OF THE EQUILIBRIUM OF FLOATATION. 285 



PROBLEM LIII. 



359. Having given the capacity or volume of the phial, the 

 whole weight of the aerometer, the specific gravity of the fluid, 

 and the radius of the wire : 



It is from thence required to determine, how much of the 

 stem or wire is immersed below the surface of the fluid when 

 the instrument rests in a state of equilibrium. 



By recurring to the equation marked (220), and separating the 

 terms, we obtain 



Trr*sl w cs; 



from which, by division, we get 



w cs 



T7T' (222). 



360. The practical rule for reducing this equation, may be expressed 

 in words at length, in the following manner. 



RULE. From the entire weight of the hydrometer, subtract 

 the capacity of the phial drawn into the specific gravity of 

 the fluid ; then, divide the remainder by the area of a trans- 

 verse section of the wire, drawn into the specific gravity of 

 the fluid, and the quotient will express how far the wire is 

 immersed below the upper surface of the fluid, when the 

 instrument floats in a state of equilibrium. 



361. EXAMPLE. The entire weight of an aerometer, when so adjusted 

 as to remain at rest in a fluid whose specific gravity is 0.5738*nr23 

 ounces ; what length of the stem or upright wire falls below the sur- 

 face of the fluid, supposing its diameter to be one twelfth of an inch, 

 and the magnitude of the immersed phial 40 inches ? 



Here, by the foregoing rule, we have 

 23 40x0.5738 

 '= 3^1416X^7x07673-8 = 15 ' 33 mcheS ** 



362. If the entire weight of the aerometer be multiplied by 1728, 

 the number of cubic inches in one cubic foot, the formulas (221) and 

 (222) become transformed into 



* The number 0.5738, by which the specific gravity is here expressed, is the 

 weight in ounces of one cubic inch, which being reduced to the standard of one 

 cubic foot, gives srr 0.57 38x1728 =98 1.5624 oz. 



