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6.System of Particles and Rotational Motion
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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
$Image$
$(A)$ $\mu_1=0 \mu_2 \neq 0$ and $N _2 \tan \theta=\frac{ mg }{2}$
$(B)$ $\mu_1 \neq 0 \mu_2=0$ and $N_1 \tan \theta=\frac{m g}{2}$
$(C)$ $\mu_1 \neq 0 \mu_2 \neq 0$ and $N _2 \tan \theta=\frac{ mg }{1+\mu_1 \mu_2}$
$(D)$ $\mu_1=0 \mu_2 \neq 0$ and $N _1 \tan \theta=\frac{ mg }{2}$
A$(B,D)$
B$(B,C)$
C$(A,D)$
D$(C,D)$
(IIT-2014)
Solution
Since rod is about to slip so both friction will be limiting$f _1=\mu_1 N _1 $
$f _2=\mu_2 N _2$
In option $(A)(D)$ $\mu_1=0$
Net torque about $A$ should be zero
$ mg \cos \theta \frac{\ell}{2}= N _1 \sin \theta \ell $
$\Rightarrow N _1=\frac{ mg \cot \theta}{2} $
$\Rightarrow N _1 \tan \theta=\frac{ mg }{2} $
$\text { and } N _2= mg $
$\text { (B) } \mu_2=0$
There is no force to balance $N _1$ so rod can not remain in equilibrium
$\text { (C) } \quad $ $N _1=\mu_2 N _2 $
$N _2+\mu_1 N _1= mg $
$N _2+\mu_1 \mu_2 N _2= mg $
$N _2=\frac{ mg }{1+\mu_1 \mu_2}$
Standard 11
Physics
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