Originally Posted by

**TriKri** Hello!

What I have understood about surface tension is that it lays a pressure on the surface in proportion to the bending of it, the angular change per meter surface , which has the unit $\displaystyle [rad/m^{-1}]=[m^{-1}]$. And pressure has the unit $\displaystyle N/m^2$. So surface tension would have the unit $\displaystyle \left[\frac{N/m^2}{m^{-1}}\right]=[N/m]$, which it has. Now, is it so that the pressure applied to the surface is given by

$\displaystyle P=S\cdot\frac{\delta^2 y}{\delta x^2}$

Where S is the surface tension, and x and y have the unit $\displaystyle [m]$, so $\displaystyle \frac{\delta^2y}{\delta x^2}$ has the unit $\displaystyle [m^{-1}]$. $\displaystyle \frac{\delta^2 y}{\delta x^2}$ is the same as $\displaystyle \frac{\delta\alpha}{\delta x}$, where $\displaystyle \alpha$ is the angle ($\displaystyle \alpha = \frac{\delta y}{\delta x}$ for small $\displaystyle \alpha$)