| Fig. 9. |
To determine the relation between the surface-tension and the pressure which balances it when the form of the surface is not spherical, let us consider the following case:—
Let fig. 9 represent a section through the axis Cc of a soap-bubble in the form of a figure of revolution bounded by two circular disks AB and ab, and having the meridian section APa. Let PQ Non-spherical soap bubble. be an imaginary section normal to the axis. Let the radius of this section PR be y, and let PT, the tangent at P, make an angle a with the axis.
Let us consider the stresses which are exerted across this imaginary section by the lower part on the upper part. If the internal pressure exceeds the external pressure by p, there is in the first place a force πy²p acting upwards arising from the pressure p over the area of the section. In the next place, there is the surface-tension acting downwards, but at an angle a with the vertical, across the circular section of the bubble itself, whose circumference is 2πy, and the downward force is therefore 2πyT cos a.
Now these forces are balanced by the external force which acts on the disk ACB, which we may call F. Hence equating the forces which act on the portion included between ACB and PRQ
πy²p − 2πyT cos α = −F (9).
If we make CR = z, and suppose z to vary, the shape of the bubble of course remaining the same, the values of y and of a will change, but the other quantities will be constant. In studying these variations we may if we please take as our independent variable the length s of the meridian section AP reckoned from A. Differentiating equation 9 with respect to s we obtain, after dividing by 2π as a common factor,
| py | dy | − T cos α | dy | + Ty sin α | dα | = 0 (10). |
| ds | ds | ds |
Now
| dy | = sin α (11). |
| ds |