91Ó°ÊÓ

The mass of the adult is\(M = 75\;{\rm{kg}}\).

The mass of the child is\(m = 25\;{\rm{kg}}\).

The mass of the board is\({m_{\rm{B}}} = 15\;{\rm{kg}}\).

The length of the board is\(l = 9\;{\rm{m}}\).

In this problem, the upward force F_P does not exert any torque about the pivot point. Also, the pivot must be kept in such a place where the net torque on the board is equal to zero.

The following is the free body diagram.

The relation to calculate the net torque can be written as:

\(\begin{array}{c}\sum \tau = 0\\\left( {Mg \times x} \right) - \left( {mg \times \left( {l - x} \right)} \right) = 0\end{array}\)

Here, \(g\) is the gravitational acceleration and \(x\) is the distance from the left end point to point P.

On plugging the values in the above relation, you get:

\(\begin{array}{c}\left( {75\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times x} \right) - \left( {25\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times \left( {9\;{\rm{m}} - x} \right)} \right) = 0\\x = 2.25\;{\rm{m}}\end{array}\)

Thus, \(x = 2.25\;{\rm{m}}\) is the position of the pivot point from the adult.

The relation to calculate the pivot point can be written as:

\(\begin{array}{c}\sum \tau = 0\\\left( {Mg \times x'} \right) - \left( {mg \times \left( {l - x'} \right)} \right) - \left( {{m_{\rm{B}}}g \times \left( {\frac{l}{2} - x'} \right)} \right) = 0\end{array}\)

On plugging the values in the above relation, you get:

\(\begin{array}{c}\left[ \begin{array}{l}\left( {75\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times x'} \right) - \\\left( {25\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times \left( {9\;{\rm{m}} - x'} \right)} \right) - \left( {15\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times \left( {\frac{{9\;{\rm{m}}}}{2} - x'} \right)} \right)\end{array} \right] = 0\\x' = 2.5\;{\rm{m}}\end{array}\)

Thus, \(x' = 2.5\;{\rm{m}}\) is the distance of the pivot point from the adult.