The Principle of Moments

A 20 N force pushes down 2.0 m to the left of a pivot. A 40 N force pushes down 1.0 m to the right. Does the see-saw turn?

1. Anticlockwise: $20 \times 2.0 = 40 \text{ N m}$.
2. Clockwise: $40 \times 1.0 = 40 \text{ N m}$.
3. Equal, so the see-saw is balanced.

A beam is in equilibrium. A 20 N force acts 0.5 m from the pivot and a 10 N force acts 1.0 m from the pivot, both turning clockwise. A force X acts 0.8 m from the pivot on the other side. Find X.

1. Clockwise total: $(20 \times 0.5) + (10 \times 1.0) = 20 \text{ N m}$.
2. Anticlockwise: $X \times 0.8$. Set them equal: $0.8X = 20$, so $X = 25 \text{ N}$.

A uniform beam is pivoted at its midpoint. A 60 kg person sits 3.0 m left of the pivot and an 80 kg person sits 3.0 m right of it. A third person of mass 45 kg sits at distance x, left of the pivot, to balance the beam. Find x. (9702/21/M/J/25 Q2(b))

1. The beam is uniform and pivoted at its midpoint, so its own weight acts at the pivot and has no moment.
2. Balance: $80g \times 3.0 = 60g \times 3.0 + 45g \times x$.
3. Every term has $g$, so it cancels: $240 = 180 + 45x$, giving $x = \dfrac{60}{45} = 1.3 \text{ m}$.

Working with weights on both sides lets g cancel. Working with masses from the start gives the same answer here, but write moments as force × distance in your working.