qz8sol_3706Hs11

# qz8sol_3706Hs11 - d 2 = 2.50 cm behind that contact point...

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TA: Tomoyuki Nakayama Tuesday, March 29, 2011 PHY 2048: Physic 1, Discussion Section 3706H Quiz 8 (Homework Set #10) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ In the figure below right, a 30.0 kg block is held in place via a pulley system. The person's upper arm is vertical; the forearm makes angle θ = 35.0° with the horizontal. Forearm and hand together have a mass of 3.00 kg, with a center of mass at distance d 1 = 18.0 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
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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.

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