A liquid having surface tension $S$ is forming a film between a horizontal surface and a cylindrical object of radius $R,$ as shown in figure. Cylindrical liquid surface has radius $r$ The contact angle is $\theta =0°.$  Find the downward force (by atmosphere + liquid) acting on unit length of the cylinder due to presence of the liquid. Height of liquid column above horizontal surface is greater than r.

From the figure
 $\cos \alpha =\frac{R-r}{R+r}$
Consider a length $L$ of the cylinder. Force due to surface tension acts tangentially and has a vertical resultant given by
$2SL.\sin \alpha$ 
 
 Pressure inside the liquid (at 1) is less than atmospheric pressure by  $\Delta P=\frac{S}{r}$
Pressure difference causes an additional downward force on the cylinder given by

$\Delta P[ \text{Rectangular area\hspace{0.17em}}2R\sin \alpha .L]=\frac{2SRL}{r}\sin \alpha$

$\therefore$  Net downward force due to presence of liquid is
$F_{net}=2SL\sin \alpha +\frac{2SRL}{r}\sin \alpha$ 
Force per unit length is
$$\begin{gathered} \frac{F_{net}}{L}=2S[1+\frac{R}{r}]\sin \alpha \\ =2S\left[\frac{r+R}{r}\right]\sqrt{1-{\left(\frac{R-r}{R+r}\right)}^2}\left[\text{Using\hspace{0.17em}}\left(\text{i}\right)\right] \\ =4S\sqrt{\frac{R}{r}} \end{gathered}$$