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How do mechanical systems transmit motion and force, and how is a moment calculated?

Mechanical systems: levers, the moment of a force and the principle of moments, gears and gear ratio.

A CCEA GCSE Engineering and Manufacturing answer on mechanical systems, covering levers, the moment of a force and the principle of moments, gears and gear ratio, with worked calculations.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

CCEA Unit 3 expects you to understand mechanical systems that transmit motion and force: levers and the moment of a force, the principle of moments, and gears and gear ratio. This includes calculations, so you must know the equations and units.

The answer

Levers and the moment of a force

The principle of moments

Gears and gear ratio

Gears are toothed wheels that mesh to transmit rotation and change speed or turning force (torque). The gear ratio compares the number of teeth:

gear ratio=teeth on driven gearteeth on driver gear.\text{gear ratio} = \frac{\text{teeth on driven gear}}{\text{teeth on driver gear}}.

A large driven gear and small driver gives a high ratio: the output turns slower but with more torque. Meshing gears turn in opposite directions. A gear train (several gears in mesh) can change speed and direction in stages, and an idler gear between two gears reverses the output direction without changing the overall ratio. Gears are a key way of matching a motor's fast, low-torque output to a load that needs to turn slowly with a large force.

Worked example: a moment and a gear ratio

Examples in context

Example 1. A spanner
A longer spanner puts the force further from the nut, giving a larger moment for the same hand force, so a tight nut is easier to undo.
Example 2. A see-saw
Children balance by the principle of moments: a lighter child sits further from the pivot so their smaller weight times the larger distance matches the heavier child's moment.
Example 3. A bicycle or drill gearing
Gears trade speed for torque: a low gear (high ratio) turns the wheel or chuck slowly but with more force, useful for climbing or drilling hard material.

The pattern is that mechanical systems let a small force or fast rotation be converted into a larger turning effect or torque, using distance (levers) or tooth ratio (gears).

Try this

Q1. Write the equation for the moment of a force and its unit. [2 marks]

  • Cue. Moment =F×d= F \times d; unit is the newton metre (N m).

Q2. A 40 N force acts 0.5 m from a pivot. Find the moment. [2 marks]

  • Cue. M=40×0.5=20 N mM = 40 \times 0.5 = 20\ \text{N m}.

Q3. A driver gear has 10 teeth and the driven gear has 40 teeth. What is the gear ratio? [1 mark]

  • Cue. 40/10=440 / 10 = 4, that is 4:1.

Exam-style practice questions

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

CCEA style4 marksA spanner is used to undo a nut. A force of 50 N is applied 0.30 m from the nut. Calculate the moment, and state how the moment could be increased without using a larger force.
Show worked answer →

The moment of a force is the force times the perpendicular distance from the pivot:

moment=F×d=50×0.30=15 N m.\text{moment} = F \times d = 50 \times 0.30 = 15\ \text{N m}.

So the moment (turning effect) is 15 N m.

To increase the moment without a larger force, increase the distance from the pivot, for example by using a longer spanner (or a tube over the handle), because a greater distance gives a greater turning effect for the same force.

Markers reward moment = F times d, the value 15 N m with the unit, and increasing the distance (longer spanner) to raise the moment.

CCEA style4 marksA see-saw (lever) is balanced. A 200 N child sits 1.5 m from the pivot on the left. A second child sits 1.0 m from the pivot on the right. Using the principle of moments, calculate the weight of the second child.
Show worked answer →

The principle of moments states that, for balance, the total clockwise moment equals the total anticlockwise moment about the pivot.

Anticlockwise moment (left child) =200×1.5=300 N m.= 200 \times 1.5 = 300\ \text{N m}.

For balance, clockwise moment (right child) =300 N m.= 300\ \text{N m}.

F×1.0=300F \times 1.0 = 300, so F=3001.0=300 N.F = \dfrac{300}{1.0} = 300\ \text{N}.

The second child weighs 300 N.

Markers reward the principle (clockwise = anticlockwise), the left moment (300 N m), and solving for the right force (300 N).

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