IGCSE Physics Turning Effect of Forces Questions and Worked Solutions (1.5.2) | 0625

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IGCSE Physics 0625 · Topic 1.5.2

IGCSE Physics Turning Effect of Forces Questions and Worked Solutions

50 topic-wise practice questions on Turning Effect of Forces (Unit 1: Motion, Forces and Energy), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions24 Core16 Extended10 Challenge144 marksTextbook §1.5 · pp. 33–54
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Turning Effect of Forces

A force can make an object turn about a pivot. The turning effect is called a moment. When an object is balanced, the clockwise and anticlockwise moments are equal — the principle of moments.

Key formulas

moment = F × dforce F (N) × perpendicular distance d (m) from the pivot; unit N mCore
Σ clockwise moments = Σ anticlockwise momentscondition for balance (principle of moments)Core
Σ upward forces = Σ downward forcesno resultant force in equilibriumExtended

Prerequisite knowledge

  • Forces and weight (1.3, 1.5.1)
  • Multiplying and rearranging
  • Idea of centre of gravity (1.5.3)

Exam tips & common mistakes

  • Use the PERPENDICULAR distance from the pivot
  • For uniform beams, include the beam's own weight at its midpoint
  • Take moments about a support to remove an unknown force from the equation

Key facts from the syllabus (1.5.2)

  • Everyday examples: door handles, spanners, see-saws, wheelbarrows
  • A longer lever arm means a smaller force is needed for the same moment
  • An object is in equilibrium when there is no resultant force and no resultant moment
  • The weight of a uniform beam acts at its centre
  • Extended describe an experiment with a metre rule and weights to show the principle of moments

Textbook: §1.5 Forces, pp. 33–54

1
Core[1 mark]

State what is meant by the moment of a force.

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Method: Moment measures turning effect.

The turning effect of a force
2
Core[1 mark]

Write the equation for the moment of a force.

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Method: Use the perpendicular distance.

Moment = force × perpendicular distance from the pivot
3
Core[1 mark]

Give one everyday example of using a moment.

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Method: Many tools use turning effects.

e.g. opening a door, using a spanner, a see-saw
4
Core[2 marks]

A force of 25 N acts at a perpendicular distance of 0.40 m from a pivot. Calculate the moment.

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Method: Moment = F × d.

M = 25 × 0.40 = 10.0 N m
5
Core[2 marks]

Explain why a long spanner makes it easier to undo a tight nut.

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Method: Larger distance gives a larger moment for the same force.

The perpendicular distance is larger, so a smaller force gives the same moment
6
Core[1 mark]

State the principle of moments.

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Method: Balance of clockwise and anticlockwise moments.

For a balanced object, sum of clockwise moments = sum of anticlockwise moments
7
Core[3 marks]

A child of weight 300 N sits 2.0 m from the pivot of a see-saw. Calculate the distance from the pivot that a 400 N child must sit to balance it.

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Method: 300 × 2.0 = 400 × d.

d = 600 ÷ 400 = 1.50 m on the other side
8
Core[3 marks]

A uniform beam is balanced at its centre. A 6.0 N weight hangs 30 cm to the left. Calculate the weight needed 45 cm to the right.

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Method: Principle of moments.

W = (6.0 × 30) ÷ 45 = 4.00 N
9
Core[2 marks]

State the two conditions for an object to be in equilibrium.

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Method: No resultant force and no resultant moment.

No resultant force and no resultant moment
10
Core[3 marks]

A 20 N force acts 0.30 m from a pivot clockwise and a 15 N force acts 0.50 m from the pivot anticlockwise. Determine the resultant moment and its direction.

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Method: Compare moments.

Clockwise = 6.00 N m; anticlockwise = 7.50 N m; resultant = 1.50 N m anticlockwise
11
Core[2 marks]

A door handle is placed far from the hinge. Explain why.

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Method: Larger distance → larger moment for the same push.

Maximises the perpendicular distance so a small force gives a large turning effect
12
Core[2 marks]

A force of 40 N is applied to a wheel at a perpendicular distance of 0.25 m. Calculate the moment.

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Method: M = F × d.

M = 40 × 0.25 = 10.0 N m
13
Core[2 marks]

A spanner of length 0.30 m is turned with a force of 60 N at its end, at right angles. Calculate the moment.

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Method: M = Fd.

M = 60 × 0.30 = 18.0 N m
14
Core[2 marks]

A moment of 12 N m is needed to loosen a bolt. A spanner is 0.40 m long. Calculate the minimum force.

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Method: F = M ÷ d.

F = 12 ÷ 0.40 = 30.0 N
15
Core[2 marks]

A force of 80 N produces a moment of 20 N m. Calculate the perpendicular distance from the pivot.

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Method: d = M ÷ F.

d = 20 ÷ 80 = 0.250 m
16
Core[2 marks]

Explain why it is easier to push a door open near the handle than near the hinge.

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Method: Larger distance.

Further from the hinge (pivot), so a smaller force gives the needed moment
17
Core[3 marks]

On a see-saw, a 450 N adult sits 1.2 m from the pivot. Calculate the distance a 270 N child must sit on the other side to balance.

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Method: 450 × 1.2 = 270 × d.

d = 540 ÷ 270 = 2.00 m
18
Core[3 marks]

A metre rule balanced at 50 cm has a 3.0 N weight at 20 cm. Calculate where a 2.0 N weight must be placed to balance it.

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Method: 3.0 × 30 = 2.0 × d.

d = 45 cm from pivot, at the 95 cm mark
19
Core[2 marks]

State the condition for an object to remain balanced (no rotation).

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Method: Principle of moments.

Total clockwise moment = total anticlockwise moment
20
Core[2 marks]

Explain why a wheelbarrow makes it easier to lift a heavy load.

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Method: Long handles.

The handles are far from the pivot (wheel), so a small upward force produces a large moment
21
Core[3 marks]

A beam pivoted at its centre has 10 N at 0.60 m left and 15 N at 0.20 m left. Calculate the force needed 0.50 m to the right.

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Method: Sum anticlockwise.

Anticlockwise = 6.0 + 3.0 = 9.0 N m; F = 9.0 ÷ 0.50 = 18.0 N
22
Core[2 marks]

Explain why a nut cannot be loosened by pushing the spanner along its length towards the nut.

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Method: Perpendicular distance zero.

The line of action passes through the pivot, so the perpendicular distance and moment are zero
23
Core[2 marks]

A pair of scissors cuts thick card more easily near the pivot. Explain why.

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Method: Load close to pivot.

The card is close to the pivot, so the force applied at the handles produces a larger cutting force
24
Core[2 marks]

Calculate the moment of a 150 N force acting 0.80 m from a pivot.

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Method: M = Fd.

M = 120 N m
25
Extended[3 marks]

A metre rule pivoted at 50 cm has 2.0 N at 10 cm and 1.0 N at 30 cm. Calculate the force needed at 90 cm to balance it.

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Method: Anticlockwise moments: 2.0 × 40 + 1.0 × 20.

Anticlockwise = 80 + 20 = 100 N cm; F = 100 ÷ 40 = 2.50 N
26
Extended[3 marks]

A uniform plank of weight 200 N and length 4.0 m is pivoted 1.0 m from its left end. Calculate the downward force needed at the left end to balance it.

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Method: The weight acts at the centre, 1.0 m right of the pivot.

F × 1.0 = 200 × 1.0; F = 200 N
27
Extended[4 marks]

A uniform metre rule is balanced at 42 cm with 0.34 N at 10 cm and 0.12 N at 80 cm. Calculate the weight of the rule.

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Method: Weight acts at 50 cm (8 cm right of pivot). Anticlockwise: 0.34 × 32. Clockwise: 0.12 × 38 + W × 8.

10.88 = 4.56 + 8W; W = 0.790 N
28
Extended[3 marks]

A see-saw pivoted at its centre has children of 250 N at 1.6 m and 150 N at 1.0 m on the left. Calculate where a 350 N adult must sit on the right to balance it.

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Method: Sum of left moments = 350 × d.

Left = 550 N m; d = 550 ÷ 350 = 1.57 m
29
Extended[4 marks]

Describe an experiment to show there is no resultant moment on an object in equilibrium.

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Method: Balance a metre rule on a pivot with known weights, measure distances, compare moments.

Balance a rule on a pivot, hang weights at different positions until level, measure distances, show sum of clockwise moments = sum of anticlockwise moments
30
Extended[3 marks]

A uniform beam of weight 80 N and length 2.0 m is supported at its two ends. A 120 N load is placed 0.50 m from the left end. Calculate the upward force at the right support.

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Method: Take moments about the left support.

R × 2.0 = 80 × 1.0 + 120 × 0.50 = 140; R = 70.0 N
31
Extended[2 marks]

For the beam in the previous question, calculate the upward force at the left support.

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Method: Total up = total down.

Left = 200 − 70 = 130 N
32
Extended[3 marks]

A crane jib has a 2000 N load 6.0 m from the pivot and a counterweight 2.4 m on the other side. Calculate the counterweight needed.

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Method: Principle of moments.

W = 2000 × 6.0 ÷ 2.4 = 5000 N
33
Extended[3 marks]

A uniform plank 3.0 m long weighing 300 N rests on a pivot at its centre. A 600 N man stands 0.50 m from one end. Calculate where a 400 N woman must stand on the other side to balance it.

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Method: Man is 1.0 m from pivot.

600 × 1.0 = 400 × d; d = 1.50 m (at the end)
34
Extended[3 marks]

A uniform metre rule of weight 1.2 N is pivoted at the 30 cm mark. Calculate the force at the 0 cm mark needed to balance it.

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Method: Weight acts 20 cm right of pivot.

F × 30 = 1.2 × 20; F = 0.800 N
35
Extended[4 marks]

A uniform beam 4.0 m long of weight 500 N is supported at each end. A 1000 N load is 1.0 m from the right end. Calculate the force at each support.

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Method: Moments about the left support.

R_right × 4.0 = 500 × 2.0 + 1000 × 3.0; R_right = 1000 N; R_left = 1500 − 1000 = 500 N
36
Extended[3 marks]

A 0.50 kg mass hangs at the 80 cm mark of a uniform metre rule pivoted at 60 cm. The rule balances. Calculate the mass of the rule (g = 9.8 N/kg; masses can be used directly).

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Method: Rule's weight at 50 cm, 10 cm left of pivot.

m × 10 = 0.50 × 20; m = 1.00 kg
37
Extended[2 marks]

Explain why an object can have no resultant force but still rotate.

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Method: Couple.

Two equal and opposite forces not acting along the same line (a couple) give zero resultant force but a resultant moment
38
Extended[3 marks]

A tower crane has a 12 000 N counterweight 5.0 m from the tower. Calculate the maximum load it can hold 15 m from the tower on the other side.

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Method: Moments balance.

W = 12000 × 5.0 ÷ 15 = 4000 N
39
Extended[3 marks]

A beam is in equilibrium with upward forces 300 N and 500 N at its ends and one downward load. State the load and explain.

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Method: No resultant force.

800 N, because the total upward force must equal the total downward force
40
Extended[4 marks]

Describe how you would use a metre rule, a pivot and a known 1.0 N weight to find the weight of an unknown object.

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Method: Principle of moments.

Balance the rule at its centre; hang the unknown at a known distance on one side and move the 1.0 N weight until balanced; measure both distances; W = 1.0 × d₁ ÷ d₂

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

A uniform see-saw is pivoted at its centre. A 300 N child sits 1.5 m from the pivot on the left; a 200 N child sits 0.75 m from the pivot on the left too. Calculate where a 450 N adult must sit on the right to balance.

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Method: Sum left moments.

Left = 450 + 150 = 600 N m; d = 600 ÷ 450 = 1.33 m
42
ChallengeCore[4 marks]

A spanner applies a moment of 45 N m. A 300 N force is applied at 90° to the spanner. Calculate the distance from the nut. Explain why applying the force at 60° to the spanner would need a larger force.

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Method: d = M/F.

d = 0.150 m; at an angle the perpendicular distance is smaller
43
ChallengeCore[4 marks]

A wheelbarrow and load weigh 800 N acting 0.40 m from the wheel. The handles are 1.6 m from the wheel. Calculate the upward force needed at the handles.

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Method: Moments about the wheel.

F = 800 × 0.40 ÷ 1.6 = 200 N
44
ChallengeExtended[5 marks]

A uniform beam of length 6.0 m and weight 400 N rests on supports at its ends. A 1200 N load sits 2.0 m from the left end. Calculate the force at each support.

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Method: Moments about left end.

Right = (400×3 + 1200×2) ÷ 6 = 600 N; left = 1600 − 600 = 1000 N
45
ChallengeExtended[5 marks]

A uniform metre rule is pivoted at 40 cm. A 0.60 N weight at 10 cm balances it. Calculate the weight of the rule.

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Method: Rule's weight at 50 cm, 10 cm right of pivot.

0.60 × 30 = W × 10; W = 1.80 N
46
ChallengeExtended[5 marks]

A 3.0 m uniform plank weighing 150 N overhangs a ledge by 1.0 m. Calculate how far along the overhang a 600 N person can walk before the plank tips.

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Method: Pivot at the ledge edge; plank's weight acts 0.50 m on the ledge side.

600 × d = 150 × 0.50; d = 0.125 m from the edge
47
ChallengeExtended[4 marks]

A crane's jib has its pivot 10 m from a 5000 N load. The counterweight of 20 000 N is 3.0 m on the other side. The jib (weight 4000 N) acts 2.0 m from the pivot on the load side. Is the crane balanced? If not, which way does it turn?

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Method: Compare moments.

Load side: 50 000 + 8000 = 58 000 N m; counterweight: 60 000 N m; not balanced — it turns towards the counterweight (net 2000 N m)
48
ChallengeExtended[4 marks]

A uniform ladder of weight 200 N and length 4.0 m is held horizontally by two people at its ends. A 100 N paint pot hangs 1.0 m from one end. Calculate the force each person exerts.

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Method: Moments about one end.

Far end: (200×2 + 100×1) ÷ 4 = 125 N; near end: 175 N
49
ChallengeExtended[4 marks]

Explain how a door can be opened by applying a force that does not pass through the hinge, and why pushing exactly at the hinge does not open it, using the idea of moments.

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Method: Moment = F × perpendicular distance.

A force with a perpendicular distance from the hinge produces a moment; at the hinge the distance is zero so the moment is zero
50
ChallengeExtended[5 marks]

Describe an experiment to verify the principle of moments using a metre rule pivoted at 50 cm and two sets of hanging masses. Include how you would check the result.

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Method: Measure distances; compare.

Hang masses on both sides, adjust until level; record each weight and distance from the pivot; calculate clockwise and anticlockwise moments; repeat for several arrangements and show they are equal within experimental error
All IGCSE Physics topics← 1.5.1 Effects of Forces1.5.3 Centre of Gravity →

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