TWO PARTS SUSTAINING EQUAL PRESSURES. 33 



being the very same expression as that which we obtained by the 

 foregoing analytical process, a coincidence which verifies the pre- 

 ceding construction. 



Returning to the equation numbered (15), and extracting the square 

 root of both sides, we obtain 



xy2=li 

 and by division, we have 



x = kW2. (16), 



48. The practical rule for determining the point of division, as 

 supplied by the above equation, is extremely simple ; it may be thus 

 expressed : 



RULE. Multiply half the length of the immersed side of 

 the parallelogram by the square root of 2, or by the constant 

 number 1.4142, and the product will express the distance 

 downward from the surface of the Jluid. 



49. EXAMPLE 7. A rectangular parallelogram, whose sides are 

 respectively 14 and 28 feet, is immersed in a cistern of water, in such 

 a manner, that its shorter side is just coincident with the surface; 

 through what point in the longer side must a line be drawn parallel 

 to the horizon, so that the pressures on the two parts, into which the 

 parallelogram is divided, may be equal to one another? 



Here, by operating according to the rule, we have 

 x i (28 X 1.4142)= 19.7988 feet. 



50. If the point through which the line of division passes, were 

 estimated in the contrary direction ; that is, upwards from the lower 

 extremity of the immersed side of the parallelogram; then, the ex- 

 pression for the place of the point will be very different from that 

 which we have given above, as will become manifest from the follow- 

 ing process. 



Recurring to the original diagram of Problem 5, and putting 

 x ed, the rest of the notation remaining, we shall have by sub- 

 traction, 



ae=il x; 



consequently, sn the depth of the centre of gravity of the rectangle 

 abfe, is 



3 J (/ x) sin. 0, 



and in like manner, it may be shown, that rm, the depth of the centre 

 of gravity of the rectangle efcd, is 



c?= (/ !#) sin. (f>. 



Now, according to the writers on mensuration, the area of the 

 rectangle abfe is expressed by b (i x}, and that of the rectangle 

 efcd by bx; consequently, the respective pressures are 

 VOL, i. i> 



