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.

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))`