Tyrolean Traverse/Static Equilibrium Problem

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SUMMARY

The discussion focuses on the Tyrolean traverse technique in mountain climbing, specifically addressing the physics involved in calculating the necessary rope sag for safety. A climber weighing 72 kg traverses a 25-meter chasm, requiring a minimum sag distance of 1.5 meters to maintain safety within the recommended limits. The tension force in the rope can reach up to 28 kN, with a safety factor of 10 applied. The equation 2Tsin(θ) = mg is critical for understanding the forces at play, where T represents tension and mg is the weight of the climber.

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  • Understanding of basic physics concepts, particularly forces and tension.
  • Familiarity with the Tyrolean traverse technique in climbing.
  • Knowledge of safety factors in climbing equipment.
  • Ability to manipulate trigonometric functions in equations.
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  • Study the principles of static equilibrium in physics.
  • Learn about tension forces in climbing ropes and their limitations.
  • Explore the application of safety factors in climbing scenarios.
  • Investigate the use of trigonometric functions in real-world physics problems.
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Climbers, physics students, and safety engineers interested in understanding the mechanics of climbing techniques and ensuring safety during outdoor activities.

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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



[tex] 2Tsin\vartheta-mg = \Sigma[/tex]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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