TWO PARTS SUSTAINING EQUAL PRESSURES. 31 



Let A E D represent a rectangular vessel filled with water, or some 

 other incompressible and non-elastic fluid, 

 of which ABEF is the surface, and ABCD 

 the fluid as exhibited in the vessel, on the 

 supposition that one of its upright sides is 

 removed. 



Let abed be the immersed parallelo- 

 gram, having its upper side a b coincident 

 with the surface of the fluid, and its plane 

 tending obliquely downwards in the given angle of inclination. 

 Bisect a b in g, and through g draw the v straight line gh parallel 

 to ad or be, the side of the given immersed rectangle, and let ef 

 parallel to ab or cd, denote the line of division; then, by the 

 problem, the pressure on the rectangle abfe, is equal to the pressure 

 on the rectangle efcd. 



Draw the diagonals dfo.ud.fa, cutting the bisecting line gh in the 

 points m and n ; then are m and n respectively, the places of the 

 centres of gravity of the spaces efcd and abfe. Through the point 

 g and in the plane of the fluid surface, draw g r at right angles to a b, 

 and from m and n demit the straight lines mr and ns, respectively 

 perpendicular to the horizontal line gr ; then are sn and rm the 

 perpendicular depths of the centres of gravity of the rectangles a bfe 

 and efcd on which the pressures are equal. 

 Put b rz ab, the horizontal breadth of the proposed rectangular plane, 



Z=n ad or be, the immersed length of ditto, or that which tends 

 downwards, 



d = rm, the vertical depth of the centre of gravity of the lower 

 portion efcd, 



$i=sn, the vertical depth of the centre of gravity of the upper 

 portion abfe, 



0z= mgr, the inclination of the plane to the surface of the fluid, 



P=r the pressure on the entire parallelogram, 



p =z the pressure on each of the portions into which the paral- 

 lelogram is divided, 



s rr the specific gravity of the fluid, 

 and x ae, the immersed length of the upper portion abfe. 



Then is edl x; gn \x, and gm # -|- \ (I x) ; con- 

 sequently, by the principles of Plane Trigonometry, we have 



57i^rS= | a sin. 0, and rm d [x -\- %(l #)J sin.0, 

 and moreover, by the principles of mensuration, the area of the upper 

 portion is expressed by b x, and that of the lower portion by b (I x) ; 



