is used to = P= tyee_ 50km/s at37°SofB
20 X 109s
Instantaneous speed is similarly defined as the denote either the vector quantity or its magni- ope
time rate of change of distance, which is the rate tude. = 2.5 km/s? at 37°S of E
of change of the magnitude of the displacement; 2-3. Example An automobile accelerates at a con- tly
hence the instantaneous speed of a particle is the ACCELERATED MOTION _ As in the case of velocity, we are frequen
magnitude of its instantaneous velocity. Unless stant rate from 15 mi/h to 45 mi/h in 10s while instantaneou s values of acceler -
concerned with
traveling in a straight line. What is the average
otherwise stated, the terms speed and velocity Objects seldom move with constant velocity. In ation. Instantaneous acceleration is defined by
refer to the instantaneous values. almost all cases the velocity of an object is con- acceleration?
Suppose that we consider the case of a body tinually changing in magnitude, direction, or The magnitude of the average acceleration, or
lim Av (25)
which is traveling in a straight line with a uni- both. Motion in which the velocity is changing the rate of change of speed in this case, is the a=
(r>0)At
formly increasing speed. A curve of displacement is called accelerated motion. The time rate at change in speed divided by the time in which it
against time for such a body is shown in Fig. 2-1. which the velocity changes is called the acceler- took place, or
Hereafter the term acceleration will be used to refet
By definition, the slope of the curve at a point ation. to the instantaneous and not the average ue,
P is determined from a line drawn tangent to the The velocity of a body may be changed by 45 mi/h — 15 m/h _ 30 mi/h
d= unless otherwise stated.
curve at P as the change in the displacement (As), changing the speed, the direction, or both speed 10s — 0 10s
the ordinate, divided by the change in the time and direction. If the direction of the acceleration = 3.0 (mi/h)/s
interval (At), the abscissa. Thus the velocity of is parallel to the direction of motion, only the
a body at the point P, is equal to the slope of 2-4
speed changes; if the acceleration is at right an- indicating that the speed increases 3.0 mi/h dur- UNIFORMLY ACCELERATED MOTION
the line drawn tangent to the curve at P,. If the gles to the direction of motion, only the direction ing each second. Since
velocity were constant, the slope would be the changes. Acceleration in any other direction pro- The simplest type of accelerated motion is uh |
same at every point and the curve would be a duces changes in both speed and direction. A 88 ft/s formly accelerated motion, defined as motion ng
30 mi/h = 30 = 44 ft/s
. Straight line. From the shape of the curve in Fig. The average acceleration of a body is defined . 60.mifr a straight line in which the direction is alway |
2-1 it is evident that the velocity of the body in by the same and the speed changes at a constant falè l
question is not constant, since the slope is steadily the average acceleration can be written also as It is important to note that the vector differen® —
increasing, as shown by the slope at the second
Average acceleration = change in velocity of velocities becomes simply the algebraic differ; |
reference point, P}. time elapsed ence, and one may work with simple algebrat |
equations. One direction along the line of moti? |
 VELOCITY AND ACCELERATION 45
46 THE PHYSICS
OF PARTICLES

Examplé How far does an automobile move

while itsspeed increases uniformly from 15 mi/h also a = “1 “0 and Up = V, — at
to 45 mi/h in 10 s?

T= 405 + 45) mi/h = 30mi/h = 44ft/s but,

v =U9+at and v= —v
Vo |30 km/s N
s = Ut = (44 ft/s)(10s)= 440ft
Therefore,
Three equations for uniformly accelerated
40 km/s E motion have now been. considered: Uy + at= 25— v4 or 2uy = 25— at
1
s=o a)
(a) Original vectors (b) Vector difference Multiplying both sides by ¢ and then dividing
v — Up = at Q) by 2, we obtain
Figure 2-2
Vector differences of two velocities. Ja mth G6) 2v =2s—at®? and uot =s — jal*
a

These three equations have been defined by Rearranging these terms,

we get
is called positive; the opposite is called negative. Vo + Vy setting certain conditions and limitations. It is
In this type of motion, the average value of the
v= ; 6)
possible to combine these fundamental equations S= Vot + $al? (4)
acceleration is the same as the constant instanta- in such a manner that two other very useful
neous value and £q. (2a) becomes for this case The distance covered at this average velocity is Example Derive the equation 2as = v? ~
equations are obtained.
given by s = Tt. It isimportant for the student to realize that U9”. Since
UV;
— vo
a= the equations used throughout this study of phys-
t ics have all been either defined or have been
or v — Up = at (2) derived from defined equations. That is, the and t= = = s[[(U + v)/AN
equations do not appear by magic but rather are
The velocity of a body in uniformly acceler- based on agreed-upon conditions and restrictions. _ 4s
ated motion changes steadily, with equak changes Inthe eyes of many physicists, a well thought out Vo + Uy i
of velocity in equal intervals of time, as shown and executed derivation of a working equation
in Fig. 2-3, where the slope is constant at all is a thing of beauty. While the authors of this then, a = v, — vo/[2s/(vo + v,)]
points. The slope taken between P, and P, and book share thisopinion toa degree, they arealso
then between P, and P, will be the same since aware that the study of physics as presented in Therefore,
the slope is Av/At, which equals the acceler- this text could become encumbered by an inordi-
ation—in this case constant. This would not be nately large number of derivations. 2as = (v, — voXv, + vo)
Conse-
true in general for a nonuniformly accelerated quently, only a few key derivations are given
throughout the book. The derivations presented and
motion, which would be represented by a curved
below are typical of thetechniques used in estab-
graph-line. The slope of the tangent to such a
curve at any point would represent the instanta- lishing working equations. 2as= v,?— v? o
neous acceleration. For the case of uniform ac-
Example Derive the equation s = yt +
hat i
celeration,
2-5
-Uo + Uy + Up +++ + Up HINTS FOR SOLVING
Time, s
v=
n \ PROBLEMS INVOLVING UNIFORMLY
Since s= pt and p= eto
Figure 2-3 ACCELERATED MOTION
Since we usually refer to an initial and a final Velocity in uniformly accelerated motion,
In the five working equations given above for
velocity in solving problems of motion, then,s = (24%), and solving problems in uniformly accelerate
E E a d mo-
tion, there are six variables: J, Vo U,
a, s, and
 48 THE PHYSICS OF PARTICLES
VELOCITY AND ACCELERATION a
sea level, the value is greater near the poles than
t. To solve problems of this type, the student Example An airplane lands on a carrier deck at the equator. Locally there may be small varia- t v 8
should first determine which of these variables at 150 mi/h and is brought to a stop uniformly, tions because of irregularities in the layers of rock sec ft/sec ft
are given and which are not. While not all six by an arresting device, in 500 ft. Find the acceler- beneath the surface. Such local variations are the 0 0 0 p0
variables will be involved in each problem, such ation and the time required to stop. basis of one type of prospecting for oil. 1@ 1” 16
an analysis will permit a judgment to be made
Since a freely falling body is uniformly accel-
about the nature of the problem and which of Uo= 150 mi/h = 220 ft/s v0 erated, the five equations already developed for
the five equations will provide the best route to 2@ ta 64
s = 500ft a=? t=? that type of motion may be applied when air
obtain the answer.
Since the equations are solved by algebraic resistance is neglected.
From Eq. (5),
addition, it is important to choose a direction to
call positive and apply it consistently to distance, Example A boy lets go of a tree limb and
2as = v? — vo? falls freely. What is his speed at the end of 0.50 s? 3 @ g 144
speed, and acceleration when values are being
inserted in the equations. 2a (500 ft) = 0 — (220 ft/s)? How far does he fall during this time?

4 _= Gon
= 220 fi/s)? |= ~ 484 fs?
Since acceleration is generally expressed in w=0 v=? s=? a= +32 ft/s?
feet per second squared and time is given in
seconds in the fps system, it is usually necessary t = 0.505
to express velocity in feet per second. Therefore, (Note: The negative sign on the acceleration indi- It is convenient in this case to call downward 4 O k 256
it is wise to change any velocity given in miles cates that the plane is slowing down.) From Eq.
per hour to feet per second. It was shown earlier quantities positive. From Eq. (2),
(2),
that 60 mi/h = 88 ft/s. This equivalency can be
used as a conversion factor.
Uy = Vo + at = 0 + (32 ft/s2)(0.50 s) = 16 ft/s
-v _ 0 — 220ft/s _

Example A car accelerates uniformly from a

tg eA Te ee From Eq. (4), Figure 2-4
Position and speed of a body falling freely from
standstill to 60 mi/h in 4.0 s. What is its acceler- rest after successive Intervals of time.
5 = Vol + fat? = 04 (32 ft/s3)(0.50 s)? = 4.0 ft eee
ation? How far does it travel during this time 2-6
interval? FREELY FALLING BODIES: Table 1 and Fig. 2-4 show the speed at the convenient to call upward positive and downward
ACCELERATION DUE TO GRAVITY |
end of time ¢ and the distance fallen during time negative.
Up =0 v, = 60mi/h = 88 ft/s The most common example of uniformly accel- t for a body that Starts from rest and falls freely. Another theory which is given for establishing
¢=40s a=? "32? erated motion is the motion of a body falling When an object is thrown with initial speed sign conventions and which gives the same results
freely, i.e., a body which is falling under the Uo, instead of falling from rest, the first term may be helpful here. An object thrown upward
of
action of its weight alone. If a stone is dropped, Eq. (4) is no longer zero. If it is thrown down- with a velocity v, will eventually stop climbing
From Eq. (2), |
it falls to the earth. If air resistance is negligible, ward, both v, and a have the same direction if it stays in the earth’s gravitational field; hence,
and
— = % _ 88 ft/s—0 _ the stone is uniformly accelerated. hence are given the same algebraic sign. If the the object must be slowing down until v, equals
a ; sha” S 22 ft/s? The acceleration of freely falling bodies is so object is thrown upward, however, Up is directed zero. As we saw earlier, an object which is slow-
important and so frequently used that it is cus- upward while a is directed downward, and it
is ing down has a negative acceleration and the
‘From Eq. (4), tomary to represent it by the special symbol g. acceleration should possess a negative sign. Con-
At sea level and 45° latitude, g has a value of versely an object which is falling picks up speed,
32.17 ft/s?, or 980.6 cm/s’, or 9.806 m/s?. For Table 1
L ignoring friction. Since the magnitude of its ve-
S = Vol + gat? = O + 4(22 ft/s*\(4.0 s)? some purposes it is sufficiently accurate to use Time 1, Speed, ft s, at Distance, ft, locity is increasing, its acceleration can be con-
= 180ft g = 32 ft/s*, or 980 cm/s?, or 9.80 m/s*. s end of time z. sidered Positive.
fallen in time ¢
The value of g is not quite the same at all
or from Eq. (1), places on the earth. We shall see later that the 1 32 16 ; Example A ball is thrown upwar
weight of a body depends upon its distance from 2 initial speed of 80 ft/s, (a) How high ddoeswith all
the center of the earth. It will also be seen that
64 64
(b) What it go?
is its speed at the end of 3.0 s? (c) How
Y ane (Vo tul Si (0 + 88 Kee 3 96
the acceleration of a freely falling body depends 144 high is it at that time?
upon this distance. At a given latitude the value 4 128 256 _ (a) Since both upward and downward :
= 180ft is greater at sea level than at higher altitude. At ties are involv quanti-
ed, upward will be called positiv
e.
 50 THE PHYSICS OF PARTICLES

VELOCITY AND ACCELERATION 49

nal speed is of much importance to us because illustrated, the velocity of car A with respect
if such objects did not behave in this manner, life the ground can be designated v,, and that of h
At the highest point the ball stops, and hence at resistance of the air depends upon the speed of would be far different in this world. For example,
that point v, = 0. the moving object. The resistance is quite small ground to car B be called vog. Then by the s
using Eqs. (1) through (5), it can be shown that called “domino rule,”
for the first one or two seconds, but as the speed
Up = 80 ft/s vu, =0 a= a raindrop falling without friction from an
of fall increasés the resistance becomes large
enough to reduce appreciably the net downward
altitude of 32,000 ft would acquire a velocity
p= —32 ft/s sane Vae + Yas = Yan
force on the body and the acceleration decreases, given by
From Eq. (5), After some time of uninterrupted fall, the body which is the velocity of A relative to B, the ela
is moving so rapidly that the drag of the air is %=0 v=? a= +32 ft/s? tive velocity. It should be noted that the velocit
2as = v? — up? as great as the weight of the body, so that there of the ground with respect to car B is negativi
5 = 32,000ft r=?
is no acceleration. The body has then reached its
2(—32 ft/s*)s, = 0 — (80 ft/s)?
terminal speed, a speed that it cannot exceed in
(Imagine B standing still, then G is movin
— (80 ft/s)? = on falling from rest, As we saw earlier in our discus- From Eq. (5), the left, or negatively.) Since vector quantities ¢
s. =a sion of equilibrium, a body falling at terminal be added algebraically as long as they are acti
speed has no unbalanced forces acting on it, so 2as = v,? — v? in a rectilinear fashion (straight line), we can $
From Eq. (2), the conditions for equilibrium have been met and that , i
the body ceases to accelerate. and
Uy = VW + at Very small objects, such as dust particles and Yan = v + Vs
water droplets, and objects of very low density Yas=v or
= 80 ft/s + (—32 ft/s)(3.0 s)
and large surface, such as feathers, have very low
v= Vas
= 80 ft/s — 96 ft/s = —16 ft/s but, since vg is negative,
terminal speeds; hence they fall only small dis-
tances before losing most of their acceleration. v, = VIX 321 X 32000
From Eq. (4), Vas = Y4 + (—Vg) = v, — Ys
The effect of-air friction on falling bodies can be
shown by a classic experiment called the “guinea = V205 KIER
Sq = Vol + gar? and feather tube” demonstration in which a coin
= (80 ft/s)(3.0 s) + $(—32 ft/s)(3.0 s)? (the guinea) and a feather are enclosed in a long = 1,430 ft/s, or 755 mi/h lute quantities; bil
tube, When the tube filled with air is inverted, inreality all velocities are relative.
= 240ft— 144ft= 96 ft
the coin falls much faster than the feather. If the The v4 abo
was designated vig, that is, veloc
air is pumped out and the tube is again inverted, ity with resp
or, from Eq, (1)
the coin and feather fall together. The same strik-
ing result can be shown by dropping a book and
$y = Ut = AM, an unfolded sheet of paper. If the paper is then
crumpled and the experiment repeated, the book Sary
= ae ie 80ft/s 4 9 = 96ft and the paper will arrive at the ground at the i Tetarding
tie force to coun terbalance the gravii
same time. Try it and see for yourself.
A man jumping from a plane reaches a termi-
Note that s is the magnitude of the displacement, nal speed of about 120 mi/h if he delays opening 2-8
not the total distance traveled. If the ball returns his parachute. When the parachute is opened, the RELATIVE VELOCITY
to the starting point or goes on past it, s will be terminal speed is reduced because of the in-
zero or negative, respectively. creased air resistance to about 14 mi/h, which is IfcarA going 50km/h overtakes carB i
about equal to the speed gained in jumping from TOL thefirstcarhas a relative Pii een
a height of 7 ft. A large parachute encounters m/h with respect’ tothesecond, It is‘this
2-7 more air resistance than a small one and hence
TERMINAL VELOCITY causes slower descent. A plane in a vertical dive
without the use of its motor can attain a speed tivevelocity ofonebod
In the preceding discussion we assumed that there of about 400 mi/h before reaching terminal respectto anoth
isthe
att vector erences of thetwovelocities, The
is no air resistance. In the actual motion of every speed. ofthevelocitie inthedifference is j
falling body this is far from true. The frictional The fact that falling bodies do reach a termi-
E J some common Teference

Kenindeterminingrelativevelocity.Taheca
Or
 VELOCITY AND ACCELERATION 54
time even though the fired bullet may land h
dreds of yards away. The reason for this can
to be that through which light waves are propa- _ 0:90 — (—0.80c) _ 1.70c _ understood if one realizes that the only fi
gated through all space. (This experiment is dis- ©) Yan =TI (0. 72c2/e2) n =99% causing both bullets to fall is the force due
cussed in detail later in the book.) If the expected
gravity. The fact that the fired bullet is mo
success had been achieved, this ether could have If Eq. (6) had been used, the relative speeds horizontally does not influence the rate of fal
been used as a kind of absolute frame of refer- would have been computed to be 0.10c and 1.70c,
ence for at least the velocities encountered in
The horizontal and vertical motions are inde
respectively.
astronomy. To the astonishment of the scientific pendent of each other. ;
world the result was a failure to find such a Another illustration of this independence
relative velocity, a result which has been con- horizontal and vertical motion of an object can
2-9 be seen by considering the situation in which a
firmed by many increasingly accurate experi-
PROJECTILE MOTION ball is thrown vertically upward by someone sit-
ments up to the present day. In the years follow-
ing the experiment many explanations were An object launched into space without motive ting in a convertible car with its roof down, first
offered for the negative result, but none that were power of its own, which travels freely under the when the car is at rest and then when the car
Figure 2-6
consistent with all the known facts. Our present action of gravity and air resistance alone, is called Path of stone thrown is moving with constant velocity. An analysis of
interpretation results from a proposal in 1905 by a projectile, or ballistic missile. While a rocket is with a speed of 50 ft/s. both cases will show that, ignoring wind resist-
Albert Einstein that the speed of light in empty self-powered for a small part of its flight, it be- EY ANA ance, both balls would be caught by the person
space is the same for all observers, regardless of comes a projectile when its fuel is shut off and in the car. The first situation where the car is at
their motions. This statement is, of course, con- travels thereafter in the same way as a bullet. For rest should prove no mystery, but perhaps an
so on, just asifit had no horizontal motion.
trary to the meaning of Eq. (6). Einstein showed flights short enough so that the curvature of the Progress during the first three seconds
Its explanation for the second ball’s being caught is
that his proposal led to the conclusion that the earth may be neglected, the motion of a projectile is illus- needed. It should be noted that both the ball and
trated inFig. 2-6. At A the stone has no
speed of light c is the limit to the possible speed is one of constant downward acceleration, but it vertical the car have the same constant horizontal veloc-
of any body relative to any observer and that Eq. speed; atB (after 1 s) itsvertical speed
differs from the uniformly accelerated motion is32ft/s;
(6) is valid only for speeds small compared with at C, 64ft/s; and atD, 96ft/s. The ity; and since the vertical motion of the ball
already discussed in that the direction of the curved line would be the same whether the car was moving
the speed of light in a vacuum, namely, ABCD in Fig. 2-6 is the path that the
acceleration is seldom the same as that of the stone fol-
lows,andthearrows at 4, B, C, and Or not, the ball thrown upward should return to
c = 186,000 mi/s, or 3.00 x 108 m/s. For two initial velocity. Hence the velocity is continually D represent the person throwing it in spite of the fact that
bodies A and B moving in the same or opposite changing in both magnitude and direction. thevelocities at those places. Note that
thehori- it is moving.
directions at speeds comparable with the speed It is convenient in studying such projectile Zontal arrows areallthesame length,
indicating Suppose that a projectile is fired with a speed
of light, their relative speed uy, is given by motion to consider it as made up of two compo- the constant horizontal speed, while
the vertical and direction such as represented by vectorYoM
nents, one vertical and the other horizontal. Since arrows increase inlength, toindicate
theincreas- Fig. 2-7. Consider the components of velocity Yo
the gravitational force is vertically downward, it ingvertical speed. Thevertical arrow
— 4 = Us at C istwice along the x (horizontal) and y (vertical) axes.
YaB = T V4U,/c? ” produces an acceleration only in that direction, aslong asthatat B, while thatat D
isthree times An object that had the simultaneous horizontal
leaving the horizontal component of the velocity as long. The resultant velocity of
the
The basis of this equation is discussed in Chap. unchanged if air resistance is neglected. The is at each point tangent to the curve projectile and vertical speeds represented by vo cos 9 and
45 along with other consequences of the special complex motion of the projectile reduces to two is constantly changing both in magn ABCD. It vo sin9)would follow exactly the same path, in
itude and in the initial direction of vo. In discussing the motion
theory of relativity. simple motions, constant horizontal velocity and direction.
uniformly accelerated vertical motion. of the projectile, one may use either the whole
{
Example Two electrons A and B have speeds Suppose that we ask ourselves how a stone will speed in the direction of v or the horizontal and
of 0.90¢ and 0.80c, respectively. Find their rela- move if it is thrown horizontally at a speed of vertical parts. The latter viewpoint simplifies the
tive speeds (a) if they are moving in the same 50 ft/s. Neglecting air resistance, the stone will problem. }
direction and (b) if they are moving in opposite travel with a constant horizontal speed of 50 ft/s The horizontal acceleration a, is zero, if aif
directions until it strikes something. At the same time it will vertical components are ind resistance is Negligible. Hence the
` epe
other can be illustrated in seve ndent of each
horizontal
execute the uniformly accelerated motion of an ral ways, component of the velocity v, remains constant
object falling freely from rest; that is, beginning If a rifle is held horizontally and at any time r
ao heat a bullet is fired from the and at the instant
U4 BT with a vertical speed of zero, it will acquire rifle another bullet is
= vy0,/c?
downward speed at the rate of 32 ft/s in each dropped from the same height, both U, = Up COS bo (8)
(2) %5 = 0.90c — 0.80c _ 0.10c _ 0.36c. second. It will fall 16 ft during the first second, hit the ground, assuming bullets will
it is level, at the same
48 ft during the next, 80 ft during the third, and The vertical acceleration of the projectile is in
 54 THE PHYSICS OF PARTICLES
VELOCITY AND ACCELERATION
53 i

ee vertical and horizontal components and the two

Using s= vot + fat? = ar’,
u = Vy = U0 C08 ĝo are considered separately.
2s 2x
39 ft
es: r= [B= aye T'S In Fig. 2-8 we note that the path may be found
by considering a uniform motion in the initial
Hence the time elapsed before the projectile direction OC and finding the distance the projec-
strikes the surface is tile has fallen from this path at each instant. In
1 s under the action of gravity, the projectile falls
VENEZ
LG s = 32's
16 ft; hence at the end of 1 s it is 16 ft below A;
in 2s it falls 64 ft and hence is 64 ft below B,
During all this time the projectile travels hori-
and so on.
zontally with a uniform speed of 86.6 ft/s. The
horizontal range R is therefore
2-10
R = v, X time in air MONKEY AND HUNTER
= vt’ = (86.6 ft/s)(3.2s)= 280 ft A classic problem, the monkey and hunter, illus-
Figure 2-7 trates the projectile-type problem. A hunter aims
Components of velocity in Projectile motion.
Dy v If the surface above which the projectile a rifle at a monkey sitting at the top of a 100-ft-
moves is not level, the time of flight will be jall tree which is 200 ft away from the hunter.
increased or decreased depending upon whether If the monkey drops from the tree the instant it
the negative (downward) direction along the y U, = u cos0 = (100 ft/s)(cos 30°) = 86.6 ft/s the striking point is below or above the firing sees the flash of the rifle being fired, will the
axis; thus the vertical component of velocity at
point. The range is correspondingly increased or bullet hit the monkey? If it does, at what eleva-
any time ż is
Using Eq. (5) as applied to the vertical motion, decreased. tion will the monkey be when hit? Assume the
Uy = vo sin Oy — gt The motion of any projectile, with air resist- muzzle velocity of the bullet to be 500 ft/s (see
(9)
v? — v? = 2as ance neglected, may be treated in this manner Fig. 2-9).
The magnitude of the velocity at any instant 0 — (50 ft/s)? = 2(—32 fi/s?)s no matter what may be the initial speed and angle The angle @ is determined by recalling that
is v = Vv,? + v,?, and the angle this resultant of projection. The initial velocity is resolved into
velocity makes with the horizontal can be found _ 2,500 ft?/s? tang = Elevation of tree _ 100 ft _ 0.5
= 39ft
from S=- AR Tange 200 ft
tang =% (10) The time required to reach the highest point Therefore, 0 = 26.6°. The horizontal and vertical
Us
is, from Eq. (2), components of the velocity of the bullet are
In projectile motion one frequently wishes to
determine the height to which the projectile rises, vy = vsin 0= 500 sin 26.6° = 224 ft/s
V, — V% = at
the time of flight, and the horizontal range. The = v cos 0= 500 cos 26.6° = 447.5 ft/s
0 — 50 ft/s= (—32 ft/s?)r
first two may be obtained by the use of Eqs. (5)
and (2), while the range is determined by multi- 50 ft/s 4g The bullet will take z= s/v, = 200ft
plying v, by the time of flight. t= nye = 16s (447.5 ft/s) = 0.4475 to reach the tree. The
bullet will have an elevation at 0.477 s of
Example A projectile is thrown with a speed Assuming that the surface above which the 1 sec 2 sec
of 100 ft/s in a direction 30° above the horizontal. projectile moves is horizontal, an equal time will S = Vt + hat? = (224 ft/s)(0.447 s)
Find the height to which it rises, the time of flight, Figure 2-8
be required for the projectile to return to the
and the ħorizontal range.
Path of a projectile fired atan
angle of 30° abov e + 4(—32 ft/s?)(0.477 s}
surface. This can be seen by realizing that vo the horizontal with an initial
Initially, (downward) is now zero at the highest point. speed of100 ft/s. 100 ft — 32 ft = 96.8 ft
oni selectiestrikeswitha
speedequalto the
Also, s = 39 ft, a = +32 ft/s? and v, and ¢ are nal speed
and at an angle of 30°
V, = v sin 0 = (100 ft/s)(sin 30°) = 50.0 ft/s unknown. horizontal. y
The monkey at the end of 0.477 s will have fallen
|
eaa ee eS ety
S = hal? =} (32 ft/s?\(0.44 s)? = 3.2 ft
 56 THE PHYSICS
OF PARTICLES
VELOCITY ‘AND ACCELERATION

Speed is distance per unit time, and velocity A freely falling body is one that is acted on _
When the range of a projectile is of the order is displacement per unit time. Their average by no forces of appreciable magnitude other than _
of 100 mi or more, account must be:taken of the _ Values are defined by the equations its weight. .
fact that the direction of the acceleration due to The acceleration of a freely falling body at sea
gravity does not remain constant but changes a level and 45° latitude is approximately 32 ft/s? J
Aee Vo s—
significantly, always pointing toward the center Om t or 980 cm/s?, or 9.80 m/s?.
of the earth. The trajectory in such a case, even The terminal speed of a falling object is the |
in the absence of air resistance, departs from the vertical speed at which the force of air resistance
parabolic and follows an elliptical path. The dis- Instantaneous velocity is the time rate of
Man change of displacement isjust sufficient to balance its weight. l
cussion of the motion of a missile or satellite for The relative velocity of one body with respect
with rifle
long trajectories is postponed to Chap. 9. to a second body is the velocity of the first minus
, . ås
the velocity of the second, vector subtraction
200 ft 2-12
KE amV3 being used.
(a) Sketch of problem AIR RESISTANCE These relations are approximations to relativ-
Velocity is a vector quantity; therefore a state- istic equations for velocities and lose accuracy —
So far in the discussion of the motion of projec- ment of velocity must specify the direction as well
wt near the speed of light.
& tiles the resistance of the air has been neglected, as the speed, forexample,25 km/h E, 30 cm/s SW.
A projectile is an object which is given an
Uy However, for high-speed projectiles this resist- Acceleration is the change of velocity per unit initial velocity and which is then allowed to move
ance is no small factor, It introduces a force time. Average acceleration is defined by under the action of gravity.
which opposes the motion, a force which varies
both in magnitude and in direction and hence In projectile motion the vertical and horizontal
produces variable acceleration in addition to the a= MENA
7 | motions may be treated separately. If air resist-
Ug constant acceleration we have assumed previ- ance is neglected, the horizontal motion is uni-
(b) Vertical and horizontal components ously. This resistance reduces the height of flight, form, while the vertical motion is uniformly ac-
of the muzzle velocity the range of the projectile, and the speed of the Instantaneous acceleration is the time Tate of celerated. Under these conditions the path is
projectile when it strikes its target. A repre- — change of velocity parabolic.
Figure 2-9
Sketches for monkey and hunter problem. sentation of these effects is given in Fig. 2-10. The range of a projectile depends upon its
.
. > ae
initial speed and the angle of projection. If aif
Tesistance is negligible, maximum range is at-
SUMMARY tained with an angle of 45°.
and will be 96.8 ft high. Therefore, the elevation
of the monkey and the bullet is 96.8ftand the Displacement is a change in position, specified by Uniformly accelerated motion is defin Air resistance decreases the speed, the maxi-
ed as mum height, and the range of a projectile.
monkey will have been hit. “a length and a direction. motion in a straight line in which the
direction
is always the same and the speed chang
es at a
constant rate.
2-11 The equations of uniformly accelerated Questions
tion for the particular case in which the
mo-
RANGE AND ANGLE OF ELEVATION direction
of motion remains fixed and the speed 1 A satellite travels with constant speed at a
The range of a projectile depends upon the angle changes fixed elevation above the earth. Show why
uniformly are the
at which it is fired as well as upon the initial velocity is not constant. What is the direction of
speed. If the angle is small, the horizontal velocity the acceleration?
is relatively high but the time of flight is so short Figure 2-10 s&u (1) 2 Show by the use of graphs why the average
that the range is small. On the other hand, if the Path of a projectile. The dotted curve represents v= Uo = at Speed of an object is 4 (Va + v,) only
angle of projection is very large, the time of flight the path that would be followed if there were no (2) of uniform acceleration and is not true
for the case
is long but the horizontal velocity is small. In the air resistance, while the solid line is an actual ble acceleration,
for varia-
absence of air resistance the maximum horizontal path. The maximum height, range, and striking
3 A man on a moving flatcar throws
Tange is attained when the angle of elevation is speed are decreased, while the striking angie is
towar a ball
increased. S= vt + far? (4) d his companion on the other end
of
45°
2as = v,?— v? car. Describe the velocity of the ball (a) relatithe
ve
(5) to the companion and (b) relative to the earth,
 58 THE PHYSICS OF PARTICLES
VELOCITY AND ACCELERATION 57
25 km/h. If he increases his average speed to long will it take to reach the ground? (c) How
when the car is moving (i) forward and (ii) back- at more than 18,000 mi/h horizontally, it won’t 40 km/h, how much time will he gain in his far will it fall from the base of the cliff?
ward. come down.” journey? Ans. (a) 103.4 ft/s; (b) 2.2 s; (c) 125.8 ft,
4 What is the average speed of a car which 14 . Describe the flying maneuver used to simu- 2 -A runnerAcan run the mile race in 4,25 min. 11 An elevator is ascending with an upward
goes at 40km/h for 20km and at 60km/h for late a condition of “weightlessness” for the train- Another runner B requires 4.55 min to run this acceleration of 4.0 ft/s*, At the instant its upward
20 km? ing of astronauts. distance. If they start out together and maintain speed is 8.0 ft/s a bolt drops from the top of the
5 Show by a vector diagram how much the 15 Why are the rear sights ofalong-range rifle their normal speeds, how far apart will they be cage 9.0 ft from the floor. Find the time until the
smekéstack on a moving train caboose would adjustable? at the finish of the race? Ans, 355 ft. bolt strikes the floor and the distance it has fallen,
have to be inclined in order for a vertically falling 16 Show clearly how the concepts expressed in 3 A train starts from rest and at the end of 90 s 12 A rocket ship is on its way to the moon at
raindrop to pass through the stack without hitting Newton’s laws of motion apply to the motion of
the sides.
has a speed of 30 km/h. What is its acceleration? 15,600 mi/h when it is alerted that a second ship `
a projectile.
4 A car going 50mi/h overtakes and passes is observed 1,000 mi away (viewed at an angle 30°
6 Cana body have a velocity without an accel- 17 .Describe the effect that an increase in the
another car moving at 45 mi/h. What length of from the line of motion of the rocket ship). The
eration? Can it have an acceleration with zero angle of elevation has on the range of a projectile
velocity? Give examples. road is required for the operation? Assume that second ship is traveling at 9,000 mi/h and will
for various angles of elevation.
7 Cite an example to show that it is possible each car is 15 ftlong and that there is a 60-ft cross the path of the first ship at right angles. Are
18 Derive an expression for the.speed of a pro-
for an object to have an acceleration without its space between them before and after passing. they on a collision path? If so how much time
jectile at time ¢ after it is fired with velocity y
speed changing. f at an angle of elevation 8.
Taking into account the approach of a car from does the command pilot of the first rocket ship
8 In a famous paradox of the Greek philoso- 19 A man stands in the center of a flatcar
the opposite direction at 50 mi/h, what clear have to take evasive action?
pher Zeno, Achilles, a fast runner, was proved moving with a uniform speed of 40 km/h. He
length of road is required? Ans. Yes, they will collide; 3.33 min,
by means of the following argument to be unable throws a baseball into the air with a speed of Ans. 1,500 ft; 3,000 ft. 13 A car travels with a constant speed of
to overtake and pass a tortoise: Achilles would 40 km/h. Compare the path ofthe ball as viewed S An object falls from a plane flying horizon- 30 km/h for 15 min. It then quickly speeds up to
first have to arrive at the place where the tortoise by the man with that as viewed by an observer tally at an altitude of 40,000 ft at 500 mi/h. How 50 km/h and maintains this velocity for 30 min.
Started, by which time the tortoise would have on the ground for the following cases: (a) ball long will it take to hit the ground? Neglecting the time needed to accelerate, what
moved on. Then he would have to arrive at this thrown vertically upward, (b) ball thrown forward 6 The initial speed of a car having excellent is the average velocity for the whole period?
second place, by which time the tortoise would- horizontally, and (c) ball thrown backward hori- brakes is 30 mi/h (44 ft/s). When the brakes are 14 An automobile has a speed of 60 mi/h.
have moved to a new place, etc. Is there anything zontally. ç applied, it stops in 2.0s. Find the acceleration. When the brakes are applied, it slows to 15 mi/h
wrong with the proof? 20. .Discuss the factors that would affect the ac- Ans, —22 ft/s?. in 4s, What is its acceleration? How far does it
9 A marble rolls with negligible friction down celeration of an Atlas rocket as it rises during the 7 A bullet accelerates from rest to a speed of travel during the fourth second?
an inclined plane. Show by means of a vector 70 s that the fuel burns. 600 m/s while traveling the 0.600-m length of a Ans. a = —16.5 ft/s?; s = 30ft
polygon how the acceleration parallel to the 21 Derive an equation for the range of a projec- rifle barrel. What is its average acceleration? 15 A sport car starting from rest can attain 4
plane may be expressed in terms of the geometri- tile as a function of the angle of its initial velocity (Treat this as a case of uniform acceleration.) speed of 60 mi/h in 8.0s. A runner can do &
cal dimensions of the plane. Galileo referred to above the horizontal, neglecting air resistance. How many times as great as the acceleration of 100-yd dash in 9.8s. Assume that the runner is
this experiment as “diluting gravity.” Show why (Hint: Write equations for x and y as functions gravity is this? moving with uniform speed and that the car starts
this is an appropriate designation. of ¢ and 9; place y equal to zero; eliminate ¢ 8 A body slides down a frictionless plane and at the instant he passes it. How far will both travel
10 Sketch rough curves to illustrate the velocity between the equations; use the trigonometric during the third second. after starting from rest until the car overtakes the runner?
as a function of time for the following cases: (a) identity for the sine of 24.) it travels 19.4 m. What is the angle of inclination 16 A stone falls from a railroad overpass which
a baseball thrown vertically upward, starting from 22 Derive an equation for y as a function of x of the plane? Ans. 52.5°. is 36fthigh into the path of a train which is
the instant in which it leaves the thrower’s hand for a projectile, treating vọ and @ as known, and 9 A body slides down a frictionless incline approaching the overpass with uniform speed. If
and continuing until the ball strikes the ground; eliminating the time £. ; 10.0m long. If the incline makes an angle of the stone falls when the train is 50 ft away from
(b) an elevator on a complete upward trip, 23, What would be the appearance of a speed- 30.0% with the horizontal, calculate (a) the time the overpass and the stone ‘hits the ground just
T1_ State two reasons why the value of g is time curve if the falling body were so light that of: descent, (b) the speed with which it reaches as the train arrives at that spot, how fast is. the
different at various places on the earth. What the effect of air friction could not be neglected?
would one expect about this value on the moon?
the bottom, and (c) the distance traversed during train moving? j Ans. 22.8 mi/h.
the second second after it starts from rest. 17 Two tall buildings are 60 m apart. With what
on the sun?
10 An object is thrown downward from a 150-ft speed must a ball be thrown horizontally from
12 If a body falls from a great height, can its cliff with a velocity of 45 mi/h so that it makes
Problems a window 150 m above the ground in one build-
Speed reach a maximum value and thereafter an angle of 30° with the horizontal. (a) How fast
decrease? Explain. 1 -A motorist has to travel 3.50km in a city ing so that it will enter a window 15 m from the
` will it be going when it hits the'ground? (b) How ground in the other?
13 Discuss the statement “If a projectile is-fired where his average speed should not exceed
 VELOCITY AND ACCELERATION

18 How high will a body rise that is projected the plane be with respect to the target when the
vertically upward with a speed of 100 ft/s? How bomb is dropped if a hit is to be made?
long will it take for the body to reach its maxi- 26 Two distant nebulae are observed to be
mum height? Ans. 156 ft; 3.1 s. moving away from the earth at 0.150 and 0.250
19 An arrow is shot vertically upward with a times the speed of light c. If they and the earth
speed of 288 ft/s, and 3.00 s later another is shot are in nearly the same straight line, how fast
up at a speed of 240 ft/s. Will they meet? If so, would one nebula appear to be moving, as seen
where? from the other? Determine their relative speed
20 A balloon which is ascending at the rate of by the classical equation first, then by the rela-
12 m/s is 80m above the ground when a stone tivistic equation. Ans. 0.100¢; 0.104c.
is dropped. How long a time is required for the 27 A ball thrown by a boy in the street is caught
stone to reach the ground? Ans. 5.4 s. 2.0 s later by another boy on the porch of a house
:21 A cannon is fired with a muzzle velocity 15.0m away and 5.0m above the street level.
of 300 m/s at an angle of 60°. What is its What was the speed of the ball and the angle
range? above the horizontal at which it was thrown?
22 A stone is dropped from a high altitude, and 28 A missile is fired with a launch velocity of
3.00 s later another is projected vertically down- 15,000 ft/s at a target 1,200 mi away. At what
ward with a speed of 150 ft/s, When and where angle must it be fired to hit the target? How long
will the second overtake the first? after it is fired will the, target be hit? (Assume
Ans. 5.70 s; 520 ft. that the accelerating power is cut off the instant Sites GEE PAREN
23 A baseball is batted into the air and caught it leaves the ground.) Ans. 32°, 248 s.
at a point 100 m distant horizontally in 4 s. If air 29 A projectile is fired at an angle of 30° above Sir Joseph John 1856-1940
resistance is neglected, what is its maximum the horizontal from the top of a cliff 600 ft high.
height in meters above the ground? The initial speed of the projectile is 2,000 ft/s. Born in Cheetham Hall, near Manchester. Profes-
24 A bomb is dropped from an airplane travel- How. far will the projectile move horizontally sor at the Royal Institute for Natural Philosophy in
ing horizontally with a speed of 300 mi/h. If the London, later master of Trinity College, Cambridge.
before it hits the level ground at the base of the
airplane is 10,000ftabove the ground, how far cliff? Discoverer of isotopy. Awarded the 1906 Nobel
from the target must it be released? Neglect air Prize for Physics for his theoretical and experi-
30 At what speed must a rifle bullet be fired so
mental investigations of the passage of electricity
friction. Ans, 2.09 mi. that it hits a monkey when the monkey is 180 ft
. 25 A bomb is dropped from an airplane 1,500 m high and at the instant of firing drops from the through gases.
above the ground when the plane is moving hori- top of a 200-ft-high tree, 350 ft away?
John Willlam Strutt, Third Baron
zontally at the rate of 160 km/h. Where should Ans. 364 ft/s.
1842-1919

_ Born in Essex, England. Chairman of the Davy-

Faraday Research Laboratory in London. Awarded
the 1904 Nobel Prize for Physics for his investi-
gations on the density of the more important
gases, and for his discovery of argon, one of the
results of those investigations.

Philipp Lenard, 1862-1947

Born in Pozony, Hungary. Professor of experi-

mental physics at Kiel, later at Heidelberg. Re-
ceived the 1905 Nobel Prize for Physics for his
work on cathode rays.
 3
62 THE PHYSICS OF PARTICLES

p
2 A turning wheel does not stop rotating as long

Force and Motion = NRA, as it is not retarded by air resistance

3 Planets and comets maintain both their pro-
gressive and their revolving motion longer in
(a) -spaces that offer less resistance

3-2
FORCE AND
ACCELERATION: NEWTON’S
SECOND LAW
Let us consider an experiment in which we have
a spring and several identical blocks of metal.
() Suppose that one metal block is placed on a
horizontal frictionless surface. The spring is at-
Figure 3-1 tached to the metal block and stretched a small
The effect of an external force on moving
objects. . known distance while the block.is held (Fig. 3-2),
We have considered the conditions under which exerting a force F} on the metal block. If the
must be exerted to start the box moving, and it
there is no change in motion of a particle. If the block is released and the stretch of the spring is
stops quickly when the force is removed. If the
net force on a particle is zero, that is, there is purely negative statement that no acceleration kept constant by pulling it along, the block is ac-
box is mounted on wheels, a smaller force is
‘No unbalanced external force, its motion will occur without a net force to cause that celerated. The acceleration a, of the block may
does not Tequiréd to start it and it continues to move
change. We have studied the motion of ‘bodies change. be determined by measuring the time required
longer. If more care is taken to reduce the fric-
that are moving with uniform acceleration, but It should be noted that this law states that in to move a known distance sstarting from rest and
tion, it "becomes easier to start thé box and it
we have not inquired about the forces that pro- order to change either the rate of motion or the then by using the equation s = uot + }.at®. Re-
continues to move more readily. We are finally
duce such acceleration. We shall now seek to direction in which an object moves, a force is peat the experiment with the spring stretched
led to the conclusion that if the friction of the
analyze the relationship between resultant forces required. Further, the direction of this applied twice as much to give a force F, = 2F,, and again
and the accélerations they produce.
floor could be entirely removed, any horizontal
force could start the box moving and once started force is important because, as Fig. 3-1 shows, the with three times the initial stretch to give
When a body is at rest, we know from experi- force in a acting in the direction in which the
it would continue to move indefinitely unless F = 3F,. The accelerations will be found to be
ence that it will remain at rest unless something a
object was moving will change only the rate of
is done to change that state. We walk without fear
force were exerted to stop it. The Property of a directly proportional to the net force F applied
body by virtue of which a net force is required Motion, while the force in b which is applied at and in the direction of the net force.
in front of a standing locomotive because we
to change its motion is called inertia. right angles to the original motion will change
know that it will not suddenly move. A heavy
the direction only. Since velocityisa vector quantity Faa
box on. the floor will stay in place unless it is
definedinterms of magnitude and direction, either
pushed or pulled. We must exert a force upon or
3-1 force can be considered as causing a change in
it to change its motion, that is, to give
it an velocity of the object. Case b will become more
acceleration. THE LAW OF Pas Site Fas const (1)
INERTIA: NEWTON’S FIRST LAW important when we consider uniform circular oi a5 Ay
We readily accept the fact that no body can
be set in motion without having a force act upon motion and central acceleration later in this book.
The conclusion which has been reached regarding There are many examples of the first law of
it. It may not be so easy to accept the equally This constant ratio of the net force to the accel-
the need of a force to change the motion of a motion, but perhaps the student would appreciate
true fact that a body in motion cannot change eration produced is a measure of the inertia of the
body was stated by Sir Isaac Newton (1642-1727), knowing the specific illustrations of this law that body being accelerated and is thus the mass of the
its motion unless a resultant force acts on it. We There is no change in the motion of a body unless
seldom if ever observe a body that has no force Newton himself gave in his book the “Principia” body.
an unbalanced external force is acting upon it, If in 1687. He observed that
acting on it. Suppose that we repeat our experiment with
the body is at rest, it will continue in motion with
A box resting on the floor has more than one spring and metal blocks, but this time we sh
constant speed in a Straight line unless there is 1. Projectiles continue in their motion until re-
force acting on it, but they do not produce a a net force acting. This law of inertia is usually keep the stretch of the spring constant and apply
change in motion. A rather large horizontal force tarded by air resistance and pulled down by grav- the same force F to one block, two blocks, threé
called Newton's first law of motion. It is the ity
blocks, etc., and measure the resultant acceler
61
 64 THE PHYSICS
OF PARTICLES
FORCE AND MOTION 63

For Newton’s second law; the proportionality ond. From the kilogram unit of mass and meter
constant k is customarily defined as 1. The value per second per second as a unit of acceleration
ofkdepends on units in which force, mass, and we can derive a unit of force that will make k
acceleration are presented. We shall see many of Eq. (3) unity. This unit of force is called the
situations in physics where proportionality con- newton. A newton (N) is the force that will give
stants are employed, e.g., Coulomb’s law and to a mass of one kilogram an acceleration of one
Newton’s universal law of gravitation. meter per second per second (m/sec?).
Since the utilization of Newton’s second law When we use a set of units, such as the mks
of motion depends greatly upon the units which system, in which one unit is defined in such a
are used, we will defer consideration of Newton's way as to make k unity, Eq. (3) reduces to
third law until we have examined in some detail
the choice of units available to us. F = ma (4)

Equation (4) can be used only when a consistent

ations. We find that the accelerationis smaller and written as the equation 3-3 set of units is employed.
when the number of blocks is increased. For two SYSTEMS OF UNITS In the cgs system the centimeter, the gram, and
blocks the acceleration is one-half that for one F=kma (3) the second are used as starting units. The acceler-
block, for three blocks one-third, etc. If m, is the Earlier we observed that three fundamental
In Eq. (3) any unit of force, any unit of mass, quantities are required to set up a system of units ation is measured in centimeters per second per
mass of one block, second. As before, we can derive a unit of force
and any unit of acceleration can be used, pro- in mechanics. The choice of these fundamental
vided that the proper valre is assigned to the quantities is rather arbitrary, and the kinds of that makes k of Eq. (3) unity. The dyne (dyn)
constant k. In general, a different value of k units set up depend upon the choice. Commonly, is the net force that will give to a mass of ont
would have to be assigned for each combination the fundamental quantities are length, mass, and gram (g) an acceleration of one centimeter per
E =m = 2m, @) chosen. For simplicity, it is customary to use a
system of units for which k has a value of 1.
time. Another system of fundamental quantities second per second (cm/s*) (Fig. 3-3). A mosquito
weighs approximately 1 dyn.
utilizes length, force, and time.
It is important to recognize the significance of Mass is independent of the place at which The cgs system was used for many years 4
= my = 3m proportionality constants, as k above. In science observation‘ismade, and hence a system of units the principal metric system. The mks system was
many physical relationships are observed as pro- based on length, mass, and time is called an adopted by an international conference to be
ete, portions, or ratios, While these ratios help to give absolute system. in the alternative choice, length, come effective in 1940. Since that time it hes
The acceleration of a body is directly propor- an understanding of the relationship of two prop- force, and time, the force commonly chosen is increasingly replaced the cgs system. An advan-
tional to the net force and inversely proportional erties, it is often desirable to express this rela- a gravitational force, or weight; and hence the tage of the mks system is that it leads tothe
to the mass being accelerated. The generalization tionship in quantitative terms by means of some system of units is called à gravitational system. “practical” electrical units in common use.
of this statement is Newton’s second law of mo- statement of equality, an equation. A typical ex- A British absolute system isbased upon the
tion. Whenever a net (resultant) force acts on a ample can be seen if we consider the geometric foot, the pound, and the second. In it the pound
body, it produces an acceleration in the direction ratio of the area of a circle to its radius. The area 3-4 isused as a unit of mass, and as before
aunit
of the resultant force that is directly proportional of a circle is proportional to. the square of the ABSOLUTE SYSTEMS OF UNITS of force isdefined to make k of Eq. (3) unity:
radius:
to the resultant force and inversely proportional to
In an absolute system of units a fundamental unit
The poundal
isthe force that will give to a mass
the mass of the body.
Aar is arbitrarily assigned to each of the fundamental of one pound an acceleration of one foot pef
According to the Newton’s second law, the second per second (ft/s). We shall not use this
following proportions may be written: quantities length, mass, and time as described in
While this can be proved to be true, we are system of units in this book.
the Introduction. Three such absolute systems are
unable to use this proportion in that form to find
commonset lyup: two metric, mks and cgs, and
ax F and eat the area of a circle. By selectingaproportionality
one British.
m
constant, however, we are able to change the
The mks system is that for which the funda- 3-5
proportion to an equation. For the particular case GRAVITATIONAL SYSTEMS OF UNITS
These proportions may be combined as mental units selected are the meter, the kilogram,
of a circle where the proportionality constant
and the second, the initials of these units forming
equals 3.14+, or 7, the following equation holds: the name of the system, In this system the unit In a gravitational system ofunits the fundamental
F unit of force is defined in terms of the pull
ax—
A=ar of acceleration is the meterper second per sec-
m
the earth upon anarbitrarily chosen body. In#
 FORCE AND MOTION 68
Table 1 _ 320lb _
m= Ast
fst = 10 slugs
System £E = m a
a = g = 32 ft/sec?
mks N = kg m/s? Table 1 lists consistent sets of mechanical
units.
_ cgs (abs) dyn = 8 cm/s?
a= 1 cm/sec? Fs] _
British (grav) 1b = slug fyè
a= 1 ft/sec? 3-8
HINTS FOR THE SOLUTION
(a) F = ma (b) F=% the central acceleration necessary to hold both OF PROBLEMS USING NEWTON'S
satellite and occupant in the orbit. Hence there SECOND LAW
Figure 3-3
is no reaction of the capsule on the occupant. This In applying the second law of motion to the
condition of zero reaction is often referred to as solution of a problem, much difficulty can be |
avoided by following a definite procedure, The
metric gravitational system the gram is used as ever the term pound is used, it will refer to force. most common source of difficulty is the failure
a unit of force. The gram force is defined as Summarizing, we shall limit ourselves to the mks, to recognize that the F of the second law always
one-thousandth the pull of the earth upon a cgs (abs), and British gravitational (fps) systems. refers to the resultant, or unbalanced, force actily
standard kilogram at a place where g has a value When a body falls freely, the only force acting on a body and the m refers to the entire mas
of 980.665 cm/s?. In this gravitational system the on it is its weight. This net force produces the of that same body. In solving any problem it
unit of mass is derived from Eq. (3) in such a acceleration g observed in freely falling bodies.
3-7 volving force and motion, the following stepsat
manner as to make k unity. No name has been From Eq. (3) we obtain
assigned to this gravitational unit of mass. It is WEIGHT: RELATION recommended:
the mass to which a gram force would give an BETWEEN MASS AND WEIGHT F = kma =
acceleration of 1 cm/s?. This system of units will 1 Make a sketch showing the conditions ofthe
The weight of a body at any point in space may
net be used in this book. be defined as the resultant gravitational force
W = kmg problem. Indicate on it dimensions or other dats
In the British gravitational system, fps, the given in the problem.
acting on the body, due to all other bodies in
fundamental unit of force, the pound (Ib), is If we use units that are consistent with Eq. (4), 2 Select for consideration the one body wht
space. When this definition is used, the body has
1/2.2046 the force with which the earth pulls on weight at all points in space except at those very the value of k is unity and motion is to be studied. Construct a force ved
a standard’ kilogram at a place where g is special points at which the resultant gravitational diagram. On this vector diagram, represent bj
32.17398 ft/s?. In this system we define a unit of force is zero. For a body near the surface of the Wa meg (5) vectors all the forces'acting on the body thathis
mass, the slug, from Eq. (3). The slug is the mass earth the gravitational forces due to outside bod- been selected. If any forces are unknown, rep
to which a force of one pound will give an accel- ies are almost negligible in comparison with that sent them also by vectors, and label them #
eration of one foot per second per second.
or AHS a
£ 6) unknown quantities.
due to the earth; the only gravitational force that
we need consider is that due to the earth, and 3 From the vector diagram, find the ren
this force is weight. Because the earth is rotating, Since weight isaforce, we use Eq. (5) toexpress force acting on the body. This resultant 18 the
observations of the gravitational force give a re- weights in newtons or in dynes when we use the of Eq. (4).
3-6 absolute mks or cgs units, respectively. In the
CHOICE OF UNITS TO BE USED sult that is slightly less than the gravitational 4 Find the unknown quantity (a, F, of m) frot
force, as we shall see later. Since this effect is British gravitational system we commonly ex- the relation F = ma. If the weight of the
We have outlined five systems of units each of small, we shall ignore it for the moment. press the mass in slugs from Eq. (6). Inany case, is given, compute m from m = W/g. If the po
which is consistent, logical, and suitable for use Some prefer to define weight as the reaction where we use units that are consistent. with Eq. lem asks for a distance, velocity, or time, 4P
in Eq. (4). It is unfortunate that in the different of a measuring instrument to the gravitational i bscan always substitute W/g for m and mg the equations of accelerated motion (Chap.
Sets of units the same word is used to designate force. According to this definition the weight for W. : required.
4 unit of mass in one set but a unit of force in depends upon the conditions under which the
another. Therefore throughout the mechanics measurement is made. For example, a body fall- Example Find the mass of an object that Example A 50-kg block rests at the wpa
section of this book we shall omit completely ing freely is accelerated by the resultant gravita- weighs 320 Ib. a smoo th
plane whose lengis2.00thmand "1
reference to the British absolute system and to the tional force, and the reaction of the measuri height is 0.50 m. How long will it take oF
cgs gravitational system. Whenever the term gram instrument as it and the body fall freely is zero. block to slide to the bottom of the plane
mal
or kilogram is used, it will refer to mass. When- Or in a satellite, the gravitational force produces g released?
 68 THE PHYSIC
OF PARTICLE
SS
FORCE AND MOTION 67

Here three forces are acting on the block. Its

weight W is 60lb downward. The force of the
plane on the block is a thrust N normal to the
plane. There is a pull P parallel to the plane.
Addition of these vectors by the polygon rule
shows an unbalanced force F acting on the block
parallel to the plane.
The weight of the block may be resolved into
components of 60.01b X cos 20° normal to the
Figure 3-4 plane and 60.0 Ib X sin 20° parallel to the plane.
Inclined plane. The normal component is balanced by the force
N. Hence the unbalanced force F parallel to the
plane and directed up the plane is
As indicated in Fig. 3-4, the forces acting on Weig
=60
ht
1b©
the block are the weight W downward and the
force N of the plane against the block. This force
F = 30.0 lb — 60.0 Ib x sin 20° @
N, called the normal, is perpendicular to the = 30.0 lb — (60.0 x 0.342) lb = 9.5 lb Figure 3-5
plane. The resultant (unbalanced) force F is par- Determination ofresultant force acting on block.
allel to the plane. The force triangle and the space
triangle are similar, and the angle @ is common
to the two triangles. From the force triangle From Eq. (4), In this time the elevator will have covered &
distance
F= Wsin 0
upward
From the space triangle
s=it= Ants +9 0.21s) =0.42ft
Note that if the angle were 30°, the component Since the initial velocity is downward, the upwar
: 9 _ 050m _ OA
sinl = 5m ; of the weight down the plane would be equal to d
acceleration will be considered negative,
the force up the plane and there would be no
The time Tequired to stop the elevator is Example A 5.0-kg block is placedon å
F = mg sin0 = (50 kg)(9.8 m/s?)(0.25) unbalanced force acting on the block. Hence it smooth horizontal surface (Fig. 3-7). A horizontal
would not be accelerated. If the angle were cord attached to the block passes over a light
= 1.2 x 10?N greater than 30°, the block would be accelerated 20 Ug' 2 0'~ 40 ft/s os
down the plane. ate OTe —19 Rates 2218
From Eq. (4),
Example A 2.0-ton elevator is supported ‘by nn ee Ec ee A
F_12x10N
ssi =o Ry T= 2.4 m/s? a cable that.can safely support 6,400 Ib. What is
the shortest distance in which the elevator can Maximum
Since the block starts from rest, the time of de- be brought to a stop, when it is descending with tension |6,400 lb }6,400 Ib
scent is determined from s = }at?. a speed of 4.0 ft/s?
The maximum net force acting on the elevator {2.400Ib= F=f
2s [2 200m _ (Fig. 3-6) is force remaining
r= [B= "Saag no to stop elevator
6,400 Ib — 4,000 Ib = 2,4001b upward Weight 4,000 lb

Example A 60.0-lb block rests on a smooth 4,000 Ib

plane inclined at an angle of 20° with the hori- The mass being accelerated is
zontal (Fig. 3-5). The block is pulled up the plane Figure 3-6 x
with a force of 30.0 Ib parallel to the plane. What — Ww _ 4,000 Ib
Ib Finding unbalanced force trom vector
m= EE = 125 slugs Figure 3-7
‘is its acceleration? addition of forces.
(aan
OT RAE Block accelerated by a falling mass.
ae
 70 THE PHYSICS
OF PARTICLES
FORCE AND MOTION 69
and ,
where m, is called the rest mass, m is the may
at speed v, and c is the speed oflight. Thischange
trictionless pulley and is attached to a 4.0-kg (0.040 kg)(9.8 m/s?) — T = (0,040 kg)a of mass with speed is negligible for speeds that
body. Find the acceleration and the tension in
the cord when the system is released.
are small compared with the speed of light and
When the two equations are added by solving ‘ence ‘is important only for the small particles
Consider first the 5.0-kg block. The forces simultaneously, we obtain
acting on this body are its weight W, downward, that can be accelerated to speeds approaching
the reaction N of the plane upward, and the that of light.
tension T of the cord to the right. Since there is
T — 0.294 N = (0.030 kg)a
no vertical acceleration, N = W, and the resultant — T + 0.392N = (0.040k Example What would be the mass of an ob-
force is T. Thus, from Eq. (4), + 0.098N = (0.070 kg)a ject having a rest mass of 1.0 g if it moves at 08
the speed oflight?
T = m,a 0.098 N
T = 1.40 m/s?
"= 000kg
T
The forces on the 4.0-kg body are its weight
me VO ~ Via (Oso
> mo = 10g Ii

W, downward and the tension T upward. The net W = més Then

downward force is W, — T and
W= mg T = 0.294 N + 0.030kg(1.40 m/s?) Then
W,-
T = ma = (0.294 + 0.042) N
Figure 3-8 1.0 EE)
Since the two bodies move together, the acceler- Unequal masses suspended
from a = 0.336 N m= Ven oA AoA
ations are equal in magnitude although the direc- pulley.
tions are not the same. If we add the two equa- ART
ad
tions by solving them simultaneously, we obtain 3-9
forces act on it, the weight m,g downward and
the upward pull 7 of the string. The resultant VARIABLE FORCE: Lo
T = m;a as = 1.67g
force on this body is T — m,g upward. From Eq. VARIABLE ACCELERATION
W,- T = ma
(4) we may write A n In our discussion thus far we have assu
W, =ma + m,a med that
W, =(m, +m,)a the forces considered have been constant 3-10
in mag-
T—mg
= m;a nitude and in direction..In general,
forces may REACTING FORCES: |
But W, = mg = 4.0kg x 9.8 m/s? = 39 N
where a is the upward acceleration of this body.
vary in any mannêr. Ifthere isany
such variation, NEWTON'S THIRD LAW |
Therefore, Newton’s second law applies at each
Now consider the body of mass m,. The forces instant and For every force that acts on one body there va
the instantaneous value of the accelerati
39N = (5kg + 4kg)a acting on this body are itsweight m,g downward Proportional to the net force at that on is Second force equal in magnitude but opposite in |
and the tension 7 upward. The resultant force instant and direction that acts upon another body. These fores
in the direction of that force.
a== Pkg
EN =i 4.4 m/s 2 is mg — T downward and from Eq. (4) are often referred to as acting and reacting ~~
The mass of abody does not neces
sarily re- |
main constant during its motion. Here the term acting force means the force th
mg
— T = m,a As a rocket
Then, substituting in T = mya, consumes fuel and expels the exhaust one body exerts on a second body, while reacting
achievthe
‘gas to force means the force that thesecond body exer |
e thrust, the mass continually decre
where a is the downward acceleration of this Other bodies may pick up material ases. on the first. There can be no force unless the
T = (5kg)(4.4 m/s?) = 22.0 N as they move,
body. Since the two bodies move together, the as in the case of growing raindrops. mutual interaction of two bodies is ne
Example Two bodies having masses m, = accelerations are equal in magnitude but opposite The principle of relativity (Cha fields arising from two different source
p. 45) states s ss
30.g and m, = 40 g are attached to the ends of in direction. that even though no material
is added, the mass
should be remembered that acting and rar
a string of negligible mass and suspended from forces, though equal in magnitude and on
a light frictionless pulley as shown in Fig. 3-8. m, = 30 g = 0.030 kg in direction, can never neutralize each other
Find the accelerations of the bodies and the ten- m, = 40 g = 0.040 kg they always act on different objects. In order pa
i
sion in the string. two forces to neutralize each other, they must
Consider the body of mass m,. Two external T — (0.030 kg)(9.8 m/s?) = (0.030 kg)a on the same object.
WMS Lactic barera > tan,KA
 72 THE PHYSICS OF PARTICLES
FORCE AND MOTION 74

a force on the ball while the two are in contact. c aaam

distance between their centers, and G is a con
During the same time the ball exerts a force of stant called the gravitational constant. The value
the same magnitude but opposite in direction on of G depends upon the system of units used in
the bat. A freely falling body is accelerated by F; Eq. (8). If the force is expressed in newtons, the |
the net force with which the earth attracts the mass in kilograms, and the distance in meters,G
body. The earth in turn is accelerated by the has the value 6.670 x 10-1! N-m*/kg?.
opposite reacting force the body exerts on the For extended bodies having a larger volume”
earth, Because of the great mass of the earth this
(a) Nozzle closed (b) Nozzle open and consisting of many particles, the gravitational |
acceleration is too small to be observed. In attraction is given in magnitude and direction oy
throwing a light object, one has the feeling that Figure 3-9 the vector sum of the attractions by the individual
he cannot put much effort into the throw, for he Forces inside an inflated balloon.
cannot exert any more force on the object thrown
than that object exerts in reaction against his mz = a
Ws _=n1.281b
ft/s? = 0.040 slug
hand. This reacting force is proportional to the one has missed seeing what happens when the
mass of the object (F œ m) and to the acceler- nozzle of an inflated balloon is released. While
ation (F œ a). The thrower’s arm must be accel- the motion quickly becomes quite random due Since
erated along with the object thrown; hence the to the aerodynamics of the balloon, the first reac-
such spheres attract as if their masses were cons
larger part of the effort exerted in throwing a light centrated at their centers. Inside a uniform spherii
tion of the balloon is to move away from the Fa = —F,
object is expended in “throwing” one’s arm. direction of the escaping air, an action-reaction cal shell the resultant force is zero. We may usi
When one steps from a small boat to the shore, phenomenon. A force diagram of the pressure on m,a, = (Mg)
—as) this property to examine the gravitational fort
he observes that the boat is pushed away as he a balloon (Fig. 3-9) reveals that when the balloon (—as)(Mma) _ 0.040 slug inside any uniform solid sphere or one that ii
steps. The force he exerts on the boat.is respon- is inflated, the internal pressure due to the forces a = poreB5 = —5.0 ft/s? 0,030slug made up of any number of concentric shells each
sible for its motion, while the force of reaction, acting on the inner surface cancel each other out. of which is uniform. Consider the sphere of Fig
exerted by the boat on him, is responsible for his This must be true or else the balloon would not = —6.7 ft/s? 3-11. A body embedded in the sphere at a radi
motion toward the shore. The two forces are be at equilibrium and would be moving around. r experiences gravitational forces due to the shel
equal in magnitude and opposite in direction, Once the nozzle is released, the equilibrium is Since the westward acceleration of B was taken between the radii r and R. The resultant fort
while the accelerations which they produce (in destroyed; and where every internal force had as positive, the negative sign for the acceleration
boat and passenger, respectively) are inversely been counterbalanced by another internal force of A indicates that its acceleration is eastward.
Proportional to the masses of the objects on (F, and F, for example), an unbalanced force is
which they act. Thus a large boat will experience ` now present because F, has escaped out the noz-
only a small acceleration when one steps from zle and F,, behaving according to the second law,
it to shore. causes an acceleration in the direction of F}. This
3-11 |
is the same principle upon which a rocket works.
UNIVERSAL GRAVITATION
A book lying on a table is attracted by the
earth. At the same time it attracts the earth, so Gases produced by a rocket engine are allowed In addition to the three laws of motion, Newton
that they would be accelerated toward each other to escape, much as the air from the balloon, and formulated a law of great importance in me-
if the table were not between them. Hence, each the unbalanced force in the chamber then causes chanics, the law of universal gravitation: Every
exerts a force on the table, and, in reaction, the the resulting acceleration. particle in the universe attracts every other particle
table exerts an outward force on each of them,
with a force that is directly proportional to the
keeping them apart. It is interesting to note that Example A 0.96-Ib ball A and a 1.28-Ib ball product of the masses of the two particles and
the table exerts outward forces on the book and B are connected by a stretched spring of negligi- inversely proportional to the square of the distance
the earth by virtue of being slightly compressed ble mass as shown in Fig. 3-10. When the two between their centers of mass. This relation may
by the pair of inward forces, which they exert on balls are released simultaneously, the initial ac-
it. be expressed symbolically by the equation
celeration of B is 5.0 ft/s? westward. What is the
Another example of the third law in action is initial acceleration of A? Figure 3-11 “a
the manner in which a rocket flies. While this will A body embedded in a solid sphere. The outer
r= Sty (8)
be treated in detail in the chapter on Space Phys- shell where theradius isgreater than r produces
ics, a simple analogy may prove helpful here. No a resultant gravitational force of zero. The i
where F istheforce of attraction, m, and m, are resultant gravitational force onthebodyis tha
due to the part of the sphere inside the radius y
the respective masses of the twoparticles, s isthe
 74 THE PHYSICS OF PARTICLES
FORCE AND MOTION 73
common to all material bodies, namely, that they and
due to this shell is zero. The body also experi- show a gravitational attraction for each other. It
ences a gravitational force due to the sphere of is not immediately apparent that the property of F's om = weight
radius r to which it is external. For the gravita- inertia is involved in gravitation. We may exam-
tional force on this body we use in Eq. (8) the ine the relationship between inertia and gravita-
mass of the inner sphere of radius r, and the tion. Suppose that we compare the masses m, and then
distance of the body from the center of the m, of two bodies by observing the accelerations
sphere: the radius r. From these considerations a, and a, produced by equal forces. From Eq.
we may conclude that the gravitational force is
a maximum at the surface of a uniform sphere.
(4), mg = ore

Inside the sphere the effective mass decreases as

the cube of the radius, while the distance de- BESS Let m, be the mass of any object on the earth's
creases only as the square of that same distance. m 44
surface and m, be the mass of the earth, Since
Outside the sphere the attracting mass does not m, appears on both sides of the equation it can:
change, but the distance increases. To a first ap- Or, if we apply forces necessary to produce equal
Figure 3-12 accelerations, cels out. Therefore,
proximation the earth may be considered as made
up of many spherical shells, and hence it approx- Cavendish apparatus.
imates the behavior just described. Near the sur- | 2em
seca
face local variations cause g to change both in F, m,
magnitude and in direction.
of the other small ball. By use of a telescope the
Newton checked his law of gravitation by minute deflection of the rod bearing the small Let us consider the gravitational force when each or
calculations based upon the orbit of the moon. balls was measured and from the interpretation of these bodies is attracted by the earth. The
With the approximate data at his disposal he still of these data the validity of the law was estab- gravitational forces W, and W, produce the same
found reasonable agreement between his calcula- lished and the vaiue of G was determined. acceleration in each body. Hence _ gs? _ (9.806 m/s?) (6.371 x 10° m}?
tions and observations. ms = G = 6610 x 10 N-m?/kg
Newton’s law of universal gravitation was not
experimentally proved until more than a century Example Two lead balls whose masses are = 5.967 X 10%kg
5.20 kg and 0.250 kg are placed with their centers
after it was first published. While few physicists
50.0cm apart. With what force do they attract
had any doubts about the truth of his hypothesis,
each other? This compares fairly well with the “best” estimate
the law was proposed in an age in which direct That is, the gravitational force is proportional to of the mass of the earth which is 5.975 x 10% kg
From Eq. (8), the mass measured by the inertial property. Thus
experimental evidence was required to prove its Similarly, the mass of the moon or of any planet
validity. It remained for Henry Cavendish = Gm
F=G-% masses may be compared by observing inertial can be approximated using certain assumption
(1731-1810) to measure in the laboratory the effects or by observing gravitational effects, and and conditions.
force of attraction between two masses. This was = 6.670 x 10-1! N- m?/kg? the results are identical, It was precisely this pro-
a formidable task because the force between any portionality between gravitational force and iner- Example Compute the mass’ of the moot:
two masses which could be conveniently handled
in the laboratory was extremely small and re-
x $20 kg x 0.250 kg tia that Newton used in accounting for planetary Since the moon is about one-fourth as large #
(0.500 m)? motion. the earth, s = (6.371 x 106 m)/4 (the value we
quired great skill to measure. Since the force A classic problem in physics is to calculate the
between the balls he used would amount to only = 3.47 x 10-°N shall use is 1.74 x 10° m). Also it is commonly
mass of the earth, or any other planet using the said that a person weighs one-sixth as much 0n
about 1/50,000,000 of their weight, it was neces-
universal law of gravity. the moon as he does on earth. Since in the equ
Sary to guard against any external factor such as
changes in temperature and air currents. Figure 3-12 tion W = mg, m is constant, then the weight and
3-12 illustrates the essential apparatus Cavendish Example At the surface of the earth g = g are proportional. Therefore gon the moon must
MEASUREMENT OF MASS 9.806 m/s?. Assuming the earth to be a sphere of
used. A 6-ft-long wooden arm was suspended by
radius 6.371 x 106m, compute the mass of the .
be approximately equal to (9.806 m/s?)/6 %
a thin wire 40 in long. Two small (2-in-diameter) We observe that we now have two equations that 1.62 m/s?. Using the equation,
balls were suspended at the ends of the arms. Two involve mass. We have defined mass from the earth.
larger balls (8 in diameter) were positioned one property of inertia that is inherent in all material
Z° _ (1.62 m/s?(1.74 x 19% my
in front of one small ball and the other in back bodies. We have observed a second property m m?/k
moon G — 6.670x 1011 N-
Mmoon = 7.39 X 102kg `
 76 THE PHYSICS OF PARTICLES
FORCE AND MOTION 75:
c If the mass is 5g and the force 10 dyn, Latitude,
This compares with the value calculated by more The slug is the mass to which a force of 1 Ib the acceleration is ____. Location deg Elevation, m g, m/st
sophisticated means for the mass of the moon, will give an acceleration of 1 ft/s?. d If the weight is 32 1b and the force 1Ib,
7.349 x 10”? kg. When the speed of a body approaches the the acceleration is ____. St. Michael, Alaska 63.5 1 9.822
speed of light, the mass of the body increases e If the weight is 320 1b and the force 20 lb, Denver, Colorado 39.7 1,638 9.796
according to the relation the acceleration is ____.
‘f If the weight is:500 1b and the force 10 1b, Portland, Oregon 45.5 8 9.806
SUMMARY
A Sl Ph the acceleration is Key West, Florida 24.6 1 9.790
The property of a body by virtue of which a VETZ g If the mass is 10.0g and the force
Canal Zone 8.9 6 9.782
resultant force is required to change its motion 9,800 dyn, the acceleration is
is called inertia. Mass is a numerical measure of The law of universal gravitation expresses the h If the mass is 6.0kg and the fore is
inertia. fact that every particle attracts every other parti- 2.0 N, the acceleration is
The relation between forces and the motions cle with a force directly proportional to the prod- i If the mass is1.0 kgand the force is9.8 N, A single object is taken to each of these stations,
produced by them was described by Newton in uct of their masses and inversely proportional to the acceleration is Compare the mass at the various places. Compare
three laws of motion. They are: ; the square of the distance between their centers 6 Explain why an empty train starts more the weights.
of mass. In equation form ;
quickly than a loaded train. 17 If a ball is thrown vertically upward by a
1 A body at rest remains at rest, and a body
7 Explain how a baseball pitcher’s windup person on a train which has a speed of 80 km/h,
in motion continues to move at constant speed = Gimme
F=G—3
in a straight line unless it is acted upon by an
enables him to throw the ball with greater speed the ball will be caught by the person if his hand
external, unbalanced force. than he otherwise could throw it. has not changed position. Explain why this hap-
2 An unbalanced force acting on a body pro- 8 Which is greater, the attraction of the earth pens.
The weight of a body at any point in space
duces an acceleration in the direction of the net may be defined as the resultant gravitational force for a pound of lead, or the attraction of the pound 18 What is the meaning of weightlessness as
force, an acceleration that is directly proportional of lead for the earth? applied to objects in space? Is there any place
acting on the body due to all other bodies in
to the unbalanced force and inversely propor- space. 9 Why would an object have greater weight at where weight is zero? If so, where?
tional to the mass of the body. the North Pole than at the Equator? 19 Derive an equation that expresses the total
3 For every force that one body exerts on a 10 If an elevator supported by'a cable is stopped downward force on a light pulley over which a
second body there is a force equal in magnitude quickly, it may oscillate up and down. Explain. pair of masses M, and M, are suspended by a
but opposite in direction that the second body Questions 11 If bullets are fired from a plane in its direc- light cord.
exerts upon the first body. tion of flight, will the velocity of the plane be 20 Describe how a satellite which orbits the earth
1 Motion and rest are referred to as being
affected? Explain. is constantly being accelerated toward the earth.
relative terms. What does this mean?
The relation expressed in Newton’s second law 12 If you found yourself on perfectly smooth
may be expressed in equation form as
2 Is the weight of a body: the same thing as
its mass? Discuss briefly. Is the weight of a body: ice in the center of a pond so that there was no
constant at all places on the earth? Is the mass? friction, how could you get off the pond?
Problems
F=kma Explain your answers. 13 If acting and reacting forces are “equal and
3 If a man standing on a scale grabs his shoe- opposite,” why can they never balance or cancel? 1 Ifa force of 2 N is applied to a 0.5-kg mass,
where F, m, and a can be in any units, provided laces and pulls up on them, will the scale reading 14 Describe how a stone which is being whirled what acceleration should result?
that the proper value is assigned to k, or be affected? If so, how? in a circular path on the end of a string will move 2 An unbalanced force of 50N acts on an
4 According to biblical account, David killed if the string breaks. Explain your answer. object weighing 100 N. What acceleration is pro-
F = ma the giant Goliath by using a sling. What scientific 15 The distance of sea level from the center of duced? Ans. 4.9 m/s*.
principle did he use to sling the stone? the earth is 3,963.34mi at the Equator and’ de- 3 How much does an object having a mass of
where one can use only those consistent sets of 5 Consider an object on a horizontal friction- creases to 3,949.99
mi at the poles. Suggest an 10 kg weigh? How much mass does a 64-1b object
less plane, acted upon by a single horizontal experiment by which this information about the have?
units in which one of the units is defined in such
a manner as to make k = 1. force. shape of the earth might be obtained. In view 4 A rope is attached to a 100-Ib object and is
A newton is defined as the force that will a If the massis 1 g and the force 1 dyn, the of it, what is meant by vertical? by horizontal? pulled upward with a force of 150 1b. What is the
acceleration is What basis is there for the statement sometimes
impart to a I-kg mass anacceleration of 1 m/s?. upward acceleration of the object? Ans. 16 ft/s’.
The dyne is defined as the force that will im- b Ifthe mass is1 g and the force 5 dyn, the made that the Mississippi River flows uphill?
acceleration is
5 An airplane in taking off from a field makes
part to a l-g mass an acceleration of 1 cm/s?. 16 Approximate values of g in various places a run of 2,300 ft and leaves the ground in 15.08
are shown in the following table. from the start. (a) What is its acceleration, as
 78 THE PHYSICS OF PARTICLES
FORCE AND MOTION 7
with the horizontal; the other makes an angle ot
sumed constant? (6) With what speed does it the frictionless plane with an acceleration of 60°, so that there is a 90° angle at the top. A
leave the ground? 8.0 ft/s, if the plane makes an angle of 30° with 6.0-kg mass and a 2,0-kg mass are attached to
6 A 10-g rifle bullet acquires a speed of the horizontal? the ends of a string which passes over a pulley
400 m/s in traversing a barrel 50 cm long. Find 16 An object of mass 8.00 kg is pulled up an at the top of the smooth double plane with the
the average acceleration and accelerating force. inclined plane, making an angle of 30° with the 2.0-kg mass on the steeper side. Find the acceler-
Ans. 1.6 X 107 cm/s; 1.6 x 108 dyn. horizontal, by a cord which passes over a pulley ation and the tension in the cord when the system
7 An electron (m = 9.11 X 10-?! kg) in empty at the top of the plane and is fastened to a 10,0-kg is released.
space experiences an upward electric force equal mass, Neglecting friction, find the acceleration 24 In Fig. 3-13 the blocksA and B of masses
to 25 times its weight. What is its acceleration? and the tension in the string. m, = 2mg, are on frictionless planes. Find the
8 Calculate the accelerating force needed to Ans, 327 cm/s*; 6,52 x 108 dyn.
change the speed of a 20-lb object from 18 ft/s 17 A 2,00-Ib hammer traveling at a speed of Figure 3-14
to 50 ft/s in a distance of 40 ft. Ans. 17 lb. 15.0 ft/s strikes a nail and drives it 0.75 in into
9 A plumb bob hangs from the roof of a rail- a block of wood. Assume that the resisting force
way coach. What angle will the plumb line make in the wood is constant. Find (a) the acceleration 29 In the arrangement of Fig. 3-15 express the
with the vertical when the train is accelerating of the hammer and (b) the constant force.
2,3 m/s”? acceleration of each block in terms of the three
18 An elevator and its load weigh 1,600 1b. Find
10 A 1.50-ton automobile crashed into a wall masses and g. Find the acceleration of each block
the tension in the supporting cable when the
at a speed of 10 mi/h. The car moved 5.00 in elevator, originally moving downward at 20 ft/s, Figure 3-13:
if m, = 3m, = 2m,. Assume the plane to be fric- |
before being brought to rest. What was the aver- is brought to ‘rest with constant acceleration in tionless and neglect the mass of the pulleys.
age force exerted on the wall by the car? a distance of 50 ft. Ans. 1,800 Ib.
Ans, 24,300 Ib 19 The reaction time of the average motorist is
11 What pull must a locomotive exert on a magnitude and direction of the acceleration
of
0.70s. Assume that by means of the brakes, a
12,000-ton train to attain a speed of 60 mi/h in retarding force equal to three-fourths the weight each block.
5.0 min? Assume uniform acceleration, and as- of the car can be applied to the car. If the car
sume that 30 percent of the applied force is used is traveling 60 mi/h, find (a) the acceleration
against friction. when the brakes are applied and (b) the distance descending with a speed of 900 ft/min.
12 A 1,000-g block on a smooth table is con- the car travels after the motorist receives the Ifthe load on the cables must not exceed 14 tons,
nected to a 500-g piece of lead by a light cord stopping signal. what is the shortest distance in which the elevator
that passes over a small pulley at the end of the 20 A 100-Ib box slides down a frictionless skid can be Figure 3-15
table. (a) What is the acceleration of the system? inclined at an angle of 60° with the horizontal. 26 A spring balance fastened to the roof of a
(b) What is the tension in the cord? f Find (a) the accelerating force, (b) the time re- Moving elevatorcar indicates 90 Ib as the weight
Ans. 327 cm/s?; 3.27 x 10° dyn. quired to travel the first 20 ft, and (c) the time of a 120-Ib body. (a) What is the magnitude and 30 Two objects of mass 500
13 A rocket has a mass of 2.00 x 10*kg of required to travel the next 20 ft. direction of the acceleration of the elevator? (b) g each are fastened
which half is fuel. Assume that the fuel is con- together by a cord and suspended over a friction-
Ans. 87 lb; 1.2 s; 0.49 s, Can one determine from these data the direction
sumed at a constant rate as the rocket is fired and 21 What force, applied parallel to the plane, is
‘less pulley at the top of a double-inclined plane.
that there is a constant thrust of 5.0 x 10°N. necessary to move a 16.0-kg object up a friction-
in which the elevator is moving? One side of theplane makes an angle of 45° with
Neglecting air resistance and any possible varia- less plane with a uniform acceleration of
Ans. 8.0 ft/s? down; no.
27 In the system shown in Fig. 3-14 find the
the horizontal,and the other side makesanangle
tion of g, compute (a) the initial acceleration and 2.00 m/s?, if the plane makes an angle of 60° with acceleration of each of the bodies at the instan of 30°,so that there is an angle of 105° at the
(b) the acceleration just as the last fuel is used. the horizontal? t top of the plane where the pulley is a
|
14 An object whose mass is 12.0 kg is acted 22 A 200-lb man stands in an elevator. What
the system is in the configuration pictured. Will
the acceleration remain the same as the motio
Calculate the acceleration of the system and the
upon by two forces that are in opposite direc- force does the floor exert on him when the eleva- n force in the cord.
progresses? If not, what will be the manner
tions: one of 540 N, the other of 1,260 N. What tor is (a) stationary; (b) accelerating upward in Ans. 102 cm/s?; 2.96 x 10° dya
which it 31 An electron
is the acceleration produced? How is the direction has a rest mass
9.11 x 10-31 kg. Find its mass. when its speed %
16.0 ft/s?; (c) moving upward at constant speed; 28 A light frictionless pulley carries a light
of the acceleration related to the forces? cord
Ans. 60 m/s?.
and (d) moving upward but decelerating at to which is attached at one end a 48-Ib
weight 0.50¢, 0.90, and 0.99c.
Ë
12.0 ft/s?? Ans. 200 1b; 300.1b; 200 Ib; 125 Ib. and at theother a 64-Ib weight. The weigh {
13 What force, applied parallel to the plane, is 23 One side of a double-inclined plane some- ts are 32 What average force is necessary to accelerslé
suddenly released. Find the accelerati
Necessary so that a 100-Ib object will slide down what like that in Fig. 3-13 makes an angle of 30° on and the a proton, rest mass 1.67 x 10-27 kg from rest © |
tension in the cord: Ans. 4.6 ft/s?; 55 Ib, one-tenth the speed of light (c = 3.0 x 10° m/9)
 FORCE AND MOTION 79
in a distance of 3.0cm? What force would be force of attraction for each other. (6) Find the
required to produce this same acceleration if the acceleration of each that is produced by this
initial speed were 0.50c? force.
Ans. 2.5 x 10-11 N; 2.9 x 10-11N, 34 What is the acceleration due to gravity on
33 The mass of the earth is approximately the surface
of the moon if the mass of the moon
5.98 x 10% kg and that of the moon is 0.0123 is 0.0127 that of the earth and the radiusof the
times as great. The mean distance between them moon is 0.25 that of the earth? Ans. 6.5 ft/s?.
is 3.84 x 10°km. (a) Compute the gravitational

Albert Abraham Michelson, 1852-103

Born in Streino, Prussia. Director of the physics

department, University of Chicago. The 1907
Nobei Prize for Physics was conferred on him for
his optical instruments
of precision and the spec
troscopic and meteorologic investigations which
he carried out by means of them.
4
82 THE PHYSICS OF PARTICLES

tance of one centimeter. An erg is 10-7 J, In the

British system the unit of work is the foot-pound
Work, Energy, and Power the work done by a force of one pound acting
over a displacement of one foot in the direction
of the force. Another unit of work, used especially |
in electrical measurements, is the kilowatthour
Work= (F cos @)s (Sec, 4-16).
Figure 4-2
Work done in dragging a sled. Example A box is pushed without acceler
ation 5.0m along a horizontal floor against a
frictional force of 180 N. How much work is
This procedure is usually denoted by the use of done?
a dot between the two vectors to indicate scalar
multiplication; thus 'W = Fs = (180 N)(5.0 m)
= 900 N-m = 900J
W=F-s (2)
A significant difference between our civilization W = (F cos 6)(s) (` Example What work is performed in drag-
and that of the ancients is our extensive utiliza- The magnitude of this scalar product is Fs cos
9, as appears in Fig. 4-2. Another kind of product ging a sled 50 ft horizontally without acceleration
tion of energy from sources other than the mus- where @ is the angle between the direction of the when the force of 60 Ib is transmitted by a rope
cles of men and animals. Many of the early ad- force and that of the displacement, and the force of two vector quantities is the vector product
described in Chap. 5. making an angle of 30° with the ground (Fig
vances in physics were made by men who were ‘F is averaged over the displacement. In the spe- 4-2)?
trying to understand and control sources of en- cial case where the force is constant and has the
The component of the force in the direction
ergy and apply them to men’s tasks. As the study same direction as the displacement, 0 = 0°, cos
of the displacement is F cos 30°.
of physics has advanced, energy has continued to 0°= 1, and the work done by the force is the
4-2
be a principal concern, playing such a crucial role product of the constant force and the distance.
Although work is the product of two vector
UNITS W = (Fcos 30°)s
that physics has been called the “science of en-
ergy and its transformations.” quantities, force and displacement, it is itself a In the mks system the unit of work is the newton- = 60 Ib x 0.866 x 50 ft = 2.6 x 103ft-lb
scalar quantity. When two vector quantities are meter, the work done by a force of one newton
multiplied so as to produce a scalar quantity, this exerted through a distance of one meter when the
4-1 process is called scalar multiplication of vectors.
WORK force is parallel to the displacement. The newton-
meter is called a joule, after the British physicist 4-3
The term work, commonly used in connection James Presçott Joule (1818-1889), whose experi- ENERGY: THE ABILITY TO DO WORK
with widely different activities, is restricted in ments contributed heavily to the acceptance of
physics to cases in ‘which there is a force and a —_— o Work = Fs That prope rty of a body or system of bodies by _
F a the relationship between heat and work. In the virtue of which work can be performe
displacement along the line of the force. In this
technical sense of the word work, a pier does no
cgs system the unit of work is the erg, which is energy (a scalar quantity). Energy d is called
work in supporting a bridge, and a man does no
the work done by one dyne exerted over a dis-
many forms and can be transformedcanfromexist in
one
work if he merely holds up a suitcase, though he
may experience muscular fatigue. But in lifting Table 1
the suitcase to a rack he would perform work. UNITS TO DESCRIBE WORK
When a force F moves through a displacement
s and the directions of these two vectors are not
Work = (F cos @)(s) ° System Work = Force Displacement
the same, the work ‘W is defined as the product
of the magnitude of the average force F and the
()
i. peo RO ane = newton a
Figure 4-1 meter =(N-m)
displacement s cos @ in the direction of the force. cgs (abs) erg = dyn
Work done on a block.
(See Fig. 4-1.) British (grav) foot-pound Soy (dyn-cm)
= pound ft (ft-lb)
81
 @4 THE PHYSICS
OF PARTICLES
WORK, ENERGY, AND POWER

F = 401b
form to another. The energy possessed by an tential energy gained, is the product of the weight
object by virtue of its motion is called kinetic W and the height A to which it is raised. This
«nergy, or energy of motion. Energy of position, increase in potential energy is given by
or configuration, is called potential energy, When
work is done on a body in the absence of fric- E, = Wh = mgh (3)
‘tional forces, the work done is equal to the sum
of the increase in kinetic energy and the increase
If W is in newtons, m is in kilograms, and A is
in potential energy. The units in which energy
is expressed are the same as the units for work, in meters, E, is given in joules. If W is in pounds,
Figure 4-3
Many problems in mechanics can be solved m is in slugs, and h is in feet, E, is given in
Work done when force is not in the direction of the displacement.
foot-pounds. If W is in dynes, m is in grams, and
by the laws of motion discussed in Chap. 3. Given
certain information about the initial status of an h is in centimeters, E, is given in ergs,
object and the forces to which it is subjected, we The gravitational potential energy is expressed
relative to a specified arbitrary reference level. to the ramp, angle @ would be zero but F would the total potential energy acquired by the body,
can predict its position and velocity at any future
This reference level may be any point that is be only 20 Ib. is represented approximately by the total area of
time, In some situations, however, as in the de-
agreed upon by those concerned, For example, the rectangles and accurately by the area under
scription of the motion of a pendulum, direct the arbitrary reference for zero gravitational po- W = Fscos0= (201b)(200 ft)(1) the curve.
application of the laws of motion would require tential energy may be chosen as sea level, or floor
complicated calculations of forces and acceler- = 4.0 x 103 ft-lb It is shown in Chap. 9 that the potential energy
level, or “at infinity,” a point so far from the earth of a body at high altitude with respect to the
ations in order to obtain a relatively simple result. as to be effectively outside the earth’s gravita- In these cases a momentary force somewhat
Consideration of the potential and kinetic energy surface of the earth is given by
tional field. greater than that discussed is needed to set the
involved and the relation between work and en-
body in motion. The work thus done on the body
ergy simplifies the solution of many problems in
mechanics. Moreover, the concept of energy leads
Example A 40-Ib stone is hoisted to the top at the beginning is recovered when the body E, = GMm(4— +) o
of a building 100ft high. How much does its comes to rest. ý
to the principle of the conservation of energy,
potential energy increase? In all three modes of (frictionless) transport,
which unifies a wide range of phenomena in the
Friction being neglected, the increase in po- the work done equals the gain of E, given by
physical sciences.
tential energy is just the amount of work done Eq. (3). In Eq. (3) we have assumed that when
in lifting the stone, so that
we elevate an object a distance h which is small
4-4 compared with the radius of the earth, the gravi-
E, = Fs = (40 1b)(100 ft) = 4.0 x 109 ft-lb tational force acting on that object remains con-
POTENTIAL ENERGY
stant. For any system in which the force is not
The energy which bodies possess by virtue of Example A 40-lb stone is carried up a ramp, constant the gain in potential energy is the prod-
their positions, configurations, or internal mecha- along a path making a 30° angle to the horizon- uct of the average force and the distance moved
nisms is called potential energy E,. Important tal, to the top of a building 100ft high. How in its direction, If an object such as a rocket is Ib
t,
forms of this type of energy are electrical, elastic, much work is done? (Neglect friction.) lifted to a height of several times the earth’s
chemical, and nuclear potential energy. The most The distance traversed is now 200 ft, as shown radius, the gravitational force is far from con- Wei;
common form of potential energy is gravitational in Fig. 4-3a. The force exerted on the stone is
stant, since this force varies inversely as the
potential energy. Since the earth attracts every equal in magnitude to its weight. The angle be-
square of the distance of the object from the
body, work is required to lift the body to a higher tween the force and the displacement is 60°.
center of the earth. This is shown in Fig, 4-4 for
level. When a brick is carried to the top of a a body which weighs 1,000 lb at the surface
building, the work done on the brick (weight of W = Fs cos@ = (40 1b)(200 ft)(0.500) of
the earth. If such a body is lifted to a high alti-
brick times vertical distance) represents energy
= 4.0 x 10° ft-lb tude, the work can be obtained by considering 3
that can be recovered. By virtue of its position
at the top of the building the brick possesses more
the operation as a large number of bits of work, Distance from center of earth in earth radii
ability to do work than it had when it was at Notice that the work is the same as for the previ- each performed with a different average force.
ground level. It has increased its potential energy. ous example (again neglecting friction). If the The area of each rectangle in Fig, 4-5 represents Figure 4-4
stone is rolled up on rollers by a force parallel the product of a force and a distance, hence Variation of weight with distance from the center
The work done on the brick, and hence the po- a
quantity of work. The total work done, and of the earth for a body of mass 31.1 slugs.
hence

i
n
 86 THE PHYSICS
OF PARTICLES
WORK, ENERGY, AND POWER

in foot-pounds. If m isin grams and v incentime- sion of one surface tothe other
and bythei
This is about equal to the work needed to lift terspersecond, Eq.(6)givesthekineticenergy locking of the irregularities of the rubh
an, object weighing 3,800 tons to a height of in ergs. surfaces, The force of frictional resistance d
1,000ftabove the earth. Although a steady force has been assumed pends upon the properties of the surfaces i
here, the result is independent of the particular upon the force keeping the surfaces in conta
manner in which a body attains its velocity, The effects of friction are often undesita
4-5 Friction increases the work necessary to oper
KINETIC ENERGY Example What is the kinetic energy of a machinery, it causes wear, and it generates het
3,000-Ib automobile which is moving at 30 mi/h which often does additional damage. To redi
Force In addition to energy of position or state, objects (44 ft/s)? this waste of energy, friction is minimized by!
may possess energy due to their motions. A car
or bullet in motion, a stream of water, or a re-
use of wheels, bearings, rollers, and lubricant
W _ 3,000Ib Automobiles and airplanes are streamlined
volving flywheel possesses kinetic energy. The ma Bina = elope
kinetic energy of a moving object can be meas- order to decrease air friction, which is large
ured by the amount of work it will do if brought E, = į mu? = į X 94 slugs x (44 ft/s)? high speeds.
Distance to rest or by the amount of work originally On the other hand, friction is desirable|
= 9.1 x 10 ft-lb many cases. Nails and screws hold boards|
Figure 4-5 needed to impart the velocity to it, in circum-
Work done by a vatying force. stances where the work cannot also go into po-
When an accelera gether by means of friction. Power may be tra
a
OTS tential energy. force isting
applied to a mitted from a motor to a machine by means
EE RE body, the work done by that force produces a
Consider a body with an initial speed v) on a clutch
or a friction belt. In walking, driving)
which a steady unbalanced force F acts as it change in the kinetic energy of the body. If a
resultant force F acts tostart a body in motion car, striking a match, tying shoes, or sewing fab
where G is the constant of universal gravitation,
M is the mass of the earth, m is the mass of the
moves a distance s. The body gains speed at a
or to stop one initially in motion, together we find friction a useful force. Sand!
rate given by a= F/m until it reaches a final Placed on rails in front of the drive wheels?
body, R is the radius of the earth, and r is the speed v,. The work done on the body by the
locomotives, cinders are scattered onicy st s
distance of the body from the center of the earth unbalanced force that accelerated it appears as Fs =ġm - (7) chains are attached to the wheels of automobil 4
(note that r is not the altitude above the surface a change in its kinetic energy. Since F = ma,
of the earth), multiplying by s gives Fs = mas and
Example What average force is necessary to and special materials are developed for use 1
stop a bullet of mass 20gand speed 250 m/s as brakes—all for the purpose of
increasing frictidl
Example.. A 200-kg satellite is lifted to an A(E,) = Fs = mas it penetrates wood to a distance of 12 cm? where it is desirable. Frictional forces are impot
(5)
orbit, of 2,20 x 104 mi radius. How much addi- The work done by the retarding force is equal tant in determining the path of a space vehi
tional potential energy does it acquire relative to From Eq. (5), Chap. 2, to the initial kinetic energy of the bullet and the heating produced when the vehicle t
the surface of the earth? enters the earth’s atmosphere.
The solution is found from Eq. (4), where 2as = v,? — v? Fs = 3mv?
or as = Xv,” — vo’) F X 0.12 m = (0.020 kg)(250 m/s)?
R = 6,37 x 106m
F = 5.2 x 103N 4-7
A(E,) = $m (0,? = U9?) = $mv,? — 4muy? SLIDING FRICTION
r= 2.20 x 104mi = 3.54 x.107m
This force is nearly 30,000 times the weigh
M = 5.98 x 10% kg
If the body was initially at rest, Vo = 0 and the t of When we slide a box across a floor, we find
the bullet.
gain in kinetic energy is the final kinetic energy. we must continue to apply a steady horiz “al
The initial kinetic energy, $ mv? = 620 J, is
force on
m = 200 kg Thus the kinetic energy of a body at any instant
is largely wasted iņ heat and in work done to cause the box to slide uniformly over &4
in de- horizontal surface. We conclude that
E, = (6.67 x 10-11 m3/kg-s?) forming the bullet. there w
X (5.98 x 10% kg)(200 kg)
E, = 4m? (6)
z (orel onies
6.37.x 108m o 3.54 x 107m
If m is in kilograms and v in meters per second,
Eq. (6) gives the kinetic energy in joules (new-
‘ton-meters). If m is expressed in slugs and v in
4-6
FRICTION ifthe applied force is greater than the frictio
= 103 x 10%J feet per second, Eq. (6) gives the kinetic energy Whenever an object moves while in force, the body will be accelerated. a
contact with The friction between solids sliding ovet 0%
another object, frictional forces oppo
se the rela- another is due to several causes acting at ont
tive motion. These forces are caused
by the adhè- If the surfaces are very rough, an interlock
 88 THE PHYSICS OF PARTICLES a

WORK, ENERGY, AND POWER §7

Here (mu) is the coefficient of friction, F the
process gives rise to large forces; but as one or frictional force, and N the normal or perpen-
both of the surfaces are made progressively dicular force.
smoother, it is found that the friction diminishes
at first, then becomes fairly constant. Thus the Example A 65-lb horizontal force is sufficient
friction between polished steel blocks remains to draw a 1,200-Ib sled on level, well-packed snow
appreciable and even increases- with further at uniform speed. What is the value of the coeffi-
smoothing. For the surfaces of the same metal cient of friction?
in contact it is found that microscopic welds are
formed and broken; and even for dissimilar Figure 4-7 2
metals strong adhesive bonds are formed. The The frictional force is independent of the area of
presence or absence of oxide coatings and water the surface of contact.
films is significant. All these processes cause vi- Figure 4-9
The frictional force is proportional to the nor- Forces acting on a body on an inclined
brations which set up waves through the materials mal force, which must include all normal compo- plane.
and produce heating. Despite the complexity and nents of forces pressing the surfaces together.
variety of the processes involved for various ma- 4 The frictional force is roughly independent of
the speed of sliding provided that the resulting Only in very special cases is the normal force the ae.
terials, certain simple observations have been
heat does not alter the condition of the surfaces. weight of the body. Example A 50-kg box is placed on an inclin
found to apply to nearly all cases of sliding fric- plane making an angle of 30° with the horizon
tion to within a few percent accuracy. 5 The frictional force depends upon the nature
of the substances in contact and the condition of Example A 1,200-Ib sled is pulled along a (Fig. 4-9). If the coefficient of kinetic friction
The observations regarding sliding friction are
these: the surfaces (i.e., on polish, roughness, grain, and horizontal surface at uniform speed by means of 0.30, find the resultant force on the box. _
wetness). a rope that makes an angle of 30° above the The weight of the box is a force acting Wa
1 The frictional force is parallel to the surfaces horizontal (Fig. 4-8). If the tension in the rope cally downward. This force may be separatedi
sliding over one another. Sliding friction is sometimes called kinetic is 100 Ib, what is the coefficient of friction? components parallel (W sin 30°) and perp
š The frictional force (parallel to the surface)
friction. These observations may be illustrated by dicular (W cos
force (parallel to30°) to the plane. A
2 The frictional force is proportional to the
force which is normal (perpendicular) to the sur- simple experiments. By the use of a spring bal- the plane) acts up thefrictior
p
faces and which presses them together. ance to pull a brick uniformly on a tabletop (Fig.
3 The frictional force is roughly independent of 4-6), the frictional force may be found to be | Ib, F = 1001bX cos 30°= 1001b x 0.866 = 86.6Ib N= W cos30°= 50 kg x 9.8 m/s?
XOM
the area of the surface of contact. whereas a force of 2 lb is required to maintain
the motion if a second brick is placed on top of | The normal force N is the weight of the sled = 430 N
the first to increase the normal force. If the sec-
downwar d minus the vertical component of the
Pakis Woka icti l iforce isis given
ond brick is tied behind the first (Fig. 4-7), the The frictiona gi by
frictional force is still 2 1b, showing its inde-
pendence of area of contact. If the bricks are F= = 3 = 130N
placed on glass or metal surfaces, one finds that
N = 1,200 lb — 100 1b x sin 30° = 1,150 lb oe ON
friction depends upon the nature of the surfaces — F _ 866lb _ R The resultant force is
in contact. h = "N= T1501b = 2.0753
DE OT S E's 8 R = Wsin30° — F
a 100Ib = 50kg x 9.8 m/s? x 0.50 — 130N
4-8
COEFFICIENT OF KINETIC FRICTION = 120 N down the plane

When one body is in uniform motion on another When two surfaces are lubricated, fii
body, the ratio of the frictional force to the per- reduced by the substitution of the internal fe
pendicular force pressing the two surfaces. to- : a lubricant for the friction between the Mf
Figure 4-6 gether is called the coefficient of kinetic friction.
Figure 4-8 Surfaces. The ratio F/N is then not 4 S
The frictional force is directly proportional to the
normal force pressing the two surfaces together.
Forces acting on a sled, i q
Constant but depends upon the properties
= E (8) : lubricant and the area and relative speed f
RET ie moving surfaces. s
 90 THE PHYSICS OF PARTICLES
WORK, ENERGY, AND POWER 89

the size, shape, and speed of the moving object,

Table 1 as well as on the nature of the fluid itself. We
COEFFICIENTS OF FRICTION have already seen fluid friction in action when
we discussed the terminal speed of a body falling —
Material My My
through air (a fluid).
Steel on steel 0.15 0.09
Metal on metal, lubricated 0.03 0.03
4-13
Leather on oak 0.4. 03
STOPPING DISTANCE
Rubber tire on dry concrete road 10 0.7
The fact that the kinetic energy of a moving
Rubber tire on wet concrete road 0.7 0.5
object is proportional to the square of its speed |
has an important bearing upon the problem of
stopping an automobile. Doubling the speed of
4-9 the car quadruples the amount of work that must
STATIC FRICTION Figure 4-10 be done by the brakes in making a quick stop.
Figure 4-11
When a body at rest on a horizontal surface is Limiting angle for friction. Rolling friction. A consideration of the equation
pushed gently sideward, it does not move, be-
cause there is a frictional force just equal to the v? ate Vo? = 2as
sideward push. If the push is increased, the fric- (Fig. 4-10) be tilted upward (gradually) until the
a steel rail there is some deformation of the two
tional force increases until a limiting friction is body B just begins to slide down the incline. „Shows that, if v, = 0 (indicating a stop), $=
Consider the weight W
of the body to be resolved surfaces. A force F (Fig. 4-11) is required to roll
reached. If the side push exceeds the limiting the wheel on the horizontal rail because the sur- —vo"/2a, so that the distance in which an auto-
friction, the body is accelerated. When there is into components Wcos@ and F’ = Wsin 8, re-
spectively perpendicular and parallel to the in- faces are continually distorted as illusttated by mobile can be stopped is likewise proportional
no relative motion between the two surfaces in the hill along line CD. The deformations of the to the square of the speed, on the assumption of
contact, the friction is called static friction and cline. The component Wcos@ = N presses the
two surfaces together, the component F’ is di- two surfaces produce internal friction in the two a constant negative acceleration. Actually, how-
the frictional force can have any value from zero bodies. This topic wilt be explored again when ever, the deceleration accomplished by the brakes
up to the limiting value. rected down the plane, and the frictional force
F is directed up the plane. Just before sliding we study the elastic properties of solids. The force is smaller at high speed because of the effect of
For limiting friction (but not for all-static of rolling friction varies inversely as the radius heat upon the brake linings, so that the increase
friction) the same laws apply as for sliding fric- occurs, there is no acceleration, and F’ is bal-
anced by the friction-F.From Fig. 4-10, of the roller, and it is less the more rigid the in stopping distance with speed is even mort
tion except, of course, that referring to the speed
of sliding. The coefficient of static (limiting) fric-
surfaces. rapid than is indicated by theoretical consid-
Rolling friction is ordinarily much smaller erations.
tion is the ratio of the limiting frictional force
to the normal force.
IN ie than sliding friction. Sliding friction on the
axle
of a wheel is replaced by rolling friction throu Example In what distance can a 3,000-lb
gh
and hence the coefficient of friction is given by the use of roller or ball bearings. automobile be stopped from a speed of 30 mi/h
W= ©) (44 ft/s) if the coefficient of friction between tires
H, = tan 0 and roadway is 0.70?
For any two surfaces the coefficient of static fric- 4-12 The retarding force furnished by the roadway
tion u, is somewhat greater than the coefficient
4-11
_FLUID FRICTION can be no greater than
of sliding or kinetic friction py.
ROLLING FRICTION The friction encountered
ingthrough fluids and re
Solid objects i Fs = uN = (0,70)(3,000 1b) = 2,100 Ib
Rolling friction is the resistance to motion caused m cdg ges
4-10 : within liquids and gases in
chiefly by the deformation produced where a motion are examples Since the work done by this force is equal to th
LIMITING ANGLE (ANGLE OF REPOSE) of fluid friction. The laws
of fluid friction differ
wheel or cylinder pushes against the surface on kinetic energy of the car, the stopping distanc?
greatly from those of sliding and
The coefficient of static friction may be found which it rolls. The deformation of an automobile
The amount
rolling friction. can be found by substituting in Eq. (7).
without measurement of forces by the following tire in contact with the pavement is readily visi- of frictional resistance enco
by an object moving thro untered
simple and convenient method. Let the plane AC ble. Even in the case of a steel wheel rolling on ugh a fluid depends on
Fs = } mv?
 92 THE PHYSICS OF PARTICLES
WORK, ENERGY, AND POWER 91

= W =
_ 3,000lb_à
Bane? = 4 slugs may receive energy of the order of 10-1? J at the
expense of a decrease in the mass of the reactants E,= min
Pee Uae 94sings (44M/s)? an which is measurable with a mass spectrometer. E,
= max
In ordinary chemical reactions the energy re- Retarding force = 200 lb
F 2 X 2,100Ib
leased per molecule is so much smaller
(10-8J) that the corresponding changes in '
4-14 mass are not apparent. All other forms of energy Figure 4-12
CONSERVATION OF ENERGY may be similarly associated with mass changes, Figure for example in Sec. 4-15.
which are in most cases unmeasurably small.
Energy is given to a body or system of bodies
when work is done upon it. In this process there The motion of a pendulum furnishes another
is merely a transfer of energy from one body ever, the velocity is directed toward the left at 0,
to 4-15 simple example of energy transformations. A
another. In such transfer no energy is created Under the constraint of the string the ball will
or TRANSFORMATIONS OF small ball (Fig. 4-13) of mass m is suspended
destroyed; it merely changes from one form to continue to move along the arc BOA, gaining
KINETIC AND POTENTIAL ENERGY from a fixed point P by a string of length /. When potential energy at the expense of its kinetic
another. This statement is known as the law of
Very frequently in mechanical systems at low the ball is pulled aside from O to position B, it energy as it approaches A. If no energy is lost `
conservation of energy. It is true that in most
speeds there is an interchange of kinetic and is raised a distance h and hence given potential to its surroundings, the ball will reach point A
processes some of the energy becomes unavail-
potential energies. If a ball is held at the top of energy mgh. When the ball is released, it moves at a height A above its lowest position. It will then
able. Work done against friction is converted into
a building, it possesses potential energy. When toward its lowest point and its energy while re- retrace its path AOB, and the motion will be
heat energy in such a form that it can seldom
it is released and falls, the kinetic energy in- maining constant changes from potential to ki- repeated.
be used. Thus, although the energy is not de-
stroyed, it is wasted as far as its usefulness in the creases as the potential energy decreases. The netic, the sum of the two forms always being
sum of E, and E, remains constant and equal equal to mgh. At point O all the energy will be
process is concerned.
kinetic. The ball will have a speed v obtained
Example The bob of a pendulum has its rest
It was mentioned in Chap. 3 that if a body to the potential energy at the top if no energy point 1.00 m below, the support. The bob is pulled
is accelerated to a speed approaching the speed is lost against air resistance. from aside until the string makes an angle of 15° with
of light its mass increases appreciably. As a con- the vertical. Upon release, with what speed does
sequence of this fact Eq. (6) for kinetic energy Example A 3,000-Ib automobile at rest at the mgh = 4mv? (11) the bob swing past its rest point?
Ceases to be accurate for such high-speed bodies, top of an incline 30ft high and 300ft long is From the geometry of Fig. 4-13,
since this equation was derived on the assumption released and rolls down the hill. What is its speed or v = V2gh (12)
that the mass remains constant. Taking the mass at the bottom of the incline if the average retard-
ing force due to friction is 200 lb? (Fig. 4-12.) This is the speed it would have acquired if it had h=PC1—
= l— lcos 15° = I (1 — cos 15°)
variation into account consists not merely in sub-
Stituting the relativistic mass in Eq. (6) but in The potential energy at the top of the hill is fallen freely through a vertical distance h. How-
available to do work against the retarding force = (1.00 m)(1.000 — 0.966) = 0.034 m
deriving the equation from the beginning with
mass variable. This yields for the kinetic energy F and to supply kinetic energy.
of a high-speed particle v= V2gh= V2 X 9.80 m/s? x 0034m
Wh = Fs + 4mv?
E, = mc? — mc? (10) = 0.82 m/s
W _ 3,000Ib_
mie Saat a

where m is the mass at the high speed, mp is the

3,000 Ib x 30 ft = 200 Ib x 300 ft 4-16
mass when the body is at rest, and c is the speed
of light. The quantity myc? is called the rest energy POWER
+ $ X 94 slugs
x v?
of the body, and the quantity mc? is called its In science and technology the word power is tê-
9.0 x 104 ft-lb — 6.0 x 10* ft-lb
total energy. This equation may be written E, = stricted to mean the time rate of doing work. The
(m — m)? = Amc? so as to associate an increase = } X 94slugs
X v? average power is the work performed divided by
of kinetic energy with an increase of mass Am. the time required for the performance. In meas- 1
2 — 3.0 X 104fi-lb _ 2 /e2
Einstein’s prediction of the equivalence of mass
and energy is readily verifiable in nuclear reac-
ide 47 slugs 640ft/s Figure 4-13 uring power P, both the work W and the elapsed
Pendulum; an example of time t must be measured
tions (Chap. 49). In certain reactions a particle v = 25 ft/s
energy transformations.
5
R = TOASUE
N
(3)
 THE PHYSICS OF PARTICLES
WORK, ENERGY, AND POWER 93
4-17 For example, if amachine is available that ena-
The same work is done when a 500-lb steel where @ is the angle between the force and the SIMPLE MACHINES bles a person to lift 500 1b by applying a force
girder is lifted to the top of a 100-ft building in velocity. of 251b, its actual mechanical advantage is
2 min as is done when it is lifted in 10 min. How- The mks unit of power is the joule per sècond, A machine is a device for applying energy to do 500 1b/25 Ib = 20. For most machines the AMA
ever, the average power required is five times as called the watt; the kilowatt (1,000 W) is also work in a way suitable for a given purpose. No is greater than unity. A machine that is designed
great in the first case as in the second, for the commonly used. The cgs unit, the erg per second, machine can create energy. To do work, a ma- to increase the force has an AMA greater than I;
power needed to do the work varies inversely as is inconveniently small. In the British system the chine must receive energy from some source, and
for example, a bench vise, a crowbar, or a block
the time. foot-pound per second and the horsepower are the maximum work it does cannot exceed the and tackle. A machine designed to increase speed
Enormous amounts of power are used in many common units of power. The latter was defined energy it receives. has an AMA less than 1; for example, a catapult,
events that last only very short times, as in the by James Watt in connection with his engineering Machines may receive energy in different a fly casting rod, the gears in a hand-operated
case of lightning. In some thermonuclear experi- studies on the steam engine. One horsepower is forms, such as mechanical energy, heat, electric
beater, or the chain drive ofa bicycle. A machine
ments moderate amounts of energy are released defined as 550 ft-Ib/s. energy, or chemical energy. We are here consid- designed to simply change the direction of the
in a.few millionths of a second to produce tre- Since work is the product of power and time, ering only machines that employ mechanical en-
mendous amounts of power. For example, in a any power unit multiplied by a time unit may. applied force has an AMA of 1. An example of
ergy and do work against mechanical forces. In
plasma experiment called Zeus, energy from a be used as a unit of work. Commonly used units this would be a machine consisting of a rope
the so-called “simple machines,” the energy is
capacitor bank is released in a brief pulse with of work formed in this manner are the watt- which is attached to an object to be lifted and
supplied by a single applied force and the machine
a power comparable with that consumed nor- second (joule), the. watthour, the kilowatthour, does useful work against a single resisting force. then passed over a beam, or a single fixed pulley.
mally in the entire United States. and the horsepowerhour. With the force being exerted downward on the
The frictional force which every machine en-
Since work is frequently done in a continuous counters in action and which causes some waste rope, the resultant force would be of the same
fashion (or energy is transported in a continu- Example By the use of a pulley a man raises magnitude but upward.
of energy will be neglected for simplicity in treat-
ous stream), another expression for power a load of 50 kg to a height of 15 m in 65 s. Find
ing some of the simple machines. Most machines,
(work/time) is useful; thus the average power required.
no matter how complex, are combinations of two
5 aia peal ht 50 kg x 9.8 m/s? x 15m or more simple machines. 4-19
F's _ ,,/As There are really two major classes of simple IDEAL MECHANICAL
Pia
At me ($) t t 65s ADVANTAGE
machines, the lever and the inclined plane. How-
= 113W ever, these usually have been modified into more In any machine, because of the effects of friction,
since As/Ar represents the velocity of the body specialized simple machines so that it may be the useful work done by the machine is always
on which the force is applied, Example What power is needed to move a considered that there are the following six simple less than the work done on the machine. The
3,000-Ib car up an 8.0° incline with a constant machines: the lever, the pulley, the wheel and input work done by the applied force F, is meas-
P=F-y (14a) speed of 50mi/h against a frictional force,of axle, the inclined plane, the screw, and the wedge. ured by the product of F, and the distance 5;
80 1b? through which it acts. The output work is meas-
or in scalar terms
By vector solution the downhill component of ured by the product of the output force F, and
the car’s weight is ¢ the distance s, through which it acts. Hence
P = Fv cos 0 (14b) 4-18
Fp = 3,000 Ib x sin 8.0° = 430 Ib
ACTUAL MECHANICAL ADVANTAGE Fs. < Fis;
Table 3
The total force needed is The utility of a machine is chiefly that it enables If we divide each member of the inequality by
UNITS OF POWER a person to perform some desirable work by F,s,, we obtain
F = 430lb + 80 lb = 510lb changing the amount, the direction, or the point
1 watt (w) = 1 newton-meter per second (N-m/s) = of application of the force. The ratio of the output
l joule per second (J/s) = 107 ergs per second The power needed is given by force F, exerted by the machine on a load
to the
Fo es
(erg/s) input force F, exerted by the operat Fs,
or on the
1 horsepower (hp) = 550 foot-pounds per second i 80 ft/mi machine is defined as the actual mechan that is, the ratio of the forces F,/F, is less than
(ft+Ib/s) = 33,000 foot-pounds per minute P = Fv = 5001b x 50 mi/h « vantage (AMA) of the machine.
ical ad-
the ratio of the distances 5,/s, for any machine.
(ft-1b/min)
= 3.7 x 10! ft-lb/s If the effects of friction are very small, the value
1 horsepower = 746 watts of the output work approaches that of the
3.7 x 10 ft:1b/5 _ 68 hp F input
T work, or the value of F,/F;
l kilowatt = 1,000 watts = 1.34 horsepower AMA = —2
= -550 (ft-1b/s)/hp becomes nearly that
of s;/s,. The ideal mechanical advantage (IMA)
 96 THE PHYSICS OF PARTICLES

WORK, ENERGY, AND POWER 95 Energy input = energy output + energy wasted rately for the input distance s; and the si
distance sọ. The ratio of these is the ideal me
is defined as the ratio of the distance s, throug
h
Fe
S eg e
SS - assuming no energy is stored in the machine. chanical advantage.
which the input force acts to the distan
ce s, The efficiency of a machine is defined as the
through which the output force acts, Example A 60-Ib sled is pulled up an
inclined)
ratio ofits output work to its input work. This ratio
is always less than 1, and is usually multiplied plane 50ft long and 30ft high by a boy why
IMA
=2 by 100 percent and expressed in percent. A ma- exerts a force of 45 Ib parallel to the plane, (j
So chine has a high efficiency if a large part of the What is the efficiency of this plane? (b) Whi
energy supplied to it is expended by the machine force would the boy have to exert if the plant
Since the forces move these distances in equal
times, the ratio s,/s, is also frequently called on its load and only a small part wasted. The were frictionless?
the efficiency (eff) may be as high as 98 percent for
velocity ratio. In a “frictionless” machine
inequality of the ratio of the forces to the
the
ratio
a large electric generator and will be less than pq = Wout _ 601b x 30 ft = 0.80 = 80%
of the distances would become an equality. Figure 4-14 X
50 percent for a screw jack. Ea 451b x 50ft
Each of the six simple machines has a predic
able ideal mechanical advantage based on
t- The human forearm as a simple
machine. Ef = output work _ F,s,
Fraon= Fotu XEff
= 451b x 0.80
=3610
the
physical structure of the machine. For exampl Sn
in the lever, which is a long rigid rod design
e, input work — Fs, An alternative solution for this force is
ed
to balance on some balancing edge, a
fulcrum,
the IMA may be predicted if we know the ratio The human forearm is thus designed to give Also since
of the length of the effort (input) arm to the mobility and speed to the hand and not lifting Fu W sin = 601b x 2°= 361b
power. The biceps, contrary to popular belief,
length of the resistance (output) arm. IMA
= especially by those who work at developing large
Eso _ E/R
length of effort arm/length of resistance.
In real-
ity although one of the most efficient machines, biceps muscle, is not designed to give strength to Fis, 3/5
the lever is not rigid, so it will bend; and there the arm. Other muscles with other mechanical _ AMA SUMMARY
may be some friction at the fulcrum; and, further arrangements, such as the triceps in back of the a IMA -
, arm, will provide the necessary lifting strength. Work is the product of force and the displaceme
the rod may not be of uniform density—all fac-
in the direction of the force:
tors which make the machine less efficient.
Yet Note that the work input times the efficiency is
the use of the rule of thumb for determining Example A pulley system is used to lift a equal to the work output
the W = Fs cosl
IMA of machines does permit the making of 1,000-Ib block of stone a distance of 10 ft by the
helpful approximations. application of a force of 150 Ib for a distance of
We saw earlier that some machines have me- (EAN(F;s,) = F,s, Energy is the capacity for doing work. 7
80 ft. Find the actual mechanical advantage and
chanical advantages less than’ 1. Using the rule- the ideal mechanical advantage. The foot-pound is the work done by 4 fort
of-thumb method we can show that the human Example What is the efficiency of the pulley of 1 lb exerted through a distance of | fl
forearm is a lever and can make some prediction system (described in the previous example) which
The joule (newton-meter) is the work
MESTF, _ io
1,0001b _=
about its IMA. Figure 4-14 shows that the biceps lifts a 1,000-1b block of stone a distance of 10 ft a force of 1 N exerted through a distance ofle
īa

muscle is attached a short distance in front of the 6-67 by the application of a force of 150 1b for a dis-
elbow on the forearm. The length of the effort A joule is 107 ergs.
s 80ft tance of 80 ft? The erg is the work done by a force of |dt
arm is the distance from the elbow to the point
— exerted through a distance of 1 cm. yi
of contact of the biceps with the forearm, This
par= Fse — (1,000 1b)(10 ft)
distance is small compared with the length of the
Tesistance arm, the distance from the
Fe = “Tsoyada. = 08 = 8% Energy is that property of a body oF ph
system of bodies by virtue of which work cati
elbow to
the hand. done.
4-20 Also
Therefore, since Potential energy is energy of position orof
EFFICIENCY
figuration. For gravitational potential energi
Because of frictional losses and other losses in
length of effort arm
~ length of resistance arm all moving machinery, the useful work done by Ef = AMA _ 667 _ 0.83 = 83%
a machine is less than the energy supplied to it. E, = Wh = mgh
and IMA < 1
From the principle of conservation of energy, To calculate the mechanical advantage of a
Kinetic energy is energy of motion.
machine, one can imagine it to have carrie
d out
a chosen motion. Expressions are written sepa-
E = 4 mv?
 98 THE PHYSICS OF PARTICLES

WORK, ENERGY, AND POWER 97

11 If increasing the speed ofa body increases Which is greater in each case, the applied force
its mass, does heating a gas cause a mass increase? or the force exerted by the machine?
For transformations between work and kinetic The ideal mechanical advantage is the ratio of 12 How can one calculate the minimum speed 27 If a block rests on a plane and the planeis
energy, the distance s; through which the input force acts gradually tilted until the block just begins to slip,
with which a projectile would have to be fired
to the distance s, through which the output force the angle of inclination of the plane is called the
Fs = 4mv?
acts: vertically in order to escape from the earth?
13 Name several types of mechanisms in which angle of limiting repose. Show why this angle?
Energy can be neither created nor destroyed, friction is essential for proper operation. isgiven by tan @ = p.
only transformed. This is the principle of the con- IMA
= 14 What becomes of the energy expended
S,
servation of energy. against friction?
Frictional force F is proportional to the nor- The efficiency of a machine is the ratio of the 15 Why should one take short steps rather than
mal force N pressing the two surfaces together Problems
output work to the input energy. long ones when walking on ice?
and is directed parallel to these surfaces. 16 An automobile is moving along a concrete 1 A loaded cart has a total mass of 227 kg.If
toad with the same speed as that of a streetcar a 312-N force acting at an angle of 30° to the
F=,pN
Questions alongside it. Which vehicle can stop in the shorter ground is applied, how much work is done in
The coefficient of friction is defined as the ratio
distance? (The coefficient of friction for rubber moving the cart 15 m? |
1 A man is supported on a ladder which is on concrete is 0.7, for steel on steel 0.2) 2 A force of 50N exerted over a distance of |
of the frictional force to the normal force. leaning against a wall. Is the wall doing any work 17 A body rests on a rough horizontal plane. 3.0m causes a block having a mass of 20 kg |
on the ladder? Show that no force, however great, applied to- move 1.0 m. Find the AMA; the IMA; the efi-
F F,
Me = and 1 = 2 If you try for 20 min to move a heavy object ward the plane at an angle with the normal less ciency. Ans. 2.5, 3.0, 83 percent |
but cannot make it move, have you done any than the limiting angle for friction, can push the 3 A safe weighing 10 tons is to be loaded on |
work on the object?
The coefficient of static friction p, may be 3 Aman rowing his boat upstream is just able
body along the plane. a truck 5.0fthigh by means of planks 20 ftlong”
calculated from the limiting angle of repose, @. to hold his position with respect to the shore. Is 18 Can a moving automobile be stopped in If it requires 350 Ib to overcome friction on tht”
Rolling friction is the resistance to motion he doing work? If so, on what? shorter distance (a) by applying the brakes to skids, find the lehst force necessary to move tht |
caused chiefly by the deformation produced 4 Account for the energy possessed by the “lock” the wheels or (b) by applying a braking safe.
where a wheel or cylinder pushes against the water stored above a dam. force just short of that which causes the tires to 4 A man weighing 450N sits on a platform
surface on which it rolls. 5 Which performs work: the hammer or the slip? Explain. suspended from a movable pulley and raisi
Fluid friction is resistance encountered by nail? the powder or the bullet? the catcher or the 19 Distinguish carefully between doing work himself by a rope passing over a fixed pulley. |
solids passing through fluids and the friction set baseball? the baseball or the bat? Explain your and the exerting force. Give examples. Assuming the ropes parallel, what force does he
up with liquids and gases. answers. 20 Show that nq work is done on a body that exert? (Neglect the weight of the platform.)
Power is the time rate of doing work. 6 Two boys are throwing and catching a ball moves with constant speed in a circle. Ans. 150N |
on a moving train. Does the kinetic energy of the 21 In a tug-of-war, team A is slowly giving 5 A box weighing 150 1b is moved acros
P= = = Fucos@ ball at any instant depend on the speed of the ground to team B. Is any work being done? If horizontal floor by dragging it by means of à 10
train? Explain. so, by what force? attached to the front end. If the rope makes #
7 Show that during the motion of a simple 22 Does the use of a lever increase one’s power?
A horsepower is 550 ft-1b/s. pendulum the work done by the tension in the
angle of 30° with the floor and if the coeffiiet
23 What kind of machine would you select if of friction between box and floor is 0.400, fit
A watt is 1 J/s. string is zero. you desired one having a mechanical advantage
One horsepower is equivalent to 746 W. 8 Show that when a body of mass m is dropped the force exerted by the man pulling the 19
of 2? of 500 or more? Which machine would be
A machine is a device for applying energy to from a heigl h, the sum of its kinetic and poten- likely to have the greater efficiency if both ma-
6 A1,000-N piano is moved 20 m across 4 fot”
do work in a way suitable for a given purpose. tial energies at any instant is constant and equals chines were as mechanically perfect as it is possi- by a horizontal force of 350 N. Find the coef
In a simple machine the energy is supplied by mgh. cient of friction. What happens to the ene
a single applied force and the machine does use- ble to make them? ~ expended? A Ans, 0%
H Trace the changes in the energy of a roller-
24 Why does a road wind upa steep hill instead
ful work against a single resisting force.
The actual mechanical advantage of a machine
coaster car.
of going directly up the slope? 7 A 350-8 block of wood on a horizontal plat!
is fastened to a cord passing over a friction
10 A 90-kg man is seated in an automobile
25 How does the IMA of an inclined plane vary
is the ratio of the output force to the input force: moving at a speed of 80 km/h. What is his kinetic
with the angle @ of the plane; that is, what trigo- Pulley and attached to a 265-g load. The c0? i
energy relative to his fellow passengers in the car? cient of kinetic friction between block and plat!
Does he have this same kinetic energy relative nometric function of @ gives the value of IMA? _
AMA =EF, to the ground? 26 Describe several machines in which the load is 0.45. (a) Determine the acceleration ° the
Moves at a greater speed than the applied force, system after it is set in motion. (b) What !
force in the cord?
 WORK, ENERGY, AND POWER
THE PHYSICS OF PARTICLES

8 A sled weighing 100 lb reaches the foot of and at point C, which is 50 ft farther along the its mass increase over its rest mass of
a hill with a speed of 40 ft/s. The coefficient of track and 20 ft higher than point B? 1.67 X 107?" kg and its relative mass gain.
kinetic friction between the sled and the horizon- Ans. 43.3 ft/s at B; 23.8 ft/s at C. 20 A loaded sled weighing 1,250 Ib is given a
tal surface of the ice at the foot of the hill is 0.030. 17 A prony brake is a device often used to speed of 25.0 mi/h while moving a distance of
How far will the sled travel on the ice? measure the power of an engine. In a small motor 140 ft from rest on a horizontal ice surface. If the
Ans. 830 ft. the brake may be a strap passing halfway around coefficient of friction is 0.105, what constant force,
9 What average net thrust must a 17-ton air- the pulley of the motor, with the ends of the strap applied horizontally, would be necessary to pro-
plane have to reach an altitude of 5,000 ft and supported by spring balances (Fig. 4-15). In one duce this motion? Ans. 320 Ib.
a speed of 600 mi/h at an airline distance of 21 An electron whose mass is 9.11 x 10-3! kg
10 mi from its starting point? is moving at 3.00 x 10° m/s. What constant force
10 An automobile traveling at 50 mi/h on a is required to bring it to rest in 1.5 x 107° cm,
level road is stopped by sliding the wheels. If the as might occur at the target of an x-ray tube?
coefficient of kinetic friction between tires and 22 The coefficient of friction between a block
road is 0.75, what is the minimum distance in and the surface on which it slides is 0.18, and
which the car can be brought to rest? the surface is inclined at an angle of 15° with
Ans. 112 ft. the horizontal. If the block weighs 300 1b, what 0.100, how far will the box travel, and how lo
11 A panelboard sheet hangs in a vertical posi- force is needed to drag it up the incline at uni- a time will elapse after it reaches the level surf
tion while gripped between the jaws of a clamp. form speed? What force is required to let it slide before it comes to rest? )
One side has a coefficient of static friction of 0.30 24 A coin lying on a meterstick starts to si
with the clamp and the other a coefficient of 0.45. down the plane at uniform speed?
Ans. 130 lb; 26 Ib. when one end of the stick is raised 36.4 cm ab
How much horizontal force must each face of the
23 A 100-Ib box slides down a skid 20.0 ft long the horizontal. (a) What is the limiting angle!
clamp exert if the sheet weighs 150 N? How much
and inclined at an angle of 60° to the horizontal. friction? (b) What is the coefficient of starti
vertical force will each face of the clamp then
exert?
At the bottom of the skid the box slides along friction? Ans. 21°; 03
Figure 4-15
12 A 3,200-lb car starts to roll from the top of A prony, or band, brake.
a level surface of equal roughness. If the co- 25 In the pulley system shown in Fig. 4-168
a 400-ft hill which is 500 ft long. How fast is it efficient of kinetic friction for the surfaces is far must the applied force move if the load mo}
going when it reaches the bottom? Ans. 160 ft/s. 1.00 ft? If the load weighs 200 Ib and the @
13 A man lifts a 100-N can of oil by pressing case a pulley had a diameter of 45.2 cm, and the ciency of the system is 90 percent, what musti
his two hands against the smooth sides, toward balances indicated a net difference in force of the applied force?
each other. If the coefficient of static friction 0.525 N, when the motor speed was 500 r/min. 26 Seven pulleys, three movable and four iX®
is 0.30, what force must he apply with each The power input was 16.7 W. What was the effi- are connected by a single cord. What resistat
hand? i ciency? will a 100-N effort move with this machine,
14 A worker can supply a maximum pull of 18 A rocket rises to a height of 20.0 mi, 200 mi,
150 Ib on a cart which has a handle set at an angle 2,000 mi, and 20,000 mi above the surface of the
of 40° above the horizontal. What is the coeffi- earth on a flight toward the moon. If it weighs
cient of friction if he can just move a total load 3,000 Ib at the surface of the earth (at a radius
of 1,500 1b? Ans. 0.082. of 4,000 mi), what is its, potential energy at each
at of these altitudes? (For which cases must the
at the end of the lever arm, move in one tui
15 A 445 x 104N_ truck is traveling
variation of gravity with altitude be taken into What is the ratio of the load to the applied fos
100 km/h. If the coefficient of friction between (called the actual mechanical advantage)?
the tires and the roadway is 0.50, what is the account?) G = 3.44 x 10-8 ft8/(slug)(s*); M, =
minimum distance the truck will go before stop- 4.10 x 1078 slugs. Ans, Taking Ep = 0 at 28 A pulley system has a mechanical advant
ping? earth’s surface: 3.17 x 108 ft-Ib; of 5and is used to lift a load of 1,000 1b. Ifi
3.00 x 10° ft-lb; 2.08 x 101° ft-lb; effort moves 10 ft, how far does the load mi
16 A roller coaster which weighs 640Ib starts
5.25 x 10" ft-lb. If this work was done in 15 s, what horsepo™
from rest at point A and begins to coast down
the track. If the frictional force is 5.0 lb, how fast
was developed? If the machine was 80 pert
will the coaster be going at point B, which is 19 A proton is accelerated to a kinetic energy efficient, how much effort would be needed toñ
100 ft down the track and 30ftbelow point A, of 1.60 x 10-®J in a modern ‘synchrotron. Find the load? Ans. 2 ft; 0.243 hp, 20
Block and tackle. 29 For the compound machine of Fig. 4-18)!
4
radii of the wheel and axle are 18 in and 60%
 WORK, ENERGY, AND POWER 101

respectively. If the load is lifted 2.00 in, through

what distance doés the applied force move? What
would be the ratio of the load to the applied force
if the efficiency were 100 percent?
30 A man raises a 500-N stone by means of a
lever 5.0 m long. If the fulcrum is 0.65 m from
the end that is in contact with the stone, what
is the ideal mechanical advantage? Ans. 6.7.
31 The radius of a wheel is 1.8 ft, and that of
the axle is 2.5 in (Fig. 4-19). What force, neglect-
Figure 4-18 ing friction, must be applied at the rim of the
A compound machine.
wheel in order to lift a load of 800 1b which is
attached to a cable wound around the axle?
32 Using a wheel and axle, a force of 30 1b
applied to the rim of a wheel can lift a load of
240 Ib. The diameters of the wheel and axle are
3 ft and 4in, respectively. Determine the effi-
ciency of the machine. Ans, 89 percent.
33 A load of 400 N is lifted by means of a screw
whose pitch is 5.0mm. The length of the lever
is 20 cm, and the force applied is 5.0 N. Deter-
mine the efficiency of the machine.
34 Out of a total of 300 ft-lb of work put into
a machine, an amount equal to 75 ft-lb is lost
in overcoming friction. What is the efficiency of
this machine? Ans. 75 percent.
35 A motor operates at its rated load of 10 hp
for 8.0 h a day. Its efficiency is 87 percent. What
Figure 4-19
is the daily cost of operation if electric energy
Gabriel Lippmann, 1845-1921
Wheel and axle.
costs 5 cents/kWh?
Born in Hollerich, Luxembourg. Director of the
physical laboratory at the Sorbonne, Paris. A
the 1908 Nobel Prize for Physics for his methodof
photographic reproduction of colors, based upon
the phenomenon of interference.
 104 THE PHYSICS OF PARTICLES

action of the force. This distance is called the

moment arm of the force. In Fig. 5-3, the momen} |
arm of the force F is indicated by the dotted line|
Torque s = OP. The line of action of the force is a mere |
geometrical construction and may be extended
indefinitely either way in order to intersect the
perpendicular OP. It has nothing to do with the
length of the force vector. The force F in Fig
5-2 produces no change in the rotation when il
is applied at C, because its line of action passes
through the axis of rotation and its moment arm
An unbalanced force produces rotational is therefore zero. The same force applied at A
acceleration if its line of actidn does not pass has a moment arm OA and, therefore, tendsto
through the axis of rotation. change the rotation.

When a body is in equilibrium, it is either at rest 5-3

1
oe
It is a common experience that the wheel can be
or moving uniformly, Forces may act either to
set into motion more easily and quickly by apply- TORQUE
change the linear motion of a body or to change
the rotation of the body, or both. If all the forces ing a force F perpendicular to a spoke at some For a fixed moment arm, the greater the fore
acting upon the body intersect at a common point (a) (b) point A far from the axis than by applying the the greater the effect upon rotational motion. The
and their vector sum is zero, creating a closed same force at a point B nearer the axis. The effect two quantities, force and moment arm, are 0
Figure 5-1
polygon, they have no tendency to change either Forces which are equal in magnitude and
of a given force upon the rotational motion of equal importance. They can be combined into4
translation or rotation. opposite in direction produce equilibrium when a body is greater the farther the line of action of, single quantity, torque (also called momentof
Since most bodies are acted upon by forces they have a common line of action (a) but do not the force is from the axis of rotation. The distance force), which measures the effectiveness of tht
that do not act through a single common point, produce equilibrium when they do not have the of the line of action from the axis is measured force in changing rotation about the chosen axis
we must consider the effect of each force in same line of action (b). perpendicular to the line of action of the force. Torque will be Tepresented by the symbol L.
changing the rotation, as well as its effect in It is not merely the distance from the axis to the The torque (moment of force) about a choset
changing the linear motion of the body. The same point of application of the force. If the same force
force applied at different places or in different axis is the product of the force and its moment arm
zero; yet it is plain that under the action of these F of Fig. 5-2 is' applied at C rather than at A,
-directions produces greatly different rotational two forces, the block will rotate. In fact, when where OC = OA, there is no effect upon
effects. Therefore, in studying equilibrium, we the vector sum of the applied forces is equal to rotation of the wheel, since the line of action of
the LasF ()
must consider the place at which a force is ap- zero, we can be sure only that the body as a whole
F passes through the axis and it merely pulls the where s is the perpendicular distance from the
plied as well as its magnitude and direction. will have no change in its linear motion; we
wheel as a whole upward. Though the magnitude
cannot-be sure that there will be no change in
of the force, its ah and the distance of its
its rotary motion. Hence complete equilibrium is
not assured. In addition to the first condition for point of application from the axis are the same
5-1
equilibrium previously stated (Chap. 1), a second in the two examples, rotation is affected when the
CONDITIONS FOR EQUILIBRIUM
condition is necessary, a condition eliminating the force is applied at A but it is not affected when
Consider an arrangement in which two opposing possibility of a change in rotational motion. The the force is applied at C.
forces equal in magnitude act on a block as in example of Fig. 5-16 indicates that this second
Fig. 5-la. It is obvious that if the block is origi- , condition is concerned with the placement of the
nally. at rest it will remain so under the action forces, as well as their magnitudes and directions.
of these two forces. We say, as before, that the In order to study the factors that determine 5-2
vector sum of the forces is zero. the effectiveness of a force in changing rotational MOMENT ‘ARM
Now suppose that the forces are applied as in motion, consider tue familiar problem of turning (a) (b)
The factor that determines the effect of a given
Fig. 5-1b. The vector sum of the forces is again a heavy wheel by pulling on a spoke (Fig. 5-2). force upon rotational motion is the perpendicula
r Figure 5-3
distance from the axis of rotation to the Measurements of moment arm.
103 line of
 x

TORQUE 105
in torque, force and distance are at right angles. dot) product and the vector (or cross) product
Any similar combination of force and distance The scalar product is the product of the magnitude
units makes a suitable unit for torque. of one vector by the magnitude of the component
of the other vector quantity in the direction of
the first. (See Fig. 5-5a.)
5-4 To find the scalar product of vector quantities
VECTOR NATURE OF TORQUE A and B in Figure 5-5a, the component of B along
A is found to be B cos8. Therefore, A+B = AB
In our discussion of the addition of vectors we cos 8. Since A is only the magnitude of A and
established the requirement that the vectors that B is the magni tude
of B and the cosine 0isa
were being added be similar in nature. That is, pure number, this product is a scalar quantity.
$ (b) BE = AB cos 30° force vectors were added to force vectors and It is seen that any relation involving the cosine
`s
L = (AB cos 30°)(DB) velocity vectors to velocity vectors. We will see, of an included angle may be written in terms of
Ny’
however, that it is possible to multiply vectors of the scalar product. For example, we have seen
C different kinds. We should be aware that since that the mechanical work ‘W done by a force F
(a) CD = DB cos 30° vectors have direction as well as magnitude, which makes an angle @ with the displacement
L = (10N)(DB cos 30°) vector multiplication does not follow exactly the s is W= Fs cos @ or, in vector notation,
algebraic rules that the multiplication of scalars W = Fes. Other scalar products appearing in
Figure 5-4 follow. There are three types of vector- a study of physics are electric power and electric
Finding the torques by two procedures. multiplication situations that may occur: (1) the potential, gravitational potential energy, and
multiplication of a vector by a scalar, (2) the electromagnetic energy density.
multiplication of two vectors in such a manner _ The vector product of two vectors A and B
axis to the line of action of the force. An axis Therefore, as to produce a scalar, and (3) the multiplication iswritten as AX B and produces another vector
must always be selected about which torques are of two vectors so that ånother vector is produced.
to be measured. The value of the torque pro- C. A x B = C.The magnitude ofC is foundto
L = (0.5 m)(10 N) = 5 The multiplication of a scalar, K, by a vector, be AB sin 8. The proof of this can be foundin
duced by a given force depends, of course, upon A, produceś another vector which has the sąme several mathematics books.! The direction of C
the axis chosen. . ne selection of an axis is quite A second procedure involves taking the verti- direction but a magnitude of KA. For example, ee ll
arbitrary; it need not be any actual axle or ful- cal component of the 10-N force. (Fig. 5-45). 2 XA = 2A. 1H, Margeneau and G., M: , “The Mathematics of
crum. In many cases, however, a wise selection From the figure it can be seen that When we multiply two vector quantities, it is
of the axis about which torques are to be calcu- necessary to distinguish between the scalar (or Physics and istry,” D Va Nostrand Company,
Inc., New York, 1957, p. 143.
lated greatly simplifies a problem, because it re- BE
cos 30° =
duces to zero the torque due to a force whose 10 N
magnitude or direction is unknown.
BE = (10 N)(cos 30°) = (10 NX(0.5) = 5N
Example Find the torque created by a 10-N
Therefore,
force acting 60°N of E at a distance of 1 m from
the axis of rotation of a lever. L = (1 m)(5N)
=5
Two approaches can be used. First, find the
perpendicular distance from the fulcrum to the Since torque is a product of a force and a
line of action of the force by extending the line distance, its unit is a force unit times a distance
of action AB and then dropping a perpendicular ` unit, such as the pound-foot, the usual unit in
to that line (Fig. 5-4a). Then from the trigo- the British system. An mks unit of torque is the
nometry of a right triangle, meter-newton. The cgs unit of torque is the centi-
meter-dyne. Because the units here formed are . Figure 5-5 č
(a) Scalar (dot) product of two vectors. (b) Vi
coss 30
30° =
CD
DB
== products of force units and distance units, as are vector quantities.
(cross) product
of two
also the units of work, the order of the units is
CD = DB cos 30° = (1 m)(0.5) = 0.5m here interchanged to call attention to the fact that
 108 THE PHYSICS OF PARTICLES

TORQUE 107
=F, =0 E)
is perpendicular to the plane of A and B. (See ZF, =0 @)
Fig. 5-5b.)
The direction of C is determined by the so- =F, =0 6)
called “right-hand rule,” actually the rule of a or =F =0
right-hand screw. If the fingers of the right hand
6)
are wrapped about an imaginary axis perpen- For an object to be in equilibrium under the
dicular to the plane of A and B, the fingers indi- action of a set of forces, the sum of the torques
cating the direction in which A will be turned into (about any axis) acting upon the body must
B through an angle of less than 180°, and the be
zero, This statement is known as the second con-
thumb kept erect, the direction of the thumb Figure 5-6 dition for equilibrium. It may be represented
gives the direction of the vector product A x B. A clockwise torque r x F. The vector representing by
the torque is perpendicular to the plane of the
the equation
From ‘this sign convention, it follows that the Figure 5-7
vector product B X A has the same magnitude Paper and directed into the paper. The magnitude
of the torque is rF sin8,where s = rsin ô is the Illustration of concurrent force
s. =L=0
as AXB but has opposite “direction, Sea (7)
Ax B= -B X A.
moment arm. EY
produced by each force of such In the first and second conditions we
Vector products are often encountered in a set is zero. If have a
any other axis is chosen in plac complete system for Solving problems
physics, for example, in angular momentum, the e of that through involving
force on a moving charge in a magnetic field, and
torque is not known from the direction of the the common point, the sum: bodies in static equilibrium. These same
force alone: of the torques will condi-
the topic we are presently considering, torque. not, in general, be zero. For tions are useful in certain problems involving
the Special case in iform motion. If the first condition is satisfi
In Fig. 5-6 we represent the two vectors in- which the resultan t of the conc ed,
volved in a torque, the force F and the length
Example A light horizontal bar is 4.0 m long.
zero, the sum of the torques urrent forces is the vector sum of the forces is zero
and no trans-
vector r from the axis to the point of application
A 3.0-N force acts vertically upward on it 1.0 m
zero. Hence a cons
about any axis is lational acceleration is produced. If the second
from the right-hand end. Find the torque about ideration of torque is not nec-
of F. The vector product is essary in the study of a set Condition is satisfied, the vector
each end. of concurrent forces sum of the
in equilibrium, torques is zero and there is no rotational accele
Since the force is perpendicular to the bar, the ation. This means not that there is r-
L=rxF (2) moment arms are measured along the bar. no motion but
only that the forces applied to the
About the right-hand end, NO change in its motion. While inbody produce
which is represented by a vector of magnitude equilibrium,
rF sin 0 perpendicular to the plane of r and F L, = 10m x 3.0N =3.0m-:N clockwise the body may have a uniform motion including
and directed into the paper. We observe that 7 both translation and
rotation.
sin ô = s is the moment arm. The consequences About the left-hand end,
of the vector nature of torque will be discussed forces. This relation
in more detail later in connection with rotary is expressed in the 5-7
L,=3.0m x 3.0N = 9.0m-N condition for equilibri second
um, j
Motion. For the present we shall confine our
counterclockwise
CENTER OF
attention to cases in which all the forces act in GRAVITY; CENTER
the same plane. For these cases the axes, and 5-6 OF MASS
The torques produced by this single force
therefore the torques, are parallel, and only the THE TWO The most common force acting upon a body is
about the two axes differ in both magnitude and
algebraic signs of the torques need be considered,
direction. CONDITIONS FOR EQUILIBRIUM its weight. For every body, no matter how irregu-
The algebraic sign of such torques is deter- laritsshape, there exists a point such that the
mined by consideration of the direction of the We have previously considered (Chap.
1) the entire
at that weight may be considered
Totation the torque tends to produce. For exam- condition necessary for equilibrium
under the point. This point is calledas concentrated
action of concurrent forces the center of
ple, the torques in Fig. 5-3 tend to produce coun- 5-5 gravity of the body. This point must be that for
terclockwise accelerations about O, while the CONCURRENT AND
vector sum of all the forces acting shall
be equal which the sum of the torques about
torque in Fig. 5-6 tends to produce a clockwise to zero. This condition
must also be fulfilled
horizontal
NONCONCURRENT FORCES the forces are not concurrent, when snes through the center of gravity produced by
acceleration. One may refer to these torques as This first condition
Positive and negative, respectively. Note that a Concurrent forces are those whose lines of action may be expressed by the statemen the weights of the Particles that make up the body
given force may produce a counterclockwise intersect in a common point (Fig. 5-7). If an axis of
t that the pen
the components in any three perpendi must be equal
are coordin to zero. If x. and y, (Fig. 5-84)
ates of the center
torque about another axis. The direction of a passing through this point is selected, the torque directions shall be zero. of gravity, the y
sie component L, of torques (about an axis parallel
to the y axis) is
 TORQUE 109

acceleration due to gravity is everywhere the i lu in the equa

same in magnitude and direction, Eq. (10) is i cone tion ob-
e
obtained by dividing Eqs. (8) and (9) by g and i X T ' for equilibri
m(x2, y2) W m,
the center of gravity and the center of mass are $
coincident. ,
A knowledge of the position of the center of finding the ce
nter of
gravity is very useful in-problems of equilibrium, strated in the follow
for that is the point of application of the vector in J
representing the weight. It is never necessary and
seldom convenient to break the weight up into
parts when considering the effect of weight in a
torque equation.

Example A uniform bar 9.0 ft long and

Figure 5-8 weighing 5.0 lb is supported by a fulcrum 3.0 ft
Coordinates of center of gravity. from the left end as in Fig. 5-9. If a 12-lb load
4
is hung from the left end, what downward pull
at. the right end is necessary to hold the bar in
L, = (x — x,)mnyg + (x2 — X,)mog + ore: equilibrium? With what force does the fulcrum forces add
push up against the bar? f 8 Upwaat
rso
dme
= 2(x — x,)mg = 0 (8) —
Consider the bar as an object in equilibrium. A. Also si
where the x’s are the x coordinates of the parti- The first step is to indicate clearly all the. forces
cles, Similarly, for the x components of. the that act on it. Since the bar is uniform, its center D +30 = 2001
torque. of gravity is at its midpoint and hence the weight
of the bar, 5.0 lb, can be considered to be concen- m atA,the foll
trated at its middle. A 12-lb force acts downward `
owing
L: = (Vy — yedmsg + (Yo — Ye)Mog + ++
= 2(y — y,jmg = 0 (9) at the
at the
left end of the bar, a force R acts upward
fulcrum, and there is an unknown down-
This procedure could be extended to a third ward force F at the right end.
dimension (z) to find the torque about an axis The first condition for equilibrium indicates
Parallel to the z axis, L,. that the vector sum of the forces applied to the
The center of gravity may be either within or bar is zero, or that
Outside the body. If a single force equal to the
weight of the body and acting vertically upward R-— 121b — 5.0lb-
F= 0
could be applied at the center of gravity, it would
Support the body in equilibrium, no matter how Without further information we certainly can-
the body might be tipped about the center of
gravity.
The center of mass is the point about which
the product of the mass and the moment arm sum
up to zero. We may then write

(x, — XM, + (X2 — X,)mg + ++

121b R
= E(x —x,)m=0 (10)
and similar equations for y and z. Figure 5-9
The forces acting on a lever.
If, over the region occupied by the ee
body, the
 112 THE PHYS
OFIC
PARTICLE
SS

isunknown. The second condition for

equi
lib- can see that a right triangle
rium isanexcellent toolto empl is formed and
situa
oy insuch a
TORQUE tion, forifweuseanaxisthrough
111 0as theaxisabout which totake thepoint Th = 1,93
Ibcos
040* = 1,930 Ib x 0.7660 =
torques equal clockwise torques, torques, the
unknown force at thehinge has zero
system consisting ofthe metal sheet and moment arm + 1,480Ib
Px = (10.0 1b)(0.5 ft) + (3.0 Ib)( tached weights, the at
1 fi) and
20x =5 4+3 tension 7 intheropewithout
knowin
Example A Tope magnitude orthedirection ofthe g either the T = 1,93
xme soap (Fig: . 5-11) helps to~ support
a uniform 200-1b beam, 20 Thetorques about anaxisthr force at 0, Ibsi0n40*= 1,9301bx 0.6430 =
"20 bere long,one‘end. of ough 0 are
which ishinged atthewal
Therefore, the of which
l andthe other end
a 1.0-ton load. The ro boone clockwisetorque
+1,2401b
fulcrum will be placed on du
0.4ftto the right of side aline an angle of50°with the makes etothe weight of the
AB. wall and the beam makes
Viewing the sheet from side an angle of 60° with the wall. Determine
CD, we see that tension in the rope. the
(200 1b)(10 ftco
Py = (10 1b)(0.5 ft) + (6.0 Ib)( s30°)
2y =546=11 1.0 ft) beam, let us consider it as the ead
object in equilib- = (200 1b)(10 ft)(0.
am 3 rium. In addition tothe200-Ib 866) = 1,730 ft-I
straight and 2,000-1b forces b ZF, = —load — we
y=} Lossy down, there arethepullof
theropeon ight of boom +7,
thebeam and theforce F whi The clockwise torque +v=0
20 j ch thehinge exerts due to the load: —2,000 Ib — 200 1b
+ 1240Ib = —y
on the beam atthe wall.
Letus-not make the
Therefore, the fulcrum will (2,00016)20ftcos30
above side AC.
be placed 0.55f is Straight up or straight alon °) u = 960Ib
g the beam. A little
This point 0.4 ft to the right thought will convince us that = (2,000 1b)(20 ft)(0. 2h = +7, -—h=0
of AB and 0.55 ft the hinge must be 866) = 34,500 ft-l
above the side AC is the center of gravity of the pushing both up and out on
the b
direction of this force, as well beam. The exact The counterclockwi h= T, = 1,480 1b
as its magnitude, se torque due to
the tension: We now have the Ver
The mome tical and horizonta
l com-
= to be 2nt0coars20
m of T isOP wh ponents of F and by trigonometry, we see that
Si Eo °= 20(0,94) = icBhan $, the angle that F make
s wit h the horizontal, can
N88 ft) or 18.87; and, Ound because
since

tang = PO
18.87 + 34,600 + l, = 06s and
1,730 =0 $ = 33°

36,400
T= “Tea = 19301p Therefore, # = (960 Ib\sin
, Sh ould benoted that 33°) = 523
Fisnot acting up 1p. It
above it. the beam
Wess A
30-ft ladder Weighi
ing its Center
ofmass One-third ofng 100 Ib hav-
the way up
that it` makes an angl
Tests against a smooth wall
so
e of 60° with the
coefficient of fr ground.
iction between
the ground
is 0.4, how
80 before the ladder high can a 150-
slips? (Fig. 5-12.) Ib man
to Me equations around point A as

= Coun
(100 1b)(10 f £95160 terclockwise torque
?) + (150 1byx)
= F(BO)
 114 a Pe
j

is the product obtained by multiplying the mag.

d rey a .

; res caeby the magnitude ofthe com.

TORQUE 113 ponent of theother vectorin the direction of the

But x isthe perpendicular dist pe vector product, or cross product, of two

ance (AE) tothe
line ofaction ofthe man’s weight.
We need to
vectors produces another vector having a mag
find itsprojon
ec theti
laddon
er (4D). tude equal to the|product of the magnitude of
Figure 5-13 the two vectors times thesine ofthe angle be
AD = (x)\(cos 60°) Forces constituting a couple. tween them and having a perpendicular
= (14)(cos 60°) = 28 ft to the plane of thetwovectors, the sense of which
{ y p ,

shown in Fig. 5-13, About the axis through is determined by the right-hand rule. a i
Therefore, the man can climb up to
fore the ladder slips. 28 ftbe- 0, the For an object tobein equilibrium, it is r
torque produced by F, is (GA)F,, and that
pro- sary (1) that the vector sum of the forces applie
duced by F, is — (OB)F,, Since F, = F, = F to it be zero and (2) that the vector sum of the
d
` total torque is the
5-9 torques acting on it be zero.
HINTS (OA)F — (OB)F = (OA — OB)F
SOLVING TORQUE PROBLEMS = (AB)F
Center of gravity
The technique used in removing This verifies the statement that
duced by a couple is the Produc the torque pro-
of ladder the unknown
force
i
about the hinge asanaxis isa stan of the forces and the Perpen t of one (either) The center ofgravity of a body is
at which its Weight may be consider the point
dard devi ce dicular distance be-
instatics. The student should alwa
ys beonthe a product independent ed as acting.
lookout for the opportunity to sides of
tari atroub
ly lesom
tep (tempo-
) e unknown force by selecting axis. A couple cannot be the loca- cent ofer
mass of or system is the
a single balanced Point about which th Product of the mass
an axis oftorques that lies onthe line
of action force, but only by the and
of the unknown force he wishes to avoid application of moment arm sum up to zero.
. in magnitude and Opposi
te A couple consists of two forces equal
in mag-
nitude, opposite in direction, and not in the sam
5-10 ine. The torque prod uced by the couple i
COUPLES SUMMARY _ 10 the magnitude of one (either) of the forces k
Figure 5-12 times the perpendicular distance between them.
llustration of ladder Problem. The motion of a bo
dy is determined by
‘ th ou
acting on it, as wel
Magnitude and dir l
ection of the forces
where BC = (30 ft)(sin 60°) = 26ftand Fis The moment arm ,
force exerted by the wall on the ladder.
the dicular distance Of a force is th
Perpen-.
from the axis to 1 How may the torqueof
F can be found by using the firstcondition
for equilibrium, EF, = 0. Theref
of the force, the fraof action
increased? 2 a en force be
ore, f, the fric- The torque Produc = ed give
tional
force between the ground and the Product of theforce
ed by a for ce i 2 What two condition
ladder, an,
equalsF.Sincep = J/N; f = uN = 04 d iits mo
mnmean
ladder + weight of man), and
(weight of ta
nt ar m,t the Object to be in equili ?
f= 0.4250) = 3 Explain why the
100 1b = F.
torque, and hence a rotational acceleration. A pair L=5F i a ladder leaning that a wall exerts on
Therefore, offorces equal in magnitude, opposite in direction, that = ladder exerts Must equal the force
or
e wall —
and not in the same line is called a couple. The . i
(100 16X5 ft) + (150 1b)(x) = (100 1b)(26 fY) plain why in
ue produced by a couple is independen
t of
exerted on the indd T problem the force
(150 1b)(x) = 2600 ft-lb — 500 ft-lb cw ara poly ae a Fhe eya
of one of the forces and the perpendicular dis- to be equal and Opposi
x== ~i
2100 ft-lb _
o = 14ft tance between them, zontal component
of the force
Consider the torque produced by the couple exerted by the —
Pend on the base of ladder.
5 Does the Center
j

Explain your answer,

 116 THE PHYSICS OF PARTICLES

TORQUE 115 and B, 6m The scaffold

weighs 200 N. by a horizontal cable fastened at the upper end,
N.
Bis250
jer ionits 150Nandthatin The boom supports a 1,000-Ib load at its upper
6 Explain why a person trying to push a stalled 0.4m from the upper end. Calculate the torque (a)Whatistheweight ofthepainter? (6) How end. Find the tension in the cable, the horizontal
automobile must lean forward to get the best due to this force about each end. far from A ishestanding? and vertical components of the thrust at the
results. 10 Asotin pols2)long ameweighs DTh hinge, and the resultant thrust there.
7 Show that if the resultant of a set of concur- Ans. 55 m-N; 14 m:N.
3 A uniform board 20.0cm x 10.0 cm has a issupported by a boy 2.0ftfrom e and a Ans. 620 Ib; 620 1b; 1,200 Ib;
rent forces acting on a point is zero, the sum of mass of 200 g.Masses of 50.0 and 80.0g are Seep heeadh Aiwhospot salda 1,300 Ib, 63° above the horizontal,
the torques about any point due to these forces attached at two corners at the ends of load of 1001b be placed so that the man will 17 A uniform ladder 8m long and weighing
must also be zero. one of the
longer sides. Locate the center of gravity. support twice as much as the boy? 350N rests against a smooth vertical wall at an
8 A man wishes to check the weight of a pur- 4 A rigid rod resting on a fulcrum has the
chase but has available only a spring balance Ans, 8.8ftfrom B. angle of 30° to the wall. A 700-N man stands
following 20-Ib forces acting on it. A is attached 11 A uniform 30-lb beam 10 ft long is carried 6m up from the bottom of the ladder. Find the
which can read only to about one-third of the 2 ft to the left of the fulcrum acting vertically
presumed weight of the object. Show how he can by two men A and B, one at each end of the horizontal force necessary at the base to keep the
upward; B is4 ft to the left of the fulcrum, acting beam, (a) IfA exerts a force of 25lb, where must ladder from slipping.
perform an accurate weighing with the spring 30°S of E; C is 6ftto the left of the fulcrum, 4 load of 50lbbe placed on the beam? (b) What 18 A uniform ladder 20.0ft long weighing
balance and a yardstick. acting 45°N of E; D is attached 2 ftto the right
9 Show how one might locate the center of force does Bexert? 30.0 1b rests on horizontal ground and leans at
of the fulcrum, acting 60°N of W; E is acting
gravity of a thin, uniform sheet of metal by using vertically downward; 4 ft to the right of the ful- 12 A bricklayer who weighs 200 Ib stands 4.0 ft an angle of 60° with the horizontal against a
only a plumb bob. crum; and F is 6 ft to the right of the fulcrum from one end of a uniform 20-ft scaffold. A 60-1b smooth vertical wall. How far up the ladder may |
10 Where is the center of gravity of a doughnut? and acting 50°N of W. Is the system in equilib-
pile ofbricks hasitscenter of gravity 7.0 ftfrom a 160-Ib man go before the ladder slips if the
Explain your reasoning. rium? If not, where can a 20-Ib force be placed thesame end. Thescaffold weighs 90Ib. Ifthe Coefficient of friction between ladder and ground |
11 In order to assist a horse pull a wagon out scaffold
issupported at the two ends, find the
to put it in equilibrium? Ans. No; 1.9 ft to is 0.433. Ans. 15.9 ft
of a rut, where on the wheel could you most force on each support. Ans. 106 lb, 244 Ib.
cause a clockwise moment. 1 A 150-1b man is able to climb up 15ft on
effectively apply a force. Why? 5 A uniform steel rod with a linear mass of 13 A nonuniform
bar weighs 400N and is4m a 20-ft ladder before it slips. What must the
12 Three unequal forces act upon a body at a 2 g/mis bent in the shape ofa letter F. The vertical long. When itissupported by a fulcrum atits
Point so that the. body is in equilibrium. If the Coefficient of friction be if the ladder is arranged
rod is 2m long, the cap is 1 m long, and the midpoint, a load of 80 N must besupplied at the So that its base makes an angle of 50° with the
magnitudes of two of the forces are doubled, how
must the third force be changed to preserve equi-
smaller arm, attached 1.5 m from the bottom, is smallendtoholdthebarina horizontal position. ground? Assume the ladder to be uniform and
0.5 m long. Find the center of mass of the object. Where isthecenter of gravity? weighing 100 Ib.
librium? Use diagrams to support your answer. 6 A letter E has the following dimensions: its n A crane boom which weighs 2001b and
13 How might one use the property of center base is 3 ft long, the middle arm is 1 ft long, and which is10 ftlong is supported sothat it makes 20 A stone having a mass of 10kg is sup-
of gravity to locate the geographic center of a the cap is2 ftlong. The vertical arm is 4 ftlong ported in the middle of a rope, which makes an
State? an angle of 50° with the su
and the middle arm is attached 1 ft from the top. angle of 150° at its point of contact with the
14 Discuss the stability of a body in equilib- If the linear weight is 20 Ib per linear foot, find Stone, Find the tension in the rope. Ans. 190N.
rium. How is stability related to potential energy? the center of mass. Ans, 0.7 ft to the right of 21 A uniform rod 30 ft long and weighing 200 Ib
Give illustrations,
15 Archimedes is reputed to have said, “Give
the vertical arm; 1.9 ft above the base. 18 pivoted at its upper end. It is drawn aside so |
7 A uniform sheet of metal 6 ftsquare weighs that it makes an angle of 60° with the vertical
me a place to stand and I can lift the earth.” |
50 Ib. If the following weights are hung on each by means ofarope fastened at the lower end and |
Comment on this statement, What are some of
the auxiliary devices that he would have needed
corner of the square in a consecutive manner: makinganangle of 90° with the rod. Find (2)
20Ib at A, 60lb at B, 801b at C, and 401b at
for this cosmic experiment?
D, find the center of gravity of the system.
ke tension in the rope, (b) the horizontal force
16 Show that a couple cannot be balanced ex- 8 The legs of a wheelbarrow are 3.0 horizontal at the pivot, (c) the vertical force at the pivot, |
cept by another couple. na (d) the resultant force at the pivot.
ft from the axle. When the wheelbarrow is un- an Pepe farm gate isSupported by two hinges
loaded, a force of 12 1b applied to the handles Tek abstthewe
4.0 horizontal ft from the axle is needed to raise ightof the
boom. isNat The gate is 12.0ftlong. Its weight
the legs from the ground. If a 120-1b box is placed 30ftlong, weighing 200 “Tey Supported by the upper hinge. If the
Problems boaa Rape Ib

tagnse int ana ae

Bay 10 ftfromthe
`“ A rope is wound around a shaft 10.0 cm in
in the wheelbarrow with its center of gravity 1.5 ft
m Laeis of uniform construction, what forces are
diameter, If a pull of 500 N is exerted
from the axle, what force must be applied to the angle of30 Withthever end and mak
on the rope, handles to raise the legs from the ground? tical, Itis
heldinposition ` y from vertical; 30 Ib, hori-
What torque is imparted to the shaft?
Ans. 57 lb. zontally.
A bar 2.0 m long makes an angle of 30° with 9 A painter stands on a horizontal uniform
the horizontal. A vertical force of 40 N is
applied scaffold hung by its ends from two vertical ropes