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 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
[pi]y^2p 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 2[pi]y, and the
downward force is therefore 2[pi]yT 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
[pi]y^2p - 2[pi]yT cos[alpha] = -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 2[pi] as a common
factor,
dy dy d[alpha]
py -- - T cos[alpha] -- + Ty sin[alpha] -------- = 0 (10).
ds ds ds