which, substituted in the equation above obtained, gives A result which might have been obtained at once without the aid of Art. 158 by supposing O to coincide with G. 165. COR. 2. If now, as in Art. 161, we suppose all the bodies equal and n in number, the last equation be Hence, in any system of n equal bodies, the sum of the squares of the lines joining their centres of gravity, two and two, is equal to n times the sum of the squares of the distances of those points from the centre of gravity of the system. 166. COR. 3. Consequently, in the case of the triangle (Art. 147), BC2 + AC2 + AB2 = 3. (AG2 + BG2 + CG3). Hence the sum of the squares of the sides of a triangle is equal to three times the sum of the squares of the dis tances of its angular points from its centre of gravity. 167. COR. 4. have In the case of the triangular pyramid we AB2 + AC2 + AD2 + BC2 + BD2 + CD2 = 4 (AG2 + BG2 + CG2 + DG3). Hence the sum of the squares of the six edges of a pyramid is equal to four times the sum of the squares of the distances of its angular points from its centre of gravity, 168. When a system of bodies is in equilibrium under the action of gravity only, the altitude of its centre of gravity is in general a maximum or a minimum. Let m1, m2, m ̧... be the particles of the system in equilibrium; 1, 2, 3... their respective altitudes above a fixed horizontal plane; the altitude of the centre of gravity above the same plane; g the accelerating force of gravity; then m1g, m2g, mg ... are the forces acting upon the particles of the system. Let now the system be disturbed in a manner subject to the same restrictions as were pointed out in the Chapter on virtual velocities, (i. e. rods must not be bent, cords must be kept of invariable length, contacts must not be broken, &c.) and let d1, dx, d... be the virtual velocities of the respective particles, then by Art. 113, m1g.dz1 + m2g.dz2 + måg.d%3+ 2 or Σ(mdx) = 0. But since Σm. z = Σ(mx); .. Σm.dz = Σ(mdx) = 0. = 0, Now dz is the differential of z, or second term of Taylor's Theorem, and this being equal to zero, it follows that is in general a maximum or minimum. It has been stated that the principle of virtual velocities extends only to quantities of the first order of smallness, that is, to the second term of Taylor's Theorem only; it does not appear therefore that (mdx) is in general equal to zero; the algebraic sign of dz will therefore decide whether z is a maximum or a minimum. 169. COR. Since the centre of gravity of the system is the point through which the resultant (mg), or g≥m of all the forces mig, meg... passes, and seeing that this resultant acts in a downward direction, it appears that, if the system be disturbed, the tendency of gravity is to make the centre of gravity descend: but if the constitution of the system be such that in passing out of a position of equilibrium the centre of gravity can only ascend, the ascent will be op posed by gravity; that is, gravity tends to bring the system back again into its position of equilibrium. But if the constitution of the system be such that in passing out of equilibrium the centre of gravity cannot but descend, it is assisted in its descent by gravity, and there is no tendency to return towards the position from which it set out. Hence it follows: (1) That if the altitude of the centre of gravity be a minimum, the system when disturbed will return towards the position from which it was disturbed. This is therefore called a position of stable equilibrium. (2) That if the altitude of the centre of gravity be a maximum, the system when disturbed will recede still farther from the position of equilibrium. This is therefore called a position of unstable equilibrium. (3) That if the centre of gravity neither ascend nor descend when the system is disturbed, it still continues in a position of equilibrium. This is therefore called a position of neuter equilibrium. 170. If a body be placed with its base upon a plane it will stand or fall according as a vertical through its centre of gravity falls within or without its base. Let AB (figs. 31, 32) be the base of the body, G its centre of gravity; draw a vertical through G meeting the plane on which the body is placed in H; H falling within the base in fig. 31, and without it in fig. 32. Every particle of the body is acted on by the force of gravity, and we have shewn that the centre of gravity is the point at which the resultant of the forces may be supposed to act this resultant is equal to their sum, that is, it is equal to (W) the whole weight of the body. We may therefore suppose the body to be without weight, and that a force acts at G equal to W. In fig. 31, we may suppose this force to be transmitted to H, which being in contact with a fixed point of the plane cannot be moved, and therefore W is counteracted, its effects being to make the body stand firm upon its base. But in fig. 32, W cannot be transmitted to a point which is in contact with the plane, and therefore as there is nothing to oppose its action, the point G will descend, thereby causing the body to turn about the point 4. 171. This reasoning applies if the plane on which the body is placed be not horizontal, provided the body be prevented from sliding by the roughness of the plane, or any equivalent cause. 172. If a body be placed on points, instead of a flat base, it will stand or fall according as a vertical through its centre of gravity falls within or without the polygon formed by passing a thread round the points. 173. If there be any case not here considered, it may be disposed of on the following principle. The whole weight of the body may be supposed to act at its centre of gravity; and as it acts in a downwards direction, its tendency is to cause that point to descend. If the geometrical arrangement of the system be such that it is impossible for it to move so as to permit the centre of gravity to descend, it will remain stationary; for in this case the tendency which gravity produces is prevented from taking effect by the construction of the machine. APPLICATION OF THE INTEGRAL CALCULUS TO FIND THE CENTRE OF GRAVITY OF BODIES. 174. To find the centre of gravity of a plane curve line. Let AB (fig. 33) be the curve line, referred to the rectangular axes Ox, Oy. P any point in AB, and Q very near to P. x = OM, dx = MN, y=MP, s= AP, ds = PQ, =3 u = the moment of the arc AP, and Su that of PQ, about Oy. = The moment of PQ about Oy is greater than it would be if PQ were all collected in a point at P; ... Su > xds; and it is less than if PQ were all collected at Q; •. Su< (x + dx) Ss. always lies between x and x + dx, consequently the integral to be taken from a OC to x = OD. = But if y be the co-ordinates of the centre of gravity of AP, we have by Art. 139, be 175. These formulæ will suffice for the determination of the point required in any given example: but it may remarked with respect to these, and other formula, which will be investigated for finding the centres of gravity of |