## CONDITIONS FOR EQUILIBRIUM

A body is said to be in equilibrium if under the action of several forces, it does accelerate or rotate.

- The sum of the upward forces must be equal to the sum of the downward forces.
- The sum of the clockwise moment above a point must be equal to the sum of anticlockwise moment about the same point

F1 + F2 = F3 + F4

(F1+F2) – (F3+F4) = 0

Clockwise moment = F_{2}X_{2} + F_{4}X_{4}

Anticlockwise moment = F_{1}X_{1}+ F_{3}X_{3}

(F_{1}X_{1}+ F_{3}X_{3}) – (F_{2}X_{2} + F_{4}X_{4}) = 0

Sum of clockwise moment = sum of anticlockwise moment

## MOMENT OF A FORCE

The moment of a force is the product of the force and the perpendicular distance

M = F x distance

Unit =Nm

## COUPLE

A couple is a system of two parallel, equal and opposite forces acting along the same line

The moment of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces

M = f x 2r

M = f x d

The distance between the two equal forces is called the *arm of the couple*

The moment of a couple is also called a *torque*

## Application of the Effect of Couples

- It is easier to turn a tap on or off by applying couple
- It is easier to turn a steering wheel of a vehicle by applying a couple with our two hands instead of a single force with one arm.

EXAMPLES

- A light beam AB sits on two pivots C and D. A load of 10N hangs at 0; 2m from the support at C. Find the value of the reaction forces P and Q at C and D respectively.

P + Q = 10N

X 2 = Q (2 + 6)

20 = 8Q

Q = 20/8 =2.5 N

OR

Taking moment about D

P x8 = 10 x6

P = 60/8

P =7.5N

Q = 10 -7.5

Q = 2.5 N

- A pole AB of length 10m and weight 600N has its center of gravity 4m from the end A, and lies on horizontal ground. Draw a diagram to show the forces acting on the pole when the end B is lift this end. Prove that this force applied at the end A will not be sufficient to lift the end A from the ground.

Clockwise moment =600 x 4 =2400Nm

Anticlockwise moment =p x 10 = 10pNm

P =240Nm

If this force of 240Nm is applied at A, we have

P= 240Nm

Taking moment about B, we have

Clockwise moment =240 x 10 =2400Nm

Anticlockwise moment =600 x 6 =3600 Nm

The anticlockwise moment is greater than the clockwise moment.

Therefore, the 240N force A will not be sufficient to lift the end A because the turning effect due to the 600N force far exceeds that due to the 240N force

Find the moment of the force of 20N in the diagram above about A and B

Taking moment about A

Cos 60 =d/3m

D= 3 cos 60

D = 1.5m

Moment about A =F x d

M = 20 x 1.5

= 30 Nm

The Moment about B= 0

- A uniform rod lm long weighing 100N is supported horizontally on two knife edges placed 10cm from its ends. What will be the reaction at the support when a 40N load is suspended 10cm from the midpoint of the rod.

R1 R2

10cm 40cm 10cm 30cm 10cm

40N 100N

R1 + R2 = 140N

Taking moment about R1

R2 x 80 = (100 x 40 ) + (40 x 50 )

80R2 = 4000 + 2000

R2 = 6000/80

R2=75N

R1 = 140 – 75

=65N

- A metre rule is found to balance horizontally at the 50cm mark. When a body of mass 60kg is suspended at the 6cm mark, the balance point is found to be at the 30cm mark, calculate:

-The weight of the metre rule

-The distances of the balance point to the 60kg mass if the mass is moved to the 13cm mark

*6cm 24cm 50cm*

* *

600N W

w x 20 = 24 x 600

w = 14400/20

= 720N

13cm xcm 37cm 50cm

600N 720N

600x(X)_{=}720(37-X)

600x = 6640 – 720x

600x+ 720x = 6640

x = 6640/1320

x = 20. 18cm

**CENTRE OF GRAVITY**

The centre of gravity of a body is the point through which the line of action of the weight of the body always passes irrespective of the position of the body. It is also the point at which the entire weight of the body appears to be concentrated.

The centre of mass of a body is the point at which the total mass of the body appears to be concentrated. Sometimes, the center of mass may coincides with the centre of gravity for small objects.

## STABILITY OF OBJECTS

There are three types of equilibrium- stable equilibrium, unstable equilibrium, and neutral equilibrium.

- Stable equilibrium: a body is said to be in stable equilibrium if it tends to return to its original position when slightly displaced. A low centre of gravity and wide base will put objects in stable equilibrium e.g. a cone resting on its base ; a racing car with low C.G and wide base; a ball or a sphere in the middle of a bowl.
- Unstable equilibrium: a body is said to be in an unstable equilibrium if when slightly displaced it tends to move further away from its original position e.g. a cone or an egg resting on its apex. High C.G. and a narrow base usually causes unstable equilibrium.
- Neutral equilibrium: a body is said to be in neutral equilibrium if when slightly displaced, it tends to come to rest in its new position e.g a cone or cylinder or an egg resting on its side.

## ONLINE WORK

- When is a body said to be in equilibrium?
- What is moment?
- Write short note on the three types of equilibrium

## ASSIGNMENT

**SECTION A**

- Two forces A and B act at a point at right angles. If their resultant is 50N and their sum is 70N, their magnitudes are: (a) 50N and 20N (b) 20N and 40N (c) 40N and 30N (d) 60N and 10N
- A uniform metre rule of mass 100g balances at the 40cm mark when a mass X is placed at the 10cm mark. What is the value of X? (a) 33.33g (b) 43.33g (c) 53.33g (d) 63.33g
- The equilibrant of a system of forces is (a) equal and opposite to the resultant of the forces (b) the force which has the same effect as the system (c) equal to resultant of the system (d) the force that makes the system unstable
- Two forces forming a couple are separated by a distance of 25cm, if one of the forces equal 40N, what is the moment of the couple? (a) 1000Nm (b) 500Nm (c) 10Nm (d) 5Nm
- Two forces each of magnitude 10N acts in opposite directions at the end of a table. If the length of the table is 50cm.Find the moment of the couple on the table (a) 0.5Nm (b) 5Nm (c) 50Nm
- A pole AB of length 5M and weigh 300N has its centre of gravity 2.0M from the end A, and lies on horizontal ground. Calculate the force required to begin to lift this end. (a) 60N (b) 120N (c) 240N
- Consider the three forces acting at a point O and is in equilibrium as shown below. Which of the equations is/are correct?

P_{3} O θ_{1} P_{1}

θ_{2}

P_{2}

- P
_{1}cos θ_{1}= P_{1}cos θ_{2}P3= P_{1}cos θ1+ P_{2}cos θ_{2}iii. P_{1}sin θ_{1}= P_{2}cos θ_{2}(a) I only (b) II only (c) III only (d) II and III only - When a body is acted upon by several forces and it does not accelerate or rotates, the body is said to be in (a) space (b) equilibrium (C) motion
- Two masses 40g and 60g respectively are attached firmly to the ends of a light metre rule. The centre of gravity of the system is: (a) at the midpoint of the metre rule (b) 40cm from the lighter mass (c) 40cm from the heavier mass (d) 60cm from the heavier mass
- A 50kg mass, suspended from a ceiling is pulled aside with a horizontal force, F, as shown in the diagram below. Calculate the value of the tension T (g=10ms
^{-2})

30^{0} T

F

15kg

- 0N (b) 173.2N (c) 30.0N (d) 17.3

** **

**SECTION B**

- (a) Explain with the aid of diagram what is meant by the moment of a force about a point (b) State the conditions of equilibrium for a number of coplanar parallel forces (c) A metre rule is found to balance at 48cm mark. When a body of mass 60g is suspended at 6cm mark, the balance point is found to be at the 30cm mark. Calculate the: (i) mas of the rule (ii) the distance of the balance point from the zero eend if the body were moved to the 13cm mark
- State the conditions necessary for a body to be in equilibrium, mention the three types of equilibrium with at least two examples each.

P

12m

) 30^{0}

10N

Use the diagram above to calculate the moment of the force of 10N about the point P

- A uniform beam HK of length 10m and weighing 200N is supported at both ends as shown below. A man weighing 100N stands at a point P on the beam. If the reactions at H and K are respectively 800N and 400N, then the distance HP is

See also

PROJECTILES AND ITS APPLICATION

DERIVATION OF EQUATONS OF LINEAR MOTION