Page:Elementary Text-book of Physics (Anthony, 1897).djvu/114

100 expressed by $$2 \pi r T \cos \theta$$. This force, for each unit area of the tube, is $$\frac{2 \pi r T \cos \theta}{\pi r^2}\cdot$$ The downward force, at the level of the free surface, making equilibrium with this, is due to the weight of the liquid column (§ 113). If we neglect the weight of the meniscus, this force per unit area, or the pressure, is expressed by $$hdg$$, where $$h$$ is the height of the column and $$d$$ the density of the liquid. We have, accordingly, since the column is in equilibrium, $$\frac{2 \pi r}{\pi r^2} T \cos \theta = hdg$$; whence $$h = \frac{2T \cos \theta}{rdg}$$, and the height is inversely as the radius of the tube.

If the liquid rise between two parallel plates of length $$l$$, separated by a distance $$r$$, the upward force per unit area is given by the expression $$\frac{2l}{lr}T \cos \theta$$, and the downward pressure by $$hdg$$; whence $$h = \frac{2T \cos \theta}{rdg}$$, and the height to which the liquid will rise between two such plates is equal to that to which it will rise in a tube the radius of which is equal to the distance between the plates.

If the two plates be inclined to one another so as to touch along one vertical edge, the elevated surface takes the form of a rectangular hyperbola; for, let the line of contact of the plates be taken as the axis of ordinates, and a line drawn in the plane of the free surface of the liquid as the axis of abscissas, the elevation