Question

Short answer

The result of the forces acting on the diver board is

(a)\({F_{\rm{A}}} = 1530.8\;{\rm{N}}\)and\({F_{\rm{B}}} = 2040.4\;{\rm{N}}\)

(b) \({F_{\rm{A}}} = 1803.2\;{\rm{N}}\) and \({F_{\rm{B}}} = 2587.2\;{\rm{N}}\)

Step-by-step solution

Given Data

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 \({m_{\rm{b}}} = 28\;{\rm{kg}}\).

Understand the torque acting on the diver

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.

Draw the free-body diagram and calculate the forces on point B

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.

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

Calculate the forces on point A

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.

Calculate the forces on point 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.

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

Calculate the forces on point A

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.