Question

A solid cylinder of $m$ is placed touching a rough step of height $\frac{R}{4}$. A horizontal force F is applied at top point so that cylinder can roll over step without sliding.

• A. Minimum force required so that cylinder climbs over step is $\frac{mg}{\sqrt{7}}$.
• B. For minimum force F, friction acting on cylinder is $\frac{mg}{\sqrt{7}}$.
• C. For minimum force F, normal reaction on cylinder is $mg$.
• D. Friction on cylinder is acting in horizontal direction.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Minimum force required so that cylinder climbs over step is $\frac{mg}{\sqrt{7}}$.
B For minimum force F, friction acting on cylinder is $\frac{mg}{\sqrt{7}}$.
C For minimum force F, normal reaction on cylinder is $mg$.


$\cos \theta =\frac{3}{4}$ and $\sin \theta =\frac{\sqrt{7}}{4}$

By torque equilibria about P

$F_m\times \frac{7R}{4}=mg\sin \theta \times R$

$F_m\times \frac{7}{4}=mg\frac{\sqrt{7}}{4}\Rightarrow F_m=\frac{mg}{\sqrt{7}}$

By force equilibria is

$F_m+fcos\theta =Nsin\theta \dots \dots \dots \left(1\right)\Rightarrow \frac{mg}{\sqrt{7}}+f\times \frac{3}{4}=\frac{\sqrt{7}}{4}N$

$Ncos\theta +fsin\theta =mg\dots \dots \dots \left(2\right)\Rightarrow \frac{3}{4}N+\frac{\sqrt{7}}{4}f=mg$

So by (1) and (2) $\Rightarrow N=\sqrt{7}f$

So $\frac{3}{4}\sqrt{7}f+\frac{\sqrt{7}}{4}f=mg\Rightarrow \frac{4\sqrt{7}f}{4}=mg$

$f=\frac{mg}{\sqrt{7}}$ and $N=mg$