## College Physics (4th Edition)

The force that the biceps muscle exerts is $130.5~N$
We can find the angle $\theta$ the force exerted by the biceps makes with the horizontal: $tan~\theta = \frac{30.0}{5.00}$ $\theta = tan^{-1}(\frac{30.0}{5.00})$ $\theta = 80.5^{\circ}$ We can consider the torques about the rotation axis in the elbow. The force from the biceps exerts a counterclockwise torque about this axis. The weight of the milk and the weight of the forearm exert clockwise torques about this axis. Since the system is in equilibrium, the counterclockwise torque is equal in magnitude to the sum of the clockwise torques. We can find the force that the biceps muscle exerts: $\tau_{ccw} = \tau_{cw}$ $(5.00~cm)~(F_b)~sin~80.5^{\circ} = (16.5~cm)(18.0~N)+(35.0~cm)(9.9~N)$ $F_b = \frac{(16.5~cm)(18.0~N)+(35.0~cm)(9.9~N)}{(5.00~cm)~sin~80.5^{\circ} }$ $F_b = 130.5~N$ The force that the biceps muscle exerts is $130.5~N$