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Tension / Equilibrium Question

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A load on which the force of gravity is 46 N hands from the centre of a frictionless clothesline, pulling its centre down so that each half of the line makes an angle of 25° with the horizontal. What is the force of tension in the line?

    2. Relevant equations
    [tex]\Sigma[/tex]Y = FT - FG
    = 2FTsin25° - FG

    3. The attempt at a solution
    I found that FT is equal to 46 N / 2sin25° [tex]\approx[/tex] 54 N. However, I don't understand if this is the force of tension in each half of the line, or through the whole line. Can someone help me understand this?


    Last edited: May 23, 2010
  2. jcsd
  3. May 23, 2010 #2
    I think you've calculated the tension in one of the lines.
    You can calculate both of them by removing the 2 from the denominator.

    I approached it like this: If the weight is 46N, then each side of the line is contributing 23N to supporting it. That means that the vertical component of the tension in one side of the line is 23N, so 23/sin(25) = F_T in one side. This is clearly the same as your expression.
  4. May 23, 2010 #3


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    The tension on both sides of the hanging weight are the same. But you can't add them to get the line tension. The line tension is an internal force. At any point on the clothesline, the line tension is___________??
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