Engineering Science A
靜力學
Topic : Statics
(a) Scalar and vector quantities, graphical method of
vector , calculations. Force system on rigid
bodies, triangle and polygon of forces
(b) Mass and force
(c) Lever, moment, principle of moments
(d) Conditions of equilibrium
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標量 向量
Scalars and Vectors
大小
A scalar quantity is defined by magnitude only.
E.g. speed, no of students, time
A vector quantity is defined by magnitude and
direction.
E.g. force, velocity, acceleration.
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Scalars and Vectors
重量
質量
Mass (symbol: m) Weight (symbol: W)
a vector
a scalar
a force
has magnitude only
has a magnitude of W=mg
unit: kg
where g is the
gravitational acceleration,
has a direction towards
the centre of earth
unit: N or kgm/s2
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力
Force
A force is a vector. It has magnitude and direction. Its
effects: changes a body’s motion and deforms the
變形
body.
A frame of reference must be applied when a force is
drawn. 10 N Line of action
Negative Positive
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Some examples of forces are involved in our
daily life:
Change motion - pushing a door
deform - pulling a rubber band
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Vector Sum (2 forces)
平行四邊形
Parallelogram Law
Sum of two vectors (Parallelogram Law)
a a+b
b
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Vector difference (2 forces)
Addition of a negative vector (Parallelogram Law) b
-b
a a
a-b
-b
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Triangle of Forces Method
Vector sums and differences
Start with any force (from the arrow end to a arrow
head)
Addition of another force
start from the arrow head of the first force, draw the
second force
Subtraction of another force
start from the arrow head of the first force, reverse &
draw the second force
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Triangle of Forces Method
The vector R is then drawn to close the triangle and
its vector direction is from F1 (arrow end) to F2
(arrow head).
Figure 5 shows F1 drawn first and Figure 6
shows F2 drawn first.
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Vector sums (> 2 forces)
Start with any force (from the arrow end to a arrow
head)
Addition of another force
start from the arrow head of the first force, draw the
second force Resultant 合力
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Vector differences (>2 forces)
Start with any force (from the arrow end to a arrow
head)
Subtraction of another force
start from the arrow head of the first force, draw the
second force
合力
Resultant
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代數的
Algebraic Method
The vectors can be added algebraically by first
resolving the forces into their rectangular
components in the x and y directions, adding the
components and then converting back into polar
coordinates.
Solution:
R x F1,x F2,x F1 cos 1 F2 cos 2
Ry F1, y F2, y F1 sin1 F2 sin 2
R Rx Ry
2 2
Ry
tan 1
Rx
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Example 1
Fy
600N
35 Fx
Calculate the components in rectangular coordinates of the 600 N force.
Solution
Fx 600.cos 35 491 N Fy 600.sin 35 344 N
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Example 2
A force vector has the components 600 kN and 300 kN
in the x and y directions respectively, calculate the
components in polar coordinates.
Solution
F 3002 6002 6708. kN
1300
tan 2656
.
600
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Example 3
Find the magnitude and direction of the resultant (i.e. in polar coordinates) of the two
forces shown in the diagram,
6kN
4kN
15 30
a) Using the Parallelogram Method
b) Using the Triangle of Forces Method
c) Using the algebraic calculation method.
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Solution (example 3)
solution to a) and b) as practice in the above examples
c) Force X-axis Y-axis
R x 6 cos 30 4 cos15 1332
. kN (kN) (kN) (kN)
4 4 cos 15 -4 sin 15
R y 6 sin 30 4 sin 15 4.035 kN
6 -6 cos 30 -6 sin 30
ΣR -1.332kN -4.035
4.035
R ( 1332
. ) 2 ( 4.035) 2 4.249 kN ; tan 1 7173
.
1332
.
or -108.26 measured from the positive x axis.
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多邊形
Polygon of forces
Addition and subtraction of a system of forces
form a polygon
The no. of sides depends on the no. of forces in the
system.
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多邊形
Polygon of forces
A closed polygon: no resultant force
A non-closed polygon: resultant force
the force = an arrow drawn from the arrow end of the
first drawn arrow to the arrow head of the last drawn
arrow.
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Class Activity Two
Draw the resultant force for the following system of
forces.
y 70N < 60 o
110N < 160 o
45N < 20 o (with
20 degree with
x the horizontal
level)
40N < 210 o
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Class Activity Three
Use analytical method, find the resultant force for
the system of forces in the Class Activity Two.
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剛
Rigid bodies
The distance between any two points remains the
same when the body is subjected to forces.
Rigid Body is a body that does not deform (change shape)
as a result of the forces acting on it.
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Free body diagrams
A body is freed from any contact and any remote
influence. R
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W
Free body diagrams
Draw the outline of the
chosen body R
Mark down all the
remote forces
(gravitational, and
electromagnetic forces)
Mark down all the
contact forces (Reaction
& Friction) W
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力矩
Moment of a force
Moment of a force about any point垂直say A: M = r x F
力臂 距離
where r (moment arm) is the perpendicular distance
from A to the line of action of F.
A
r
F
Line of action
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力偶
Couples
非一至
A couple is two equal and opposite parallel, but non-
coincident forces.
Moment of a couple M = d x F.
d F
F
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Principle of Moments
槓桿 支撐了
• A lever AB, simply supported at C,
負荷
loads P
反作用力
and Q are at D and E. The
reaction at C is R.
D C E
A B
a b
P Q
R
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Principle of Moments
• If a body is at rest under the action of
several forces, the total clockwise
moment (CM) of the forces about any
axis is equal to the total anticlockwise
moment (ACM) of the forces about the
same axis.
• CM = ACM
• or CM -ACM = 0
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Principle of Moments
• Examples:
• Moment about C:
• (Q . b) = (P . a)
• or (Q . b) - (P . a)= 0
• Moment about D:
• Q . (a + b) = R . a
• or Q . (a + b) - R . a = 0
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Principle of Moments
• Examples:
• Moment about E:
• R . b = P . (a + b )
• or R . b - P . (a + b ) = 0
• For equilibrium,
• downward force = upward force
• i.e. P + Q = R
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Principle of Moments
• Examples:
• If P is 50N , a and b are 3m and 2m
respectively, find Q and R if the lever
is in equilibrium.
• Take moments about C:
• 50N x 3m = Q x 2m
• Q = 75N
• and R = 50N + 75N = 125N
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Degrees of freedom (dof)
1 dof = a way of moving (in a line) or rotating freely
3 translational dof
+
3 rotational dof
= 6 dofs
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均衡
Equilibrium of plane rigid bodies
Conditions for static equilibrium
F 0 F xi 0
F
i
yi 0
M
OR
i 0
M i 0
F is a vector. Component form
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Application of Moment
Consider a person trying to open a door, by applying a force, of magnitude, F, as
shown below.
The perpendicular distance, d, between the line of action of the force and
the pivot (the hinge of the door) is r sinα
Therefore, the moment of the force is given by
Moment = F r sinα
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The two obvious changes the person could make in order to open the door more easily
are
(i) he/she could increase the distance "r"
(ii) he/she could push at 90° to the door.
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桿
Example: Find the reaction supports at RA and RB as shown in the loaded beam.
20 kN
35 kN
RA RB
Engineering Science A MMI6801
3.5m 7m 3.5m
Solution:
Taking moments about A,
RB 14 35 10.5 20 3.5 0
RB 31.25 kN
From vertical equilibrium
R A RB 35 20
R A 23.75 kN
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