Question

(I) Approximately what magnitude force, \({F_{\bf{M}}}\)must the extensor muscle in the upper arm exert on the lower arm to hold a 7.3-kg shot put (Fig. 9–71)? Assume the lower arm has a mass of 2.3 kg and its CG is 12.0 cm from the elbow-joint pivot.

Short answer

The magnitude of force that the extensor muscle in the upper arm must exert on the lower arm to hold the shot put is 967.6 N.

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The force due to bicep muscle is\({F_{\rm{M}}}\).
• The mass of the lower arm is\({m_l} = 2.3{\rm{ kg}}\).
• The mass of the shot put is\(m = 7.3{\rm{ kg}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Understanding the forces acting on the forearm

The weight of the ball held in the palm acts in the downward direction. The force is due to the bicep muscle acting in the downward direction. The force acts due to the lower mass of the arm.

At equilibrium, balance the torques about the elbow joint. The torque acting in the clockwise direction will be equated to the torque acting in the anti-clockwise direction.With the help of these torque balance equations, the force acting due to the biceps muscle can be evaluated.

Determination of the magnitude of force the extensor muscle in the upper arm must exert on the lower arm to hold the shot put

Let mg be the weight of the ball held in the hand. The mass that the person holds in his hand is m.CG is the center of gravity. The force acting due to the lower mass of the arm is\({F_{\rm{l}}}\).

The torque due to the force\({F_{\rm{M}}}\)acts in the anti-clockwise direction.The torque due to the force\({F_{\rm{l}}}\)acts in the clockwise direction. The torque due to the weight of the ball acts in the clockwise direction. The force\({F_{\rm{l}}}\)acts at the CG of the hand, and the distance from the center of gravity to the elbow joint is 12 cm.

The forces acting on the forearm can be represented as:

At equilibrium, the net torque acting on the forearm about joint (J) of the elbow becomes zero.

From the above figure, the torques equation can be expressed as:

\(\begin{array}{c}\sum {{T_{net}}} = 0\\{F_{\rm{M}}} \times AJ = JB \times {F_{\rm{l}}} + JC \times mg\\{F_{\rm{M}}} \times 2.5{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) = 12{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) \times {F_{\rm{l}}} + 30{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) \times mg\\{F_{\rm{M}}} \times 0.025{\rm{ m}} = 0.12{\rm{ m}} \times {m_l}g + 0.30{\rm{ m}} \times mg\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{\rm{M}}} \times 0.025{\rm{ m}} = 0.12{\rm{ m}} \times 2.3{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} + 0.30{\rm{ m}} \times 7.3{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2}\\{F_{\rm{M}}} = \frac{{24.19{\rm{ kg}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}}}{{0.025{\rm{ m}}}}\\{F_{\rm{M}}} = 967.6\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\{F_{\rm{M}}} = 967.6{\rm{ N}}\end{array}\)

Thus, the magnitude of force that the extensor muscle in the upper arm must exert on the lower arm to hold the shot put is 967.6 N.