What is the minimum angle for a ladder to not slip against a frictionless wall?

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To determine the minimum angle for a ladder leaning against a frictionless wall, the discussion focuses on the relationship between static friction and the angle of inclination. The coefficient of static friction is given as 0.43, leading to the equation tan(θ) = 0.43. Participants suggest using torque equations to analyze the forces acting on the ladder, specifically referencing the maximum angle formula θ = tan⁻¹(1/(2μ)). The calculations involve balancing forces and moments to find the angle at which the ladder will not slip. Understanding these principles is essential for solving the problem effectively.
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Homework Statement


A uniform ladder of mass m and length L leans at an angle θ against a frictionless wall, Fig. 9-61. If the coefficient of static friction between the ladder and the ground is 0.43, what is the minimum angle at which the ladder will not slip?

Pic:
http://www.webassign.net/giancoli5/9_61.gif


Homework Equations



?? tan(theta) = .43?

The Attempt at a Solution



23.26
 
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Hi tigerwoods99! :smile:

(have a theta: θ :wink:)
tigerwoods99 said:
?? tan(theta) = .43?

The Attempt at a Solution



23.26

I don't think so.

What equations did you use?
 
Hi! I am really not sure its more or less a wild guess using trigonometric functions.
 
(just got up :zzz: …)
tigerwoods99 said:
Hi! I am really not sure its more or less a wild guess using trigonometric functions.

thought so! :rolleyes:

ok … call the normal reaction force (from the ground) N, and the horizontal reaction force H …

how much is N? :smile:
 
I think this is right.


τorque = 1/2mgx - FNx + Ffy

at max angle
μmgy = 1/2mgx

the maximum angle is θ = tan−1(1/(2μ))
 
tigerwoods99 said:
τorque = 1/2mgx - FNx + Ffy

at max angle
μmgy = 1/2mgx

the maximum angle is θ = tan−1(1/(2μ))

Yup! :biggrin:

Easy, isn't it? :wink:
 
haha yea i guess.. though i haven't learned this yet so i had to figure it out.
 
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