FLUID PRESSURES.] MECHANICAL PHILOSOPHY. HYDROSTATICS. 



753 



its own pressure, increased by the weight of the mass 

 above it ; and so on. 



PROBLEM. A hollow cone rests in a vessel with its 

 base on a smooth horizontal plane, and water is 

 poured in at the top of the cone. How high will the 

 water rise before it lifts the cone off its support and 

 escapes ? (See Fig. 165). 



As the water rises in the cone, it exercises an upward 

 pressure on its slant sides, which increases as the per- 

 pendicular height of the fluid increases ; and as soon as 

 this upward pressure becomes equal to the downward 

 pressure or weight of the conical shell, the equilibrium 

 is j ust maintained, and no more water ; that is, no more 

 upward pressure can be sustained by the cone, which 

 will therefore be lifted up, and give egress to the water : 

 we have therefore to find the height of water, the 

 upward pressure of which is just equal to the weight of 

 the cone. 



The upward pressure, thus just balanced by the weight 

 of the cone, would be equally balanced if a cylinder on 

 the same base, and of the same altitude as the water, 

 were to surround the cone, and the vacant space to be 

 filled with the fluid. Thus if 

 B E (Fig. 165) were the height 

 of the water, and if a cylinder 

 H D were to surround the cone, 

 and the space H F C to be filled 

 with water, the downward pres- 

 or weight of the water in 

 space would replace the 

 weight of the cone, so that the 

 equilibrium would remain un- 

 disturbed if the cone were 

 withdrawn ; we have therefore 

 to di-tcrmine the height B E of the cylinder, so that the 

 weight of water contained between it and the cone may 

 be exactly equal to the weight of the cone. 



Let A B = , BC = 6, AE = a;: then the volume of 

 the cone A C D ia i-6 2 o ; and the volume of the cone 

 A F G, being to that of A C D as z 3 to a 3 , is \irWa, x 



3? 



$ ' hence the volume of water in the frustum F D ia 



volume of water in F D : 



weight of the same 



--,)?. . .(1) 



the specific gravity of the fluid being 1. 



Also, if the cone were removed, the volume of water 



in the cylinder H D would be 

 volume of water in cylinder irV (a, >a) 

 weight of the same = irb 3 (a x) g . . . (2) 



Hence, subtracting (1) from (2), we have for the 



weight of water between the cylinder and cone that is, 



for the upward pressure of the water in the cone on its 



sides, 



z 3 ) 

 upward pressure on cone = irtfy { (a x) Ja(l -. ,) j 



If, therefore, the weight of the cone be w, we shall have 

 to solve the cubic equation 



or 



The upward pressure of the fluid compels an equal 

 pressure downwards on the base ; the water in the cavity 

 between the cone and cylinder is just sufficient to balance 

 the upward pressure, or to replace the resistance of the 

 sides, the pressure on the base remaining undisturbed ; 

 ami it is thus that the base of the cone supports not 

 only the water in it, but also an amount of pressure 

 equal to the weight of the additional water between the 

 cone and cylinder. 



VOL. I. 



Fig. 106. 



In any vessel containing fluid, where all the vertical 

 pressures are downwards that is, where the sides do 

 not any of them incline inwards the sum of the vertical 

 pressures must be equal to the weight of the fluid. For 

 every vertical line of particles presses downwards with 

 the weight of those particles ; so that the whole vertical 

 pressure is the whole weight of the fluid. 



The pressure is, therefore, the same as it would be if 

 the fluid were to become solid, and the sides of the vessel 

 to be removed ; and the effect is the same as if the entire 

 pressure or weight were concentrated in the centre of 

 gravity. 



The horizontal pressure at any depth, is of course the 

 same all round the vessel at that depth, the pressure of 

 every horizontal line of particles being equal to the 

 pressure of a vertical line of particles reaching from the 

 surface to the horizontal line. The downward vertical 

 pressure on the bottom of a vessel can exceed the 

 weight of the fluid contained in the vessel, only when 

 one or more of its sides, or a portion of a side, inclines 

 inwards, occasioning an upward pressure, which reacts 

 downwards on the base, the excess of the pressure on 

 which, above the weight of the fluid, is just equal to 

 this additional pressure ; and the vertical pressure on 

 the bottom can fall short of the weight of the fluid in 

 the vessel, only when a side, or portion of a side, in- 

 clines outward, occasioning a downward pressure on 

 that side. The whole downward pressure on the sides, 

 together with that on the base, makes up the weight of 

 the fluid. 



Take, for example, the case of a common decanter : 

 the pressure at any point C (Fig. 166) is in the direc- 

 tion of the straight line perpendicular to the surface. 

 This may be decomposed into two 

 pressures, of which one C E is hori- 

 zontal, and the other C F vertical : this 

 last pressure being directed upwards in 

 the figure, if the point C had been near 

 to the bottom, it would have been di. 

 rected downward, on account of the 

 curvature there being in a contrary di- =i 

 rection. If we conceive the pressure at \ 

 every point of the interior surface of 

 the decanter to be decomposed in like 

 manner into a horizontal and vertical pressure, there 

 will be a series of horizontal pressures like C E, and a 

 series of vertical pressures like C F. Tho horizontal 

 components mutually destroy one another ; otherwise 

 the decanter would tend to move horizontally, which is 

 not the case. Of the vertical components, some are up- 

 ward pressures and the others downward pressures : they 

 may, therefore, be replaced by a single vertical force, 

 which will act upward or downward, according as the 

 component vertical forces upward or downward prevail.* 

 If this simple resultant pressure act upward, the pressure 

 on the bottom of the decanter will exceed the weight of 

 the liquid, because the downward pressure on the bottom, 

 minus this upward pressure, must be equal to the weight : 

 on the contrary, if the resultant be a downward pressure, 

 then because the downward pressure on the bottom, phis 

 this other downward pressure, is equal to the weight, 

 the pressure on the bottom will be less than the 

 weight. 



PROBLEM. To determine the resultant of all the 

 pressures of a fluid upon the surface of a body immersed 

 in it. 



Instead of the immersed body C (Fig. 167), conceive 

 the fluid it displaces to become solidified : the surround- 

 ing pressures will keep the solidified fluid at rest, just as 

 if it were in its original state. But for these pressures, 

 the mass would fall downwards in virtue of its weight : 

 the resultant of the pressures, therefore, just balances 

 the weight, and acts in a direction opposite to that of 

 gravity ; that is, vertically upwards through the centre 

 of gravity of the mass. And as the surrounding fluid 

 exerts the same pressures, whatever be the body whose 

 surface is pressed, it follows that the resultant of the 

 pressures on the surface of any solid body, is equal in 

 Sec Statics; ante, p. 720. 



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