What horizontal distance can a painter climbing a ladder before it slips

[zdickz]

This is for an online homework assignment that apparently is not crediting the proper answer. I'm just curious to know if I got this right or if I'm way off. Thanks in advance.

Homework Statement

If the coefficient of static friction is 0.362, and the same ladder makes a 61.0° angle with respect to the horizontal, how far along the length of the ladder can a 70.2-kg painter climb before the ladder begins to slip?

Homework Equations

$$\mu$$s = $$\frac{P}{N}$$
P = force of wall on ladder
N = weight of ladder + painter

dx = [P(dy)sin$$\theta$$a]/(Nsin$$\theta$$b)

theta a = the angle between the ladder and the floor
theta b = the angle between the ladder and the wall

The Attempt at a Solution

Force of wall on ladder

P = 0.362(743.66) = 268.84N

Horizontal distance of painter

dx = [268.84(10.6)sin 61]/[(743.66)sin 29] = 6.91 m

 

[Delphi51]

Welcome to PF, zdickz. I don't follow your solution so can't comment directly, but I think it should include some torque calcs. Here are some tips for a similar ladder problem:
https://www.physicsforums.com/showthread.php?t=269592

 

[zdickz]

Thank you! I've been poking around here and there. I've almost posted a couple of other questions, but in the process of trying to input the problem on here, I've solved the problem. I attached a FBD I drew up.