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I've got two trees and between them is a rope. When someone about 100kg sits at the center of the rope, what are the forces on both parts of rope (to the left-hand side from the person and right-hand side)? When no-one is sitting on the rope, the angle between rope and ground is 0 degrees. When he sits, it is about 15 to 20 degrees.

And how can I compute that?

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    $\begingroup$ Hint: A tension $T$ acting at an angle $\theta$ has a vertical component of $T\sin(\theta)$. $\endgroup$
    – lemon
    Apr 25, 2015 at 15:54
  • $\begingroup$ I'm sorry but I still don't know. I suppose I will use sin(15°). But how to combine that with weight? $\endgroup$
    – Martin Heralecký
    Apr 26, 2015 at 5:06
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    $\begingroup$ Forces must balance. So the vertical downward force (your weight) must equal the vertical upward force... $\endgroup$
    – lemon
    Apr 26, 2015 at 11:40

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Since the length of the rope must remain constant the trees must bend towards the center on both sides .

If the mass of the person is M , then by balancing forces in vertical

Mg = 2.T.sin(theta) : theta being angle ~20 degrees

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