Moments

Key ideas

• Moment = force × perpendicular distance from the pivot to the force line.
• For rotational balance, clockwise and anticlockwise moments about the same point are equal.
• A force through the pivot has zero moment about it.
• A lower centre of gravity and a wider base usually improve stability.

The method

1. Identify the pivot.
2. Convert distances to metres. Measure from the pivot to the force line at a right angle.
3. Calculate each moment as force × perpendicular distance.
4. For balance, equate total clockwise and anticlockwise moments.
5. For a beam at rest, also set total upward force equal to total downward force.

Worked examples

Mistakes that lose marks

• Measuring from the end of a beam instead of its pivot.
• Leaving distances in centimetres when the answer uses N m.
• Forgetting the weight of a uniform plank at its midpoint.
• Treating two forces with the same turning direction as a balanced pair.

EXAM-STYLE questions written by Apex tutors

Practise with 12 original questions and full solutions. The suggested marks are a guide from our tutors. M means method, A means accuracy and B means an independent point. Try each question, then open its solution.

1. Apex original core A 60 N force acts perpendicular to a spanner, 0.30 m from the nut. Find the moment. [2 marks]
   Show full solution

   1. Moment = force × perpendicular distance. M1
   2. 60 × 0.30 = 18 N m. A1

   Answer: 18 N m

2. Apex original extended A door needs a moment of 24 N m. Find the perpendicular force at 0.80 m from the hinges. Then find the perpendicular distance needed for a force of 48 N. [4 marks]
   Show full solution

   1. Force = 24 ÷ 0.80 = 30 N. M1
   2. Distance = 24 ÷ 48 = 0.50 m. A3

   Answer: 30 N; 0.50 m

3. Apex original extended A 500 N load acts 1.2 m left of a seesaw pivot. A 750 N load acts on the right. Ignore the seesaw weight. Find its distance for balance. [4 marks]
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   1. Left moment = 500 × 1.2 = 600 N m. M1
   2. 750d = 600, so d = 600 ÷ 750 = 0.80 m. A3

   Answer: 0.80 m right of the pivot

4. Apex original extended A uniform metre rule is pivoted at 50 cm. A 3.0 N weight hangs at 15 cm. At which mark on the other side must a 2.5 N weight hang for balance? [4 marks]
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   1. Left distance = (50 − 15) ÷ 100 = 0.35 m. M1
   2. Left moment = 3.0 × 0.35 = 1.05 N m. The rule weight acts through the pivot. M1
   3. Right distance = 1.05 ÷ 2.5 = 0.42 m = 42 cm. M1
   4. Position = 50 + 42 = 92 cm. A1

   Answer: 92 cm mark

5. Apex original extended A uniform 3.0 m plank weighs 60 N. Its pivot is 1.0 m from the left end. A 90 N load hangs at the left end. Find the downward force at the right end needed for balance. [5 marks]
   Show full solution

   1. Plank weight acts at 1.5 m from the left end, so its pivot distance is 1.5 − 1.0 = 0.50 m. M1
   2. Left moment = 90 × 1.0 = 90 N m. M1
   3. Right moment = 60 × 0.50 + F × (3.0 − 1.0) = 30 + 2.0F. M1
   4. F = (90 − 30) ÷ 2.0 = 30 N. A2

   Answer: 30 N downwards

6. Apex original extended A light 4.0 m beam is supported at its ends A and B. A 120 N load acts 1.0 m from A. Find both upward support forces. [4 marks]
   Show full solution

   1. Take moments about A: R_B × 4.0 = 120 × 1.0 = 120 N m. M1
   2. R_B = 120 ÷ 4.0 = 30 N. M1
   3. Vertical balance: R_A = 120 − 30 = 90 N. A2

   Answer: A: 90 N; B: 30 N

7. Apex original extended Why does placing a door handle far from its hinges make the door easier to open? [3 marks]
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   1. The hinges form the pivot. M1
   2. Moment = force × perpendicular distance. M1
   3. A greater distance needs a smaller force for the same moment. A1

   Answer: A larger perpendicular distance reduces the force needed.

8. Apex original extended Describe how to find the centre of gravity of an irregular piece of card. [4 marks]
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   1. Suspend the card freely from a hole near its edge. M1
   2. Hang a plumb line from the same support. When the card is still, mark the vertical line. M1
   3. Repeat from a different hole and draw another vertical line. M1
   4. The intersection locates the centre of gravity. A third suspension can check it. A1

   Answer: Find the intersection of the vertical lines.

9. Apex original extended Why is a tall narrow bookcase less stable than a low wide one? [3 marks]
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   1. The tall narrow bookcase has a higher centre of gravity and a narrower base. M1
   2. On tilting, the vertical line through its centre of gravity reaches the base edge sooner. M1
   3. It topples once this line passes outside the base. A1

   Answer: A smaller tilt can move its weight line outside its base.

10. Apex original extended A light wheelbarrow carries a 300 N load 0.40 m from the wheel axle. The handles are 1.20 m from the axle. All distances are perpendicular to the forces. Find the total upward handle force, then repeat with the load at 0.20 m. [5 marks]
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    1. First balance: F × 1.20 = 300 × 0.40 = 120 N m. M1
    2. F = 120 ÷ 1.20 = 100 N. M1
    3. New force = (300 × 0.20) ÷ 1.20 = 60 ÷ 1.20 = 50 N. A3

    Answer: 100 N; 50 N

11. Apex original extended A uniform metre rule is pivoted at 50 cm. A 2.0 N weight hangs at 20 cm. A spring balance pulls DOWN at 90 cm to keep it horizontal. Find the reading. Repeat when the balance moves to 70 cm. [4 marks]
    Show full solution

    1. The left weight gives an anticlockwise moment: 2.0 × ((50 − 20) ÷ 100) = 0.60 N m. M1
    2. The downward pull on the right gives the balancing clockwise moment. M1
    3. At 90 cm: F = 0.60 ÷ ((90 − 50) ÷ 100) = 1.5 N. M1
    4. At 70 cm: F = 0.60 ÷ ((70 − 50) ÷ 100) = 3.0 N. A1

    Answer: 1.5 N; 3.0 N

12. Apex original extended A 300 N load acts 2.0 m left of a seesaw pivot and a 200 N load acts 1.5 m right. Ignore the seesaw weight. Which side falls? Where should the 200 N load move for balance? [4 marks]
    Show full solution

    1. Left moment = 300 × 2.0 = 600 N m anticlockwise. M1
    2. Right moment = 200 × 1.5 = 300 N m clockwise. The left side falls. M1
    3. For balance, right distance = 600 ÷ 200 = 3.0 m. A2

    Answer: Left side falls; move the right load to 3.0 m from the pivot.

Where this is in the syllabus

Explore the matching courses: Cambridge IGCSE Physics 0625; Pearson Edexcel International GCSE Physics 4PH1.

For the official syllabus, use the board links: Cambridge IGCSE Physics 0625; Pearson Edexcel International GCSE Physics 4PH1.

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