Practice Exam 4 - practicework 04 BAUTISTA, ALDO Due: May 1...

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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 free-body 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 two-thirds 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.

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Practice Exam 4 - practicework 04 BAUTISTA, ALDO Due: May 1...

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