Question

The minimum coefficient of friction ${\mu }_{min}$ between a thin homogeneous rod and a floor at which a person can slowly lift the rod from the floor without slippage to the vertical position, applying to it a force perpendicular to it is $\frac{1}{x\sqrt{x}}$. Find the value of $x$.

Answer 1

We consider the equilibrium of the rod when it is inclined at an angle $\alpha$ with the horizontal. The forces acting on the rod are shown in the figure.
We consider torque about the point of intersection of the force of gravity, mg; and the force F applied by the person perpendicular to the rod. Since the torque of these forces about this point is zero, it eliminates the unknown force F.
$\sum \tau$ = $Nl\cos \alpha -F_{fr}l\left(\frac{1}{\sin \alpha }+\sin \alpha \right)$ = 0 - (i)
Moment arm for $N$ is $l\cos \alpha$, while for friction is
$\frac{l}{\sin \alpha }+l\sin \alpha$
From equation (i), we get
$F_{fr}$ = $N$ $\frac{\cos \alpha \sin \alpha }{1+si{n}^2\alpha }$
$=N\frac{\cos \alpha \sin \alpha }{2si{n}^2\alpha +co{s}^2\alpha }$
$=N\frac{1}{2\tan \alpha \cot \alpha }$
As the force of friction cannot exceed $\mu N$, we have
$\mu$ $\ge$ $\frac{1}{2\tan \alpha +cot\alpha }$
$\mu$ is minimum when $2\tan \alpha +\cot \alpha$ is maximum
$\frac{d}{d\alpha }(2\tan \alpha +\cot \alpha )=0$
$\Rightarrow$ $2{\sec }^2\alpha -cose{c}^2\alpha =0$
$\tan \alpha$ = $\frac{1}{\sqrt{2}}$
Thus the required coefficient of friction of ${\mu }^{}$ = $\frac{1}{2\sqrt{2}}$