Tension in a rope hanging between two trees

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Homework Help Overview

The problem involves a uniform rope of weight W hanging between two trees, with both ends at the same height and making an angle θ with the trees. Participants are tasked with finding the tension at the ends and in the middle of the rope.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss splitting the tension into horizontal and vertical components and balancing forces to find the tension at the ends. There is also an exploration of the tension in the middle of the rope, with references to the catenary shape of the rope.

Discussion Status

Some participants have drawn free-body diagrams to analyze the forces acting on sections of the rope. There is an ongoing exploration of how the tension components change along the rope, with hints provided to simplify the analysis.

Contextual Notes

Participants are considering the implications of the rope's shape and the forces acting on it, including the need to integrate over the curve, which is identified as a catenary rather than a simpler shape.

Buffu
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Homework Statement



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

Homework Equations

The Attempt at a Solution



[/B]
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 ?
 
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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?
 
Orodruin said:
If you draw a free body diagram of an arbitrary section of the rope, what is the total horizontal force component?

Ok let me try.
 
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## ?
 
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?
 
Orodruin said:
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?

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 ?
 
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?
 
Buffu said:
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 ?
Yes.
 

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