What is the Tension at the Ends of a Hanging Rope?

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The discussion centers on calculating the tension at the ends of a uniform rope hanging between two trees. Participants clarify that the vertical component of the tension at each end is W/2, but the total tension is not simply W/2. The correct approach involves using a free body diagram and applying Newton's second law to analyze the forces in both the x and y directions. It is established that the tension T can be derived from the equation 2*T*cos(θ) = W, leading to a clearer understanding of the forces at play. The conversation emphasizes the importance of distinguishing between vertical and total tension in the context of the rope's weight and angle.
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Homework Statement



A uniform rope of weight W hangs between two trees. The ends of the rope are the same height, and each make angle \theta 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.

Homework Equations



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 F = (W/2) \cos \theta, but the book answer says otherwise. Is the force at each end W or W/2?
 
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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 :/
 
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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 T_x = T \sin \theta. Using this equation and T_y = (W/2) \cos \theta I can find the tension at an end for a given \theta (for theta = 45 degrees, the tension is W/\sqrt{2}), but I cannot see the forces at work here.
 
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.
 
Then how can I know the horizontal tension in the rope in order to find the total tension at an end?
 
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

\sum F_y=0=2*T\cos\theta - W

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.
 
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Okay, thanks :smile:. Some sort of brain block there!
 
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