Tyrolean Traverse/Static Equilibrium Problem

[ConstableZiM]

Homework Statement

Please read the WHOLE post before replying, because I don't want the answer, but how to obtain one of the equations.

A mountain-climbing technique called the "Tyrolean traverse," a rope is anchored on both ends (to rocks or strong trees) across a deep chasm, and then a climber traverses the rope while attached by a sling as in the figure (Intro 1 figure) . This technique generates tremendous forces in the rope and anchors, so a basic understanding of physics is crucial for safety. A typical climbing rope can undergo a tension force of perhaps 28 kN before breaking, and a "safety factor" of 10 is usually recommended. The length of rope used in the Tyrolean traverse must allow for some "sag" to remain in the recommended safety range.

Consider a 72-kg climber at the center of a Tyrolean traverse, spanning a 25-m chasm. To be within its recommended safety range, what minimum distance x must the rope sag?

Homework Equations

$$2Tsin\vartheta-mg = \Sigma$$F

The Attempt at a Solution

I don't have a problem with the question itself, I have a problem with understanding one of the equations, 2TSin(Theta) = mg... Why is the TSin(Theta) multiplied by 2? I can't conceptually comprehend that... if the line can take only 2900 Newtons, then multiplying it by two means we are putting a load of 4800 Newtons on the whole line?...

The answer is 1.5 meters

 

[rock.freak667]

The tensions in the two sides would be the same if the weight is at the center.