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: (D)

$\mu_1=0, \mu_2 \ne 0$ and $N_1 \tan \theta=\frac{mg}{2}$

Answer 2

Condition of translational equilibrium
$N _{1}=\mu_{2} \,N _{2}$
$N _{2}+\mu_{1}\, N _{1}= Mg$
Solving $N _{2}=\frac{ mg }{1+\mu_{1} \mu_{2}}$
$N _{1}=\frac{\mu_{2} mg }{1+\mu_{1} \mu_{2}}$
Applying torque equation about corner (left) point on the floor
$mg \frac{\ell}{2} \cos \theta= N _{1} \ell \sin \theta+\mu_{1} N _{1} \ell \cos \theta$
Solving $\tan \theta=\frac{1-\mu_{1} \mu_{2}}{2 \mu_{2}}$