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Unformatted text preview: alexander (jra2623) – homework 17 – Turner – (92510) 1 This printout should have 11 questions. Multiplechoice 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 θ 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 xaxis will the resultant force point? Let counterclockwise be positive, within the limits 180 ◦ to 180 ◦ . Correct answer: 29 . 6657 ◦ . Explanation: The angle θ shown is θ = 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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 Summer '08
 Kaplunovsky
 Force, Mass, Work, General Relativity, Correct Answer, Celestial mechanics

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