OCR GCSE Physics A: Moments, levers and gears (J249)

This guide teaches statements P2.3k, P2.3l and P2.3m of the OCR GCSE (9-1) Gateway Science Physics A specification (J249, Version 5.0, July 2026), within topic P2.3 Forces in action. The specification prints all three in bold, which marks them as Higher Tier only. Foundation Tier papers do not test them. Turn to the revision notes when you need fast recall, and the practice set for questions. The rest of topic P2 is in the forces guide. Course pages are at the OCR GCSE Physics hub and the printable checklist. Not sure what you already know? Try the free 10-minute diagnostics first.

What you must be able to do

Statement	Outcome	Tier
P2.3k	Describe examples in which forces cause rotation, including the location of pivot points and whether the resultant turning force is clockwise or anticlockwise	Higher Tier only
P2.3l	Define and calculate the moment of a force, and apply the principle of moments to objects that are balanced	Higher Tier only
P2.3m	Explain how levers and gears transmit the rotational effects of forces, using ratios, so that they work as force multipliers	Higher Tier only

P2.3k: forces that cause rotation (Higher Tier only)

A force can change how an object moves in a straight line. It can also turn an object about a fixed point. That fixed point is the pivot. A turning effect needs a force whose line of action does not pass through the pivot.

Everyday examples:

• A door turns about its hinges when you push the handle.
• A spanner turns a nut about the centre of the nut.
• A see-saw tips about the support under its middle.
• A wheelbarrow tips up about the axle of its wheel when you lift the handles.

To decide the direction, imagine the pivot is held fixed and ask which way the object would turn if only that force acted. Looking at the object from a given side, it turns clockwise (like clock hands) or anticlockwise. If two forces turn the object in opposite directions, the larger turning effect wins. That gives a resultant turning effect in one direction. If the two effects are equal, there is no resultant turning effect and the object does not start to rotate.

Moving the pivot changes the answer. A downward push to the right of a pivot is clockwise. The same push to the left is anticlockwise.

Worked example. A uniform plank rests on a support under its middle. A weight hangs on the left side and a weight hangs on the right side.

Force	Side of pivot	Direction of turning
Weight on left, pulling down	Left	Anticlockwise
Weight on right, pulling down	Right	Clockwise
Upward support force at the pivot	At the pivot	None, because its line of action passes through the pivot

A force acting through the pivot has no turning effect about it. This idea returns in the next section.

P2.3l: the moment of a force (Higher Tier only)

The moment of a force is the size of its turning effect about a pivot. It depends on how big the force is and how far from the pivot it acts:

moment (N m) = force (N) × perpendicular distance between the pivot and the force’s line of action (m)

or M = F × d.

Measure the distance at 90 degrees to the force’s line of action. If a question gives a distance along a slanted bar, check whether it is the perpendicular distance. If the force pushes straight along the line towards the pivot, the perpendicular distance is zero and so is the moment.

You can rearrange the equation: F = M ÷ d, and d = M ÷ F. Keep distances in metres. If a distance is given in centimetres, divide by 100 first.

Worked example 1: finding a moment. A mechanic pulls with a force of 65 N on a wrench, at right angles to it, 0.28 m from the centre of a bolt.

M = F × d = 65 × 0.28 = 18.2 N m

The answer is 18 N m to 2 significant figures.

Worked example 2: finding a distance. The same bolt needs a moment of 24 N m. The mechanic can push with 80 N. Rearranging, d = M ÷ F = 24 ÷ 80 = 0.30 m. A wrench at least 0.30 m long is needed.

Worked example 3: why handles sit far from hinges. A door handle is 0.85 m from the hinges. A force of 12 N is applied at right angles to the door.

M = 12 × 0.85 = 10.2 N m

Pushing at 0.20 m from the hinges to get the same moment needs F = 10.2 ÷ 0.20 = 51 N. The force is about four times larger, because the distance is about four times smaller.

The principle of moments

An object is balanced (in equilibrium, not turning) when, about one chosen pivot, the clockwise moments add up to the same total as the anticlockwise moments:

sum of clockwise moments = sum of anticlockwise moments

Method for balance problems:

1. Choose the pivot and mark it.
2. List every force that is not at the pivot, with its distance from the pivot.
3. Sort the forces into clockwise and anticlockwise.
4. Work out each moment, using M = F × d.
5. Add the moments in each direction and set the totals equal.
6. Solve for the unknown.

A force of weight is found from W = m × g. In the examples below, g is taken as 10 N/kg, which would be given to you in a question as data. The weight of a uniform object, such as a regular plank or metre rule, acts at its midpoint.

Worked example 4: a plank. A uniform plank is pivoted at its centre, so its own weight acts through the pivot and has no moment. A girl of weight 380 N sits 2.1 m to the left of the pivot. A man of weight 840 N sits on the right. How far from the pivot must he sit?

Anticlockwise moment = 380 × 2.1 = 798 N m. Clockwise moment = 840 × d. For balance, 840 × d = 798, so d = 798 ÷ 840 = 0.95 m.

The plank is not accelerating up or down, so the upward force from the support equals the total downward force: 380 + 840 = 1220 N, plus the weight of the plank itself. If the plank weighs 300 N, the support force is 1520 N.

Worked example 5: a metre rule. This matches the practical suggestion in the specification, an investigation of moments using a metre ruler, a pivot and balancing masses. A uniform metre rule of mass 0.120 kg is pivoted at the 40 cm mark. A 0.250 kg mass hangs at the 10 cm mark. Where must a 0.200 kg mass hang to balance the rule?

Weights: rule 0.120 × 10 = 1.2 N, at the 50 cm mark. Left mass 0.250 × 10 = 2.5 N. Right mass 0.200 × 10 = 2.0 N.

• Anticlockwise: the 2.5 N weight is 0.40 − 0.10 = 0.30 m from the pivot, so its moment is 2.5 × 0.30 = 0.75 N m.
• Clockwise from the rule: its weight is 0.50 − 0.40 = 0.10 m from the pivot, so its moment is 1.2 × 0.10 = 0.12 N m.
• Clockwise from the 2.0 N weight: 2.0 × x.

Balance: 0.12 + 2.0x = 0.75, so 2.0x = 0.63 and x = 0.315 m. The mass hangs 0.315 m to the right of the pivot, at the 71.5 cm mark.

P2.3m: levers and gears (Higher Tier only)

Levers

A lever is a rigid bar that turns about a pivot. You apply an effort force at one point, and the lever applies a load force at another. When the lever is balanced or moving steadily:

effort × distance of effort from pivot = load × distance of load from pivot

So the moments are equal, and the ratio of the distances is the ratio of the forces:

force ratio = load ÷ effort = distance of effort from pivot ÷ distance of load from pivot

If the effort is further from the pivot than the load, the effort can be smaller than the load. The lever is a force multiplier. A crowbar, a wheelbarrow, a nutcracker and a long spanner all work this way.

Worked example 6. A wheelbarrow carries a load of 480 N. The load acts 0.60 m from the axle of the wheel, which is the pivot. The gardener lifts the handles 1.20 m from the axle.

Effort × 1.20 = 480 × 0.60 = 288, so effort = 288 ÷ 1.20 = 240 N.

The ratio 1.20 ÷ 0.60 = 2, so the lever halves the force needed. There is a cost. The handles move twice as far as the load does, so you do not get extra energy for free.

If the effort is nearer the pivot than the load, the effort must be larger than the load, although the load moves further. Tweezers are like this. The specification asks about force multipliers, so concentrate on the longer effort arm.

Gears

Gears are toothed wheels that mesh, so that turning one makes the other turn in the opposite direction. The gear you turn is the driver. The gear it turns is the driven gear. At the point where the teeth touch, the two gears push on each other with equal and opposite forces. The moment of a force is force × distance, so the larger gear has a larger distance from its centre to the contact point, and it receives a larger moment from the same force.

The sizes of gears are compared by counting teeth.

gear ratio = number of teeth on driven gear ÷ number of teeth on driver gear

• Moment on the driven gear = moment on the driver × gear ratio.
• Speed of the driven gear = speed of the driver ÷ gear ratio.

A driver with few teeth turning a driven gear with many teeth gives a ratio greater than 1. The output moment is larger, which is a force multiplier, but the output turns more slowly. This is the choice when a cyclist needs to climb a hill. If the driven gear is the smaller one, the ratio is less than 1. The output turns faster but with a smaller moment.

These calculations assume no energy is lost to friction.

Worked example 7. A driver gear with 15 teeth turns at 240 revolutions per minute and applies a moment of 2.5 N m. It meshes with a driven gear with 60 teeth.

• Ratio = 60 ÷ 15 = 4.
• Output moment = 2.5 × 4 = 10 N m.
• Output speed = 240 ÷ 4 = 60 revolutions per minute.

Worked example 8: two pairs of gears. The 60-tooth gear is fixed to the same shaft as a 20-tooth gear, which drives a 50-tooth gear. The second ratio is 50 ÷ 20 = 2.5. Ratios in a train multiply, so the overall ratio is 4 × 2.5 = 10.

• Output moment = 2.5 × 10 = 25 N m.
• Output speed = 240 ÷ 10 = 24 revolutions per minute.

Common errors

• Using the length of the whole bar instead of the distance from the pivot to the force.
• Forgetting the weight of the bar itself when the pivot is not at its midpoint.
• Treating a moment as a force. A moment is measured in N m, not N.
• Inverting the gear ratio, so that the moment goes down when it should go up.
• Saying a force multiplier gives more energy. It gives a bigger force over a smaller distance.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249). Specification Version 5.0, July 2026; first assessment 2018. Topic P2.3, statements P2.3k to P2.3m.