Advanced 2014 | JEE 2014 | Discussion Page

In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with the horizontal floor. The coefficient of frictional between the wall and 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 N₁ and that of the floor is N₂. If the ladder is about to slip, then

$\mu_{1}=0,\mu_{2}\neq0$ and

$N_{2}\tan\theta=\frac{mg}{2}$

$\mu_{1}\neq 0,\mu_{2}=0$ and

$N_{1}\tan\theta=\frac{mg}{2}$

$\mu_{1}\neq 0,\mu_{2}\neq0$ and

$N_{2}=\frac{mg}{1+\mu_{1}\mu_{2}}$

$\mu_{1}= 0,\mu_{2}\neq0$ and

$N_{1}\tan\theta=\frac{mg}{2}$

Answer:

Option C,D

Explanation:

$\mu_{2}$ can never be zero for maximum equilibrium.

when $\mu_{1}$ = 0 we have

$N_{1}=\mu_{2}N_{2}$ ..........(i)

$N_{1}=m_{2}g$ ...........(ii)

$\tau_{B}=0\Rightarrow mg\frac{L}{2}\cos\theta=N_{1}L\sin\theta$

$\Rightarrow$ $N_{1}=\frac{mg \cot \theta}{2}$

$\Rightarrow$ $N_{1}\tan \theta=\frac{mg }{2}$

when $\mu_{1}\neq 0$ we have

$\mu_{1}N_{1}+N_{2}=mg$ ........(i)

$\mu_{2}N_{2}$ = N .......(ii)

$\Rightarrow$ $N_{2}=\frac{mg}{1+\mu_{1}\mu_{2}}$

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