# a0_001 - 2 ) = 7 . ( 2 1 + 0 . 50 1 . 327-1 2 ) = 4 . 9 m ....

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PHY 2053, Section 3794, Fall 2009, Quiz 10 1. One end of a uniform 7.0 m long rod of weight w is supported by a cable. The other end rests against the wall, where it is held by friction. The coeﬃcient of static friction between the wall and the rod is μ s = 0 . 50. Determine the minimum distance, x , from point A at which an additional weight w (the same as the weight of the rod) can be hung without causing the rod to slip at point A. Let T be the tension in the cable and N the normal force at the wall, with friction given by F R μ s N . Then: F x : N - T cos( θ ) = 0 F y : T sin( θ ) + F R - 2 × w = 0 , and τ : T × L sin( θ ) - w × L 2 - w × x = 0 , where torque has been calculated around point A. From the ﬁrst two equations we have: 2 × w = T sin( θ ) + F R T sin( θ ) + μ s × T cos( θ ) = T sin( θ ) 2 × w 1 + μ s × cot( θ ) , while from the third equation we ﬁnd T sin( θ ) = w × ( 1 2 + x L ) . Putting these together we have x L ( 2 1 + μ s × cot( θ ) - 1
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Unformatted text preview: 2 ) = 7 . ( 2 1 + 0 . 50 1 . 327-1 2 ) = 4 . 9 m . 2. A planet of radius 8 . 00 10 3 km spins with an angular velocity of 20 . 10-5 rad/s about an axis through the North Pole. What is the ratio of the normal force experienced by a person at the equator to that experienced by a person at the North Pole? Assume a constant gravitational acceleration of 1.60 m/s 2 and that both people are stationary relative to the planet and are at sea level. At the North Pole we have N NP-mg = 0 = N NP = mg , while at the equator we have N eq-mg =-m 2 R = N eq = mg-m 2 R . Thus the ratio N eq /N NP is given by: N eq N NP = 1- 2 R g = 1-(2 . 00 10-4 ) 2 8 . 00 10 6 1 . 60 = 1 . 000-. 200 = 0 . 800 ....
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## This note was uploaded on 12/15/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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