min friction is ?????

well this is a small problem in statics

u see force is applied such that it is moved slowly

=> At every instant it is in equillibrium

let the angle of the rod be theta

then as F is applied perpendicular to the rod, Fsinθ=f

for coefficient of friction to be minimum F should be minimum(as it will pass through all angles)

so to produce the maximum torque F should be applied at the end

at an instant

F*2L=mgcosθL

F=mgcosθ/2

further for horizontal and vertical equillibrium

Fsinθ=f
and Fcosθ+N=mg
as f<=μN

solving we get μ>cosθsinθ/(1+sin²Î¸)

now for no slipping μ should be greater than this value for all angles upto 90

If we make μ greater than the maximum possible value of the expression then it will take care of all angles ..

so maximum value of cosθsinθ/1+sin²Î¸

= sinθ/cosθ/(sec²Î¸+tan²Î¸)

=(1/((1/tanθ)+2tanθ))

applying A.M>=G.M to denominator
we get minimum value as 1/2√2 not 1/√3 as u mentioned in the question

I am very sure that this is the answer .. check your question

oherwise maybe i understood your problem wrongly - but that will be evident from my method above so please check ..