244 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES, 



290. EXAMPLE. Suppose that a cone and cylinder of copper, whose 

 specific gravity is nine times greater than that of water, are immersed 

 in a fluid, whose specific gravity is 1.85, and placed under the con- 

 ditions specified in the problem ; how much must be cut from the 

 lower part of the cone to restore the equilibrium, the diameter of the 

 bases and the altitude of the cone and cylinder being respectively 

 2 and 5 inches ? 



Here, by operating according to the rule, we shall have 



5.236* X 9 47.124 

 = 6(9=L85)= : 1^ 



the part to be cut off from the cone, in order that the remainder may 

 equipoise the cylinder; or we may calculate the magnitude of the 

 equipoising cone by equation (190), in the following manner. 



5.236(5X9 6X1.85) 



m" ^ ~ i4.135 cubic inches; 



o(y l.oo) 



which, by subtraction, gives 1.098 cubic inches for the part to be 

 cut off. 



Put d =. the diameter of the base of the cone and cylinder, and 

 h ~ the common height or altitude. 



Then, the equations (190) and (191) will become transformed, in 

 terms of the dimensions, into those that follow, viz. 

 ,,__.2618<fA(5s' 6s) 



~6(s'- 5 ) (192). 



This equation expresses the magnitude of the cone which restores 

 the equilibrium, and the following one expresses the magnitude of the 

 frustum which has to be deducted, in order that the equilibrium may 

 obtain ; that is, 



.2618^5' 



6(s f s)' (193). 



The above is a better mode of expressing the quantities, than that 

 exhibited in equations (190) and (191) ; since it is not probable, that 

 the magnitudes or solid contents of the bodies will be proposed, with- 

 out having previously stated their linear dimensions. 



It would be superfluous to propose an example, for the purpose 

 of illustrating the reduction of the equations in their modified form ; 

 for since the expression .261 8e? 2 A, is equivalent to the magnitude of 



* The number 5.236 is that which expresses the magnitude of the cone, for 

 2 2 X. 7854 X 5X$=5.236. 



