(I) Suppose the point of insertion of the biceps muscle into the lower arm shown in Fig. 9–13a (Example 9–8) is 6.0 cm instead of 5.0 cm; how much mass could the person hold with a muscle exertion of 450 N?

The given data can be listed below as:

• The force due to bicep muscle is\({F_{\rm{M}}} = 450{\rm{ N}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

The forces on the forearm act in the vertical direction. Take the torques about the elbow joint. The net torque in the counterclockwise direction is equal to net torque in the anti-clockwise direction.

The force is due to the combined mass of the forearm and the hand acting in the clockwise direction. The force is due to the ball's weight acting in the clockwise direction. The force is acting in the anti-clockwise direction due to the bicep muscle.

The diagram of the forearm can be shown as:

Here, mg is the weight of the ball that the person holds in the hand,\({F_{\rm{J}}}\)is the force exerted by the bone at the joint of the elbow, m is the mass that the person holds in his hand, \({F_{\rm{M}}}\) is the force due to the bicep muscle, and CG is the center of gravity.

At equilibrium, the net torques acting on the forearm about the joint of the elbow or the point at which\({F_{\rm{J}}}\)is acting becomes zero.

From the above figure, the net torque equation can be expressed as:

\(\begin{array}{c}\sum {{T_{net}}} = 0\\{F_{\rm{M}}} \times 6{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) = 15{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) \times {\rm{2 kg}} \times g + 35{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) \times mg\\{F_{\rm{M}}} \times 0.060{\rm{ m}} = 0.15{\rm{ m}} \times {\rm{2 kg}} \times g + 0.35{\rm{ m}} \times mg\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}450{\rm{ N}}\left( {\frac{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1{\rm{ N}}}}} \right) \times 0.060{\rm{ m}} = 0.15{\rm{ m}} \times {\rm{2 kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} + 0.35{\rm{ m}} \times m \times 9.81{\rm{ m/}}{{\rm{s}}^2}\\27{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2} = 2.943{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2} + 3.43m{\rm{ m/}}{{\rm{s}}^2}\\{\rm{24}}{\rm{.057 kg}} \cdot {\rm{m/}}{{\rm{s}}^2} = 3.43m{\rm{ m/}}{{\rm{s}}^2}\\m = 7.01{\rm{ kg}}\end{array}\)

Thus, a mass of 7.01 kg can be held by the person with a muscle exertion of 450 N.