SUSTAINING EQUAL PRESSURES 



Let A ED represent a cistern filled with 

 fluid, of which ABEF is the surface, sup- 

 posed to be perfectly quiescent, and con- 

 sequently, parallel to the horizon ; and let 

 ABCD be a vertical section of the cistern, 

 exhibiting the fluid with the immersed 

 rectangle abed. 



Draw the diagonal ac, and in a d take 

 any point e ; through the point e thus 

 assumed, draw the straight line ef parallel to a c the diagonal of the 

 parallelogram; then is edf the triangle, on which the pressure is 

 equal to that upon the polygonal figure e a b cf. 



Take dn and dt respectively equal to one third of de and df, and 

 through the points n and t, draw nm and tm respectively parallel to 

 ab and ad, the sides of the parallelogram, and meeting one another 

 in the point m ; then, according to problem B, m is the place of the 

 centre of gravity of the triangle e df. 



Produce tm directly forward, meeting ab the upper side of the 

 parallelogram perpendicularly in s ; then, through the point s, and in 

 the plane of the fluid surface, draw the straight line sr also at right 

 angles to a b, and from m, the centre of gravity of the triangle edf, 

 demit the line mr perpendicularly on sr ; then is rm the perpendicular 

 depth of the centre of gravity of the triangle edf, and msr is the 

 angle of inclination of the plane to the horizon. 



Put b=^ab, the horizontal breadth of the given parallelogram, 

 I = ad, the length of the immersed plane tending downwards, 

 d rm, the perpendicular depth of the centre of gravity of the 



triangle efd, 



p ~ the whole pressure perpendicular to its surface, 

 <t> msr, the angle which the immersed plane makes with the 



horizon, 



s =the specific gravity of the fluid, 

 and x ed, the perpendicular of the triangle edf, of which the base 



isrf/. 



Then, by reason of the parallel lines ac and ef, the triangles adc 

 and ec?/are similar to one another, and consequently, by the property 

 of similar triangles, we have 



ad : dc : : ed : df, 

 which, by restoring the symbols, becomes 



I : b : : x : df, 

 and from this analogy we have 



D2 



