ON DIFFERENT SECTIONS OF PARALLELOGRAMS. 27 



And in like manner it may be shown, that if the side BC were 

 horizontal, the perpendicular depth of the centre of gravity would be 



d=b; 



therefore, take T.m and BW respectively equal to one third of BC and 

 BA, and through the points m and n, draw mG and no parallel to BA 

 and BC, meeting each other in the point G ; then is G the position of 

 the centre of gravity. 



The intelligent and attentive reader will readily perceive that the 

 above determination is not legitimate, since it supposes the pressure 

 upon the triangle ABC to be given ; now, this pressure depends 

 entirely upon the position of the centre of gravity, and consequently, 

 the problem supposes the position of the centre of gravity of the 

 triangle ADC to be known; the principle, however, will be more 

 distinctly indicated when applied to other figures, where the above 

 determination may be admitted, without infringing on the precepts of 

 scientific propriety. 



4. OF THE PRESSURE OF INCOMPRESSIBLE FLUIDS ON DIFFERENT 

 SECTIONS OF PARALLELOGRAMS PARALLEL TO THE HORIZON. 



PROBLEM V. 



43. A rectangular parallelogram is obliquely immersed in an 

 incompressible and non-elastic fluid, in such a manner, that one 

 side is just coincident with the surface : 



It is required to compare the pressure on the upper and the 

 lower portions, supposing the parallelogram to be bisected by 

 a line drawn parallel to the surface of the fluid. 



Let A ED represent a vessel full of water, or some other non-elastic 

 and incompressible fluid, of which ABEF 

 is the surface, and suppose one side of the 

 vessel to be removed, exhibiting the fluid 

 and the immersed rectangle as represented 

 by A B c D and abed. 



Bisect the parallelogram abed by the 

 straight lines ef and gh respectively parallel 

 to a b and ad\ then are abfe and efcd, 

 the portions on which the pressures are to be compared, and gh is the 

 line in which the centres of gravity occur. 



