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Work done by lowering a piano

  1. Apr 16, 2010 #1
    The two ropes seen in Figure Ex11.9 are used to lower a 235 kg piano 5.5 m from a second-story window to the ground. How much work is done by each of the three forces? (T1 = 1820 N andT2 = 1110 N)

    t1 is 45degrees north of east
    t2 is 60degrees north of west

    w = J
    T1 = J
    T2 = J


    w=fd

    I got work done by weight to be 12666.5j which is correct, how do i find the tensions?
     
  2. jcsd
  3. Apr 16, 2010 #2

    kuruman

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    To get help, you need to show Figure Ex11.9. You have it, but we don't.
     
  4. Apr 16, 2010 #3
  5. Apr 16, 2010 #4

    kuruman

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    Assuming that the piano is lowered at constant speed (as is usually the case with pianos being moved), what is the net force acting on the piano?
     
  6. Apr 16, 2010 #5
    0? but its looking for work
     
  7. Apr 16, 2010 #6

    kuruman

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    Zero is correct. Although the net force does zero work and the net work done on the piano is zero, the individual forces acting on the piano do do work. The sum of all the works done by all the forces (i.e. the net work) must be zero.

    First things first. How did you calculate the works done by T1 and T2? Please show what equation you used and what numbers you put in.
     
  8. Apr 16, 2010 #7
    Those are the ones i need to figure out, i figured out the force in the y direction with trig functions then used w=fd

    I got t1 = 4316.886899
    t2 = 8668.914292

    but they were wrong
     
  9. Apr 16, 2010 #8

    kuruman

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    I asked you to show what equation you used and what numbers you put in. You say W = Fd, OK. What numbers did you put in W = fd to get t1 = 4316.886899 and t2 = 8668.914292 ?

    By the way, lose the extra significant figures. You don't need more than three.
     
  10. Apr 16, 2010 #9
    to get t1 I used w= (1100*sin45)*5.5
    to get t2 I used w= (1820*sin60)*5.5

    *If I dont keep the sigfigs the webquestions get counted wrong
     
  11. Apr 16, 2010 #10

    kuruman

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    You mean "to get w1 and w2" because t1 and t2 are given. Minor mistake. The most serious mistake is in you application of the work equation.

    The work done by a constant force is W = f d cosθ, where θ is the angle between the displacement vector and the force. Since the displacement vector is down, and t1 is 60o above the horizontal, what is the angle between the displacement and the t1? How about t2?
     
  12. Apr 16, 2010 #11
    I have no idea, what that means t1s angle 150? t2s 135?
     
  13. Apr 16, 2010 #12

    kuruman

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    Correct. So what are the cosines of 150o and 135o? You can put it together now.
     
  14. Apr 16, 2010 #13
    cos(135) = -sqrt(2)/2 and cos(150) = -sqrt(3)/2

    Where do I plug them into? what would be the equation to find T1?
     
  15. Apr 16, 2010 #14

    kuruman

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    Are you looking for T1? Look at your initial posting. It says T1 = 1820 N and T2 = 1120 N. You are looking for the work done by T1 and T2. As I said before, the work done by each force is F d cosθ. Just put it together.
     
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