416 OF THE CENTRE OF PRESSURE. 



This is the general form of the equation, corresponding to a fluid oi 

 any density whatever denoted by s; but when the fluid is water, of 

 which the specific gravity is unity, the above equation becomes 



(306). 



The method of reducing this equation, may be very simply expressed 

 in words as follows. 



RULE. Divide six times the momentum of the given force,* 

 by the horizontal breadth of the side to which it is applied, 

 and the cube root of the quotient will be the height to which 

 the vessel must be filled. 



EXAMPLE. The horizontal breadth of one side of an oblong pris- 

 matic vessel, is 30 inches ; now, supposing this side to be loose and 

 moveable about a Hinge at the bottom ; how high must the vessel be 

 filled with water, in order that the pressure of the water, and a force 

 of 400 Ibs. applied at the distance of 12 inches from the bottom, may 

 exactly balance each other ? 



Here we have given 



b = 30 inches ; 3 zr 12 inches ; and /zz 400 Ibs ; 

 consequently, we. obtain 



z=f/ 6x4QQ Xl2 

 30 



9.865 inches. 



But the centre of pressure, at which the weight of the fluid is sup- 

 posed to be applied in opposition to the given force, is situated at 

 one third of the above distance from the bottom of the vessel ; hence 

 we have 



cm zzi of 9.865 3.288 inches. 



Therefore, the pressure of the fluid acting on a lever of 3.288 inches, 

 must be equal to a force of 400 Ibs. acting on a lever of 12 inches ; 

 that is 



400 X 12 = 30 X 9.865 s -:- 6. 



* The momentum of the given force, is equivalent to its magnitude, drawn into 

 the distance above the bottom of the point at which it is applied. 



