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Unformatted text preview: Overton, Mays Oldquiz 4 Due: May 3 2005, 2:00 pm Inst: Turner 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The system shown in the figure is in equilib rium. A 17 kg mass is on the table. A string attached to the knot and the ceiling makes an angle of 68 with the horizontal. The coeffi cient of the static friction between the 17 kg mass and the surface on which it rests is 0 . 41. The acceleration of gravity is 9 . 8 m / s 2 . 17 kg m 6 8 What is the largest mass m can have and still preserve the equilibrium? Correct answer: 17 . 2514 kg. Explanation: Let : M = 17 kg , m = 17 . 2514 kg , and = 68 . Basic Concepts: The net force exerted on an object in equilibrium equals zero. Solution: For the system to remain in equi librium, the net forces on both M and m should be zero; and thus the tension in the rope has an upper bound value T max T max cos = M g , so (1) T max = M g cos = (0 . 41)(17 kg)(9 . 8 m / s 2 ) cos(68 ) = 182 . 341 N . For m to remain in equilibrium T max sin = m max g , so (2) m max = T max sin g = (182 . 341 N)sin(68 ) (9 . 8 m / s 2 ) = 17 . 2514 kg . Alternative Solution: Equations 1 and 2 come directly from the freebody diagram for the knot. Dividing Eq. 2 by Eq. 1, we have tan = m max M , so m max = M tan (3) = (0 . 41)(17 kg) tan(68 ) = 17 . 2514 kg . 002 (part 1 of 1) 10 points A uniform ladder of mass 20 kg and length 6 m leans against a smooth wall at an angle 72 with respect to the ground. A person of mass 88 kg climbs up the ladder to a point twothirds of the way to the top. Assume that there is no friction between the top of the ladder and the wall. The acceleration of gravity is 9 . 8 m / s 2 . Compute the minimum coefficient of static friction between the ladder and the ground that would prevent the ladder from slipping. Correct answer: 0 . 206584 . Explanation: Consider the freebody diagram for the lad der shown below. f W =mg W =Mg O 1 2 x y w f f Overton, Mays Oldquiz 4 Due: May 3 2005, 2:00 pm Inst: Turner 2 The three conditions of equilibrium are 0 = X F x = f x f w , 0 = X F y = f y Mg mg, 0 = X = f w L sin  Mg 1 2 L cos  mg 2 3 L cos , where the torques are with respect to the point where the ladder contacts the ground. Solving these equations, we find f x = f w = 1 2 M g + 2 3 mg tan = 1 2 (20 kg)(9 . 8 m / s 2 ) tan(72 ) + 2 3 (88 kg)(9 . 8 m / s 2 ) tan(72 ) = 218 . 649 N and f y = M g + mg = (20 kg)(9 . 8 m / s 2 ) + (88 kg)(9 . 8 m / s 2 ) = 1058 . 4 N ....
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This note was uploaded on 03/20/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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