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Question about tension?

  1. Jul 1, 2010 #1
    1. The problem statement, all variables and given/known data

    A 20 kg loudspeaker is suspended 2.0 m below the ceiling by two cables that are each 30 degrees from the vertical. What is the tension in the cables

    2. Relevant equations


    3. The attempt at a solution

    Here is how I solved it but I am not sure if it is right. Also, I don't understand how to use the 2.0 meters given in the problems statements. Anyways, here goes:

    Fnetx=Wx + T1x + T2x =max=0
    0 + T1 - T2=0
    T1sinθ -T2sinθ=0

    Fnety=wy + T1y + T2y=may=0
    -w + T1 + T2=0
    (20 * -9.8) + T1cosθ + T2cosθ=0
    -196 + (T1 + T2)(cosθ)=0
    (T1 + T2)(cosθ)=196
    (T1 + T2)=196/cos(30)
    (T1 + T2)=226.3

    Since T1=T2, then 226.3/2=113 N. Therefore 113 N is the tension for both ropes?

    Is this correct?
  2. jcsd
  3. Jul 1, 2010 #2
    somebody please help!
  4. Jul 1, 2010 #3


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    Yes! The 2 meters is not relevant to the solution of this problem.
  5. Jul 1, 2010 #4
    So the answer is 113 N for both cables? And my work is correct?
  6. Jul 1, 2010 #5


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    Yes, the tension is 113 N in each cable. In your work, though, you noted
    Fnetx=Wx + T1x + T2x =max=0, correct, but then you said
    0 + T1 - T2=0, instead of saying 0 +T1sintheta -T2 sin theta = 0, which you later corrected. You did the same thing in the y direction, looks like just a typo.
  7. Jul 1, 2010 #6
    Thank you. I'll make sure not to make that mistake again.
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