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hw 28 - homework 28 HETTIGER CHRISTOF Due Apr 8 2008 4:00...

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homework 28 – HETTIGER, CHRISTOF – Due: Apr 8 2008, 4:00 am 1 Question 1, chap 14, sect 2. 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 one-fourth the way parenleftbigg d = 4 parenrightbigg 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. b 4 W 2 + b 2 W 1 = Fh correct 2. h 4 W 2 + h 2 W 1 = Fb 3. h 2 W 2 + hW 1 = Fb 4. ( W 1 + W 2 ) h 2 = Fb 5. b 2 W 2 + bW 1 = Fh 6. ( W 1 + W 2 ) b 2 = Fh Explanation: Pivot F T N f W 2 W 1 θ summationdisplay F x : T F = 0 , (1) summationdisplay F y : N f W 2 W 1 = 0 , and (2) summationdisplay τ P : W 2 d cos θ + W 1 2 cos θ (3) Fℓ sin θ = 0 , where d is the distance of the person from the bottom of the ladder. Therefore 2 Fℓ 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 = Fh . (4) Question 2, chap 14, sect 2. 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 = 3 4 W 2. F = 5 12 W correct 3. F = 5 6 W
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homework 28 – HETTIGER, CHRISTOF – Due: Apr 8 2008, 4:00 am 2 4. F = 2 3 W 5. F = 1 3 W 6. F = 1 12 W 7. F = 1 4 W 8. F = 1 2 W 9. F = 1 6 W 10. F = 7 12 W Explanation: For W 2 = 3 W 1 = W and h = b , Eq. 4 gives b 4 W + b 2 W 3 = Fb, so that F = W 4 + W 6 = (3 + 2) 12 W = 5 12 W . Question 3, chap 14, sect 2. 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 = 1 4 b 2. y = 5 6 b 3. y = 5 12 b 4. y = 2 3 b 5. y = 1 12 b 6. y = 1 3 b correct 7. y = 1 6 b 8. y = 3 4 b 9. y = 1 2 b 10. y = 7 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 = Fh = 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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