A uniform ladder of length $5\text{m}$ and mass $100\text{kg}$ is in equilibrium between a vertical smooth wall and a rough horizontal surface as shown below. Find the minimum friction co-efficient between floor and ladder for this equilibrium.

The correct option is B $\frac{2}{3}$
Point of contact of the ladder with the horizontal floor will tend to slip in the forward direction, hence static friction will act along backward direction.

So, the F.B.D of ladder can be drawn as


For vertical equilibrium of ladder

$N_2=mg.....\left(i\right)$

For horizontal equilibrium,

$N_1=f_s$

Since coefficient of friction is minimum, $f_s$ will act at limiting value to prevent slipping.

$\Rightarrow N_1=\mu N_2....\left(ii\right)$

Balancing clockwise and anticlockwise moment of forces about the point $\text{A}$

$({\tau }_f=0,{\tau }_{N_2}=0)$

$\Rightarrow N_{1}l\sin {37}^{\circ }=mg\times \frac{l}{2}\cos {37}^{\circ }.....\left(iii\right)$

From equations $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$:

$\mu mg\times \frac{3}{5}=mg\times \frac{1}{2}\times \frac{4}{5}$

$\Rightarrow 3\mu =2$

$\therefore \mu =\frac{2}{3}$