260 OF THE EQUILIBRIUM OF FLOATATION. 



315. EXAMPLE 3. A cubic mass of oak, whose specific "gravity is 

 0.872, is placed in a cistern of water, and when it has attained a state 

 of equilibrium with its sides vertical, it stands 20 inches above the 

 surface of the fluid; what is the magnitude of the solid, the specific 

 gravity of the water being represented by unity ? 



In order to resolve this example, we shall first investigate a general 

 formula, which will apply to every case of a similar nature, when the 

 specific gravity of the fluid and that of the solid are given ; for which 

 purpose, 



Put x nr the length of one side of the solid ; then is 



x a* the length of that portion which is below the plane of 

 floatation. 



But by the principles of mensuration, the magnitude of the whole 

 body is 



razz a? 8 , 



and that of the part immersed, is 



m'= (x a) Xs'zz a: 8 ax* ; 



therefore, by the Proposition, we obtain 



m ' m! \ '. s' i s ; 

 and this, by substitution, becomes 



a; 8 : x* a a? 2 : : s' : s ; 



from which by equating the products of the extremes and means, 

 we get 



sz'zz s'O 8 as 2 ); 

 and by separating the terms, we obtain 



sx 9 zz 5' x 3 as' x 9 ; 

 consequently, by transposition and division, it is 



as' 



~7^7)' (202). 



And the practical rule supplied by this equation, may be expressed 

 in words at length in the following manner. 



RULE. Multiply the specific gravity of the fluid by the 

 height of the body above its surface, and divide the product 

 by the difference between the specific gravity of the fluid 

 and that of the solid, and the quotient will give the side 

 of the cube required. 



* The quantity a, is here put to denote the height of the body abovelhe fluid. 



