hw17 - alexander (jra2623) homework 17 Turner (92510) 1...

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Unformatted text preview: alexander (jra2623) homework 17 Turner (92510) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points Three 2 kg masses are located at points in the xy plane as shown. 40 cm 53 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? The universal gravita- tional constant is 6 . 6726 10 11 N m 2 / kg 2 . Correct answer: 1 . 91978 10 9 N. Explanation: Let : m = 2 kg , x = 40 cm = 0 . 4 m , y = 53 cm = 0 . 53 m , and G = 6 . 6726 10 11 N m 2 / kg 2 . The force from the mass on the right points in the x direction and has magnitude F 1 = G mm x 2 = Gm 2 x 2 = (6 . 6726 10 11 N m 2 / kg 2 ) (2 kg) 2 (0 . 4 m) 2 = 1 . 66815 10 9 N . The other force points in the y direction and has magnitude F 2 = (6 . 6726 10 11 N m 2 / kg 2 ) (2 kg) 2 (0 . 53 m) 2 = 9 . 50174 10 10 N . F 2 F 1 F The magnitude of the resultant force is F = radicalBig F 2 1 + F 2 2 = bracketleftBig (1 . 66815 10 9 N) 2 + (9 . 50174 10 10 N) 2 bracketrightBig 1 / 2 = 1 . 91978 10 9 N . 002 (part 2 of 2) 10.0 points At what angle from the positive x-axis will the resultant force point? Let counterclockwise be positive, within the limits- 180 to 180 . Correct answer: 29 . 6657 . Explanation: The angle shown is = arctan parenleftbigg f 2 f 1 parenrightbigg = arctan parenleftbigg 9 . 50174 10 10 N 1 . 66815 10 9 N parenrightbigg = 29 . 6657 ....
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas at Austin.

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hw17 - alexander (jra2623) homework 17 Turner (92510) 1...

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