Tyrolean Traverse/Static Equilibrium Problem

[SMA83]

Homework Statement

In 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 75-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?

I attatched the picture from masteringphysics.

Homework Equations

\sum{F}=0

The Attempt at a Solution

T=2800 (this is 28 kN/10 for the safety factor)
m=75 kg

\sum{F}=2Tsin\vartheta-mg=0

\vartheta=sin^{-1}\frac{mg}{2T}=7.54 degrees

cos\vartheta=\frac{x}{12.5}

x=12.39 m

Obviously, this answer is too large, and I've verified that masteringphysics won't accept it, but I'm still pretty confused as to where I went wrong. Thanks in advance for the help!

 

[rl.bhat]

x/12.5 is tan(theta), not cos (theta)

 

[SMA83]

doh! I knew it would be something simple that was throwing me off. Well thanks for the help...I got 1.7 m now, and masteringphysics approves, so all is right with the world.