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Simple tension problem

  1. Feb 2, 2005 #1
    Hi... I just stumbled upon this forum trying to find a reasonable answer to an introductory physics problem I am working on. Maybe someone here could help me. If you have a mass hanging from three wires, how do you compute the tension in each of the wires? It's a conceptual question, I only need to derive the formula for T1, which is actually given to me, but I still can't figure it out... any help would be greatly appreciated.
     
  2. jcsd
  3. Feb 2, 2005 #2
    You could evaluate vectorally in three dimensions.
     
  4. Feb 2, 2005 #3
    Why three dimensions? Wouldn't it just be in the x-y plane? So I just resolve each of the three vectors into their x and y components and then add them? I didn't mention also that the two wires are different lengths, at different angles from the ceiling.
     
  5. Feb 2, 2005 #4
    Ok, I assumed three dimensions, but it can be in the x-y plane. In that case you have the right approach. Trigonometry can be used, alternatively.
     
  6. Feb 2, 2005 #5
    Yeah, the equation they want me to derive gives
    T1 = {mg cos(angle 2)}/{sin(angle 1 + angle 2)}
    *sorry* I don't know how to type in equations like this properly

    T1 and T2 are the two wires supporting the wire holding the mass, T3. I get that the tension for T3 is mg, but the tension in T1 and T2 aren't equal if the mass isn't centered between the two wires, right? I'm sorry, this is probably so simple...
     
  7. Feb 2, 2005 #6
    No need to apologize.

    Hm, from what I understand of your description, you have a mass supported by a wire which splits into two separate wires of unequal lengths before reaching the ceiling, or whatever the mass is hanging from. You are on the right track, and you are correct in assuming unequal tension if the lengths of T1 and T2 are different. Draw a vector diagram showing the forces present and set up a force-vector triangle. Then use trigonometry to find an expression for T1.
     
  8. Feb 3, 2005 #7
    Thanks sirus... I appreciate all your help. Still haven't figured it out though. I think it's not even so much I don't understand physics, but more so I don't get trigonometry. Maybe I should go to a trig forum. Thanks though!!
     
  9. Feb 3, 2005 #8

    Doc Al

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    Start by describing the orientation of your three wires. (Draw a picture.) At any point (such as the junction of two wires) the net force is zero.
     
  10. Feb 3, 2005 #9
    OK, so the net force is zero when a mass is just hanging from 3 wires... two wires attached to a support and the third attached to those two with the mass hanging from it. So, does that mean the x-component of the net force is equal to the y-component of force? or that each are equal to zero?
     
  11. Feb 3, 2005 #10
    Here's the diagram... can anyone still help me?
     

    Attached Files:

  12. Feb 3, 2005 #11

    Doc Al

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    Both the x and y components of the net force must equal zero.

    In your diagram, T3 must balance the weight of the object (since the object is in equilibrium). So now you know T3!

    Now consider the forces at the junction of wires: Find the vertical components of each tension: they must add to zero. Same for the horizontal components. You'll have two (easy) equations and two unknowns (T1 and T2), so you can solve for T1 and T2.
     
  13. Feb 3, 2005 #12
    All right, I guess I've done that... but now I can't seem to derive the equation (from above). Can you tell me if this is right so far? (please?)

    for the y-component: T1sin(angle1) + T2sin(angle2) - mg = 0
    for the x-component: T2cos(angle2) - T1cos(angle1) = 0

    I don't know how to combine these to derive the formula they're asking for. Any hints? Thanks for helping me, by the way.
     
  14. Feb 3, 2005 #13

    Doc Al

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    So far, so good!

    First step: In the second equation, solve for T2 in terms of T1. Then substitute that into your first equation and solve for T1.

    To simplify the resulting expression, you'll need a trig identity. ([itex]sin(A + B) =[/itex] ? Look it up!)
     
  15. Feb 3, 2005 #14
    Ha! thanks so much... I think I got the right answer now. I knew it was a problem I was having with trig. Haven't taken it in years. Anyway, thanks a lot for all the help.
     
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