Given,
The distance from the elbow joint to the point where the muscle is attached to the radial bone, d=3.6 cm=0.036 m
The mass of the forearm and the hand, M=2.7 kg
The combined length of the forearm and the hand, L=36 cm=0.36 cm
The mass of the ball. m=0.146 kg
From the diagram, the centre of mass of the forearm and the hand is at a distance of L/2.
For the arm to be in the rotational equilibrium, the net torque on the arm must be equal to zero.
That is,
[tex]F_{\text{bicep}}\times d-Mg\times\frac{L}{2}-mg\times L=0[/tex]Where g is the acceleration due to gravity and F_bicep is the force applied by the bicep muscle.
On rearranging the above equation,
[tex]F_{\text{bicep}}=\frac{Mg\times\frac{L}{2}+mg\times L}{d}[/tex]On substituting the known values,
[tex]\begin{gathered} F_{\text{bicep}}=\frac{2.7\times9.8\times\frac{0.36}{2}+0.146\times9.8\times0.36}{0.036} \\ =\frac{4.76+0.52}{0.036} \\ =146.67\text{ N} \end{gathered}[/tex]Thus the force that the bicep muscle must exert in order to keep the arm in rotational equilibrium is 146.67 N
