A solid cylinder of mass 3 kg rests on a rough horizontal plane. A horizontal force F is applied to the top most point of the cylinder as shown in the figure. Coefficient of friction between the cylinder and the surface is 0.6. Then the maximum value of F for which the cylinder will roll without slipping is

Frictional force f is given by

$\mathrm{f}=\mathrm{F}\left(1-\frac{\mathrm{h}}{\mathrm{rp}}\right)$

Where ${\mathrm{r}}_{\mathrm{P}}={\mathrm{r}}_{\mathrm{C}}+\frac{{\mathrm{I}}_{\mathrm{C}}}{{\mathrm{mr}}_{\mathrm{C}}}$

${\mathrm{r}}_{\mathrm{C}}=$ Distance of centre of mass from instantaneous axis of rotation O.

For solid cylinder, ${\mathrm{r}}_{\mathrm{P}}=\mathrm{R}+\frac{\frac{1}{2}{\mathrm{mR}}^2}{\mathrm{m}.\mathrm{R}}=\frac{3\mathrm{R}}{2}$

$\therefore \mathrm{f}=\mathrm{F}\left(1-\frac{\mathrm{h}}{{\mathrm{r}}_{\mathrm{P}}}\right)=\mathrm{F}\left(1-\frac{2\mathrm{R}}{3/2\mathrm{R}}\right)=-\frac{\mathrm{F}}{3}.$

Since ‘f’ is negative, it must act in the forward direction. Now ${\mathrm{f}}_{\max }=\mathrm{\mu mg}$

No slipping takes place if $\left|\mathrm{f}\right|<\mathrm{\mu mg}$

i.e., if $\frac{\mathrm{F}}{3}<\mathrm{\mu mg}$

$\therefore {\mathrm{F}}_{\max }=3\mathrm{\mu mg}=3\times 0.6\times 3\times 10\mathrm{N}=54\mathrm{N}$