Question

A cylinder of weight $W$ and radius $R$ horizontal step of height $h$ as shown in fig. A rope is wrapped around the cylinder and pulled horizontally with force $F$. Assuming the cylinder does not slip on the step, the minimum force $F$ necessary to raise the cylinder is given by $\frac{W\sqrt{xRh-h^2}}{(xR-h)}$. Find the value of $x$.

Answer 1

When its just about to lift , normal at the bottom surface is zero .

Let the normal at the point of contact at a height h be N and let the angle made by it with horizontal be $\theta$ (the force passes through the centre ) .

Writing vertical force equation :

$Nsin\left(\theta \right)=W$

$N\frac{R-h}{R}=W$

$N=\frac{RW}{R-h}$

Writing horizontal force equation :

$F=Ncos\left(\theta \right)$

$F=N\frac{\sqrt{R^2-(R-h{)}^2}}{R}$

$F=\frac{RW}{R-h}\frac{\sqrt{R^2-(R-h{)}^2}}{R}$

$=W\frac{\sqrt{2Rh-h^2}}{R-h}$

$\therefore x=2$