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Unformatted text preview: Alvarado, Patrick – Homework 8 – Due: Nov 2 2006, 10:00 am – Inst: Andrei Sirenko 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider a uniform ladder leaning against a smooth wall and resting on a smooth floor at point P . There is a rope stretched horizon tally, with one end tied to the bottom of the ladder essentially at P and the other end to the wall. The top of the ladder is at a height is h up the wall and the base of the ladder is at a distance b from the wall. The weight of the ladder is W 1 . Jill, with a weight W 2 , is onefourth the way µ d = ‘ 4 ¶ up the ladder. The force which the wall exerts on the ladder is F . P W 1 W 2 ‘ d T θ h b F Note: Figure is not to scale. The torque equation about P is given by 1. h 4 W 2 + h 2 W 1 = F b 2. ( W 1 + W 2 ) h 2 = F b 3. ( W 1 + W 2 ) b 2 = F h 4. h 2 W 2 + hW 1 = F b 5. b 4 W 2 + b 2 W 1 = F h correct 6. b 2 W 2 + bW 1 = F h Explanation: P ivot F T N f W 2 W 1 θ X F x : T F = 0 , (1) X F y : N f W 2 W 1 = 0 , and (2) X τ P : W 2 d cos θ + W 1 ‘ 2 cos θ (3) F ‘ sin θ = 0 , where d is the distance of the person from the bottom of the ladder. Therefore 2 F ‘ sin θ = 2 W 2 d cos θ + W 1 ‘ cos θ . Since sin θ = h ‘ and cos θ = b ‘ , the torque equation about P is given by b 4 W 2 + b 2 W 1 = F h . (4) 002 (part 2 of 3) 10 points Given: W 2 = 3 W 1 = W , h = b . Determine the force F the wall exerts on the ladder. 1. F = 1 4 W 2. F = 1 3 W 3. F = 3 4 W Alvarado, Patrick – Homework 8 – Due: Nov 2 2006, 10:00 am – Inst: Andrei Sirenko 2 4. F = 1 6 W 5. F = 5 12 W correct 6. F = 2 3 W 7. F = 1 12 W 8. F = 7 12 W 9. F = 1 2 W 10. F = 5 6 W Explanation: For W 2 = 3 W 1 = W and h = b , Eq. 4 gives b 4 W + b 2 W 3 = F b, so that F = W 4 + W 6 = (3 + 2) 12 W = 5 12 W . 003 (part 3 of 3) 10 points Given: W 2 = 3 W 1 = W , h = b. When Jill has climbed up the ladder such that the rope tension reaches T = W 2 deter mine Jill’s height y from the floor. 1. y = 3 4 b 2. y = 5 6 b 3. y = 2 3 b 4. y = 5 12 b 5. y = 1 4 b 6. y = 7 12 b 7. y = 1 2 b 8. y = 1 3 b correct 9. y = 1 6 b 10. y = 1 12 b Explanation: For rope tension T = W 2 , and by applying Newton’s 2nd law in the horizontal direction (giving T F = 0), we have F = T = W 2 . With the origin at P , denote Jill’s coordinates on the ladder by ( x,y ). Then the torque equation gives xW 2 + b 2 W 1 = F h = bW 2 . (5) For W 2 = 3 W 1 = W , Eq. 5 gives xW + b 2 W 3 = bW 2 . Finally, since h = b , Jill is at height y , where y = x = 1 2 b 1 6 b = 1 3 b ....
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 Spring '08
 OPYT
 Force, Work, Sin, Cos, Andrei Sirenko

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