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?

Solution: 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

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