The greatest angle at which a ladder won't slip

  • #1
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[EDIT: Never mind about this problem, I found my mistake.]

1. Homework Statement

A uniform ladder of mass m and length L rests against a wall. The coefficients of static friction between the floor and the ladder and between the wall and the ladder are equal to each other (μ). What is the maximum value of angle θ that the ladder can make with the wall without sliding? Solve in terms of m, g, L, μ, and θ.


Homework Equations



Newton's Second Law.

The maximum magnitude of static friction = μN, where N is the normal force.

The magnitude of torque is rF.

The Attempt at a Solution



I drew a free-body diagram, treating the floor as a pivot point. I labeled the center of the mass of the ladder occurring at L/2 along the ladder, exerting a downward force of mg. At the point where the ladder touches the wall I draw a normal force Nw directed away from the wall and a static force μNw directed up from the contact point. At the point of contact with the floor I draw a normal force Nf and a frictional force μNf.

From here I know that, since the ladder isn't falling, the vertical forces balance, that is to say

mg = Nf+μNw
Since the ladder isn't translating right or left then

Nw = μNf
Since the ladder isn't experiencing any spin then

(mg cosθ)L/2 = L( μNwcosθ + Nwcos(θ+π/2) )
I'm solving for θ theta. I start by canceling L on both sides. I can also make the trig identity cos(θ+π/2) = sin( π/2 - [θ+π/2] ) = -sinθ. Also, I want to eliminate Nw so from the earlier equations I infer

mg = Nw/μ+ μNw
= Nw(1/μ + μ)
∴ Nw = mg/(1/μ + μ)
And so I can write

(mg cosθ)/2 = (mg)( μcosθ - sinθ )/(1/μ + μ)
∴ (1/μ + μ)/2 = μ - tanθ

∴ tanθ = 2μ/2 - (1/μ + μ)
= (μ-1/μ)/2
= (μ2 - 1)/(2μ)
There are two problems with this, though. One is that usually μ<1 so that this would be a negative value of θ. The other is that I pulled this problem off of the EdX Physics Review and their answer was

θ = arctan( 2μ / (1-μ2) )
That's really close to mine but the argument of the arctan is the negative reciprocal of my answer. I just can't see where I went wrong. Any help would be appreciated.
 
Last edited:

Answers and Replies

  • #2
20
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Sorry, never mind, I found my mistake--I was decomposing the components of force in the wrong way at the contact point with the wall.
 

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