ON IMMERSED RIGHT ANGLED PARALLELOGRAMS. 25 



40. The practical rules for calculating the pressures on the triangles, 

 as deduced from the equations (11) and (12) are as follows. 



1. When the base of the triangle is coincident with the surface. 



RULE. Multiply the square of the immersed length, or the 

 perpendicular of the triangle, by the base drawn into the 

 specific gravity of the fluid, and again by the natural sine of 

 the plane's inclination, and one sixth part of the product will 

 express the whole pressure upon the triangle. (Eq. 11). 



2. When the vertex of the triangle is coincident with the surface. 



RULE. Multiply the square of the perpendicular of the 

 triangle, by the base drawn into the specific gravity of the 

 fluid, and again by the natural sine of the plane's inclination, 

 and one third of the product will express the whole pressure 

 on the triangle. (Eq. 12). 



41. EXAMPLE 5. A rectangular parallelogram, whose sides are 

 respectively 26 and 14 feet, is immersed in a cistern of water, in such 

 a manner, that its shorter side is coincident with the horizontal sur- 

 face ; what will be the pressure on each of the triangles, into which 

 the parallelogram is divided by its diagonal, supposing its plane to be 

 inclined to the surface of the fluid in an angle of 56 35' ? 



Here, by the rule, we have 



p= 26 2 X 14 X .83469 X i= 1316.58436 cubic feet 

 of water; but one cubic foot of water weighs 62 J Ibs. ; therefore, to 

 express the pressure in Ibs., we have 



p= 1316.58436 X 62J = 82286.5225 Ibs. 



The pressure which we have just obtained, refers to that portion of 

 the parallelogram which has its base coincident with the surface of 

 the fluid; that is, to the triangle abd, and the pressure on the other 

 portion, or the triangle bdc, is determined as follows. 



/z=:26 2 X 14 X .83469 X i = 2633. 16872 cubic feet of water; 



or to express the pressure in Ibs. we have 



p'= 2633. 1687 X 62.5= 164573.045 Ibs. 



If the plane of the immersed parallelogram were perpendicular to 

 the surface of the fluid, the pressures on the triangles abd and bdc 

 would be respectively as follows. 

 p=:26 2 X 14 X 62 J X i = 98583ilbs.,andy 26* X 14 X 62|X 



i= 197 166| Ibs. 



COROL. The circumstance of the aggregate pressure on the paral- 

 lelogram, being equal to the sum of the pressures on the constituent 

 triangles, furnishes a very simple and elegant method of determining 



