OF THE EQUILIBRIUM OF FLOATATION. 277 



consequently, by division, we obtain 



_mns' m(s' s) * 



mn (215). 



345. And from this equation we deduce the following rule. 



RULE. Multiply together, the magnitude of the body, the 

 number which expresses what part of it is immersed in the 

 lighter, and the specific gravity of the heavier fluid ; then, 

 from the product subtract the difference between the specific 

 gravities of the fluid drawn into the magnitude of the solid 

 body, and divide the remainder by the magnitude of the body t 

 multiplied by the number which expresses what part of it is 

 immersed in the lighter fluid ; then shall the quotient express 

 the specific gravity of the body. 



346. EXAMPLE. The specific gravities of two unrnixable fluids are 

 respectively 1.24 and 0.716, that of water being unity ; now, sup- 

 posing that when these fluids are poured into the same vessel, a body 

 of 2000 cubic inches which is in equilibrio between them, has one 

 seventh part of its magnitude immersed in the lighter fluid ; what is 

 the specific gravity of the body ? 



Here, by proceeding according to the rule, we have 



mns / = 2000x7Xl.24 17360 

 7M(5's):= 2000 (1.24 0.716) zz 1048 subtract 



difference zr 163 12; 

 consequently, by division, we shall obtain 



mns' m(s' s) 16312 



x s = r- - zz: 1.17 nearly. 



mn 2000X7 J 



347. From what has been done above, it is easy to ascertain what 

 will be the specific gravity of the body, when equal portions of it are 

 immersed in the lighter and in the heavier fluids ; for in that case, we 

 have n equal to 2, which being substituted in equation (215), gives 



*~s' J(*' s). (216). 



And the practical rule for reducing this equation may be expressed 

 in words at length, in the following manner. 



* This equation is susceptible of a simpler form, for by casting out the common 

 factor m, it is 



