If the drop is cut in half the force imposed by surface tension on the half which is remaining will be 2πRS. Now consider a spherical surface, specifically the spherical drop. Note that it makes no difference whether the liquid is on the outside or on the inside. ![]() ![]() ![]() Thus the surface tension causes greater pressure inside the surface and the difference is the surface tension divided by the radius of curvature (so that a flat surface yields no pressure difference). Opposing this will be the surface tension components that are known to force S acting on the two ends of the section of surface which yield a force from the core, equal to 2 sin dθS. By the result described in static forces the net force in the outward direction due to forces near and at a fluid surface, pO and pI will be 2R sin dθ(pI − pO). Consider all the forces in the direction normal to the center of the fluid element.
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