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Forces and Tension

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A 14 kg loudspeaker is suspended 2.0m below the ceiling by two 3.0m long cables that angle outward at equal angles. What is the tension in the cables?


    2. Relevant equations

    F=ma

    3. The attempt at a solution

    I drew the free body diagram, and I broke the tension of the two wires into their components. I'm unsure about how to tackle the problem.
     
  2. jcsd
  3. Sep 22, 2007 #2

    learningphysics

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    Where are you getting stuck? What are the forces acting on the suspended mass vertically?
     
  4. Sep 22, 2007 #3
    The forces acting are the tension of the two cables and the weight of the object. I just don't know how to find the angle so I can break up the tensions into components.
     
  5. Sep 22, 2007 #4

    learningphysics

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    Find the angle... you know the y displacement is 2.0m. You know the length of the cable is 3.0m. You can find the angle.
     
  6. Sep 23, 2007 #5
    Would the angle be 48.19 degrees? And if that's the case, where do I go from there?
     
  7. Sep 23, 2007 #6

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    I'm getting 41.81 degrees. Well... if T is the magnitude of the tension... what's the vertical component of tension? What are the forces in the vertical direction?
     
  8. Sep 23, 2007 #7
    The forces in the vertical direction are T1x and T2x
     
  9. Sep 23, 2007 #8

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    Oops... sorry, this angle is fine... I just got the other angle of the right triangle... 90-48.19 = 41.81
     
  10. Sep 23, 2007 #9

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    The tensions are equal... So what is Tx in terms of T?
     
  11. Sep 23, 2007 #10
    Tx = Tcos(theta)
     
  12. Sep 23, 2007 #11

    learningphysics

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    So what is the F=ma equation in the vertical direction?
     
  13. Sep 23, 2007 #12
    Is it Tcos(theta) = m(ax)?
     
  14. Sep 23, 2007 #13

    learningphysics

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    accleretation is 0... you have 2 tensions... and also gravity.
     
  15. Sep 23, 2007 #14
    I put horizontal anyway. In the vertical direction it would be Fnety = T1y + T2y - mg = m(ay).
     
  16. Sep 23, 2007 #15

    learningphysics

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    Can you put in the angles, sin/cos etc... ay = 0... and T1=T2... so just use T.
     
  17. Sep 23, 2007 #16
    I don't understand.
     
  18. Sep 23, 2007 #17

    learningphysics

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    Did you find T?
     
  19. Sep 23, 2007 #18
    No I didn't. I don't understand how you can find T.
     
  20. Sep 23, 2007 #19

    learningphysics

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    Why are you using two different variables T1 and T2?
     
  21. Sep 23, 2007 #20
    Am I not breaking the tension into components?
     
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