In the figure, a 11 kg block is held in place via a pulley system. The person\'s

Question

In the figure, a 11 kg block is held in place via a pulley system. The person's upper arm is vertical; the forearm makes angle ? = 32° with the horizontal. Forearm and hand together have a mass of 1.8 kg, with a center of mass at distance d1 = 17 cm from the contact point of the forearm bone and the upper-arm bone (humerus). The triceps muscle pulls vertically upward on the forearm at distance d2 = 2.8 cm behind that contact point. Distance d3 is 38 cm. Take the upward direction to be positive. What are the forces on the forearm from (a) the triceps muscle and (b) the humerus?

Explanation / Answer

a) What is the force on the forearm from the triceps muscle?
Since the block is in equilibrium, the tension in the string is equal to the gravitational force Mg on the block. Therefore, the force exerted at the person’s hand is Mg and it is directed upward. We take counterclockwise as positive. The balance of torques equation around the contact point yields
-Ftd2cos – mgd1cos + Mgd3cos = 0
Ft = (Mgd3 – mgd1)/d2 = 1356 N,
where m is the mass of the forearm. The force
is directed upward.

b) What is the force on the forearm from the humerus?
We take our positive direction upward. The balance of forces equation yields
Mg + Ft – Fh – mg = 0 Fh = mg – Mg – Ft = -1446.16 N
The negative sign indicates it is directed downward

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