# Surface tension of a torus raindrop

by peripatein
Tags: raindrop, surface, tension, torus
 P: 818 1. The problem statement, all variables and given/known data When calculating the difference in pressure inside a spherical raindrop, the force exerted by the surface tension is calculated to be 2pi*R*gama, where R is the radius of the drop and gamma is dE/dS (dyne/cm). When the shape of the raindrop is said to be that of a torus, the force exerted by the surface tension is calculated to be 2pi*R*gamma + 2pi(R+2r)*gamma (please see attachment). My question is simply why does the force in the case of the torus have two components, whereas in the case of a sphere it has only one? 2. Relevant equations 3. The attempt at a solution Attached Thumbnails
 Mentor P: 11,840 In both cases, the expression is the length of the boundary (as seen in a 2D-projection) multiplied by gamma. A circle has one boundary, the torus has 2 (inner+outer).
 P: 818 But isn't the circle's circumference considered a boundary? Or shouldn't it be? Is it not a thin layer of fluid verging on air?
Mentor
P: 11,840
Surface tension of a torus raindrop

 But isn't the circle's circumference considered a boundary?
Of course. That should be the origin of 2pi*R*gamma.
You get a layer of fluid/air in contact there.
 P: 818 Okay. Thank you.

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