Question

A 70 kg father and a 50-kg mother are at opposite ends of a seesaw 6 m long. Where should their 25-kg son sit so that the seesaw is balanced at the center?*

Answer 1

$0 .5m_1l=0.5m_2l+xm_3,$

$x=\frac{0.5l(m_1-m_2)}{m_3}=2.4$ m to the mother.