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Unformatted text preview: P1112 Fall 2008 Homework 6, Solution 1. a) The forces acting on the barbell are the gravity m b g and F a b , the force exerted by the athlete, while the forces acting on the athlete are the gravity m a g , the normal force exerted by the ground, and F b a , the reaction force exerted by the barbell. b) For the barbell, F a b − m b g = m b a and for the athlete N − F b a − m a g = since the athlete is at rest. F a b and F b a are a pair of action/reaction forces so they have the same magnitude. So we have N = m a g m b g m b a . The mass of the barbell is m b = W b / g = 490 N / 9.8 ms − 2 = 50 kg . a is determined by a = 2d / t 2 = 0.47m / s 2 . So finally, N = 90.0 kg × 9.8 ms − 2 490 N 50 kg × 0.47 ms − 2 = 1.4 × 10 3 N . 2. a) The box is moving at a constant velocity so its acceleration = a . Apply Newton’s 2 nd law along the y direction: ) sin( = = + y ma w N F θ ....
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This note was uploaded on 08/28/2009 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell.
 Fall '07
 LECLAIR,A
 Force, Gravity, Work

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