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Help finding Tension on a hanging bag!

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data
    FBD: http://i560.photobucket.com/albums/ss45/ZimmyGFX/photo.jpg
    A bag of cement of weight Fg hangs from three fires as shown in the figure. Two of the wires make angles theta1 and theta2 with the horizontal. If the system is in equilibrium, show that the tension in the left hand wire is T1 = Fgcos(theta2)/sin(theta1 + theta2)


    2. Relevant equations
    T1cos(theta1) - T2cos(theta2) = 0
    T1sin(theta1) + T2sin(theta2) = Fg (aka T3)


    3. The attempt at a solution
    Using the equation T1sin(theta1) + T2sin(theta2) = Fg I was able to single out T2 to plug into the other equation (since I'm solving for T1).
    T2 = (Fg - T1sin(theta1))/sin(theta2) is what I got
    Then I substituted T2 for the equation found into T1cos(theta1) = T2cos(theta2) because the system is at equilibrium in the x direction.
    So I get T1cos(theta1) = ((Fg - T1sin(theta1))/sin(theta2)) * cos(theta2)
    Simplifies to T1cos(theta1) = (Fgcos(theta2) - T1sin(theta1)cos(theta2))/sin(theta2)
    I then divide off cos(theta1) and get the following
    T1 = (Fgcos(theta2) - T1sin(theta1)cos(theta2))/(sin(theta2)cos(theta1))

    This isn't what I'm supposed to get though. I still have a T1 on both sides of the equation but if I divide it over then it will cancel off.
    I'm supposed to be getting Fgcos(theta2)/sin(theta1 + theta2)
    and sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
    I have all the elements for the denominator on the right side, but two are on top and should be on the bottom.
    I obviously messed up big time somewhere, but I'm stuck and don't know where to go from here.
     
  2. jcsd
  3. Sep 25, 2012 #2

    Ibix

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    Science Advisor

    You're doing fine. On the left hand side, you have one term that is a multiple of T1. On the right hand side, you have a term that is not a multiple of T1 minus a term that is a multiple of T1. In other words, you've got something that looks like 5x=2-3x. How would you solve that?
     
  4. Sep 25, 2012 #3
    Yeah I took a look back right before you posted that and I realized I wasn't done simplifying. I didn't take a deep enough look at it and at first glanced I figured that I would end up with a 2T1 or something, not that it would factor out. I guess I should have worked it out instead of being lazy next time. Thanks anyway though
     
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