70 OF THE PARALLELOGRAM AND ITS INSCRIBED PARABOLA. 



Draw the diagonal AF intersecting CD, the vertical axis of the 

 parabola in the point n ; then is n the centre of gravity of the rectan- 

 gular parallelogram ABFE ; and because, as we have stated in the 

 construction of the preceding case, the centre of gravity of a parabolic 

 plane is situated in the axis, at the distance of three fifths of its length 

 from the vertex ; it follows, that if the axis DC be divided in the point 

 m, into two parts such, that Dm is to me in the ratio of 2 to 3 ; then 

 shall m be the centre of gravity of the parabola ACB. 



Put 6 = AB, the horizontal breadth of the rectangular parallelogram 



ABF,E, or the base of its inscribed parabola ACB, 

 / DC or AE, the vertical axis of the parabola, or the depth of 



its circumscribing rectangle, 



d D n , the depth of the point n, below A B the surface of the fluid , 

 nr DW, the depth of the point m as referred to AB, 

 P the pressure on the surface of the rectangle ABFE, 

 j9=:the pressure on the parabolic surface ACB, 

 Az=the area of the rectangular parallelogram, 

 a the area of its inscribed parabola, and 

 s zn the specific gravity of the fluid. 



Now, it is manifest from the principles of mensuration, that the 

 area of a rectangular parallelogram, is equal to the product that 

 arises when the two dimensions of length and breadth are multiplied 

 into one another ; that is, 



A=bl, 



and according to the writers on conic sections, the area of a parabola 

 is equal to two thirds of the area of its circumscribing rectangle ; 



therefore, we have 



oz=f6Z. 



Referring to the construction of the figure, we find that the axis 

 DC is divided at m, into the two parts Dm and me, having to one 

 another the ratio of 2 to 3 ; it therefore follows, that 



Dmzi: -:/; 

 consequently, the pressure on the parabolic surface ACB, is 



p ^blX^lXs ^b l*s. (36). 



Again, since AF the diagonal of the parallelogram, bisects DC the 

 axis of the parabola in the point n ; it follows, that 



Dw = d|/; 



therefore, the pressure perpendicular to the surface of the rectangular 

 parallelogram ABFE, becomes 



P blX i/X s= 



