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)

From the condition of equilibrium of bodies:
$ \begin{array}{l} \Sigma F _{ x }=0 \\ \therefore- N _{1}+\mu_{2} N _{2}=0 \\ \therefore N _{1}=\mu_{2} N _{2} \\ \text { 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}\left(1+\mu_{1} \mu_{2}\right)= mg \\ \therefore N _{2}=\frac{ mg }{1+\mu_{1} \mu_{2}} \end{array} $
Applying torque equation about corner (left) point on the floor:
$mg \frac{1}{2} \cos \theta= N _{1} 1 \sin \theta+\mu_{1} N _{1} 1 \cos \theta$ 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: $N_{1} \tan \theta=\frac{ mg }{2}$