62 OF FLUID PRESSURE ON THE SIDES AND BASE OF CUBICAL VESSELS. 



In the solution of the present problem, it will be unnecessary to 

 exhibit the construction of a separate diagram ; because, the bound- 

 aries when considered individually, being rectangular parallelograms, 

 the investigation for each would be similar to that required in Pro- 

 blem 4, and the resulting formulae would coincide in form with that 

 exhibited in equation (8). 



Therefore, put b = the horizontal breadth of the greater opposite 



sides, 



/3 n: the horizontal breadth of the lesser ditto ditto, 

 I z= the perpendicular depth of the fluid, whose den- 



sity is uniform, 

 P= the aggregate, or total pressure on the upright 



surface, and 

 p zz the pressure on the bottom. 



Then, according to equation (8) under Problem 3, the pressure on 

 each of the narrower sides is expressed by |/3/ 2 s, and that on each of 

 the broader sides by \bl*s\ consequently, the entire pressure on the 

 upright surface, is 



P=l'04..&)< 



and by the third inference to Proposition (A), the pressure sustained 

 by the bottom of the vessel, is 



consequently, by analogy, we obtain 

 P:p::l(P + b):pb. 



Therefore, when the opposite sides of the rectangular vessel only, 

 are equal to one another, 



The total pressure on the upright surface, is to the pressure 

 on the bottom, as half the area of the former is to the area of 

 the latter. 



If b in /3, or if all the sides of the vessel are equal to one another ; 

 then, the entire pressure on the upright sides, is 



and that on the bottom, is 



p=b*ls; 



therefore, by analogy, we obtain 

 P :p :: 2/ : 6. 



