Two children of mass \({\rm{20}}\;{\rm{kg}}\)and \({\rm{30}}\;{\rm{kg}}\) sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of \({\rm{3}}\;{\rm{m}}\), at what distance from the pivot point is the small child sitting in order to maintain the balance?

Figure \({\rm{9}}{\rm{.1}}\)

The overall movement of an object, regardless of direction, is called distance.

The mass of the heavier child is \({{\rm{m}}_{\rm{1}}}{\rm{ = 30}}\;{\rm{kg}}\).

The distance between the children is \({\rm{r = 3}}\;{\rm{m}}\).

The mass of the small child is\({{\rm{m}}_{\rm{2}}}{\rm{ = 20}}\;{\rm{kg}}\)

Let, the small child is at a distance\({\rm{a}}\)from the pivot.

The torque on the pivot point due to the small child is,

\(\begin{array}{c}{\rm{t = mgr}}\\{\rm{ = 20 \times ag}}\end{array}\)

The torque on the pivot point due to the big child is,

\(\begin{array}{c}{\rm{\tau = }}{{\rm{m}}_{\rm{2}}}{\rm{g(3 - a)}}\\{\rm{ = 30g}}\left( {{\rm{3 - a}}} \right)\end{array}\)

For the equilibrium, both the torque will be the same. So,

\(\begin{array}{c}{\rm{20ga = 30g}}\left( {{\rm{3 - a}}} \right)\\{\rm{2a = 3}}\left( {{\rm{3 - a}}} \right)\\{\rm{2a = 9 - 3a}}\\{\rm{5a = 9}}\\{\rm{a = }}\frac{{\rm{9}}}{{\rm{5}}}\\{\rm{a = 1}}{\rm{.8}}\;{\rm{m}}\end{array}\)

Thus, the distance is\({\rm{a = 1}}{\rm{.8}}\;{\rm{m}}\).