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Rope Tension question

  1. Oct 1, 2007 #1

    klm

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    a 500 kg piano is being lowered into position by a crane while 2 people steady it with ropes pulling to the sides. bob's rope pulls to the left, 15 degrees below horizontal, with 500 N of tension. ellen's rope pulls toward the right, 25 degrees below horizontal. what tension must ellen maintain in her rope to keep the piano descending at a steady speed?

    ok so what i understood from this, is that bobs rope is in the 3rd quad and ellens is in the 4th. and i guess there is a main cable from the crane going straight up. and then since is says it is descending at a steady speed that means that there is no net force or acceleration. so i figured i should get the components of bob's rope, Bx=-482 By=-129.
    and the i thought that
    Fnetx = Bx+Ex =0
    -482 + Ecos25 =0
    Ecos25=482
    E=532
    Fnety = By+Ey=o
    -129+Esin25=0
    E=306
    and then i found the magnitude of the two components of E and 614 N. but that is incorrect. so i dont know what to do anymore!
     
  2. jcsd
  3. Oct 1, 2007 #2

    Doc Al

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    Staff: Mentor

    It's the total force on the piano that must be zero, including the force of the crane (not just Bob and Ellen's force). But you know (or can assume) that the crane is just pulling vertically. So the only horizontal forces are exerted by Bob & Ellen, so the horizontal components of their forces must add to zero. And that's all you need to find Ellen's force. (You've already done the calculation.)
     
  4. Oct 1, 2007 #3

    klm

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    so is ellen's force 532 N? b/c i am not understand why you do not need bob and ellens vertical component force?
     
  5. Oct 1, 2007 #4

    Doc Al

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    Staff: Mentor

    That's correct. Since they are just steadying the crane, their total horizontal force must be zero. That's all you need to solve the problem. Their vertical force isn't zero, because the crane add its force to the piano. (The crane pulls vertically.)
     
  6. Oct 1, 2007 #5

    klm

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    thank you doc al
     
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