Question

(II) If 25 kg is the maximum mass mthat a person can hold in a hand when the arm is positioned with a 105° angle at the elbow as shown in Fig. 9–74, what is the maximum force\({F_{{\bf{max}}}}\) that the biceps muscle exerts on the forearm? Assume the forearm and hand have a total mass of 2.0 kg with a CG that is 15 cm from the elbow, and that the biceps muscle attaches 5.0 cm from the elbow.

Short answer

The magnitude of the maximum force that the biceps muscle exerts on the forearm is 1849.58 N.

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The maximum force due to the biceps muscle is\({F_{{\rm{max}}}}\).
• The total mass of the forearm and the hand is\({m_l} = 2{\rm{ kg}}\).
• The mass of the ball is\(m = 25{\rm{ kg}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Understanding the forces acting on the forearm

The person holds the ball whose weight acts in the downward direction. The force is due to the biceps muscle acting at an angle in the upward direction. The force due to the total mass of the arm acts in the downward direction.

Apply the equilibrium conditions; the net torque about the elbow joint can be equated to zero.

On solving these equations, the maximum force exerted by the biceps muscle can be estimated.

Determination of the magnitude of the maximum force that the biceps muscle exerts on the forearm

The forces acting on the forearm can be represented as:

At equilibrium, the net torques acting on the forearm about the elbow joint (A) becomes zero.

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

\(\begin{array}{c}\sum {{T_{net}}} = 0\\{F_{{\rm{max}}}}\sin \phi \times AB = AC \times {m_l}g + AD \times mg\\{F_{{\rm{max}}}}\sin \left( {180^\circ - 105^\circ } \right) \times 5{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) = 15{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) \times {m_l}g + 35{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) \times mg\\{F_{{\rm{max}}}}\sin 75^\circ \times 0.05{\rm{ m}} = 0.15{\rm{ m}} \times {m_l}g + 0.35{\rm{ m}} \times mg\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{{\rm{max}}}}\sin 75^\circ \times 0.05{\rm{ m}} = 0.15{\rm{ m}} \times 2{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} + 0.35{\rm{ m}} \times 25{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2}\\{F_{{\rm{max}}}} = \frac{{88.78}}{{0.048}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ = 1849.58{\rm{ N}}\end{array}\)

Thus, the magnitude of the maximum force that the biceps muscle exerts on the forearm is 1849.58 N.