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Tension in Cables of Differing Angles

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data
    α=30°
    β=45°
    m=10kg
    a=0
    Tα=unknown
    Tβ=unknown


    THE ANSWERS SHOULD BE Ta=72N, Tβ=88N

    2. Relevant equations
    mg=10kg*9.8m/s2=98N
    ∑Fx=ma(x)=0
    ∑Fy=ma(y)=0

    A Free Body Diagram would show three forces: 98N facing downwards, Ta facing 30° up and to the left, and Tb facing facing 45° up and to the right.


    3. The attempt at a solution

    I tried to isolate the horizontal and vertical components of the forces with
    Tαcos30°+Tβcos45°=0→Tαcos30°=-Tβcos45°
    Tαsin30°+Tβsin45°-98N=0→Tαsin30°+Tβsin45°=98N

    I then tried to solve for T with
    Tα=-Tβcos45°/cos30°

    And substitute that into the vertical component equation
    (-Tβcos45°/cos30°)sin30°+Tβsin45°=98N

    But I seem to be walking in circles when I attempt to go any further.
     
  2. jcsd
  3. Mar 1, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    The x components of the tension forces have opposite signs. Write

    Tαcos30°-Tβcos45°=0

    Correct.

    That is two equation with two unknown, just go ahead. Why do you think that you move in circles? :smile:

    ehild
     
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