Problem — Seesaw in equilibrium

Problem

A seesaw is in equilibrium. On one side sits a child of $30\;\mathrm{kg}$ at $2{,}0\;\mathrm{m}$ from the pivot. On the other side sits a second child of $40\;\mathrm{kg}$. At what distance from the pivot must they sit?

Solution

At equilibrium the sum of the torques about the pivot is zero: the two torques (in opposite directions) balance. The weight forces are $m g$, but $g$ cancels: $m_1\,g\,d_1 = m_2\,g\,d_2 \quad\Rightarrow\quad 30\cdot 2{,}0 = 40\cdot d_2$ $d_2 = \frac{30\cdot 2{,}0}{40} = \ev{1{,}5\;\mathrm{m}}$ The heavier child must sit closer to the pivot to have the same torque as the lighter one.

Links

Topics: Statics and equilibrium Concepts: Torque · Static equilibrium Skills: Torque balance