Question

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 $\mu$. The angle between two surfaces is $\theta$.

• A. the magnitude of force on the spool at B in order to maintain equilibrium is $mg\sqrt{(\frac{r}{R}{)}^2+(1-\frac{r}{R}{)}^2\frac{1}{ta{n}^2\theta }}$
• B. the magnitude of force on the spool at B in order to maintain equilibrium is $mg(1-\frac{r}{R})\frac{1}{tan\theta }$
• C. the maintain value of $\mu$ for the system to remain in equilibrium is $\frac{cot\theta }{\left(R/r\right)-1}$
• D. the maintain value of $\mu$ for the system to remain in equilibrium is $\frac{tan\theta }{\left(R/r\right)-1}$

Answer 1

Answer: (A), (D)

the maintain value of $\mu$ for the system to remain in equilibrium is $\frac{tan\theta }{\left(R/r\right)-1}$
the magnitude of force on the spool at B in order to maintain equilibrium is $mg\sqrt{(\frac{r}{R}{)}^2+(1-\frac{r}{R}{)}^2\frac{1}{ta{n}^2\theta }}$

$mgr=fR$ ....(i)

$N_{1}sin\theta +f=mg$ ....(ii)

$N_{1}cos\theta =N_2$ ....(iii)

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

From (ii) and (iii),

$N_2=\frac{mg-f}{tan\theta }=\frac{mg}{tan\theta }[1-\frac{r}{R}]$

Net force at B: $F_B=\sqrt{f^2+N_2^2}$

$F_B=mg\sqrt{(\frac{r}{R}{)}^2+(1-\frac{r}{R}{)}^2\frac{1}{ta{n}^2\theta }}$

For minimum value of $\mu :f\le \mu N_2$

$\Rightarrow \frac{mgr}{R}\le \frac{\mu mg}{tan\theta }[1-\frac{r}{R}]\Rightarrow \mu \ge \frac{tan\theta }{R/r-1}$