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Tension in a rope hanging between two trees

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  1. May 21, 2017 #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 same height, and they each make an angle ##\theta## with the trees. Find :

    a): The tension at the either end of the rope.

    b): The tension in the middle of the rope.

    upload_2017-5-21_19-33-11.png


    2. Relevant equations


    3. The attempt at a solution


    For the tension at the end, We split the tension in horizontal and vertical compoenents. ##T \sin\theta## and ##T \cos \theta## respectively.

    Since the vertical forces balance the weight, ##2T \cos\theta = W \iff T = \dfrac{W}{2\cos\theta}##,

    I am stuck at second part.
    I know the tension would be tangent to the curve. I thought I will integrate over the curve to find the total tension but the curve is not a semi circle instead it is a catenary which has a locus ##y = \alpha \cosh(x/\alpha)##, nevertheless this locus looks a complete mess to.

    Is there a way apart from this mess ?


     
  2. jcsd
  3. May 21, 2017 #2

    Orodruin

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    If you draw a free body diagram of an arbitrary section of the rope, what is the total horizontal force component?
     
  4. May 21, 2017 #3
    Ok let me try.
     
  5. May 23, 2017 #4
    path42016.png


    So the total horizontal force would be ##dT \cos(da/2) \cos a## is this correct ? Now should I integrate from ##0 \to \pi/2## ?
     
  6. May 23, 2017 #5

    Orodruin

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    It is much (much) simpler than that. You just need to draw the free-body diagram of the correct part of the rope. Hint: Try drawing the free-body diagram for half the rope. What forces act on that half in the different directions?
     
  7. May 23, 2017 #6
    Ok I did that, the horizontal force toward right is ##T\sin \theta## and towards left is ##T^\prime##, since rope is at rest, ##T^\prime = T\sin \theta = W\tan\theta /2##.

    Is this correct ?
     
  8. May 23, 2017 #7

    scottdave

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    As you move along the rope, how would you expect the vertical component of tension to change? How would you expect horizontal component to change along the rope?
     
  9. May 23, 2017 #8

    Orodruin

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    Yes.
     
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