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

Answer 1

Answer: (C), (D)

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

Consider of translational equilibrium
$N_1={\mu }_2{N}_2$ (i)
$N_2+{\mu }_1{N}_1=Mg$ (ii)

Solving $N{N}_2=\frac{mg}{1+{\mu }_1{\mu }_2}$
$N_1=\frac{{\mu }_{2}mg}{1+{\mu }_1{\mu }_2}$
Applying torque equation about corner (left) point ont he floor
$mg\frac{l}{2}cos\theta =N_{1}lsin\theta +{\mu }_1{N}_{1}lcos\theta$
Solving $tan\theta =\frac{1-{\mu }_1{\mu }_2}{2{\mu }_2}$