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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 through. For regular shaped bodies the C.G is at the geometric centre of the body. For irregular bodies their weight still acts at the centre of the gravity and the law of moments can be used to determine the weight of the body.



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.


Taking moments about the pivot, clockwise moments = W × 20 N cm.

Anticlockwise moments = 20 × 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. a) The sum of upward forces must be equal to the sum of downward forces.
  2. b) The sum of clockwise moments equals the sum of anticlockwise moments.

The two are called the first and second condition of equilibrium respectively.

Examples 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

  1. b) Reading of the spring balance.


  1. 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 × 0.2 Nm = 0.2W Nm.


Anticlockwise moments = 1.6 × 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


  1. 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

  1. 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 2kg is hung from the rod as shown; determine,
  2. a) Reading of the spring balance
  3. b) Reaction force, F, from the table.



  1. 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 × 10) × 0.4 + (5 × 0.5) = 10.5 Nm. Anticlockwise moments = S × 1 (reading of the spring balance)

1S = 10.5, hence S = 10.5 N. b) Upward forces = downward forces

Downward forces = (2 × 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. a) Base area– the bigger the base area the more the stability.
  2. b) 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.


Applications of stability

Neutral equilibrium It is used mainly in the design of motor vehicles i.e.

  1. a) Racing cars – they have a low and wide wheelbase to increase their base area.
  2. b) 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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