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Tension- Hanging Sign

  1. Dec 18, 2007 #1
    [SOLVED] Tension- Hanging Sign

    A 20.0 kg sign is being held up by 2 ropes. Each rope makes a 60 degree angle with the sign. So there is an upright triangle, each angle 60 degrees. I have to calculate the tenstion in the two ropes.



    So far, I have
    [tex]\Sigma[/tex] F[tex]_{}y[/tex]=ma[tex]_{}y[/tex]
    +Fa[tex]_{}1[/tex]sin60 + Fa[tex]_{}2[/tex]sin60 = 20 kg (9.81 m/s[tex]^{}2[/tex])
    I divided by sin60
    Fa[tex]_{}1[/tex] + Fa[tex]_{}2[/tex] = 226.55 N
    Each rope has 112.78 N


    I'm not sure if i have to do the sum of the forces in the x direction or if the work i already did is correct.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 18, 2007 #2

    stewartcs

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    Draw a FBD then break it down into components and use Newton's second law...

    x: -T1cos(theta) + T2cos(theta) = 0

    y: T1sin(theta) + T2sin(theta) - Mg = 0

    Now solve the equations simultaneously.
     
  4. Dec 18, 2007 #3
    when you say solve them simultaneously do you mean set them equal to each other since they both = 0?
     
  5. Dec 18, 2007 #4

    stewartcs

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    The easiest way is to substitute.

    Hint: Solve the x component equation for T1 and substitute into the y component equation.
     
  6. Dec 18, 2007 #5
    OK! i did this and got each rope had 113.28 N of tension.
     
  7. Dec 18, 2007 #6

    stewartcs

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    Looks about right.

    You should have come up with T1 = T2 = (Mg)/(2sin(theta)) = (20*9.8)/(2*sin(60)) = 113.16 N.
     
  8. Dec 18, 2007 #7
    yep! thanks so much!
     
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