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Finding The Tension force with different angles

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data
    I have done all the physics for this problem which I'll detail below. I am only having trouble in doing the simple math to single out the t1 or t2.

    A mass is hanging from a ceiling. It is supported by two strings which both are attached above to the ceiling. String 1 is at an angle of 30. String 2 is at an angle of 65. While the
    mass is 20 lb. ( in the center)





    2. Relevant equationsMa = 0 in both the x and y directions.



    3. The attempt at a solution
    left T (T1) =30 degrees

    right T (T2) = 65 degrees.

    ok, here are my equations that I get to:

    X: T2cos(65) - T1cos(30) = 0

    Y: T2sin(65) + T1sin(30) = 0

    now I just need to solve for, say, T1 in the x dierection and use that new eq. to
    pluge into Y. Right?
     
  2. jcsd
  3. Feb 28, 2009 #2

    PhanthomJay

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    The 20 pounds is the value of the weight of the hanging mass; it is not the mass. In your last equation in the y direction, you are missing a force.
     
  4. Feb 28, 2009 #3
    right. I am missing -w. Sorry about the mistake.

    Now what of the T that I need to solve for. I have Solved for the T in my
    x and y equation and get bad nasty answers.
     
  5. Feb 28, 2009 #4

    PhanthomJay

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    First plug in the value for cos 65, sin 65, etc. Then solve the 2 equations with the 2 unknowns as you see fit.
     
  6. Feb 28, 2009 #5
    Do you mean this:

    I'll take the x eq. and solve for t1.

    I get T1 = T2cos(65) / cos (30)

    now I can take this eq. for T1 and plug it in for T1 in the y eq?
     
  7. Feb 28, 2009 #6

    PhanthomJay

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    It will be a lot easier if you rewrite your first equation as T1= T2(.423)/.866 = 0.49T2. Rewrite the 2nd equation in the same manner, then do the substitution.
     
  8. Feb 28, 2009 #7
    I see.

    T1 = T2(.49)

    T2 = 21.74lb + T1(.54) I divided .92 into lbs., think you can do that.

    now ill put T2 in the T1 eq. Right?
     
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