92 OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES. 



or multiplying both sides of the equation by 3, we get 



3czzl.5708D 3 ; 

 consequently, by equation (50), we have 



or if s denote the specific gravity of the fluid, we shall obtain 



COROL. Hence we infer, that if a hollow sphere or globe be filled 

 with an incompressible and non-elastic fluid : 



The whole pressure sustained by the internal surface of the 

 sphere is equal to three times the weight of the fluid which it 

 contains. 



102. EXAMPLE 21. A hollow spherical shell or vessel, whose inte- 

 rior diameter is equal to 30 feet, is completely filled with water ; what 

 weight is equivalent to the pressure sustained by its internal surface ? 



Here, by operating according to the process indicated in equation 

 (50), we have 



P=: 1.5708 X 30 3 zr42411.6cub.ft. 



Now, since the fluid with which the vessel is filled, is water, giving a 

 weight of 62 \ Ibs. to a cubic foot, we have 



P =. 4241 1.6 X 62 J =: 2650725 Ibs. ; 



but 2240 Ibs. are equal to one ton; therefore, the pressure on the 

 internal surface of a hollow spherical vessel whose diameter is 30 feet, 

 when completely filled with water, is 



P = 2650725 H- 2240 1 183 Hi tons. 



PROBLEM XVI. 



103. Suppose a sphere or globe to be immersed in an incom- 

 pressible and non-elastic fluid, in such a manner, that the upper 

 extremity of the vertical diameter is just in contact with its 

 surface : 



It is required to determine through what point of the axis 

 a horizontal plane must pass, so to divide the sphere, that the 

 pressure on the convex surface of the lower segment, may be 

 equal to the pressure on the convex surface of the upper. 



Let AD BE represent the sphere in question, so placed, that A the 

 upper extremity of the vertical diameter, is just in contact with FG 

 the surface of the fluid. 



