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What is the turning effect of a force, and how do moments, levers and gears work?

The moment of a force as a turning effect, the principle of moments for a balanced object, levers and gears as force multipliers, and applications such as spanners and seesaws.

A focused answer to OCR Gateway GCSE Physics A topic P2 on moments, covering the turning effect of a force, the moment equation, the principle of moments for a balanced object, and how levers and gears act as force multipliers.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. The moment of a force
  3. The principle of moments
  4. Levers and gears
  5. Try this

What this topic is asking

OCR wants you to define the moment of a force, apply the principle of moments to a balanced object, and explain how levers and gears multiply forces. This is part of topic P2.3 (Forces in action) of the OCR Gateway Physics A (J249) specification.

The moment of a force

Two things increase a moment: a larger force and a larger perpendicular distance from the pivot. This is why a longer spanner undoes a stiff nut more easily, why door handles are placed far from the hinges, and why a crowbar with a long handle lifts a heavy load. The distance must be measured at right angles to the force.

The principle of moments

So on a balanced seesaw, a heavy child sitting close to the pivot can balance a lighter child sitting far from it, because moment depends on both force and distance. To use the principle, work out each moment (F×dF \times d) on each side, set the clockwise total equal to the anticlockwise total, and solve for the unknown.

Levers and gears

A lever is a force multiplier. A force (the effort) applied at a large distance from the pivot produces a large moment, which can balance or overcome a larger force (the load) acting at a small distance. Crowbars, scissors, wheelbarrows and bottle openers all use this principle to let a small effort move a large load.

Gears also transmit and multiply the turning effect of a force. A small gear driving a larger gear turns it more slowly but with a greater moment (turning effect), while a large gear driving a smaller one gives a faster but weaker turn. This is how the gears on a bicycle or in a machine match the turning effect to the job.

Try this

Q1. A force of 20 N20\,\text{N} acts at a perpendicular distance of 0.50 m0.50\,\text{m} from a pivot. Calculate the moment. [2 marks]

  • Cue. moment =F×d=20×0.50=10 N m= F \times d = 20 \times 0.50 = 10\,\text{N m}.

Q2. State the principle of moments. [1 mark]

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

Exam-style practice questions

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

OCR 20183 marksA spanner is used to undo a nut. A force of 40 N40\,\text{N} is applied at the end of the spanner, 0.25 m0.25\,\text{m} from the nut, at right angles to the spanner. Calculate the moment of the force about the nut, and state one way to increase the moment.
Show worked answer →

A P2 Calculate question on the moment equation, moment =F×d= F \times d. Substitute: moment =40×0.25=10 N m= 40 \times 0.25 = 10\,\text{N m} (2 marks for the equation and the answer with units). To increase the moment you could use a longer spanner (a greater distance from the pivot) or apply a larger force (1 mark). Markers reward the correct calculation, the unit newton metres, and a valid way to increase the moment. A common error is to omit the unit or to use the wrong distance.

OCR 20214 marksA uniform beam is balanced on a pivot. A 30 N30\,\text{N} weight is placed 0.40 m0.40\,\text{m} to the left of the pivot. Calculate the distance to the right at which a 20 N20\,\text{N} weight must be placed to balance the beam, using the principle of moments.
Show worked answer →

A P2 question applying the principle of moments. For a balanced beam, the total clockwise moment equals the total anticlockwise moment (1 mark). The anticlockwise moment from the 30 N30\,\text{N} weight is 30×0.40=12 N m30 \times 0.40 = 12\,\text{N m} (1 mark). Setting the clockwise moment equal: 20×d=1220 \times d = 12, so d=1220=0.60 md = \dfrac{12}{20} = 0.60\,\text{m} (2 marks for the rearrangement and answer). Markers reward stating the principle of moments, calculating one moment, and solving for the unknown distance. A common error is to add rather than balance the moments.

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