298 Hydrostatics. 



416. As the pressures exerted upon the several points of the 

 same plane surface are perpendicular to the surface, and conse- 

 quently parallel among themselves, the resultant or whole pres- 

 sure must be parallel to the components. Now as we know how 

 to find the resultant as well as that of each of the partial pres- 

 sures, it will be easy to determine, as we have occasion, through 

 what point the resultant passes ; it evidently cannot pass through 

 the centre of gravity, but must pass through some point lower 



Fig.201. down. It is only in the case where the surface is infinitely small 

 that the whole pressure passes through the centre of gravity of 

 this inclined surface. 



417. In order to find the resultant of all these pressures both 

 in a vertical and in a horizontal direction, any body, whatever 

 its figure, may be considered as composed of an infinite number 

 of strata, parallel among themselves, and the surface of the pe- 

 rimeter of each stratum may be represented by a series of trap- 

 ezoids, of which the number is infinite, when the surface is a 

 curve. So that in order to estimate the resultant of the pressure 

 exerted by a fluid either upon the interior sides of a vessel, or 

 upon the exterior surface of a solid immersed in it, we must de- 

 termine the pressure exerted upon a trapezoid of an infinitely 

 small altitude. 



Accordingly, let us suppose a trapezoid ABCD, of which the 

 Fig.2C2. two parallel sides are AB, C/), and the altitude infinitely small 

 compared with these sides ; and let there be applied at the cen- 

 tre of gravity G of the trapezoid perpendicularly to its plane, a 

 force p equivalent to the product of the surface of this trapezoid 

 into the distance of its centre of gravity from the horizontal 

 plane XZ. 



To determine the effect of this force, as well in a horizontal 

 as in a vertical plane, suppose through the line CD a vertical 

 plane CDFE, and through the line AB, considered as horizontal, 

 a horizontal plane AFEB. Having drawn the vertical lines 

 CE, DF, meeting this latter plane in E and F, we join BE, AF-^ 

 and through the direction Gp of the force />, suppose a plane 

 A7/I cutting CD at right angles, HGK and HI being the inter- 

 sections of this plane with the two planes ABCD, FECD, res- 

 pectively. The plane KIH will be perpendicular to each of 



