276 OF THE EQUILIBRIUM OF FLOATATION. 



343. In this example, a greater portion of the body is immersed in 

 the lighter fluid than what is immersed in the heavier ; but this cir- 

 cumstance manifestly depends upon the nature of the immersed body, 

 and the relation of the specific gravities ; for an instance may readily 

 be adduced, in which exactly the reverse conditions will obtain : 

 Thus, let the magnitude of the body and the specific gravities of the 

 fluids remain as above, and suppose the specific gravity of the body 

 to be 1.17 instead of 0.872 ; what then are the parts immersed in the 

 respective fluids ? 



The numerical process is represented as below. 



2000(1.241.17) 

 * = - (1.24- 0.716) = 267.175 cub.c. nc hes. 



Here then, we have 267.175 cubic inches for the portion which is 

 immersed in the lighter fluid, while that immersed in the heavier is 

 2000-267.175~1732.825 cubic inches; this agrees with the case 

 represented in the diagram, for there the body displaces a greater 

 quantity of the heavier than it does of the lighter fluid. 



PROBLEM L. 



344. Having given the specific gravities of two unmixable 

 fluids, and the magnitude of a solid body which floats in equili- 

 brio between them : 



It is required to determine the specific gravity of the solid, 

 so that any proposed part of it may be immersed in the 

 lighter fluid. 



Put m zr the magnitude of the immersed solid, the same as in the 



preceding Problem, 



s ~ the specific gravity of the lighter fluid, 

 s' the specific gravity of the heavier fluid, 

 x = the specific gravity of the solid body, being the required 



quantity, and 

 n n= the denominator of the fraction which expresses the part 



of the body immersed in the lighter fluid. 



Then, according to the principle demonstrated in Proposition VIII., 

 we shall obtain 



ms , (m m)s' 



mx= \-- , 



n n 



and from this, by a little farther reduction, we get 

 mnxmmns' m(s r s) ; 



