Question

In the figure, a ladder of mass $m$ is shown leaning against a wall. It is 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 ${\mu }_2$. The normal reaction of the wall on the ladder is $N_1$ and that of the floor is $N_2$. If the ladder is about to slip, then

• A. ${\mu }_1=0,{\mu }_2\ne 0\text{and}{N}_2\tan \theta =\frac{mg}{2}$
• B. ${\mu }_1\ne 0,{\mu }_2=0\text{and}{N}_1\tan \theta =\frac{mg}{2}$
• C. ${\mu }_1\ne 0,{\mu }_2\ne 0\text{and}{N}_2=\frac{mg}{1+{\mu }_1{\mu }_2}$
• D. ${\mu }_1=0,{\mu }_2\ne 0\text{and}{N}_1\tan \theta =\frac{mg}{2}$

Answer 1

The correct option is C ${\mu }_1\ne 0,{\mu }_2\ne 0\text{and}{N}_2=\frac{mg}{1+{\mu }_1{\mu }_2}$

${\mu }_2$ can never be zero for equilibrium.


When ${\mu }_1=0,$ we have

$N_1={\mu }_2{N}_2;N_2=mg$

${\tau }_B=0$
$\Rightarrow mg\frac{L}{2}\cos \theta =N_{1}L\sin \theta$

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

When, ${\mu }_1\ne 0$ we have

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

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

$\Rightarrow N_2=\frac{mg}{1+{\mu }_1{\mu }_2}$

Hence, option $\left(\text{C}\right)$ is correct.