A massless spool of inner radius r, outer radius R is placed against vertical wall and tilted split floor as shown. A light inextensible thread is tightly wound around the spool through which a mass m is hanging. There exists no friction at point A, while the coefficient of friction between spool and point B is $μ$ . The angle between two surfaces is $θ$ .

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the minimum value of $μ$ for the system to remain in equilibrium is $\frac{\tan θ}{(R / r) - 1}$

the magnitude of force on the spool at B in order to maintain equilibrium is $mg (1 - \frac{r}{R}) \frac{1}{\tan θ}$

the minimum value of $μ$ for the system to remain in equilibrium is $\frac{\cot θ}{(R / r) - 1}$

the magnitude of force on the spool at B in order to maintain equilibrium is $mg \sqrt{{(\frac{r}{R})}^{2} + {(1 - \frac{r}{R})}^{2} \frac{1}{\tan^{2} θ}}$

answer is A, D.

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Detailed Solution

$mgr = fR$ …(i)

$N_{1} \sin θ + f = mg$ …(ii)

$N_{1} \cos θ = N_{2}$ …(iii)

From (i), $f = \frac{mgr}{R}$

From (ii) and (iii),

$N_{2} = \frac{mg - f}{\tan θ} = \frac{mg}{\tan θ} [1 - \frac{r}{R}]$

Net force at $B : F_{B} = \sqrt{f^{2} + N_{2}^{2}}$

For minimum value of $μ : f \leq {μN}_{2}$

$\Rightarrow \frac{mgr}{R} \leq \frac{μmg}{\tan θ} [1 - \frac{r}{R}] \Rightarrow μ \geq \frac{\tan θ}{R / r - 1}$

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