HYDROSTATICS. 



29 



lumn of water whose base is A B C and 

 altitude G M : that is, equal to C P x 



M 



C* AT 



A B + j- x g ; g being the specific 



gravity of water. 



Draw E F parallel to A B , and let W O 

 - x, C W= a, C P= 6, AB = c, G M = m. 

 Then C O = .r -a; E F= (by similar tri- 



angles) - > r/), and the pressure 



upon ABC= x m ^- And by the 



property of the centre of percussion, the 

 distance of the centre of percussion of 

 C E F below W R is equal to the fluent 



- xO a)x- x; and finding the 



fluent, we hare - ( - -- ^ ) + C, 



b \. 4 8 / 



a constant quantity; which constant 

 quantity (as the fluent vanishes at C, or 



when x a - 0, that is x = ) is - - ; 

 so that the whole fluent is x (^~ 



- + ~T~) ' an d if we take this for 



the whole triangle ABC, where x = a 

 + b ; we have to substitute thisforrr, and 

 to divide by AB C x G M, that is, by 



- . Let a + b = d ; the expres- 



sion becomes 



2 /</ 4 _ a d 3 a * \ 



^~b~* X V 4 3 h TT ) ' 



Therefore the centre of percussion, 

 is in the line found E F. But if 

 the triangle were to revolve round 

 the intersection of its plane with the 

 plane of W R, and strike against a 

 plane passing through E F, its mo- 

 tion would be destroyed, the motions 

 on the opposite sides of that line 

 balancing each other; therefore the 

 pressure of the fluid on the triangle 

 being counteracted by an equal and 



opposite pressure on the line E F, the 

 triangle must be supported; that is, 

 the centre of pressure is in E F. 



If the vertex of the triangle be situ- 

 ated at W, the surface of the water, 



g 



then a= 0, d b, and m = 6, and the 



O 



expression becomes - b ; or the cen- 

 tre of pressure is at three-fourths of 

 the depth of the water. 



The point of the line E F, which is 

 the centre of pressure, that is, the dis- 

 tance of that centre from C P, may be 

 found by a similar process. It is equal 

 to 

 2 tan. <p /d* _ 2 3 a 2 d 2 a 4 



~^T^~ x \T ~ ~s ad + ~~2~ T 



9 being the angle P C G. When, 

 therefore, the triangle is isosceles, and 

 the centre of gravity is in C P, this 

 quantity vanishes, and the centre of 

 pressure is in O, the intersection of C P 

 and E F ; and when the vertex of the 

 triangle touches the surface W R, the 



expression becomes - 6 x tan. <p. 



Of the two expressions, for the dis- 

 tance of E F from W R, and for the dis- 

 tance of the centre from C P, it may 

 be remarked that, in the former, C 

 vanishes at one of the steps of the 

 analysis. Wherefore the distance of 

 E F 'from W R does not depend on the 

 line A B : in other words, all triangles 

 of the same altitude, and in the same 

 depth of water, have the centre of pres- 

 sure at the same depth below the sur- 

 face, whatever be their bases. But the 

 latter expression involving the angle <p, 

 the horizontal distance of the centre of 

 pressure from the vertical line depends 

 upon the base of the triangle, because 

 that determines the position of G, the 

 centre of gravity. 



To apply this analysis to other figures, 

 the line E F = n must be found in terms 

 of x, or of the given quantities, and 

 substituted in the fluxional expression 



nx 2 J : and the fluent being found, it 



must be divided by the expression for 

 the figure. This ives the depth of the 

 line E F below W R. 



To find the distance of the centre from 

 C P on E F, we must substitute for n 

 (or E F) its value in terms of x as 

 before, in the expression 



