# Homework Help: Surface tension of a torus raindrop

1. Nov 25, 2012

### peripatein

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 Files:

• ###### Torus.JPG
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2. Nov 25, 2012

### Staff: Mentor

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).

3. Nov 25, 2012

### 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?

4. Nov 25, 2012

### Staff: Mentor

Of course. That should be the origin of 2pi*R*gamma.
You get a layer of fluid/air in contact there.

5. Nov 25, 2012

### peripatein

Okay. Thank you.