A ladder leans against a frictionless wall. If the coefficient of friction with the ground is \(\mu\), what is the smallest angle that the ladder can make with the ground and not slip?


Figure showing the forces acting

Let the ladder have mass m and length \(l\). We have three unknown forces: (i) the frictional force F
(ii) the normal forces \(N_1\) and \(N_2\). And we fortunately have three equations that will allow us to solve for these three forces:
\(\Sigma F_{vert}=0\), \(\Sigma F_{horiz}\) and \(\tau=0\).
Now, from the figure we can see that \(N_1=mg\).And then when we look at the horizontal forces, we see that \(N_2=F\).

We will now use \(\tau = 0\) to find \(N_2\) (or F).

So the number of unknowns are reduced frook three to one.
There are two forces acting at the bottom end of the ladder.
Balancing the torques due to gravity and \(N_2\), we have $$N_2lsin\theta=mg\Bigr(\frac{l}{2}\Bigr)cos\theta$$

( The factor 1/2 comes into play because the ladder behaves like a point mass located halfway up)

$$\Rightarrow N_2=\frac{mg}{2tan\theta}$$
This is also the value of the frictional force F. The condition
$$ F\mu N_1=\mu mg$$

therefore becomes

$$\frac{mg}{2tan\theta}\leq \mu mg$$ $$\Rightarrow tan\theta\geq\frac{1}{2\mu}