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

Explanation

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 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 thee fores about this point is zero, it eliminates the unknown force `F`.
`sumtau=Nlcosalpha-F_("fr")l(1/(sinalpha)+sinalpha)=0`.......i
moment arm for `N` is `l cosalpha,` while for friction is
`l/(sinalpha)+lsinalpha`
From equation i we get
`F_("fr")=N(cosalphasinalpha)/(1+sin^(2)alpha)=N(cosalphasinalpha)/(2sin^(2)alpha+cos^(2)alpha)`
`=N1/(2tanalpha cotalpha)`
As the force on friction cannot exceed `muN`, we have `muge1/(2tanalpha+cotalpha)`
`mu` is minimum when `2tanalpha+cotalpha` is maximum,
`d/(dalpha)(2tanalpha+cotalpha)=0implies2sec^(2)alpha-cosec^(2)alpha=0`
`tanalpha=1/(sqrt(2))`
Thus the required coefficient of friction is `mu_("min")=1/(2sqrt(2))`