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

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Homework Help Overview

The problem involves determining the minimum angle at which a uniform ladder can lean against a frictionless wall without slipping, given the coefficient of static friction between the ladder and the ground.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of trigonometric functions and the relationship between forces acting on the ladder. Some express uncertainty about their calculations and the equations used, while others propose equations related to torque and friction.

Discussion Status

The discussion is ongoing, with participants exploring different approaches and questioning the validity of their assumptions and calculations. Some guidance has been offered regarding the forces involved, but there is no explicit consensus on the correct method or solution.

Contextual Notes

Participants mention varying levels of familiarity with the concepts involved, indicating that some are still learning the relevant material.

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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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