Ladder on rough ground against rough wall

[Yoonique]

1. Homework Statement
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?

Homework Equations

∑F=ma
∑τ = F⊥ r
f_s ≤ μ_sN

The Attempt at a Solution

[/B]

∑F_y = 0
N₂ + f₁ - W = 0

∑F_x = 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?

 

[ehild]

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

 

[Yoonique]

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

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

 

[Yoonique]

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

 

[ehild]

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

 

[Yoonique]

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?

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

 

[ehild]

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

It is correct now. Nice work!