Question

A ladder of mass m is leaning against a wall. lt 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 $N_{1}tan\theta =\frac{mg}{2}$

Balancing forces in horizontal direction,

$N_1=\mu N_2$

Balancing forces in vertical direction,

$N_2+\mu N_1=mg$

Solving we get,

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

$N_1=\frac{{\mu }_{2}mg}{1+{\mu }_1{\mu }_2}$

Applying rotational equilibrium,

$mg\frac{l}{2}\cos \theta =N_{1}l\sin \theta +\mu N_{1}l\cos \theta$

Solving, $\tan \theta =\frac{1-{\mu }_1{\mu }_2}{2{\mu }_2}$

If ${\mu }_1=0,{\mu }_2\ne 0$

$\tan \theta =\frac{1}{2{\mu }_2}$

$N_1={\mu }_{2}mg$

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