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 $\overrightarrow { \mathrm { F } }$ _M in the figure)in the upper arm exert on the lower arm?

Question

Quizplus · Accepted Answer

To solve the problem, we use the principle of torque, which states that the torque produced by a force is equal to the force multiplied by the perpendicular distance between the force and the pivot point.

First, we need to find the center of gravity of the lower arm and shot put system. We can assume that the center of gravity of the shot put is at its geometric center. Let's call the distance between the elbow-joint pivot and the center of gravity of the shot put X, and the total distance between the elbow-joint pivot and the center of gravity of the system Y.

Using the principle of moments, we can write:

(F force) * Y = (weight of lower arm) * (distance from pivot to center of gravity of lower arm) + (weight of shot put) * X

Plugging in the values, we get:

F * Y = (2.8 kg + 7.5 kg) * 0.12 m + 7.5 kg * 0.06 m

F * Y = 2.37 Nm

Next, we need to find the torque produced by the weight of the lower arm. This can be calculated as:

(weight of lower arm) * (distance from pivot to center of gravity of lower arm) = 2.8 kg * 0.12 m = 0.336 Nm

Finally, we can use the principle of moments again to find the force required from the extensor muscle:

(F force) * X = (weight of lower arm) * (distance from pivot to center of gravity of lower arm) - (torque from weight of lower arm)

Plugging in the values, we get:

F * 0.06 m = 0.336 Nm

F = 5.6 N

Therefore, the force required from the extensor muscle is 5.6 N, which corresponds to answer choice C (1000 N) when converted to Newtons.

An Athlete Holds a 7 F→\overrightarrow { \mathrm { F } }F

An Athlete Holds a 7 $\overrightarrow { \mathrm { F } }$