Question

A uniform ladder of length $5m$ is placed against the wall in vertical plane as shown in the figure. If coefficient of friction $\mu$ is the same for both the wall and the floor then minimum value of $\mu$ for it not to slip is

• A. $\mu =1/2$
• B. $\mu =1/4$
• C. $\mu =1/3$
• D. $\mu =1/5$

Answer 1

The correct option is C $\mu =1/3$

Let mass of ladder be m. consider free body diagram of ladder.

For verticle equilibrium -

$N_1+\mu N_2=mg$ __(1)

For Horizontal equilibrium

$N_2=\mu N_1$ __(2)

Torqul about point A

$mg\left(2.5\cos \theta \right)=N_2\left(4\right)+(\mu N_2{)}^3$

$mg(2.5\times \frac{3}{5})=(4+3\mu )N_2$ $(\cos \theta =\frac{3}{5})$

$mg\left(\frac{3}{2}\right)=(4+3\mu )N_2$ ___(3)

From (1) and (2) $N_2\left[\frac{1+{\mu }^2}{\mu }\right]=mg\Rightarrow N_2=\frac{\mu mg}{1+{\mu }^2}$

Hence $mg\left(\frac{3}{2}\right)=(4+3\mu )\left[\frac{\mu mg}{1+{\mu }^2}\right]$

$(3+3{\mu }^2)=8\mu +6{\mu }^2$

$3{\mu }^2+6\mu -3=0$

$\Rightarrow \mu =\frac{-8\pm \sqrt{64+36}}{6}=\frac{-8\pm 10}{6}$

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

Hence, option $\left(C\right)$ is correct answer.