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Suspended Sign

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Attached is a diagram of the problem.

    A 28kg sign is suspended by two massless cables. Find the tension in each.

    2. Relevant equations



    3. The attempt at a solution

    I'll call the cable at 57 degrees T1 and the other cable T2.

    T1∙cos 57 + T2∙cos 36 = 9.8m/s2∙28kg
    T1∙cos 57 + T2∙cos 36 = 274.4N

    T1∙sin 57 + T2∙sin 36 = 0N

    This is where I'm getting confused - rusty on my algebra with simultaneous equations. Am I on track so far though, then just solve for T1 and T2?
     

    Attached Files:

  2. jcsd
  3. Sep 29, 2009 #2
    T1 = (-T2∙sin 36)/sin 57

    Substitute that into the first equation...

    (-T2∙sin 36∙cos 57)/sin 57 + T2∙cos 36 = 274.4N
    T2 = 143.05N

    Then solve for T1.

    T1∙cos 57 + (143.05N)∙cos 36 = 274.4N
    T1 = 291.33N
     
  4. Sep 29, 2009 #3

    rl.bhat

    User Avatar
    Homework Helper

    From third equation you can write
    T1 = - T2*sin36/sin57.
    Substitute this value in eq. 2 and solve for T2.
     
  5. Sep 29, 2009 #4
    Thanks - that's exactly what I ended up doing. Not sure why I was over-complicating it. Could anyone confirm my solutions are correct?

    Thanks!
     
  6. Sep 29, 2009 #5
    Can anyone verify that my work is correct? Thanks!
     
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