(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A uniform ladder of mass m and length L rests against the wall as shown. 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? (Use m for m, L for L, mu for μ, and arcsin, arccos, and arctan for the inverse trig functions.)

2. Relevant equations

f(friction)≤μN(normal force)

t(torque)=r(moment arm)Fsinθ

G=mg

3. The attempt at a solution

FBD is attached

because the ladder is not moving

t=-mgLsinθ/2+N_{f}Lsinθ-μN_{f}Lcosθ=0

F_{x}=N_{w}-f_{f}=0

F_{y}=N_{f}+f_{w}-mg=0

Solve t for N_{f}

N_{f}Lsinθ-μN_{f}Lcosθ=mgLsinθ/2

N_{f}tanθ-μN_{f}=mgtanθ/2

N_{f}=mgtanθ/(2(tanθ-μ))

When θ is at its maximum without the ladder slipping, f=μN.

f_{f}=μmgtanθ/(2(tanθ-μ))

Solve F_{x}for N_{w}

N_{w}=μmgtanθ/(2(tanθ-μ))

Also f_{w}=μ^{2}mgtanθ/(2(tanθ-μ))

Solve F_{y}for θ.

mgtanθ/(2(tanθ-μ))+μ^{2}mgtanθ/(2(tanθ-μ))=mg

tanθ/(2(tanθ-μ))+μ^{2}tanθ/(2(tanθ-μ))=1

tanθ=2(tanθ-μ)/(1+μ^{2})

(tanθ-μ)/tanθ=(1+μ^{2})/2

1-μ/tanθ=(1+μ^{2})/2

μ/tanθ=1-(1+μ^{2})/2

tanθ=μ/(1-(1+μ^{2})/2)

θ=arctan(μ/(1-(1+μ^{2})/2))

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# Ladder problem (statics) with 2 sources of friction

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