A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its center' O' perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is :
Option: 1 Mg\sqrt{1-\left ( \frac{R-a}{R} \right )^{2}}
 
Option: 2 Mg\sqrt{\left ( \frac{R}{R-a} \right )^{2}-1}
Option: 3 Mg\frac{a}{R}  
Option: 4 Mg\sqrt{1-\frac{a^{2}}{R^{2}}}

Answers (1)

For minimum force to topple

Let the net torque about point P tends to zero

i.e

\begin{array}{l} (\tau)_{P}=0 \\ F_{min}.R. -\mathrm{mgx}=0 \end{array}

And x=\sqrt{R^{2}-(R-a)^{2}}

So

 \begin{array}{l} \mathrm{F_{min}}=\mathrm{mg} \frac{\mathrm{x}}{\mathrm{R}} \\ \\ \mathrm{F_{min}}=\mathrm{mg} \sqrt{1-\left(\frac{\mathrm{R}-\mathrm{a}}{\mathrm{R}}\right)^{2}} \end{array}

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