Seesaw and fulcrum

Problem

Two children of masses $m_1 = 30\;\text{kg}$ and $m_2 = 45\;\text{kg}$ sit at the two ends of a seesaw. Explain qualitatively where the centre of mass of the two-children system is, and where the fulcrum must be placed for the seesaw to be balanced.

Solution

The centre of mass is the weighted average of the positions: the heavier child “pulls” it towards themselves. Letting $L$ be the distance between the two children, and taking the origin at the light child, the distance of the CM from $m_1$ is $d_1 = \frac{m_2}{m_1 + m_2}\,L = \frac{45}{75}\,L = \frac{3}{5}\,L$ The CM is therefore located at $3/5$ of the distance from the $30\;\text{kg}$ child, i.e. closer to the $45\;\text{kg}$ child.

For the moments to balance, the fulcrum must be placed exactly at the CM: there, the moments of the two weights balance each other ($m_1 g\,d_1 = m_2 g\,d_2$, with $d_1/d_2 = m_2/m_1$).

Links

Topics: Centro di massa Concepts: Centro di massa · Equilibrio statico Skills: Calcolo del centro di massa