All subjects · Mathematics · WME01 (M1)

Beams on supports and hanging rods

A typical M1 moments question has a plank on two supports (or a rod held by two vertical strings) with loads on it. You find the reactions using moments and vertical forces.

Method

1. Draw the beam, all weights (a uniform beam's weight at its midpoint) and both reactions.

2. Take moments about one support to find the other reaction.

3. Resolve vertically (or take moments about the other support) to find the first.

Equal reactions and given conditions

Questions often say "the reactions are equal" or "the reaction at C is twice the reaction at D". Write R and R (or 2R and R), then use the two equilibrium conditions to find the unknown distance or mass.

Rods on strings

A rod held horizontally by two vertical strings works the same way: the tensions play the role of the reactions.

Worked example

A uniform plank AB of length 6 m and mass 30 kg rests on supports at C (1 m from A) and D (2 m from B). A man of mass 80 kg stands 2 m from A. Find the reactions.

  1. Moments about C: R_D × 3 = 30g × 2 + 80g × 1.
  2. R_D = 457 N.
  3. Vertically: R_C = 110g - R_D = 621 N.

Answer: R_C = 621 N, R_D = 457 N

Key idea: Moments about one support for the other reaction; then forces up = forces down.

Check yourself

  1. A uniform beam AB of length 4 m and mass 20 kg is supported at A and B. A 50 kg load is placed 1 m from A. Find the reaction at B, to 2 significant figures.
    Show answer

    220 N. M(A): 4R_B = 20g × 2 + 50g × 1 = 90g, so R_B = 22.5g = 220.5 N, which is 220 N to 2 s.f.

  2. For the same beam, find the reaction at A, to 3 significant figures.
    Show answer

    466 N. R_A = 70g - 22.5g = 47.5g = 466 N.

  3. A uniform rod AB of length 2 m and weight 40 N is held horizontally by vertical strings at A and B. A 60 N weight hangs at x m from A, and the tensions are equal. Find x.
    Show answer

    1 m. Equal tensions: each is 50 N. M(A): 50 × 2 = 40 × 1 + 60x, so x = 1 m.

  4. A uniform plank AB of length 5 m and weight 200 N rests on supports at A and B. A child of weight 300 N stands on it so that the reaction at A is twice the reaction at B. Find the distance of the child from A.
    Show answer

    1.11 m. R_A + R_B = 500 and R_A = 2R_B, so R_B = 500/3. M(A): (500/3) × 5 = 200 × 2.5 + 300x, so 300x = 333.3, x = 1.11 m (3 s.f.).

  5. In the worked example, the man walks to 3.5 m from A. Find the new reaction at C, to 3 significant figures.
    Show answer

    229 N. M(C): 3R_D = 30g × 2 + 80g × 2.5 = 260g, so R_D = 849 N and R_C = 110g - R_D = 229 N.

  6. Where do you draw the weight of a uniform plank?
    • At each support
    • At the end nearer the load
    • At its midpoint
    • At the point where the person stands
    Show answer

    At its midpoint. Uniform means the centre of mass is at the midpoint.

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Last updated 7 November 2026. Spotted a mistake? Tell us.