solution10

# solution10 - fullwood(tf5349 Homework Static Equilibrium...

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fullwood (tf5349) – Homework Static Equilibrium – Catala – (49033) 1 This print-out should have 21 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 523 N window washer is standing on a uniform scaffold supported by a vertical rope at each end. The scaffold weighs 158.4 N and is 4.54 m long. Assume the window washer stands 1.20 m from one end. What is the force on the farther rope? Correct answer: 217 . 438 N. Explanation: Let : F g,w = 523 . 0 N , F g,s = 158 . 4 N , d s = s 2 = 2 . 270 m , d w = 1 . 20 m , and d 2 = 4 . 54 m . Applying the second (rotational) condition of equilibrium (axis of rotation at T 1 ), F T, 2 d 2 - F g,w d w - F g,s d s = 0 F T, 2 = F g,w d w + F g,s d s d 2 = (523 N)(1 . 2 m) 4 . 54 m + (158 . 4 N)(2 . 27 m) 4 . 54 m = 217 . 438 N . 002 (part 2 of 2) 10.0 points What is the force on the closer rope? Correct answer: 463 . 962 N. Explanation: Applying the first (translational) condition of equilibrium, F T, 1 + F T, 2 - F g,w - F g,s = 0 F T, 1 = F g,w + F g,s - F T, 2 = 523 N + 158 . 4 N - 217 . 438 N = 463 . 962 N . 003 10.0 points A uniform 6.0 m tall aluminum ladder is lean- ing against a frictionless vertical wall. The ladder has a weight of 288 N. The ladder slips when it makes a 49.0 angle with the horizon- tal floor. Determine the coefficient of static friction between the ladder and the floor. Correct answer: 0 . 434643. Explanation: Let : F g = 288 N L = 6 . 0 m θ = 49 . 0 F n = F g , so applying the first (transla- tional) condition of equilibrium vertically, F y = F n - F g = 0 and horizontally, F x = F s - F wall = μ s F n - F wall = 0 F wall = F s = μ s F n = μ s F g . Applying the second (rotational) condition of equilibrium (with axis of rotation at the base of the ladder), F wall L sin θ - F g parenleftbigg L 2 parenrightbigg cos θ = 0 2 F wall L sin θ = F g L cos θ 2 μ s F g L sin θ = F g L cos θ μ s = 1 2 parenleftbigg cos θ sin θ parenrightbigg = 1 2 parenleftbigg cos 49 sin 49 parenrightbigg = 0 . 434643 .

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fullwood (tf5349) – Homework Static Equilibrium – Catala – (49033) 2 004 (part 1 of 3) 10.0 points A ladder with a length of 12.9 m and weight of 512.0 N rests against a frictionless wall, making an angle of 52.0 with the horizontal. Find the horizontal force exerted on the base of the ladder by Earth when a firefighter weighing 817.0 N is 2 . 45 m from the bottom of the ladder. Correct answer: 321 . 239 N. Explanation: Let : L = 12 . 9 m , W = 512 . 0 N , θ = 52 . 0 , W f = 817 . 0 N , and d f = 2 . 45 m . Applying the second (rotational) condition of equilibrium (with axis of rotation at the base of the ladder), F wall L sin θ - W parenleftbigg L 2 parenrightbigg (cos θ ) -W f d f cos θ = 0 2 F wall L sin θ = W L cos θ + 2 W f d f cos θ Applying the first (translational) condition of equilibrium horizontally, F x = F x,E - F wall = 0 F x,E = F wall F x,E = ( W L + 2 W f d f ) cos θ 2 L sin θ = W cos θ 2 sin θ + W f d f cos θ L sin θ = (512 N) cos 52 2 sin 52 + (817 N)(2 . 45 m) cos 52 (12 . 9 m) sin 52 = 321 . 239 N .
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