Moments, levers and gears (physics only) — AQA GCSE Physics

SPEC 4.5.4 • TOPIC 5 FORCES — PAPER 2

Moments, Levers & Gears

The turning effect of a force. One equation, one principle, and the unit conversion that costs a mark a year.

⚡ PHYSICS ONLY — NOT IN COMBINED SCIENCE

What You'll Learn

Moments are the physics of turning forces — door handles, spanners, seesaws, gears. The equation is short. The problems are short. The trap is the unit conversion.

Lesson Goals

Describe everyday examples in which forces cause rotation (door handles, spanners, levers, gears)

Define a moment, a pivot, and the principle of moments in mark-scheme language

Apply M = F d (and its rearranged forms) with correct unit conversion (cm → m) at substitution

Use the principle of moments to find an unknown force or distance at equilibrium

Explain how levers transmit force and how gears change the size of a moment

1

Equation to recall

⭐

High-priority spec point

24%

Lost a mark on Jun 2024

Key Definitions

Mark-scheme wording. Learn these word-for-word — examiners are strict on the word "perpendicular" and on "clockwise = anticlockwise".

Definition

Moment of a force: the turning effect of a force about a pivot. Calculated using M = F d, where d is the perpendicular distance from the pivot to the line of action of the force.

Definition

Pivot: the fixed point about which an object can rotate.

Definition

Principle of moments: for an object in equilibrium, the sum of the clockwise moments about a pivot equals the sum of the anticlockwise moments about that pivot.

The Equation — Recall List

M = F d is a recall equation — it is not on the Physics Equations Sheet. You must know it from memory.

Moment of a forceM = F × dM — moment, N mF — force, Nd — distance, m

Rearranged forms — for "find F" or "find d" questions

F = M ÷ d | d = M ÷ F

The spec specifically requires you to "calculate the size of a force, or its distance from a pivot, acting on an object that is balanced" — so both rearrangements are examinable.

Critical: d is perpendicular distance

d is the perpendicular distance from the pivot to the line of action of the force — not the straight-line distance from the pivot to where the force is applied. If the force is applied at an angle, you must use the perpendicular component.

Why "Perpendicular"?

The further from the pivot — and the more directly the force pushes at right angles — the bigger the turning effect. Pushing a door at the handle takes less force than pushing near the hinge for exactly this reason.

Intuition

Same force, applied twice as far from the pivot → twice the moment. Same distance, twice the force → twice the moment. M = F × d is just that relationship as an equation.

When the force isn't perpendicular

If a force is applied at an angle, only the perpendicular component contributes to the turning effect. A force pulling along the spanner (towards or away from the pivot) produces zero moment — it doesn't turn the bolt at all. This is exactly why the word "perpendicular" appears in every mark scheme definition.

ESUA Worked Example

A spanner exerts a force of 80 N at a perpendicular distance of 25 cm from a bolt.

Calculate the moment of the force on the bolt. 3 MARKS

E

Equation: M = F × d

S

Substitute (convert first!): d = 25 cm = 0.25 m → M = 80 × 0.25

U

Unit: newton metres (N m)

A

Answer: M = 20 N m

Principle of Moments

When something is balanced — a seesaw, a pair of scales, a balance beam — it isn't rotating. So the turning effects on either side of the pivot must cancel out.

In equilibrium

Σ clockwise moments = Σ anticlockwise moments

F₁ × d₁ = F₂ × d₂

Worked Example — Balance

A seesaw is balanced. A child of weight 300 N sits 1.6 m from the pivot. A second child sits 1.2 m from the pivot on the opposite side.

Calculate the weight of the second child. 3 MARKS

E

Equation: at equilibrium, clockwise moment = anticlockwise moment → F₁d₁ = F₂d₂

S

Substitute and rearrange: 300 × 1.6 = F₂ × 1.2 → F₂ = (300 × 1.6) ÷ 1.2

U

Unit: newtons (N)

A

Answer: F₂ = 400 N

Levers

A lever lets a small force at a large distance from the pivot balance a large force at a small distance. That's how a crowbar lifts a paving slab and how a wheelbarrow lets you carry a heavy load.

Mark-scheme phrasing

A simple lever transmits the rotational effect of a force. Because the input force acts at a greater distance from the pivot than the load, its moment can be much greater — so a small input force balances a large output force.

The Trade-off — Force vs Distance Moved

A lever multiplies force, but at the cost of distance moved. You push the end of a crowbar through a long arc to lift a heavy load by only a few centimetres. There's no free energy here — work done in equals work done out.

✓ What multiplies

ForceSmall input force × large input distance = large output force × small output distance.

✗ What's traded

Distance movedYou move the input end a long way; the load only moves a short way.

Examples to know

Crowbar — lifting paving slabs. Wheelbarrow — carrying heavy loads with the wheel as the pivot. Spanner — loosening a tight bolt by applying a moment at the end of a long arm.

Gears — Transmitting Rotation

Gears transmit the rotational effects of forces. When two gears mesh, the teeth of the driving gear push the teeth of the driven gear — so one moment causes another moment about a different pivot.

Direction rule

When two gears mesh, they rotate in opposite directions. (If gear A turns clockwise, gear B turns anticlockwise.)

Why the moment changes size

At the meshing teeth, the force one gear exerts on the other is the same (Newton's 3rd law). But each gear pivots about its own centre, so M = F × radius. The gear with the larger radius experiences the larger moment. That's the whole reason gears change the size of a moment.

Small Gear ↔ Large Gear

Gears change the size of the moment in a transmission system. The direction of the change depends on which gear drives which.

SetupMomentRotational speedWhere you see it

Small → Large (small gear drives a large one)Increases ↑Decreases ↓Low gear on a bike — climbing a hill

Large → Small (large gear drives a small one)Decreases ↓Increases ↑High gear on a bike — going downhill fast

Memory aid

"Big wheel = big moment, slow turn." The larger gear always carries the larger moment but turns more slowly. The smaller gear always turns faster but carries a smaller moment.

Exam-Style Multi-Part Question

A clock mechanism contains two meshed gears. Gear A is small. Gear B is larger. A force of 4.0 N is applied to the rim of gear A at a perpendicular distance of 6.0 cm from the pivot of gear A.

(a) Write down the equation linking moment of a force (M), force (F) and distance (d). 1 MARK

(b) Calculate the moment of the force about the pivot of gear A. 3 MARKS

(c) Describe what happens to gear B when gear A rotates. 2 MARKS

(a)

M = F × d (1 mark)

(b)

Convert: 6.0 cm = 0.060 m (M1)

Substitute: M = 4.0 × 0.060 (M1)

Answer: M = 0.24 N m (A1)

(c)

Gear B rotates in the opposite direction to gear A (1)

Gear B rotates more slowly than gear A / gear A exerts a force on gear B causing a moment about its pivot (1)

Examiner Alert

Lost-marks alert — recurring examiner finding

On the most recent moment calculation set by AQA on Paper 2 (a multi-mark M = F d question with a cm value), around 75% of students scored every mark. The remaining ~24% lost a mark because they substituted the cm value directly into M = F d without converting to metres.

The rule is absolute: distance in M = F d is in metres. Always convert at the substitution step, before multiplying. Do not multiply first and "convert later" — that's where the slip happens.

Convert at substitution

7.5 cm → 0.075 m | 25 cm → 0.25 m | 4 cm → 0.04 m

Weak vs Strong Answers

Same question. Two students. One mark.

✗ Weak

"M = 80 × 25 = 2000 N m"Distance left in cm. Magnitude is wrong by a factor of 100, and "N m" is being claimed when the substituted value was in cm — examiner cannot award the answer mark.

✓ Strong

"25 cm = 0.25 m; M = 80 × 0.25 = 20 N m"Conversion shown at substitution. All three marks awarded: equation (M1), substitution (M1), answer with unit (A1).

✗ Weak

"Gear B turns."No direction stated. Mark scheme needs "opposite direction" OR "clockwise/anticlockwise". One mark lost.

✓ Strong

"Gear B rotates in the opposite direction to gear A because gear A exerts a force on gear B."Direction stated; mechanism (force/moment) explained. Both marks.

Common Errors

Students often writeCorrect versionWhy it matters

Distance in cm substituted directly into M = F dConvert cm → m at the substitution stepWrong magnitude and wrong unit — loses the answer mark

"Distance from the pivot" (no qualifier)" Perpendicular distance from the pivot to the line of action of the force"Mark schemes specifically require "perpendicular"

"Clockwise moment > anticlockwise" at balanceAt equilibrium: clockwise = anticlockwiseWrong setup gives wrong answer for unknown F or d

"The gear turns" (no direction)Driven gear turns in the opposite direction to the driverDirection is a separate mark in 2-mark gear questions

Answer given as "20" with no unitAnswer given as "20 N m""Calculate" questions always require a unit

Checkpoint

You can now

Define a moment as the turning effect of a force about a pivot

Apply M = F d with confident unit conversion at substitution

State the principle of moments (clockwise = anticlockwise at equilibrium)

Use the principle of moments to find an unknown force or distance on a balance

Explain how a lever transmits force (and trades distance moved)

Explain how gears change the size of a moment in a transmission system

One thing to take away

Convert cm to m at the substitution step. Every year, this single habit separates the 75% who get full marks on a moment question from the 24% who don't.

Next lesson

5.07 — Pressure in a fluid; atmospheric pressure (physics only). Two equations: p = F/A (recall) and p = h ρ g (HT, equation sheet). The HT version is the high-value one.