hw3_sol

# hw3_sol - 2.33 The homogeneous gate shown in Fig. P183...

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Unformatted text preview: 2.33 The homogeneous gate shown in Fig. P183 consists of one. quarter of a circular cylinder and is used to maintain a water depth of 4 m. That is. when the water depth exceeds 4 m. the gate opens slightly and 13L: the water {law under it. Determine I11: weight of the gate per meter oriength. EFIGURE 92.88 Comider Hm free boa/y diagrams of H19 gafc and a pal‘ﬁoﬂ of {he wafer anhm. ZMoT-O J or Mum ~ 5,4, — mm; Were F” Maﬁa/4 = Mm“ Jr3,"‘*3/{3»35rr.v) (MN/9,) 33523 k/V qafc since for "H76 Verffaal side} [16 = 4m -' a 5,9; :35»? 29/50, M F} = 3%)? = ﬂaw/033,15, (4m) (mum) : 39.2 luv ﬁlm} a” w: = “my? — K'(§(!m)1)(/m): id’XKOEEL-l —;’?]m3: 2,1!» M’ MM; [4 = 0:5»; and [3 z: 0'5“ + 0er = 0.5;); + 1;“ BEUMJUmf Ya” :05”, + 3,5,» (man) ﬁzz/m‘%2/— 20.576,” To defermme 3,} consider a viii/swam 71/951 6017535 01" a qmrfer circle and Has reminder MIhOWn in we figure. 77%: 06199013.: Maren: 6') and é) are as ind/bqfsd. r9 T5053 (0.5 15mg : (0.5 24% 50525917 ( 60/? ’1‘) _ Ccan‘ﬂ \$0 #161“ {AH-ﬂ? 192 =¥OJZ =' £2 Md A); if”??? 7951.} WW; Mame:J by combining £7: (U Mun/9}) (5"): (69-5765») W +(o. 123m) (2.“; M) - (3mm (0.52%) - (am mug.st = a OI‘ w = 66% AN 2.39 The concrete (speciﬁc weight = 150 lbfft’) seawall of Fig. P18“! has a curved surface and restrains seawater at a depth of 24 ft. The trace of the surface is a parabola as illustrated. Determine the moment of the ﬂuid force (per unit length) with respect to an axis through the toe (point A). FIGURE P2.8q The campanrn-ls oac 7}): fluid {Sr-re new; an 7%: all” are 5' and W 05 Show» on ﬁt: fly“!!! wéere g: aged/l :— (M.0;’i,)(l~’;—_{*){zy,q,mj :- f‘gtfooa and jir°%-’ct: ao-Fz.‘ 24/59) l5W: 3’47; 73 dcﬁ’r‘mme '5‘ 14nd area 5'60, ThusJ (Sec 43w: 24w- nym‘) ’56 [fay-0.3x“) ﬁx 0 x5 x9 :4: {ﬁg-EJJKL' 0 go : :20 3 n : [34X " ‘9'ng l) (ﬂak: All fen-37h: m ft) 6014' an??? 9": V; 24- =-' 1‘75 Hz 5o ﬁaé ) +L= Ax 1H :- 175—161“; Th 3 u) %=(¢%0ﬁ3){175{r’)= f42m/A 7; 5647.4: (Pd/I'an of A -‘ «‘0 "e 9‘" 3 z 9' XCA : fa: 004 = fie—ija’x -/(Mx—a2x «‘c/x = ’ﬂo"i§° (J 0 0 m. (Ir—1, 2- 0.2 fun)“ and X: = n ) fr 2 4.“ 1'75 I :~ goo At, ){f’itj - [/4 Zoo A )0; «ﬁt-5M £25) -'-' .2 €200 [IL/é) J14» g :9} -— wfxr—xc) 2.15 Three gates of negligible weight are used to hold back water in a channel of width b as shown in Fig. P235. The force of the gate against the block for gate (b) is R. Determine {in terms of R) the force against the blocks for the other two gates. - 5:. ages-355:; 5r Cast-20,) I FIGURE 92.15 Hf Mrs/4 -‘ J’C‘Q'X/MLJ = 3321117 i2=(-§U(L§fé) _ 34211, ’e‘ ‘5“ F5? (’35:? (a) 0!? ﬁre-ho}- drefma Sign)»: l 5;: {if} (6%)” 45am?) d/m’ :.- 3. 5A 3 i ﬁnd W : Xx "Hal M1132 If. If ﬁﬂf 55 .— M‘A {49. 37‘?) FM”, 57, (0 (WA :53, 2‘4“; 55 : /./7£ ﬂy- Gauge {C'JJ 5r the {4195- My— abjmm Shot-aw) 7%: far“. 7;. a»; 7hr.“ darted SPGfJa-n [Deﬁes 770mg}: 7‘74: hlﬂfe 601d 77733.6”? 0/035 #025 Eaafhéwk 1‘: 7'71! mammf ﬁrmmd H. 0n éaﬂam Par-25 0,5 5d: g _. MM = Myyiéxw z E53515 1. ﬁm \$3.0) M1535": 75”“ @sgg: 1.. H32 An inverted test tube partially ﬁlled with air ﬂoats in a plastic water-ﬁlled soft drink bottle as shown in \"ideu \-’2.7 and Fig. PZJOZII‘he amount of air in the tube has been adjusted 59 that itjusl floats. The bottle cap is securely fastened. A slight squeezing of the plastic bottle will cause the test tube to sink to the bottom of the bottle. Explain this phenomenon. WA”: 7778 49.519 “£1411! Lt ffaaéxéj 77;: weary” of— ‘ﬁr éalgl‘l’d; is éefanced 5:, 7‘3: éueymi 74k”; a! 4: 5:99:10: 1;! 73'! 14er . 71g éualwrxé Aer-c: i3 due :51, 732 dispiawa’ with”! MUJgé’F 45 Fhﬁwﬁ. 7:913 Ira/um; {is dire 22‘: 7‘71: 1m“- pwsswg ﬁg WeﬁWK H? 77!! 11:335.? faker: ?: ﬁ-f-ajédlk When “ﬁfe if s‘fueegen" 7739' mi- ﬁres”: m 774: We 1%) 2‘3 Margaux ﬁft‘gé‘f/y (Md The)»: time increases 72, 7%: pres”: Compresw'ng The arm n 73:: éesé-eate, 72“; 12c div/dud VD/nrm: Ls {karma-9d m'ﬁ A Jubﬂgﬂﬁsi decrees: an F5. ' Sine: '11} ’3 59051571315) 4 deal/9:192 in F8 19H! (has: "ﬁve {(524 iuét 4» Sin/f. Fesl lube a Ptasnc name. I FIGURE P2.102 dnfleud I 2J0!» When a hydmmeter {see Fig. P2105 and Vida" V2.9} hm- ing a stern diam-later of 0.30 in. is placed in water. the stem pm- n'udes 3.15 in. abcwe 111:: water surface. If the water is replaced Hydrometer with a liquid having a speciﬁc gravity of LID. how much of the stem would protrude above the liquid surface? The hydmmeter weighs 0.042 lb. IFIGUFIE P2.105 Man '11): hydrameérr :5 Hamilhg, H‘s wer'girtj 7d} 13 balanced by 17” buquné éfce)F;3. Fay fguihlmhm, Z FVW/Tcdl :0 77M; 24:? haul-EV 53 =w (a; yr = 1,.) m 1.0 when 1%; )3 171g sulmergeol volume. er'h The new It‘ :3 3” 66)(ljm)ﬁ=—ﬂu a) (maﬁa/hm} 555,0) Mara) wl'f'h ‘71) darlj‘bmt (3220)“??? :: (5007,”)?2. GM“ .lV4' +1: 56 73135 _1‘(. l = OJgIXJO 3 7; obi-ma 777:3 dfFfermce 1%: chant): n? fragmA—f, LS {{JKMMYM = (o.ux;o“‘H-‘){f728 I . I A}: “Min. “"751 7'7" 49'” 1‘23ch "ﬁat Shem woe/d Profmde 3.!5'1'n. + 1.9:; in. z. (71,494}: . about The Juwéce 2.10? (See Fluids in the News article titled “Concrete canuc.” Section 2.11.1.) Haw much extra water does a 1474b concrete ca~ no: displace compared to an ulu‘alightwcight 38-lb Kevlar canoe of the same size carrying the same load? Fay QﬁuIHbrl‘um‘, 2 FVer'J-id :0 a _ '1“ ch) = F3 = ‘3‘“ JV— aml va [.5 (JuplaCfA Vaiume. 2 For Canard-z Canoe) Itn Lbrﬂ-M‘ﬁQ ‘5’: JV; = 2. 3L 42:53 F3 Fay Kevlar came, 32 u... =(bz. 1+ wk 491 = o. m 9&3 3 EMT-a, bowl-tr dLSPlac-c meat ‘= 2. 3L 'Fts- 0.1203 R = L196 2.108 An ice berg (speciﬁc gravity 0.917) floats in the ocean (Spe- ciﬁc gravity 1.025). What percenl of the volume of the iceberg is under water? For eqyi lr'bm'vm, W = weigh)’ of iceberg = Fé c ﬁlm/Mi farce OJ" ﬁe dfce r 5% £53m. , WIN” a!” 5 VOW/W76 m“ 1.69 Ja'ﬁ’wf/‘yed. "Thus! 32:01: 5 a?“ __— ﬂ. 5 Gigs 0.9/7 86.069631 ‘ '_‘ = f‘" = 0.8%“ 30?“ £59m I'm“; 2.'l l3 An Open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m. what is the maximum horizontal acceleration (along the long axis of the tank] that can develop before the gas- oline would begin to spill? 76': prevent! sf: “\$375) (#0716: 4ecelemlwe Could be ei'l'her 4-0 The Flﬁhi or 1H”; leLt) For ﬁve surﬁme ) 4L: LIIB An opcn. 2-ft-diameter tank contains water to a depth of 3 fr when at rest. If the tank is rotated about its vertical axis with an anguiar velocity of 180 revfmin. what is the minimum height of the tank walls to prevent water from spilling over the sides? wzk: 2} +12, (53 2.32) 4.1.244 71¢ volumg arr Ifmld :3! “ﬁle minim? Hat is 31.?” by R ‘bf; = [gm-4 dr- 0 2 F431)?“ *3- 7,—{15’0 ‘f- “has: R1 v re“ _. may“! ﬁll"! Rr 5c (Raﬁ?) 5'!— 1r (2.7; +19.) £223 5mm 1'11: Mini! volume I DY 2nd MM 592 ﬁrm ? 0+)dr +w£0Hf am.” #0 :5: Ft) ﬁ-zrﬂ'vﬂL- = 7:" ((451)1(31ct) -' hart3 (Md We. 191m! “Mum: Mus»t 1’? 63ml, +3 = +2- 73"[2.7L 17?.) acts-— 37";{3 {0: 0.29041“ Thus, Fir-am The A131“ Eﬁuahén (E15131) and “’1”: 1- 0.1% H1 .ﬁ: 2? rev 'r‘ I‘mn'a H. = (’3‘3 ‘ﬁi‘n I‘ 11'- v“ (ms "1‘" 1632.2 ﬂ) 5‘ )1“ ([4901 + amﬂ 2' 51bit 1D. Archimedes, it is saicL discovered the buoyancy lows when asked by King Hioro II ot'Sg..rI:a,I:-1.Ise1 to determine his new crown was pure gold {speciﬁc gravity SG=IQ.3}. Archimedes weighed the crown in air to he 11.3 N and weight in water to he 1.13.9 N. “in: the crown pure gold? Solution: The buoyancy is the diﬂ'erenee between an'weight and underwater weight: B=Wﬁ —Wm =ll_8—lﬂ.9=ﬂ_9 N: ymumﬂ But also War =(SGjymmUm, so Wmm = B(SG— l) Solve for SGM :1 + Wmma’B = 1+ IUBJ’US = 13.1(not pure gold) Ans. ...
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## This note was uploaded on 04/03/2012 for the course EE 199 taught by Professor Liu during the Spring '10 term at UCLA.

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hw3_sol - 2.33 The homogeneous gate shown in Fig. P183...

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