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Unformatted text preview: hiarker (srh959) – 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 A 13 m ladder whose weight is 253 N is placed against a smooth vertical wall. A person whose weight is 880 N stands on the ladder a distance 5 . 1 m up the ladder. The foot of the ladder rests on the floor 3 . 77 m from the wall. 880 N 253 N 5 . 1 m 13 m b 3 . 77 m Note: Figure is not to scale. Calculate the force exerted by the wall. Correct answer: 142 . 945 N. Explanation: Let : ℓ = 13 m , d = 5 . 1 m , s = 3 . 77 m , W ℓ = 253 N , and W p = 880 N . Pivot b N w F f N f W p W ℓ θ θ = arccos s ℓ = arccos parenleftbigg 3 . 77 m 13 m parenrightbigg = 73 . 142 ◦ 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 (880 N)(5 . 1 m) + (253 N)(13 m) 2 (13 m) × cos 73 . 142 ◦ sin73 . 142 ◦ = 142 . 945 N . 002 (part 2 of 2) 10.0 points Calculate the normal force exerted by the floor on the ladder. Correct answer: 1133 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 = 880 N + 253 N = 1133 N . hiarker (srh959) – homework 27 – Turner – (58220) 2 003 (part 1 of 3) 10.0 points An 11 . 5 m , 552 N uniform ladder rests against a frictionless wall, making an angle of 61 . 2 ◦ with the horizontal. b 884 N 552 N 4 . 63 m 11 . 5 m 61 . 2 ◦ Note: Figure is not drawn to scale. Find the horizontal force exerted on the base of the ladder by Earth when an 884 N fire fighter is 4 . 63 m from the bottom. Correct answer: 347 . 393 N. Explanation: Let : L = 11 . 5 m , δ = 4 . 63 m , α = 61 . 2 ◦ , W ℓ = 552 N , and W p = 884 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 p δ...
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 Spring '09
 NUI
 Physics, Force, Mass, Work, Sin, Trigraph, Correct Answer, Ry

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