38 CENTRE OF GRAVITY OF MIXED SPACE 



6. METHOD OF FINDING THE POSITION OF THE CENTRE OF GRAVITY 

 OF ANY MIXED SPACE OF RECTILINEAR FIGURES IMMERSED IN 

 NON-ELASTIC FLUIDS. 



54. Since the pressure on the entire parallelogram, is equivalent to 

 the sum of the pressures on the two parts into which it is divided by 

 the line ef; it follows from thence, that the position of the centre of 

 gravity of the figure abcfe can be determined, as is shown in what 

 follows. 



Let ABCFE be the figure, of which the centre of gravity is required 

 to be found, the angles at A, B and c being right 

 angles ; join the points A and F by the straight line 

 AF, dividing the figure into two parts, one of which is 

 the triangle AFE and the other the trapezoidal space 

 ABCF. 



Now, almost every writer on mechanical science 

 has given the method of finding the centre of gravity 

 of those figures separately, from which that of the compound space 

 may easily be determined ; but we are not aware of any method that 

 has been proposed, for the direct discovery of the centre of gravity 

 of the mixed space ABCFE, and that is what we are now about to 

 investigate. 



Produce the sides AE and CF till they meet in D; then, because 

 the angles at A, B and c are right angles, the angle at D is also a right 

 angle; from the point D, set off Da and D&, respectively equal to 

 one third of DE and DF, and through the points a and b, draw am 

 and bm parallel to DA and DC, which produce directly forward to d 

 and c ; then are cm and dm respectively, the perpendicular depths of 

 the centre of gravity of the triangle EDF, according as the side AB or 

 BC is supposed to be coincident with the surface of the fluid. 



Put b m AB, the breadth of the rectangular parallelogram A BCD, 

 I AD, the length of ditto, 



/3= DF, the base of the right angled triangle EDF, 

 Z'nz DE, the corresponding perpendicular, 

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



triangle EDF, when the side AB is horizontal, 

 3 dm, the perpendicular depth of the centre of gravity m, when 



the side BC is horizontal, 



jazr the pressure perpendicular to the surface of the triangle EDF, 

 7>'z= the pressure on the irregular figure ABCFE, 



