Question

Two small kids weighing $10kg$ and $15kg$, respectively, are trying to balance a see saw of total length $5.0m$, with the fulcrum at the centre. If kid of mass $10kg$ is sitting at an end, at what distance from fulcrum on other side (in m) should the other kid sit?

Answer 1

Answer: 1.7

If the kid of mass $15kg$ sits at the end, he will produce maximum torque about fulcrum. The see-saw will not be balance for any position of kid of mass $10kg$. It is clear that the $10kg$ kid should sit at the end and the $15kg$ kid should sit closer to the centre. Suppose his distance from the centre is $x$.

As the kids are in equilibrium, the normal force between a kid and the see-saw equals the weight of that kid. Considering the rotational equilibrium of the see-saw, the torque of the forces acting on it should add to zero. The forces are as follows:
(a) $(15kg)g=150N$ downwards by the $15kg$ kid,
(b) $(10kg)g=100N$ downwards by the $10kg$ kid,
(c) weight of the see-saw, and
(d) the normal force by the fulcrum.
Taking torques about the fulcrum, $150\times x=100\times 2.5$
or $x=1.7m$