OF THE EQUILIBRIUM OF FLOATATION. 265 



and this, by division, becomes 



*=%> 



and finally, by extracting the square root, we get 



(204). 



325. The practical rule deduced from this equation is very simple ; 

 it may be expressed in words at length in the following manner. 



RULE. Divide the specific gravity of the solid, by that of 

 the fluid on which it floats ; then, multiply the square root 

 of the quotient by the axis of the body, and the product will 

 give the height of the part below the plane of floatation. 



Therefore, by retaining the data of the foregoing example, we shall 

 obtain as under, 



c i x = 29 V0.76 = 25.251 inches; 



being a difference of 10.46 inches, in the depths of immersion, for the 

 two cases of the question. 



PROBLEM XLVII. 



326. When a body floats in equilibrio on the surface of a 

 homogeneous fluid, it is a necessary condition, that the centre 

 of gravity of the body, and that of the fluid displaced, shall be 

 in the same vertical line. Supposing, therefore, that the equili- 

 brium is disturbed by the addition or subtraction of a certain 

 weight : 



It is required to determine, how far the body will be 

 depressed or elevated in consequence of the extraneous weight? 



The above problem will obviously admit of two cases, for a weight 

 may be added to a body, and it may be subtracted from it ; in the 

 one case the body will descend, and in the other it will ascend ; the 

 following general solution, however, wiH 

 answer both cases. 



Let AFBC represent a vertical section 

 of the solid body, floating in a state of 

 equilibrium on a fluid of which the hori- 

 zontal surface is K L ; and suppose, that 

 when the body is acted on by its own 

 weight only, the straight line D E is coin- 



