Forces
Turning effects and stability
The turning effect of a force (moment)
The moment of a force about a pivot is the turning effect of that force. It depends on the size of the force and how far from the pivot it acts.
\[ M = F \times d \]
| Symbol | Quantity | Unit |
|---|---|---|
| \( M \) | Moment | newton-metre, N·m |
| \( F \) | Force | newton, N |
| \( d \) | Perpendicular distance from the pivot to the line of action of the force | metre, m |
The distance \( d \) must be the perpendicular distance from the pivot to the line of action of the force — not the distance along the object. If the force is applied at an angle, you must resolve it or find the perpendicular distance geometrically.
Clockwise vs anticlockwise moments:
A force acting to cause rotation clockwise produces a clockwise moment; a force that tends to rotate the object anticlockwise produces an anticlockwise moment.
A force acting to cause rotation clockwise produces a clockwise moment; a force that tends to rotate the object anticlockwise produces an anticlockwise moment.
Principle of moments
For an object in rotational equilibrium (not rotating):
\[ \text{Sum of clockwise moments} = \text{Sum of anticlockwise moments} \]
Example 1 — balanced beam
A beam is balanced on a pivot. A 6 N force acts 0.4 m to the left of the pivot. What force \( F \) must act 0.3 m to the right to maintain equilibrium?
Anticlockwise moment = \( 6 \times 0.4 = 2.4\,\text{N·m} \)
For equilibrium: clockwise moment = 2.4 N·m
\( F \times 0.3 = 2.4 \Rightarrow F = \dfrac{2.4}{0.3} = 8\,\text{N} \)
Example 2 — find unknown distance
A 10 N weight acts 0.5 m left of the pivot. A 25 N weight is on the right. How far from the pivot must the 25 N weight be placed for balance?
Anticlockwise: \( 10 \times 0.5 = 5.0\,\text{N·m} \)
Clockwise: \( 25 \times d = 5.0 \Rightarrow d = 0.20\,\text{m} \)
Centre of gravity
The centre of gravity of an object is the point through which its entire weight appears to act (the point at which the gravitational force effectively acts on the whole object).
Finding the centre of gravity of a flat lamina (irregular shape):
- Suspend the lamina freely from one point near its edge. It will hang in equilibrium.
- Hang a plumb line from the same pin. Draw a line on the lamina along the plumb line.
- Repeat from a different point on the edge.
- The centre of gravity is at the intersection of the two lines.
Stability
An object is more stable when:
- Its centre of gravity is lower (closer to the ground).
- Its base area is wider.
| Stability type | Displaced slightly… | Example |
|---|---|---|
| Stable equilibrium | Returns to original position | Ball in a bowl, racing car (low CoG) |
| Unstable equilibrium | Continues to move away (topples) | Pencil balanced on its tip |
| Neutral equilibrium | Stays in new position | Ball on a flat surface |
Double-decker buses, racing cars, and cargo ships are designed with low centres of gravity (heavy components at the bottom) and wide bases to maximise stability. Questions often ask you to explain how a design feature improves stability — always link it to the lower centre of gravity or the wider base.