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How does a lever multiply force, and what keeps it balanced?

Levers and the moment of a force, calculating a moment, and applying the principle of moments to a balanced lever.

An SQA National 5 Engineering Science answer on levers and moments, covering the moment of a force as force times perpendicular distance, the principle of moments for a balanced lever, and how a lever provides a mechanical advantage to lift a large load with a smaller effort.

Generated by Claude Opus 4.89 min answer

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

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  1. What this key area is asking
  2. Moments
  3. The principle of moments
  4. Levers and mechanical advantage
  5. Why moments matter
  6. Try this

What this key area is asking

The SQA wants you to calculate the moment (turning effect) of a force and apply the principle of moments to find an unknown force or distance on a balanced lever.

Moments

The distance must be measured at right angles to the force. Doubling the distance doubles the moment for the same force, which is the whole point of using a long lever.

The principle of moments

To use it, work out the moment on each side of the pivot (force times its perpendicular distance), then set the clockwise total equal to the anticlockwise total and solve for the unknown.

Levers and mechanical advantage

Because a moment depends on distance as well as force, a lever can let a small effort balance a large load. If the effort acts far from the pivot and the load sits close to it, a small effort produces the same moment as the large load. This force-multiplying effect is the lever's mechanical advantage (covered in detail in its own key area). Everyday levers include a crowbar, a wheelbarrow, scissors and a see-saw.

Why moments matter

Moments appear throughout engineering: in spanners and tools, in cranes and loading arms, and in the analysis of structures, where the principle of moments helps find the reaction forces at supports. Being confident with M=F×dM = F \times d and the balance condition is essential for the structures work later in this area.

Try this

Q1. A force of 25 N25 \text{ N} acts 0.40 m0.40 \text{ m} from a pivot, at right angles. Calculate the moment. [2 marks]

  • Cue. M=F×d=25×0.40=10 NmM = F \times d = 25 \times 0.40 = 10 \text{ Nm}.

Q2. State the principle of moments. [1 mark]

  • Cue. For a balanced lever, the total clockwise moment equals the total anticlockwise moment about the pivot.

Q3. A 60 N60 \text{ N} load is 0.50 m0.50 \text{ m} from a pivot. What effort, applied 1.0 m1.0 \text{ m} on the other side, balances it? [2 marks]

  • Cue. Anticlockwise =60×0.50=30 Nm= 60 \times 0.50 = 30 \text{ Nm}; effort ×1.0=30\times 1.0 = 30, so effort =30 N= 30 \text{ N}.

Exam-style practice questions

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

SQA N5 style3 marksA spanner is 0.25 m long. A force of 60 N is applied at right angles to the end of the spanner. Calculate the moment about the bolt.
Show worked answer →

Use the moment relationship.

Relationship: M=F×dM = F \times d, where dd is the perpendicular distance from the pivot.

Substitution: M=60×0.25=15 NmM = 60 \times 0.25 = 15 \text{ Nm}.

Markers reward selecting moment equals force times perpendicular distance, correct substitution, and a final answer in newton metres (Nm). A longer spanner would give a larger moment for the same force.

SQA N5 style4 marksA uniform beam is balanced on a central pivot. A 40 N load sits 0.60 m to the left of the pivot. Calculate the force needed 0.30 m to the right of the pivot to keep the beam balanced.
Show worked answer →

Apply the principle of moments: for balance, clockwise moments equal anticlockwise moments.

Anticlockwise moment (the 40 N load on the left): 40×0.60=24 Nm40 \times 0.60 = 24 \text{ Nm}.

Clockwise moment (the unknown force FF on the right): F×0.30F \times 0.30.

For balance: F×0.30=24F \times 0.30 = 24, so F=240.30=80 NF = \dfrac{24}{0.30} = 80 \text{ N}.

Markers reward writing both moments, setting clockwise equal to anticlockwise, and solving for the force in newtons. The shorter distance needs a larger force.

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