- #1

Dustgil

- 42

- 0

## Homework Statement

A uniform ladder leans against a smooth vertical wall. If the floor is also smooth, and the initial angle between the floor and the ladder is theta(0), show that the ladder, in sliding down, will lose contact with the wall when the angle between the floor and the ladder is

[tex]sin^{-1}(\frac {2} {3} sin \theta_{0})[/tex]

## Homework Equations

[tex] \frac {dL} {dt}=N [/tex]

where L is the angular momentum and N is the net external torque

## The Attempt at a Solution

I understand why the ladder loses contact with the wall at some point. There is a normal force on the ladder exerted by the wall that is not balanced by anything. So as the ladder falls, it sort of gains length in the x direction while at the same time being pushed away from the wall. Eventually there comes a point where the normal force exerted by the wall is zero and then negative, which will be at the angle I'm looking for.

Let's say that the angle that the ladder makes with the floor is theta. The translational motion equations are

[tex] m \ddot y=F_{n} - mg[/tex]

[tex] m \ddot x=F_{w}[/tex]I start to get a bit confused when trying to define the rotational motion. The first question is where's the best place to define the rotation at? For now I choose the point where the ladder makes contact with the ground. This is where I run into my second problem. For whatever reason I'm having trouble confidently expressing the torques exerted from both gravity and the normal force from the wall. I'm just kinda getting mixed up by it. I make a guess:

[tex]\frac {dL} {dt}= N[/tex]

[tex]I \ddot {\theta} = mgcos(\theta)(\frac {l} {2}) - F_{w}sin(\theta)(\frac {l} {2})[/tex]

Are these the correct equations? It took me a long while to get to this point so It'd be cool to be confident before I soldier on.