Surface tension of a torus raindrop

[peripatein]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[mfb]

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).

 

[peripatein]

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?

 

[mfb]

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.

 

[peripatein]

Okay. Thank you.