Moments: balancing a seesaw

Forces

GCSE Physics (8463)

ORIGINAL

Total anticlockwise moment = total clockwise moment, with every force paired with its own distance.

Board and spec code confirmed against AQA 8463 · registry checked 2026-07-17How this checking works

Question

A uniform seesaw is pivoted at its centre. Child A, weight

320\,\text{N}

, sits

1.5\,\text{m}

to the left of the pivot, and a bag of weight

80\,\text{N}

rests

0.50\,\text{m}

to the left of the pivot. Child B, weight

400\,\text{N}

, sits on the right. How far from the pivot must Child B sit for the seesaw to balance?

Every step worked, with the reasoning.

1
Balanced: total anticlockwise moment $=$ total clockwise moment
The principle of moments. A uniform seesaw pivoted at its centre has its weight acting through the pivot, so its own weight has no moment.
2
Anticlockwise: $M = Fd = (320 \times 1.5) + (80 \times 0.50)$
Each moment is force × perpendicular distance from the pivot; both loads on the left turn the seesaw anticlockwise.
3
$= 480 + 40 = 520\,\text{N m}$
Add the moments — not the forces or the distances.
4
Clockwise: $400 \times d = 520$
Child B's weight provides the only clockwise moment.
5
$d = \dfrac{520}{400} = 1.3\,\text{m}$
Sense check: Child B is heavier than Child A, so sits closer to the pivot than Child A's $1.5\,\text{m}$ .

Answer: $1.3\,\text{m}$ to the right of the pivot