418 OF THE CENTRE OF PRESSURE. 



but by Problem XVII. Chapter V. equation (56), the whole pressure 

 upon the side ACB is expressed by / 8 \/2 ' consequently, the pressure 

 upon the triangle CDB, becomes 



This is manifest, for by Proposition I. Chapter I. the pressure upon 

 the triangle ACB, is equal to its area drawn into the perpendicular 

 depth of the centre of gravity, the specific gravity or density of the 

 fluid being expressed by unity ; but by Problem XVII. Chapter V. 

 the perpendicular depth of the centre of gravity of the side of a 

 tetrahedron below the vertex, is 



and according to the writers on mensuration, the area of an equilateral 

 triangle, is expressed by one fourth of the square of the side, drawn 

 into the square root of 3 ; therefore, we have 



where a denotes the area of the triangle ACB; consequently, by 

 multiplication we obtain 



from which, by division, we get 



p-=L -rW's. (307). 



519. This equation satisfies the first demand of the problem, and the 

 second manifestly requires the determination of the centre of pressure ; 

 for by the definition, the point of application of a force, equal in 

 intensity to the pressure of the water, must occur at the centre of 

 pressure of the plane CDB, on which that pressure is exerted. 



Now, by the principles of Plane Trigonometry, the length of the 

 perpendicular c D is thus determined, 



rad. : I : : sin. 60 : CD, 

 from which, by reduction, we get 



but sin.60=r %\/3, and consequently, by substitution, we get 



Therefore, since the triangles CDB and CEF are similar, by the 

 property of similar triangles, we have 



JJ/3 : ' x : y, 

 and by reduction, we get 



