HYDROSTATICS. 1850. (A). Find the pressure referred to a unit of area at any depth below the surface of a heavy incompressible fluid. (B). If from every point in the vertical side of a rectangular vessel containing fluid a horizontal line be drawn proportional to the pressure at that point; find the locus of the extremities of such lines; and thence deduce the amount of the whole pressure upon one of the vertical sides of a cube filled with fluid, and the point of application of the resultant of the pressures. We obtain from (4) that pressure at any point of a fluid = σ%, where σ specific gravity of fluid, ≈ = the surface. z = depth of the point below Let ABCD (fig. 51) be a vertical section of the rectangular vessel. If from a point P in AB, a line PQ be drawn proportional to the pressure at P, PQ must, by the proposition, be proportional to AP. Therefore and therefore the locus of Q is the straight line AQ; and therefore the locus required is a plane making a constant angle QAD with the surface of the fluid. Let now the vessel be a cube of which AB a is one vertical edge. Since σ is a numerical quantity, we may take PQ = σAP, in which case PQ will represent the pressure at P. Hence the whole pressure on the line AB is equal to the sum of all the lines similar to PQ which lie between A and BR, i.e. = area ABR. Therefore whole pressure on a vertical side of the cube volume of the solid of which ABR is a vertical section, volume of rectangular box ABRL, = {AB2. BR, = α. σα, = 10a3. Again, to determine the point of application of the fluid pressures on AB is nothing more than to determine the point of application of a system of parallel forces proportional to lines such as PQ; which problem is exactly the same as that of finding in which of the lines parallel to BR, of which the triangle ABR is made up, the centre of gravity of the triangle ABR lies. We know that it lies in that line which is at a distance AB from A; the point therefore (in the vertical line bisecting the side of the cube) which is at a distance a below the surface of the fluid, is the point of application of the resultant of the fluid pressures upon the side. 1849. (A). Determine the whole pressure on a surface immersed in a heavy fluid of uniform density. (B). What must be the vertical angle of a conical vessel, in order that when it is placed with its vertex upwards, and filled with heavy fluid through a hole at the vertex, the pressure on the curved surface may be to the pressure on the base as 4 to 3? Prove that the ratio above mentioned cannot for any cone be less than 2 : 3. The whole pressure on a surface whose area is S, which is immersed in a fluid of specific gravity σ, and whose centre of gravity is at depth z below the surface of the fluid, is oz S. If h, c, r are respectively the height of the cone in (B), the length of the slant side, and the radius of the base, area of curved surface = πrC, and depth of its centre of gravity – 3h; therefore whole pressure on curved surface (P1) = }σ.πrch. Again, area of base and depth of its centre of gravity = h, therefore whole pressure on the base (P) = orrh. But if be the semi-vertical angle of the cone, 1849. (4). Determine the conditions of equilibrium of a floating body. (B). A cylindrical vessel, the radius of the base of which is one foot, contains water: if a cubic foot of cork (sp. gr. 24) be allowed to float in the water, find the additional pressure sustained by the curved surface, and by the base, respectively. That a body may float, its weight must be equal to the weight of the fluid displaced, and the centres of gravity of the body and of the fluid displaced must lie in the same vertical line. The latter condition shews that the cubic foot of cork mentioned in (B) will float in stable equilibrium with a side horizontal; and then the former condition gives us (if x is the depth to which the cork will sink, expressed as a fraction of a foot), (sp. gr. of water) x (1 sq. ft.). x = (sp. gr. of cork). (1 cub. ft.), or × X x = '24. Again, if y be the height through which the water rises when the cork is put in, the fact that the volume of the water remains the same gives or y (area of base of cylinder) = volume of cork immersed, 24 (. radius of the base of the cylinder = 1), Now the increased pressure on the curved surface will clearly be equal to the pressure on a strip of the cylinder of length y supposed added on to the bottom of the cylinder; i.e. (if h is the original height of the water) 2π.y. (h+y), sp. gr. of water being 1, = ·24 ( 2h + •24 -24). The increase of pressure on the base π.(h+y) — π.h, TY, = '24. Obs. These numerical results give the ratios of the to the weight of a unit of volume of water. pressures (B). The specific gravity of coal is about 1.12, that of water being 1, and a cubic foot of water weighs 1000 oz.; find the edge of a cubical block of coal which weighs 2000 tons. The specific gravity of a substance is the ratio of the weight of any portion of its volume to that of an equal volume of some standard substance whose specific gravity is taken as unity. Hence, water being taken as the standard substance, therefore weight of 2000 × 32 cubic feet of coal = 2000 tons. Thus the volume of the block of coal is 64000 cubic feet, and the edge is therefore 40 feet. 1850. (4). Define specific gravity. Given weights of substances of known specific gravity are compounded; find the specific gravity of the compound. (B). Eleven ounces of gold (sp. gr. 19.3) are mixed with one ounce of copper (sp. gr. 8.8), find the specific gravity of the compound, supposing its volume to be the sum of the volumes of the two metals. The result of the investigation in (A) is sp. gr. of compound σσ'. W + W' σW' + σ'W' assuming that the volume of the compound is the sum of the volumes of the components. |