The Physics Question Bank

Course: Separate Physics only  |  Tier: Foundation + Higher  |  Equation sheet: given in the exam

The equation

moment of a force = force × distance (normal to direction of force)    M = F d

Quantity	Symbol	Unit
moment	M	newton-metres (N m)
force	F	newtons (N)
perpendicular distance from pivot	d	metres (m)

Rearranged: F = M ÷ d  |  d = M ÷ F  |  Balanced: clockwise moments = anticlockwise moments

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

Foundation: aim for Q1–8.   Higher: aim for Q4–10. Most questions need rearranging.

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Questions

Q1 (F) A mechanic pushes on a spanner with a force of 40 N, 0.25 m from the nut. Calculate the moment.

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F: M = F d
I: M = 40 × 0.25
F: M is already the subject – no rearranging needed
A: 10 N m

Q2 (F) A door handle is 0.8 m from the hinges. A moment of 16 N m is needed to open the door. Calculate the force needed.

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F: M = F d
I: 16 = F × 0.8
F: F = 16 ÷ 0.8
A: 20 N

Q3 (F) A child with a weight of 300 N sits on a see-saw and produces a moment of 450 N m. How far from the pivot is the child?

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F: M = F d
I: 450 = 300 × d
F: d = 450 ÷ 300
A: 1.5 m

Q4 (F) A bottle cap needs a moment of 3.0 N m to open. The bottle opener handle is 12 cm long. Calculate the force needed at the end of the handle.

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Convert first: 12 cm = 0.12 m
F: M = F d
I: 3.0 = F × 0.12
F: F = 3.0 ÷ 0.12
A: 25 N

Q5 (F) A gardener lifts wheelbarrow handles with 150 N, producing a moment of 180 N m about the wheel. How far are the handles from the wheel, in cm?

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F: M = F d
I: 180 = 150 × d
F: d = 180 ÷ 150 = 1.2 m
A: 1.2 × 100 = 120 cm

Q6 (F/H) A child weighing 250 N sits 2.0 m from the pivot of a see-saw. Where must a parent weighing 500 N sit on the other side to balance it?

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Child’s moment: M = F d = 250 × 2.0 = 500 N m
Balanced, so parent’s moment = 500 N m
F: M = F d
I: 500 = 500 × d
F: d = 500 ÷ 500
A: 1.0 m from the pivot

Q7 (F/H) A tower crane lifts a 12 kN load 15 m from the tower. Its 60 kN counterweight balances the load. How far from the tower is the counterweight?

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Load’s moment: M = 12 000 × 15 = 180 000 N m
F: M = F d
I: 180 000 = 60 000 × d
F: d = 180 000 ÷ 60 000
A: 3.0 m

Q8 (F/H) A crane can safely take a maximum moment of 3.6 × 10⁶ N m. What is the largest load it can lift 40 m from the tower? Give your answer in standard form.

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F: M = F d
I: 3.6 × 10⁶ = F × 40
F: F = 3.6 × 10⁶ ÷ 40
A: 9.0 × 10⁴ N

Q9 (H) A 30 kg child sits 1.4 m from the pivot of a see-saw. Her brother sits 1.2 m from the pivot on the other side and it balances. Calculate the brother’s mass. (g = 9.8 N/kg)

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Step 1 – her weight and moment
W = m g = 30 × 9.8 = 294 N
M = F d = 294 × 1.4 = 411.6 N m
Step 2 – brother’s weight
F: M = F d
I: 411.6 = F × 1.2
F: F = 411.6 ÷ 1.2 = 343 N
Step 3 – brother’s mass
m = W ÷ g = 343 ÷ 9.8
A: 35 kg

Q10 (H) A car wheel nut needs a moment of 110 N m to undo it. A mechanic can push with a maximum force of 320 N. Calculate the minimum length of wrench she needs. Give your answer to 3 significant figures.

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F: M = F d
I: 110 = 320 × d
F: d = 110 ÷ 320
A: 0.34375 = 0.344 m (3 s.f.)