OF THE CENTRE OF PRESSURE. 



417 



PROBLEM LXVII. 



518. A vessel in the form of a tetrahedron is entirely filled with 

 water, and has one of its planes bisected by a line drawn from 

 the vertex to the middle of the opposite side ; now supposing 

 one half of the bisected plane to be loose, and moveable about a 

 hinge at its lower extremity : 



It is required to determine the magnitude of the force, the 

 point of application, and the direction in which it acts with 

 respect to the horizon, when the moveable half of the contain- 

 ing plane is just retained in a state of quiescence. 



Let ACB be one side of the vessel, divided by the line CD into two 

 parts, which are equal and similar to 

 one another; and let the part BCD 

 be moveable about the hinges at e 

 and/. 



Suppose the centre of pressure of 

 the loose part B c D to be at the point 

 m, and through m draw the straight 

 lines mb and mn, respectively paral- 

 lel to CD and AB; in CD take any 

 other point E, and through E draw 

 EF perpendicular to CD, or parallel 

 to DB, making the triangles CDB and CEF similar to one another. 



Put I =z AB, BC or AC, the side of the containing plane, or the 



edge of the tetrahedron, 

 d r= CD, the length of its perpendicular, 

 5 =z the perpendicular depth of its centre of gravity below the 



vertex at c, 



p = the pressure of the water on the loose triangle CDB, 

 ar^: CE, any variable distance, 

 y zz EF, the corresponding co-ordinate, and 

 (f> z= the angle which the direction of the retaining force makes 



with the horizon. 



Then it is manifest from the nature of the .problem, that in the case 

 of an equilibrium, the magnitude of the retaining force must be equal 

 to the whole pressure of the water upon the moveable triangle CDB ; 



VOL. i. 2 E 



