Set07S - Page of w ENGINEER 2P04 ENGINEERING MECHANICS “A” PRA CTICE PROBLEM SET 7 SOL U T I ONS Faculty of Engineering McMaster University

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Unformatted text preview: Page: / of w ENGINEER 2P04 ENGINEERING MECHANICS “A” PRA CTICE PROBLEM SET: 7 SOL U T I ONS Faculty of Engineering, McMaster University, Hamilton, Ontario, CANADA ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: 2 'of2é Practice Problem Set Solutions [ Problem Set: 7] ‘ ' P /r W Tihe reganiular cross-sectional links BC and» DE are both made b’fj’siéé'l’ E (E=200GPa , and are 12mm wlde and 6mm thick_. If a lOkN force P 1_s ap 11ed to t e rigid member AF shown 1n F1gure:/ , determlne (a) the force in link C, and the force 1n link_DE (b) Determine the reactions at support F. (c) Determine the deflection of Pomt A. Figure:/ Mmex— AF: V; a m‘gcd «AW .2 Ft 4.3“ NOT ch—W. 12% cm €\‘z§|~©\ Z (i)r‘ot<:\2%, (ii) fif‘amfikoxte Chi) F‘c‘CQéQ & ém$\abe, __ v) & member DI: com WWW AD gawk AP 1 _ Free Bcdy DngPW 0+ Riva em ‘_ .. _,...____ A >\G\<I\l loom“ :60 _ _ -C ‘4.“ C W 7k— - p D 9:05 W 4/- _ - F —‘>pr PF A6 CICr\ toe/96m W the, F.%-D 7 Pk“ almikm isr‘l‘makag : AD : Ac : AA 6‘ > AD Ac: ° . '56 {CO 200 .> z :5 o . AC: 2A0 = F '* . Ac : __L‘3¢_C°_-‘fi_____~__ t Céack4x¢wmova DP béc Age C CO at“? k g. 0, 7<\ 6‘) Ccoc\l~\) Cc-ccém\) E be; :05 WE ; _ FOE (0415“) = C8~é>8(7‘(<;fiDFOE ‘ a ........ . .EQEAQE. ...... . .(2???>[email protected]~>§ ....................... .. 5 ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: ‘of20 Practice Problem Set Solutions [Problem Set: 17’] ' E a :icm eta-O) a : be: 2&9 (é’qzflz’l M04) Fee ; 1 C 8.68! x 6;“) Foe Fm : 2,55 F05 W (L) eke «emcee meflw (M m Pea) at)th 915-? C°~°5->Coe Jr (04%)ch 3 (\CKN>Q°-l-) : Foe“ Jc 292,0: AGKN wcfi) algebhévtywxfl 6%. (2,) :«b: a? C?>> ' F05 «L (1.?) Doe) : («OK/\J 0° Fog: éae‘FW“ Fee: \é»6‘+V\’\‘ éca)’fi«€ W lm hr\\/\ E>c, {s \e,é;=,L KN <§QN6,‘Q\> ' W i“ KKK be {8 60(31l/V\l CCWP/esfip‘cm3 €03) fig Teocfim a6 p25. i: ‘_ 5:39:43 '0 ;y-TaL-fi— \OKIN! = [636174451 4‘“ é°6_+l/“\l prv (53% KM _ ZR =01 l:':"( :6 Eons-the magma elf/~06; [1 we 1 ' FL : \03.’%>[’U/\’\l R ; omvl Cc) Defleotycm 5‘: Pé’A' I PM 6%., C\> AA:— 2&0 = 2,(é,6k4\41 74 6“ page 3 2 (équ‘z’k 79(7‘0‘ > 06.51% (0,51%) s 00 ENGINEER. 2PO4 - Engineering Mechanics ‘A’ Page: 4 o ‘ Practice Problem Set Solutions [Problem Set:?—] 3 or”, , , , _, , _ ,, o -, ,,, 7, w, ., r _\ A _. The steel rods BE and CD each have a 16-mm diameter T k E =200 . . " : Z ends of the rod are single threaded with a pitch of 2.5(rrlz1.eK1-SfowinGt1h:t;) a3; 1 hem snugl fitted, the not at C IS tlghtened one full turn, determine (a the stress 1n rob €11,618 the stress in rod BE, and (c) the deflection of point C of the rigid mem er . 0 Figure: Mewxlov AEZ>C i9 Nanci .n 76 will J\lc‘/T <fi©~PC~r~M , Cm el'zslzxe/r‘ ‘ (if) WSlaée/ 6m) r‘c‘éca‘ée <6 Misuse, the Ave is awed we ~RH W § axe, Rafi Q(\\ m 1.5%“ Ar‘eoesc5'r‘ccj‘b €£§CC>1 T‘:8.mm§ _ c , L z : AEE“ Aco- (\CW) : 20La47-m : 5 2.0 ice 7‘ (0‘1"ng the, m:& cm} C; :5 tighbe-xeci CK‘Q‘TVH , T; It ovqrféE 6° W 2,6M m 'Q‘Fé). w ('06 $255 M5 “€21; phebem‘é W bd‘ ' ,2 rod $5 is dere‘mero‘ 197 Ag, - Pct) DC, is Wed 57, AC. :3 _...——-’ 250 i550» greed g5 : A93 —:; ng (bag 5 C1fl$m58§fie E (2507‘ lch’AD Q2 3 ClOéfi to““ n3) 2 39660“ Ac : (2comco‘l:>é::?oev<\c'fio~"> : (ARCH— filo‘89Tco $524: ) “662/al olefieczsrox a? (>25. c; :3 CC = weer-3% - (euca- 7<\C>~8 >19 ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: 5 0120 Practice Problem Set Solutions [Problem Set: ] 4 : : Gho‘é» \o‘é (c-26M> (C- \Sm) 0‘60 25%— Cpl>q¥7<\o‘$>Tc.-> 7'<LZL\Z'>><KC‘1>TO>E V ~1 ——-— ‘8 ("zhbxto > \e’5 f CLI°Q1xiO >TCD : C.QCL§~\ W (L) latmm ,— Ker) 2 B $3 (‘97 base C2,) : on \65: (6.0.le . o —— b C “TV” .0 \CO ; Ci,qr$\/\I\4 ,/(- (—3 TCO : Cabefiw’éia Rec} CD 1 a) ' a : TC‘Q/ : Cq'?bf\0'h\> : D A A9 (Lowe filo‘“ n3- _ P - ( ’ s > 5 a 6% - {A - ec/A ., (Léfllnclc N) - 80“??me (lactoéflc“\m7’ Cc) DQRCC/é‘cx 5‘— P6. C: T o.cc,0f5 m —— C‘flach-7‘l :78) (<2 ‘?‘$$lO§N> Laé’vww [to tke, \64125] ~ ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: '63f Practice Problem Set Solutions [ Problem Set:7-] ' ' i :73 . A Bin connected structure is loaded and su ported as shown in Fi ure: 3. Member 5 C IS RIGID and )8 horizontal before the aged P = 150kN is app ied. Member A IS an aluminum allo w1th a modulus ofelast1c1 of 75GPa, and a cross-sectional area of 1000mm2. ember B 1.5 a structural stee bar with a modulus of elasticity of 200G_Pa, and a cross-sectional area of 500mm2. Determine, (a) the ax1al stresses in bars A, and B, (b) the deflection of pomt D. Figure: 3 CQ'MCQ: W490 GD ;5 “had he 9.1;“ MOT deficMfl _ Rig; c‘ bmifegs Cm We, srx 7;) “cast 6?) Ra‘boxfiim (ii) ‘ér‘oné\oq§;m (Hi) F’Obcxéim + M\¢\bf0m Fr‘cz/ ‘0ch diagm g— Y'N‘gr‘cl bm 03 2 FE; FA T T i——> ! n I 42-h§m+ jag/w _—;‘4|M‘7£ c7 \anKN Ccneiciehlmfl mlj/ axe nWmfi cg),;\ibr\;m M ()6. (\.c5m)Fv>+ (l.§~r\+ 2,6“) ;A _ C, E Cu‘fwfifig + cams) 9A C: 96%;? 155% +\)W§ :1 O : C55M>(\5<;N\I) Fr‘crw 5'.rv~i\cr\ firfic -‘ M66 ) é2~ AA—AD v: 2F “ ‘5— Wm 3 “We, defi'wfiu «5 ; 6km or? K dembicmfi 0V”! (QC so‘g$z\:ijc\flfiw7§ed. % : an” 96m 193 \H‘AWQH7 h \2 66". ‘bkoaS' (\iaid 1:» CD ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: ‘EZon; Practice Problem Set Solutions [Problem Set: 9—] E ' Fhm ‘66, thf5\ A: PLA : bA A99: FE) L$ 2-, F72; (hsfirx 3 C‘.5><\O‘ 8) F 55 AG; C lco 7c \GQPQDCBX (c‘é‘ ML) E> A : PA LA FA (QaCM> PK : — -————-—————————— :, 2, we ~ cg AA (%$\o‘\ QA>CO‘OCIM1> C xko > PA _ , «8 [36/ : AA/ —9 (“57“6 >F€> : (l,éé‘:lb fiko~8)FI ‘36 4‘ {05 _—————_ A Ll Pg; ; Oaéééj‘SP/A‘ : Or") 7" ‘05 28b6‘éi‘éjv‘birfl K160 ea}. 0); ' Lgpg -\- ZACA : ¥5c:\/u\l \.5Fe;+ 40.5»:ng = 3%ch 009 z Fm: (éoKm gsflmsce 1m bof‘fi A4 6°. W616”: F/A c Woofi mam) ‘ 0a., in E>c(\ A 2? :— Ffi/ 5 ~ \BCMPCA AA (o‘er/Marx 519%? in 13;» e) 1 _ ._, Te) _' Pg/ -_—, ((66% {oaN Z ‘ A49 —’ 2C0 0\ ‘96 S [30: 5/3?) = 5C|.S7\\C:5>C\ccfi\c’b/V> la'b LS ; 5W a 5‘ Pt: D 5W demo I ENGINEER. 2P04 — Engineering Mechanics ‘A’ Page: 5 Practice Problem Set Solutions [Problem Set:¥] “ ‘ 43? A brass tube 300 mm long, outside diameter 32 rnm and wall thickness 3 m, ‘V; is placed between thejaws of a Vice, Wthh 15 ad usted until the 'aws ust touch : each end of the tube. Two forces, P = 190 . and Q = 166 , are then apghed to the tube as shown 1n the figure. Knowmg that Young's modulus (E) is 03 GPa determine, _ a The forces exerted by the Vice at the ends A and D of the tube. b The change 1n length of the portion BC of the tube. F;flufie: 4.. Crbe§~€>ecfiiom <53: bar-2:95 tvbe; ° 3mm Area 6‘: tube) Abbe ° At:ch = «Cléwbz - fi (‘23 M)?— ' " 2%.?)2.m~ml : cmooozqu‘bl ML . A : D wmtokdpxfigdve the, mtibilltj Wibio’é’: 57 9%??th Wifiimf‘) 456/ lilo/“M 6‘) 4Q 5402, bebfibfi lmQfi/gih‘brxim: EQUILIBRIUM CUN‘D’TM,“ ' FA + \é£\/u\\ 3 7:04— \CLO.K,\. ' 0 00 PA“ FOZEOVV-xl wok) o CQ\‘<SC>\ i ‘évbe. 0 9 792 de‘Plecjsichg @ pt. A & 6;: o ; 5-D : O Eflt/Mbfifl%_ CoNCDITIa/x} ; , ~ . . . . . . . . ‘ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ENGINEER. 2P04 — Engineering Mechanics ‘A’ Pagezgjof Practice Problem Set Solutions [Problem Set: 9} ' .Flf‘S‘b) WC). W6 WWW mwemts 0rd ~F—u‘rfl the, SQ: lac; KN. [email protected]%€<5F& mm) éafléy ¥O\ the W) mm>A T5 -‘Ff:s.ecl <§ (375—0 T9 m ‘QGKN léfil’kl‘c A‘ 6—] ‘——> i ’WNE is 75:; 4?ch ‘éhe, imtomaq 100% E lccw C a % lrfi W SCOIgI‘C/fi cavscci ' be PtlochM ) & Qv\éc\/v~l. (0 Wham A8) 1 l l I .. I ~e_u____. 'l’Ab~\%K/~l~\' \éCKl'xl :0 I l ( *co D film) = ’50qu i FAfi : *3cl/V4 O ' l «7an or; bow/4 0'“: Mrtsgrm w <— '————«»———«»"a PEG C D (9,) Seoéicx 8-20 1, l | ":50 + (QCVN :0 ‘ MOW“ lécKN F66: \éCKINJ . (—— ~—> a %' ——O———-———o————-—-c FA& 9) C D “0 Léc KN *5: Wim (5) SCCQM GD 1 : C0:). = C IAN E Find the, de¢\eo$icr\ at Pb, duet exilemafi £57:in E z p), ' a NOTE; C‘O-W‘O\ Me, {9 “egcéixae becéwse, {~5— ; (5/9 /I::/>\ ) Wficx‘3 a W‘ M— + Uéomg’NXCW) C we; x koqpcx>C27+5527< (5“ «3 C \cfi 7‘qu 960(24632 »< to‘“ A3 fiecvsr‘m A9: Scosim 93c; : 'Go 10% + c.568mm : Oeaélb MM 00’: (36°13 Cal“ o-QélEDMW 60 6‘6 Plgfl'\to M W % cl ' . 2n ‘ Cofidl‘eim we KF‘OW {she} dc—lHQC/fii‘m (7‘75 D “AL/9U ‘06 m_ WW) cqe, will 0x9pr the wp%§"om ~¥o"cg ‘10 91> pfiD 75c ml flee; 69%merm (5313(ngin A | ’4 $0 .hbfiywm D ENGINEER. 2P04 — Engineering Mechanics ‘A’ Page: [00f Practice Problem Set Solutions [ Problem Set::]r] : 5: PLEA *> Pow = my ‘> Men—:1 Fe NM «a «a w 1 L_ Cm‘éif‘e, \Mfik‘. PD 1 C %~bxno’“ ~>ac3>7< \o“ Paczsififizx (om/J) (0.73m) defie Fm; “Cd ~ team-4 f— -—'-——oj <—————~4\°.> 25m F’bo C D ‘Ffio-Hec v aafiy/w :0 F53; we»? KM 00° a W § \\é:q‘Kl—l $50 : 2: Clké\q‘X\c$ N\>Co. ‘ M) ( (C?) 7< \oq Pa) (2.130327g(o"" ML) My; Ayah/Mam” 5mm” W g¢Mz/&%_ C9 __‘.—- [fig/A /g4€/6/M C’flflfi/Wam/ [,7 -—- /7/ .y— /%9 3 0 "7 [See P52 myéa/c Jfl i ' 5, 5 —-0 3 : Qmflflf/M/f/ dim/WWW“ fl " . : : m g/a/o’ Mvmtzanr 5I-[Xs~5¢)+/§~¢Q+Ké—&J:OM52””¢k ’6 5 E ’ ' ’aM/flV/é/A/V 3 A48. [is + Asa .f A45 4"” z 5,. Cam/Z/r/an/ - ,, flm 22 it» ' 46' fig fie mam; M %r. m #5 2”“fl: M7 Q. m8 2 .. 52 (am Wm» My, gar/90w- = ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: / offlb Practice Problem Set Solutions [Problem Set: 7 ] g 3 5 A 250mm of 15 X SO-tnm rectangular cross. sectioh eonsists of two . I - aluminum layers, S—mm thick, brazed to a center brass layer of the same thick- . P w .ness. ll" it is subjected to centric forces -of magnittlde P- 30 RN, and knowing . that E. = 70 GP: and E], = 105 CPI, determine the normal stress (a) in the alu- _ 5 mm minum layers, (12) in the brass layer. V r . . 1 1 5 mm ‘ hf: Cam 1:2 aggm tkca 5C le—l I Aluminum fg‘RCe a“ W “17% the, boxdecl ~37 Wimvtw> W Cbnecécmg go? W MBA/0 Wm be 13.6 some” £0 5 ‘FM tke/ Cox~yc>oxz§v‘bl'lr‘£~—7 Caxch‘éu‘cx‘ 2 AN :- A65- NC?) gmmce 5:19p EA 000 A o 3 m pm LAX & Ag» Pep. Lw/ 594° AAA .. 9 M, C x = r55- C c.1S A,_> ‘90 7‘ \C“ Ptfl> (ccc‘sfi‘DLC.C3M> : ‘ P734, 7: C7- ngx / ' 3J0 fkié; i~tc C‘> 2(0-6;P&,> -\— P80. :gnscKlq o". \1.82K/\; :; p a BA, (CWI‘Q\) no: YA‘_: \-.:’L~\8YI\ Cm ' i W .> 829‘ KM 90“ N. S‘é'fi'o '2 : A ' Nwtrxw (bath \c‘flzfib : C"j’\8NCEN>/Cl>¢§Wch/M~> : "‘ 5:. a 5 Brae”; ’ 6 9 ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: Practice Problem Set Solutions [ Problem Set: 7] 2 ~ A polystyrene rod consisting of two cylindrical portions AB and BC is restrained at both ends and supports tw0 26-kN loads as shown. Knowing that E = 3.1 GFa, determine (a) the reactions at A and C, (b) the normal stress in each /, portion of rod. / / "/ to k :0 \ \\ \\ Practice Problem Set Solutions [Problem Set: 7] Fromflfl“ - '* QMb' L’Mb *?%C'L'%Q : O (fi/kél‘kqa)‘; (fi/A’A‘L‘x‘zc) E RAILMD Ar 1“_"\?\C_,\_O=C F O Mai-A C759: &LKV§°)Z “1/4,; avg-(3y? ' AM Skfixoiv ’QC’XQA as Q-o’sl)L+ * QC$1.gng~——_————-(>b) “Put :5: ‘m UA we. 06cc QA*”>'E€>QA~—>§)_ ' Qp‘g: \\,\§O\KN [\‘k RM RC: 40.?)4 \m '\+ fig? as; \gkxvm “M MA MB. \Xmsm age-,X 6’ kg —-, @5423 z’X-RA ~ \V-Ngqxxfi3 ._ (3 SCEC —, ©ebg;r<lc_;fil«o$eqx\€> A‘s; R m 'R/Axxmssx‘fg GJRQ : =—-—10;%M(\30\j 1 VS (.4747? If??? ........ ENGINEER. 2P04 — Engineering Mechanics ‘A’ Page; #0pr57 Practice Problem Set Solutions [Problem Set: 7 ] E r , ; 7c The assembly shown consists of an aluminum shell (El,I = 70 GPa. a. = 2 ’ 23.6 x lO“/°C) fully bOnded to a steel core (E, = 200 Ch, a. $11.7 x 10"/-'C) ' and is unstressed at a temperature of 20°C. Considering only axial deformations, determine the stress in the aluminum shell when the temperature reaches 180°C. ,\‘va’v__fi,_.___’ r ' gtv‘C'Q— wwM 'V‘ \°°""3‘Q—A V {QRC’ &°w~fi§\ aowe \ Aumvwx \A‘VKN , 1&1)? 1§OW¥QW~ Leg I ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page/50mg? Practice Problem Set Solutions [Problem Set: 7] w/wigi~irwam«a~~eé k . I \ [é gm gym E “\wac \XL‘xg‘J‘Q" L“ /%Ve 9'13 (oxMaL’ {\o-l'fl \Kao '?*b "l— ___. ;§\o 'oglr— O Q’Lz\‘/‘-)V°*\b ~Ac> A l a} R/A akfis'o‘) f)»th ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: /60% Practice Problem Set Solutions [Problem Set: 7 ] 1 Cdcv‘d fie Pods ton/cl“ each , g Rod AB is made ofbrwas (15,, = 105 CPn, ab = 20.9 x 10"“PC) and rod CD C t 3-} WKQ/\ m fieewsw I of aluminum (E, = 70 GPa, a“ = 23.6 X 10‘°/°C). Knowing that at 15°C 30.5—mm 34:; (ficfige : L 65 8., _ KP . ' gap exists between the ends of the two rtxls. detennine (a) the normal stress in each 8 C’ ‘ rod after the temperature has been raised to 85°C. (1)) the defimnation of rod AB at QT : 8%° A \‘g" —_— qo°c that time. 0.5 mm In 91%00“) W.f\m\¢\l efiDONéiQr“ 32:: Z Emi = 04» ATo L rthe 1 Z 50-min diameter 75-mm diameter A fax C 6 E E > t MM ETVYFMGA ©<pcr\$i¢:\ ir\ W Z WAC/3*") : £141 \j’qMMZJ E 5%“ - ‘ ; mi - CZC-q7‘\GG/°C/>C‘:}c°c)(?>c~cw> : Q4458qu Qrpmsim if\ HWimK/M ’1 Z ‘6 o mad (ZOD’Q rkc / : G.Q\5W 2%3RC6 % ‘éc‘écx‘ ewe/elm +O.A\fbw\m\ : C)8<:_>\o\ > Egrtocém then m 0.5% : 3“? five "‘CC\$ ‘\\ ' e ‘65» . cnce ‘ka was ficvck eack 6% 6‘ cmd 3Aducc the reacyérm ‘FCF‘CQ/ 73437 WC“ CW -é:, Q79 (A ' / ix“ 1*\ dw to AT. Wnl‘ \l(\ 3 dV‘L ‘60 Rafi - ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page: / Practice Problem Set Solutions [Problem Set: 7 ] : ; FM 6% diéP‘Qmea—Né 5‘03”:va <2 gains Cmdé' Ct?“ be egta’o‘igked. Mg 7 I Icvx E<€><pomsfo~ in AL due 15:: AT * Swmtzmerxfl in A! due 7%; R) + E 900 (0' __ J 5 ‘ T K M f <5T‘Jfi>A—lvv¢\w 2 C'BMM KC.) T=29~ex~-m 3 R> Peao‘éicn D Ebechéu’cexg (ax/Sch b7 43% Fec‘Acéim %\ce ) b be» Ash 0 CE) GP«>(\G<>?:- S MAE) = (kl/15.35 xko‘o‘ > P : Q : (ansQ >< \o “6) R cm AM CTR: CP¢k>C4141l1‘c\ MAJ) : oxg 64%:de C \kacoéG F S 1 of, E E «Aves; \vxto (er/l. cl ) ‘ / a a o {W > G” 5 \4‘ ~ 9‘ ‘ E 7‘\C./ M> {C3 + (Qflgr‘o Li R) .CSZ/VX cfimég (fiz‘zéaé‘ K‘C‘a)gl (69:35»? xto\6”\\> 3 Q': ‘BBan—Vw ‘ on “A ‘ ' 6 30 (253—7\ NWAk/W ~— g : (\SSsQ—V‘WLC Ki) : $5.10! [\qu m in W85 ~ R K : (\S$>é‘?—7<lcsi\l> :2 tpvune ixqu 6(- T‘ C Z€M3L (C<~\;=>fes§fc~) EDe¢c-<uémx a: fled A6 [ion—:25. (tel) : ’ 3099 «w ‘2 v a (c N --r 00 erbm D ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page:// Practice Problem Set Solutions [Problem Set: 7] ' The material properties associated with the block shown in the. figure are E=SOGPa, and Poisson’s . Ratio v=0.33. Knowing that a‘= -180MPa, determine (a) the magnitude of a, for which the change in the heightof the block will be zero, (b) the VcorreSponding change in the area of face ABCD, (c) the corresponding. change in the volume of the block. - 2 : - t 6296‘ Ed. E ~ 0311 “ 0T2) V‘C‘) 1“ 9 E ‘ E-x ems 986% i \(— E C‘QUDL+ 0:71 ~QTZ> "“13 (T3,: —\8CMF>0‘ 83 =JéC—vcr1 ~ 002, 1‘ 0‘5) «(3) (Int? 9‘; COMP“ __ l O __ (5%an < lam Pa + a» + on can W7 : -gqaé‘ floépffi a b MPQ .. with (5‘ : - => ~ I a \( gqéz Ivn at the ckmfle m FQI'gkjf (57) Wm be 2mg 3 (b) Ckaae ir\ arm Agco 1 AA: Lafit+ei>L8<l+ezE> «Lazy/9 Eek: J’CQ‘J'L50T ‘3? v I - E 7 2) m E (~\%cmf%> ‘5CI -élMgflbl (Wax) 2: c.0:;[ 58 L v offifitecmqu) ~ofi>$ (—56.9. ENGINEER. 2P04 - Engineering Mechanics ‘A’ Page/got??? 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This note was uploaded on 02/19/2010 for the course ENG 2P04 taught by Professor Sivakumaran during the Spring '10 term at McMaster University.

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Set07S - Page of w ENGINEER 2P04 ENGINEERING MECHANICS “A” PRA CTICE PROBLEM SET 7 SOL U T I ONS Faculty of Engineering McMaster University

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