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 ${\mu }_2$. 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. ${\mu }_1=0,{\mu }_2\ne 0$ and $N_{2}tan\theta =\frac{mg}{2}$
• B. ${\mu }_1\ne 0,{\mu }_2=0$ and $N_{1}tan\theta =\frac{mg}{2}$
• C. ${\mu }_1\ne 0,{\mu }_2\ne 0$ and $N_2=\frac{mg}{1+{\mu }_1{\mu }_2}$
• D. ${\mu }_1=0,{\mu }_2\ne 0$ and $N_{1}tan\theta =\frac{mg}{2}$

Answer 1

Answer: (C), (D)

The correct options are
C ${\mu }_1\ne 0,{\mu }_2\ne 0$ and $N_2=\frac{mg}{1+{\mu }_1{\mu }_2}$
D ${\mu }_1=0,{\mu }_2\ne 0$ and $N_{1}tan\theta =\frac{mg}{2}$

From the condition of equilibrium of bodies:
$\Sigma F_x=0$
$\therefore -N_1+{\mu }_2{N}_2=0$
$\therefore N_1={\mu }_2{N}_2$
Also $\Sigma F_y=0$
$\therefore N_2+{\mu }_1{N}_1-mg=0$
$\therefore N_2+{\mu }_1{\mu }_2{N}_2-mg=0$
$\therefore N_2(1+{\mu }_1{\mu }_2)=mg$
$\therefore N_2=\frac{mg}{1+{\mu }_1{\mu }_2}$
Applying torque equation about corner (left) point on the floor:
$mg\frac{l}{2}cos\theta =N_{1}lsin\theta +{\mu }_1{N}_{1}lcos\theta$

Solving: $N_{1}tan\theta =\frac{mg}{2}-{\mu }_1{N}_1=\frac{mg}{2}-\frac{{\mu }_1{\mu }_{2}mg}{1+{\mu }_1{\mu }_2}$
Solving we get: $N_{1}tan\theta =\frac{mg}{2}$