Page:Scientific Papers of Josiah Willard Gibbs.djvu/327

Rh equilibrium. But it is evident from the points of application and directions of these forces that they cannot have a zero resultant unless each force is zero. We may therefore introduce a fourth mass $$D$$ without disturbing the parts which remain of the surfaces $$\text{A-B, B-C, C-D}$$.

It will be observed that all the angles at $$a, b, c$$, and $$d$$ in figure 14 are entirely determined by the six surface-tensions $$\sigma_{AB}, \sigma_{BC}, \sigma_{CA}, \sigma_{DA}, \sigma_{DB}, \sigma_{DC}$$. (See (615).) The angles may be derived from the tensions by the following construction, which will also indicate some important properties. If we form a triangle $$\alpha \beta \gamma$$ (figure 15 or 16) having sides equal to $$\sigma_{AB}, \sigma_{BC}, \sigma_{CA}$$, the angles of the triangle will be supplements of the angles at $$d$$. To fix our ideas, we may suppose the sides of the triangle to be perpendicular to the surfaces at $$d$$. Upon $$\beta \gamma$$ we may then construct (as in figure 16) a triangle $$\beta \gamma \delta '$$ having sides equal to $$\sigma_{BC}, \sigma_{DC}, \sigma_{DB}$$, upon $$\gamma \alpha$$ a triangle $$\gamma \alpha \delta $$ having sides equal to $$\sigma_{CA}, \sigma_{DA}, \sigma_{DC}$$ and upon $$\alpha \beta$$ a triangle $$\alpha \beta \delta '$$ having sides equal to $$\sigma_{AB}, \sigma_{DB}, \sigma_{DA}$$. These triangles are to be on the same sides of the lines $$\beta \gamma, \gamma\alpha, \alpha \beta$$, respectively, as the triangle $$\alpha \beta \gamma$$. The angles of these triangles will be supplements of the angles of the surfaces of discontinuity at $$a, b$$, and $$c$$. Thus $$\beta\gamma\delta ' = dab$$, and $$\alpha\gamma\delta '' = dba$$. Now if $$\delta '$$ and $$\delta $$ fall together in a single point $$\delta$$ within the triangle $$\alpha\beta\gamma$$, $$\delta '$$ will fall in the same point, as in figure 15. In this case we shall have $$\beta\gamma\delta + \alpha \gamma\delta = \alpha \gamma \beta$$, and the three angles of the curvilinear triangle $$adb$$ will be together equal to two right angles. The same will be true of the three angles of each of the triangles $$bdc, cda$$, and hence of the three angles of the triangle $$abc$$. But if $$\delta ', \delta , \delta '$$ do not fall together in the same point within the triangle $$\alpha \beta \gamma$$, it is either possible to bring these points to coincide within the triangle by increasing some or all of the tensions $$\sigma_{DA}, \sigma_{DB}, \sigma_{DC}$$, or to effect the same result by diminishing some or all of these tensions. (This will easily appear when one of the points $$\delta ', \delta , \delta$$ falls within the triangle, if we let the two tensions which determine this point remain constant, and the third tension vary. When all the points $$\delta ', \delta , \delta$$ fall without the triangle $$\alpha \beta \gamma$$, we may suppose the greatest of the tensions $$\sigma_{DA}, \sigma_{DB}, \sigma_{DC}$$—the two greatest, when these are equal, and all three when they all are equal—to diminish until one of the points $$\delta ', \delta , \delta$$ is brought within the triangle $$\alpha \beta \gamma$$.) In the first case we may say that the tensions of the new surfaces are too small to be represented by the distances of an internal point from the vertices of the triangle representing the tensions of the original surfaces (or, for brevity, that they are too small to be represented as in figure 15); in the second case we may say that they are too great to be thus represented. In the first case, the sum of the angles in each of the triangles $$adb, bdc, cda$$ is less than two right angles (compare figures 14 and 16);