KinexiaLEARN PHYSICS

Torque — Balancing a Seesaw (Worked Example)

Medium
Problem

A 30 kg child sits 2.0 m left of a seesaw pivot. Where must a 45 kg child sit on the right for the seesaw to balance horizontally?

Given

  • m₁ = 30 kg at x₁ = 2.0 m

  • m₂ = 45 kg

  • g = 9.81 m/s²

Find

The distance x₂ of the heavier child from the pivot.

Principle

Balance means zero net torque: clockwise torques equal counterclockwise torques about the pivot (weights act downward at each seat).

Formula

Solution

  1. Set torques equal: m1gx1=m2gx2m_1 g x_1 = m_2 g x_2

  2. Cancel g from both sides: m1x1=m2x2m_1 x_1 = m_2 x_2

  3. Solve: x2=m1x1/m2=30×2.0/45x_2 = m_1 x_1 / m_2 = 30 \times 2.0 / 45

  4. Compute: x2=60/45≈1.33 mx_2 = 60/45 \approx 1.33\ \mathrm{m}

1.33 m from the pivot

Unit check

kg·m = kg·m on both sides; g cancels — the balance point doesn't depend on gravity at all.

What this means physically

The heavier child sits closer (1.33 m vs 2.0 m) — torque balance is why seesaws pair big kids near the middle. On the Moon the same seats would balance identically.

Common mistake

Forgetting that the pivot itself can exert a force but contributes zero torque (its lever arm is zero). Only the two weights matter for the balance equation.