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Homework Help: Forces and Tension in cable

  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

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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?
     
  22. Sep 23, 2007 #21

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    I don't understand... is T1 and T2 different components? I thought they were for each rope on either side of the mass...

    Both of those ropes have exactly the same tension (magnitude)... As written in the question, "two 3.0m long cables"... both of those cables have the exact same tension...
     
  23. Sep 23, 2007 #22
    Are the ropes straight up and down or something?
     
  24. Sep 23, 2007 #23
    No, cause it says outward.
     
  25. Sep 23, 2007 #24

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    Actually you don't have to assume they're equal tensions...

    [tex]\Sigma{F_x} = 0[/tex]

    T2cos(41.8) - T1cos(41.8) = 0

    T1 = T2

    so from the sum of forces in the x-direction, we get that the two tensions in the two ropes are the same.

    Then [tex]\Sigma{F_y} = 0[/tex]

    T1sin(41.8) + T2sin(41.8) - mg = 0

    sub in T1=T2

    2T1sin(41.8) = mg

    and you can solve for T1.
     
  26. Sep 23, 2007 #25
    I think I got it now. Since the acceleration in the x direction is zero, the tensions in the x direction are zero. Since you know they're equal in the x direction you can substitute them in for the y direction. I just don't think I knew that you could substitute.
     
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