Mountain climber Equilibrium Question

[simondsaid]

Homework Statement

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.

Homework Equations

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

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

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

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

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°

 

[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.

 

[simondsaid]

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

the acceptable answer was 26.

Thanks for your help!