Question

A uniform ladder of mass $10\text{kg}$ leans against a smooth vertcal wall making an angle ${53}^{\circ }$ with it. The other end rests on a rough horizontal floor. Then, the friction coefficient just necessary for the ladder to be at rest is approximately

Answer 1

FBD of the ladder is given below:


For translational equilibrium, in horizontal direction:
$N_2=\mu N_1$ for limiting value of friction ("just necessary")
and in vertical direction:
$N_1=mg$

For rotational equilibrium, net torque about point of contact of ladder and vertical wall
${\overrightarrow{\tau }}_{net}=0$
$\Rightarrow N_1\times \frac{4}{5}l=mg\times \frac{4}{5}\times \frac{l}{2}+\mu N_1\times \frac{3}{5}l$
where $l$ is the length of the ladder.
$(\because N_1=mg)$
$\Rightarrow mg\times \frac{4}{5}\times \frac{l}{2}=\mu mg\times \frac{3}{5}l$
$\therefore \mu =\frac{2}{3}=0.67$