In order that the fluid in a canal leading from A to a point in the surface of the exterior fluid, may be in equilibrium, we must have K-H+g.AC = K, ·. H = gbc; If a be the angle between a tangent to APB at B and AN, V the volume of the fluid in the tube elevated above the surface of the exterior fluid, depends only on the nature of the fluid, and of the substance of which the tube is formed. When the fluid is water, and the tube of glass, V = (0,023444) πa, a being expressed in linear inches, and V in cubic inches. Also when a is small compared with c, the surface of the fluid is a concave hemisphere, ... V = wа3c+}πα3, ··· ac +}a2 = 0,023444. If day = tan 0, and a be very small compared with c, fyx is small compared with ca2, and 48. To determine the surface of a fluid between two parallel vertical plates. Let D'APD (fig. 24.) be a section of the surface of the fluid and of the parallel plates, made by a plane perpendicular to their surfaces; AC equidistant from the parallel plates, meeting the surface of the interior fluid in A, and the plane of the surface of the exterior fluid in C. Then, the rest of the construction and the notation being the same as in Art. 47.) we have d2y g 2 = y + b = { 1 + (d_y)2 } } ' H When the fluid is water, and the plates of glass, and very close to each other, B'AB is a semicircle, ... area D, B" ABD = 2αe +2α2 — —πα2 = 2ac + 3a2 nearly, H and = g It appears that the elevation of a fluid between two parallel plates, is nearly half the elevation in a tube whose diameter is equal to the distance between the plates. have When a single plate is immersed vertically in a fluid, we This is the differential equation to the elastic curve. The investigation of the form of the surface of the fluid when it is depressed, leads to precisely the same equations as when it is elevated, the sign of y being changed. If V be the space between the surface of the mercury in a vertical glass tube and the plane of the surface of the mercury on the outside, and a the radius of the tube, V = (0,01) π α. 49. To find the tension of a flexible cylindrical vessel containing fluid. Let MK, PQ, HL (fig. 25.) be equidistant sections of the cylinder made by planes perpendicular to its axis. Draw PE, QE normals at the extremities of the small arc PQ; MPH, KQL perpendicular to PEQ; and let p be the pressure of the fluid at P, t. MH the tension of MH or KL, r the radius of curvature of PQ. Now ML is kept at rest by the pressure of the fluid, and the tensions of its edges; the tensions of MH, KL, and the pressure of the fluid, are the only forces that act in the plane PEQ; the tensions act perpendicular to PE, QE respectively, and the pressure of the fluid acts perpendicular to PQ, therefore ultimately EP = PQt. MH p. (area ML)=t: p.PQ; . t pr. 50. To find the tension of a vessel of any form containing fluid. = Let PCP', QCQ' (fig. 26.) be the normal sections of least and greatest curvature of the vessel at C; PC PC, QC = QC; PE, PE, QF, QF normals at the extremities of the small arcs PCP, QCQ; MPK, HPL sections of the vessel made by planes perpendicular to PEP'; MQH, KQL sections of the vessel made by planes perpendicular to QFQ. Let p be the pressure of the fluid at C, t. QQ' the tension of HL or MK, v. PP' the tension of MH or KL; r, s the radii of curvature of PCP, QCQ. ML is kept at rest by the pressure of the fluid, and the tensions of its edges; therefore the resultant of the tensions must be equal and opposite to the pressure of the fluid. The resultant of the tensions of MK, HL. And the resultants of the tensions act in the direction CE; the pressure of the fluid = p. PP'. QQ', and it acts in the direction EC, When the tensions are the same in every direction, or v = t, When the vessel is immersed in fluid, p is the difference of the pressures of the interior and exterior fluids. E SECTION V. ON THE MOTION OF FLUIDS. ART. 51. WHEN an incompressible fluid flows through a tube, the velocities of the fluid at any two points, are inversely proportional to the areas of the perpendicular sections of the tube at those points; supposing the tube to continue always full, and the velocities at all points in the same section, to be equal to one another, and perpendicular to the section. For equal volumes of fluid must pass through each section in the same time; and if u, v be the velocities at the two sections; H, K the areas of the sections; uHt, v Kt will be the volumes of the fluid that passes through the two sections in the small time t; and these are equal, 52. .. UH = vK, and .. u: v=K : H. When a fluid is in motion acted on by any forces, to determine the effective accelerating force in the direction of its motion at any point. Draw the curve APQR (fig. 27.) so that a tangent to it at any point may be in the direction of the motion of the fluid at that point. The motion of the fluid will not be altered if we suppose a portion of it, of the form of a very small cylinder having PQ for its axis, to become solid for an instant. Let Ρ be the pressure of the fluid at P, S the accelerating force at P resolved in the direction of a tangent to APR at P, AP = s; then p + d.p. PQ will ultimately be the pressure at Q; and if p be the density of the fluid at P, κ the area of the base of the cylinder PQ; the mass of PQ = pk. PQ, and the moving force K |