OF THB EQUILIBBIUM OF A SYSTEM. 187 



products of all the separate bodies, into their distances 

 from this plane, is equal to nothing: for the distances must 

 be in some given proportion to all the coordinates x, y, 

 and z, [depending on the properties of similar triangles 

 (117) and therefore ** linear functions", not involving their 

 squares; for example nx, ny, or n'^z: but when Swz^zrO, 

 it is obvious that I.mnx^.O, since n is constant;] -whence 

 the property of the plane passing through the centre of 

 gravity is evident. 



In order to determine the position of the centre of 

 gravity of any body, we may suppose X, Y, and Z to be 

 its coordinates with respect to any given origin, x, y, andz 

 being those of m, x\ y\ and / of »/, . . . , with respect to 

 the same point. We shall then have, from the equations 

 (o), OzzStw (x— X)[the X of those equations being supposed 

 to begin at the centre of gravity, and therefore answering 

 to a:— X here]; now S/wXzzXSm, Sm being the mass of 



?/Ynv 



the system; we have therefore Xzz-— — ; and in the same 



manner Y=---^, and Z= . It is also evident that 



2m Sm 



the coordinates X, Y", and Z, being thus completely deter- 

 mined by the magnitude and position of the separate bodies 

 of the system, they can only belong to a single point for 

 any one system of bodies at the same time. For the direct 

 distance of the centre of gravity we have the equation 



X2+1^ + Z2=i^ \. o -\ which may be 



transformed into 



^mm'\ {x'^xy^{y'^yf+{z'-^zY } 



