# Ladder on rough ground against rough wall

1. Apr 15, 2015

### Yoonique

1. The problem statement, all variables and given/known data
A ladder on the rough floor is leaning against a vertical rough wall. The ladder has length l and mass m. The coefficients of friction are μ for both contact surfaces. What is the smallest angle between the ladder and the floor?

2. Relevant equations
∑F=ma
∑τ = F⊥ r
fs ≤ μsN

3. The attempt at a solution

∑Fy = 0
N2 + f1 - W = 0

∑Fx = 0
f2 - N1 = 0
f2 = N1

∑τ about the ground = 0
N1lsinθ + f1lcosθ - W(l/2)(cosθ) = 0

For θ to be the smallest angle, what is the condition? Is it f1 = μN1 or f2 = μN2 or both of them need to happen at the same time?

Last edited: Apr 15, 2015
2. Apr 15, 2015

### ehild

Can the ladder move if the friction is static at some of the contact surfaces at the ends of the ladder?

3. Apr 15, 2015

### Yoonique

I think the ladder can't move if one the friction is static. So both must happen at the same time.

Last edited: Apr 15, 2015
4. Apr 15, 2015

### Yoonique

Okay, I finally managed to solve it. θ = tan-1[1-μ2/(2u)]. Thanks ehild!

5. Apr 15, 2015

### ehild

Are you sure that $\theta= \tan^{-1}\left(1 - \frac{ \mu^2 }{2 \mu }\right)$ ? Why you do not simplify with μ? Or some parentheses missing?

6. Apr 15, 2015

### Yoonique

Oh it is a typo! Should be θ = tan-1[(1-μ2)/(2μ)]. Thanks anyway!

7. Apr 15, 2015

### ehild

It is correct now. Nice work!