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qz9sol_3885f10

# qz9sol_3885f10 - direction to be positive a What is the...

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TA: Tomoyuki Nakayama Monday, November 1, 2010 PHY 2048: Physic 1, Discussion Section 3885 Quiz 9 (Homework Set #10) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ In the figure below right, a 15.0 kg block is held in place via a pulley system. The person's upper arm is vertical; the forearm makes angle θ = 25° with the horizontal. Forearm and hand together have a mass of 2.50 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 d 2 = 2.00 cm behind that contact point. Distance d 3 is 40.0 cm. Take the upward
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Unformatted text preview: direction to be positive. 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 -F t d 2 cos θ – mgd 1 cos θ + Mgd 3 cos θ = 0 ⇒ F t = (Mgd 3 – mgd 1 )/d 2 = 2720 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 + F t – F h – mg = 0 F h = mg – Mg – F t = -2840 N The negative sign indicates it is directed downward....
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