Question

Determine the minimum coefficient of friction between a thin uniform rod and a floor at which a person can slowly lift the rod from the floor without slipping, to the vertical position applying to its end and a force always perpendicular to its length.

Answer 1

From FBD,

$Fl=\frac{mgl}{2}$ $\Rightarrow F=\frac{mg}{2}$

$Fcos\theta =N+mg$

$\frac{mg}{2}cos\theta +N=mg$

$N=mg(1-\frac{cos\theta }{2})$ eq(1)

$Fsin\theta =f$

$\frac{mg}{2}sin\theta =\mu N$ eq(2)

From eq(1) and eq(2)

$\mu =\frac{sin\theta }{2-cos\theta }$

Maximum of above function is $\frac{1}{\sqrt{3}}$

So, minimum coefficient of friction required is $\frac{1}{\sqrt{3}}$