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Finding tension

  1. Nov 7, 2009 #1
    I am unsure of how to go about this problem, because whatever I tried wasn't right. So if anyone could help me by going step by step through it, showing all equations and numbers used, and then showing me the answers so I can make sure I get the same thing, I would appreciate it.


    A uniform beam of length x = 1.0 m and mass 10 kg is attached to a wall by a cable at angle Θ = 30° to the horizontal, as shown in the figure. The beam is free to pivot at the point where it attaches to the wall. What is the tension in the cable?

    http://session.masteringphysics.com/problemAsset/1013774/7/jfk.Figure.P08.08.jpg
     
  2. jcsd
  3. Nov 7, 2009 #2

    rock.freak667

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    Well no one at PF will directly give you the answers, we will help you however, arrive it. After you show some effort.

    If that beam is in equilibrium in the diagram, what does that mean about the forces on the beam?
     
  4. Nov 7, 2009 #3
    It means they will equal zero.
     
  5. Nov 7, 2009 #4

    rock.freak667

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    So if the sum of the forces in the y direction is zero and the sum of forces in the x-direction is zero, and the sum of the moments about any point is zero.

    Can you use one of those conditions to find the tension?
     
  6. Nov 7, 2009 #5
    Well, what I did when I first attempted the problem was I used the equation:

    "Net Torgue = Tsin(30) - mg" which gave me 196 N as my tension, but this was wrong. Now I don't know where I went wrong.
     
  7. Nov 7, 2009 #6

    rock.freak667

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    If you are taking moments about the end, then the net torque is zero.

    so

    0= (Tsin30)*1-(A)mg

    what the value of A (the distance from the force to the pivot point)?
     
  8. Nov 7, 2009 #7
    1 m? Cause the board is 1 m long.
     
  9. Nov 7, 2009 #8

    rock.freak667

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    The board is uniform, where is the weight acting?
     
  10. Nov 7, 2009 #9
    The weight of the board is acting on the center of the board.
     
  11. Nov 7, 2009 #10

    rock.freak667

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    and the center is how far away from the end?
     
  12. Nov 7, 2009 #11
  13. Nov 7, 2009 #12

    rock.freak667

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    so in 0= (Tsin30)*1-(A)mg, what is A?
     
  14. Nov 7, 2009 #13
    0.5 m? So just plug that into the equation?
     
  15. Nov 7, 2009 #14

    rock.freak667

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    which gives T as?
     
  16. Nov 7, 2009 #15
    I got 98 N.
     
  17. Nov 7, 2009 #16

    rock.freak667

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    Is that the correct answer?
     
  18. Nov 7, 2009 #17
    Yes. Thanks a lot dude.
     
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