26 EXAMPLES. 



2. Find the centre of pressure of an isosceles triangle 

 whose base is horizontal and opposite vertex in the surface 

 of the fluid. 



Let a be the altitude of the triangle and 5 its base. Take 

 the intersection of the plane of the triangle with the surface 

 of the fluid for the axis of y and the vertex for the origin, 

 the axis of x bisecting the triangle. Then from (7) we 

 have, 



na r&a pa 



I I x*dxdy I x*dx 



t/o t/o t/o 



X = 



J Q J Q xdxdy 



3. A quadrant of a circle is just immersed vertically in a 

 fluid, with one edge in the surface. Find its centre of 

 pressure. 



Take the edge in the surface for the axis of y, and the 

 vertical edge for the axis of x, and let a be the radius. 

 Then, from (7) and (8), we have 



pa p |V x* pa 



J J tfdxdy I x*(a z x*)* dx 



> Va'x 



pa /> Va' na 



J J xdxdy J x(a* 



'0 



atn a* 3 

 = 16*3 = 16 a7r; 



na /> v a' * \ f*a 



J J xydxdy -J ' x(a*-x*)dx 



and y = - = - 



/* r> VaPx* /xj . 



J J xdxdy J x (a 2 x*)* dx 



_ a< _._ a 8 _ 3 

 ~ 8 : 3 : ~ 8 a> 



(See Besant's Hydromechanics, p. 41.) 



