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Tension in a cable

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data

    An 88kg construction worker sits down 2.0 m from the end of a 1430 kg steel beam to eat his lunch. The cable supporting the beam is rated at 15,000 N. Should the worker be worried? To answer this, determine the tension in the cable.

    2. Relevant equations



    3. The attempt at a solution

    First, I have no idea how to start this problem. Second, I don't know what they mean by "the beam is rated."
     
  2. jcsd
  3. Oct 19, 2007 #2

    learningphysics

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    do they give a picture?
     
  4. Oct 19, 2007 #3
  5. Oct 19, 2007 #4

    learningphysics

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    Take the torque about the point where the beam is attached to the wall... take clockwise positive just like the other problem...

    so you have these forces causing torque: the weight of the beam... the weight of the man... and the tension...

    divide the tension up into the horizontal and vertical component.... think about the torque due to each component separately...

    the net torque about the point is 0...

    you should be able to solve for the tension...
     
  6. Oct 19, 2007 #5
    I tried solving this like the other one. The torque of the beam alone is 42042. Then the T with the man is T = (88)(9.8)(6) + 42042. For the tension, does the tension in the x-direction play a part?
     
  7. Oct 19, 2007 #6

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    no tension in the x-direction doesn't exert a torque about that point... one way to see this is that torque = r X F... when two vectors are parallel, their cross product is 0. r is in the x-direction...Tx is in the x-direction... so torque due to Tx is 0.

    so you just need to consider Ty.
     
  8. Oct 19, 2007 #7
    Ty = 7500. Was the first part right?
     
  9. Oct 19, 2007 #8

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    No, that's not what I'm getting...

    Also here:

    T = (88)(9.8)(6) + 42042

    it should be 88*9.8*4 since the man is 2m away from the end...

    how did you get the 7500?
     
  10. Oct 19, 2007 #9
    Oh. I did Ty = Tsin(theta) = 15000sin(30).
     
  11. Oct 19, 2007 #10

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    oh! yeah, don't use the 15000N... that's just a safety rating... means that beyond that the cable might snap...

    Use the sum of the torques about the pin = 0

    Tsin(theta) would appear in the above equation... letting you solve for T.
     
  12. Oct 19, 2007 #11
    So T = 45491.6/sin theta?
     
  13. Oct 19, 2007 #12

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    You forgot to multiply Tsin(theta) times the distance to give torque due to Tsin(theta)...
     
  14. Oct 19, 2007 #13
    T = 45491.6/(6m x sin30)?
     
  15. Oct 19, 2007 #14

    learningphysics

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    exactly.
     
  16. Oct 19, 2007 #15
    Got it. Finally. T = 15163.867. Thanks!!! :)
     
  17. Oct 19, 2007 #16

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    no prob.
     
  18. Oct 19, 2007 #17
    My english is ok but I have hard time comprehending some questions.

    How did you know that it meant he was sitting 2m from the right end, and not the left end?

    The picture doesn't help me determine this either, looks like he was sitting in the middle there. thanks.

    If it was from the left end i got 14.606kN, therefore the man should not be worried.
     
  19. Oct 19, 2007 #18

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    Yeah, I assumed it was from the right end... I think I looked at the diagram and thought he was leaning a little to the right so the center of mass would be slightly to the right...

    But you're right it could be either way from the wording of the question.
     
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