Suggested_Ch4_sol - i - T 4.9 The 40-fi boom AB weighs 2...

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Unformatted text preview: i - T 4.9 The 40-fi boom AB weighs 2 kips; the distance from the axle A to the center of ' roblem 4.9 I. I I I __ gravity G of the boom is 20 ft. For the position shown, determine the tension T in the cable and the reaction at A. I 1/ .. ' ‘. .1: '_v_ o_ I - RD: gimzoIB . :(uommmg AD: mm ; ___________ Ill-e” 6 5r ' .. .. I I l— —|— I CF — - III- ‘05 — J____L_| .__;_! W _Zk'i?‘ 1 'T(HD)—2(20£0$30') 1 !._!_!___l . . .! ; | _ “ . .. a -5(‘/oc,os-30')=0 J 7 ! m __ _'_| i “9”” . _ 5;; T(6.’IS)“Z/l7.32)-:S‘(3%6Q)=0 _ ..__ . . - _ __ __j_j_I_|_'_I ___ILL 6.9532013.- ' 7': 2mm” 4_ i: i :T— “'Cét’itzt~‘*)w‘20°+%=0 ‘ 32:2&"‘.‘i? _ _ 34% [ _W‘21350:j(“24.4)sinZD’My-arszo flit/125K333" ______ ' : ' ‘- 5:53.0Kl'fidt3I-50 - l : I I I I raised by the cable BC. _5_ Determine the tension T required to raise end B just off the floor (a) in terms of W i and6,(b)ifh=8ft,L=10fl,andW=351b. #1 ! l [1—H] i 'l iLf emiafi . __ It .i .e i | r. l. J -;—|— 4.10 The ladder AB, of length .L and weight W, can I i I . A; -'-——n. L C I i(¢) Trianfie ABC; z I __' ii rec-51:4: its- Th“ . _ 1"“.‘7 = . - L ’0“ E: in _i_ HQ»: 3. m5:- . $ III! - --—i-- +3 th rd: - .. _ ___ I ___I. . _. B . I LI _ _I __ . g._i_ lifts; _.__- T(L€o§%> —IIII\:/(§ £059) :0. __i_:_._;_. . i . ' -.—- _ £0 a a. a ; . . ___i_I_: . i . g ——!—; - T— IECDEE; ‘ - _ _ ' . (b) Sine = if: : 0.5 5* : 53,1" ' i. _ 4.12 The lever‘AB is hinged at C and attached to a control cable atA. If the lever is - _ subjected at B to a 500-N horizontal force, determine (a) the tension in the cable, (b) the reaction at C. Problem 4.12 1 L Thar-n5}: [5 li'EDECff-ES 255mm .5 I I! _ . m‘figcrdi'fl-l-‘BU =35 ‘_ _ E1 200 m g H z {I} = Hgggimu‘j =3e._ C . Thus 1m Farm mgr ar!- 15' flan/HI? hurr'éan . .22” l 3-. i . I _ _l' E ._ L: l ._| L | '_ - I | |_ Wt ESDFW fig: fhh WFwfl-afl‘ Rims {QB-find fermeAr‘Lufnr TC: l: zapsm 30“ = 100m E, 590 N '1') ' "_=(b)g,zF=o': . 1_ . . (400 N)cos 40‘ +62 afoo/s/ :0 III-fin -3: 32ch =0: (7:9 SrnZU‘X250mm) — (500 N)(100mm) =0 f2; : #OUM €=+300M 7t. " —'i- J Problem 4,19 I“ 4.19 A 160-lb overhead garage door consists of a uniform rectangular panel AC, 84 E - _ . . -. in. high, supported by the cable AE attached at the middle of the upper edge of the ! I | | l | . l door and by two sets of fiictionless rollers at A and B. Each set consists of two 1— rollers located on either side of the door. The rollers A are free to move in ' horizontal channels, while the rollers B are guided by vertical channels. If the door is held in the position for which BD = 42 in., determine (a) the tension in cable AE, (b) the reaction at each of the four rollers. . +3 EMA”! (23); — loo/5M :0. _ F (25)(42;n.)—(léo/l,)(3s.e;n.)= o ET-Efi‘lflflb _ T: 128.0/1‘2? <- !_Problem 422 .._. .. 4.22 The rig Shown consists of a 1200-lb horizontal member ABC and a vertical ' i i _ member DBE welded together at B. The rig is being used to raise a 3600-lb crate . - i at a distance x = 12 it from the vertical member DBE. If the tension in the cable is ' —i— "‘ 4 kips, determine the reaction at E, assuming that the cable is (a) anchored at F as shown in the figure, (b) attached to the vertical member at a point located 1 it above E. : ” it“: 131;. imam + moms“): o M5375 7—36001 4200 a) Gil/tn Data: fizz/Z .fii‘ 7’: woo/£5 ; H; r 3.7s(4w¢—3eo‘o(iz)-7eoo :46 000 Wow? :0 ; fz : 0 f; ,.' 5.: : jaw/200+ 1/000 :gaow; ;:5,wkipstyg=3e,0k¢.igl {- ‘ +3 2 ME = o: : Mé+(3‘600)(l2)+ (/100)(é.<3)=o _ ME = .“57 000 /é. HF Elite: 2:30 2‘ :0: %:fsoo + aoosgzgoo ,3 _ g: 4-, go hyst; Hg 51-0“? ‘Q ‘1 _ i I I l | I l I l I I I ' I ’ I V I ' ' 4.25 A trues me}r he supported in eight different ways as shown. All connections - consist of frictionless pins. rollers. and short links. In each case, determine whether {a} the truss is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the truss is maintained in the position shown. Also, wherever possible, compute the reactions, assuming that the magnitude of the force P is 12 kips. |Iil:|.:llil,i.|llll|lll. . 3 hon-mourrenle',non-Eunild reactions. 0». Truss com‘gid'd'f constrained. 1:. Reactions si‘a'h'saii‘tdehrminal‘e. C. Bauiiihn‘um mthtm‘vmai‘. H :8 kites, Avy=)gk,'l,; x . {I fi.=llh'f2kl't?545'éu33§:fi -ei't4«ert-i —' fl WW” l!- EMF-s i'} (2) Comorrcnt’ read-thus (fhrnosh Fl)- a‘. Truss improper! ' consf‘r'al‘mafl. - 5‘ Read-bus staid Indeterminate. Q c. No ectol‘i‘ilorium {ZMAR'EQQ (3} grim-tonch enf, hawtgnrauei medians. . anTrots mmwt i211 emetmtmof. i3. Reeeh'ons thh‘milul daemon-hi. E. EfiUl‘l'IIEI'I‘Um Main haittu't. fl=g36kif3¥ (it) Linonvwvseurrcni} nonvpamiiei tranche“ s . ‘ a.Truss wmplei‘eitl congi‘mi‘nfd; 5. Rent New: shit. ivxde terminate . C.- Egltl'l‘ll‘i‘th‘dkM mainhnineoi. 5:3 I5—‘1E:3_k‘— ’ (PHI 1- Etf'r' 52: in; I}; ‘ (5} if Cantuflifii" rEtJ-L Hens. — (times “‘1 0.. TNSS Em ‘ o r1 consimmeol . b. chh'onsfiign ‘ihde krmtna'te . C. No equiifbriuwx I .. _ . , ._ . . l i _ . . ! ._a_t;o_e.-i_i_m_e_iai_l .|____.. _ I I '. |___| .- :rrit-4*'-fiwi-_fi-'—_ I[I .'>i mull”- _'_ “ll air . I I I _r. I l l l I I I - l ‘ ‘ Z reaoHuns. 'a. TM; 92: Hah- Consfiamd. E. Reacfg‘onfi- Ae'f'ermimi‘e. c. Eqvm‘bfi'um Winfafinfidi under gm“ Yoadmv a: g 2 Kips (7) 3 non- concwrm I" nqn~\9ara “el- VCad'x'cMS. ’ a. TESS tom PlfieH mnahmnedfi- fl Reachms Achrm'z‘mfa- 6' chdlllbvmvm Main'fiuhmfi- fl = 6 k‘u‘rsf,‘ E: 8 kip» 3:. J0 kirk}; 36‘” (3) Lfnuvx- (om urrewfl now- {mm “PL redch'ows- a. Tru‘s} COMWMT Wham. J: -. Rea: him; Sfa Whack-{MINA l‘e. .e. E1u§War€uM ma in {ZN-.4035. 2M0“): :4; = 6 Mr; f -- 4.36 Determine the reactions at A and E when a'= 0. \ i _l_.}- ' 5;".5‘ 3 f'm'tffi act 9’1- "5'25! I . i- —i Pro blem 4.315 — J I_ - T b”""‘[; 5. Must P155 +£mu3h. H d I W11“: 1% arm-f _E_ intersect. Than .1 My fl: - "in! a fling i dried-hr. of fill r'E drtcrml‘nrd a: fill'l'uwil _ fern qb :15"— : 0.375 H00 4: : 20.51;“ 5 We draw #1: farce Mans/e 1 ' "E— M Write A =‘_flN—° L: 539 N 60520.56 fl: 539M zi6‘M’" <1- 15 :[500 NHan 20,562 187.3 N 4.37 Determine (a) the value of a for which the reaction at A is vertical, (b) the corresponding reactions at A and E. ' ;_i i_|.. I_ i _!_I :_:_- ; 5 .._ H (a) 5mm 3 ram; dd am Hm: r‘r'flfd’ bad E must 1:. {hmgh H Whjl‘i fl and E Fn‘fErJEr—‘t. The angle: 0C 55 Ofiid'lned b7 Wrih‘nj -'..,‘._.S:9-- ': fan as 200 . cat/:35? 4’ V/e' JVaN’ ihe {one frianale and write ,4. :(509 meosj’éfl”: 400 N- fl' :Ho’oN ‘i E :.(soo~)s.=n 35.13300»; . Problem 436 . . 4.86 A 60-kg cabinet is mounted on casters that can be locked to prewnt their . rotation. The coefficient of static friction between the floor and each caster is 0.35. Assuming that the casters at both A and B are locked, determine (a) the force ' P required to move the cabinet to the right, (b) the largest allowable value of h if the cabinet is not to tip over. IJIJIJI’, I l 4.4 .L—lm—Jr—4—4 - - - . We is” a; Cabinet = W: (60k3){7,8/M/s*):55’6.6 N [fl)HNCRSh‘Z Loch Hit-3:0: M+’V8"W=O NHHV 2535.5 N 6+ F; =M+mg= Mow/e) : 0.35(5‘88.6;N):206,0N 0: 79- f; — : 0 Pg _P: 206 N—I> " Check {$1M cabrnctdcies no:L HP: . +5 2/76 :0: —( 206.0N10iém )+ (553.5 N)(o,‘25m) * NR [0, m) = o All, :+ #7.I.N>o 0k ' W T I 1 " [1311332373 N/owaé/e Valuc of h When Cabu‘net is aimed: to tip, them: is no Lm‘t‘aet at H! N :0 . ‘74 +§ZM6=M W{0.Z5'm)—P/7=_O~ h lgfo 5m) WMJS'm) h raj/Hm) 5:7”th ‘_ i; F 2;: fin“ 4.103 The 300-lb beam AB carries a 500-lb load at B. The beam is held by a fixed ' support at A and by the cable CD which is attached to the counterweight W. (a) If W = 1300 lb, determine the reaction at A. (b) Determine the range of values of W for- which the magnitude of the couple at A does not exceed 1500 lb - fl. iiillllllllllllll (4) F“ W=l3oolb: i>ZEfi=OI /. , _' Ax—‘%(/300/6)_0 A awe/A.» 4 '1: t”? =0; 43 - 3 00 —500 ( lane) : go > A; :300 fly :joolafrfi 1+5 '3. W20: MA —(soo)fa)—(soo)(ls) + 7352(13ooX'22.).=-0 MA = 2 WM 8000 — 6 now = #00 fiflfltg/oo/Méd ' wives oi’ Which é £500 /6_#: +52 M,of :0: MA —(3oo)(&) —(soo)(l é) +{7531W)(/.2)'= i} _ _ 60W- _ _6oW ' Cmdr‘b'ln is Sat ltf§rcimi An" -1500 5 /o'.400?é/%1/5 [3-00 ""900 i ~60W/I3 5 — 8.qu 3:900 _< éow/Ia $14700: 1928/6 s W 5 2,580 is 4% - Problem 4_1o5 - - -- 4.105 The overhead transmission shaft A5 is driven at a constant speed by an . electric motor connected by a flat belt to pulley B. Pulley C may be used to drive a machine tool located directly below C, while pulley D drives_a parallel shaft located at the same height as AE. Knowing that T3 + T13 = 36 lb, To = 40 lb, T ’c = 16 lb, To = 0, and T’D = 0, determine (a) the tension in each portion of the belt driving pulley B, (b) the reactions at the bearings A and E caused by the tension in the belts. :m w innit "infest—ii 1-; : +0 41.5%.9'.‘fiaqfliq'syfificrfifin-wzr :03; 3; :qncl. Tammi. 'Jt‘a'vicé -'1:0'UFJ'E_' ajstewe ad: .13 r I}: I: fatal“ 11°12 2*- .cue 9:11“- _ + T31 [5—1'r1n 30° '- {95 30:31 swam]: én'i; ' ' me i mm 3: 1M; entity; J This Fences: w «techie: new :c-vi “Hue telnet-H I eye :15.- Elam-uni AH =31'un EHEIH'JF'S in‘ inches. r I T! .Cin-inued -P~M'|nJ€-M “L155 Euh‘TinUeal. M = 0: ~ am Irang-cwxmumnm. _ + (a; x Mum”? - M2234 I2g' x Q'mnso‘g-cossey} + $144.65)) +72; x(Eafi+ = o (— BT§+8T5 ~95)g ' f‘ f'2(T,3_su'nZ?.'1-T;5 5m 30°+ e 533‘ ._ ‘2 (Te c6511} +1} cuSaHlfiB-G if)? :- o m set the coefficient 0F i, (qua! tozercv: ‘. - _'- _ I) T,3 Tb _ :2, ( TB anl1°+TéS€h30° +3 Ego (7A Tf5 coutH-T'ficmd’ ~ 6 57 +168 = o '(3) am) veoall {that T513115 = 56 (4) (a) films (Wand (1+): 7g=Z#/A;T1;=IZIB 1 (b) from B)? 651:-(2Q)Cu522°+(1z)cos3O°+I68 . V : 22.2% I0.37‘+168 :Zoo.é I £{=+352.#/b 4 : . Ema)!éfiar-(ztfjsrvazF-(Izfiwa’ J . 3- "‘ 8| y" :" . . . ~— ‘7/ M‘ 47 :_Z’30/b ‘ :on: (1-3 °—'l a—' ': 3 r J 7360927, 7344530 5é+€7 o 7 RJ —22,2%’-/0.3‘i'$‘é +53.s‘=o _ ‘ F9 : +552/15 4 if: :0! Ha +7; Sin71°+TéSfih300+Ei :0 92 +51% +6 - 2.50 =0 _ ,4;)( and fix are I‘wdcfcr’mina'k. "‘ " Problem 4.107 4.107 A force P is applied to a bent rod AD which may be supported in four different ways as shown. In each case determine the reactions at the supports. -(a) -i--§Zf"‘i’,,f 30': (chslrgja 4.135.349st —— Pa = 0a , gout-twain?” <1 EMA-=0? -(Dcoswfk +(Ds;n\rs°)za 432: o- Dfiz :1? PHJHWDAHS'“ fl 1%; ‘(Acostrfh — (A-sm‘ts‘jja— . E: awwzlas’ 4; 1: *Efizorbe-U‘szrr —.-z D. “'3 I p? Firm: :3 3* v3 3" H2355 :0: DJ mfg—9:0, DJ: P—g = 13 :0,74§Pi‘63,# 2-. 3 1:1 “ +7 Elm/2:0: Pa -.(R sin 45°)2a = 0‘ 2233—: = P 19:0.707P545? 4 i; 22:30: stésa— Cws‘rS“: oo- c:n Q:0‘7D7PJ745 4 =0." HSEHLFSg-CSI'PHS‘ED‘P:Q 3"??9 D: F’ .£= Pt 1 C ...
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Suggested_Ch4_sol - i - T 4.9 The 40-fi boom AB weighs 2...

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