the bodies have the same effect on the line af, to turn it about the point c, whether they are placed at the points a, b, d, e,f, or in any part of the perpendiculars Aa, Bb, Dil, Ee, Ef. 102. PRop. If there be three or more bodies, and if a line be drawn from any one body D to the centre of gravity of the rest c; then the common centre of gravity E of all the bodies, divides the line cD into two parts in E, which are reciprocally proportional as the body D to the sum of all the other bodies. That is, cE : Ed : : D : A + B, &c. For, suppose the bodies A and B to be collected into the common C &c. TX Corol. Hence we have a method of finding the common centre of gravity of any number of bodies; namely, by first finding the centre of any two of them, then the centre of that centre and a third, and so on for a fourth, or fifth, &c. 103. Prop. If there be taken any point P, in the line passing through the centres of two bodies; then the sum of the two products, of each body multiplied into its distance from that point, is equal to the product of the sum of the bodies multiplied into the distance of their common centre of gravity c from the same point P. 104. Corol. 1. Hence, the two bodies A and B have the same force to turn the lever about the point P, as if they were both placed in c their common centre of gravity. Or, if the line, with the bodies, move about the point P; the sum of the momenta of A and B, is equal to the momentum of the sum s or A + n placed at the centre c. 105. Corol. 2. The same is also true of any number of bodies whatever, as will appear by cor. 4, art. 100. namely, PA. A + PB. B+ Pd . D, &c. =rc. (A+B+D, &c.) where P is in any point whatever of the line Ac. And, by cor. 5, art. 101, the same thing is true when the bodies are not placed in that line, but any where in the perpendiculars passing through the points A, B, D, &c.; namely, Pa. A + Pb. b + Pd. D, &c. = PC. (A + B + D, &c.) 106. Corol. 3. And if a plane pass through the point P perpendicular to the line cF ; then the distance of the common centre of gravity from that plane, is Pa . A + Pb. B  pal . D, &c. A + b + D, &c. T of all the moments divided by the sum of all the bodies. Or, if A, B, D, &c. be the several particles of one mass or compound body; then the distance of the centre of gravity of the body, below any given point P, is equal to the forces of all the particles divided by the whole mass or body, that is, equal to all the Pa. A, Pb. B, Pd. D, &c. divided by the body or sum of particles A, B, D, &c. 107. Prop. To find the centre of gravity of any body, or of any system of bodies. Through any point p draw a plane, and let Pa, Pb, Pd, &c. be the distance of the bodies A, B, D, &c. from the plane; then, by the last cor. the distance of the common centre of gravity from the plane, will be Pa . A + Pb . B  pp. D, &c. A + B + D, &c. 108. Or, if b be any body, and QPR any plane; draw PAh &c. perpendicular to QR, and through A, B, &c. draw innu. merable sections of the body b parallel P to the plane QR. Let's denote any one Q of these sections, and d=PA, or PB, &c. its distance from the plane Q.R. Then will the distance of the centre of gravity of the body from the plane be _ sum of ! the ds. And if the distance be thus found for two intersecting planes, they will give the point in which the centre is placed. 109. But the distance from one plane is sufficient for any regular body, because it is evident that, in such a figure, the centre of gravity is in the axis, or line passing through the centres of all the parallel sections. , that is, equal to the sum Wol. II, 24 Thus, if the figure be a parallelogram, or a P R cylinder, or any prism whatever ; then the i axis or line, or plane Ps, which bisects all the sections parallel to att, will pass through the T centre of gravity of all those sections, and E. consequently through that of the whole figure f c. Then, all the sections s being equal, and the body b = Ps. s, the distance of the cen tre will be po = S middle of the axis of any figure whose parallel sections are equal. 110. In other figures, whose parallel sections are not equal, but varying according to some general law, it will not be easy to find the sum of all the PA s, PB . so, PD. s.", &c. except by the general method of Flūxions; which case therefore will be best reserved till we come to treat of that doctrine. It will be proper, however, to add here some examples of another method of finding the centre of gravity of a triangle, or any other rightlined plane figure. 111. Prop. To find the centre of gravity of a triangle. From any two of the angles draw lines AD, ce, to bisect the opposite sides; so will their intersection G be the centre of gravity of the triangle. For, because AD bisects BC, it bisects also all its parallels, namely, all the parallel sections of the figure : therefore AD passes through the centres of gravity of all the parallel sections or component parts of the figure; and consequently the centre of gravity of the whole figure lies in the line AD. For the same reason, it also lies in the line ce. Consequently it is in their common point of intersection G. 112. Corol. The distance of the point G, is AG = 3AD, and cq = 3cF : or AG = 2GD, and cq = 2GE. For, draw Bf parallel to AD, and produce ce to meet it in F. Then the triangles AEG, BEF are similar, and also equal, because AE = BE; consequently AG = BF. But the triangles cDG, car are also equiangular, and cb being == 2cd, therefore BF = 2GD. But BF is also = AG ; consequently AG = 2Gd or 4AD. In like manner, cq = 2GE or #CE. 113. Prop. To find the centre of gravity of a trapezium. Divide the trapezium Abcd into two A. triangles, by the diagonal BD, and find E, F, the centres of gravity of these two o: : then shall the centre of gravity of the trapezium lie in the line EF connecting them. And therefore if EF be divided, in G, in the alternate ratio of the two triangles, namely, Eg: Gr:: triangle BCD : triangle ABD, then G will be the centre of gravity of the trapezium. 114. Or, having found the two points E, F, if the trapezium be divided into two other triangles BAc, DAC, by the other diagonal Ac, and the centres of gravity H and 1 of these two triangles be likewise found ; then the centre of gravity of the trapezium will also lie in the line H1. So that, lying in both the lines, EF, HI, it must necessarily lie in their intersection G. 115. And thus we are to proceed for a figure of any greater number of sides, finding the centres of their component triangles and trapeziums, and then finding the com: mon centre of every two of these, till they be all reduced into one only. ProbleMS For exeRCISE. 1. Find, geometrically, the centre of gravity of a trapezoid. 2. Find, geometrically, the centre of gravity of a triangular pyramid. 3. Infer, thence, the centre of gravity of any pyramid. 4. Find, algebraically, the centre of gravity of the frus. tum of a pyramid. 5. Let a sphere whose diameter is 4 inches, and a cone whose altitude is 8 inches, and diameter of its base 3 inches, be fastened upon a thin wire which shall pass through the centre of the globe and the axis of the cone ; let the vertex of the cone be toward the sphere, and let its distance from the sphere's surface be 12 inches. Required the place of their common centre of gravity. 6. Demonstrate 1st, That the surface produced by a plane line or curve by revolving about an axis in the plane of that curve, is equal to the product of the generating line or curve into the path described by the centre of gravity. And 2dly. That the solid produced by the revolution of a plane figure about an axis posited in the plane of that figure, is equal to the product of the generating surface into the circumstance described by the centre of gravity. ON THE EQUILIBRIUM OF ARCHES. 116. A very interesting department of the science of Statics, is that which relates to the stability of arches, as introduced in the construction of bridges, powdermagazines, &c. Every such structure is a system of forces, and the examination of its firmness, therefore, requires the application of the general principles of equilibrium. We shall here present a #. useful propositions in elucidation of the more received theories. 117. Prop. The force of a voussoir depending on the magnitude of the angle formed by its sides, the impelling force, and the resistance to be overcome, is on the first account directly as the radius of curvature of the arch at that point, on the second as the square of the sine of the angle included between the tangent of the curve at the given point and the vertical passing through that point, and on the third, as the sine of the same angle. 1. Let EAFF, eabf, be two similar concentric curves, and AB, ab, two voussoirs similarly situated, whose sides perpendicular to the curve converge to the centre c. The forces of these voussoirs considered as portions of wedges, are inversely as the sines of the half vertical angles (schol. art. 82.) or, because each wedge occupies an equal portion of its respective arch, directly as the radii of curvature. 2ndly, Let Hh be the invariable breadth of the voussoirs on the arch calf, Gghh the incumbent weight, which, since GH is supposed given, is as the breadth hk, or as the sine of the angle hilk : by the resolution of the force gh into two him, *k, the latter is the force impelling the voussoir to split the
