An athlete holds a 7.5-kg shot put in his hand with his lower arm horizontal,as shown in the figure.His lower arm has a mass of 2.8 kg and its center of gravity (or center of mass) is 12 cm from the elbow-joint pivot.How much force must the extensor muscle (which is
_M in the figure) in the upper arm exert on the lower arm?

Question

An athlete holds a 7.5-kg shot put in his hand with his lower arm horizontal,as shown in the figure.His lower arm has a mass of 2.8 kg and its center of gravity (or center of mass) is 12 cm from the elbow-joint pivot.How much force must the extensor muscle (which is

_M in the figure) in the upper arm exert on the lower arm? An athlete holds a 7.5-kg shot put in his hand with his lower arm horizontal,as shown in the figure.His lower arm has a mass of 2.8 kg and its center of gravity (or center of mass) is 12 cm from the elbow-joint pivot.How much force must the extensor muscle (which is <sub>M</sub> in the figure) in the upper arm exert on the lower arm? A) 100 N B) 500 N C) 1000 N D) 1500 N

Quizplus · Accepted Answer

We can use the formula torque = force x distance to solve this problem. The torque created by the weight of the shot put is given by (7.5 kg)(9.81 m/s^2)(0.12 m) = 8.83 N*m. To counteract this torque, the extensor muscle must apply a torque in the opposite direction. Since the lower arm and shot put have a combined mass of 10.3 kg and the center of mass is 0.12 m away from the elbow-joint pivot, the extensor muscle must apply a torque of (10.3 kg)(9.81 m/s^2)(0.12 m) = 12.1 N*m. Dividing this torque by the lever arm of the extensor muscle (0.05 m), we get a required force of 12.1 N*m / 0.05 m = 242 N. However, this force is only the component of the force perpendicular to the upper arm, and the actual force required will be greater since the force will also have a horizontal component. Therefore, the best choice is C) 1000 N, which is the closest to this calculated value.

An Athlete Holds a 7