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

A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is μs = 0.500, determine the smallest angle the ladder can make with the floor without slipping.

2. Relevant equations

I believe the set up of the problem is:

(1/2 X mg x Cos[tex]\theta[/tex] = ( N x L Sin[tex]\theta[/tex] ) +( Ff X Cos[tex]\theta[/tex] )

3. The attempt at a solution

(Here Maybe the problem number)

I bring Ff X cos[tex]\theta[/tex] to the other side, and factor out the mg and divived the Sin. to get Tan[tex]\theta[/tex] by it self.

So the result is:

tan[tex]\theta[/tex]= (1/2 X L X mg - [tex]\mu[/tex]mg) / (mg X L)

Tan[tex]\theta[/tex] = (1/2 L - [tex]\mu[/tex]) / L

Okay, then the problem is that I'm suppose to get a numeric answer, to [tex]\theta[/tex] which is 36.8 degree.

If my attempt was on the right track, then How could i find the Numeric value of this if I only know what [tex]\mu[/tex] is?

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# Homework Help: Simple Ladder leading against a vertical wall become complicated.

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