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Unformatted text preview: practicework 04 BAUTISTA, ALDO Due: May 1 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. The system shown in the figure is in equilib rium. A 12 kg mass is on the table. A string attached to the knot and the ceiling makes an angle of 66 with the horizontal. The coeffi cient of the static friction between the 12 kg mass and the surface on which it rests is 0 . 41. The acceleration of gravity is 9 . 8 m / s 2 . 12 kg m 6 6 What is the largest mass m can have and still preserve the equilibrium? Correct answer: 11 . 0505 kg (tolerance 1 %). Explanation: Let : M = 12 kg , m = 11 . 0505 kg , and = 66 . 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) (12 kg) (9 . 8 m / s 2 ) cos(66 ) = 118 . 544 N . For m to remain in equilibrium T max sin = m max g , so (2) m max = T max sin g = (118 . 544 N) sin(66 ) (9 . 8 m / s 2 ) = 11 . 0505 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) (12 kg) tan(66 ) = 11 . 0505 kg . Question 2 Part 1 of 1. 10 points. A uniform ladder of mass 30 kg and length 3 . 8 m leans against a smooth wall at an angle 52 with respect to the ground. A person of mass 59 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 . 476964 (tolerance 1 %). Explanation: Consider the free body diagram for the ladder shown below. f W =mg W =Mg O 1 2 x y w f f practicework 04 BAUTISTA, ALDO Due: May 1 2006, 4:00 am 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  M g 1 2 L cos  m g 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 m g tan = 1 2 (30 kg) (9 . 8 m / s 2 ) tan(52 ) + 2 3 (59 kg) (9 . 8 m / s 2 ) tan(52 ) = 416 . 008 N , and f y = M g + m g = (30 kg) (9 . 8 m / s 2 ) + (59 kg) (9 . 8 m / s 2 ) = 872 . 2 N . Since the friction force f x cannot exceed s times the normal force f y , we must have s f x f y = . 476964 ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Mass, Work

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