Question

A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a distance L from the pivot. On the other side of the pivot point are two smaller children each of mass M/2. One is sitting a distance L/6 from the pivot. How far from the pivot must the other small child be sitting in order for the system to be balanced?

Answer 1

Answer:$\frac{5L}{6}$

Explanation:

Given

Wooden board is pivoted at center and

Older child of mass $M=32\ kg$ is sitting at a distance of L from  center

if two child of mass $\frac{M}{2}$ is sitting at a distance $\frac{L}{6}$ and $x$(say) from pivot then net torque about pivot is zero

i.e.

$\Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx$

as $\tau_{net}=0$

Therefore

$MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx$

$L-\frac{L}{6}=x$

$x=\frac{5L}{6}$

Therefore another child is sitting at a distance of $\frac{5L}{6}$