# hw28_s - homework 28 YOO HEE Due Apr 8 2008 4:00 am...

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homework 28 – YOO, HEE – Due: Apr 8 2008, 4:00 am 2 4. F = 1 2 W 5. F = 1 6 W 6. F = 5 6 W 7. F = 2 3 W 8. F = 3 4 W 9. F = 1 3 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 12 b 3. y = 2 3 b 4. y = 3 4 b 5. y = 1 6 b 6. y = 5 6 b 7. y = 7 12 b 8. y = 1 3 b correct 9. y = 1 2 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 = 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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