Tension in a rope hanging between two trees

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SUMMARY

A uniform rope of weight W hangs between two trees, creating tension at both ends and in the middle. The tension at either end of the rope is calculated using the formula T = W/(2 cos θ), where θ is the angle the rope makes with the trees. The tension in the middle of the rope requires understanding the catenary shape of the rope, which is described by the locus y = α cosh(x/α). A free-body diagram approach reveals that the horizontal force components balance out, confirming that T' = T sin θ = W tan θ / 2 for equilibrium.

PREREQUISITES
  • Understanding of basic physics concepts such as tension and equilibrium
  • Familiarity with free-body diagrams and their application in mechanics
  • Knowledge of catenary curves and their mathematical representation
  • Ability to perform trigonometric calculations involving angles and forces
NEXT STEPS
  • Study the properties of catenary curves and their applications in physics
  • Learn how to draw and analyze free-body diagrams in static equilibrium scenarios
  • Explore the derivation of tension in various configurations of hanging ropes
  • Investigate the effects of varying angles θ on tension and stability in hanging systems
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Students studying physics, particularly those focusing on mechanics, engineers designing cable systems, and anyone interested in the mathematical modeling of hanging structures.

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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