HW3_SOL - 2.84 The 18~fHong gate at Fig. 132.9443 :4...

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Unformatted text preview: 2.84 The 18~fHong gate at Fig. 132.9443 :4 quarter circie and is hinged at H. Determine the horizontal force. P, required to hold the gate in place. Neglect friclion at the hinge and the weight ofthe gate. +11. F F.“ P FEW egurfiiwém (all-gm M-‘ds—dtfiiflm l a; 457m}! Ira-nan)J ‘L G Hf [1% 2—1: :0 31 5, 5'0 int-5 ‘lf o H“ v Fri-r KM: = £2,4fi3y§§)/éféxflft) zflqzoafla Similar-13) :0 50 fligf 1 F‘y: W :3 £20: {wafume or" £403) = (62le fia)[¥(5;t)y {gfijzfl’goafl AL”) 2c. —"-' 4(69") 2: E 361‘ (See 53.2.1?!) I 3?:— H' at”: a} : %_£" ___. 276* 5r {Jug-flirth {fim fire—éoog-dzéynm 5f junk) Z M, =0 .30 P(Mt) = E, (g,)+ F} (a) 9 9 r {201290l5)(2-F£)+(3IJ8001L)('1'i-'Ft) p : -'-' 29,200 H: 643'! 238 The homogeneous gale shown in Fig. P188 consists of one quarter of a circular cylinder and is used to maintain a water depth of4 to. Thai is. when the water depth exceeds 4 m. lhe gale Opens slightly and lets the water flow under il. Determine the weight of the gate per meter of length. Comic/er H76 free body diagram: of He gafe and a pof‘Hon all {he wafer aufiauw. qm‘e ZMOZO J 01‘ (l) [ZWJrfiM-Effa ~FV£¢:0 where (2) F” :d‘ficfl = flex/a" £35m) (/my/m) s 34.3 W since for #79 Verfiml side) f1; = 4m “9.5m =35»: 19/50, M (3) FY :dflfl‘fl = mun}? (4m) {MN/m) #39,}. kW Iqde 3/‘/ (“J wI = 5'UmJ3 — Hiram") (for) = WW 5,3[1 afjfz 2,10 M! r : 0.51,»; d 5) NOW, 34‘ +620; )zog + I“ has +éflmmms 5052* “J [3 " 0'5” )5" y“ ' ”’ m ' ' ”’ 3,5mflm-XM) ' ’” (7) MM 1’2 = M — 541% =/ — 4321’”) x 0,575.»: To defermz’ne 2, J consider a Why/are Mal comfis of a qmm’er aria/c and Me rewalhder I MJDOWH in {be Wm. 7728 cem‘mfd: sigma: (D and Q are a: I‘fi/licallsd '9 T/WSJ 1 (0.5 -5’%)fl2 : (o.s—£,Jfl, _ (60/250 0.5-1??— (8') 50 #er wffb fig #3702 = a? mid fl; =/-%7 79% 9/1/63: (0.5 3%); =(M -£)(!—23‘) or I, 5 0-223!» Hence) by combrbr'ng 57: (I) Happy/1(8): (0.576m3w +(0.133m)(2.l0kA/) -(36¢.3/:M(0.526’m)-(3?.2 More.st :0 or w = 5 6w luv 2.89 The concrete (specific weight = 150 lbi'ft’) seawall of Fig. P23? 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 P139 The Compam’m‘s mt fie {Ir/I'd 16H: tic/15:; a»; 7%: an” or! 5 and W as shown on fire #13“: wfiere E": Kit/1 = {é¥.0fi3)(;‘gt)(2¥{£x#i) = llfi’flbo a and 5;? ‘11?!- : (3°47? a lav/5:9) 9w: w 7; de-z‘frmme "V 14216! mm 3'50. 77m:J [Sec 143:: in nyfif) ’10 '4 :/(-24—5)a’x zfm'woxlfl’x 0 a gag = Ma 02%.? x1: I '7 [39X —' 1) (Movie: All ftngfiis m If) and iwfiz 3,: Va J ,4- =' {7515252 J:- fliaé +’-= Ax Me = 275—159 Thus J “W: (Lake £3 )(2’75'152‘3) = L5 200 I1, 7; food: (Ed/mg}! of A .‘ 1‘s )‘o X‘, 3 2. Xg/i : [x 5'14 = /{a?LI-5)x.cfx :fl7¥X-D,2x )4; = szo—fl.:,g d 0 0 ~— 5* m fir-z, 2- 6.2 (no? and X = fl) 1,. = 4.“ If C. 175 Thus, _ MA .- L; y, ~ why—Xe) : fig; 40,, Ir. More) — m, 200 ,2 )0; AL ~59” H) = .2552” J2.sz / qu A closed tank is filled with water and has a 4—fl- K") diameter hemispherical dome :15 shown in Fig. P192 A U—luhc 1/ A" manometer is connected to the tank. Determine the vertical r4itt1tametcr . " force of the water on the dome if the differential manometer reading is 7 ft and the air pressure at the upper end of the ma- nometer is |2.6 psi. 20 PSI -- Gage flutd DU - 311t— Pa- 3T3 Far Eja't/iérléml 7— Fl’ftr'hia/ :° 50 flta’é T 12A FE): " 20 (I) MAM-e F; 13 7'71: 749mg. 7?): clam: aft—{rs an 711:;th find ? 1'; "five Wafer PV€SSur€ 41L 771g «9&5? 67‘ 774640913 Ham 777.2 mane/rte is)”; Jo 7,7241- .. [£2 In r—fizwttte )*(5‘°)(‘2-*}'333W)‘M§a)/m 2.33.9 9—2:; Tim'sJ from 53.0) MW: ten/am: of sphere = gammzérrjj _ ii! 2 F5 /Zm fley—gfttfl) — %[%(H£)flazlffia) : 3532M “9 me etch/g 71ml: 7kg Vehénidl 79ft: 114M“ The Mr EXEFB a» The dons; 1’3 55:100/5 IT .. 2.9-5L A 3—mdiameter open cylindrical tank contains water and has a hemispherical bottom as shown in Fig. P238" Determine the magni» tude, line of action, and direction of the force of the water on the curved bottom. 1—3rn—vi FIGURE P2.‘34 Err: = wed?“ 9,5 waérr Juflzarée’d Jr; hemat‘spkewéd infirm- : X [{b’ofumc if Cqfim’tr )-—- (Valqmc 0F den-‘Jph€fl)] #10 2 (3M)1C8m) '— T—‘i (3m?) 2 "fSS'EN TA; 2(5er :1: wreath! verifier}: downward, ind due. to Symmef'rg H: act: on “Hie hemg‘sphere. aloqg T71; varth axis a-F The Cylinder. #354.” 2. H1 A S-gal. cylindrical open container with a bottom area of [20 in.2 is filled with glycerin and rests on the floor of an elevator. (3) Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of 3 ft/sz. (b) What resultant force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (Note: l gal = 231 ln.3) {3. a (at 9g; = 70 (3142*) (Eagle) Thus-JP db 3 —F( 1‘4) di '9. (120 llH.L):(5-fifll)( l 1° 3 * it 0 Mr] E :- (0 (?+q§)‘fl [is] From ere—beclg-dtrtgmm of con-EmeerJ e—- t, A = (4M fi,)(xzatn?) L151" ) b lwltt'n.‘ = 57.9- lb Thus} fierce 61C Cam-Lemar an Fla“. 1'; 57.1% lb downward-1. 2..| l '4- If the tank of Problem 2. I I3 slides down a frictionless plane that is inclined at 30° with the horizontal, determine the angle the free surface makes with the horizontal. From Neuflon’s Jan! law, Shite, "/71: ion/j érre m 771: j-Jdueed-Jéu 1.5 The CoMfonpa-é a! weriphf {ngSJHE‘J (M5)5}m9 =. M £5. 50 775415 l' : I. 6 a5 fli SM and 'fl’iffihgl"! - f -' l ‘ a5 - a5 C056 0% - — d3 512:9 A150) CE H _ “5 (E? z 23) :1 3*“; I _. _ as C1559 = _ 2m}; 69519 3 — 5195:”? g __ 3 SJ'nQSiJ’lE a , M mama i—mfie = "W view d; HenceJ a? = "fflh 9J .90 11/76 {r68 dz I; surface is a‘l H78 some Meg/8 as #78 MW. 2.116 The open U-tube of Fig. P2.|l6 is par- tially filled with a liquid. When this device is ac- celerated with a horizontal acceleration, a. a difierential reading. it, develops between the ma- nometer legs which are spaced 3 distance l’. apart. Determine the relationship between a. l‘. and h. FIGURE P2. i If: 01—? = -— a” (fig. 2.2.?) d5 grit; "Since .3 2 de- .. __. _ 8—5 "- IE and flg—o "file?! ‘12 _ a E _ g-ro 0]" ,Q at = L 5 2.120 A closed, 0.4-m-diameter cylindrical tank is completely filled with oil (36 = 0.9) and rotates about its vertical longitudinal axis with an angular velocity of 40 radis. Determine the dif- ference in pressure just under the vessel cover between a point on the circumference and a point on the axis. Pressure :3: a mid {My f/um' varies 1}; accordeyce mm #1: figurine», 2 15 _— [92552; a"? + waskmf [1.53, 2.33) Slflflfl at" "-' £8 2.121 (See Fluids in the News article titled “Rotating mercury flecewer _ mi" 01' MESCOPE,” Section 2.12.2.) The largest liquid mirror Lele~ Llum rays scope uses a 6-ft-diarneter tank of mercury rotating at 7 rpm to pro- duce its parabolic-shaped mirror as shown in Fig. 132.121. Deter- mine the difference in elevation of the mercury. Ah. between the edge and the center ofthc mirror. Mercury _ fl FIGURE P2121 Far {Cree Surface 0F rel-e:ng “$01!”; ,3: dig-2 + wnsfqn'f (E3 2.32) Let Z-=0 at r=a and dherm‘ure Cons-bani- :0‘, Thus) WI“ . co = ('7 WNW = 0733 figs’ H: "Follows 'Tka“: Ah __ (0.733 rat“ 7'(3 Lb) = o. 0751 3+ 2 (32,7. 232:) r; ...
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This note was uploaded on 04/07/2010 for the course MAE MAE 103 taught by Professor Mae103 during the Fall '09 term at UCLA.

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HW3_SOL - 2.84 The 18~fHong gate at Fig. 132.9443 :4...

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