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A ring of radius '1 m' is touching the rough vertical wall and is supported by a massless rope whose one end is attached to the wall. If the distance between the points where the rope is touching the wall and the ring is '1 m'. Then what should be the minimum value of coefficient of friction, so that the ring can remain in equilibrium?

Option: 1
$\quad \mu=\frac{1}{2}$

Option: 2
$\mu=\frac{1}{\sqrt{2}}$

Option: 3
$\quad \mu=\frac{2}{3}$

Option: 4
Not possible

Answers (1)

best_answer

$\tan \theta =\frac{Q R}{P Q} \\$

$\tan \theta =\frac{1}{1}=1 \\$

$\theta =45^{\circ}$

for horizontal equilibrium

$T \cos \theta+f=m g \\$ -------(i)

$T \sin \theta=N$

for rotational equilibrium, taking torque about centre of mass -

$N \times O+m g \times O+T \times O+f \times R=C \\$

$C=f R \\$

$\alpha=\frac{f R}{I}$

from equations (i) & (ii)

$\left(\frac{N}{\sin \theta}\right) \cos \theta+f=m g \\$

$f=m g-N \cot \theta$

As, $f \neq 0, \alpha \neq 0$

$\Rightarrow$ ring will rotate about its centre of mass.

Posted by

jitender.kumar

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