This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 = 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 = 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 = s F n F wall = 0 F wall = F s = s F n = s F g . Applying the second (rotational) condition of equilibrium (with axis of rotation at the base of the ladder), F wall L sin  F g parenleftbigg L 2 parenrightbigg cos = 0 2 F wall L sin = F g L cos 2 s F g L sin = F g L cos s = 1 2 parenleftbigg cos sin 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 = 512 . 0 N , = 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 W parenleftbigg L 2 parenrightbigg (cos )W f d f cos = 0 2 F wall L sin = W L cos + 2 W f d f cos Applying the first (translational) condition of equilibrium horizontally, F x = F x,E F wall = 0 F x,E = F wall...
View Full
Document
 Spring '10
 PRO.KIROSHIMA
 Static Equilibrium, Work

Click to edit the document details