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h27 - hernandez(ejh742 homework 27 Turner(58120 1 This...

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hernandez (ejh742) – homework 27 – Turner – (58120) 1 This print-out should have 13 questions. Multiple-choice 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 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 N w F f N f W p W θ θ = arccos s = arccos parenleftbigg 8 . 82 m 14 m parenrightbigg = 50 . 9499 In equilibrium summationdisplay vector F = 0 and summationdisplay vector τ = 0 . summationdisplay τ : W p d cos θ + W 2 cos θ N w sin θ = 0 , where d is the distance of the person from the bottom of the ladder. Therefore 2 F w sin θ = 2 W p d cos θ + W cos θ F w = 2 W p d + W 2 cos θ sin θ = 2 (486 N)(5 . 4 m) + (297 N)(14 m) 2 (14 m) × cos 50 . 9499 sin 50 . 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 .
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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. 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 , δ = 5 . 25 m , α = 64 , W = 489 N , and W p = 825 N . L sin α L cos α L 2 cos α δ cos α Pivot N w f N g W p W Applying rotational equilibrium with the pivot at the point of contact with the ground, summationdisplay τ = W p δ cos α + W L 2 cos α −N w L sin α = 0 2 W p δ cos α + W L cos α = 2 N w L sin α N w = W p δ cos α L sin α + W cos α 2 sin α = W p δ L tan α + W 2 tan α = (825 N) (5 . 25 m) (18 . 3 m) tan 64 + 489 N 2 tan 64 = 234 . 687 N .
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