## Centre of Gravity

- Centre of gravity (COG) of a body is the point of application of the resultantforce due to earth’s attraction. It is the point where the whole weight of the body appears to act from. The resultant force is the weight (W=mg) of the body.

### Centre of Gravity of Regular Shapes

- The centre of gravity of a uniform body(body with weight evenly distributed) lies at the body’s geometrical centre. For example, a uniform meter rule balances at the 50cm mark when suspended.

- The centre of gravity of regular shapes can also be determined by construction e.g.
- For square and rectangular plates, diagonals are constructed.The point of intersection is the centre of gravity.

- For triangular plate, perpendicular bisectors of the sides are constructed. The point of intersection is the centre of gravity.

- For circular plate construct diameters. The point of intersection, which is the centre of the circle, is the centre of gravity.

- For square and rectangular plates, diagonals are constructed.The point of intersection is the centre of gravity.

**Examples**

- A uniform meter rule is balanced at 20cm mark when a load of 1.2N is hung at the zero mark.
- Draw a diagram of meter rule showing all the forces acting on it.

- Calculate the weight and mass of the meter rule.
**solution**

At equilibrium (balance), Sum of clockwise moment =Sum of anticlockwise moment

W × 0.3m =1.2N x 0.2m

0.3W =0.24

∴W =^{0.24}/_{0.3}=0.8N - Determine the reaction on the pivot.

Solution

total upward forces = total downward forces.

R=1.2+W

R=1.2+0.8

R=2.0N

- Draw a diagram of meter rule showing all the forces acting on it.
- The diagram below shows a metalplate 3m long, 1m wide and negligible thickness. A horizontal force of 100N applied at pointD just makes the plate tilt. Calculate the weight of the plate.

Solution

At equilibrium (balance), Sum of clockwise moment = Sum of anticlockwise moment

100N x 3m=W ×0.5m

300=0.5W

∴W =^{300}/_{0.5}=600N

**Exercise**

- A uniform half-meter rule is pivoted at the 10cm mark. Find the position of a 2.0N weight that will balance the rule horizontally if the weight of the rule is 0.4N.
- The figure below shows a uniform plank of length 6.0m acted upon by the forces shown. If the plank has a weight of 300N, draw the diagram showing all the forces acting on the plank. Calculate the tension T in the string and the reaction at the pivot.

- Define the centre of gravity of a body.
- The figure below shows a diagram, of mass 150kg and radius 0.5m being pulled by horizontal force F against a step 0.1m high.What initial force,F, is just sufficient to turn the drawn so that it rises over the step.If the diagram below shows spherical balls placed at different positions on a surface.

## Equilibrium States

- State of equilibrium refers to state of balance of a body. There are three states of equilibrium:

### 1. Stable equilibrium

- A body is said to be in a stable equilibrium if it returns to the original position after being displaced slightly. The funnel does not topple over since the line of action of weight still falls inside the base of the funnel.

### 2. Unstable equilibrium.

- A body is in unstable equilibrium if on being displaced slightly, it does not return to its original positions but occupies a new position. The funnel below topples over because the line of action of weight falls outside the base of the funnel.

### 3. Neutral equilibrium

- A body is said to be in neutral equilibrium if on being displaced it occupies a new position which is similar to the original position.

### Conditions for Equilibrium

- The sum of forces on the body in one direction is equal to the sum of forces acting on the body in the opposite direction.
- The sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.