A man is running along a road with speed u.  On his chest there is a paper of mass m and area  S. There is a wind blowing against the man at 
speed V. Density of air is $\rho$. Assume that the air  molecules after striking the paper come to rest  relative to the man. Find the minimum coefficient 
of friction between the paper and the chest so that  the paper does not fall?

Volume of air striking the paper in unit time is = S (V + u)  
Mass of air striking the paper in unit time = $\rho$S (V+ u) 
In reference frame of man, the air molecules strike at a speed (V + u) and come to rest.
$\therefore$Rate of charge of momentum for air particles $=\mathrm{\rho S}(\mathrm{V}+\mathrm{u})(\mathrm{V}+\mathrm{u})$
$\therefore$Force on paper due to air $=\mathrm{\rho S}(\mathrm{V}+\mathrm{u}{)}^2$
For the paper to not fall, friction on it must balance its weight
$$\begin{gathered} {\mathrm{f}}_{max}\ge \mathrm{mg} \\ \mathrm{\mu \rho S}(\mathrm{V}+\mathrm{u}{)}^2\ge \mathrm{mg} \\ \mathrm{\mu }\ge \frac{\mathrm{mg}}{\mathrm{\rho S}(\mathrm{V}+\mathrm{u}{)}^2} \end{gathered}$$