hydrostatic force on a plane surface

2.8 Hydrostatic Force on a Plane Surface 61

(a)

I

I1

l

Diaphragmstop

Beam (strain gages deposited on beam)(b)

Hydrostatic Force on a Plane Surface

. FIGURE 2.15(a) Two different sizedstrain-gage pressuretransducers (Spectra-med Models PIOEZand P23XL) com-monly used to mea-sure physiologicalpressures. Plasticdomes are filled withfluid and connected toblood vessels througha needle or catheter.(Photograph courtesyof Spectramed, Inc.)(b) Schematic diagramof the P23XL trans-ducer with the domeremoved. Deflection ofthe diaphragm due topressure is measuredwith a silicon beam onwhich strain gagesand an associatedbridge circuit havebeen deposited.

When a surface is submerged in a fluid, forces develop on the surface due to the fluid. Thedetermination of these forces is important in the design of storage tanks, ships, dams, andother hydraulic structures. For fluids at rest we know that the force must be perpendicular tothe surface since there are no shearing stresses present. Wealso know that the pressure will

vary linearly with depth if the fluid is incompressible. For a horizontal surface, such as thebottom of a liquid-filled tank (Fig. 2.16), the magnitude of the resultant force is simplyFR = pA, wherep is the uniformpressureon thebottomandA is the areaof thebottom.Forthe open tank shown, p = yh. Note that if atmospheric pressure acts on both sides of thebottom, as is illustrated, the resultant force on the bottom is simply due to the liquid in thetank. Since the pressure is constant and uniformly distributed over the bottom, the resultantforce acts through the centroid of the area as shown in Fig. 2.16.

For the more general case in which a submerged plane surface is inclined, as is illus-trated in Fig. 2.17, the determination of the resultant force acting on the surface is moreinvolved. For the present we will assume that the fluid surface is open to the atmosphere. Letthe plane in which the surface lies intersect the free surface at 0 and make an angle e with

I

I

62 Chapter 2 / Fluid Statics

. FIG UR E 2. 1 6 Pressure and resultanthydrostatic force developed on the bottom of anopen tank.

i

:!~jij3iij,1

If,,:iiI!1.~:!~;j

j]j~

~~

(

j.'

this surface as in Fig. 2.17. The x-y coordinate system is defined so that 0 is the origin and yis directed along the surface as shown. The area can have an arbitrary shape as shown. Wewish to determine the direction, location, and magnitude of the resultant force acting on oneside of this area due to the liquid in contact with the area. At any given depth, Ii, the forceacting on dA (the differential area of Fig. 2.17) is dF = 'Yh dA and is perpendicular to thesurface. Thus, the magnitude of the resultant force can be found by summing these differentialforces over the entire surface. In equation form

FR = L 'YhdA =)L 'YYsin e dA

where h = y sin e. For constant 'Yand e

FR = 'Ysin e L y dA(2.17)

!

I

t.~

I

,,,,,,x ', '

XR

~~"'/"/ //

//

Location ofresultant force

(center of pressure, CP)

. FIGURE 2.17Notation for hydro-static force on an in-cIined~plane surfaceof arbitrary shape.

- -1

if


The integral appearing in Eq. 2.17 is the first moment of the area with respect to the x axis,so we can write

L Y dA = YeA

where Yc is the Y coordinate of the centroid measured from the x axis which passes through O.Equation 2.17 can thus be written as

FR = yAyc sin ()

or more simply as

I FR = yhcA I

(2.18)

where he is the vertical distance from the fluid surface to the centroid of the area. Note thatthe magnitude of the force is independent of the angle () and depends only on the specificweight of the fluid, the total area, and the depth of the centroid of the area below the surface.In effect, Eq. 2.18 indicates that the magnitude of the resultant force is equal to the pressureat the centroid of the area multiplied by the total area. Since all the differential forces thatwere summed to obtain FRare perpendicular to the surface, the resultant FRmust also beperpendicular to the surface.

Although our intuition might suggest that the resultant force should pass through thecentroid of the area, this is not actually the case. The y coordinate, YR' of the resultant forcecan be determined by summation of moments around the x axis. That is, the moment of theresultant force must equal the moment of the distributed pressure force, or

FRYR= L Y dF = LYSin () y2 dA

and, therefore, since FR = yAy c sin ()

L y2 dA

YeAYR =

The integral in the numerator is the second moment of the area (moment of inertia), Ix, withrespect to an axis formed by the intersection of the plane containing the surface and the freesurface (x axis). Thus, we can write

IxYR = -

YeA

Use can now be made of the parallel axis theorem to express Ix as

Ix = Ixc + Ay~

where Ixc is the second moment of the area with respect to an axis passing through its centroidand parallel to the x axis. Thus,

- Ixc + YcYR - YeA

(2.19)

Equation 2.19 clearly shows that the resultant force does not pass through the centroid but isalways below it, since IxclYcA > O.

--~---~ - ~ --- --~ --

--- lI


The x coordinate, XR' for the resultant force can be determined in a similar manner bysumming moments about the Y axis. Thus,

F~R = L '}' sin 8 xy dA

and, therefore,

XR =L xy dA

YeA= [xy

YeA

where [xyis the product of inertia with respect to the x and Y axes. Again, using the parallelaxis theorem,l we can write

[xye + xeXR = YeA

(2.20)

{:Ix, = 112 ba3

I - 1 b3

ye - I2a

A =ba

Ixy,= 0(a)

rrR2A=- 2

Ixc = 0.1O98R4

lye = 0.3927 R4

Ixy,= 0

(c)

4R1 13rr

4RI3rr

A =rrR2

Ixc = lye = rrR44

Ixy,= 0

(b)

A=ab I=ba32 xc 36

ba2Ixye= n(b - 2d)

(d)

A = rrR24

Ixc = lye = 0.05488R4

IXyc= -0.01647R4

. FIGURE 2.18

(e)

Geometric properties of some common shapes.

]Recall that the parallel axis theorem for the product of inertia of an area states that the product of inertia with respect to an orthogonalset of axes (x-y coordinate system) is equal to the product of inertia with respect to an orthogonal set of axes parallel to the originalset and passing through the centroid of the area, plus the product of the area and the x and y coordinates of the centroid of the area.

Thus, Ixy = Ixy, + Axcy c'

I - --- -~ ~ ~~~~.-~---~- ~~~ --'-~--- -


where Ixye is the product of inertia with respect to an orthogonal coordinate system passingthrough the centroia of the area and formed by a translation of the x-y coordinate system. Ifthe submerged area is symmetrical with respect to an axis passing through the centroid andparallel to either the x or y axes, the resultant force must lie along the line x = xc' since Ixyeis identically zero in this case. The point through which the resultant force acts is called thecenter of pressure. It is to be noted from Eqs. 2.19 and 2.20 that as Ye increases the center ofpressure moves closer to the centroid of the area. Since Ye = hclsin e, the distance Ye willincrease if the depth of submergence, he' increases, or, for a given depth, the area is rotatedso that the angle, e, decreases. Centroidal coordinates and moments of inertia for some com-mon areas are given in Fig. 2.18. .

by

leI

The 4-m-diameter circular gate of Fig E2.6a is located in the inclined wall of a large reservoircontaining water (y = 9.80 kN/m3). The gate is mounted on a shaft along its horizontaldiameter. For a water depth of 10 m above the shaft determine: (a) the magnitude and location'of the resultant force exerted on the gate by the water, and (b) the moment that would haveto be applied to the shaft to open the gate.

~O)

0

/"x 1

00

;,Y /I I

I II I

I l {j°-

I I a0';' / "'.$

~/ vj)'~ ~)Centerof

pressure

(a)

. FIGURE E2.6

SOLUTION

F,.,k~~MOx

ow(c1

(a) To find the magnitude of the force of the water we can apply Eq. 2.18,

FR = yheA

and since the vertical distance from the fluid surface to the centroid of the area is 10mit follows that

FR = (9.80 X 103 N/m3)(10 m)(41T m2)

= 1230 X 103 N = 1.23 MN (Ans)

To locate the point (center of pressure) through which FRacts, we use Eqs. 2.19and 2.20,

alaJa. Ixye + Xc

XR = YeA

I

- Ixe + YeYR - YeA

I

~---~ ~ ~ ~----


For the coordinate system shown, XR = 0 since the area is symmetrical, and the center

of pressure must lie along the diameter A-A. To obtain YR' we have from Fig. 2.18nR4

I =-xc 4

and Ye is shown in Fig. E2.6b. Thus,

(n/4)(2 m)4 10 mY = +-

R (10 m/ sin 60°)(4n m2) sin 60°

= 0.0866 m + 11.55 m = 11.6 m

and the distance (along the gate) below the shaft to the center of pressure is

YR - Ye = 0.0866 m(Ans)

. We can conclude from this analysis that the force on the gate due to the water has amagnitude of 1.23 MN and acts through a point along its diameter A-A at a distance of0.0866 m (along the gate) below the shaft. The force is perpendicular to the gate surfaceas shown.

(b) The moment required to open the gate can be obtained with the aid of the free-bodydiagramof Fig. E2.6c.In this diagram"W is the weight of the gate and Ox and Oy arethe horizontal and vertical reactions of the shaft on the gate. We can now sum momentsabout the shaft

2: Me = 0

and, therefore,

M = FR(YR - yJ

= (1230 X 103 N)(0.0866 m)

= 1.07 X 105 N.m (Ans)

EXAMPLE2.7

A large fish-holding tank contains seawater (y = 64.0 lb/ft3) to a depth of 10 ft as shownin Fig. E2.7a. To repair some damage to one corner of the tank, a triangular section is replacedwith a new section as illustrated. Determine the magnitude and location of the force of theseawater on this triangular area.

SOLUTION

The various distances needed to solve this problem are shown in Fig. E2.7b. Since the surfaceof interest lies in a vertical plane, Ye = he = 9 ft, and from Eg. 2.18 the magnitude of theforce is

FR = yheA

= (64.01b/ft3)(9ft)(9/2 ft2) = 2590 lb (Ans)

:er

IS)

,aofce

jyIreIts

IS)

vn~dhe

cehe

IS)


(a)

~(c) . FIGUREE2.7(b)

Note that this force is independent of the tank length. The result is the same if the tank is0.25 ft, 25 ft, or 25 miles long. The y coordinate of the center of pressure (CP) is found fromEq.2.l9,

- Ixc + YeYR - YeA

and from Fig. 2.18

I = (3 ft)(3 ft)3 = ~ f4xc 36 36 t

so that

81/36 ft4y= +9ft

R (9 ft)(9/2 ft2)

= 0.0556 ft + 9 ft = 9.06 ft (Ans)

Similarly, from Eq. 2.20

- Ixyc + XcXR - YeA

and from Fig. 2.18

= (3 ft)(3 ft)2 (3 ft) = ~ ft4Ixye 72 72

68 Chapter 2/ Fluid Statics

so that

81/72 ft4

XR = (9 ft)(9/2 ft2) + 0 = 0.0278ft

Thus, we conclude that the center of pressure is 0.0278 ft to the right of and 0.0556 ftbelow the centroid of the area. If this point is plotted, we find that it lies on the median linefor the area as illustrated in Fig. E2.7c. Since we can think of the total area as consisting ofa number of small rectangular strips of area 8A (and the fluid force on each of these smallareas acts through its center), it follows that the resultant of all these parallel forces must liealong the median.

2.9 Pressure Prism

(Ans)

An informative and useful graphical interpretation can be made for the force developed by afluid acting on a plane area. Consider the pressure distribution along a vertical wall of a tankof width b, which contains a liquid having a specific weight y. Since the pressure must varylinearly with depth, we can represent the variation as is shown in Fig. 2.19a, where thepressure is equal to zero at the upper surface and ,equal to yh at the bottom. It is apparentfrom this diagram that the average pressure occurs at the depth h/2, and therefore the resultantforce acting on the rectangular area A = bh is

FR = PavA = Y (~) A

which is the same result as obtained from Eq. 2.18. The pressure distribution shown in Fig.2.19a applies across the vertical surface so we can draw the three-dimensional representationof the pressure distribution as shown in Fig. 2.19b. The base of this "volume" in pressure-area space is the plane surface of interest, and its altitude at each point is the pressure. Thisvolume is called the pressure prism, and it is clear that the magnitude of the resultant forceacting on the surface is equal to the volume of the pressure prism. Thus, for the prism of Fig.2.19b the fluid force is

The 'pressure prismis a gl!°metric rep-'lisen/ation of thehydrostatic force on'a plane surface. .

FR = volume = ~ (yh)(bh) = y G) A

where bh is the area of the rectangular surface, A.

1h

1h3

~yh~

(a)

I

. FIGURE.2.19Pressure prism forvertical rectangulararea.(b)

...

IS)

,ftneofalllie

-'a[lkryhentnt

g.me-IS

;e

g.

9

I

2.9 Pressure Prism 69

yhj

SMA

TIT1Yl YA

21. FIGURE 2.20Graphical representa-tion of hydrostaticforces on a vertical rec-tangular surface.

E

(a) (b)

The resultant force must pass through the centroid of the pressure prism. For the volumeunder consideration the centroid is located along the vertical axis of symmetry of the surface,and at a distance of hl3 above the base (since the centroid of a triangle is located at hl3above its base). This result can readily be shown to be consistent with that obtained fromEqs. 2.19 and 2.20.

This same graphical approach can be used for plane surfaces that do not extend up tothe fluid surface as illustrated in Fig. 2.20a. In this instance, the cross section of the pressureprism is trapezoidal. However, the resultant force is still equal in magnitude to the volumeof the pressure prism, and it passes through the centroid of the volume. Specific values canbe obtained by decomposing the pressure prism into two parts, ABDE and BCD, as shown inFig. 2.20b. Thus,

FR = FI + F2

where the components can readily be determined by inspection for rectangular surfaces. Thelocation of FRcan be determined by summing moments about some convenient axis, such asone passing through A. In this instance

FRyA = FIYI + F2Y2

and YI and Y2can be determined by inspection.For inclined plane surfaces the pressure prism can still be developed, and the cross

section of the prism will generally be trapezoidal as is shown in Fig. 2.21. Although it isusually convenient to measure distances along the inclined surface, the pressures developeddepend on the vertical distances as illustrated.

TIh2

l . FIG UR E 2. 2 1 Pressure variationalong an inclined plane area.

..


The res~ltant fluidforce acting on asubmerged area is

affecte~}Y ~hep~~s-sure at the free sur-face.

EXAMPLE2.8

I

The use of pressure prisms for determining the force on submerged plane areas isconvenient if the area is rectangular so the volume and centroid can be easily determined.However, for other nonrectangular shapes, integration would generally be needed to determinethe volume and centroid. In these circumstances it is more convenient to use the equationsdeveloped in the previous section, in which the necessary integrations have been made andthe results presented in a convenient and compact form that is applicable to submerged planeareasof any shape. .

The effect of atmospheric pressure on a submerged area has not yet been considered,and we may ask how this pressure will influence the resultant force. If we again consider thepressure distribution on a plane vertical wall, as is shown in Fig. 2.22a, the pressure variesfrom zero at the surface to yh at the bottom. Since we are setting the surface pressure equalto zero, we are using atmospheric pressure as our datum, and thus the pressure used in thedetermination of the fluid force is gage pressure. If we wish to include atmospheric pressure,the pressure distribution will be as is shown in Fig. 2.22b. We note that in this case the forceon one side of the wall now consists of FRas a result of the hydrostatic pressure distribution,plus.the contribution of the atmospheric pressure, PatmA,where A is the area of the surface.However, if we are going to include the effect of atmospheric pressure on one side of thewall we must realize that this same pressure acts on the outside surface (assuming it is exposedto the atmosphere), so that an equal and opposite force will be developed as illustrated in thefigure. Thus, we conclude that the resultant fluid force on the surface is that due only to thegage pressure contribution of the liquid in contact with the surface-the atmospheric pressuredoes not contribute to this resultant. Of course, if tne surface pressure of the liquid is differentfrom atmospheric pressure (such as might occur in a closed tank), the resultant force actingon a submerged area, A, will be changed in magnitude from that caused simply by hydrostaticpressure by an amount PsA, where Ps is the gage pressure at the liquid surface (the outsidesurface is assumed to be exposed to atmospheric pressure).

Patm Patm

Th

1Palm A Patm A

. FIGURE 2.22Effect of atmosphericpressure on the resul-tant force acting on aplane vertical wall.

f--rh--j(a) (b)

A pressurized tank contains oil (SG = 0.90) and has a square, 0.6-m by 0.6-m plate boltedto its side, as is illustrated in Fig. E2.8a. When the pressure gage on the top of the tank reads50 kPa, what is the magnitude and location of the resultant force on the attached plate? Theoutside of the tank is at atmospheric pressure.

SOLUTION

The pressure distribution acting on the inside surface of the plate is shown in Fig. E2.8b. Thepressure at a given point on the plate is due to the air pressure, Ps' at the oil surface, and the

n~:

I...

IS

:d.

aeas1d1e

'd,1e

es

al

1e

'e,~e

n,e.Ie~d

Ie

Ie

reat

19IC

Ie

2c1-a

dse

~

I

2.9 Pressure Prism 71

yhjr-+-P,-1

/1/

//

//

/

LOi I surface

I2m

~0.6 mt

(a) (b)

. FIG URE E2. 8

pressure due to the oil, which varies linearly with depth as is shown in the figure. The resultantforce on the plate (having an area A) is due to the components, FI and Fz, with

FI = (Ps + '}'hl)A

= [50 X 103N/mz + (0.90)(9.81 X 103N/m3)(2 m)](0.36 mZ)

- 24.4 X 103N

and

(hz - hi

)Fz = '}' 2 A

= (0.90)(9.81X 103N/m3) (O.~m) (0.36 mZ)= 0.954 X 103N

The magnitude of the resultant force, FR, is therefore

FR = FI + Fz = 25.4 X 103 N = 25.4 kN (Ans)

The vertical location of FR can be obtained by summing mOments around an axisthrough point 0 so that

FRyo = FI(0.3 m) + Fz(0.2 m)

or

(25.4 X 103 N)yo = (24.4 X 103 N)(0.3 m) + (0.954 X 103 N)(0.2 m)

Yo = 0.296m (Ans)

Thus, the force acts at a distance of 0.296 m above the bottom of the plate along the verticalaxis of symmetry.

Note that the air pressure used in the calculation of the force was gage pressure. At-mospheric pressure does not affect the resultant force (magnitude or location), since it actson both sides of the plate, thereby canceling its effect.

..


2.10 Hydrostatic Force on a Curved Surface

V2.4

The development ofa free-body diagramof a suitable volumeof fluid can be usedto determine the re-

sultant fluid forceacting on a curvedsurface. '

The equations developed in Section 2.8 for the magnitude and location of the resultant forceacting on a submerged surface only apply to plane surfaces. However, many surfaces ofinterest (such as those associated with dams, pipes, and tanks) are nonplanar. Although theresultant fluid force can be determined by integration, as was done for the plane surfaces, thisis generally a rather tedious process and no simple, general formulas can be developed. Asan alternative approach we will consider the equilibrium of the fluid volume enclosed by thecurved surface of interest and the horizontal and vertical projections of this surface.

For example, consider the curved section BC of the open tank of Fig. 2.23a. We wishto find the resultant fluid force acting on this section, which has a unit length perpendicularto the plane of the paper. We first isolate a volume of fluid that is bounded by the surface ofinterest, in this instance section BC, the horizontal plane surface AB, and the vertical planesurface AC. The free-body diagram for this volume is shown in Fig. 2.23b. The magnitudeand location of forces Fl and F2can be determined from the relationships for planar surfaces.The weight, OW,is simply the specific weight of the fluid times the enclosed volume and actsthrough the center of gravity (CO) of the mass of fluid contained within the volume. Theforces FHand Fv represent the components of the force that the tank exerts on thefluid.

In order for this force system to be in equilibrium, the horizontal component FH mustbe equal in magnitude and collinear with F2,and the vertical component Fvequal in magnitudeand collinear with the resultant of the vertical forces Fl and OW. This follows since the threeforces acting on the fluid mass (F2,the resultant of Fl and OW,and the resultant force that thetank exerts on the mass) must form a concurrent force system. That is, from the principlesof statics, it is known that when a body is held in equilibrium by three nonparallel forcesthey must be concurrent (their lines of action intersect at a common point), and coplanar.Thus,

FH = F2

Fv = Fl + OW

and the magnitude of the resultant is obtained from the equation

FR = Y(FH)2 + (Fv)2

__nn__n_nn-----. --n_nnn_n_n- _--nnnnn. nnn

A

c=vB

c

(a)

II FIGURE 2.23(b) (c)

Hydrostatic force on a curved surface.

-'ce

of

hel1S

<\she

sh

tar

of

ne

je

:s.

ts

1e

st

Ie~e

Ie

~s

~s

r.

2.10 Hydrostatic Force on a Curved Surface 73

The resultant FR passes through the point 0, which can be located by summing momentsabout an appropriate axis. The resultant force of the fluid acting on the curved surface BC isequal and opposite in direction to that obtained from the free-body diagram of Fig. 2.23b.The desired fluid force is shown in Fig. 2.23c.

The 6-ft-diameterdrainageconduitof Fig. E2.9ais half full of waterat rest. Determinethemagnitude and line of action of the resultant force that the water exerts on a l-ft length ofthe curved section BC of the conduit wall.

A B

1.27 It

AH1\-- - - - --I FR= 523 Ib I

32.5°~

1ft[:

B

Fj FH

c

(a) (b)

III FIGURE E2.9

(c)

SOLUTION

We firstisolatea volumeof fluidboundedby the curvedsectionBC, the horizontalsurfaceAB, and the vertical surface AC, as shown in Fig. E2.9b. The volume has a length of 1 ft.The forces acting on the volume are the horizontal force, FI, which acts on the vertical surfaceAC, the weight, W, of the fluid contained within the volume, and the horizontal and verticalcomponents of the force of the conduit wall on the fluid, FHand Fy, respectively.

The magnitude of FI is found from the equation

FI = ')!hcA = (62.4 lb/ft3)(! ft)(3 ft2) = 28llb

and this force acts 1 ft above C as shown. The weight, W, is

W = ')!vol = (62.4 lb/ft3)(97T/4 ft2)(l ft) = 44llb

and acts through the center of gravity of the mass of fluid, which according to Fig. 2.18 islocated 1.27 ft to the right of AC as shown. Therefore, to satisfy equilibrium

FH = FI = 281 lb Fy = W = 441 lb

and the magnitude of the resultant force is

FR = V(FH? + (Fy)2

= V(28l lb? + (441 lb)2 = 523 lb (Ans)

The force the water exerts on the conduit wall is equal, but opposite in direction, to the forces

FH and Fy shown in Fig. E2.9b. Thus, the resultant force on the conduit wall is shown in Fig.E2.9c. This force acts through the point o at the angle shown.

,


An inspection of this result will show that the line of action of the resultant force passesthrough the center of the conduit. In retrospect, this is not a surprising result since at eachpoint on the curved surface of the conduit the elemental force due to the pressUreis normalto the surface, and each line of action must pass through the center of the conduit. It thereforefollows that the resultant of this concurrent force system must also pass through the centerof concurrence of the elemental forces that make up the system.

This same general approach can also be used for determining the force on curvedsurfaces of pressurized, closed tanks. If these tanks contain a gas, the weight of the gas isusually negligible in comparison with the forces developed by the pressure. Thus, the forces(such as FJ and F2 in Fig. 2.23b) on horizontal and vertical projections ofthe curved surfaceof interest can simply be expressed as the internal pressure times the appropriate projectedarea.

2.11 Buoyancy, Flotation, and Stability

The resultant fluidforce acting on abody that is com-pletely submergedor floating 'in afluid is called thebuoyant force.

V2.5

2.11.1 Archimedes' Principle

When a body is completely submerged in a fluid, or floating so that it is only partiallysubmerged, the resultant fluid force acting on the body is called the buoyant force. A netupward vertical force results because pressure increases with depth and the pressure forcesacting from below are larger than the pressure forces acting from above. This force can bedetermined through an approach similar to that used in the previous article for forces oncurved surfaces. Consider a body of arbitrary shape, having a volume ¥, that is immersedin a fluid as illustrated in Fig. 2.24a. We enclose the body in a parallelepiped and draw afree-body diagram of the parallelepiped with the body removed as shown in Fig. 2.24b. Notethat the forces FJ, F2' F3, and F4 are simply the forces exerted on the plane surfaces of theparallelepiped (for simplicity the forces in the x direction are not shown), "Wis the weight ofthe shaded fluid volume (parallelepiped minus body), and FBis the force the body is exertingon the fluid. The forces on the vertical surfaces, such as F3 and F4, are all equal and cancel,so the equilibrium equation of interest is in the z direction and can be expressed as

FB = F2 - FJ - "W

If the specific weight of the fluid is constant, then

(2.21)

F2 - FJ = y(h2 - hdA

where A is the horizontal area of the upper (or lower) surface of the parallelepiped, and Eq.2.21 can be written as

FB = y(h2 - hJ)A - y[(h2 - hJ)A - ¥]

Simplifying, we arrive at the desired expression for the buoyant force

I FB = y¥ I

(2.22)

where y is the specific weight of the fluid and ¥ is the volume of the body. The directionof the buoyant force, which is the force of the fluid on the body, is opposite to that shownon the free-body diagram. Therefore, the buoyant force has a magnitude equal to the weight

hydrostatic force on a plane surface

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