CENTER. 



x for the center of gravity of the part 

 D B E F, consequently the center of gra- 

 vity of the cone A B C G is distant from ' 

 the vertex B of the side B G, in a circle 

 parallel to the base. 



To find the center of gravity in a paral- 

 lelogram and parallelepiped, draw the 

 diagonal A D and E G (fig. 3,) likewise 

 C B and H F ; since each diagonal A I) 

 and C B divides the parallelogram A C D B 

 into two equal parts, each passes through 

 the center of gravity : consequently the 

 point of intersection, I, must be the cen- 

 ter of gravity of the parallelogram. In 

 like manner, since both the plane C B F H 

 and A D G E divide the parallelepiped 

 into two equal parts, each passes through 

 its center of gravity, so that the common 

 intersection I K is the diameter of gravi- 

 ty, the middle whereof is the center. 

 After the same manner may the center 

 of gravity be found in prisms and cylin- 

 ders, it being the middle point of the 

 right line that joins the center of gravity 

 of their opposite bases. 



The center of gravity of a parabola is 

 found as in the triangle and cone. Thus, 

 let B F in the parabola ABC (fig. 4) be 

 equal to x, D E = y, then will y x be the 

 fluxionary weight, and yxx the fluxion of 

 the momenta ; but from the nature of the 

 curve we have y = x$ ; whence yx = x$ 

 " 



x, and y xx = x*xx, whose fluent -x% 

 divided ^.rf the fluent of x*x will give - 



O 3 



3 



x = B F for the distance of the center of 



gravity from the vertex B in the part of 

 D B E ; and so J of B G is that center in 

 the axis of the whole parabola ABC 

 from the vertex B. 



The center of gravity in the human 

 body is situated in that part which is call- 

 ed the pelvis, or in the middle between 

 the hips. For the center of gravity of 

 segments, parabolics, conoids, spheroids, 

 &c: we refer to Wolfius. 



CENTER of gravity of two or more bodies, 

 a point so situated in a right line joining 

 the centers of these bodies, that, if this 

 point be suspended the bodies will equi- 

 ponderate and rest in any situation. In 

 two equal bodies it is at equal distances 

 from both : when the bodies are unequal, 

 it is nearer to the greater body, in pro- 

 portion as it is greater than the other ; or 

 the distances from the centers are in- 

 versely as the bodies. Let A (fig. 5,) be 

 greater than B, join A B, upon which 

 take the point C, so that C A : C B : : B : 



A, or that A X C A = B xCB; then is* 

 C the center of gravity of the bodies A 

 and B. If the center of gravity of three 

 bodies be required, first find C the center 

 of gravity of A and B ; and supposing a 

 body to be placed there equal to the sum 

 of A and B, find G the center of gravity 

 of it and D ; then shall G be the center of 

 gravity of the three bodies A, B, and D. 

 In like manner the center of gravity of 

 any number of bodies is determined. 



The sum of the products that arise by 

 multiplying the bodies by their respective 

 distances, from a right line or plane given 

 in position, is equal to the product of the 

 sum of the bodies multiplied by the dis- 

 tance of the center of gravity from the 

 same right line or plane, when all the 

 bodies are on the same side of it : but 

 when some of them are on the opposite 

 side, their products, when multiplied by 

 their respective distances from it, are to 

 be considered as negative, or to be sub- 

 ducted. Let I L, (fig. 6.) be the right 

 line given in position, C the center of 

 gravity of the bodies A and B ; A a, B b t 

 C c } perpendiculars to I L in the points 

 a, b, and c : then if the bodies A and B be 

 on the same side of I L we shall find A -j- 

 Aa-f6xB6 = A-f B X C c. For draw- 

 ing through C, the right line M N" paral- 

 lel to I L meeting A a in M, and B b in N, 

 we have A : B :: B C : A C by the proper- 

 ty of the center of gravity, and conse- 

 quently A : B :: B N : A M, or A x A M 

 = BxBN ; but A x A a + B X B 6= 

 AxCc-f A x AM-f-BxCc B xB 



N= AxCc-r-BxCc = A-r-BxCc. 

 When B is on the other side of the right 

 line I L (fig. 7,) and C on the same side 

 with A, then AxAa B X B =: A X 

 Cc+Ax AM B xBN-fBxCc== 

 A -}- B X C c -, and when the sum of the 

 products of the bodies on one side of I L, 

 multiplied by their distances from it, is 

 equal to the sum of the products of the 

 bodies multiplied by their distances on 

 the other side of I L, then C c vanishes, 

 or the common center of gravity of all 

 the bodies falls on the right line 1 L. 



Hence it is demonstrable, that when 

 any number of bodies move in right 

 lines with uniform motions, their com- 

 mon center of gravity moves likewise in 

 a right line with an uniform motion ; and 

 that the sum of their motions, estimated in 

 any given direction, is precisely the same 

 as if all the bodies in one mass were car- 

 ried on with the direction and motion of 

 their common center of gravity. 



CEVTETI of an hyperbota, a point in the 

 middle of the transverse axis. 



