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su [chapter Peay ant at mcr Solution |The agora its eos dp aps The age per nae se i te ae Sethi be denne Asepioar We aes lhe ey an ual ape sale spar Tact ie ee 2 Schhg fr 9 ard wth hes cay _ Nop ite pan tena pn nll Salton the avuice pes Is war a sete eth gem the fo sre ese ae the ‘Moots pesese's coed esate eee st Asap Ths snd yak ae anes, rata Me ti ery ie. S038, Wea be ely wa be DK Thum eet fn eerie ie pe ei tn ry pS pge = DMI IU I) gen ie psa, the amen este = 117540 “Looby 9.8L ) Teele emu tad ae T9dsRkPs Ceo Myre: 0 eas Noe iene ue hide Dis oreo gel Saluian —iierosessows ot gon? = 1522p Asis Nou bys 980605, N= 025814 a = 2k an elas Lg — ws N= oan | OIE 2.20063 ot sedge! =s20tscmten = suey at amin? —1z2pss cussion is Hong nab ue es cesvern fc (Chapter Pressure aud Mid Statics ‘Th change i pe peesine ft be eetrmined Fora gen yom, Initia Param ryt 0 Aor he pressure is applied Prope rthasiens Gs ab ry sho (On ths other han, fro the contin swe oman = sey,sh0.0-,AiU9 (1.o1<1.26—.01) ap -867 Pa ‘Chapter 3 Pressure and Flu States aa Solution The poss ins pcan wae i i emaeavn! hy a ml Ml ysomuimer The gage prs e inthe ark sta be estermined Assumptions The ar pressure inthe ais ufo (ie, veresen with elevaten Is negligible ots ow dent), Sad thus we can determine te seesce athe at-ater nerface Properies The densiios of matcury, water, and ol ars glven 1 be 13.600, 1690 and 850 in", respect ty niin the le yoshi i 2. al ig Analg Sarg a he pss pHa ae sete ne gunn esas (see rp eye Ions wi we ech ‘eae tle anesilee x Fi + Poacthy ~ Cathe ~Pcsan sls ~Pne clung for 7, H PPoy Pon Bh Paths Pee =F Paras Poe ong that ge 2) Bae ated stun, wor LY), Poe (SL AEY CHOU} OHH) OMT nF Lira iig-ns*liecon vsrtocn( -978kPa Discussion Note that juan hoizetlly Eom ane tube tthe nex a sealzing char pressure remains he = sane Sui simplifies ts analysis areal. inthe Sokition The beret be catenin Properties The density of morcuy is given 9 be 13.600 kg ‘Avalale The cmaophene presse s deexmined dcety trom 2, ng aa locaton fs glen. In elght af mercury columa, The stmasperke pressure Isto aaah Linon) = 18.kgin vast vs nT] ES eres =98.1KP3 Discussion We round off che fina! answer to tee significant ists. 190 KPa is a fly type value af amosphoric pressure orld sight above sea lve (Chapter 3 Pressure and Fluid Staves 236 [A manometer s designed to measure pressures, A cerain geometric reo i the manumeiet fr Keeping samrundors specified vale In he eerie vss Pare rin8. ; the (4) 7 re D L weft «sna leer nd fiero date th ees er Ia “E", ve fete Bue Za low sine P19) sya] Poa! Shay 0.025. Sin 00/981 010194 0.194 man wake gen eta PY ne am vwzs=|' oD 0.8, Pip +050] "> area {D 0.0985.+0.10 [Chapter 3 Pressure aud Ml Sates a6 Solution sesh and seawater flowing in parallel horizona,ppalines are com anuicler, xe essa diene ost ie pipelines Bt becca, tod ta each eter bya double Utube Assumpilans 404k ligula facar pest. [Properdes ‘The denskies of seawater and mercury’ ate giver 1 be A, ~ 1025 igi? and, 13.600 kgm. We eke the dorsty oPwrte ta he ay =i0 kgm! The spete gravy fol gon we 7. and ts deny is 770 kgm sess he hs pie vin 1 aa ving in ht yang Gs he eh termes wh. yar nest Un se water pe (pu 2, a ela te real sal Analesis Sori, with ‘dv subtract is sw oP) gives Ht pethewrgths Pathe ~Peathes FP eythy +easShg Pathan Aon: Sse ty Pata Pally Ponltead Subseuting, B=, = L981 62 1(13.600 kgm M0. am) 20 Kgy¥® MT rm) — HUN Kae. (es UsSkgin' Us9) — log wis 20 Shia" =10.5 kPa “Thesefve the pressure in he fesh water pige is 10.2 KPa higher thas the pressure inthe seawater poe. Pseusioy —_The rl pre tha oI 2th ei Chapter 3 Pressure and Fluld Stales 2.48R, Solution The pressure ina natural gay pipeline fy mmeascul by ‘en 0 the atmosphere. The absiute pressure Inthe fpeline 4s to be de:ermined. ube manonueser wis one of Uw ae Assumptions 1 All she liquids are incompressible, 2 Ihe pressure throughout che natural gas (including the te is Sas sine ts density is law. Properties We ake the density of sate to be 9, = U2 omit the specific gravity of mercury is given tobe 136, 1s ens 8 = 12. 6362.4 — B48 ha The spre gravity af a is pen nb, an ths it density 6 0.69402 4= 42.1 fof Analysis Stnting with the prosame at point | inthe mata’ gas pipeline and ~oving along the mie hy adn {Uwe or sublraeing (os wo up) he ph ws wl we reach ive sac ool lene diel abe fs exes andi ‘aumosphere, and scting the ccsultoqual to Pea. gives PigShing | Patthea ~ Ogee Bhgcas ~ Pon Solving for, P.~ Pou Onging acm ~ Bu th Substitting P= 14 2poin + (2.2. (848 the?) (2H) = (62 Athan (27H12%) =u bm )a512%9 eared 2b tus ) =17psia Liscussion Note that jumping horizentally from one tube to the next anc realizing that pressure remains the same inthe iid inplies ce analys's realy (Chapter 3 Pressure and Fuld Stades am Solution Ts jssane fsa mia ys pts fs ows ye ble UE bs envoy wie me he mas spe asta Th bate pts tt pt Be astm Assmann 1 AI ihe bays fnoam yee 2 The effet fi ea “roughout “he ncural gas (Iaslding the eb) Is untfor since ts dens 11. ssa bs igi & Th Properes We ake the centy af wate'o be gy ~ 524 Toit The specie aly of marca 8 gben 0 Be 125, ul Cus is Jost is y= 1524 = 8886 Ta “Analyse Stating with he prssire at oir the matural ga pplin, ar moving lor the abe by ading as we ancien or suhracting (sep pte gh terms url ace rch he fon see af where the ol res exposed ro ‘Beatnosptere and sezng he esl equal © i 05 a Ba = =P 4.eh, Selving for? Fi =Fag = Pages Pasay Subst, . 8 oath we ane) Po162 pan (02.254 [845.6 tba 6/128) + (2Atbre?} (aur ny] —-ME_ | 1 P Me vores) we sertimwie Li me 8.0 psia Discusion — eke Usa juni ually rm nl Ue ent an ein essa ems Une san ae “ari uit siopliies the analysts grear'y Iso, can he sewn thatthe Lin igh re nina sith a denst'y of 075 Tm’? corresponds ton presatre d erence of 000068 ps! Therefore seer nthe pressure éifference heroen the nr pies is negizble Chapter 3 Pressure and Fluid States 349 Solution The gaye prossae of aie in a gressuraed welet lank is neasued sinulascously by bull a praise gaye anda manometer. The differential height ff the mercury column isto be cetersined, Assumpdous Toe ls pussure in he tank ib wlfor Geis vsiation wid slevatin bs msgligible due tis Jaw ders and this the pressure athe ai-water Interface isthe same as the inccated gage pressure. Properios Wie take the densey of water they 1000 Kg The sper grits af and meciny ate given to WeO.72 ad 195, sepesiely Analysis Staringswith the presse cr subtracting (as we gous) the f@h terms unl we seach the free surface ofc: atmosphere, and soring the result equal Pa v8 ie he an (pine 1, and ~ving along the nb by sling fas wr ge es Shere the oll tube is expesed to the Bi eash, uth, Pasha A Ban Pasha, * PrgSlng~ Auth Ale Foe h 2k rene Suhstistng i Whee (im Hom og tsom,-03m (goo egén?)(astavs?) | Laka a Solving for uy gives fy = 0.47 m. Therefore. she ciffererial height of the mercury column must be cm. Discassian Double ieisumentaies Uke dhs lows one tw verily the messusement ol wae of dhe isumetats Ly Uae teasureanoal ofan teu. (Chapter 3 Pressure and Flu States ast Solution te tit ols wl fv! tbs manne fang sa eh sans ep Syne Bea given prs hyd ks ang hse en ob evo Assumprions 1 All the lguds ate Sncempressibe. 2 Pressure Inthe brine pipe enmins sonvart, 2 The vaso of ressure in the tapped ay space i oaible Properties The specific ras ac vent be 19.5 fer mescury and I ferrin, We take the tar dansk of ‘eater 0 be 9, -1069 gm Avatjs 11s sea fromthe peslem stent ard the figure Cu the bene pressure 5 much Migr te che alr [essure, and en te ar pressure dros by 0.9 KPa, th posse cifeence betven the rine andthe ai space also Increases bythe sane nour Satng with the at pressure (poi 8} anc msviag aor the tube by adding (as we g0 dwn) er saoaecting a we ge up se opt ters url we reac ch ene pipe pot Bar setng te esl ecu. fe and after he pressure change of give Bafne: Py +P, ch. ~ Oe Bhaga~ Bins “Pe Mor Bae thy Pegtayas hs Subdting Pas sPugiing Pobthy 2 + ATE san, Soh -9 ise Shy el Sig ae Ue ges eer omcany a ci als igh espectvely de die op Ina ser Had af tine sv pn ks seas enon vs ef he ler lil oi ‘rbot mercy end rie Iacease Neng also na te volume of ete scnsant, Ne BIE ins ~ 42 hig Pup y DRL mb Nee? antigen 2h, -000Em Mg hate “Agr Nig # Aly = Age 4 /y) Tag ay 13 550.0054 44) 11200051 m Taives 24 0838 x Ae rine pre corn oa fe sem a. Disansions —_Inaiinn to He spatial indie, easaba necomnens ona mess tim (Chapter 3 Pressure ard Fuld Sackes 259 Solution Wert deormine ef gid te earth yale art wn Airset eas. Asuna 1 Tie il fsemssiie 2 je fs rest rng th nails thjerostls, te When = 0. che pressure at the ‘odor of etc piston mascbe the sare Tis, a i Ana £88" j 18) apo * Rowers” Tae ‘Auth beglaring, when.) dhe requlrg fee thus F, ~260N. (©) Vien 2. te yes pressure duet be ection dererce mast beaker Io account. ame. Regn Ee ah on 0.0008 ~cnnnn OMNES ye nt?) ean tn? kg mi -274N "Tus, sir the ea has bean ase 2 tes, the egule Fc 27.4 N, Comparing the 8 resus Maes more fore 1o Keep the ear elevated that does te hol Ie at =O. Ths makes sense plysscall because che elevation dilference generates a higher pressure (and thus a higher required force) at the lower piston ane ride Lisension hen f= 6 be see godly Gases oft kale Ml ds ones the clase problem imps to seting te fo preeur aul However when Oar is ahyscotee bead ane cece the ‘dens f the fh enters th cale.aton Chapter 2 Pressure and Fd Statics ask Solution Two wus tanks are eommeciod ws wach oiler Uhcoayh a mescury manmneter wilh nein bes, Fora give pressure difercice beso te tn aks, ie paractens nl @ abe te be dela, ui, Assumptions Both wake sal Properies Tho sovelic gravity of me-riry is pv 10 he 13.6. We take the stoma density nf water to ar AN yi’, seury ate incompressiie Analysis String. with the pressune in the tank A. and moving slong the nine by adding las w> pr dein) or -suhuroetng (ase 219) the agh tems rl we cree tank P and setting he result equal tive a a ee 6 Recttanging and substincng the koe values, =e 20k Zemee 21185) 10001 (see |-0.9750m-7.5.6m jain kN From geometrteconsiéertions PR Awna— 2a (om) Thuselare, Discussion _Note tat vertical distances are use i manometer analysis Harizantal distances are of na ennseqnenee (Chapter 3 Pressure and Fluid Scr 268 Solution Ihe feigh fw ates ser veda yee ge ng ie envy “he tye i foxce on the "He welght of te elinder pert lng are tobe dtermined. Assumptions Viking lioness 2 Amr essa tls lid ne eal Le be aed in calculates for convensnce Properties Wo'ako the density af water tobe G2. Ibm hroustout Analysis _(o) Wie consst te fee ody lagram ef the lguldbiock enclose Ry the clues sutfce af the eylndst fr te verical ané sorizeta, projections The hycrosatie Fores acing on the verieal aad horzoral ple surfaces ae ‘velar the weg ofthe lquld block pr length ofthe eyinder ate ovina fore om verte sta Ty FPA Pack PEE + RIDA = 62.4 Tom? 1(32 252 p13 2/2 MYCE Ax | we) albm we annie > ‘Vor ee on hates anc 7 Fy~ Bagh PEA Phan 224 2 ” ee raf 2b = (oedumat sez nat yoo nee 9 1 bees ( u y M 32.2 Ibm.’ ieee ‘Welght of ld lock prt length (es. nearh = P= ges? as QED ibe cers Weme = eel = ei ~ ea rmnvazanis yan) U-xnUR ‘Therefore, the net upwasd wertcl farce is Wy etwr=81 <1 “Thea the mages aa drt ofthe hyo ace acting cathe Gini suttce become Feo Rh FE
wea eae 2351 N 3 3 ‘Live distance of ihe centroid ul a iriample Crom @ side is LS of Ue bright of Ube triamyle (or tal side, “Lekiny te moment Sethe Fan ssn nd ong ah he sna be sie be ee eat yin tigate aoe oe _(0 AT) 5 Pel OTRO Ss 66108 Eu-v + on zoom Themis map cae oe DN Ox TG Sev tn Foaling sig a2 ee i ae PoP B20 aon Discussion ‘The tension farce here ts «factr of ebous 3.5 sole han hat ofthe previews problem, eve though the looagh is une al Cll (Chapter 3 Pressure and Fluid Statics 3-96 Solution 4 retaining wal! agains: mit sie isto he ennstructes hy recrangular eanceetehioeks, The ‘hich exe blocks will sar sliding, are the blocks will ip ove ae to be determines height at Assamprians Atmos fe pressure acts am heh sides of che wl, and thus i ean be ignored in calulations for ‘Properties ‘Te density i sven to be LAOU kg for the mud, znd 200 kan for concrete blocks Analysis) The weight af the concrete wall peri length [= mand the Hesian Faroe etwven the wal an! the ‘round ace 5 {1 Wena = Pe = 700g" .R1 my 751.248 2 | ras ig mis Fsoion = ag =VAUTSAE N} = B78 sm “The rast force exered hy the mu tothe walls Fig oF, Pag A~ othe A~ paths wong enim. A Tiga \ Fe uri? r Sseting the hycratatic and friction forces equal to eachother aves Frc + curio +h 630m Ya = cen (8) The tine of accion ofthe hysmasatie foe prsses themigh the pressure center. which is 5 fram the free sonface. The Line of action of the weight of tie wall asses throug the midplane ofthe wal. Laking the mome=t sbeut paixt A and sorng Is equal to zere gives THO Mya 2 API Hyg 2) 80H Fad substituting the mud height fr tip over is determined a he 0.157 m Discusston Chapter 8 Pressure and Fhikt Stnes on ® Solution A quarer-crevia gate hinged shan ts ype ege cones ke ow af water owe he lege at where the ale Bs ese by a spi, ie mu spa force reuured kee tie gale lane et tue wale Kv ies (92 ire yr ao thee bs a der Aswumptions The hinge sfictioness.2 Anmospherie pressure ces or both ses ofthe gate, an ths it caa be ignored iret iows feasts THe weigh othe te iy We ae the density af wate oe 1000 gn oho We corsder ti fie body dara ofthe qu Hoc closed by the teu surface ofthe gu andi. See a ere ; scene fe Np nh ney tm aie d= py). 1 AM -100tgnhissimishe/amtimcam{ HN] A Tie a vse Tu (ten ieaineeemitn «anf tL cena ‘Tho yin affuad Block por 2 ange (Cowart: Wa patm ing] we eB /A| (1000 kgm") (48t me" fm) crema ‘Therefoce, the net upward vercal force is ‘Thon the magnitude and direction of te estate force acting othe surface ofthe d-m long quatr-ciruler section of the ate becuse y-\id or} ~lareown® -as.63007 ~1922 1 158k tan pam. + 2-232 Fy Te BKN “Therefore, the magncade af the hydrostatic fores acing c= the gfe 192.2 KN, and is line af action pases throug: the comer ef te quare-crcula:gae making an angle 232° upsands fom che Rorizarl “The maw spring force needed ts determined by teking a moment about the pein 4 where the hinge I. and MeO Fp Rsia09 61 Figg P 9 Solving for ya and subsutng the spring fore is determined tobe ag — Fasiuih-6) (192 24M) i008 73.29) 177 kN Discussion Several variations ofthis design ete possible, Can you thin of som af ther? 261 cre Finan nay ener Ths med a he eid er, ped Pwned teed, oe CChapler'$ Pressure anal Fil States 87 Solution retaining wall agains: mud side sro be consucreé by rectangular concrete blocks. The sé helght at ‘which che biceks wil stat sliding, andthe blecks will do evar are co be etermlned. Assumprians Arwasperic pressure acts on both sides of the wil, al ths It een Se ignored in calculations fe Properties The density sglvea wo be 1400 kg/m forthe mud, and 2700 Annlysis (a) The weight of the concrete wall er unit lrg (F = Lm and she fet Fores he-wren re wall and the ‘ud ate exec = FEV = (TON gn" (O81 -V47y [0 AOR Fevaon ~ HT igos “OAIBATEN) = 8990 N vats “The yea ce cere by th md oe wa Bye h 2B An pthc d= pe(hV4 ca wa = vrng'yasiyaer0 sf} IK. freer a ~a8eTH? W f 7 Setting the hydrate and fricsianfonreseqial seach eter gives by -Fece > veNTIE SK) + -0.708m (®) The line of action of ee hyiostate force passes through the pressure line of acon of the weipst of the wall passes arough cae midplane ofthe wall, Taxing the stom setting it equal to zere ives yw ester whlch 4s 2/3 from the fee surface, The about point and 0 alt= Fy Ci/9 + Tyg lt/2)= 080789 Solving for hand substiuting, the mud height fr tip over is determined 1 be Wimst |" _(SaeaTex0d zeusur) ~~ bi 0.905 m te wall will side Lefuse pping, Three, sliding fs amore uri! daa ping Discussion — Note dha Gs cu (Chapter 3 Pressure and J fuid Statics Solution A quarcer etreular gate hinged bout Is upper edge contro the flow of watcr oer th ledge a: B where she sis prosody u sprig, The rnin spring once required lo Seep dhe gals clusud wien dhe waler love rises ly a ‘upper edg: of she gate i ta be determined. Assumpttons 1 The hing® ts ficaness, 2 Aumospherte pressure acts on bath siccs of the g: inealeulations for eonvenionce, $ Ihe weigh of lie gale is negligible and thus can be Ygnersd Properties Wie take te density of water tobe 1M p73 throughout. Analysis We censicer the free horly ingam af the liq block encased hy the circular sare of the gate ene its ‘vorcal and hovtzoncal projections. The Fyrostade Forces acting on the versal and hastzonca plane surfzces as well asthe ‘weight ofthe liquid block are determine a allows Dp aly Pi gd = ethic d= aK DA 4 trang nstmcverrmiansan{ y ig, og > -a1s94N La, oalforcotertoe ance rad thes thera Ti AN } ry | roots wa 1 pe ming eed Wem og = peiwe nk? JA) 1000 hyn?) (9.84 ms") (wx (A)* 4) \Tocokg mis” = 198.141 “Therefone the net upwenrd vertical fare Fy oF, -W = 827 8-193.1 = 134.7 NS ‘Then the magnitude and direction of the hydeostatie force acting on the surface of the 4-m long 9 gate necome retreuler section of Fy ~ Via +8 ~ (@1s9n00’ + 0347 07 -311.6 kN tang FE TAIN ggg, @-25.2° F, S1S9kN ‘herefore, the magnitude cf the hydrostatic force acting on the gate is 41.0 KN, and is line of action passes through the center af the qunte= eller gore malng an angle 25.22 upmenn fram tho Portznntal hou the point where the hinge i, and ‘The minlmum spring force necéed Is detcrrincd by taking a mome setting i equal to ze, Yao ve tsinoo—o1 1 on Solving for Fes and substhuting the spring Zee 's determines to 56 ya —Fisi2(90-8) — (241.8 KN)sin(90* 23.25) 314 kN Discussion —_ Ue previous pullin iy solved usin a pr nike EES, iL stp w seca with different values. (Chapter 3 Pressure and Pld Sas Solution An nnd ye sts wt foe tho geen aston tn Sterne Wa hl Te cha sw hwy ea Ascunpoos 1 Aumesphat pressure act: on boi sce ofthe ate, and aus W canbe geared fa cleo fr Comestenee 2 ie weit degli ropes we alah ye ae ob 109 ago Ine pe tet ene m ei wa Se hme Ara " 25m 1 Asal & zane! The herapphosty cates eee iar = 1.974 = P8688 a 28) * seo ° ipnesbesindds )Apllg gab, veln® Ficlis-toabFrsltg-ror]_ side) Se 4O- 180g pgp ap nSefely rs) aR =m ne] Chapter 3 Prescre and Fuld Staues aim Solution The lng of Ue portion of a cable iow block that extenls abuve the water suka i ‘of de ie black below de sueface is Le determine sae, The vit Assimptions the buoyancy fares in aici nepligihle 2 The top surface of the ice blocs parallel ta the surface of the nies The specific pravites af ice and sea ‘omresponding densities are 92) kg/m! anc 102% kgm Amaysis The weigh ofa body ating in Maids wu Hho Daya one ating ont fa eusequece uf yeni Tore bance Fram svi equilibrium). Therefore in tit cae re average ent f the Rey mst he eq he dens ‘ofthe Fact snes er are given to be 0.92 and 1.025, respectively, and thus the WH Fs > Oraty£V oat ~ Pats €Vorrmeracd Venaged Phat leek ee = sa foot Pe © ‘The emss sectional of cube is cans, ad thus the “volume clo" ean be iy replaced by “height ratio", Then, 0.92 L025 ia the surface, Solving far gives Discussion Note that 97/095 —0 89758, s9approximate'y 509% of the volume of an ies black main ror symszetrical sce biceks this alsa represents the fraction of height that remains under water inde= water 390 Solution se censity f aliquid isto be desermined by 2 hyd-ometer by establishing division marks in water are ia ‘the liquid, and meesuring che cistance berwsen thesa marks. Properties Anatysis A hydromster floating in water isin atic equilibrium, ane tte buayact force J's exerted by the iguld must slays be equal tothe weight W ofthe hydrometer, = 1 ‘We take the density of pure water to be ILD kg. BPM ag: ~ ee, ‘where isthe height ofthe submerged portion ofthe hydrometer and isthe ‘ones srcrinn ama hich fs canta Inpursvawr = p.2h A. Inthe higuish T= aguas Ling the relaens abu xual we vale oll (sine bo ey dre weigh! of the hyceometer} gives Plc he = Prqua Rl quar Solving forthe liquid density and sutstituting, incrcal hydrometer, the product of the fluid deasky: and the height of the Ssubmergee portion of tse kyeromets- i constant in ay fluid Chapter 8 Pressure and Ful Staies 3.100F, Solution 4 concro block Is Iawered imo sho Sea. The trslon inthe ope Ista be detesmired bstore and block is immersed in was Assumpttons 1 The buoyancy force in art negligible. 2 The w: Properties The density of steel boc’ i given zo he 696 Ibmsit. Analysis (o) The forces acting om the concrete block in air ace cs downward weight and the upward pull action (tension) by the rose. These two forces rust belznce ecch other, and chus the terston in the rope must be equal to the weight of te black: Vedra! (art 5 Ay ia 137 the 1 of xe ropes neg FWP = (496 tv 9 fo’) va 6841 = 6980 tht jn Iam Fa | (8) Wren Bn eter in ate, sho eh “cuagupmas Ile ors lance is ene ys aue)(uarar 1M 522th Fasag “WF, 09848620102 if = 6100 Hot tonal fare of tayaney 2 be yn pyald—(62 A toms Discussion Nets thatthe vieight of the concrete block anc thus the tensioa of the rope decreases by [GRA U2) 4698 = 12.1% in water. a0 Solution An ivegulety shaped bocy is weighed ia alr and thea in water with a sping scale, he volume and he seems eens othe by are ae determined Properties Wetake the density of water to 100 kaa Assumptions The buoy Fors fn arb lige 2 Tha bay fs co ely subengol in wales Anniysis The mnssof the by 6 By TaNN | Ukg-ass®) - Fay, WEN (TBs!) ays oyg ew 2 oeims"| IN) sane ‘The dTerence betwee: the weights in air and in ware is due to the buoyancy "” fone ner I, WPM ATE 24108 a= 1 Ta-ra0% oting that 4 ~ Pyyur the volume ofthe bodys determined to be ve FON ines yy! 20.288 mn? [R000 Kgim®)iast mas") ‘he ney wh oy bes LIBS cor int 290 kg Discursion The volume uf the butly ca use be mesma by wlceving Ge ange ks de volume of the euntaer vin lo bud is dropped i i fasumig he Lay fs 1 paren). (Chapter 3 Pressure and Flu Staes Solution nvered cone laced ina water ank, The ela he cd comeing the cone ote ota of he Asermptins The hinynney force inci gh, Propertice The Eons of wot 4009 kg? sonra Desai wal 7 vee maar wel Fox tescemiein en / Pak 2 Pressure and Fd Stade Solution wae tanks song towed ya muck 20a el ocd ne he ang che fee sue mekos va che ausal a astm Sueded of eb te denn _Assunpdons 1 1 ds ertzrtl ota ection i ra vaca compan (0 = ‘Thttee of epleting. bear, bers oe: bumps. ane lenge eur ee De conan, end ono! conieed $ ne celeron reals Sra » ‘io tte se a tot th deco of rota, ae zal te bee ep ‘teva decals Thebes of gh saa hes ae Sobng fora a saa, ey Uae me = (98 eS ae 2.08 custo Nowe tht he arable val or rfl wit connt cea since we wee mo torent rab Solution Two water tanks fled ih Tanks rune ‘et che stor and be ebier mins 2 Sovards af coman secthauian he is tee Amunpiins Le tan sos : 2 Goneane 2 Wane isan lncomgresiie \ Biases > 1. Properdes We tke te deraty of wae a] Beira Anafds—Theprasur dleranca brwwen to pols {and nan incamprste fad s gen by PicReruinainele alsa) we RReety ak sins y= Toki 2 ah fs saa pi Lhe takai, ine B= Pp tl Poe Foam area and tha the prs eae Br Porm athe tabi i@sim |_| santa? “Tank Wa have o,~-S t atdskas the presste tebe vam 8] ; Been 20g nonin ivan sme?nem| ES) ees eur * nee mje a | rk age: preset Ue ton Diss Wi can css see hs problem qeky by owemrng he elton Ryya= ele a) Accel or fk isan Les on nf Th AIM ST 8! nent ft ye ne ah a “Tafre, tne ts th he hry tesa id igh pokes Fook nce wl hows ger esoreatths hoon (Chapter 3 Pressure and Fluid Statics 1. sl of te bal etabe dewrnlned. fo Tae mms wo soa a ey vt an he ‘Asumprioae 1 The dynamic stfect ofthe waves are disrgardec. 2 Lhe buoyancy force i iris neliibe Properties The dersty of se ws hae Anabsis ‘The weight ofthe urloed boats bs glven tobe 1.031000 — 1030 ign’, We tae the dena of water tobe 1000 ig Frog ~mg~ (9550 Ne BL mis?}| LAN eo Baie TON « “Tho moxeney fore heeemeca mevtmurs when she cae the a Bog in ale, al steams be a2 ~ (460 igi. 81 vs" ane) ES _ zs Toobg- mi HIRAI eee eee {1000ig-mi wal edgitof a oan a ge bua tse Bs ea 10 eta Lore “Thocecre she weight ofthe maximum lad = “Tho enmespaniing mass aed ae 168240 1050455") 94 grag Saimst TN 170.9008 TAN Hs (eng Piscussiat Nowe ha this bua ext cant meaty 30sec es ae: Puy teed ‘bons ea water should expect ic sink lnc ter deeper sehen ty eer tes vate, suchas ariver bers 3 Frente ud ai Stats Soltion Awake ath bsg aed unas ophil wad ena cn Ube sah de suai of wae These onl 0 be deemed, 2 elton 0S eps ferhe dnb ren we, a vem = comin 1 Rot ling hon diving er gs ang hen aairdoesr, Anais We ake dee atl ane sl i Un gue owe pate! consti, Ue aad nel Ue angle fa sea eal come Grete eo - angie) gana 4a.” poesia Rimi Sins l= Wine the deton of maton severe hha cd acorn mgive sand Aetion wecct' an hs eeone rage cus ‘Thor ce wrset ofthe ane he te uae mks wt the hele beseres cone fm na78) + o-~208" Bra Frese BRtmi? Gms" vation ote tae se ans is wll fr any Aid wth esas Eeny, et jnt itr, a we lntanacan hsprn ose te ioe, (Chapter 3 Pressure and Fld Stas 3436 Solution A seui-irvular gle is hinged. The required force at he cemr of gravy wo keep the gate closed so be letersi Properties The speilc paves ut oi and ulyeerin are given in We figure Analysis Pa ab anal oil, SG-091 £ Ciyeortn SG=126 Force applied by glycerin 20.5 aaa? xi toga 2° Hyd 2B eB “The gage pressure of ai alrapped on Be wp of theo surface B 80-100 -20.LPa (gage) ‘This egative pressure would rel ian imayary reduction in dhe ol level by 20000 ‘Tore asi 224.0 Ure dhe iainany oi level would be Af — 4.74 2242.50, fm ylycerin surface, The force applied by el is Uns pit 0.9n 9810] 254475], 7x05" _gsog av wel a a le0s? 03025» Jy. — 0.10988! — 0.1088 <0.3° —0.,0068625 ar! Hoag! TS 0.2122 0.0068625 | _ 9.2915 0 M08 oatee 0.0068625 - 27124 A 2.712 «0.39267 ‘Mament ahout hinge would give Fo tgeg “Fry Tere Faltr-e25) 0 F< 0.2122 +1030% 0.2945 —9508(2.784—2.5)-0 F-11296 N=11.3kN ey (Chapter 3 Pressure and Fuld Sides 3.51 Solution An claste air balloon submerged in water ss attached to the base of the tank. The change in the tension free of the cable isto he determined when che tank pressure is increased anc! the ballon diameter is deceensed in fcenrdance with the relation P = C2 Ascumprions 1 Arsesphvie preseason al sees, an tus 9 Be ignated in calculations foc convenience. 2 Water i an incempressible fia. 3 Pr=100 kPa “heat of theo and he as nage Properties We take the density of water to be 1000 kg/m’, ‘Analysis eens oe onto cable ole he balloon sdetrmnes fom rc balance on balloon Fae Fr Mica =F "he anya erceaetngn hs san ay Fas PatVeneas~ — (000 pny?) Z0S0*{ AN pag y é which is equivalent to D ~ VOTE. Then the Anal diameter Trg “The variation of pressure with ameter is glven as 2— CD" cf he ball becomes p, Jere Th socom EDMPE oy DB Jen VP . at ONT guppy OT “The buuyancy force weting om the balloon in tis case Faso Meatana ~ bu8 2% ~ 1000 kin?) 281m (cal 5 enn ‘hon he percsr change nh cal fo cams Fressiot Fenn ® Fay Fas AT Change nao F2L Fae epg HRI22 soo o6.4ne, "herefore, Increasing the tank pressure in thls case resus In 4% reduction In cable tenskon Discussion — We can obtain a relation for dhe change i cable wasn as follows: Visa Po 8Veacnt Fai -Fas = changers F84=F 8S 199 Fs 100 1001 ‘Chapter 3 Pressure and Pula States 5.153 Solution 4 gasoline ine is cone Wo & pressure goye Urungu double-U nanan, For given reading of the ‘ressuze gage. the gage pressure ofthe gasoline he sco be determined. Asvumplioes 1 A he liguids ure inconpessile, 2The fle of ir cols on pressure is eligi Properies The specific graves of oll. mercury, and gasoline are given to be 0.79, 13.6, and 0.70. respectively. We lake dhe density water bey = 1000 hgh Analysis Steting withthe pressure indicated by the pressure gage and moving along the tbe by adding, wr 20 down} subtracting [as we grap the pgH terms nll we each the gasoline pipe, ad setting the esult equal 19 Py Frage eB Puce PegBhheg PysamSyacine — Fost Rearanging. 2, PARC, SCs | Gage! SCorainl gain) Subtituig, Prac = 261 KPa ~ (NO Rp}. is”UD A) ~ 0.79405 om) = 13.40.10, 7010-22.99) vey Yatra (TooO kg sas? kN 245 kPa ‘Therefre, the pressure in the gasoline pipe is 15 KPa lower dua the pressure reading adhe pressure wage (No B0tte “Jig Discussion Note that sometimes the use of specific gravity afers great canvenienre Inthe soluttan of peoblems that invelve several Disida, Chapter 3 Pressure and Plaid States aan Solution A gasoline line is connec lua pressuy gage zug x dauble-U masmurcwter, Far a given reading of the pressure gage, tae gaye pressure u Une gusoline Line yo be ele, Assuimpelans 1 All the liquids are nonmpressible, 2 The offect of airealuinn an pressing Properties "The specific pravities af ol, mercury, aed gasoline are glven robe 0.70, V6, ant 1.70, eespetively. We lake te density of war be a, L000 hy Analysis Staring with the pressure indcated hy’ che pressure page aad moving along the whe by adding (er gO fossa subtracting (as we go up) the ei terms ual we reach the gasoline pipe, and seing the result equal 10 Py. ives Page ~ Pn hy Pasian ~ Pag ~ Panstc pete “Pete Reatranging Frater Page Crllhy SgsitMaatstar Suche 5 Substning, 80 kPa (1000 ig sf tte Yt «00K mF J Tk =mkPa 81 mts) 450) ~0.7940.5 m} = 12.6(0.1 49) + 0.701022 my) "hore. the pressure in the gasoline pp Is 13 KPa lower than the pressure cea of the pressute gage Discussion Nowe that sometimes the use of spell gravry aess great convenience Inthe salutlan of problems that inte several lids, (Cluapter 3 Pressure aun Haid States 2.155F Solution A water pipe is eannected to a double U manomesee whase foo arm ss open to the atmosphere. The Atal press ar the estes the ps he dred, Assumprions 1 XI que are ncoenpocsile. 2 The solubility afte liquids in each hors nti Properties The specie graves of mucry tl il ae gives Wo be 13.8 and 0.40, espectively, We take te dey of eater ibe p= 62.4 Tome Analysts Stang wth the pressure at the center of the water pipe and moving along. the tube by ating (as we 20 Clow} subtracting (as we go up the gh terns unl we reac the Fee surface af all where the ol rahe Isexpmsed othe -axmosphere, and seing the est eal t0 Pay Ives Pramesse~ Pnuathoaee~ Basin Maat Pie Paistar Fae Saling for Ppa ys Feowenie Fer +P ues ~SGahnseat ~SGuc hie SGauh) ‘Subottig, Pong ~Hpsa | (G2Albw TD ILLTAMHOSHZK O8OIGHLZ R 16Usiz0) senscanr2tgjs{ OEY an cre pa ee aetis Li “zefere, de use pressure i the wae pipe is 22. pa Diceussion Nove that jumping horzoatal'y from ane be a the est and realiing ha pressure resins he am the same fi stopifes tho aay rea. ‘Chapter 3 Pressure and Fluid Statics 2456 Solution The pressure of water flowing though pe is measured by an arangesest that volves bth a pressure _Bep and 9 manometer. Fr the values gven the presse i the pipe Ista he determine, Assumpldons 1 Allth Uquids ate incomtesible, 2 The effet of cola on poesure is neil, Properdes The specific gravity of gage uid is ghven to be 24. We take te sasdatd density of water tobe = 1000 ig Arps Sening with bs pss nice bythe presse eel ening be ey aig is ‘dvs ar subratng las gous the gh ers atl we Zech he weer pp oi eng te ces Gua 02a ave Loge tate ~Pgeelyy, “Pat 2 -Reacanging, Pau —Pan Fate Suaphge Reb Rag 12th SC Esin® Eysnd} Nosing tht tn bl sein Prous — 040s (OOO ky DLAI A=]. ANNO IONE? OBIE ‘cto, de pessun in asl pes. KP ver ean Ue pressure ae. age hid Sood Discussion ‘Note has even wou a manonee, the eadng of a presse gage canbe in ear if ts. nt placed atthe Sr feel the pis len te i pi (Chapter 3 Pressure aad Pid Stasis aas7 Solution A U-tube Milled ith mercury except he 12cxm high portion at the op. Oils poured Sate die left arn, Torin some mercury fom Uae left aru in ie right ume, The mains amount uf wil at can be add ina he et arn Isto be determined helene Pr oy UPR hand GADD yi 81 eas), 1 NOU ‘fee bale on the eyliner ia tbe veri retin yields Solving for and sebstiating, 2708 sy ana PP OW = 980 Ni Discusslon We could also solve this prableo hy conshesing the atninpheric pressure, but we wold oben the same result since almuspheric pressure wuuld cancel eu. Chapter 8 Pressure and Pluld States aan 2170 Solution The danity of 4 veo! log is wo Le messured by yng ead weighs co i unl bude og a he hts Solution An ievberg loaing in seawater fs considered. The volume fraction uf the iceberg submerged in sewwatr is sat completely subserged at then weighing chem epartely Ina. The average cen of a glvea log so be determned tobe cetermine. andthe rasan foe tei tnnaver ta be explained Assumptions A The buoyancy love in aes eligible, 2 The Uensity of ieberg und seuater are uniform, Properties the densides of iceberg und vewyuater are given \o be 917 Ay ad LOZ? kyl’, respectively Analysis (a) Ths weight of a body Mating in v Dud fs eu « the buoyant force acting om i (a consequence of ‘eres fore hala from static elitr, Thesefre, woi's Proeet Mana sa ost 88% V, ut Pith Plas ‘Therefore. 88% of the volume of te teoberg is submerged inthis case (6) Heat ransfor tothe iceberg ue tothe temperature ditference between the seanyater snd an iebeng cases uneven meting uF Ue sregubury shaped ister, "he resulting shift In the center of mass causes the keeheng to wurn aver Discussion The submerged fraction depends on the deusity of seawatet, an ths fraction can dif in diffrent seas. bys approach Properties ‘Uh tei o le weigh give oe 1.900 kg We take he deri a ater be LODO Ky Analysis ‘The weight ofa bady Is equal tthe ayant foo whew the hod Is nang in 2 fad we Beg Compieiely merged ini (2 comyeeaet af atta free ble: Fone sae lib. In tis case dhe average deny of he boy mst equal he lens of he dss WE > erat Maat Py Paws ‘Thecke 140. Subse, the volume and str af he log ane deci ta he Mac tM yA 1ST.ON KE 5 oogg 10° a has I an 0.18802 a 1000 i 197081 KE gas Mea 1ST OBEN 5.174 kg/m? 835 kgm? PT 0 18800 " ‘s Diversity es ely meng oe Bs ly to vai Tay th eo aso be completely submerged, bu tl bot Very cal because of the soll valure ofthe lead weights. [Chapter 3 Presse and Fuld Stas [Chapter 3 Preesure and Flald States wn Solution A recnguar gate tat Jews ain We ver wise angle uf 45° wh Ue burizonl fo be opened Ena ts ler enge hy applying a ninnal foree at its ener, The minimums farce F requleed to pen the wear gate is ta be eter, Assumptions 1 Atmospheric pressure acts on both sides of the gate, and hus it can be ignores in calculations for ‘onvenionce, 2 Triclin a he hinge is epi Properties We take the density a water to he 1000 kin? shroghour Analysis Tho Tength oF the grip and the distance of the uppor eg of Pee gate {poi 3) Leon de fre sure in he plane of the gte are am am 24am ands nis Sind atm 059 a8m=20 2 ‘The average pressure ona surface is dhe pressure atthe certo (mdi) of the ‘utfce, and mulipling hy the pte arma gives the rela hyeasrare-s she ware, Poh cA By Ped = foogn (asim ame> 424907 =S744KN the distance of the pressure he gales From the fee surface of water along the plane of gait 12431 1 4248/2) oe ‘a tae 87a adm ‘The disc ofthe pressure center Som the hinge a poi D is aaml4-1131=: 583m Taking the moment abowt point 2 and sectg st equal to zero elves Poin Dyson we Selvin for F ana substi, he regi force wo avercone dhe pressure pn Rel _ UOTANNNZSNI Moo guy @ 1243 In axtdition eo this. theres the weight ofthe gare self, aehich must be added, Ta the 45 dri, v onstts| =mgeastase) = anagvastouséy) TSN leogiairy 1.042 0N ron gov? | "hus, the total force required nthe 45" lietion Is the un of these tn vals /01-3KN = TOTKN in th 45° diwcrion vi Discussion The applied force is inversely proportional ( Uhe distance uf Ue pia of application frum the hinge, and the requlved Force eam he reduce by applying the force ata leer polat en the gate. "The welghr ofthe gate Fe early ‘glib compared lo the pressure force inthis example: fn swlity. «beaver gate vould probably be require, 
