Mathematics · OCR H240 · Topic 3.04

Moments

Moments: the turning effect of a force, and rigid bodies in equilibrium.

  • 1 sub-topic
  • Taught in Year 13
  • 29 practice questions
  • 1 animated lesson
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 3.04a-cMomentsY13

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 29 questions in this topic, free, no account needed.

Moments

Not needed for Year 12Easy

Balancing a see-saw · Year 13 · 2 marks

  1. A 30 kg child sits 2 m from the pivot of a see-saw. Where must a 40 kg child sit to balance it?
    (2)
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  1. Clockwise moments = anticlockwise moments.30g×2=40g×d30g \times 2 = 40g \times d
  2. d=1.5d = 1.5 m on the other side

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Moments

Not needed for Year 12Medium

A non-uniform rod · Year 13 · 4 marks

A non-uniform rod ABAB, 3 m long and of mass 6 kg, rests on supports at AA and BB. The reaction at AA is twice the reaction at BB.
  1. Find the distance of the centre of mass from AA.
    (4)
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  1. RA+RB=6gR_A + R_B = 6g and RA=2RBR_A = 2R_B
  2. RB=2gR_B = 2g
  3. Moments about AA: 6g×x=2g×36g \times x = 2g \times 3
  4. x=1x = 1 m

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Moments

Not needed for Year 12Hard

A ladder · Year 13 · 7 marks

A uniform ladder, 4 m long and of mass 15 kg, rests against a smooth wall with its foot on rough ground. It makes 60∘60^\circ with the ground and is about to slip.
  1. Find the coefficient of friction between the ladder and the ground.
    (7)

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  1. The smooth wall gives only a horizontal force SS.Vertically: R=15g=147R = 15g = 147 N
  2. Horizontally: F=SF = S
  3. Use perpendicular distances.Moments about the foot: S×4sin⁡60∘S \times 4\sin 60^\circ=15g×2cos⁡60∘{}= 15g \times 2\cos 60^\circ
  4. S=42.4S = 42.4 N
  5. About to slip, so friction is at its limit.μ\mu=FR{}= \dfrac{F}{R}=42.4147{}= \dfrac{42.4}{147}
  6. μ=0.289\mu = 0.289

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