Question

A uniform ladder of length $5m$ is placed against the wall as shown in the figure. If coefficient of friction $\mu$ is the same for both the walls, what is the minimum value of $\mu$ for it not to slip ?

• A. $\mu =\frac{1}{2}$
• B. $\mu =\frac{1}{4}$
• C. $\mu =\frac{1}{3}$
• D. $\mu =\frac{1}{5}$

Answer 1

The correct option is C $\mu =\frac{1}{3}$

$N_1=\mu N_2$ ...(1)
$\mu N_1+N_2=mg$ ...(2)

Torque about A: ${\tau }_A=0$
$\Rightarrow 3N_2-4N_1-\frac{5}{2}\times \frac{3}{5}mg=0$ ....(3)

From (1) and (2)

$N_2=\frac{mg}{1+{\mu }^2}$ and $N_1=\frac{\mu mg}{1+{\mu }^2}$

Substituting $N_1$ and $N_2$ in (3) we get
$\frac{3mg}{1+{\mu }^2}-\frac{4\mu mg}{1+{\mu }^2}=\frac{3}{2}mg$
$\Rightarrow \frac{3-4\mu }{1+{\mu }^2}=\frac{3}{2}$
$\Rightarrow 3{\mu }^2+8\mu -3=0$
$\Rightarrow (\mu +3)(3\mu -1)=0$
$\Rightarrow \mu =-3$ or $\mu =\frac{1}{3}$ but $\mu =-3$ is not physically possible.
$\therefore \mu =\frac{1}{3}$