26 OF THE AGGREGATE PRESSURE OF FLUIDS 



the centre of gravity; which method, in so far as respects plane 

 figures of particular forms, may, in many instances, be very advan- 

 tageously applied. 



It would be foreign to our present purpose to enter into a detail of 

 the method alluded to in this place ; nevertheless, for the satisfaction 

 of our readers, we shall briefly introduce it, not being aware that it 

 has been suggested by any other writer in ancient or modern times. 



PROBLEM B. 



42. The base and perpendicular of a right angled triangle 

 being supposed known : 



It is required to determine the position of its centre of 

 gravity, or that point on which, if the surface were supported, 

 it would remain at rest in any position. 



Let ABC be the triangle given, of which it is required to determine 

 the centre of gravity. 



Complete the parallelogram A B c D, by drawing the 

 dotted lines AD and DC; then, because the entire 

 pressure on the parallelogram ABCD, is equal to the 

 sum of the pressures on the triangles ABC and ACD; 

 it follows, that the pressure on the triangle ABC, is 

 equal to the difference between the entire pressure on 

 the parallelogram, and that on the triangle ACD; consequently, we 

 have, by retaining the foregoing notation, 



p = P p r ; that is, 



But it has been elsewhere demonstrated, that the pressure on any 

 surface, is expressed by the area of that surface, drawn into the 

 perpendicular depth of its centre of gravity ; consequently, the per- 

 pendicular depth of the centre of gravity, must be equal to the 

 pressure divided by the area of the surface. 



Now, in the present instance the pressure is known, and since by 

 the problem, the base and perpendicular of the triangle are given, its 

 area can easily be found. 



Thus, the writers on mensuration have shown, that the area of a 

 triangle is equal to half the product of the base drawn into the per- 

 pendicular altitude ; consequently, if a be put to denote the area of 

 the triangle ABC, we shall have 



therefore, by division, the perpendicular depth of the centre of gra- 

 vity, is 



