Question

A 15-kg uniform beam is 3.0 m long and is pivoted 1.0 m from the left end. A 35 kg child sits on the left end of the beam 0.60 m from the pivot. How far from the pivot must a 20 kg child sit in order for the beam to be in equilibrium?

Answer 1

The red arrows show the force caused by the mass of the beam B, b, the green arrows show the force caused by the the mass of kids M, m. For the beam to be in equilibrium, the torques must be equal:

$$\frac13m_{beam}g\cdot0.5+Mg\cdot0.6=\frac23m_{beam}g\cdot1+mgx,\\\space\\ x=0.8\text{ m}.$$