Unformatted text preview: d 2 = 2.50 cm behind that contact point. Distance d 3 is 40.0 cm. 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 joint yields -F t d 2 cos θ – mgd 1 cos θ + Mgd 3 cos θ = 0 ⇒ F t = (Mgd 3 – mgd 1 )/d 2 = 4.49 ×10 3 N, where m is the mass of the forearm. Note this 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 + F t – F h – mg = 0 F h = mg – Mg – F t = -4.76 ×10 3 N The negative sign indicates it is directed downward....
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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
- Spring '08