Torque and Equilibrium of a rope climber

[geah]

Homework Statement

A climber with a weight of 533.8 N is held by a belay rope connected to her climbing harness and belay device; the force of the rope on her has a line of action through her center of mass. The indicated angles are 40 degrees and 30 degrees. If her feet are on the verge of sliding on the vertical wall, what is the coefficient of static friction between her climbing shoes and the wall. (40 degrees is the angle between her feet and the wall, 30 degrees is the angle between the rope and the wall)

Homework Equations

torque = rxF

The Attempt at a Solution

First off I drew a free body diagram of the situation. Because torque must equal to zero we know that the torque counterclockwise = torque clockwise. Knowing this I put the information into the equation and got r Ff sin(110) = r/2 (533.8) (cos 70) what's stumping me is the fact that I have 2 variable and I can figure out how to move from here.


If you could offer a clean explanation that would be wonderful,
Thanks
~geah

 

[Doc Al]

Because torque must equal to zero we know that the torque counterclockwise = torque clockwise. Knowing this I put the information into the equation and got r Ff sin(110) = r/2 (533.8) (cos 70) what's stumping me is the fact that I have 2 variable and I can figure out how to move from here.

It's unclear to me what point you are using as a reference in calculating torques. (In any case, the "r" drops out of this equation, leaving you with one variable.)

I suggest you solve this in multiple steps. First take torques about the point where her foot contacts the wall. This will allow you to find the tension in the rope.