420 OF THE CENTRE OF PRESSURE. 



522. EXAMPLE. A vessel, in the form of a tetrahedron has the length 

 of its edges equal to 15 inches; now, supposing it to be filled with 

 water, and placed upon its bottom with the axis vertical ; conceive 

 one of its sides or containing planes to be bisected by a line drawn 

 from the vertex to the middle of one of the bottom edges, and let one 

 half of this plane be considered as loose, and moveable about a hinge 

 at the bottom ; what must be the magnitude of a force that will just 

 retain it in its place, and at what point must it be applied ? 



By equation (307), the magnitude of the required force is precisely 

 equal to the pressure of the fluid upon the moveable plane ; therefore, 

 by substituting the datum of the above example, we have 



p z= T VX 15 s X -/2 397.74 cubic inches of water ; 



which being reduced to Ibs. avoirdupois, becomes 



p = 397.74 X62.5 -4-1728 14.38 Ibs. 



523. Again, for the point at which this force must be applied, in order 

 to counteract the pressure of the water, we have by equation (308), 



cw~|7^/3" of 15 X -v/3 9| inches nearly. 

 And in like manner, for the corresponding co-ordinate nm, we have 

 by equation (309), 



nm ^l -?^ of 15 2.8125 inches, 

 from which the position of the point m can easily be ascertained. 



524. If the vessel should be placed with the bottom upwards and 

 parallel to the horizon, then we shall have 



j9 1=7.19 Ibs. ; DW 6.495 inches, and nm~ 1.875 inches. 



A great variety of useful and interesting problems similar to the 

 preceding, might be proposed in this place ; but as we have already 

 overleaped the limits of this subject in the present volume, we must 

 defer their consideration till another opportunity. 



