Moments, levers and gears (physics only) – GCSE Physics exam questions with mark scheme

1Grade 1–3FoundationTriple only3 marks

A mechanic uses a spanner to tighten a wheel nut.

(a) Write down the equation that links distance, force and moment of a force. [1]

(b) The mechanic applies a force of 40 N to the spanner. The perpendicular distance from the line of action of the force to the centre of the nut is 0.20 m.

Calculate the moment of the force. [2]

(a) moment = force × distance allow $M = Fd$; allow 'distance' to be 'perpendicular distance'; do not accept an equation written only as units(1)
(b) $M = 40 \times 0.20$(1)
(b) 8.0 N m allow 8 N m; allow Nm; an answer of 8 (N m) with no working scores 2 marks(1)
Answer: (a) moment = force × (perpendicular) distance (b) 8.0 N m
Examiner tip: The unit of moment is the newton metre (N m) — writing 'N/m' or 'Nm²' loses the answer mark.

2Grade 4–5Triple only3 marks

The diagram shows two children, A and B, sitting on a seesaw. The seesaw is balanced and horizontal.

The weight of the seesaw acts through the pivot.

Child A weighs 300 N and sits 1.6 m from the pivot. Child B weighs 400 N.

Calculate the distance $d$ of child B from the pivot. [3]

(clockwise moment = anticlockwise moment so) $400 \times d = 300 \times 1.6$ allow moment of A = 480 (N m) for this mark(1)
$d = \dfrac{480}{400}$(1)
1.2 m allow 120 cm; an answer of 1.2 (m) with no working scores 3 marks(1)
Answer: $d$ = 1.2 m
Examiner tip: Set the total clockwise moment equal to the total anticlockwise moment before rearranging — don't just divide the two weights.

3Grade 4–5Triple only5 marks

A gardener lifts the handles of a wheelbarrow. The wheelbarrow pivots about the axle of the wheel.

The weight of the wheelbarrow and its load is 600 N. The diagram shows the horizontal distances from the axle.

(a) Calculate the upward force $F$ the gardener must apply to the handles to just lift the wheelbarrow. [3]

(b) The gardener moves the load in the wheelbarrow closer to the wheel.

Explain how this changes the force needed to lift the handles. [2]

(a) moment of load $= 600 \times 0.40 = 240$ (N m)(1)
(a) $F \times 1.20 = 240$ allow $F = 240 \div 1.20$(1)
(a) $F$ = 200 N an answer of 200 (N) with no working scores 3 marks(1)
(b) (perpendicular) distance of the weight from the pivot/axle decreases so the moment of the load decreases(1)
(b) (so) a smaller force is needed (at the handles, which are the same distance from the axle) do not award this mark without a reason(1)
Answer: (a) 200 N (b) smaller moment of the load, so smaller force needed
Examiner tip: In (b) say what happens to the moment of the load, not just 'it is easier to lift'.

4Grade 1–3FoundationTriple only5 marks

The diagram shows a spanner on a nut. Three different forces, P, Q and R, can be applied to the spanner, one at a time. Each force is 20 N.

(a) Which force, P, Q or R, produces the largest turning effect on the nut? [1]

(b) Calculate the moment of force Q about the centre of the nut. [2]

Explain why. [2]

(a) Q(1)
(b) $M = 20 \times 0.25$(1)
(b) 5.0 N m allow 5 N m; an answer of 5 (N m) with no working scores 2 marks(1)
(c) (the line of action of) force R passes through the pivot / centre of the nut allow R acts along the spanner towards the nut(1)
(c) (so) the perpendicular distance (from the pivot) is zero, so the moment is zero allow there is no perpendicular distance(1)
Answer: (a) Q (b) 5.0 N m (c) its line of action passes through the pivot, so perpendicular distance and moment are zero
Examiner tip: The distance in $M = Fd$ is the perpendicular distance from the pivot to the line of action of the force — for R it is zero even though R acts at the end.

5Grade 6–7Triple only5 marks

The diagram shows two gears in a machine. A motor turns gear A clockwise. Gear A turns gear B.

(a) State the direction in which gear B turns. [1]

(b) Gear A makes 6 complete turns.

How many complete turns does gear B make? [1]

Explain why. [3]

(a) anticlockwise(1)
(b) 2(1)
(c) the force between the teeth of the two gears is the same (size) allow the teeth of A push on the teeth of B with the same force as B pushes back on A(1)
(c) gear B has a larger radius / the teeth of B are a larger (perpendicular) distance from its axle/pivot(1)
(c) moment = force × distance, so the moment on gear B is larger allow (gear B gives a) moment 3 times bigger(1)
Answer: (a) anticlockwise (b) 2 (c) same force at the teeth acts at a larger distance from B's axle, so larger moment
Examiner tip: Gears pass on the same force at the teeth; the moment changes only because the radius (distance from the axle) is different.

6Grade 8–9HigherTriple only6 marks

A small electric winch is used to lift loads. A motor turns gear A, which turns a larger gear B. Gear B is fixed to the same axle as a drum. A rope is wound around the drum and the load hangs from the rope.

radius of gear A = 2.0 cm
radius of gear B = 6.0 cm
radius of drum = 4.0 cm

The motor exerts a moment of 0.30 N m on gear A. Assume there is no friction.

(a) Calculate the force that the teeth of gear A exert on the teeth of gear B. [2]

(b) Calculate the moment that this force exerts on gear B. [2]

(a) $0.30 = F \times 0.020$ allow $F = 0.30 \div 0.020$; distance must be converted to metres(1)
(a) 15 N an answer of 15 (N) with no working scores 2 marks(1)
(b) $M = 15 \times 0.060$ allow their (a) × 0.060(1)
(b) 0.90 N m allow 0.9 N m; allow a correct answer using an incorrect value from (a) (ecf)(1)
(c) $0.90 = W \times 0.040$ allow their (b) = W × 0.040(1)
(c) 22.5 N allow 23 N; allow ecf from (b)(1)
Answer: (a) 15 N (b) 0.90 N m (c) 22.5 N
Examiner tip: Convert cm to m before using $M = Fd$; an answer of 0.15 N in (a) shows the conversion was missed.

7Grade 6–7Triple only4 marks

A student balances a uniform metre rule on a pivot at the 50 cm mark. She hangs three weights from the rule, as shown in the diagram. The rule is balanced and horizontal.

a 2.0 N weight at the 10 cm mark
a 3.0 N weight at the 30 cm mark
an unknown weight W at the 90 cm mark

Calculate W. [4]

distances from pivot: 0.40 m, 0.20 m and 0.40 m allow 40 cm, 20 cm, 40 cm(1)
total anticlockwise moment $= (2.0 \times 0.40) + (3.0 \times 0.20) = 1.4$ (N m) allow 140 N cm(1)
$W \times 0.40 = 1.4$ allow total clockwise moment = total anticlockwise moment stated or used(1)
W = 3.5 N an answer of 3.5 (N) with no working scores 4 marks; ignore the weight of the rule (it acts through the pivot)(1)
Answer: W = 3.5 N
Examiner tip: Measure each distance from the pivot (50 cm mark), not from the end of the rule — using 10 cm, 30 cm and 90 cm gives the wrong answer.

8Grade 6–7Triple only6 marks

A student investigates the force needed to keep a lever balanced. A light beam is pivoted at one end. A 6.0 N weight hangs 0.20 m from the pivot.

The student uses a newtonmeter to pull upwards on the beam at a distance $d$ from the pivot. She records the force $F$ needed to keep the beam horizontal.

$d$ in m	0.10	0.15	0.20	0.25	0.30	0.40
$F$ in N	12.0	8.0	6.0	5.6	4.0	3.0

(a) Calculate the moment of the 6.0 N weight about the pivot. [2]

(b) One of the results is anomalous. Identify the anomalous result. [1]

(d) Predict the force needed when $d$ = 0.60 m. [1]

(a) $M = 6.0 \times 0.20$(1)
(a) 1.2 N m an answer of 1.2 (N m) with no working scores 2 marks(1)
(b) 5.6 N (at 0.25 m) allow the result at $d$ = 0.25 m(1)
(c) calculates $F \times d$ for at least three (non-anomalous) results, all equal to 1.2 allow shows that when $d$ doubles, $F$ halves, eg 0.10 → 0.20 m, 12.0 → 6.0 N(1)
(c) $F \times d$ is constant (so $F$ is inversely proportional to $d$) allow a second correct doubling/halving pair, eg 0.20 → 0.40 m, 6.0 → 3.0 N(1)
(d) 2.0 N allow 2 N(1)
Answer: (a) 1.2 N m (b) 5.6 N at 0.25 m (should be 4.8 N) (c) $F \times d$ = 1.2 for each result (d) 2.0 N
Examiner tip: To show inverse proportion, calculate $F \times d$ for several pairs and say it stays constant — 'as d goes up F goes down' is not enough.

9Grade 8–9HigherTriple only6 marks

The diagram shows the horizontal arm (jib) of a tower crane. The jib can turn about point P at the top of the tower.

A counterweight of 60 000 N is fixed 8.0 m from P. The weight of the jib is 20 000 N and acts 6.0 m from P. A load of weight L hangs 24 m from P.

(a) Calculate the value of L when the jib is balanced. [4]

(b) The crane operator needs to lift a heavier load.

Explain why the load must be lifted closer to the tower to stop the crane toppling. [2]

(a) anticlockwise moment of counterweight $= 60\,000 \times 8.0 = 480\,000$ (N m)(1)
(a) clockwise moment of jib $= 20\,000 \times 6.0 = 120\,000$ (N m)(1)
(a) $480\,000 = 120\,000 + (L \times 24)$ allow total clockwise moment = total anticlockwise moment with both clockwise moments included(1)
(a) L = 15 000 N an answer of 15 000 (N) with no working scores 4 marks; an answer of 20 000 N (jib weight ignored) scores 1 mark(1)
(b) a heavier load at the same distance would produce a larger (clockwise) moment (than the anticlockwise moment)(1)
(b) moving it closer reduces the (perpendicular) distance so the moment of the load is reduced so the moments stay balanced allow so the resultant moment is (close to) zero(1)
Answer: (a) 15 000 N (b) smaller distance keeps the load's moment small enough to balance the counterweight's moment
Examiner tip: There are two clockwise moments here — forgetting the weight of the jib gives 20 000 N.

10Grade 8–9HigherTriple only6 marks

On a bicycle, the cyclist pushes down on a pedal. The pedal is on the end of a crank that turns the front gear wheel (chainring). A chain connects the chainring to a smaller gear wheel (sprocket) fixed to the rear wheel.

To ride up a steep hill, the cyclist changes to a larger rear sprocket.

Explain how the turning effect of the cyclist's force is transmitted to the rear wheel, and why changing to a larger rear sprocket makes it easier to ride up the hill. [6]

Pedal and crank (lever): the downward force on the pedal produces a moment about the crank axle; moment = force × perpendicular distance from the axle; the moment is largest when the crank is horizontal (and zero when it is vertical)(1)
Chainring and chain: the moment turns the chainring; this pulls the chain with a force (tension); force in chain = moment ÷ radius of chainring(1)
Rear sprocket: the same force in the chain acts on the teeth of the rear sprocket; this produces a moment on the rear wheel = chain force × radius of the rear sprocket(1)
Larger rear sprocket: larger radius / larger perpendicular distance from the rear axle; so for the same pedal force the moment on the rear wheel is larger; so the driving force (the forward push of the road on the tyre) is larger (allow 'the tyre pushes harder on the road'); this can overcome the extra force needed to go uphill (ignore 'gears change the energy'; do not credit 'the gear produces more force' without reference to radius / perpendicular distance)(1)
Trade-off: the rear wheel makes fewer turns for each turn of the pedals; so the bicycle moves a shorter distance per pedal turn / goes more slowly unless the cyclist pedals faster(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically linked explanation that follows the turning effect from the pedal through the chain to the rear wheel, uses force, perpendicular distance and moment correctly at each stage, and explains why the larger radius of the rear sprocket gives a larger moment on the rear wheel and so a larger driving force.
Level 2	3–4	A partly linked explanation. Moments are used for at least two parts of the system, and there is a relevant reason why the larger sprocket helps, but some steps are missing or unclear.
Level 1	1–2	Simple, separate statements about levers, gears or moments, with little or no linking to climbing the hill.
No relevant content	0	Nothing worthy of credit.

Answer: Moment on crank → chain force = moment ÷ chainring radius → same chain force × larger rear-sprocket radius = larger moment on rear wheel → larger driving force; but fewer wheel turns per pedal turn (slower).
Examiner tip: Top-band answers follow the force step by step and say why a larger radius gives a larger moment — the chain force itself does not change.

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