90 Mr. J. Kam on Molecular Attraction and 



and every molecule in the " stationary motion " of Clausius 

 and let OL be the radius of the " sphere of attraction "" 

 for 0. 



Every molecule on the spherical surface RFL will be just 

 the last to attract 0, and each radius through a molecule of 

 that surface is the direction of a force attracting 0. 



Suppose k to be such a force, and some function of distance, 

 of the radius r, 



« =/(r). 



The horizontal components compensate each other, but the 

 vertical components of all forces k form the normal resultant 

 to which is subjected. 



If <f) is the angle between such a force k and OF, the normal 

 component of k is 



k = k . cos </> (29) 



All forces at the angle <p have this same component h and 

 lie in the surface of the cone with the angle </> in and go 

 through the curve of intersection with the sphere which is 

 the circle with the radius x. 



If n is the number of molecules per unit of distance, 



K = %k = 2ir . x .n . k . cos <£, . . . (30) 



or, as t i' = sin cf>, 



K = 2it .n . k . sin$ . cos 4> (31) 



A second cone with the angle (cj)-\-d(f>) cuts the sphere 

 a distance dh above the first. We obtain a second curve of 

 intersection with the radius (x-\-dx) parallel to the first one. 

 The surface between the two planes is for the height dh the 

 surface of a cone-frustrum, and equals 



d<t 

 dO =2II{x+(x+dx)}~= 2.Tl.xds. . . (32) 



The number of molecules in this surface is 



Z 1 = 2U.x.n 2 .ds (33) 



Through each of these molecules a force is directed, and 

 the vertical component is (ds = d(f>) 



dF = 2II.n 2 .sin(/>.cos</>.d(/> (34) 



It is evident that 



sin 2 <j) 



F = \dF = 2n.nK\d- 



