mae_101a_hw2s

mae_101a_hw2s - MAE101A HW2 SOLUTION Problem 1 Problem...

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Unformatted text preview: MAE101A - HW2 SOLUTION Problem 1 Problem 3.23 [2] Problem 2 Problem 3.27 [2] Problem 3 Problem 3.33 [3] s = L/∆he = L/(SG h) = 5/SG Problem 4 Problem 3.51 !"# $ Given: Find: ,-./-)01$.($23)4.05-$46$(.0$-789:9;098/ h H$%$&'$()$ A R$%$*+$()$ y B FA y z x Solution: <3=95$-783)9.> ! " 4? $ " @ A6 # A@ $ ρ% 2 AB 4? $ @5% 6 ΣCD $ + GHH 1F $ 15 & 6 % 15 IB-0-$1$I.8:A$;-$/-3=80-A (0./$)B-$(0--$=80(35- .[email protected])9>2$-783)9.>= 6==8/@)9.>=J$=)3)95$(:89AK$!$%$5.>=)3>)[email protected])/$.>$.)B-0$=9A-K$A..0$9=$9>$-789:9;098/ G>=)-3A$.($8=9>2$-9)B-0$.($)B-=-$3@@0.35B-=E$I-$>.)-$)B-$(.::.I9>2E$8=9>2$1$3=$9>$)B-$=L-)5B ΣCD $ + 46 $ *! " % 1% ρ% 2% B A6 ?" # ! " 46% ? $ " 1% @ A6 # I9)B I9)B @ $ ρ% 2% B 1 $ 0% =9> M θN M,32-$@0-==80-E$=9>5-$@$% @3)/$.>$.)B-0$=9A-N B $ O'1 A6 $ 0% A0% Aθ 3>A π O->5- ! π? " P *! ! ρ% 2 " ( O % ? ? &+ 46 $ % " " ρ% 2% 0% =9> M θN % M O ' 0% =9> M θN N % 0 A0 Aθ $ % =9> M θN ' % =9> M θN , Aθ %" ) ? #+ #+ ? " P #+ * 4? $ " P ( &% O% ?& π% ?" + ρ% 2 ( &% O% ? π% ? + , $ ρ% 2% ) , %) ' ' ?* " QQ*" R=9>2$29S->$A3)3 4? $ *TUP% =:82 () " . "&T&% & &π "2 :;( % = . / . &'% () . M *+% ()N ' . M *+% ()N 3 . 0 & 1" Q 4 =:82% () = () & 4? $ VTUW . *+ % :;( P Problem 5 Problem 3.53 !"# Given: Find: )*+&*,-.$+/$0123*$42,* 5636&7&$8*649,$,+$:**0$6,$;1+<*= $ L$%$"$& h y dF W w$%$($& L'( Solution: >2<6;$*?72,6+3 ! " @A $ " 0 =B # =0 $ ρ% 4 =9 @A $ 0; % B Σ5C $ D GHH .F $ .; & B % .; +-E$7<*$;+&07,634$*?72,6+3< B<<7&0,6+3<I$<,2,6;$/176=J$!$%$;+3<,23,J$02,&$+3$+,9*-$<6=*J$=++-$6<$63$*?7616K-67& G3<,*2=$+/$7<634$*6,9*-$+/$,9*<*$200-+2;9*<E$8*$3+,*$,9*$/+11+8634E$7<634$.$2<$63$,9*$<:*,;9 Σ5C $ D L*$21<+$92P* =@ $ 0% =B 86,9 ! M " L% % ;+< N θO $ " . =@ ( # 0 $ ρ% 4% 9 $ ρ% 4% .% <63N θO N)24*$0-*<<7-*E$<63;*$0$%$02,&$+3$+,9*-$<6=*O Q*3;* L$ ! ! ( ( " " % " .% 0 =B $ % " .% ρ% 4% .% <63N θO % 8 =. M% ;+< N θO # M% ;+< N θO # M ! ( (% ρ% 4% 8% ,23N θO ! ( ( ( " " . =. $ % ρ% 4% 8% M % ,23N θO L$ % " .% 0 =B $ % #D M% ;+< N θO # M " R<634$46P*3$=2,2 L$ ( :4 & W %< ( % SDDD% ' TUVS% ' (% & ' N "% & O ' ,23N "D% =*4O ' " ( " :4% & & < ( L $ XV% :W Problem 6 Problem 3.76 !"# Given: 89,($*(/2(,0: Find: ;/01($/5$-,/3$< $ Solution: <9-)1$(=49,)/5>3 # ρ" * >+ Σ?@ # A x D FV A R FB WGate y’ R6& W% FH y " R67! W& x '()*+,-$./0$1/234,)5* FV F% @--423,)/5-B$-,9,)1$.C4)>D$!$E$1/5-,95,D$39,2$/5$/,+(0$-)>( ;/0$)51/230(--)FC($.C4)> 3 # ρ" *" + G+(0($3$)-$*9*($30(--40($95>$+$)-$2(9-40(>$>/G5G90>- '($5((>$,/$1/234,($./01($H)51C4>)5*$C/19,)/5I$>4($,/$G9,(0$/5$140J(>$-40.91($95>$45>(05(9,+K$$;/0$140J(>$-40.91($G($1/4C>$)5,(*09,( 30(--40(L$F4,$+(0($G($4-($,+($1/51(3,-$,+9,$;M$H-(($-N(,1+I$)-$(=4)J9C(5,$,/$,+($G()*+,$/.$.C4)>$9F/J(L$95>$;O$)-$(=4)J9C(5,$,/$,+($./01($/5 9$J(0,)19C$.C9,$3C9,(K$$P/,($,+9,$,+($-N(,1+$/5C:$-+/G-$./01(-$,+9,$G)CC$F($4-(>$,/$1/234,($,+($2/2(5,$9,$@ ;M # '% $ '& N* 2 P" '% # ρ" *" G" Q" R # %AAA" % SKT%" % 7" 2 % "KU" 2 % 7" 2 % 7 & N*" 2 2 '& # ρ" *" G" π" R " & & ;/0$;M G),+ '% # 7SV" NP # %AAA" N* 2 7 % SKT%" 2 & % 7" 2 % π " % H 7" 2I % & P" - & N*" 2 '& # &AT" NP ;M # '% $ '& R "" R ;M" W # '%" $ '&" & 7" π W# ;M # %TS" NP '% R '& "" R "$ " ;J & ;J 7" π G),+$W$*)J(5$F: /0 W# 7SV 7" 2 &AT " % $ % % 7" 2 %TS & %TS 7" π ;O # 31" @ W # %KVU 2 ZWW :Y # :1 & @ " :1 ;/0$;O X/234,)5*$(=49,)/5- !"#$" *' $ %! " &$! ' " ρ! (! % ) # ( ! ,! * +) & %! " -...! /( 0 1 * 234-! 0 5 + * % 637! 0 # $ & 1! 0 ' ( * 1! 0 * 1! 0 * +) /(! 0 8! 5 + %! " +97! /8 :;"<=>[email protected]>#<>B<@;A5<B>C$"<A5 1 + FGG * ' ,! * * * DE " D$ + " $) # ( + * " )# + % ' ! D$ & +) -+ *' + *' , ! *! $ ) # ( -+! $ ) # ( % % +) +) & & DE " 637! 0 # 1! 0 + + H 1! 0I + 1! 0 ' -+ * $ 637! 0 # % ( +) & DE " 13+7 0 :;"<B>C$"<%-<>#<@;"<J>@@>0<>B<@;"<(?@"<A5 %- " &! ' " ρ! (! )! ,! * %- " -...! /( 0 1 * 234-! 0 5 + * 637! 0 * 1! 0 * 1! 0 * 8! 5 /(! 0 + %- " 12K! /8 %>C<@;"<$>#$C"@"<(?@"<HLM<N<+36<BC>0<:?J="<'3+I π! * /( 0 π + 8! 5 OM?@" " LM! ρ! (! ,! " +36! -...! * 234-! * 1! 0 * * H 1! 0I * 1 + 6 6 /(! 0 0 5 * 6! * # OM?@"! # %S! G # %!! T DE # H ) # *I U " . + 1! π + + OM?@" " 622! /8 !"#$"P<@?/A#(<0>0"[email protected]<?J>Q@<' %R! * + %-! %R " 6 G T DE # H ) # *I U !O + !% + ! %! # ! %1! π M?@" * S * + 6 -3K7 T 13+7 # H 637 # 1I U * 622! /8 + * -42! /8 + * +97! /8 # * 12K! /8 1! π 1 1 + %R " %R " +K4! /8 Problem 7 Problem 3.71 [3] Part 1/2 Problem 8 Problem 3.83 [3] FV acts down at (4R/3") # 0.849 m to the left of C. Sum moments clockwise about point C: ' MC # 0 # (2 m)P $ (58740 N)(0.667 m) – (92270 N)(0.849 m) # 2P $ 117480 Solve for P # 58,700 N # 58( 7)kN Ans. Problem 9 2.87 The bottle of champagne (SG # 0.96) in Fig. P2.87 is under pressure as shown by the mercury manometer reading. Compute the net force on the 2-in-radius hemispherical end cap at the bottom of the bottle. Solution: First, from the manometer, compute the gage pressure at section AA in the champagne 6 inches above the bottom: 154 Chapter 2 ! Pressure Distribution in a Fluid 123 Fig. P2.87 Solutions Manual ! Fluid Mechanics, Fifth Edition "2 # "4 # Solution: $ (0.96 % 62.4) ( is ft ) & (13.56 % 62.4) ( 1 fluid ' patmosphere in3, (gage),12 fl. oz. " p AA First, how high the container? Well, ft oz. " 1.805 ' 0 hence * fat, ) + 12 + 21.66 in3 " #(1.5 in)2h,* or h $ 3.06 in—It is a 12 nearly square little glass. Second, determine the acceleration or: PAA the center of (gage) toward ' 272 lbf/ft 2 the merry-go-round, noting that the angular velocity is % " (12 rev/min)(1 min/60 s)(2# rad/rev) " 1.26 rad/s. Then, for r " 4 ft, Then the force on the bottom end cap is vertical2only (due to symmetry) and equals the 2 a " ght " the champagne below AA: 2 force at section AA plus the xwei% r of(1.26 rad/s) (4 ft) " 6.32 ft/s Then, for steady rotation, the W F ' F ' p (Area) $ water surfaceW the glass will slope at the angle & in V AA AA 6-in cylinder 2-in hemisphere a 6.32 , " " 0.196, or: ' (6/12) & (0.96 % 62.4)(2, /3)(2/12)3 tan & " x (4/12)2 $ (0.96 % 62.4), (2/12)2 h left to center " (0.196)(1.5 in) " 0.294 in ' (272) g ( a z 32.2 ( 0 4 ' 23.74 $ 2.61 & 0.58 - 25.8 lbf Ans. Thus the glass should be filled to no more than 3.06 ) 0.294 $ 2.77 inches This amount of liquid is * " # (1.5 in)2(2.77 in) " 19.6 in3 $ 10.8 fluid oz. Ans. 2 ircular-arc P.88 Cabout point roblem 10 pivots Tainter gate ABC O. For the position 2.139 determine (a)liquidhydrostatic force The tank of the in the figure P2.139 shown, onaccelerates (per the right with the fluid in the gate to meter of width into the 2 rigid-body motion. (a) Compute ax in m/s paper); and (b) its line of action. Does the . (b) pass through point O? force Why doesn’t the solution to part (a) depend upon fluid density? (c) Compute gage pressure horizontal the fluid Solution: The at point A if hydrostatic is glycerin at on vertical projection: force is based20+C. Fig. P2.139 Fig. Solution: (a) The slope of the liquid gives us the acceleration: P2.88 FH ' . h CG A vert ' (9790)(3)(6 % cm' 176220 N at 4 m below C a 28 ) 15 1) tan& " x " " 0.13, or: & " 7.4+ 100 cm The vertical force is upward and g equal to the weight of the missing water xin the segment thus a " 0.13g " 0.13(9.81) " 1.28 m/s 2 Ans. (a) ABC shown shaded below. Reference to a (b) handbook solution y (a) is geometric good Clearly, thewill give toou thepurely geometric and does not involve fluid density. Ans. (b) 3 (c) From Table A-3 for glycerin, , " 1260 properties of a circular segment, and you kg/m . There are many ways to compute pA. For example, that the straight area is may compute we can go segment down on the left side, using only gravity: 3.261 m2 and its centroid is 5.5196 m from 2 p 0.3235 z from vertical 3 )(9.81 m/s point O, or A " , g'm " (1260 kg/m line AC, )(0.28 m) " 3460 Pa (gage) Ans. (c) as shown in the figure. The vertical (upward) Or we can start on the right s thus hydrostatic force on gate ABC iside, go down 15 cm with g and across 100 cm with ax: p width) ' ( ,a x 'x " (1260)(9.81)(0.15) ( (1260)(1.28)(1.00) FV ' . A ABC(unit A " , g'z (9790)(3.2611) " 1854 ( from B ' 31926 N at 0.4804 m1607 " 3460 Pa Ans. (c) ...
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This note was uploaded on 02/03/2010 for the course MAE mae150 taught by Professor Alisonmarsden during the Fall '10 term at UCSD.

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