(I) What is the mass of the diver in Fig. 9–49 if she exerts a torque of \({\bf{1800}}\;{\bf{m}} \cdot {\bf{N}}\) on the board, relative to the left (A) support post?

The torque is\(\tau = 1800\;{\rm{m}} \cdot {\rm{N}}\).

The distance between points A and B is\(d = 1.0\;{\rm{m}}\).

The distance between the diver and point B is\(d' = 3.0\;{\rm{m}}\).

The diver’s torque is her weight times the distance between her and the support of the plank.

The relation for mass is shown below.

\(\begin{array}{l}\tau = {F_{\rm{W}}} \times \left( {d + d'} \right)\\\tau = mg \times \left( {d + d'} \right)\end{array}\)

Here,\(g\)is the gravitational acceleration, and\({F_{\rm{W}}}\)is the weight of the diver.

Put the values in the above relation.

\(\begin{array}{c}\left( {1800\;{\rm{m}} \cdot {\rm{N}}} \right) = m\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {1\;{\rm{m}} + 3\;{\rm{m}}} \right)\\m = 45.9\;{\rm{kg}}\end{array}\)

Thus, \(m = 45.9\;{\rm{kg}}\) is the mass of the diver.