# Mountain climber Equilibrium Question

1. Jun 21, 2009

### simondsaid

1. The problem statement, all variables and given/known data
Picture: http://www.webassign.net/grr/p8-34.gif

A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber's waist 15 cm to the right of his center of gravity and makes an angle of $$\theta$$= 19° with the wall. The climber weighs 744 N.

(a) Find the tension in the rope=680N
(b) Find the magnitude and direction of the contact force exerted by the wall on the climber's feet.

Magnitude=240N
Direction=$$\theta$$=? above the horizontal.

2. Relevant equations
T=$$\frac{W*.91m}{1.06m*cos19}$$

W=Tcos$$\theta$$+F$$_{v}$$

T*sin$$\theta$$=F$$_{h}$$

Magnitude F$$_{w}$$=$$\sqrt{F^{2}_{v}+F^{2}_{h}}$$

3. The attempt at a solution
The equations basically explain my attempt at the problem. I just can't seem to be able to find the direction of the contact force exerted by the wall.

I've tried:
arctan$$\frac{Fv}{Fh}$$=25°

2. Jun 21, 2009

### LowlyPion

Welcome to PF.

I get roughly what you are getting. What seems to be telling you it is wrong? (I get 244 N at 25.5 degrees above the horizontal.)

The wall is pushing up - countering the weight not made up in the tension, and out against the compression of the feet.

3. Jun 21, 2009

### simondsaid

Ah I got it. It was a significant digits thing.