Tension in a rope hanging between two trees

AI Thread Summary
A uniform rope of weight W hangs between two trees, forming an angle θ with each tree. The tension at the ends of the rope is calculated using the balance of vertical forces, resulting in T = W/(2cosθ). The discussion shifts to finding the tension in the middle of the rope, which involves understanding the catenary shape of the rope. A free-body diagram is suggested to analyze the forces acting on half the rope, leading to the conclusion that the horizontal tension components balance out. The final tension relationships are confirmed as correct, emphasizing the importance of visualizing the forces involved.
Buffu
Messages
849
Reaction score
146

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 ?
 
Physics news on Phys.org
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.
 
Back
Top