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## Homework Statement

[/B]

A 3.0-m-long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.32.

What is the minimum angle the ladder can make with the floor without slipping?

## Homework Equations

With the axis of rotation at the bottom of the ladder:

ΣT = 0 <- torque

ΣF = 0

T = LF

T = LFsinφ

ƒ = μN

## The Attempt at a Solution

X to the right is positive, Y up is positive

Nw = Normal of wall

Nf = normal of floor

fs = static friction force

Fg = gravity

ΣFx = 0

Nw - fs=0

Nw = μNf

ΣFy = 0

Nf-Fg = 0

Nf = mg

So,

Nw=μmg

Now I'm still having trouble with torques so bear with me. Since these two forces go through the axis then they produce no torque correct?

TNf = 0

Tfs = 0

and

TFg = (Fg)(L/2) = mg(L/2)

TNw = (Nw)(L)

Am I on the right path? I think you sum the two torques but I don't have the mass so that wouldn't work.

Also, how do I know if the torque is negative or positive? and which way do I make torque positive? CW or CCW?

Appreciate any and all advice :)

link to the picture: http://session.masteringphysics.com/problemAsset/1385677/5/K-P12.60.jpg