Determine the minimum coefficient of friction `mu_("min")` between as thin homogeneous rod and a floor at which a person can slowly lift the rod the floor without slipage to the vertical position, applying to it a force perpendicular to it.

Answer

`1/(2sqrt(2))`

Explanation

Consider a general position of the rod as shown in the figure. Let the applied force be `F`. As the end is moved slowly, some rod has no acceleration. Applying Newton's law in `x`-direction, `f-Fsintheta=0`

`f=Fsintheta`
In `y`-direction `Fcostheta+n=mg` ......iii
Making `tau_(A)=0impliesFl-mg1/2costheta=0`
`F=(mgcostheta)/2`.........iii
For no slipping `flemuNimpliesmugef/N`
`muge(Fsintheta)/(mg-fcostheta)impliesmuge(sinthetacostheta)/(2-cos^(2)theta)`
`mu_("min")=((sinthetacostheta)/(2-cos^(2)theta))=1/(2sqrt(2))`