This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Overton, Mays Oldquiz 4 Due: May 3 2005, 2:00 pm Inst: Turner 1 This print-out should have 23 questions. Multiple-choice 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 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)(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 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 . 206584 . Explanation: Consider the free-body 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 ....
View Full Document
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