If the axes are rectangular, the direction-ratios become direction-cosines, so that λ2 + μ2 + ν2 = 1, whence
R2 = X2 + Y2 + Z2.
(6)
The conditions of equilibrium are X = 0, Y = 0, Z = 0.
§ 2. Statics of a System of Particles.—We assume that the mutual forces between the pairs of particles, whatever their nature, are subject to the “Law of Action and Reaction” (Newton’s Third Law); i.e. the force exerted by a particle A on a particle B, and the force exerted by B on A, are equal and opposite in the line AB. The problem of determining the possible configurations of equilibrium of a system of particles subject to extraneous forces which are known functions of the positions of the particles, and to internal forces which are known functions of the distances of the pairs of particles between which they act, is in general determinate. For if n be the number of particles, the 3n conditions of equilibrium (three for each particle) are equal in number to the 3n Cartesian (or other) co-ordinates of the particles, which are to be found. If the system be subject to frictionless constraints, e.g. if some of the particles be constrained to lie on smooth surfaces, or if pairs of particles be connected by inextensible strings, then for each geometrical relation thus introduced we have an unknown reaction (e.g. the pressure of the smooth surface, or the tension of the string), so that the problem is still determinate.
| Fig. 6. |
| Fig. 7. |
| Fig. 8. |
The case of the funicular polygon will be of use to us later. A number of particles attached at various points of a string are acted on by given extraneous forces P1, P2, P3 ... respectively. The relation between the three forces acting on any particle, viz. the extraneous force and the tensions in the two adjacent portions of the string can be exhibited by means of a triangle of forces; and if the successive triangles be drawn to the same scale they can be fitted together so as to constitute a single force-diagram, as shown in fig. 6. This diagram consists of a polygon whose successive sides represent the given forces P1, P2, P3 ..., and of a series of lines connecting the vertices with a point O. These latter lines measure the tensions in the successive portions of string. As a special, but very important case, the forces P1, P2, P3 ... may be parallel, e.g. they may be the weights of the several particles. The polygon of forces is then made up of segments of a vertical line. We note that the tensions have now the same horizontal projection (represented by the dotted line in fig. 7). It is further of interest to note that if the weights be all equal, and at equal horizontal intervals, the vertices of the funicular will lie on a parabola whose axis is vertical. To prove this statement, let A, B, C, D ... be successive vertices, and let H, K ... be the middle points of AC, BD ...; then BH, CK ... will be vertical by the hypothesis, and since the geometric sum of BA>, BC> is represented by 2BH>, the tension in BA: tension in BC: weight at B
as BA : BC : 2BH.
The tensions in the successive portions of the string are therefore proportional to the respective lengths, and the lines BH, CK ... are all equal. Hence AD, BC are parallel and are bisected by the same vertical line; and a parabola with vertical axis can therefore be described through A, B, C, D. The same holds for the four points B, C, D, E and so on; but since a parabola is uniquely determined by the direction of its axis and by three points on the curve, the successive parabolas ABCD, BCDE, CDEF ... must be coincident.