distance from that point, is equal to the product of the sum of the bodies multiplied by the distance of their common centre of gravity C from the same point P. That is, PA. A + PB. B = PC . A + B. FOR, by the 38th, CA. A = CB. B, that is, - PB. B, therefore, = PC PA PC. A by adding PA. A+ PB. B = PC. A + B. A Corollary. 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 + B placed at the centre C. PROP. XVI. C Corollary. 2. The same is also true of any number of bodies whatever, as will appear by cor. 4, prop. 38, namely, PA. A + PB . B + PD . D, &c= PC. A + B + D, &c., where P is any point whatever in the line AC. And, by cor. 5, prop. 38, 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 that is, equal to the sum of all the forces Corollary. 3. And if a plane pass through the point P perpendicular to the line CP; then the distance of the common centre of gravity from that plane, is PC= Pa. A+ Pb. B+ Pd. D, &c. A+B+D, &c. 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 the particles A, B, D, &c. 62. 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 lies also in the line CE. of intersection G. B B A And consequently it is in their common point Corollary. The distance of the point G, is AG = AD, and CG = {CE; or AG 2GD, and CG = 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, CBF are also equiangular, and CB being = 2CD, therefore BF = 2GD. But BF is also = AG; consequently AG = 2GD or AD. In like manner, CG = 2GE or 2CE. PROP. XVII. 63. To find the centre of gravity of a trapezium. DIVIDE the trapezium ABCD into two triangles, by the diagonal BD, and find E, F, the centres of gravity of these two triangles; 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: GF: triangle BCD : triangle ABD, then G will be the centre of gravity of the trapezium. B PROP. XVIII. A 64. 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 I of these two triangles be also found; then the centre of gravity of the trapezium will also lie in the line HI. So that, lying in both the lines, EF, HI, it must necessarily lie in their intersection G. AG: GI. AB: KI:: AH: KH:: 3:1 .. AG: AI:3:4 or AG AI. E 65. And thus we are to proceed for a figare of any greater number of sides, finding the centres of their component triangles and trapeziums, and then finding the common centre of every two of these, till they be all reduced into one only. 66. To find the centre of gravity of a triangular pyramid. Let ABCD be a triangular pyramid, and to the point of bisection of DC draw AH, BH. Take HK=HA and HI+HB; then K and I will be the centres of gravity of the surfaces ACD and BCD respectively. Join KI, AI, BK. Now if the pyramid be resolved into elements, by means of planes parallel to BCD, it is evident that the line AI must pass through the centre of gravity of the pyramid, since I is the centre of gravity of BCD. For the same reason, the centre of gravity of the pyramid is in the line BK, and because AI and BK are in one plane, the centre of gravity of the pyramid ABCD must be at G, the point of intersection of AI and BK. By similar triangles AGB and KGI, we have H K G Q C B Cor. 1. Bisect AB in P and join HG and GP; then if PQ be drawn parallel to AI, we have BQ=QI=IH; but AI=2PQ and AI=4GI; hence PQ=2GI; and therefore HI : HQ : : IG : PQ; whence HGP is a straight line. Cor. 2. Hence the centre of gravity of a triangular pyramid is the middle of the line joining the points of bisection of any two edges that do not meet. Cor. 3. A solid bounded by plane surfaces may be divided by planes into a number of triangular pyramids; and if a plane be drawn parallel to the base, at a distance equal to 4 of the altitude of the pyramid, then the centre of gravity of the whole pyramid must be in this plane, for that of each of the triangular pyramids is in this plane. Hence, the line joining the vertex of the pyramid and the centre of gravity of its base will cut the plane in the centre of gravity of the whole pyramid. 67. To find the centre of gravity of any body, or system of bodies. Let v1, V2, V3, &c., denote the volumes of the material particles which compose the body or volume V; and x1, Y1, Z1; X2, Y2, Z2, &c.; their co-ordinates in reference to three rectangular axes; then if X, Y, Z denote the co-ordinates of the centre of gravity, we have (by Prop. XV., Cor. 3.) PROP. XIX. But to adapt these expressions to computation, we shall introduce the principles of the Differential and Integral Calculus, and then the preceding expressions will take the form X = fxdv Y = Sydv ; 2= Jdv' where x, y, z denote the distances of the centre of gravity of du from the three rectangular planes. fzdv By means of these three equations, the determination of the centre of gravity may be effected; and when the figure is a plane surface, two of these equations are only required, since the centre of gravity is in the plane. I. When the figure is a plane curve. Here dv differential of the arc = √dx2 + dy2 = ds Y = Jyds S and if the arc be symmetrical on each side of the axis of x, we have y = 0, and then we have simply X .. X = fxds; S II. When the figure is a plane surface. X = √xds S differential of the arca = ydx; hence And if the area is symmetrical on each side of the axis of x, we need only one equation, viz. III. For a surface of revolution round the axis of x. Here one equation only is necessary; and since dv = differential of the surface 2 yds, we have IV. For a solid of revolution round the axis of x. Here dv differential of the volume = ry2dx, and hence X = xd We shall now apply these formulæ to a few examples. EXAMPLES. 68. Ex. 1. To find the centre of gravity of a circular arc. Here the curve is symmetrical on each side of the axis of x, and the equation is y2 = 2ax - x2; hence we have X = X = fxds S xdx √2ax-x2 ay ( (s y) = a S S that is, the distance of the centre of gravity of a circular arc from the vertex is a ay, and therefore the distance of the centre of gravity from the 8 centre = ay S = = = 3 4 = S = = Ex. 2. To find the centre of gravity of a cone. Let x1, x2 represent any two parts of the axis of the cone, measuring from the vertex, and y1, y2 the radii of the circular sections of the cone corresponding to the altitudes x1, x2; then, if x denote any variable part of the axis, and y the radius of the corresponding circular section, we have xx1y: Y1 y2. 2 xda y❜dx Y1⁄23 fx dx + y2 fx dx 2 3 fx3dx 4x4 3 = 4 3 X = X = Syxdx X = fxyds (x2 + x22) (x + x2) x2 + x1x2 + x22 3 4 X = x1 = = x= distance from vertex. 2 x12 + x ̧x2 + x 22 This expression gives the distance of the centre of gravity of the frustum of a cone from the greater end; hence, if R, r represent the radii of the greater and less ends of the frustum, and h its altitude, we have the distance h R2+2Rr + 3r2 ;; and when 70, we have 4 R2+ Rr + r2 of the centre of gravity: = · h the distance of the centre of gravity of a cone from its base = fourth of the altitude as found above. Ex. 3. Four bodies, whose weights are w1, w2, wз, w, pounds, are placed at the successive angles of a square whose side is 2a inches; required the position of their common centre of gravity, the square being considered without weight. W4 y Take O, the centre of the square, as the origin of co-ordinate reference, and the two rectangular axes parallel and perpendicular to the sides; then we have (w2+ W3 W4) -Wi X= and? a, y= w1 + w2 + w3 + W4 Thus if w=3, w2=4, w3=5, w1 = 6; 4 a. = w1 + wz W3 w1 + w2 + W3 + W4 and 2a = 12 inches; .. X=0 and y = —: 6=- ·14 = OG = distance of 18 centre of gravity below O on the axis of y. PROBLEMS FOR EXERCISE. 1 = one 69. Ex. 1. Find the centres of gravity of (1.) The common parabola and the paraboloid. (4.) The sector of a circle, and a spheric sector. WI G W2 Ex. 2. Two cones are placed with their equal bases in contact, and the altitude of the one is three times that of the other; find the position of their common centre of gravity. Ex. 3. The surface generated by a plane line or curve revolving about an axis in the plane of the figure, is equal to the product of the generating line or curve, and the path described by its centre of gravity. Ex. 4. The volume of the solid generated by the revolution of a plane figure about an axis in the plane of the figure, is equal to the product of the generating surface, and the path described by its centre of gravity. Ex. 5. From a given rectangle ABCD of uniform thickness, to cut off a triangle CDO, so that the remainder, ABCO when suspended at O, shall hang with AB in a vertical position. |