Lecture6 Hydrostatic Force on curved Surfaces

1 Hydrostatic Force on a curved SurfacesHenryk Kudela1 Hydrostatic Force on a curved SurfaceOn a curved surface the forcesp Aon individual elements differ in direction, so a simplesummation of them may not be made. Instead, the resultant forces in certain directionsmay be determined, and these forces may then be combined vectorially. It is simplest tocalculate horizontal and vertical components of the total component of Hydrostatic forceAny curved surface may be projected on to a vertical plane. Take, for example, the curvedsurface illustrated in Fig. 1: Hydrostatic Force on a curved surfaceIts projection on to the vertical plane shown is representedby the trace AC. LetFxrepresent the component in this direction of the total forceexerted by the fluid on thecurved act through the center of pressure of the vertical projection andis equal in magnitude to the Force F on the fluid at the any given direction, therefore, the horizontal Force on any surface equals the forceon the projection of that surface on a vertical plane perpendicular to the given line of action of the horizontal Force on the curved surface is the same as that of theforce on the vertical component of Hydrostatic forceThe vertical component of the Force on a curved surface may bedetermined by consideringthe fluid enclosed by the curved surface and vertical projection lines extending to the freesurface.

2 ThusFH=F2= gzs(1)FV=F1= gV(2)whereVis the volume of the liquid between the free surface liquid and solid curved magnitude of the resultant is obtained form the equationFR= F2H+F2V(3)Example 1A sector gate, of radius 4 m and length 5m, controls the flow of water in ahorizontal channel. For the (equilibrium) conditions shown in Fig. 2 , determine the totalthrust on the 2: Hydrostatic Force at a curved the curved surface of the gate is part of a cylinder, thewater exerts no thrust along itslength, so we consider the horizontal and vertical components in the plane of the horizontal component is the thrust that would be exertedby the water on a verticalprojection of the curved surface. The depth d of this projection is4 sin 30o= 2m, (R=4m) and its centroid is1 +d/2 = 2mbelow the free surface.

3 Therefore horizontal forceFHis equalFH= gzsA= 1000 2 (5 2) = 105 NIts line of action passes through the center of pressure of the vertical projection, that is,at a distanceIx/Azsbelow the free surface, given by:IxAzs=Is+Az2sAzs=bd312(b d)zs+zs=212 2+ 2 + vertical component of the total thrust is equal of weightof imaginary water (4 4 cos 30o) = Force is equalFV= gV. ThenFV= 1000 5[( 1) + 42 30360 12 2 4 cos 30o]= 104 NThe horizontal and vertical components are co-planar and therefore combine to give asingle resultant Force of magnitude|F|= F2H+F2V= ( )2+ ( )2= 105 Nat angle equal toarctg(6180196200) = the horizontal dam in Fig. 3 is a quarter circle 50 m wide into the paper. Determine the horizontaland vertical components of the Hydrostatic Force against the dam and the point CP wherethe resultant strikes the 3: Hydrostatic Force at a curved dam( , FV= )Problem the horizontal and vertical components of the Hydrostatic Force on the quarter-circle panel at the bottom of the water tank in Fig.

4 43 Figure 4: Hydrostatic Force at the bottom of water tank2 BuoyancyBecause the pressure in a fluid in equilibrium increases withdepth, the fluid exerts aresultant upward Force on any body wholly or partly immersedin it. This Force is knownas thebuoyancy. This buoyancy Force can be computed using the same principles usedto compute Hydrostatic forces on Surfaces . The results are the two laws of buoyancydiscovered by Archimedes in the third century :1. a body immersed in a fluid experiences a vertical buoyant Force equal to the weightof the fluid it a floating body displaces its own weight in the fluid in whichit g(4)In fig. 5 the vertical Force exerted on an element of the body inthe form of vertical prismof cross section Ais F= (p2 p1) A= gh A= g Vin which Vis the volume of the prism. Integrating over the complete body gives theformula (4).

5 Weighting an odd-shaped object suspended in two different fluids yields sufficient datato determine its weightW, volume V, specific weight = g(or density ) and specificgravitySG= water. Figure 6 shows two free-body diagrams for the same object suspendedand weighed in two the weight when submerged and 1= 1gand 2= 2gare the specific weights of the , the weight and volume of the4 Figure 5: Vertical Force componets on element of bodyobject, respectively, are to be found. The equations of equilibrium are writtenF1+V 1g=W,F2+V 2g=Wand solvedV=F1 F2g( 1 2),W=F1 1 F2 2 2 1 The hydrometr uses the principle of buoyant Force to determine specific gravities of 6: Free-body diagrams for body suspended in a fluidFigure 7 shows a hydrometr in two liquids. It has a stem of prismatic cross the liquid on the left to be distilled water,SG= , the hydrometr floats inequilibrium whenV0 g=Win whichV0is the volume submerged, g= is the specific weight of water, andWis theweight of the hydrometr.

6 The position of the liquid surface is marked as on the stem5to indicate unit specific gravitySG. When the hydrometr is floated in another liquid theequation of equilibrium becomes(V0 V)SG g=Win which V=a h. Solving for husing above equations gives h=V0aSG 1 SGfrom which the stem can be marked off to read specific gravitiesor 7: Hydrometr in water and in liquid of specific [1] F. M. White, mechanics , McGraw-Hill.[2] B. R. Munson, Young and T. H. Okiisshi, of fluid mechanics ,John Wiley and Sons, Inc..[3] B. S. Massey and J. Ward-Smith, 2006, mechanics of Fkuids, Taylor and Francis6