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Unformatted text preview: sanchez (ds28677) – homework 27 – Turner – (58220) 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 An 11 m ladder whose weight is 337 N is placed against a smooth vertical wall. A person whose weight is 342 N stands on the ladder a distance 4 . 4 m up the ladder. The foot of the ladder rests on the floor 4 . 62 m from the wall. 342 N 337 N 4 . 4 m 11 m b 4 . 62 m Note: Figure is not to scale. Calculate the force exerted by the wall. Correct answer: 141 . 292 N. Explanation: Let : ℓ = 11 m , d = 4 . 4 m , s = 4 . 62 m , W ℓ = 337 N , and W p = 342 N . Pivot b N w F f N f W p W ℓ θ θ = arccos s ℓ = arccos parenleftbigg 4 . 62 m 11 m parenrightbigg = 65 . 1654 ◦ In equilibrium summationdisplay vector F = 0 and summationdisplay vector τ = 0 . summationdisplay τ ◦ : W p d cos θ + W ℓ ℓ 2 cos θ − N w ℓ sin θ = 0 , where d is the distance of the person from the bottom of the ladder. Therefore 2 F w ℓ sin θ = 2 W p d cos θ + W ℓ ℓ cos θ F w = 2 W p d + W ℓ ℓ 2 ℓ cos θ sin θ = 2 (342 N)(4 . 4 m) + (337 N)(11 m) 2 (11 m) × cos 65 . 1654 ◦ sin65 . 1654 ◦ = 141 . 292 N . 002 (part 2 of 2) 10.0 points Calculate the normal force exerted by the floor on the ladder. Correct answer: 679 N. Explanation: Applying translational equilibrium, summationdisplay F y : N f −W p −W ℓ = 0 N f −W p −W ℓ = 0 . N f = W p + W ℓ = 342 N + 337 N = 679 N . sanchez (ds28677) – homework 27 – Turner – (58220) 2 003 (part 1 of 3) 10.0 points A 17 . 8 m , 481 N uniform ladder rests against a frictionless wall, making an angle of 68 . 3 ◦ with the horizontal. b 767 N 481 N 5 . 82 m 17 . 8 m 68 . 3 ◦ Note: Figure is not drawn to scale. Find the horizontal force exerted on the base of the ladder by Earth when a 767 N fire fighter is 5 . 82 m from the bottom. Correct answer: 195 . 505 N. Explanation: Let : L = 17 . 8 m , δ = 5 . 82 m , α = 68 . 3 ◦ , W ℓ = 481 N , and W p = 767 N . L sin α L cos α L 2 cos α δ cos α Pivot b N w f N g W p W ℓ Applying rotational equilibrium with the pivot at the point of contact with the ground, summationdisplay τ = W p δ cos α + W ℓ L 2 cos α −N w L sin α = 0 2 W p δ cos α + W ℓ L cos α = 2 N w L sin α N w = W p δ cos α L sin α + W ℓ cos α 2 sin α = W...
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This note was uploaded on 04/07/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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