Question

In the figure, a ladder of mass $m$ is shown leaning against a wall. It is in static equilibrium making an angle $\theta$ with the horizontal floor. The coefficient of friction between the wall and the ladder is ${\mu }_1$ and that between the floor and the ladder is ${\mu }_2$.

Answer 1

The normal reaction of the wall on the ladder is $N_1$ and that of the floor is $N_2$

If ladder is about to slip then

According to the figure

$N_1={\mu }_2{N}_2$

$N_2=mg+{\mu }_1{N}_1$

By torque balance equation, we get

$\frac{mg\cos \theta }{2}={\mu }_1{N}_1\cos \theta +N_1\sin \theta$

if ${\mu }_1=0$ then

$\frac{mg\cos \theta }{2}=N_1\sin \theta$

$N_1\tan \theta =\frac{mg}{2}$

if ${\mu }_2=0$ then

$N_1=0$