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How does a force make an object turn, and when is a balanced object in equilibrium?

Define the moment of a force, apply the principle of moments, and state the conditions for equilibrium

A focused answer to the O-Level Physics outcome on moments. The moment of a force as force times perpendicular distance, the principle of moments for a balanced beam, the conditions for equilibrium, and centre of gravity.

Generated by Claude Opus 4.88 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to define the moment of a force, to use the principle of moments to solve balanced-beam problems, to state the two conditions for equilibrium, and to understand centre of gravity and stability. The big idea is that a force can make an object turn, and the turning effect depends on both the force and how far it acts from the pivot.

The answer

The moment of a force

The moment of a force is its turning effect about a pivot. It equals the force times the perpendicular distance from the pivot to the line of the force:

moment=F×d\text{moment} = F \times d

Force is in newtons and distance in metres, so the unit of a moment is the newton metre (N m\text{N m}). A larger force or a longer distance gives a bigger turning effect, which is why a long spanner loosens a tight bolt more easily.

The principle of moments

When an object is balanced (in equilibrium) about a pivot, the total clockwise moment equals the total anticlockwise moment about that pivot:

total clockwise moment=total anticlockwise moment\text{total clockwise moment} = \text{total anticlockwise moment}

This lets you find an unknown force or distance on a balanced beam such as a seesaw or a metre rule.

Conditions for equilibrium

An object is in equilibrium when it has no resultant force and no resultant turning effect. The two conditions are:

  1. The resultant force is zero (forces balance in every direction).
  2. The resultant moment about any point is zero (clockwise moments equal anticlockwise moments).

Centre of gravity and stability

The centre of gravity is the single point through which the whole weight of an object seems to act. An object is more stable when its centre of gravity is low and its base is wide, because it can tilt further before the centre of gravity passes outside the base and it topples.

Examples in context

Example 1. A wheelbarrow. The load sits near the wheel (the pivot), so its weight has a small distance and hence a small moment, while you lift the handles far from the wheel. Your smaller lifting force acting at a large distance balances the load's larger weight acting at a small distance, letting you lift heavy loads easily.

Example 2. A racing car. Racing cars are built low and wide so their centre of gravity is low and their wheelbase is broad. This means the car can corner hard and tilt without the centre of gravity passing outside the wheels, so it is far less likely to topple than a tall, narrow vehicle.

Try this

Q1. Calculate the moment of a 25 N25\ \text{N} force acting 0.40 m0.40\ \text{m} from a pivot. [2 marks]

  • Cue. Moment =F×d=25×0.40=10 N m= F \times d = 25 \times 0.40 = 10\ \text{N m}.

Q2. State the principle of moments. [2 marks]

  • Cue. For a balanced object, the total clockwise moment about a pivot equals the total anticlockwise moment about that pivot.

Q3. Explain why a double-decker bus is designed with its heaviest parts low down. [2 marks]

  • Cue. It keeps the centre of gravity low, so the bus can tilt further before toppling, making it more stable.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA uniform metre rule is balanced at its centre. A 2.0 N2.0\ \text{N} weight hangs 40 cm40\ \text{cm} from the pivot on the left. (a) Find its moment about the pivot. (b) Where must a 4.0 N4.0\ \text{N} weight be hung on the right to balance the rule?
Show worked answer →

(a) Moment =force×perpendicular distance=2.0×0.40=0.80 N m= \text{force} \times \text{perpendicular distance} = 2.0 \times 0.40 = 0.80\ \text{N m} (anticlockwise).

(b) For balance, clockwise moment equals anticlockwise moment: 4.0×d=0.804.0 \times d = 0.80, so d=0.804.0=0.20 m=20 cmd = \dfrac{0.80}{4.0} = 0.20\ \text{m} = 20\ \text{cm} from the pivot.

Markers reward moment as force times perpendicular distance with the distance in metres, applying the principle of moments (clockwise equals anticlockwise), and the correct balancing distance.

Original4 marks(a) State the two conditions for an object to be in equilibrium. (b) Explain why a bus is more stable when its heavy luggage is stored low down rather than on the roof.
Show worked answer →

(a) For equilibrium: the resultant force is zero (forces balance), and the resultant moment about any point is zero (the principle of moments, clockwise moments equal anticlockwise moments).

(b) Storing luggage low keeps the centre of gravity low. A lower centre of gravity makes the bus harder to topple, because it can tilt further before its centre of gravity passes outside the wheelbase, so it is more stable.

Markers reward both equilibrium conditions, and linking low stored mass to a low centre of gravity and hence greater stability.

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