- #36

Hak

- 709

- 56

I don't think so. If we use Laplace-Young equation, we have ##\Delta P = \alpha \left(\frac{1}{R_1} +\frac{1}{R_2} \right)##, where ##\Delta P## is the pressure difference across the interface and ##R_1 = \frac{\mathrm{d}x}{\mathrm{d}(\sin \theta)}## and ##R_2 = \frac{x}{\sin \theta}## radii of curvature normal to the surface.haruspex said:Aren't they both the radius of the cut disc?

Am I wrong in thinking that, and is the matter simpler than that? Perhaps yes, it is as you say, ##F = \pi x \alpha##.