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Unformatted text preview: hernandez (ejh742) homework 27 Turner (58120) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A 14 m ladder whose weight is 297 N is placed against a smooth vertical wall. A person whose weight is 486 N stands on the ladder a distance 5 . 4 m up the ladder. The foot of the ladder rests on the floor 8 . 82 m from the wall. 486 N 297 N 5 . 4 m 14 m b 8 . 82 m Note: Figure is not to scale. Calculate the force exerted by the wall. Correct answer: 272 . 54 N. Explanation: Let : = 14 m , d = 5 . 4 m , s = 8 . 82 m , W = 297 N , and W p = 486 N . Pivot b N w F f N f W p W = arccos s = arccos parenleftbigg 8 . 82 m 14 m parenrightbigg = 50 . 9499 In equilibrium summationdisplay vector F = 0 and summationdisplay vector = 0 . summationdisplay : W p d cos + W 2 cos N w sin = 0 , where d is the distance of the person from the bottom of the ladder. Therefore 2 F w sin = 2 W p d cos + W cos F w = 2 W p d + W 2 cos sin = 2 (486 N)(5 . 4 m) + (297 N)(14 m) 2 (14 m) cos 50 . 9499 sin50 . 9499 = 272 . 54 N . 002 (part 2 of 2) 10.0 points Calculate the normal force exerted by the floor on the ladder. Correct answer: 783 N. Explanation: Applying translational equilibrium, summationdisplay F y : N f W p W = 0 N f W p W = 0 . N f = W p + W = 486 N + 297 N = 783 N . hernandez (ejh742) homework 27 Turner (58120) 2 003 (part 1 of 3) 10.0 points An 18 . 3 m , 489 N uniform ladder rests against a frictionless wall, making an angle of 64 with the horizontal. b 825 N 489 N 5 . 25 m 18 . 3 m 64 Note: Figure is not drawn to scale. Find the horizontal force exerted on the base of the ladder by Earth when an 825 N fire fighter is 5 . 25 m from the bottom. Correct answer: 234 . 687 N. Explanation: Let : L = 18 . 3 m , = 5 . 25 m , = 64 , W = 489 N , and W p = 825 N . L sin L cos L 2 cos cos Pivot b N w f N g W p W Applying rotational equilibrium with the pivot at the point of contact with the ground, summationdisplay = W p cos + W L 2 cos N w L sin = 0 2 W p cos + W L cos = 2 N w L sin N w = W p cos L sin + W cos 2 sin = W p L...
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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