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Tension angle

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Weight suspended by 3 cords. 1 from the weight goes up vertically to the knot. The next goes left 30° below horizontal and the last goes to the right 45° above horizontal.

    2. Relevant equations
    f=ma


    3. The attempt at a solution
    I dont think I've ever worked a problem with a cord below horizontal. So I'm stumbling a bit.

    Since there's only 1 cord above horizontal (T1) it should carry the whole y force. So, in y T1=T2+T3
    And, in x the left and right cords must be equal T1=T2


    T1sin(45)=T2sin(30)+mg

    Not sure about the rest. Any help?
     
  2. jcsd
  3. Apr 22, 2012 #2

    OldEngr63

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    Gold Member

    The problem statement is somewhat confusing. A figure would really help.
     
  4. Apr 22, 2012 #3
    Quick picture I just made.
     

    Attached Files:

  5. Apr 22, 2012 #4
    As the object not going up or down
    it's too not going left or right, equal forces must be there to make it in equilibrium.
     
  6. Apr 23, 2012 #5

    OldEngr63

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    Gold Member

    You simply need to sum forces, both horizontal and vertical, at the junction point, and require that both sum to zero. That will give you everything you need.
     
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