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Unformatted text preview: fullwood (tf5349) – Homework Static Equilibrium – Catala – (49033) 1 This printout should have 21 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 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 = ℓ 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 . ◦ 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 ◦ sin49 ◦ parenrightbigg = . 434643 . 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 ℓ = 512 . 0 N , θ = 52 . ◦ , 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 θW ℓ parenleftbigg L 2 parenrightbigg (cos θ )W f d f cos θ = 0 2 F wall L sin θ = W ℓ 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...
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This note was uploaded on 06/10/2010 for the course PHY 100 taught by Professor Pro.kiroshima during the Spring '10 term at Adelphi.
 Spring '10
 PRO.KIROSHIMA
 Static Equilibrium, Work

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