Calculate the moment of a 60 N force acting 0.5 m from a pivot. [2 marks]

AQA AQA 2021-style practice: A pulley system lifts a load of 600 N using an effort of 250 N. To raise the load by 0.5 m the effort moves 1.5 m. Calculate the mechanical advantage, the velocity ratio and the efficiency of the system.

A good answer takes the three quantities in order with units and a final percentage. Mechanical advantage = loadeffort = 600250 = 2.4. Velocity ratio = distance moved by effortdistance moved by load = 1.50.5 = 3. Efficiency = MAVR × 100\% = 2.43 × 100\% = 80\%. The system is 80\% efficient; the lost 20\% goes to friction in the pulleys and the weight of the moving parts. Markers reward MA = 2.4, VR = 3, and efficiency = 80\%, each shown clearly with the correct formula.

EnglandEngineeringSyllabus dot point

How do we calculate forces, moments and the advantage a machine gives?

Forces, moments and the principle of moments, mechanical advantage and velocity ratio, and efficiency in simple machines.

A focused answer to AQA GCSE Engineering on forces, moments and the principle of moments, mechanical advantage, velocity ratio and the efficiency of simple machines such as levers and pulleys.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
Forces and moments
The principle of moments
Mechanical advantage, velocity ratio and efficiency
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What this dot point is asking

AQA wants you to calculate forces and moments, apply the principle of moments to a balanced system, work out mechanical advantage and velocity ratio for a simple machine, and calculate efficiency. These are calculation marks, so show the formula, substitute clearly, and state the unit or percentage.

Forces and moments

The principle of moments

For a seesaw: if a $300 \text{ N}$ child sits $2 \text{ m}$ from the pivot, that side has a moment of $600 \text{ Nm}$ , so balance needs an equal $600 \text{ Nm}$ on the other side.

Mechanical advantage, velocity ratio and efficiency

A machine that lets a small effort lift a heavy load gives a high MA, but you must move the effort a greater distance, which is the velocity ratio. The two are linked by efficiency: in a perfect machine MA would equal VR, but friction always makes MA smaller, so real efficiency is below $100\%$ . This is the conservation-of-energy idea behind simple machines: a lever, pulley or gear train cannot create energy, so the work you put in (effort times the distance the effort moves) always equals the useful work out (load times the distance the load moves) plus the energy lost to friction and to lifting the machine's own parts. That is why a machine that multiplies force must, in return, move its input through a larger distance, and why no real machine reaches $100\%$ efficiency.

Worked example: balancing a loaded lever using the principle of moments

A uniform lever pivots at one point. A $50 \text{ N}$ load sits $0.8 \text{ m}$ from the pivot on one side. Find the effort needed $1.0 \text{ m}$ from the pivot on the other side to balance it.

step State the principle

At balance, clockwise moment $=$ anticlockwise moment about the pivot.

step Write the load (anticlockwise) moment

Load moment $= F \times d = 50 \times 0.8 = 40 \text{ Nm}$ .

step Set the effort moment equal

The effort acts at $1.0 \text{ m}$ , so effort moment $= F_{effort} \times 1.0$ . Setting moments equal: $F_{effort} \times 1.0 = 40$ .

step Solve and interpret

$F_{effort} = \dfrac{40}{1.0} = 40 \text{ N}$ . An effort of $40 \text{ N}$ balances a $50 \text{ N}$ load because it acts at a greater distance from the pivot, giving a mechanical advantage of $\dfrac{50}{40} = 1.25$ . This shows how the principle of moments turns a balance condition into a single equation to solve.

Try this

Q1. Calculate the moment of a $60 \text{ N}$ force acting $0.5 \text{ m}$ from a pivot. [2 marks]

Cue. $60 \times 0.5 = 30 \text{ Nm}$ .

Q2. State the formula for mechanical advantage. [1 mark]

Cue. Mechanical advantage = load divided by effort.

Exam-style practice questions

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

AQA 20184 marksA spanner is

0.25 \text{ m}

long and a force of

40 \text{ N}

is applied at its end to undo a nut. Calculate the moment, and state how to increase it without using more force.

Show worked answer →

A good answer uses the moment formula and suggests a longer lever.

A moment is force times perpendicular distance from the pivot: $\text{moment} = F \times d = 40 \times 0.25 = 10 \text{ Nm}$ .

To increase the moment without a bigger force, increase the distance from the pivot, for example by using a longer spanner or a tube extension on the handle. Doubling the length to $0.5 \text{ m}$ would double the moment to $20 \text{ Nm}$ .

Markers reward the correct moment of $10 \text{ Nm}$ with units and the point that a longer lever increases the moment.

AQA 20216 marksA pulley system lifts a load of

600 \text{ N}

using an effort of

250 \text{ N}

. To raise the load by

0.5 \text{ m}

the effort moves

1.5 \text{ m}

. Calculate the mechanical advantage, the velocity ratio and the efficiency of the system.

Show worked answer →

A good answer takes the three quantities in order with units and a final percentage.

Mechanical advantage $= \dfrac{\text{load}}{\text{effort}} = \dfrac{600}{250} = 2.4$ .

Velocity ratio $= \dfrac{\text{distance moved by effort}}{\text{distance moved by load}} = \dfrac{1.5}{0.5} = 3$ .

Efficiency $= \dfrac{\text{MA}}{\text{VR}} \times 100\% = \dfrac{2.4}{3} \times 100\% = 80\%$ .

The system is $80\%$ efficient; the lost $20\%$ goes to friction in the pulleys and the weight of the moving parts. Markers reward MA $= 2.4$ , VR $= 3$ , and efficiency $= 80\%$ , each shown clearly with the correct formula.

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Sources & how we know this

AQA GCSE Engineering (8852) specification — AQA (2017)