OF THE EQUILIBRIUM OF FLOATATION. 291 



PROBLEM LV. 



370. Having given the specific gravity of distilled water, 

 equal to 1000 ounces per cubic foot : 



It is required to determine the specific gravity of a solid 

 body that is wholly immersed in it. 



It is manifestly implied by the total immersion of the body, that its 

 specific gravity exceeds the specific gravity of the fluid in which it is 

 immersed ; therefore, attach the body to the hook in the bottom of 

 the scale E by a very fine and light thread, and balance it exactly by 

 weights put into the other scale at D ; then, immerse the body in the 

 water, and find what weight is required to restore the equilibrium, the 

 weight thus required will measure the specific gravity of the body. 

 Put w z= the weight of the body when weighed in water, 

 w' ~ the weight when weighed in atmospheric air, 

 s zn the specific gravity of water, and 

 s' the specific gravity of the body sought. 



Then is w w' equal to the weight which must be put into the scale 

 E to restore the equilibrium; consequently, by the fifth proposition, 

 we have . w' w : w' : : s : s' ; 



from which, by reduction, we get 



,'-_^i- 



~w' w (225). 



371. The following is the practical rule in words at length for 

 reducing the above equation. 



RULE. Multiply the weight of the body when weighed in 

 air, by the specific gravity of the fluid, and divide the product 

 by the weight which it loses in water for the specific gravity 

 of the body. 



This rule determines the specific gravity of the body when it 

 exceeds that of the fluid in which it is weighed ; but when the body 

 is specifically lighter than the fluid, the method of finding its specific 

 gravity is shown in Problem XLIV., it is therefore unnecessary to 

 repeat it here. 



372. EXAMPLE. If a piece of stone weighs 20 Ibs. in air, but in water 

 only 13J Ibs.; required its specific gravity, that of water being 1000 ? 



Here, by the rule, w=: 13J, w'= 20, s=z 1000, 



20X1000 20000 

 therefore s = T^T ~ 3076.923 rz the specific gravity 



Z(j ~ Aug O.O 



of the mass when it is wholly immersed in water. 



u 2 



