solution-hw18

solution-hw18 - lai (yl8859) HW18 Ross (15200) 1 This...

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Unformatted text preview: lai (yl8859) HW18 Ross (15200) 1 This print-out should have 9 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 A 0 . 095 kg meter stick is supported at its 40 cm mark by a string attached to the ceiling. A 0 . 702 kg object hangs vertically from the 6 . 29 cm mark. A second mass is attached at another mark to keep it horizontal and in rotational and translational equilibrium. If the tension in the string attached to the ceiling is 21 . 6 N, find the value of the second mass. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 40708 kg. Explanation: Let : m = 0 . 095 kg , m 1 = 0 . 702 kg , and T = 21 . 6 N . Because of vertical equilibrium, T = mg + m 1 g + m 2 g m 2 = T g- m 1- m = 21 . 6 N 9 . 8 m / s 2- . 702 kg- . 095 kg = 1 . 40708 kg . 002 (part 2 of 2) 10.0 points Find the mark at which the second mass is attached. Correct answer: 56 . 1429 cm. Explanation: Let : L = 100 cm , = 40 cm , and 1 = 6 . 29 cm . Applying torque about the 0 cm mark, T = mg L 2 + m 1 g 1 + m 2 g 2 2 T = mg L + 2 m 1 g 1 + 2 m 2 g 2 2 = T m 2 g- m 1 1 m 2- mL 2 m 2 = (21 . 6 N)(40 cm) (1 . 40708 kg)(9 . 8 m / s 2 )- (0 . 702 kg)(6 .....
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solution-hw18 - lai (yl8859) HW18 Ross (15200) 1 This...

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