Answer to Question #151450 in Physics for Jonalyn Catolico

Question

1. A 15.0 kg child and a 25 kg child sits at opposite ends of a 4 m see saw pivoted at its center. Where should a third child whose mass is 20 kg sit in order to balance the see saw?

Answer 1

$$m_1g(0.5L)+m_3g(0.5L-x)=m_2g(0.5L)\\m_1(L)+m_3g(L-2x)=m_2g(L)\\ \\15(4)+20(4-2x)=25(4)\\x=1\ m$$

Answer: 1 m from a 15.0 kg child.