..

>asses:eachormalrefore;enter

urved

gasISforceslrfacejected

-rtiallyA net:orcesanbeesonlersedrawaNote)f theghtof~rtingmcel,

:2.21)

dEq.

2.22)V2.6

:ctionhown

'eight

I

2.11Buoyancy,Flotation,andStability 75

11"'-'.: :--: ::::::-::_-:::::~ -:: :-:--:-:'=:-:::::::::~~~::::::::::::::I ~1

h2l

[~ - --- B

;L,I --I III Y~ I xJ

--

Centroidof displaced

volume

(d)

D c

(a)Centroid

A

.FIGURE 2.24Buoyantforceonsubmergedandfloat-ingbodies.

F3-+-

D(e)

c

(b)

of thefluid displacedby thebodyandis directedverticallyupward.This resultis commonlyreferredto asArchimedes'principle in honorof Archimedes(287-212B.C.),a Greekmech-anicianandmathematicianwho firstenunciatedthebasicideasassociatedwithhydrostatics.

The locationof theline of actionof thebuoyantforcecanbedeterminedby summingmomentsof theforcesshownon thefree-bodydiagramin Fig. 2.24bwith respectto someconvenientaxis.For example,summingmomentsaboutanaxis perpendicularto thepaperthroughpointD wehave .

FBYe =F2YI - FIYI - 'WY2andonsubstitutionforthevariousforces

VYe= VTYI - (VT - V)Y2 (2.23)

whereVT is thetotalvolume(h2- hI)A.Theright-handsideofEq.2.23isthefirstmomentof thedisplacedvolumeV withrespecttothex-zplanesothatYcis equaltotheYcoordinateof thecentroidof thevolumeV. In asimilarfashionit canbeshownthatthex coordinateof thebuoyantforcecoincideswiththex coordinateof thecentroid.Thus,weconcludethatthebuoyantforcepassesthroughthecentroidofthedisplacedvolumeasshowninFig.2.24c.Thepointthroughwhichthebuoyantforceactsis calledthecenterofbuoyancy.

Thesesameresultsapplyto floatingbodieswhichareonlypartiallysubmerged,asillustratedin Fig.2.24d,if thespecificweightof thefluidabovetheliquidsurfaceis verysmallcomparedwiththeliquidin whichthebodyfloats.Sincethefluidabovethesurfaceisusuallyair,forpracticalpurposesthisconditionis satisfied.

In thederivationspresentedabove,thefluidis assumedtohavea constantspecificweight,y.If abodyisimmersedin afluidinwhich'Yvarieswithdepth,suchasinalayeredfluid,themagnitudeof thebuoyantforceremainsequaltotheweightof thedisplacedfluid.However,thebuoyantforcedoesnotpassthroughthecentroidof thedisplacedvolume,butrather,it passesthroughthecenterof gravityof thedisplacedvolume.

76 Chapter2 / Fluid Statics

EXAMPLE2.10

A sphericalbuoyhasadiameterof 1.5m,weighs8.50kN, andis anchoredtotheseafloorwithacableasis showninFig.E2.l0a.Althoughthebuoynormallyfloatsonthesurface,atcertaintimesthewaterdepthincreasessothatthebuoyis completelyimmersedasillus-trated.Forthisconditionwhatis thetensionof thecable?

Pressureenvelope

(a) (b) (c) .. FIGURE E2.10

SOLUTION

Wefirstdrawa free-bodydiagramof thebuoyasis shownin Fig.E2.1Ob,whereFEis thebuoyantforceactingonthebuoy,OWis theweightof thebuoy,andT is thetensionin thecable.Forequilibriumit followsthat

T =FE - OW

FromEq.2.22

FE = ')IV

and for seawaterwith ')I = 10.1kN/m3andV = 1Td3/6then

FE = (10.1 X 103N/m3)((1T/6)(lSm)3] = 1.785X 104N

Thetensionin thecablecannowbecalculatedas

T =1.785X 104N - 0.850X 104N =9.35kN (Ans)

Notethatwereplacedtheeffectofthehydrostaticpressureforceonthebodybythebuoyantforce,FE,Anothercorrectfree-bodydiagramof thebuoyis showninFig.E2.1Oc.Theneteffectof.thepressureforcesonthesurfaceof thebuoyis equaltotheupwardforceof magnitude,FE(thebuoyantforce).Donotincludeboththebuoyaritforceandthehydro-staticpressureeffectsin yourcalculations-useoneortheother.

2.11.2 Stability

Anotherinterestingandimportantproblemassociatedwithsubmergedorfloatingbodiesisconcernedwiththestabilityofthebodies.A bodyissaidtobeinastableequilibriumpositionif, whendisplaced,it returnsto its equilibriumposition.Conversely,it is in anunstableequilibriumpositionif,whendisplaced(evenslightly),itmovestoanewequilibriumposition.Stabilityconsiderationsareparticularlyimportantforsubmergedorfloatingbodiessincethecentersof buoyancyandgravitydonotnecessarilycoincide.A smallrotationcanresultineithera restoringor overturningcouple.For example,for thecompletelysubmergedbody

)r

v,

s-

hehe

1S)

theDc.rcero-

;isionbleon.thetin)dy

I

~ -~~_u_--~-_u_~--------

V2.7

:=-=-=-=-==-~:~=:=====:-~-:-=-=-=-=-~cY=-:-=-:-=-=-:==:=:::.:=:~:=:

Stable

Restoringcouple

. FIGURE 2.25Stabilityof a completelyim-mersedbody-centerof gravitybelowcentroid.

2.11 Buoyancy,Flotation,andStability 77

-:-:~~-= =-:-=-:-:-~c-:-=:.:_=-:--_'!:-=::====:=====:=:::==:=c

Unstable

Overturningcouple

. FIGURE 2.26Stabilityofa completelyim-mersedbody-centerof gravityabovecentroid.

shownin Fig.2.25,whichhasacenterof gravitybelowthecenterofbuoyancy,a rotationfromitsequilibriumpositionwillcreatearestoringcoupleformedbytheweight,OW,andthebuoyantforce,FE,whichcausesthebodytorotatebacktoitsoriginalposition.Thus,forthisconfigurationthebodyis stable.It is tobenotedthataslongasthecenterof gravityfallsbelowthecenterofbuoyancy,thiswill alwaysbetrue;thatis, thebodyis in astableequi-libriumpositionwithrespecttosmallrotations.-However,asis illustratedinFig.2.26,if thecenterof gravityis abovethecenterofbuoyancy,theresultingcoupleformedbytheweightandthebuoyantforcewillcausethebodytooverturnandmovetoanewequilibriumposition.Thus,acompletelysubmergedbodywithitscenterof gravityaboveitscenterofbuoyancyis in anunstableequilibriumposition.

Forfloatingbodiesthestabilityproblemis morecomplicated,sinceasthebodyrotatesthelocationof thecenterof buoyancy(whichpassesthroughthecentroidof thedisplacedvolume)maychange.As isshowninFig.2.27,afloatingbodysuchasabargethatrideslowinthewatercanbestableeventhoughthecenterofgravityliesabovethecenterofbuoyancy.Thisistruesinceasthebodyrotatesthebuoyantforce,FE,shiftstopassthroughthecentroidof thenewlyformeddisplacedvolumeand,asillustrated,combineswiththeweight,OW,toformacouplewhichwill causethebodytoreturntoitsoriginalequilibriumposition.How-ever,fortherelativelytall,slenderbodyshowninFig.2.28,asmallrotationaldisplacementcancausethebuoyantforceandtheweighttoformanoverturningcoupleasillustrated.

It is clearfromthesesimpleexamplesthatthedeterminationof thestabilityof sub-mergedorfloatingbodiescanbedifficultsincetheanalysisdependsinacomplicatedfashionontheparticulargeometryandweightdistributionof thebody.Theproblemcanbefurthercomplicatedbythenecessaryinclusionofothertypesofexternalforcessuchasthoseinducedbywindgustsorcurrents.Stabilityconsiderationsareobviouslyof greatimportancein thedesignof ships,submarines,bathyscaphes,andsoforth,andsuchconsiderationsplayasig-nificantroleintheworkofnavalarchitects(see,forexample,Ref.6).

c =centroidoforiginaldisplacedvolume

n . FIGURE 2.27Stabilityof a float-ingbody-stableconfiguration.

c' =centroidofnewdisplacedvolume

Restoringcouple

Stable

I"

78 Chapter2 / Fluid Statics

()c =centroidoforiginal c' =centroidof new Overturning

displacedvolume displacedvolume coupleUnstable

. FIG U R E 2.28 Stabilityofa float-ingbody-unstableconfiguration.

2.12 PressureVariationin a Fluid withRigid-BodyMotion

Althoughin thischapterwehavebeenprimarilyconcernedwithfluidsatrest,thegeneralequationofmotion(Eq.2.2)

- Vp - -yk =pa

wasdevelopedfor bothfluidsatrestandfluidsin motion,withtheonly stipulationbeingthattherewerenoshearingstressespresent.Equation2.2in componentform,basedonrectangularcoordinateswiththepositivez axisbeingverticaUyupward,canbeexpressedas

- apax= pax

ap-ay=pay - ap= l' + pazaz

(2.24)

Eventhoughafluid.maybein.motion,if..' ..j. .,. , ,it moves'asa rigidbodytherewillbe

,noshearlngstressespresent.

A generalclassof problemsinvolvingfluidmotionin whichthereareno shearingstressesoccurswhenamassoffluidundergoesrigid-bodymotion.Forexample,if acontainerof fluidacceleratesalongastraightpath,thefluidwill moveasarigidmass(aftertheinitialsloshingmotionhasdiedout)witheachparticlehavingthesameacceleration.Sincethereisnodeformation,therewill benoshearingstressesand,therefore,Eq.2.2applies.Similarly,if afluidiscontainedinatankthatrotatesaboutafixedaxis,thefluidwill simplyrotatewiththetankasarigidbody,andagainEq.2.2canbeappliedtoobtainthepressuredistributionthroughoutthemovingfluid.Specificresultsforthesetwocases(rigid-bodyuniformmotionandrigid-bodyrotation)aredevelopedin thefollowingtwosections.Althoughproblemsrelatingtofluidshavingrigid-bodymotionarenot,strictlyspeaking,"fluidstatics"problems,theyareincludedin thischapterbecause,aswewill see,theanalysisandresultingpressurerelationshipsaresimilartothoseforfluidsatrest.

2.12.1 Linear Motion

Wefirstconsideranopencontainerofaliquidthatis translatingalongastraightpathwithaconstantaccelerationa asillustratedin Fig. 2.29.Sinceax=0 it followsfromthefirstofEqs. 2.24thatthepressuregradientin thex directionis zero (ap/ax = 0). In they andzdirections

ap-ay- - pay

apaz = -peg + az)

(2.25)

(2.26)

,a float-

-meral

gthat19u1ar

[2.24)

~aringtainerinitialtereis

Harly,~with}utionlotionblemsJlems,~ssure

withairstofandz

(2.25)

(2.26)

I

--

2.12PressureVariationinaFluid with Rigid-Body Motion 79

Freesurfaceslope=dzldy a'L~:

ayPI Constantpz pressureP3 lines . FIGURE 2.29

Linearaccelerationofa liquidwitha freesurface.

Thechangeinpressurebetweentwocloselyspacedpointslocatedaty,z,andy + dy,z + dzcanbeexpressedas

ap apdp =- dy + - dzay az

or in termsof theresultsfromEqs.2.25and2.26

dp = - paydy - p(g + az)dz (2.27)

Alonga lineof constantpressure,dp = 0,andthereforefromEq.2.27it followsthattheslopeof thislineis givenbytherelationship

dz = -~dy g + az

(2.28)

'ressuredistri-

inafluidatisaccel-alonga

htpathisnot'static.

Alongafreesurfacethepressureis constant,sothatfortheacceleratingmassshowninFig.2.29thefreesurfacewillbeinclinedif ay0;6O.In addition,alllinesofconstantpressurewillbeparalleltothefreesurfaceasillustrated.

Forthespecialcircumstanceinwhichay= 0,az0;60,whichcorrespondstothemassof fluidacceleratingin theverticaldirection,Eq.2.28indicatesthatthefluidsurfacewillbehorizontal.However,fromEq.2.26weseethatthepressuredistributionis nothydrostatic,butis givenbytheequation

dpdz = - p(g + az)

For fluidsof constantdensitythisequationshowsthatthepressurewill varylinearlywithdepth,butthevariationis duetothecombinedeffectsof gravityandtheexternallyinducedacceleration,p(g + az),ratherthansimplythespecificweightpg.Thus,for example,thepressurealongthebottomof aliquid-filledtankwhichis restingonthefloorof anelevatorthatis acceleratingupwardwill beincreasedoverthatwhichexistswhenthetankis atrest(ormovingwitha constantvelocity).It is tobenotedthatfor afreelyfalling fluidmass(az=- g),thepressuregradientsin all threecoordinatedirectionsarezero,whichmeansthatifthepressuresurroundingthemassiszero,thepressurethroughoutwill bezero.Thepressurethroughouta "blob" oforangejuicefloatinginanorbitingspaceshuttle(aformoffreefall)is zero.Theonlyforceholdingtheliquidtogetheris surfacetension(seeSection1.9).

80 Chapter2 / Fluid Statics

EXAMPLE2.11

Thecrosssectionfor thefueltankof anexperimentalvehicleis shownin Fig. E2.ll. Therectangulartankis ventedtotheatmosphere,andapressuretransduceris locatedin itssideasillustrated.Duringtestingof thevehicle,thetankis subjectedtoaconstantlinearaccel-eration,ay.(a)Determineanexpressionthatrelatesayandthepressure(in lb/ft2)atthetransducerforafuelwithaSG =0.65.(b)Whatisthemaximumaccelerationthatcanoccurbeforethefuelleveldropsbelowthetransducer?

ay .ZL

Y

I--0,75ft---+--- 0,75ft--! . FIGURE E2.11

SOLUTION

(a) Foraconstanthorizontalaccelerationthefuelwill moveasarigidbody,andfromEq.2.28theslopeof thefuelsurfacecanbeexpressedas

dz ay

dy g

sinceaz =O.Thus,forsomearbitraryay,thechangein depth,z\,of liquidontherightsideof thetankcanbefoundfromtheequation

z ay1-- = --0.75ft g

or

z\ = (0.75ft) (~)Sincethereis noaccelerationin thevertical,z, direction,thepressurealongthewallvarieshydrostaticallyasshownbyEq.2.26.Thus,thepressureatthetransducerisgivenbytherelationship

p =yhwhereh is thedepthof fuelabovethetransducer,andtherefote

p = (0.65)(62.4lb/ft3)[0.5ft - (0.75ft)(ay/g)]

ay= 20.3- 30.4-g

forz\ :S0.5ft.As written,p wouldbegivenin lb/ft2.

(Ans)

Theside

ccel-t theIccur

Eg.

ght

allen

s)

2.12 PressureVariationin a Fluid withRigid-Body Mot/on 81

(b) Thelimitingvalueforay(whenthefuellevelreachesthetransducer)canbefoundfromtheequation

0.5ft = (0.75ft) [(aY~max]or

2g(~)max=3

andforstandardaccelerationof gravity

(ay)max=1(32.2ft/s2) = 21.5ft/s2 (Ans)Notethatthepressurein horizontallayersis notconstantin thisexamplesinceap/ay=- pay' O.Thus,forexample,PI 'P2'

2.12.2 Rigid-BodyRotation

Afteraninitial"start-up"transient,a fluidcontainedin atankthatrotateswithaconstantangularvelocitywaboutanaxisasis showninFig.2.30will rotatewiththetankasarigidbody.It is knownfromelementaryparticledynamicsthattheaccelerationofa fluidparticlelocatedatadistancer fromtheaxisofrotationisequalinmagnitudetorw2,andthedirectionof theaccelerationistowardtheaxisofrotationasis illustratedin thefigure.Sincethepathsof thefluidparticlesarecircular,it isconvenienttousecylindricalpolarcoordinatesr, e,andz,definedin theinsertin Fig.2.30.It will beshowninChapter6thatin termsofcylindricalcoordinatesthepressuregradientV'pcanbeexpressedas

t'7 ap A 1apA apAvp =- e + - - Ce+ - Car r r ae azZ

Thus,in termsof thiscoordinatesystem

(2.29)

ar = - rw2er ae=0 az =0andfromEq.2.2

apar = prw2 ap= 0ae

ap-az- - 'Y (2.30)

I Axisof, rotation

I,I,

~a,= r(JJ2

y

e'"

r i?/e

'\'"e,

x . FIGURE 2.30Rigid-bodyrotationof a liquidin a tank.

ap apdp = - dr + - dzar az

82 Chapter2 / Fluid Statics

Theseresultsshowthatforthistypeof rigid-bodyrotation,thepressureis afunctionof twovariablesrandz,andthereforethedifferentialpressureis

or

dp =prul dr - 'Ydz (2.31) ~

Alongasurfaceof constantpressure,suchasthefreesurface,dp = 0,sothatfromEq.2.31(using'Y = pg)

w2r2

z =2g + constant(2.32)

dz rw2---dr g

urfaceinliquidis'herthan

and,therefore,theequationfor surfacesof constantpressureis

Thisequationrevealsthatthesesurfacesof constantpressureareparabolicasillustratedinFig.2.31.

IntegrationofEq.2.31yields

I dp =pW2 I r dr - 'YI dz

or

pw2r2p =~ - 'YZ+ constant

(2.33)

wheretheconstantof integrationcanbeexpressedin termsof aspecifiedpressureatsomearbitrarypointro,zoThisresultshowsthatthepressurevarieswiththedistancefromtheaxisofrotation,butatafixedradius,thepressurevarieshydrostaticallyintheverticaldirectionasshowninFig.2.31.

PI

Constant P2'pressure

lines P3

P4

r~

P3- I~

(j)2r2

2g

ry . FIG U R E 2. 3 1 Pressure

distributionin a rotatingliquid.x

I

2.12 PressureVariationin a Fluid with Rigid-Body Motion 83

}ftwo XAMPLE2.12

It hasbeensuggestedthattheangularvelocity,w,ofarotatingbodyorshaftcanbemeasuredbyattachinganopencylinderof liquid,asshowninFig.E2.l2a,andmeasuringwithsometypeofdepthgagethechangein thefluidlevel,H - ho'causedbytherotationof thefluid.Determinetherelationshipbetweenthischangein fluidlevelandtheangularvelocity.

~R---1

(2.31)

1111

dr

1.2.31

(2.32)

(a)

SOLUTION

(b) . FIGURE E2.12

atedin Theheight,h,of thefreesurfaceabovethetankbottomcanbedeterminedfromEq.2.32,andit followsthat

w2r2h =- + ho

2g

Theinitialvolumeof fluidin thetank,Vi, is equalto

Vi = l7R2H

(2.33)

Thevolumeof thefluidwiththerotatingtankcanbefoundwiththeaidof thedifferentia]elementshowninFig.E2.l2b.Thiscylindricalshellis takenatsomearbitraryradius,r, aneitsvolumeis

dV = 21T1"hdrIt some'omtheirection

Thetotalvolumeis,therefore

IR

(w2r2 )

l7W2R4

V =217 r - + ho dr =- + l7R2ho 2g 4g

Sincethevolumeof thefluidin thetankmustremainconstant(assumingthatnonespilloverthetop),it followsthat

l7W2R4l7R2H =- + l7R2h(l

4g .

orw2R2

H-ho=- 4g

Thisis therelationshipwewerelookingfor.It showsthatthechangein depthcouldinde(beusedtodeterminetherotationalspeed,althoughtherelationshipbetweenthechangedepthandspeedis notalinearone.

(An:

Pressureliquid.

I

84 Chapter2 / Fluid Statics

References

1. TheU.S.StandardAtmosphere,1962,U.S.GovernmentPrintingOffice,Washington,D.C.,1962.

2. TheU.S.StandardAtmosphere,1976,U.S.GovernmentPrintingOffice,Washington,D.C.,1976.

3. Benedict,R.P.,FundamentalsofTemperature,Pressure,andFlowMeasurements,3rdEd.,Wiley,NewYork,1984.

4. Dally,J. W., Riley,W. F., andMcConnell,K. G., Instrumentationfor EngineeringMeasurements,2ndEd.,Wiley,NewYork, 1993.

5. Holman,J. P.,ExperimentalMethodsforEngineers,4thEd.,McGraw-Hill,NewYork,1983.

6. Comstock,J. P.,ed.,PrinciplesofNavalArchitecture,SocietyofNavalArchitectsand,MarineEngineers,NewYork,1967.

7. Hasler,A. F., Pierce,H., Morris,K. R., andDodge,J., "MeteorologicalDataFields'In Perspective'", Bulletinof theAmericanMeteorologicalSociety,Vol. 66,No.7,July 1985.

ReviewProblems

Note:Problemsdesignatedwith(R)arereviewproblems.Thephraseswithinparenthesesrefertothemaintopicstobeusedin solvingtheproblems.Complete,detailedsolutionstothesereviewproblemscanbefoundin thesupplementtitledStudentSolutionManualfor Fundamentalsof Fluid MechanicsbyMunson,Young,andOkiishi(JohnWileyandSons,NewYork,1997).

2.1R (Pressurehead) Comparethecolumnheightsofwater,carbontetrachloride,andmercurycorrespondingto apressureof50kPa.Expressyouranswerinmeters.(ANS:5.10m;3.21m;0.376m)

2.2R(Pressure-depthrelationship) A closedtankis par-tiallyfilledwithglycerin.If theairpressurein thetankis 6Ib/in.2andthedepthof glycerinis 10ft,whatis thepressurein Ib/ff atthebottomof thetank?(ANS:1650Ib/ft2)

2.3R(Gage-absolutepressure) OntheinletsideofapumpaBourdonpressuregagereads600Ib/ft2vacuum.Whatis thecorrespondingabsolutepressureif thelocalatmosphericpres-sureis 14.7psia? .(ANS: 10.5psia)

2.4R(Manometer) A tankisconstructedofaseriesofcyl-indershavingdiametersof0.30,0.25,and0.15masshowninFig. P2.4R.Thetankcontainsoil, water,andglycerinanda

mercurymanometerisattachedtothebottomasillustrated.Cal-culatethemanometerreading,h.(ANS:0.0327m)

t0.1m

+-0.1m

t-0.1mt-0.1m.t

Ih

~Mercury

.FIGURE P2.4R2.5R(Manometer) A mercurymanometerisusedtomea-surethepressuredifferencein thetwopipelinesof Fig.P2.5R.Fuel oil (specificweight = 53.0 Ib/ft3) is flowing in A andSAE 30lubeoil (specificweight= 57.0Ib/ft3)is flowinginB. An airpockethasbecomeentrappedin thelubeoil asindi-cated.DeterminethepressureinpipeB if thepressureinA is15.3psi.(ANS: 18.2psi)

i!j

Ij

ij,i

IIjj!

J

'

j

I

I

I' J

.,;~1

I ~

I

11i ~Ii

I j

Airbubble-m,

m, SAE36 oil

rd

ng

ok,P2.5R

lld 2.6R (Manometer) Determine theangle8 of theinclinedtubeshownin Fig. P2.6R if thepressureatA is 1 psi greaterthanthatat B. .

ds

7,

B.Air

. FIGURE P2.6R

- 2.7R(Forceonplanesurface) A swimmingpoolis18mlongand7 m wide.Determinethemagnitudeandlocationoftheresultantforceof thewaterontheverticalendof thepoolwherethedepthis 2.5m.(ANS:214kN oncenterline,1.67mbelowsurface)

11-

2.8R(Forceonplanesurface) Theverticalcrosssectionofa7-m-longclosedstoragetankis showninFig.P2.8R.Thetankcontainsethylalcoholandtheairpressureis 40kPa.De-terminethemagnitudeoftheresultantfluidforceactingononeendof thetank.

(ANS:847kN)

r-2 m1

IAir

a-{.IdIII]-is

2m

t4m

1~4m~.FIGURE P2.8R

I

ReviewProblems 85

2.9R(Centerofpressure) A 3-ft-diametercircularplateislocatedin theverticalsideofanopentankcontaininggasoline.Theresultantforcethatthegasolineexertsontheplateacts3.1in. belowthecentroidof theplate.Whatis thedepthof theliquidabovethecentroid?(ANS:2.18ft)

2.10R(Forceonplanesurface) A gatehavingthetrian-gularshapeshowninFig.P2.lORis locatedintheverticalsideof anopentank.Thegateis hingedaboutthehorizontalaxisAB. Theforceof thewateronthegatecreatesamomentwithrespectto theaxisAB. Determinethemagnitudeof thismo-ment.

(ANS:3890kN.m)

T8m

I

T6m

11--6 m-1-7 m l

P2.10R

2.11R(Forceonplanesurface) TherectangulargateCDof Fig P2.11Ris 1.8m wideand2.0m long.Assumingthematerialof thegatetobehomogeneousandneglectingfrictionatthehingeC, determinetheweightof thegatenecessarytokeepit shutuntilthewaterlevelrisesto2.0mabovethehinge.(ANS: 180kN)

. FIGURE P2.11R

2.12R(Forceoncurvedsurface) A gatein theformofapartialcylindricalsurface(calleda Taintergate)holdsbackwaterontopof adamasshownin Fig.P2.12R.Theradiusofthesurfaceis 22ft,andits lengthis 36ft.ThegatecanpivotaboutpointA, andthepivotpointis 10ft abovetheseat,C.

86 Chapter 2/ FluidStatics

Determinethemagnitudeof theresultantwaterforceon thegate.Will theresultantpassthroughthepivot?Explain.(ANS: 118,000Ib)

Taintergate '>..

A

. FIGURE P2.12R

2.13R(Forceoncurvedsurface) A conicalplugislocatedin thesideof atankasshowninFig.2.13R.(a)Showthatthehorizontalcomponentoftheforceofthewaterontheplugdoesnotdependonh.(b)Forthedepthindicated,whatis themag-nitudeof thiscomponent?(ANS:735Ib)

. FIGURE P2.13R

2.14R(Forceon curvedsurface) The9-ft-Iongcylinderof Fig.P2.14Rfloatsinoil andrestsagainstawall.Determinethehorizontalforcethecylinderexertsonthewallatthepointof contact,A.

(ANS:2300Ib)

II FIGURE P2.14R

I

2.1SR(Buoyancy) A hot-airballoonweighs500Ib, in-cludingtheweightof theballoon,thebasket,andoneperson.Theairoutsidetheballoonhasatemperatureof80of, andtheheatedairinsidetheballoonhasa temperatureof ISOof. As-sumetheinsideandoutsideairtobeatstandardatmosphericpressureof 14.7psia.Dete~minetherequiredvolumeof theballoonto supporttheweight.If theballoonhada sphericalshape,whatwouldbetherequireddiameter?(ANS:59,200ft3;48.3ft)

2.16R(Buoyancy) An irregularlyshapedpieceof asolidmaterialweighs8.05lb in air and5.26Ib whencompletelysubmergedinwater.Determinethedensityof thematerial.(ANS:5.60slugs/ft3)

2.17R(Buoyancy,forceonplanesurface) A cube,4ftona side,weighs3000Ib andfloatshalf-submergedin anopentankasshownin Fig.P2.17R.For a liquiddepthof 10ft,de-terminetheforceoftheliquidontheinclinedsectionAB ofthetankwall.Thewidthof thewallis 8 ft. Showthemagnitude,direction,andlocationof theforceonasketch.

(ANS: 75,000Ib oncenterline,13.33ft alongwallfromfreesurface)

1-4It-I

A

II FIGURE P2.17R

2.18R(Rigid,bodymotion) A containerthatis partiallyfilledwithwateris pulledwithaconstantaccelerationalongaplanehorizontalsurface.Withthisaccelerationthewatersur-faceslopesdownwardatanangleof 40withrespectto thehorizontal.Determinetheacceleration.Expressyouranswerinm/s2.(ANS:8.23m/s2)

2.19R(Rigid-bodymotion) An open,2-ft-diametertankcontainswaterto a depthof 3 ft whenatrest.If thetankisrotatedaboutitsverticalaxiswithanangularvelocityof 160rev/min,whatis theminimumheightof thetankwallstopre-ventwaterfromspillingoverthesides?(ANS:5.18ft)