Question

Two small kids weighing $10\text{kg}$ and $15\text{kg}$ respectively, are trying to balance a seesaw of total length $5.0\text{m}$, with the fulcrum at the centre. If one of the kids is sitting at one end, where should the other one sit?

• A. $1.5\text{m}$ away from fulcrum
• B. $1.2\text{m}$ away from fulcrum
• C. $1.7\text{m}$ away from fulcrum
• D. None of these.

Answer 1

The correct option is C $1.7\text{m}$ away from fulcrum

If the kid of mass $15\text{kg}$ sit at the end, he will produce maximum torque about fulcrum. The seasaw will not be balanced for any position of kid of mass $10\text{kg}$. It is clear that the $10\text{kg}$ kid should sit at the end and the $15\text{kg}$ 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 seesaw equals the weight of that kid. Considering the rotational equilibrium of the seesaw, the torque of the forces acting on it should add to zero.


The forces are as follows:
a. $15g=150\text{N}$ downwards by the $15\text{kg}$ kid,
b. $10g=100\text{N}$ downwards by the $10\text{kg}$ kid,
c. Weight of the seesaw, and
d. The normal force by the fulcrum.

Taking torques about the fulcrum O, we get
$150\times x=100\times 2.5$
or $x=1.7\text{m}$