7 Statics
By Liew Sau Poh
1
Objectives
7.1 Centre of Gravity 7.2 Equilibrium of particles 7.3 Equilibrium of rigid bodies
2
Learning Outcome
(a) define centre of gravity (b) state the condition in which the
centre of mass is the centre of gravity (c) state the condition for the equilibrium
of a particle (d) solve problems involving forces in
equilibrium at a point (e) define torque as = r F
3
(f) state the conditions for the equilibrium of a rigid body
(g) sketch and label the forces which act on a particle and a rigid body
(h) use the triangle of forces to represent forces in equilibrium
(i) solve problems involving forces in equilibrium.
4
7. Statics
Statics equilibrium is a state where balanced forces acting on a rigid body or a particle so that it remains at rest.
5
7.1 Centre of Gravity (C.G.)
The Centre of Gravity (C.G.) of a body is the point through which the whole weight of the body appears to act For a regular object, the C.G is always at the centre
6
7.1 Centre of Gravity (C.G.)
7
Determine the C.G of a lamina (Experimental method) Line of action of its weight must pass through the centre of gravity.
8
hung freely at A
at rest, draw vertical line
hung freely at B
at rest, draw vertical line
G is centre of gravity
© Manhattan Pre ss (H.K.) Ltd.
The Centre of Gravity of a Lamina (Mathematical method)
Assumption: Lamina is made up of n particles
coordinate of each particle
coordinate of CG
The Centre of Gravity of a Lamina
Taking moments about the y-axis, Xavg = Mg( ) = (m1g1)x1 + (m2g2)x2 + (m3g3)x3 + . . . + (mngn)xn
Mg
xgmx
xgmx
n
iiii
n
iiii
1
1
)(
)(
The Centre of Gravity of a Lamina
If g = constant (in a uniform gravitational field), g1 = g2 = g3 = ... = gn = g, then Similarly, by taking moments about the x-axis,
M
xmx
n
iii
1)(
M
ymy
n
iii
1)(
The Centre of Gravity of a Lamina
Coordinates of centre of gravity: Note: These are in fact the coordinates of the centre of mass.
M
xmx
n
iii
1)(
M
ymy
n
iii
1)(
Centre of Mass (C.M.) vs. Centre of Gravity (C.G.)
Centre of mass of a body is defined as the point where the entire mass of the body acts. Centre of gravity of a body is defined as the point where the weight of the body acts.
13
Example of C.G.:
If a body is uniform and regular in shape, then the c.g. is at the geometrical centre.
c.g. c.g.
c.g. c.g. Centre of cylinder
7.2 Equilibrium of particles
A body is balanced if pivoted at a point which passes through c.g..
c.g.
The uniform ruler above is balanced.
7.2 Equilibrium of particles
The condition for a particle in static equilibrium state.
(a) Resultant force is zero. If F1 = - F2
Then the resultant force, Fnet = F1 + F2
= 0 16
F1 F2
7.2 Equilibrium of particles
Thus, F1 + F2 = 0 The vector sum of the forces must be zero.
17
F1 F2
Example 1
Let three forces F1, F2, and F3 acting on a particle O, which is in equilibrium, as shown:
Then, (components x) F1(x) + F2(x) + F3(x) = 0, and (components y) F 1(y) + F2(y) + F3(y) = 0
18
F1 F2
F3
O F1(x)
F3(x)
F2(x)
F1(y )
F3(y)
F2(y)
Alternative method
Vector sum of the forces must be zero. F1 + F2 + F3 = 0
19
F1 F2
F3
O
F1
F2 F3 A
B
C
Closed polygon
Consider three forces acting on a particles is in equilibrium. The forces can be joined to form a polygon. The resultant force = 0 20
F2
F1 F3
Closed polygon
If the forces acting on a particle is not in equilibrium, a resultant force is existing.
21
The length of the sides representing the magnitude of the forces. The direction of force is represented by its arrow.
F1 F3
F2
F4 Resultant, F 22
F1
F2 F3
F1
F2
F3
Resultant F
Resultant Force, F = F1 + F2 + F3
7.3 Equilibrium of a Rigid Body
The following shows a rod in balance, which is so called in static equilibrium. The weight of the rod is supported by the reaction, R
23
Support
weight
R
7.3 Equilibrium of a Rigid Body
Two conditions are necessary: No resultant force (R = weight) No resultant torque (Moment = 0)
24
weight
R
Toque
If a force, F is applied on the one end of the rod, it will start to rotate about G. A torque is produced by the force, where torque, = F r
25
weight
r
F
G
Toque Torque can be produced by two parallel forces of the magnitude, but opposite direction of a distance from the rotating axis. The pair of forces = a couple
26
F1
F2
d = 1 m d = 1 m
Torque The a couple (torque), if exists, will cause a moment (torque) about the rotating axis, which is given by = Fr = F(2x) 0 Hence the body is not in equilibrium.
27
F (external force) x = 1 m x = 1 m
F 28
More about Torque
Torque is the tendency of a force to rotate an object about an axis. Torque, , is a vector quantity. Consider an object pivoting about the point P by the force F being exerted at a distance r.
F
d
P r
Moment arm
Line of Action
29
More about Torque
The line that extends out of the tail of the force vector is called the line of action. The perpendicular distance from the pivoting point P to the line of action is called Moment arm.
F
d
P r
Moment arm
Line of Action
30
More about Torque Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm.
21
2211 dFdF
sinrF Fd F
d
P r
Moment arm
Line of Action
F Sin
31
More about Torque When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise.
21
2211 dFdF
sinrF Fd
F1
d1
P
d2
F2
7.3 Equilibrium of a Rigid Body
If a mass of 1 kg is put on one end, the rod will rotate in clockwise direction. While the force, F1 = 10N The torque, = Fr = (10)(1) = 10 Nm
32
Support F1 =10 N
r = 1 m
7.3 Equilibrium of a Rigid Body If another mass of 1 kg is applied on the other end, the system will be in static equilibrium again, where R = F1 + F2, or R + F1 + F2 = 0.
33
Support F2 = 10N F1 =10N
1 m 1 m R = 20N
7.3 Equilibrium of a Rigid Body
In this case, the sum of moment/torque = F1r1 F2r2 = (10)(1) (10)(1) = 0.
34
Support F2 = 10N F1 =10N
1 m 1 m
Direction of Rotation
Example 1
The figure shows a rod of length 2 m and weight 1 kg resting on supports at end A and B. Find the reaction at A and B. (g = 10 ms-2)
35
A B
0.5 m
Solution
Let R1 = reaction at point A, R2 = reaction at point B, F = force due to weight of the rod.
36
F = 10N A B
0.5 m R1 R2 0.5 m
Solution
Since the system is in equilibrium, Resultant Force = 0
37
F = 10N A B
0.5 m R1 R2 0.5 m
Solution
Sum of moment about A, M = 0 R2 (0.5 + 1) 10(1) = 0 R2 = 10/1.5 =6.67 N From (1), R1 = 10 6.67 =3.33 N
38
F = 10N A B
0.5 m R1 R2 0.5 m 1 m
Solution
OR, Sum of moment about B, M = 0 R1 (0.5 + 1) 10(0.5) = 0 R1 = 5/1.5 = 3.33 N R2 = 10 3.33 = 6.67N
39
F = 10N A B
0.5 m R1 R2 0.5 m 1 m
Example 2 A uniform horizontal beam with a length of l = 8.00 m and a weight of Wb = 200 N is attached to a wall by a pin connection and supported by a cable as shown. A person of weight Wp = 600 N stands a distance d = 2.00 m from the wall.
40
Example 2
41
Find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam.
Solution
Analyze Draw a free body diagram Use the pivot in the problem (at the wall) as the pivot
This will generally be easiest
Note there are three unknowns (T, R, q)
42
Solution The forces can be resolved into components in the free body diagram.
43
Solution
Apply the two conditions of equilibrium to obtain three equations Solve for the unknowns
Taking moment about R,
44
Nm
mNmNl
lWdW
T
lWdWlT
bp
bpz
31353sin)8(
)4)(200()2)(600(sin
)2
(
0)2
())(sin(
Solution
Considering sum of Fx and sum of Fy,
45
0sinsin0coscos
bpy
x
WWTRFTRF
NNT
R
TTWW
TTWW
RR
bp
bp
5817.71cos
53cos)313(coscos
7.71sin
sintan
sinsin
tancossin
1
Example 3
A safe whose mass is M = 430 kg is hanging by a rope from a boom with dimensions a = 1.9 m and b = 2.5 m as shown in the figure. The uniform beam has a mass m of 85 kg; the mass of the cable and rope are negligible.
46
a
b
Example 3
a) What is the tension Tc in the cable; i. e., what is the magnitude of the force Tc on the beam from the horizontal cable?
b) What is the force at the hinge?
47
a
b
Solution
Consider the sum of Fx and sum of Fy,
48
; 0,, chchxextxnet TFTFFFMgTmgMgFFF rvyextynet ; 0,,
Tc
mg
Mg
Fh
Tr
Fv
Solution
Taking moment about the hinge
49
Tc
mg
Mg
Fh
Tr
Fv
NabgmM
abmg
abMgT
MgbaTbmg
bTaTbmg
c
c
rc
TrTcmgvhzextznet
6093)2
(2
02
02
00
,,
Solution
50
Tc
mg
Mg
Fh
Tr
Fv
NFFF
NmgMgFNTF
vh
v
ch
7912
50476093
22
Summary
Centre of Gravity Centre of Mass
Statics
Resultant Force = 0 Triangle of forces
Equilibrium of Particles
Torque, = r F Resultant force = 0 Resultant torque = 0
Equilibrium of Rigid Bodies
51