91Ó°ÊÓ

The force on A is\({F_{\rm{A}}}\).

The force on B is\({F_{\rm{B}}}\).

The mass of the person is\(m = 52\;{\rm{kg}}\).

The mass of the board is \({m_{\rm{b}}} = 28\;{\rm{kg}}\).

In this problem, the net torque is equivalent to zero because the board is in equilibrium position. To calculate the net torque at the left side of the board, use the counter-clockwise direction as positive.

The following is the free-body diagram.

The relation for net torque is written below:

\(\begin{array}{c}\sum \tau = 0\\\left[ {\left( {{F_{\rm{B}}} \times 1\;{\rm{m}}} \right) - \left( {mg \times 4\;{\rm{m}}} \right)} \right] = 0\\{F_{\rm{B}}} = 4mg\end{array}\)

Here,\(g\)is the gravitational acceleration.

Put the values in the above relation.

\(\begin{array}{c}{F_{\rm{B}}} = \left( {4\;{\rm{m}}} \right)\left( {52\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\\{F_{\rm{B}}} = 2040.4\;{\rm{N}}\end{array}\)

The relation for the forces in the y-direction is given below:

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{B}}} - {F_{\rm{A}}} - mg = 0\end{array}\)

Here,\(g\)is the gravitational acceleration.

Put the values in the above relation.

\(\begin{array}{c}\left[ {\left( {2040.4\;{\rm{N}}} \right) - \left( {{F_{\rm{A}}}} \right) - \left( {52\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)} \right] = 0\\{F_{\rm{A}}} = 1530.8\;{\rm{N}}\end{array}\)

Thus, \({F_{\rm{A}}} = 1530.8\;{\rm{N}}\) and \({F_{\rm{B}}} = 2040.4\;{\rm{N}}\) are the force exerting on points A and B.

The relation for net torque is given below:

\(\begin{array}{c}\sum \tau = 0\\\left[ {\left( {{F_{\rm{B}}} \times 1\;{\rm{m}}} \right) - \left( {{m_{\rm{b}}}g \times 2\;{\rm{m}}} \right) - \left( {mg \times 4\;{\rm{m}}} \right)} \right] = 0\end{array}\)

Here,\(g\)is the gravitational acceleration.

Put the values in the above relation.

\(\begin{array}{c}\left[ {\left( {{F_{\rm{B}}} \times 1\;{\rm{m}}} \right) - \left( {28\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times 2\;{\rm{m}}} \right) - \left( {52\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2} \times 4\;{\rm{m}}} \right)} \right] = 0\\{F_{\rm{B}}} = 2587.2\;{\rm{N}}\end{array}\)

The relation for the forces in the y-direction is given below:

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{B}}} - {F_{\rm{A}}} - mg - {m_b}g = 0\end{array}\)

Put the values in the above relation.

\(\begin{array}{c}\left[ {\left( {2587.2\;{\rm{N}}} \right) - \left( {{F_{\rm{A}}}} \right) - \left( {52\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) - \left( {28\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)} \right] = 0\\{F_{\rm{A}}} = 1803.2\;{\rm{N}}\end{array}\)

Thus, \({F_{\rm{A}}} = 1803.2\;{\rm{N}}\) and \({F_{\rm{B}}} = 2587.2\;{\rm{N}}\) are the force exerting on points A and B.