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Equilibrium and Centre of Gravity | High School Physics Form 2

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Centre of Gravity

- Centre of gravity or C.G is the point of balance of a body in which the total weight of the body seems to act.

Practice Example 1

The figure below shows a uniform bar of weight 'W' and length 80 cm. If a force of 20 N keeps it in balance, determine the weight 'W' of the bar.

Centre of Gravity Example - Physics Form Two

Taking moments about the pivot, clockwise moments = W x 20 N cm.
Anticlockwise moments = 20 x 30 N cm.
Clockwise moments = anticlockwise moments.
20 W = 600, therefore W = 30 N

Parallel Forces and Equilibrium

- For a body to be in equilibrium (neither moving nor rotating), under the action of parallel forces, the following conditions will be satisfied;
  1. The sum of upward forces must be equal to the sum of downward forces.
  2. The sum of clockwise moments equals the sum of anticlockwise moments.
- The two are called the first and second condition of equilibrium respectively.

Practice Example 1

1. A uniform rod of length 1.0 m is hung from a spring balance as shown and balanced in horizontal position by a force of 1.6 N. Determine:
a) The weight of the rod.
b) Reading of the spring balance.

Centre of Gravity Example - Physics Form Two

a) Let the weight of the rod be 'W'. W acts at 50 cm mark, therefore taking moments about point of suspension, clockwise moments = W x 0.2 Nm = 0.2W Nm.
Anticlockwise moments = 1.6 x 0.3 = 0.48 Nm.
Using the law of moments, then
Anticlockwise moments = clockwise moments
0.48 = 0.2 W, hence W = 2.4 N

b) Upward forces = downward forces
Downward force = W + 1.6 N
= 2.4 + 1.6 = 4.0 N
Upward force = reading of the spring balance = 4.0 N

Practice Example 2

2. A uniform rod is 1.0 m long weighs 5 N. It is supported horizontally at one end by a spring and the other end rests on a table as shown below. A mass of 2 kg is hung from the rod as shown; determine:
a) Reading of the spring balance
b) Reaction force, F, from the table.

Centre of Gravity Example - Physics Form Two

a) The 2kg mass and the weight of the rod (5 N) gives clockwise moment while the spring balance provides anticlockwise moments.
Clockwise moments = (2 x 10) x 0.4 + (5 x 0.5) = 10.5 Nm.
Anticlockwise moments = S x 1 (reading of the spring balance)
1S = 10.5, hence S = 10.5 N.

b) Upward forces = downward forces.
Downward forces = (2 x 10) + 5 = 25 N.
Therefore F + 10.5 = 25, hence F = 14.5 N.


-This is a term which explains how easy or difficult it is for an object to topple over when a force is applied to it.

Factors affecting stability

  1. Base area – the bigger the base area the more the stability.
  2. Position of the centre of gravity – the higher the centre of gravity the less stable the body will be.

States of equilibrium.

  1. Stable equilibrium – if a body is displaced by a small amount of force it returns to its original position.
  2. Unstable equilibrium – if a body is displaced by a small amount of force it toppled over and does not return to its original position.
  3. Neutral equilibrium – a body is at rest in whichever position it is placed in i.e. it does not rise or fall when displaced.

Centre of Gravity Example - Physics Form Two
State of Equilibrium

Applications of Stability

It is used mainly in the design of motor vehicles i.e.
  1. Racing cars – they have a low and wide wheelbase to increase their base area.
  2. Double decker buses – they are manufactured with a low centre of gravity by mounting their chassis and engines as low as possible.

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