FORCES IN EQUILIBRIUM  
Moment of a Force  
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Is the product of the force applied and the perpendicular distance from a fixed  
point or pivot  
OR  
M
oment of a force is the turning effect of the force about a point  
Moment of force = force x perpendicular distance from pivot  
∴ 푴 = 푭 풙 풅  
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The SI unit of moment of force is Newton meter (Nm)  
See the figure below  
NB:  
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The point where the object rotates after turning force is called pivot or fulcrum  
Moment of force is applied in different activities such as opening bottle caps  
,
door opening and tightening nuts etc  
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The moment of a force depends on the following  
a. Size of a force  
b. Perpendicular distance  
Application of Turning Effect in daily life  
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It is applied by a hand to unscrew a stopper on the bottle  
It is applied by a spanner to unscrew a nut on a bottle  
It is applied when turning a steering wheel of a car.  
A
force is applied to a door knob and the door swings open about its hinge  
It is applied when closing and opening the water tapes  
.
This gives some reasons as to why:-  
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It is easier to open a nut with a long spanner opener than with short spanner  
fingers (This is due to the high moment as a result of perpendicular distance  
arise from the long spanner)  
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It is easier to open the cap of the bottle with a bottle opener than with your  
fingers(This is due to the addition perpendicular distance arise from the opener)  
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Knob on a door is placed as far as possible from the hinges (This is due to the  
addition perpendicular distance arise from the hinges to the knob)  
Example,  
1. A line of action of a force of 90 N acts at a perpendicular distance of 2.5 m,  
from a point. Find the moment of the force  
Solution  
Data given  
Force applied, F = 90 N  
Perpendicular distance, d = 2.5 m  
Moment, M =?  
From  
M = f  
×
d
M = 90 N  
×
2.5 m  
∴
Moment of force = 225 Nm  
Individual Task – 1  
1. The line of action of a force 48N is at perpendicular distance of 1.5m from the  
point. Find the moment of the force about the point.  
(ANS: Moment of the force = 72Nm)  
2. The moment of a force about a point is 1120Nm. If the magnitude of a force is  
5600N, find the perpendicular distance between the point and the line of action  
of the force. ( ANS: Perpendicular distance = 0.2m)  
3. The moment of a force is 1000 Nm. If the line of the force is at perpendicular  
distance of 100m, find the magnitude of a force.(ANS: F = 10N.)  
4. If 150 N of force is applied on a spanner of 10 cm to open a nut. What is the  
length on a spanner when a force of 60 N is applied? (ANS: L = 25 cm)  
Principle of Moments  
Consider the ruler below  
Now: The principle of moments states that  
“When a body is in equilibrium, the sum of the anticlockwise moments about  
any point is equal to the sum of the clockwise moments”  
OR  
“When a system is in equilibrium the sum of clockwise moments is equal to the  
sum of anti-clockwise moments  
”
CM = AM  
That is:  
Whereby:  
M = F x d  
CM = Clockwise moment  
(F1d1)  
AM = Anticlockwise moment (F2d2)  
∴
F1d1 = F2d2  
Example,  
1. A 100 g weight is suspended 45 cm from the pivot, of a light rod. If a weight w  
suspended 20 cm from the pivot balances the 100 g weight. Find weight w.  
Solution  
See the diagram below:  
From: The principle of Moments, C.M = A.M (M1 = M2) , and M = F × d  
But:  
F = m × g  
Now: m1 × g × d1 = m2 × g × d2  
→
→
100 × 10 × 45 = w × 10 × 20  
100 × 10 × 45 = 200 w  
∴
w = 225 N (m = 22.5 g  
)
Individual Task – 2  
1. A uniform meter ruler is pivoted at its centre. A 20 g mass is placed at the 10  
cm mark and a 50 g mass at the 40 cm mark. At what mark must a second 50  
g mass be placed for the system to be in rotational balance? (ANS: d =26 cm)  
2. A uniform rod with a mass 120 g and a length of 130 cm is suspended by a  
wire from a point 80 cm from the rod’s left end. What mass must be hang from  
the right end of the rod for it to be in equilibrium? What will be the tension of  
the wire? (T = 1.56 N)  
3. David and his father are sitting at the end of a seesaw 2 m from the pivot while  
David's mother is sitting at a distance d from the pivot. The seesaw balances  
as shown in the figure below. Determine d.  
(ANS: Therefore, David's mother is sitting 1m from the pivot)  
Application of Principle of Moments  
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Used to unscrew a stopper on the bottle  
Used to unscrew a nut on a bolt  
Used to open a metal cap from a bottle of soda, etc  
Turning a steering wheel of a car  
When the door is opened, the force on the handle exerts a turning effect about  
the hinges  
Center Of Gravity  
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The weight of body is due to the attraction of the earth for its particles.  
This attracting force is the weight of each individual particle. Since the body  
consists of many particles then the weight is the resultant of all the parallel  
forces acting on the individual particles as shown below.  
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For a rigid body, there is one point at which the resultant force appears to act,  
this point is known as the center of gravity G of the body  
.
∴
The center of gravity:  
Is the point through which the resultant of the  
weight of all the particles of the body acts  
OR  
OR  
Is the point where the force of gravity can be  
considered to act  
Is the point on a body along which all the weight  
of the body is likely to act  
Difference between centre of gravity and centre of mass  
Centre of gravity  
Centre of mass  
Is the point at which the whole weight  
of the body is likely to act  
Is the point at which the whole mass of  
the body is assumed to be concentrated  
Weight distribution of the body around  
centre of gravity is uniform  
It changes with the change in the  
force of gravity  
The mass distribution around the  
centre of mass is uniform  
It remains unchanged with the change  
in the gravitational field  
Couple  
A couple Is a pair of equal but opposite parallel forces applied to the same  
body but not acting in the same line  
Characteristics of a couple  
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Comprises a pair of forces  
The forces must be equal  
The forces must be parallel  
The forces must act in opposite directions  
Moment (torque) of a couple  
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Is the product of one of the anti – parallel forces, F by the perpendicular  
distance between them.  
Equilibrium  
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Is a state achieved by a body when all the forces which act upon it are balanced  
OR  
Is the state of a body to balance  
NB: The force which brings a body into mechanical equilibrium is called “Equilibrant”  
Conditions for a body to be in equilibrium  
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The net force on the object must be zero  
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The system (body) must have an acceleration of 0 m/s2  
OR (also you can state in this way)  
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The sum of the forces in one direction must be equal to the sum of the forces in  
the opposite direction  
The Sum of clockwise moment should be equal to the sum of anticlockwise  
moment  
Conditions for equilibrium depends on three factors  
o
o
o
Magnitude  
Direction of forces  
Point of application  
Factors which affecting the stability of a body  
A stable body should have:-  
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Wide base  
Low centre of gravity  
Types of Equilibrium  
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Stable equilibrium  
Unstable equilibrium  
Neutral equilibrium  
Stable Equilibrium  
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This occurs when a body is slight displaced and then it returns to its original  
position after displacement  
Unstable Equilibrium  
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It occurs when a body is slight displaced and the body it does not returns to its  
original position after displacement  
Neutral Equilibrium  
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This occurs when a body is slight displaced and the body does not alter the  
position of the center of gravity  
Application of Equilibrium  
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Vehicles used in a car race have tyres wide apart to increase stability  
Cambered wheels: The wheels of a car are slightly tilted to increase stability.  
That is why when a car is tilted to the side, it only skids  
Used in designing of structures like bridges, Aeroplane, furniture, machines, car  
boats, ships etc  
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Our bodies muscles are always kind of equilibrium that is why we can walk, seat,  
eat, run, squat, jump etc  
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Tall structures such as buildings and pylon, they have wide base and low centre of  
gravity hence provide stability  
Bus (car) with seated passengers and loading the lower compartments is more  
stable than one with standing passengers and loaded at the top  
Ships have long and wide projecting plates extending from their bases into the  
water to increase stability  
Beam balance - used for measuring masses of different objects by comparison  
with known masses.  
Steel yard - is a machine used for weighing heavy objects. It uses the principle of  
moments by balancing heavy objects with lighter objects on longer arm.  
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Seesaw – is a long plank balanced at the fulcrum so that an increase in weight in  
one side causes it to go down while the other side goes up  
Class Activity  
1. A spanner of length 40 cm is used to tighten a bolt. A force of 400 N is used.  
Calculate the moment of the force ( ANS: 160 NM  
)
2. Abuu has a mass of 60 kg and he is sitting on a see – saw at a distance of 2.5  
m from the pivot .Calculate the moment due to his weight  
3. If a 100g weight is used to balance the weight determine the distance of the  
300g weight from the point.  
(
ANS: x = 15 cm)  
4. A heavy uniform beam AB of weight 500 N is supported at its ends. The beam  
carries a weight of 3000 N at a distance of 1.5m from the end A. if the beam is  
4m long. Find the thrust (tension) / reaction at A and B (ANS: 1375 N, 2125 N)  
5. Explain what is meant by stable, unstable and neutral equilibrium. Give one  
example of each  
6. Explain what is meant by  
(i) Moment of a force about a point  
(ii) Centre of gravity of a body  
(iii) Equilibrium  
(iv) Equilibrant  
7. A metallic rod of 2 m long has a mass of 500 g. The rod is balanced on a  
wedge when a 50 g solid is hung 40 cm from one of its ends. The wedge is 85  
cm from the same end  
a) Sketch a diagram of the arrangement  
b) How far is the wedge from the centre of the metallic rod?  
8. The diagram below shows a 150g rod balanced at its centre of gravity. A 20g  
mass is placed 120cm from the pivoted point  
a. Find the value of x  
( ANS: 48 cm)  
b. What upward force does the pivot exert on the rod  
?
( ANS: 0.7 N)  
9. From the diagram below calculate reaction M and N (ANS: M = 40 N, N = 50 N)  
10. From the diagram below calculate  
i. Reaction A and B  
(ANS: A = 1 N, B = 5 N )  
ii. Additional weight at C that will just tilt the beam about B? (C = 2 N)  
11. A metre rule is pivoted at its mid – point. If two objects of weight 1.0 N and  
2.0N are suspended at 30 cm and 90 cm respectively from one end ,calculate  
the position where an upward force of 3.0 N must be applied in order for the  
metre rule to balance horizontally (ANS: 20 cm from pivot or 70 cm mark)  
12. A Uniform wooden plank with a mass of 75 kg and a length of 5 m is placed on  
top of a brick wall so that 1. 5 m of the plank extends beyond the wall’s edge.  
How far beyond the edge of the wall can 100 kg woman walk before the plank  
began to rotate ?( Let the plank’s axis of rotation be at the wall’s edge)  
13. A pole AB of length 10.0 m and weight 800 N has its centre of gravity 4.0 m  
from the end A and lies on a horizontal ground. The end B is to be lifted by a  
vertical force applied at B. Calculate the least force that is required to do this  
14. Does an object have to be at rest to be in a state of equilibrium? Explain your answer  
15. A tree trunk of length 44 m is pivoted 12 m from one of its ends. It is balanced  
when a 1500 N solid is hung 8 m from the pivot. Calculate the weight of the  
tree trunk ( W = 1200 N )  
16. It is found that a uniform wooden lath 100 cm long and of mass 95 g can be  
balanced on a knife – edge when a 5 g mass is hung 10 cm from one end .How  
far is knife – edge from the centre of the lath? ( ANS: 2 cm )  
17. State the conditions of equilibrium when a body is acted upon by a number of  
parallel forces. A uniform metal tube of length 5 m and mass 9 kg is suspended  
horizontally by two vertical wires attached at 50 cm and 150 cm respectively  
from the ends of the tube. Find the tension in each wire ( ANS: 30 N, 60 N)  
18. A uniform rod with a mass of 120 g and a length of 130 cm is suspended by a  
wire from a point 80 cm from the rod’s end  
What mass must be hang from the right end of the rod for it to be in  
equilibrium? What will be the tension in the wire? ( m = 36g (w =0.36 N) ,T= 1.56 N)  
19. A uniform metre rule is pivoted at the centre is balanced by four suspended  
forces as shown in the figure below. Calculate the force exerted in the 60 cm  
mark ( ANS: 60 N )  
20. Mention the four properties (features) of a couple  
21. Two parallel and opposite forces acted on the handle of a bicycle. Each of the  
forces had a magnitude of 45 N. The distance between them was 50 cm .What  
is the torque produced?  
22. A uniform wooden bar AB of length 120 cm weighing 1.2 N rest on two sharp  
edged supports C and D placed 10 cm from its either ends .A 0.2 N load hangs  
from a loop of a string 30 cm from A and a 0.9 N load hangs at 40 cm from B.  
Find the:  
(i) Reaction at C  
(ii) Reaction at D  
23. A uniform metre rule of weight 0.9 N is suspended horizontally by two vertical  
loops of thread A and B placed at 20 cm and 30 cm from its ends respectively.  
Find the distances from the centre of the rule at which a 2 N weight must be  
suspended  
(a) To make loop A become slack  
(ANS: 29 cm)  
(b) To make loop B slack  
(ANS: 43.5 cm)  
24. A uniform bar AB of height 5m weights 60N. The bar is supported at a  
horizontal position by two vertical strings X and Y. If string X is 0.6m from A  
and string Y is 1.8m from B. Find the tension in the string. (ANS: 16.5 N)  
25. A uniform half metre rule is freely pivoted at the 15 cm mark and it balances  
horizontally when a body of mass 40 g is hung from the 2 cm mark.  
a) Draw a clear force – diagram of the arrangement  
b) Calculate the mass of the rule  
26. Explain the following:-  
a) Why a loaded test tube floats upright?  
b) It is more difficult to balance a nail on its tip than on its base  
c) A bus carrying a very big load on its carrier can easily overturn  
27. A metre rule of weight 1.0 N is supported horizontally on two knife edges each  
placed 10.0 cm from its ends. If the weight of 1.5 N is placed at its mid – point ,  
calculate the reaction at the supports.  
28. A simple weighing machine is made of a uniform bar 125 cm long and mass 5  
kg and pivoted 2.5 cm from one end .Find the mass that must be suspended at  
the end of the short arm. (ANS: m = 4.08 kg)  
29. A 2.0 N weight placed on a 10 cm mark of a meter rule just balances an object  
hanging from the 60 cm mark. Calculate the weight of the object. (ANS: W =8N)  
30.  
How can a metre rule be balanced on a knife edge?  
31.  
Briefly explain why the handle of a door is near its outside edge?  
32. Two spheres of mass 3.0 kg and 2.0kg are joined by a light rod so that their  
centers are 0.45 m apart. Locate the center of gravity of the system  
.
33. Define centre of gravity. Hence outline the main difference between the centre  
of gravity and centre of mass.  
34. List the factors that affect the stability of a body  
35. Explain why racing cars should have wide wheel tracks.  
36. Define turning effect of force and give its SI unit  
37. Why should a mechanic choose a long spanner to undo a tight nut?  
38. A uniform half meter rule is pivoted at its 30 cm mark. A mass of 50 g hung at the 45  
cm mark keeps the rule horizontal. Determine the mass of the half meter rule.  
39. Force applied by a lady is 2 N and moment of force is 16 Nm, distance of pivot  
from effort would be --------------  
40. (a) What is meant by balanced beam?  
(b) A uniform rod AB of mass 6.0 g is balanced horizontally about a knife edge  
at a distance of 3 cm from end A where a mass of 8.0g is hanging. Find the  
length of the rod (ANS: L= 14 cm)  
41. Moment of force applied on a door is 15 Nm and force applied is 3.75 N,  
distance of the handle from the pivot is -------------  
42. Door hinge is about 1.5 m away from handle, and a boy applies a force of 4 N.  
What will be the moment of force applied?  
43. Bilqees has a weight of 300 N and sits 2.0 m from the pivot of see – saw. Asia  
has a weight of 450 N and sits 1.5 m from pivot. Who will move down  
44. A uniform meter rule of weight 16 N is pivoted at the 60 cm mark. A 4.0 N  
weight is suspended from one end .At the instant when the rule is horizontal,  
what is the value of the resultant turning moment about the pivot.  
45. A 2.0 N weight placed on the 10 cm mark of a meter rule just balances an  
object hanging from the 60 cm mark. Calculate the weight of the object. (w= 8 N)  
46. Study the diagram below and determine the value of X and hence the length of  
the bar (ANS: L = 115 cm)  
.