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Tension In Cables

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data
    In the figure below, the left-hand cable has a Tension T1 and makes an angle of 58° with the horizontal. The right-hand cable has a Tension T3 and makes an angle θ3 with the horizontal. A 97.6N weight is on the left and an 84N weight is on the right. The cable connecting the two weights is horizontal.

    a) Find the Tension T1.
    b) Find the Tension T2
    c) Find the angle θ3.

    Image If It Helps: http://i.imgur.com/TTlwqhB.jpg?1
    2. Relevant equations
    ƩF = 0
    T1*sin(58°) - 97.6 = 0
    -T1*cos(58°) + T3*cos(θ3) = 0
    T3*sin(θ3) - 84 = 0

    3. The attempt at a solution

    a) I used the first equation: moved the -97.6 to the right then divided by sin(58°) to get 115.088.

    c) I used the third equation: moved 84 to the right and divided by sin(θ3) to get T3 on its own. Then, I substituted this into equation 2, and ended up getting: -T1*cos(58°) + 84/(sin(θ3))*cos(θ3) = 0. Then, I moved the -T1*cos(58°) to the right, plugged in 115.088 for T1, then simplified it down to cot(θ3) = .7260. I did arccot(.7260) and got 54.0188° for θ3.

    b) This is where I'm stuck. I never considered T2 in any of my equations, and I can't figure out if I'm forgetting some essential part of Tension. Any help to guide me would be very appreciated!
  2. jcsd
  3. Sep 29, 2013 #2


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    You did well. For c), you essentially took a free body diagram of both nodes that support the loads. When doing it this way, T2 force is internal and is thus is not included in your equation. To find T2, take a free body diagram of one of the nodes to solve. It's worth taking a free body diagram of the other node also as a check on your work.
  4. Sep 30, 2013 #3
    Ok, so I would end up just taking the x-component of T1 or T3 because the Tension T2 will essentially be the same. So T2 would be 60.9873?
  5. Sep 30, 2013 #4


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    Yes, call it 61 N.
  6. Sep 30, 2013 #5
    Alright, I get it. Thank you!
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