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 of 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 $\mu =\frac{1}{3}$

Let mass of ladder be $m$

consider free body diagram of ladder

For verticle equilibrium

$N+\mu N_2=mg⟶1$

For horizontal equilibrium

$N_2=\mu N_1⟶2$

Torque about point $A$

$mg\left(2.5\cos \theta \right)=N_2\left(4\right)+\mu N_2\left(3\right)mg(2.5\times \frac{3}{5})=(4+3\mu )N_{2}mg\left(\frac{3}{2}\right)=(4+3\mu )N_2⟶3$

Form $1$ and $2$

$N_2\left[\frac{1+{\mu }^2}{\mu }\right]=mg{N}_2=\frac{\mu mg}{1+{\mu }^2}$

Hence

$mg\left(\frac{3}{2}\right)=(4+3\mu )\left(\frac{\mu mg}{1+{\mu }^2}\right)(3+3{\mu }^2)=8\mu +6{\mu }^{2}3{\mu }^2+8\mu -3=0$

By solving this

$\mu =\frac{1}{3}$