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

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The discussion revolves around a physics homework problem involving a painter on a ladder, focusing on the static friction coefficient and the ladder's angle. The user calculated the force of the wall on the ladder and attempted to find the horizontal distance the painter can climb before slipping occurs. They initially struggled with the solution but later indicated they solved the problem after drawing a free body diagram (FBD). Additional advice was offered regarding the inclusion of torque calculations for similar ladder problems. The conversation highlights the importance of understanding forces and friction in static equilibrium scenarios.
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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


\mus = \frac{P}{N}
P = force of wall on ladder
N = weight of ladder + painter

dx = [P(dy)sin\thetaa]/(Nsin\thetab)

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
 
Last edited:
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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.
 

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