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Rope hung between two trees

  1. Jun 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A uniform rope of weight W hangs between two trees. The ends of the rope are the same height, and each make angle [tex] \theta [/tex] with the trees (the angle is between the rope and the lower part of the tree). Find the tension at either end of the rope.

    2. Relevant equations

    3. The attempt at a solution

    I'm having troubles understanding that the tension at the rope ends. The rope's weight is W, so to me it seems that there is a force of W/2 on each tree to keep the rope up. Then the tension at the rope end would be [tex] F = (W/2) \cos \theta [/tex], but the book answer says otherwise. Is the force at each end W or W/2?
     
  2. jcsd
  3. Jun 21, 2009 #2
    The vertical component of the tension at each end is indeed (W/2)*cos(theta)

    But that is not the total tension in each end.

    Draw a Free body diagram and sum the forces in both x and y directions.

    Edit: T_y=W/2 sorry, I answered too quickly :/
     
    Last edited: Jun 21, 2009
  4. Jun 21, 2009 #3
    Do you mean that the total tension is W/2? I cannot see anything when it comes to ropes, but I have read in a book that the horizontal tension is constant and it makes sense that [tex] T_x = T \sin \theta [/tex]. Using this equation and [tex] T_y = (W/2) \cos \theta [/tex] I can find the tension at an end for a given [tex] \theta [/tex] (for theta = 45 degrees, the tension is [tex] W/\sqrt{2} [/tex]), but I cannot see the forces at work here.
     
  5. Jun 21, 2009 #4
    I'll see if I can help. Each tree holds up half the rope. By symmetry, each holds up half the weight. Each tree only knows about the end of the rope, not what the middle does. It knows 1) the direction the rope is pulling and 2) how much downward force there must be.
     
  6. Jun 21, 2009 #5
    Then how can I know the horizontal tension in the rope in order to find the total tension at an end?
     
  7. Jun 21, 2009 #6
    You do not need to. If T is the tension built up in the rope at some angle theta, then by applying Newtons 2nd Law in the y-direction you have that

    [itex]\sum F_y=0=2*T\cos\theta - W[/itex]

    so what is T?

    EDIT: Also, I answered to hastily in post #2. The vertical component Ty is NOT (W/2)*cos(theta), it simply Ty=W/2

    But Ty=T*cos(theta) which when solved for T will give you the same results as above.
     
    Last edited: Jun 21, 2009
  8. Jun 22, 2009 #7
    Okay, thanks :smile:. Some sort of brain block there!
     
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