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

These notes condense statements P2.3k, P2.3l and P2.3m of OCR GCSE (9-1) Gateway Science Physics A (J249, Version 5.0, July 2026), topic P2.3 Forces in action. The specification prints all three in bold, so everything here is Higher Tier only. Look in the study guide; test yourself afterwards with the practice set. Course pages: the OCR GCSE Physics hub and the checklist. Not sure where to begin? Use the free 10-minute diagnostics.

Key definitions

Term	Meaning
Pivot	The fixed point an object turns about
Moment	The turning effect of a force about a pivot
Clockwise, anticlockwise	The two directions of turning, seen from one side
Balanced	No resultant turning effect, so the object does not start to turn
Lever	A rigid bar that turns about a pivot
Effort, load	The force you apply, and the force the lever applies to the thing being moved
Force multiplier	A device that gives an output force larger than the input force
Driver, driven gear	The gear that is turned, and the gear it turns

Formulas

Quantity	Equation	Units
Moment	M = F × d	N m
Force from a moment	F = M ÷ d	N
Distance from a moment	d = M ÷ F	m
Weight	W = m × g	N
Principle of moments	total clockwise moment = total anticlockwise moment	N m
Lever balance	effort × effort distance = load × load distance	N m
Gear ratio	teeth on driven gear ÷ teeth on driver gear	no unit
Output moment	input moment × gear ratio	N m
Output speed	input speed ÷ gear ratio	rev/min

In M = F × d, d is the perpendicular distance between the pivot and the force’s line of action. Convert centimetres to metres first. Gravitational field strength is given as data when needed, for example g = 10 N/kg.

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

• A force turns an object only if its line of action does not pass through the pivot.
• Direction: hold the pivot fixed, picture the object moving under that force alone, and decide clockwise or anticlockwise.
• Moving the pivot can reverse the direction.
• Opposite turning effects compete. The bigger one gives the resultant turning effect.
• Examples to quote: door about its hinges, spanner about a nut, see-saw about its support, wheelbarrow about its wheel axle, scissors about the screw joining the blades.

P2.3l: moments and balance (Higher Tier only)

Method for a balance problem

1. Mark the pivot.
2. List each force and its distance from the pivot. Ignore a force acting at the pivot, because its moment is zero.
3. Label each force clockwise or anticlockwise.
4. Work out each moment.
5. Add them in each direction and set the totals equal.
6. Solve, and check the answer is sensible.

Worked reminders

Finding a moment. A force of 45 N acts at right angles to a handle 0.32 m from the pivot. M = 45 × 0.32 = 14.4 N m.

Finding a distance. A moment of 21 N m is needed from a 150 N force. d = 21 ÷ 150 = 0.14 m.

Units. A 30 N force at 25 cm from a pivot: 25 cm = 0.25 m, so M = 30 × 0.25 = 7.5 N m. Writing 30 × 25 = 750 and calling it N m is wrong.

Balance. A clockwise force of 12 N acts 0.50 m from the pivot. An anticlockwise force acts 0.20 m from the pivot on the other side. Clockwise moment = 12 × 0.50 = 6.0 N m. For balance, F × 0.20 = 6.0, so F = 30 N.

Uniform bar. The weight of a uniform bar acts at its midpoint. If the pivot is at the midpoint, the bar’s weight has no moment about it. If the pivot is elsewhere, include the bar’s weight at the midpoint, at its own distance from the pivot.

Support force. For a balanced bar, the upward force at the pivot equals the total downward force, including the weight of the bar.

Practical suggestion

The specification suggests investigating moments with a metre ruler, a pivot and balancing masses. Typical readings: the mass on each side, and its distance from the pivot. Multiply for each side and compare the two moments.

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

Levers

• Balanced lever: effort × effort distance = load × load distance.
• Force ratio = effort distance ÷ load distance.
• A longer effort distance means a smaller effort for the same load. The lever is a force multiplier.
• Trade-off: the effort moves through a larger distance than the load. The lever does not create energy.

Worked reminder. A crowbar has its load 0.15 m from the pivot and its effort 0.90 m from the pivot. The load force is 1800 N. Effort × 0.90 = 1800 × 0.15 = 270, so effort = 300 N. Force ratio = 0.90 ÷ 0.15 = 6.

Gears

• Meshed gears turn in opposite directions.
• The tooth count gives the ratio, so you do not need to measure radii.
• Ratio greater than 1 (small driver, large driven): bigger moment, slower turning.
• Ratio less than 1 (large driver, small driven): smaller moment, faster turning.
• In a train of gear pairs, multiply the ratios.
• Calculations assume no energy loss to friction.

Case	Driver teeth	Driven teeth	Ratio	Moment	Speed
Multiplies force	24	36	1.5	8.0 N m becomes 12 N m	90 rev/min becomes 60 rev/min
Multiplies speed	40	10	0.25	12 N m becomes 3.0 N m	50 rev/min becomes 200 rev/min

Must-know distinctions

• Moment and force. A moment is force × distance (N m). A force is a push or pull (N).
• Balanced moments and balanced forces. Moments can be balanced while the support force is large. A see-saw with equal moments still has an upward force from the pivot.
• Force multiplier and energy multiplier. A lever or gear pair can multiply force. It cannot multiply energy. Force goes up and the distance moved, or the speed, goes down.
• Driver and driven. The ratio is driven ÷ driver. Reverse them and the answer is the wrong way round.

Quick self-test

1. State the equation for the moment of a force and give the unit.
2. A force of 90 N acts at right angles to a spanner at 0.40 m from the nut. Calculate the moment.
3. A moment of 36 N m is needed and the force available is 120 N. How far from the pivot must it act?
4. A 50 N force acts 8.0 cm from a pivot. Calculate the moment in N m.
5. A weight of 18 N hangs 0.30 m to the left of a pivot. What weight hangs 0.45 m to the right if the bar balances?
6. Explain why a force whose line of action passes through the pivot has no turning effect.
7. A load of 600 N is 0.25 m from a pivot. The effort is 1.50 m from the pivot. Calculate the effort.
8. A driver gear has 12 teeth and a driven gear has 48 teeth. State the gear ratio and say what happens to the speed.
9. The driver in question 8 applies a moment of 1.5 N m. Calculate the output moment.
10. A gear train has two pairs, with ratios 3 and 2.5. State the overall ratio.
11. Explain in terms of distance why a lever that multiplies force cannot also multiply energy.
12. A uniform 2.0 m plank weighing 100 N is pivoted 0.50 m from one end. State where its weight acts and give its distance from the pivot.

Answers

1. Moment = force × perpendicular distance between the pivot and the force’s line of action. Unit: newton metre (N m).
2. 90 × 0.40 = 36 N m.
3. d = 36 ÷ 120 = 0.30 m.
4. 8.0 cm = 0.080 m. M = 50 × 0.080 = 4.0 N m.
5. Anticlockwise moment = 18 × 0.30 = 5.4 N m. W × 0.45 = 5.4, so W = 12 N.
6. The force’s line of action goes through the pivot, so d is zero and force × distance is zero.
7. Effort × 1.50 = 600 × 0.25 = 150, so effort = 100 N.
8. Ratio = 48 ÷ 12 = 4. The driven gear turns four times more slowly.
9. 1.5 × 4 = 6.0 N m.
10. 3 × 2.5 = 7.5.
11. The effort moves through a larger distance than the load. The extra force is traded for a longer distance, so the work done on the load does not exceed the work put in.
12. At its midpoint, 1.0 m from the end, so 0.50 m from the pivot on the longer side.

Where marks are usually lost

• Using a distance in centimetres in M = F × d, so the answer is 100 times too big.
• Measuring the distance to the end of the bar instead of to the pivot.
• Leaving out the weight of a uniform bar when the pivot is not at its midpoint.
• Giving the unit as N instead of N m, or writing Nm without a space or with the wrong case.
• Inverting the gear ratio and getting a smaller output moment from a larger driven gear.
• Saying “balanced forces” when the question is about balanced moments.
• Writing that a lever “gives you extra energy” instead of “a smaller force over a larger distance”.
• Not showing the clockwise and anticlockwise sums separately, so a slip cannot be traced.
• Giving no direction, clockwise or anticlockwise, when the question asks for the resultant turning effect.

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.