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Calculate the tension in the string

  1. Sep 28, 2007 #1

    ssj

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    1. Calculate the tension in the string T in each of the following situations.
    In all calculations ignore the mass of the beam.


    [​IMG]


    2. Moment=F X Perpendicular Distance and Basic Trigonometry



    3.I have tried 200N X 0.5m=100 100/Cos30

    Help is much apperchated.
     
  2. jcsd
  3. Sep 28, 2007 #2

    Doc Al

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    Staff: Mentor

    I understand the left side of this equation: the clockwise torque due to the 200 N force. The right side should be the counter-clockwise torque due to the tension of the cable.

    Correct this equation, rewriting the right hand side in terms of T (tension). Then solve for T.
     
  4. Sep 28, 2007 #3

    ssj

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    I think i have found the solution by calculating the "opp" tan30 X 0.5 = 0.2886 from there onward we do 200 X 0.5= 100/0.2886 =346.5/cos 30 = 400 N

    However now I am stuck on an similar equation of [​IMG]
     
    Last edited: Sep 28, 2007
  5. Sep 28, 2007 #4

    Doc Al

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    Don't really understand what you did there with tangents and cosines. All you need is:
    0.5*200 = 0.5*T*sin(30)
    which gives you T = 400.

    Solve it in the same way. The only difference is the moment arm associated with the 200 N force.

    I strongly suggest you adopt a systematic approach, like the one I used.
     
  6. Sep 28, 2007 #5

    ssj

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    But why would you use sine30 wouldnt that work out the "opp"?. Due to the fact the T is on the hypoteanus.
     
  7. Sep 28, 2007 #6

    Doc Al

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    You want the component of T perpendicular to the beam, which in this case means [itex]T\sin {30}[/itex].
     
  8. Sep 28, 2007 #7

    ssj

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    But why would T need to become perpendicular ? Also for question 2 how am I ment to work it out with out the given length ?
     
    Last edited: Sep 28, 2007
  9. Sep 28, 2007 #8

    Doc Al

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    Because you are finding the torque (moment) it generates about the pivot point. Do you understand the definition of moment?
     
  10. Sep 28, 2007 #9

    ssj

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    The moment is the turning force right ?
     
  11. Sep 28, 2007 #10

    Doc Al

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    You could call it that, yes.
     
    Last edited: Sep 28, 2007
  12. Sep 28, 2007 #11

    ssj

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    Im stuck on the second question I posted because they havent stated the length what do I do ?
     
  13. Sep 28, 2007 #12

    Doc Al

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    You are given all the lengths you need. (The force is at the midpoint of the beam.)
     
  14. Sep 28, 2007 #13

    ssj

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    So the length is the mid-point between 0 and 0.5 ? meaning its 0.25
     
  15. Sep 28, 2007 #14

    Doc Al

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    That's correct. But realize that you don't need the length of the beam: You can just use L or L/2. (The actual length doesn't matter.)
     
  16. Sep 28, 2007 #15

    ssj

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    But doesnt the formular state Motion=Force * perpendicular distance ? This would mean the formula is 200*L = L*T*sin30 correct ?
     
    Last edited: Sep 28, 2007
  17. Sep 28, 2007 #16

    Doc Al

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    Sure. Notice that the length cancels from both sides of that equation.
     
  18. Sep 28, 2007 #17

    ssj

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    So your left with 200/sin30= T meaning 400= T ?
     
  19. Sep 28, 2007 #18

    Doc Al

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    Right, as I stated in post #4. (This is for your first problem.)
     
  20. Sep 28, 2007 #19

    ssj

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    This is for the second problem though, wouldnt the Motion be different due to the force being applyed to the mid-point? I seem to of gotten T=400 in my previous post.
     
  21. Sep 28, 2007 #20

    Doc Al

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    These calculations apply to the first problem.

    For the second problem, the method is exactly the same: Set clockwise and counter-clockwise torques equal. Of course the torque exerted by the 200 N force is different in the second problem, since it's applied at the midpoint instead of at the end.
     
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