A bowler holds a bowling ball (M = 5.8 kg) in the palm of his hand (see the figu

Question

A bowler holds a bowling ball (M = 5.8 kg) in the palm of his hand (see the figure below). His upper arm is vertical, his lower arm (1.9 kg) is horizontal. Lower arm (forearm plus hand) center of mass (a) What is the magnitude of the force of the biceps muscle on the lower arm? XN (b) What ?s the magnitude of the force between the bony structures at the elbow contact point? N Do you see that we have two unknown forces and that if we take the application point of one of them, then we can eliminate that force [rom a torque calculation balance-of-torques equation about the chosen point to solve for the other unknown force. Then use a balance-of-forces equation for vertical force components. Additional Materials Section 12.2

Explanation / Answer

A.) torque about elbow point should be Zero ,

F* 4 = 1.9 *9.8 * 15 + 5.8*9.8*33 ,

F = 538.75 N (here g = 9.8 m/s^2) ,

B.) System is in equilibrium, so

F net = 0 ,

f = 538.75 - ( 1.9 + 5.8)*9.8 ,

f = 463.29 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---