SECTION VI. ON RESISTANCES. ART. 67. THE resistance of a fluid on a solid moving in it, is the resultant of the excess of the pressure of the fluid on the solid in motion, above the pressure of the fluid on the solid at rest. Let APB (fig. 32.) be a solid, moving in a fluid with the velocity V in the direction BA. Now if we communicate to the fluid and the solid a velocity V in the direction AB, the pressure of the fluid on APB will not be altered; and the solid will be at rest in a fluid moving in the direction AB with a velocity V. Hence the force with which a fluid in motion impels a solid immersed in it, is equal to the resistance of a stagnant fluid on a solid in motion, the velocity of the fluid, in one case, being equal to the velocity of the solid in the other. So also, when both the solid and fluid are in motion, the resistance on the solid, is equal to the force with which the solid at rest would be impelled by a stream moving with the relative velocity of the fluid and solid. Let an enveloping cylinder parallel to AB touch the solid in the curve PQR. The pressure on the surface RAPQ, will upon the whole be greater, and the pressure upon PBQR, less, than when the solid and fluid are relatively at rest. the following Articles we shall consider that part only of the resistance, which arises from the increased pressure on RAPQ. In It must be observed that the theory of resistances is very imperfect, and that it is useless to expect any close agreement between the results deduced from it, and those obtained by experiment. 68. To find the force with which a stream impels a plane, the plane being perpendicular to the direction of the stream. Let P (fig. 33.) be a point in the plane; EP, the direction of the stream, perpendicular to the plane; p the pressure at P; p the pressure of the fluid at P before the plane was immersed, or, the pressure of the fluid at the point P in a plane moving with the same velocity, and in the same direction as the fluid; the density of the fluid; K the area of the plane. Ρ Then p' -p will be the resistance on an unit of the plane and after the plane is immersed, the velocity at P = 0, and the resistance on the plane = (p' − p) K = 1⁄2 pv2 K. pv2K is the weight of a column of fluid having the given plane for its base, and whose altitude is the space due to the velocity of the fluid. If the plane be made to move in a direction perpendicular to EP, it is manifest that the force with which the stream impels the plane, will not be altered. Hence the force with which a given fluid impels a given plane, depends only on that part of the relative velocity of the fluid and plane, which is perpendicular to the plane. Also, since the resistance, or impelling force of the fluid arises from the pressure of the fluid on the plane, it must act in a direction perpendicular to the plane. 69. A stream impinges obliquely on a plane; to find the force with which the stream impels the plane. Let P (fig. 33.) be a point in the plane; AP the direction of the stream; EP perpendicular to the plane; v the velocity of the stream; R the resistance, or the force with which the stream impels the plane. The velocity of the stream estimated in the direction EP = v.cos APE, .. R = 1⁄2 pv2. (cos APE)2 K. COR. 1. The resolved part of the impelling force estimated in the direction of the stream = R.cos APE = pv2. (cos APE)3 K. COR. 2. The resolved part of the impelling force estimated in a direction perpendicular to the stream, and in the plane APE, = R.sin APE = pv2. (cos APE)2. sin APE. K. 70. A cylinder having the curve BPC (fig. 34.) for its base, is immersed in a stream flowing in the direction AE; to find the force with which the stream impels the cylinder in the directions AE and MA. Draw AN perpendicular to AE; PM, QN parallel to AE; PE a normal to BP at P. Let a be the altitude of the cylinder, MP = x, AM=y, MN = dy, R the impelling force, or resistance, on that part of the cylinder which stands on BP, estimated in the direction AE, therefore ultimately d1R.dy resistance on that part of the cylinder which stands on PQ = · 1 pv2 (cos AEP)3a. PQ = 1⁄2 pv2 (cos AEP)2ady, α and tan AEP = − d„a, ..d ̧R = 1⁄2, pv2 1 + (d„a)2 * = So also, if S be the resistance on the part BP of the cylinder, estimated in the direction MA, dyS. Sy = pv2 (cos AEP) sin AEP.a.PQ = 1 pv2. cos AEP. sin AEP.a.dy; dyx .. dyS = 1 pv2 a 1 + (d,x)2 and S = 1 pv2 a ↓↓ dyx The integrals must be taken between the limits corresponding to B and C. 71. A solid is generated by the revolution of the curve BPC (fig. 34.) round AE; to find the force with which it is impelled by a stream moving in the direction AE. Let R be the resistance on that part of the solid which is generated by the revolution of BP round AE; then, retaining the construction and notation of the preceding Article, we have ultimately, d, R. dy = 1 v2 уду 1 + (d ̧x)2 ; 72. To find the resistance on a sphere. Let the centre of the sphere be the origin of the co-ordinates, a the radius of the sphere, and, therefore, x2 + y2 = a2 the equation to its generating circle. Then, y + xd,x = 0, a2 = x2 + y2 = x2 {1 + (d ̧x)2} = (a2 − y3) {1 + (d,x)2}, and the resistance on the sphere = p22πа2. Αρυπα. a2 COR. 1. The resistance on a circular plate, the radius of which is a, = p2πa2, therefore the resistance on a sphere is half the resistance on a circular plate of the same radius as the sphere. COR. 2. If the density of the sphere = σ, its mass =σπa3, and the retarding force arising from the resistance of the fluid = (resistance) ÷ (mass of the sphere) G 3 P v2 = 16 σα SECTION VII. DESCRIPTION OF INSTRUMENTS. METHODS OF FINDING ART. 73. ALMOST all bodies expand by heat, and contract by cold. This property furnishes the only known mode of comparing and recording the temperatures to which any body is exposed. The expansions of mercury, or air, combined with that of the glass vessel in which they are contained, are usually employed for this purpose. 74. The common mercurial thermometer is a glass tube of uniform bore, having a bulb at one end, which, with part of the tube, is filled with mercury; the other end is usually sealed, the space between it and the mercury being a vacuum. To fill the thermometer with mercury, the air must be partly expelled from the bulb by holding it over the flame of a lamp, and then, the other end, which is open, immersed in mercury. As the bulb cools, the mercury will be forced into it by the pressure of the atmosphere. If a paper funnel be now tied round the open end, and filled with mercury; and the mercury in the bulb be heated till it boils, the remainder of the air will be driven out, and its place supplied by mercurial vapour: this condenses on cooling, and the mercury will descend from the funnel and fill the instrument completely. When it has cooled down nearly to the highest temperature intended to be measured by it, the open end must be sealed; and as it continues to cool, the mercury will descend leaving a vacuum in the upper part of the tube. |