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# a9_001 - a How far can the person climb up the ladder...

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PHY 2053, Section 3794, Fall 2009, Quiz 9 G = 6 . 67428 × 10 11 m 3 / kg / s 2 . M Earth = 5 . 9742 × 10 24 kg . 1. An artiﬁcial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.30 m/s 2 . a) What is the orbital period of the satellite? First, we compute r from a = GM/r 2 . We ﬁnd r = GM/a . It is also possible to compute v from v 2 /r = a , which gives v = ar = ( GMa ) 1 / 4 . Now we can compute T from v = 2 πr/T , which gives: T = 2 πr v = 2 π × ± GM a × 1 ( GMa ) 1 / 4 = 2 π × ( GM a 3 ) 1 / 4 = 87 . 9 min = 1 . 46 hr . 2. A 4.25 m, 235 N uniform ladder leans against a smooth wall. The coeﬃcient of static friction between the ladder and the ground is 0.425, and the ladder makes an angle of 53 . 13 with the ground. A 645 N person begins to climb the ladder.
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Unformatted text preview: a) How far can the person climb up the ladder before it begins to slip? We have F x : F R = N W , where F R is the force of friction and N W is the normal force exerted by the wall, and F y : N G = W P + W L , where N G is the normal force exerted by the ground, W P is the weight of the person and W L is the weight of the ladder. We can put these together: F R = μN G = μ ( W P + W L ) = N W . Now we calculate torque about the base of the ladder. τ : L N W sin( θ ) − L/ 2 W L cos( θ ) − x W P cos( θ ) = 0 . Thus: x = Lμ ( W P + W L ) sin( θ ) − L/ 2 W L cos( θ ) W P cos( θ ) = 2 . 51 m ....
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