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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 (allgm M‘ds—dtﬁiﬂm l
a; 457m}! Iranan)J ‘L G Hf [1%
2—1: :0 31 5,
5'0 int5 ‘lf o H“
v
Frir KM:
= £2,4ﬁ3y§§)/éféxﬂft) zﬂqzoaﬂa
Similar13) :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 {Jugﬂirth {ﬁm ﬁre—éoogdzéynm 5f junk)
Z M, =0
.30 P(Mt) = E, (g,)+ F} (a)
9 9
r {201290l5)(2F£)+(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) [ZWJrﬁMEffa ~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 ' ' ”’ 3,5mﬂmXM) ' ’”
(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 0.51??— (8') 50 #er wffb ﬁg #3702 = a? mid ﬂ; =/%7 79% 9/1/63: (0.5 3%); =(M £)(!—23‘)
or
I, 5 0223!» 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; dez‘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?LI5)x.cfx :ﬂ7¥XD,2x )4; = szo—ﬂ.:,g
d 0
0 ~— 5* m firz, 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) MAMe 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 .. 2.95L 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 aF The Cylinder.
#354.” 2. H1 A Sgal. 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) ThusJP
db 3 —F( 1‘4) di '9. (120 llH.L):(5ﬁﬂl)(
l 1° 3 * it 0 Mr]
E : (0 (?+q§)‘ﬂ [is] From ere—beclgdtrtgmm of conEmeerJ
e— t, A = (4M ﬁ,)(xzatn?) L151" ) b lwltt'n.‘
= 57.9 lb
Thus} ﬁerce 61C CamLemar an Fla“. 1'; 57.1% lb downward1. 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: jJdueedJéu
1.5 The CoMfonpaé 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 Utube 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 _ gro
0]"
,Q
at = L
5 2.120 A closed, 0.4mdiameter 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 6ftdiarneter tank of mercury rotating at 7 rpm to pro
duce its parabolicshaped 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 rele:ng “$01!”; ,3: dig2 + wnsfqn'f (E3 2.32) Let Z=0 at r=a and dherm‘ure
Consbani :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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 Fall '09
 MAE103
 Force, open cylindrical tank, place. Neglect friclion, H76 free body, m. lhe gale, er wffb ﬁg

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