MECHANICAL PHILOSOPHY. HYDROSTATICS. [FLCID PRBBUHI 



rectangle, and if tliu rectangle be divided by a diagona 



its. 



two triangle*, then the pressure 

 on one triangle will bo double 

 that on the oth.T. 



Kor let A BCD (Fig. 102) be 

 the rectangle, and AC a dia- 

 gonal. Lot A E bUfct D C, and 

 OF, AB 2 Mid take Em- JK A, 

 and F-JFC : then m and n 

 are the centre* of gravity of the 

 two triangles. The depths of 

 thews centres are to one another 

 as Em : On ; that is, as J : I, or 

 as 1 : 2 ; and as the areas of the 

 two triangles are equal, there- 

 fore the pressures upon them are as 1 : 2. 



4. PROBLEM. To divide the preceding rectangle, by 

 lines parallel to the base, into rig. 163. 



rectangles, so that the pressure 

 on each may be the same. 



Let EB (Fig. 163) be the 

 lowest rectangle ; then, since 

 the centre of gravity of a rec- 

 tangle is its middle point, the 

 depth of the centre of gravity 

 of AC is ADA, and of that of 

 EC, iDE. The pressures, 

 therefore, on these rectangles 

 are as 

 DAx JDA:DExiDE, 



. ! A, I > A : 1 > K . 

 But the pressure on A C is divided into n equal pres- 

 sures, and the pressure on A C into n 1 equal pressures: 

 hence 



DA 1 :DE*::n :n 1. 



DA. 



.'. EA-DA DE-DA {1 y/ *^ 



If, for instance, the rectangle is to be divided into two 

 rectangles, so that the pressure on each may be the same 

 that is, half the whola pressure then since n = 2, D E 



It appears from this proposition, that in constructing 

 a flood-gate, or an upright embankment, to resist the 

 pressure of water against it, it is unnecessary to make 

 the material equally strong throughout ; the proper 

 degree of strength being secured at the bottom, where' 

 the pressure is greatest, the strength upwards may be 

 diminished in the proportion of 1) A ! to D E* ; that is, 

 the strength may decrease as the square of the depth 

 decreases. 



PROBLEM. To determine the pressure upon the in- 

 ternal surface of a hollow sphere filled with fluid. 



The centre of gravity of the surface pressed, is at the 

 centre of the sphere, so that the distance of the centre 

 of gravity from the surface of the fluid is the radius r. 

 The area of the surface of the sphere is 4nT*, where v 

 stands for 3' 141 6: hunce, the pressure is the same as 

 that of a cylindrical column of the fluid of base lirr* and 

 altitude r. 



If S be the speci6c gravity, the weight of this column 

 of the fluid is 4-r 1 S ; / ; but the volume of the sphere 



is T, and therefore the weight of the contained fluid 

 3 



Sy. Hence the pressure on the internal surface 

 3 

 of the sphere is three times the weight of the contained 



If a cone have its base eqnal to the surface of the 

 inhere, and iU altitii'li: w|iial to the radius of the sphere, 

 toe pressure on the bam of the cone will be the same as 

 that on the surface of the sphere, when both are filled 

 with the same fluid (\w 7 



pRoBLfM. To determine the pretwure on the hori- 

 zontal base of a vessel containing different fluids. 



Let E F, G H, J K (Fig. 104), bo the surfaces of the 



different fluids ; these 

 surfaces are all horizontal 

 (p. 751). Let ;., i 

 p,, be the perpen.lu-ular 

 il> pilis of the several 

 layers of fluid J C, G K, 

 EH, AF, andS, 8,, 8,, 

 8, tlieir respective specific 

 gravities : then the pres- 

 sures of these several 

 layers on the base DO 

 will be, 



. -7, DC.p.S.o, DO.p.S,?, CD.p^ 

 and consequently the whole pressure on the base will be 



that is, the pressure on the base is found thus : multiply 

 the area of the base by the sum of the products of the 

 perpendicular thickness of each fluid into its specific 

 gravity : the number of cubic feet, in the result, will be 

 the number of 1,000 ounces of pressure. 



If the depths of the several layers of fluid are all equal, 

 then the pressure on the base will be 



so that if we multiply the area of the base by the per- 

 pendicular height of one of the layers of the fluid, and 

 the product by the sum of the specific gravities, the 

 result will be the number of 1,000 ounces of pressure. 



Example. A cylindrical vessel is filled with mercury 

 to half its height, and the remainder is then tilled with 

 water. Supposing the specific gravity of mercury to be 

 14, required the pressure on the base, and on the con- 

 cave perpendicular surface. 



Let a be the height of the cylinder, and r the radius 

 of its base : then the area of the base of the cylinder 

 will be vr'* ; and since S = 14, and 8, = 1, we shall have 

 for the pressure on the base 



Jo (14 + 1) 1000 oz. = 



1000 oz. 



The pressure on the concave surface of the lower half 

 of the vessel is the same as if the water in the upper 

 half were removed, and a column of mercury ,ijth as 

 high substituted in its stead, making the height of the 

 entire column of mercury (J + .^a. Hence, the area 

 of the lower half of the concave surface being 2irrAa, 

 and the depth of the centre of gravity of it Jo + jya fa, 

 we have for the pressure on that surface 



a . 14j = 4jrra 2 x 1000 oz. (=> g) 



The pressure on the upper half of the surface is 

 2jrrjo . %ag \Trr<i-g : therefore the whole pressure on 

 the concave surface is lirra-g + JsTa 2 <j \ l *ra 3 X 

 1000 oz. (- g). 



Or the pressure on the concave surface may be found 

 thus : remembering the fundamental principle, that a 

 pressure exerted on the surface of a fluid is transmitted 

 in all directions ; 



Pressure on lower half of cylinder by the mercury 

 alone, 



1irr\a. Jo X 14</ 



Pressure of the water on the surface of the mercury, 

 2irria . $a X lg 



Pressure of the water on the upper half of the 

 cylinder, 



2>rrja . Ja X 1? 



Sum of the pressures, 



17 17 



-j- mcPg -^ wra' X 1000 oz. (= g). 



In a similar manner may the united pressures of layers, 

 equal or unequal, of different fluids be always ascer- 

 i Hie bottom layer exercises its own pressure, and 

 this is increased by the weight of the whole superin- 

 mass : the second layer in like manner exorcises 



