revolve through a very small angle 0 in the plane ADB, so that the quantity of fluid displaced may be the same as before; and let ADB meet the surface of the fluid in a Cb. Draw MF vertical through the centre of gravity of the fluid displaced by the solid in its new position, and mp, nq vertical through the centres of gravity of the wedges ACa, BCb. Then if the plane of floatation be symmetrical with respect to the plane ADB, mp, nq and consequently MF will be in the plane ADB. Draw HFE parallel to ab. Then from the two ways of making up the solid ADb, = (vol. a Db). FE+ (wedge AC a). Cm : (vol. ADB). HE – (wedge BCb). Cn. And if x, y be the co-ordinates of any point in the boundary of the plane of floatation, ACB being the axis of x, and CY (perpendicular to ACa) the axis of y, (wedge ACa). Cm=20 fxy, from C to A, (wedge BCb). Cn=20, x2y, from C to B. But 2y from C to A+ 2y from C to B=2,ay from A to B = k2 A, k2 A being the moment of inertia of the plane of floatation round CY, .. (wedge ACa). Cm + (wedge BCb). Cn=0k2 A. And wedge ACa wedge BCb, therefore = 20xy from C to A=20xy from C to B, therefore C is the centre of gravity of the plane of floatation. Also if the volume of the fluid displaced = V, (vol. ADB). HE — (vol. a Db). FE=V.HF, HF=HM.0, .. V.HM=k2 A. The point M, in which FM ultimately cuts HG is called the metacentre. A force acting in the direction FM will tend to diminish or increase the angle HMF according as M is above or below G, therefore the equilibrium of the solid is stable or unstable according as M is above or below G. If the plane of floatation be not symmetrical with respect to ADB, let a Yb (fig. 16.) be the section of the solid made by the surface of the fluid; and let H, G, &c. be the projections of H, G, &c. in (fig. 15.) on the plane a Yb. Draw pr, q8, MN perpendicular to ab. It may be proved as before, that the centre of gravity of the plane of floatation lies in CY, and that V.HNk 40, Also V.MN+ (wedge Ya Y').pr= (wedge YbY').q8; (wedge Ya Y').pr=0. » Sy xy, from C to a; 0.ffy xy from C to b−0. f. fyxy from C to a V.MN=TO. And the equilibrium will be stable or unstable according as H and G lie on the same or opposite sides of MN. 27. To determine the small oscillations of the solid DC (fig. 15.) when left to itself, after its equilibrium has been slightly disturbed, the solid being symmetrical with respect to the plane ADB. Let the figure represent the position of the solid at the end of the time t from the beginning of the motion; and let be P the density of the fluid, K the radius of gyration of the solid revolving round G in the plane ADB, HG=c; then, retaining the notation of Art. 26., the moment of the pressure of the fluid tending to turn the solid round G in the direction FMG and the moment of inertia of the solid round G in the plane ADB=KpV, SECTION III. ON THE EQUILIBRIUM OF ELASTIC FLUIDS ACTED ON BY ART. 28. To measure the pressure of the atmosphere. Let a glass tube ABC (fig. 17.) closed at the end A, be bent at B, so that the branches AB, BC may be parallel and AB about thirty one inches longer than BC. Then if AB and part of BC be filled with mercury, and placed in a vertical position, the mercury will rise in BC, and sink in AB, (leaving a vacuum in the upper part of the tube,) till the pressure of the mercury at the common surface of the air and mercury in BC is equal to the pressure of the atmosphere. Let P, Q be points in the upper and lower surfaces of the mercury; through P, Q draw horizontal planes cutting a vertical HK in H and K. Let II be the pressure of the atmosphere, σ the density of the mercury; then (Art. 12. Cor. 1.) the pressure of the mercury at Q=go.HK; and this must be equal to the pressure of the atmosphere at Q when the mercury is at rest, .. Ilgo.HK. An instrument of this description furnished with a scale for measuring HK, is called a barometer. 29. The expansion of mercury between the temperatures 10 of melting snow and boiling water is of its volume at the 555 former temperature, and the increment of its volume is very nearly proportional to the increment of its temperature. Hence if oo, o be the densities of mercury at 0°, t° (Centigrade) If HK (fig. 17.)=h, t the temperature of the mercury in ABC, II=go1.h=go,h÷ (1 + nearly. t 5550)=go,h. {1-(0,00018)} At the level of the sea in latitude 50° the mean value of h is 30,035 inches, the temperature of the air being 12o.2. 30. The pressure of air at a given temperature varies inversely as the space it occupies. (I.) Let ABD (fig. 18.) be a glass tube having an open capillary termination at A, and bent at B so that the branches AB, BD may be parallel, and PC the axis of AB vertical. Pour mercury into DB till it rises to P, cutting off the communication between AB and BD, and then seal the aperture at A. Let mercury be now poured in till the horizontal plane through the surface of the mercury in BD meets PC in any point_C'; and let the surface of the mercury in AB meet PC in M. Then if h be the altitude of the mercury in the barometer, and σ its density at the time of making the experiment; u, v the capacities of the portions AM, AP of the tube; M, II the pressures of the air in the tube when occupying the spaces u, v: we have (Arts. 29. 12.) II= pressure of the exterior air=goh, M=goh+go.CM. But if u, v be measured, it will be found that v: u=h+MC: h, .. M : II=v : u. (II.) Let AB (fig. 19.) be a glass tube having an open capillary termination at A; P, C any two points in its axis. Immerse AB vertically in mercury till the surface of the mercury meets PC in P, and seal the aperture at A. Elevate the tube till the plane of the surface of the mercury outside meets PC in C; and let the surface of the mercury within the tube meet PC in M. Then as before if u, v be the capacities of the portions AM, AP of the tube; M, II the pressures of the air in the tube when occupying the spaces u, v: we have см II=pressure of the exterior air=goh, M=goh-go.MP. And if u, v be measured, it will be found that v: u=h― MP: h, : II = v u. Hence the pressure of air at a given temperature varies inversely as the space it occupies, when the pressure is less than that of the atmosphere, as well as when it is greater. 31. Let po be the density of atmospheric air under the pressure II, at the temperature of melting snow; then since the pressure of air at a given temperature varies inversely as the space it occupies, and therefore directly as its density, II=μ pos is constant. where If h be the altitude of the mercury in the barometer, o its density, II=goh, therefore goh upo Το It is found that =10467, σ, being the density of mercury Po at 0°, when h=0,76 metres = 29,9218 inches, and g=9,8088 metres 386,18 inches, = .. (u)=279,33 metres=916,46 feet. 32. The expansion of air between the temperatures of melting snow and boiling water, under a constant pressure, is equal to 0,375 of its volume at the temperature of melting snow, and the increment of its volume is proportional to its temperature above 0°, as indicated by a mercurial thermometer. Hence if be the densities of air at the temperatures 0°, T° Роз Рт under the same pressure, p= {1+(0,00375)T}p, therefore II=μ. {1+(0,00375) T} pr 33. When a given mass of atmospheric air is suddenly compressed or dilated, its temperature is increased or diminished according to the following law. Let p, To; P1, T° be the corresponding densities and temperatures of a given mass of air: then, T, — T = (111,25). (1 If, in making the first experiment described in Art. 30, the mercury be suddenly poured into the tube DB, the temperature C |