FLUIDS IN ROTATION.] MECHANICAL PHILOSOPHY. HYDROSTATICS. 



761 



that is, the surfaces are at the same height above a hori- 

 zontal plane, or are level. Suppose above A and C, two 

 other fluids of specific gravities S", S'", to be introduced 

 into the tube, and that the equilibrium remains undis- 

 turbed ; then the new fluid in C' C communicates to the 

 horizontal plane A the only upward pressure it receives, 

 and this by hypothesis is balanced by the downward 

 pressure of the new fluid in A' A. Hence, as before, 



A'E'xS" = C'F' x S" .'. AE. S + AE*. S" 

 C F . S' + C'F' . S"' 



and so on for any number of superposed fluids in equi- 

 librio. Hence, if any number of fluids be in equilibrio 

 in a bent tube, the sum of the products of the vertical 

 heights into the specific gravities of the fluids in one 

 branch of the tube, will be equal to the sum of the 

 corresponding products in the other branch. The vertical 

 height of each fluid is estimated from the horizontal 

 plane on which it rests ; the height of the lowest fluid 

 H B C is estimated from the lowest point B ; the vertical 

 distance of B from H is the height in one branch of the 

 tube ; and the vertical distance of B from C, is the height 

 in the other branch. 



In illustrations such as the above, the tubes are always 

 represented as circular ; but, as the reasoning shows, 

 the shape of the channel which connects the two upper 

 surfaces of a fluid together does not enter into considera- 

 . A pool of water connected by underground open 

 passages, however tortuous, with an adjacent river, will, 

 on the above principle, have its surface on the same 

 level aa that of the river. Also, water conveyed by 

 pipes from a reservoir can never deliver a supply to any 

 place, on a higher level, than the surface of the source, 

 without the application of mechanical force or pressure. 



The student must be apprised, that in discussing the 

 circumstances of the equilibrium of fluids in tubes, 

 though the shape of the tube is of no moment, nor are 

 the sectional area of the branches at any height, provided 

 only that these exceed a certain amount of smallness ; 

 yet, if the section at any part be so small as to bring into 

 operation what is called capillary attraction, the fluid 

 surfaces in the two branches will not stand at the same 

 level. If the section be circular, the diameter of it 

 should exceed ^th of an inch, in order that capillary 

 attraction may not oppose that of gravity, and thus 

 lessen downward pressure. We shall devote a short 

 article to this curious subject presently. 



PROBLEM. Equal lengths of two fluids which do not 

 intermix, and whose specific gravities are as m to 1, are 

 poured into a uniform circular tube ; required the posi- 

 tion in which they will rest. 



Let P C H (Fig. 181) be the heavier fluid, and f H the 

 lighter, the lengths of the arcs P H, P' H being equal ; 



Fig. 181. 



let a represent each of 

 these lengths ; and 

 draw P M, P' M' each 

 perpendicular to the 

 vertical diameter, as 

 alsoHNR. 



Then the lighter fluid 

 P' H is upheld entirely 

 by the pressure of the 

 portion P R of the 

 in ;i\ i>r fluid ; hence 

 the downward pressure 

 of the lighter fluid mi 

 the horizontal plane H 

 must be equal to the upward pressure of the portion P R 

 of the heavier fluid, transmitted to that plane. The 

 heights of these equilibrating portions of fluid are re- 

 pctively M' N, and M N ; and as these are inversely as 

 their spociBc gravities, from what is shown above, we 

 have 

 M'N:MN: : m : 1 /. M'N = m . MN (1) 



Let the arc C P be represented by a; .'. C H = a x, 

 and C P' = a x + a = 2a x, then M' N and M N 

 may be found in terms of a and x as follows, observing 

 that 



VOL. I. 



M' N = M' C N C 

 and M N = M C N C. 

 M' C = ver. sin. (2a x), N C = ver. sin. (a x), 



M C = ver. sin. x ; 

 .'.M'N = ver. sin. (2a x) ver. sin. (a x), 



M N = ver. sin. x ver. sin. (a x); 

 .'. (1) ver. sin. (2a x) ver. sin. (a x) = m ver. sin. 



x m ver. sin. (a x) . . (2). 



Or, since ver. sin. = R cos. , where R is the radius of 

 the circle, we may use cosines instead of versed sines ; 

 and reduce the geometrical to the ordinary trigonome- 

 trical cosines ; that is, to the scale R = 1 ; we shall only 

 have to regard and x as standing, not for lengths of 

 arc, but for the degrees in those lengths ; we shall thus 

 have from (2) 



1 cos. (2a x) 1 + cos. (a x) = m m 



cos. x m -\- mcos. (a x) ; 



that is, cos. (a x) cos. (2a x) = m cos. (a x) 

 mcos. x; 



.'.cos. (2a x) + (m 1) cos. (a x) = mcos. x; 

 that is, cos. 2a cos. x + sin. 2a sin. x + (m 1) 

 (cos. a cos. x -f- sin. a sin. x) = m cos. x ; 



.'. cos. 2a + sin. 2a tan. x + (m 1) 



(cos. a + sin .a tan. x) = m ; 

 m cos. 2a (m l)cos. a 



sin. 2a -f- (m 1) sin. a 

 From this expression the trigonometrical tangent of 

 the arc x may be found from the tables ; and thence the 

 number of degrees in that arc, or the angle which C P 

 subtends at the centre of the circle. 



Let a = 2 , then cos. a = 0, cos. 2a = 1, sin. a = 1, 



sin. 2a = 0, and the expression in these circumstances 

 becomes 



m+1 



.". tan. x 



.(3) 



tan. x 



(4) 



m 1 



If m = 1, that is, if the two fluids be the same, then 

 (3) gives 



1 cog. 2o 1 (1 2 sin. 2 a) 

 sin. 2a 2 sin. a cos. a 



sin. a 

 = = tan. a. 



tan. x 



that is, the arcs a and x are equal, as we otherwise know 

 they must be. 



EQUILIBRIUM OF A ROTATING FLUID. In 

 all the preceding investigations, the only force supposed 

 to act on the fluid is the force of gravity ; we shall give 

 an illustration of a case in which the equilibrium is due 

 to the united effects of gravity and centrifugal force. 



PROBLEM. A cylindrical vessel, containing water or 

 any other liquid, revolves round its axis, perpendicular 

 to the horizon, with a given angular velocity ; to de- 

 termine the form assumed by the internal surface of the 

 fluid. 



The rotation of the vessel necessarily puts the con- 

 tained fluid in motion, and this motion continues till all 

 the forces balance and the fluid attains a fixed form 

 and position ; it is in this state of equilibrium that we 

 have to consider it. 



Suppose A H D (Fig. 182) to be the form assumed by 

 the surface of the fluid when the state of equilibrium is 

 attained, and P any particle on that surface. From P 

 draw P M perpendicular to the 

 axis of the cylinder, and P N per- 



Fig. 182. 



pendicular to the curve at P. The 

 forces which keep P in equilibrio * 

 are 1. Gravity, acting in the 

 vertical direction N M ; 2. The 

 centrifugal force, acting in the 

 horizontal direction M P ; and 3. 

 The pressure or reaction of the 

 particle in the direction P N per- 

 pendicular to the surface ; for it is 

 plain that the vertical and hori- 

 zontal forces acting on P are the only forces which causo 



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