Recently, my professor in *IE 5513 - Engineering Safety* presented the class with the task of being able to predict whether or not a ladder will slip or not when it's leaned towards a wall. It's been a while since I've had physics, so I found some notes on ladders and walls available online by Richard Fitzpatrick from the University of Texas at Austin which I'm going to go through.

### Problem statement

A ladder of length $l$ with neglible mass leaned towards a wall with an angle $\theta$. A person of mass $M$ climbs a distance of $x$ up the ladder. Given the static friction $\mu$, will the ladder slip from the ground?

### Resolving forces

There are four forces involved in this problem: The weight of the person $Mg$, the normal force at the wall $S$, the normal force at the ground $R$ and the friction force from the surface $f$. The sum of these forces needs to equal zero. For the horizontal forces we have:

$
S = f

$

Analogously, for the vertical forces:

$
R = Mg

$

Remember, these values are scalars and not vectors.

### Considering torque

The next step in this exercise is to consider the torque acting on the ladder. To simplify, the point where the ladder touches the ground is chosen as the basis for torque calculation since the displacement vector for both $R$ and $f$ is zero. Decomposition gives a contributing value of $M~g~\cos(\theta)$ for the man on the ladder and $S~\sin(\theta)$ for ladder top. Multiplying with the respective displacement values, we get the torques, which should equal to zero.

$
M~g~x~\cos(\theta) = S~l~\sin(\theta)

$

Which further can be written as:

$
S = {M~g~x \over l~\tan(\theta)}

$

### Putting it all together

The condition for the ladder not to slip is $f < \mu~R$. We also know the following:

$
f = S = {M~g~x \over l~\tan(\theta)}

$

Hence:

$ {M~g~x \over l~\tan(\theta)} < \mu~R $

Substituting $R = M~g$, removing it, as well as multiplying by $l$ and $tan(\theta)$ gives:

$
x < \mu~l~\tan(\theta)

$

If this inequation holds, the ladder will not slip. To find out if the whole ladder can be used, substitute $x$ for $l$.

A huge thanks to Tormod Haugland for reviewing and suggesting improvements for this blog post!