Tyrolean Traverse/Static Equilibrium Problem

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The discussion centers on the Tyrolean traverse, a climbing technique involving a rope anchored across a chasm, where understanding the physics is essential for safety. A climber's weight creates tension in the rope, and the equation 2Tsin(θ) = mg is questioned for its reasoning, particularly why the tension is multiplied by two. This multiplication accounts for the tension in both sides of the rope when the climber is at the center. The climber's weight and the required sag distance to maintain safety are also highlighted, with the minimum sag calculated to be 1.5 meters. Understanding these forces is crucial for ensuring the safety of climbers using this technique.
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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



<br /> 2Tsin\vartheta-mg = \Sigma<br />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
 
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The tensions in the two sides would be the same if the weight is at the center.
 

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