sol... II - cap12-dynamics - f beer & e russel - 5th edition solution bo, Manuais, Projetos, Pesquisas de Engenharia Mecânica

Baixe sol... II - cap12-dynamics - f beer & e russel - 5th edition solution bo e outras Manuais, Projetos, Pesquisas em PDF para Engenharia Mecânica, somente na Docsity! az pozsw = 00259 = 0,025 (322/95) a = 0,805 Hs” «4 Weng- F=ma (2800 lb)zing *— (500 tp = 2800 q 1953 —/500 = ge.g6a a-- s00 ftfs* Va = SS Mi/h = P0,67H)s a>0 vemtrrar: os(B66 Protcisso)x z= 217 ft «4 12.9] We nte that ag=ta,. Block a ob ao, Na $ ZE=ma,: T- (too lsinao' = 200 4, (1) Bluk Bp27 Ma e 1 a- 300 (144) tool 20 -B 373 NEEM: fool)engo-a7 = EG a) (9 Moltiply (1) by É and add (3); q aê 2.2 2004 4300 2 (200 bo)sin 20 + (Soo hsm 2 de0 a, + O (r do) Going op: Truck in eguilibrica IZÃO] We first find the valve cf Mk required Since Speed is constant, Maintnin eouilibriom! z KIT =6: f ' O ando “ 20h 1 Bb P . Pe lbsinças Wiand = 0025W (N) AZR=o: hp-do asi =o Ng = 187.44 db M ta DE =0 TA, (8249)- 200 sm 6-0) Fe 2a À * ta = cê Votar eme go Tt lt)-corco () o = F. 7 E ! ty Eta f2E=o: NgrJoocindi'so 5Zf=mas P= ma N a o Mg = 02,61 b Sobstitote for P fem (1): So dom 5 ei das) doentio + My (102,61) - 2B691= B02SW = ma 4 Mottigho (1) by 2 and subtract Leco (2) Mpeg (07,49 ) = 148/36 Pre; = 0,305 Sine Meg > A, dobocts mijl move (À q 8 dum) 5 nl 4, = ED0 (4 Eu NA as to Me Es 322 (4 af) sr joo b =030(102,60) = 20.52 , . , ATH= ma Smacii-Z7-20,58 cd (2%) E) Moltipk, 6) by 2 and adé (4): - Alo Jeinié -2 (8759) panos 20 - Zns7c 2/26 as ea 46%0 cÉR A, GEIMÃO, ans Ag fogo 4 (287) MTE (6) Fm (2): Tedunsinad +37 574 EBURO), To (24,0 lb ZA Ge Gkm/h= 1/26 =25 04 23 Both ; ms m&z e E cars ttsmoda (Ponbke AcrEdS = (A -CED SON FRON ) sTE-Ema: conop =(4Sxw ta As 4833 m/s! — verter: ops Z (-h333)A a=434 m «f = EST = »y : sto = 552 dp, ATE, Gas tgp ra |) CACO, (amido soremas ME dani (een), agoristi q) CESO = ES sorosragona (6) Fem (1): Te(gokb)engo + 200 (8,405) Sonia Pelos xi h383) 30x? 3, Ccpiib q Pes an PEIN T 4h 00000000000000000000000000000000000000000000000600  IZIZ] = 70km/h = Po/86 = 25 mf (23 Both cars Ma” “ma =(asuoa = (207% =— totisa Beto Bl J6xdN Find P for impendiny motion t Both olets are in eguilibrivm EN STA BAN rã B E 12.13 conTINVED T 33.352,35 -(Wig)sinzs Sá T- 4885 = /So (2) Lo) dd Ego (1) ara (2)! rHsê= 40 a a. =+ MEB6 mf (E) Sutetihk fera ia Bo); 3203 T= 45 (orBes) TEZBN Check We Shovld verify tunar bis achally move by dei delererning the valve 0É the fine P for uhrek atm |emofion 28 imppêncting CCONTINVED) A ThzTma Jumin (esa kyda 4 Sun F AOL mb mts 2 from ZF=o, we find a W=n'raro 0=(25)4 2(-06667)z g70 3 PD Ne Z223N W= 3557 (b) Car B mea E=M 120 (2273 N) = qué N ? >TED [E Fx = punto 0,20(3357N)= TMN , Free Body: Block A iso. , º : +aGama) Po ga, WXÊ=d: Priusassints- meme -T=o (3) =(o x toh o 6667 ma) Te Borty; Block b P-4 1233 N, Poiss3eN Ca] 7 ZE=o T-lyntSsinZs maus Tu DO (4) [213] Aesvrr melico, with block À movins Aida Es Cora (1) E] remo Pe (Rysas= tan tabeio 15 Una se)-ny O Elori À macia =H8.6N mj=205, 288 Z A He Since acha! velve c Piz5o N)is beger ttom valve for ido - ER aos impendicg motion (118.6 4), motion atos place as assume tha ! sr Ha (25 tsfastmjo) [2.14 | Assume motivo, vrth Block A moving vp. É ABM =ÉW2sN BORA (motas Na See Colubim of dk RR TO Ma Pob.I2I3 br nIg= W-f(2u5,15n)cosasto, N=222,8N A. corpo of = Mp No 0.i5(222,8N)= 383409 Au Pa card Mg 44 ZF=mas 250(2u5.2)smestaa3m Tezsa) | =MaN O AZT=0: N-(Usas)emas' =0 ? ss r=25a po tt Vo wo 22234 Bu T=M Mo 0,5(222,3N)= 33,34 N oo EN so W=m, , a . ma =!5 e Nida ) irzE-ma: Ty 25b6ins'- 33,34 = 254 nO = <Usam di T- 136,99 = 25a (1) = IIS Btx B Ms peso “a Cu p FÉ 7 E ce KZ NON (18TIS Jtnsas'= O Pu ira = REZA Nczzz 84 V3ah = 35570 G pin +5 NLM -/I47ISERAS= 0 , , . =0; NONc(ig7 isto = o Ma N'=015 (3557) = 53354 t ed Eltema T-R-Fi-Msnisema a Pan cançoso n)= 5235 N Hr ER=ma: 250-E, Fis hpsmasTe isa 250-33,34 - “sado +(UTIS) SNIS To Isa 225,50 —T = Isa (2) Mol Eos (0) omilo!: 225,50-136.90 = 0a a=+22/3 nr 4=221 mta A É) Sobstitute tr a in E (1) TEM =25(2283 =I32.3N Check. We shovld verify that slots actual, move js assume by determinmy He valve of P fo which motor of A is impending vpward. — (CONTINVED) 4 5  IZ.z2 Dr A (4), Where (= vel. of Shop Before cha brenks” Ship is im equilibriva, mia Sa bl, S2f=o Tkgeo Edy A=6/6 Ager chaim breadest £EBcma: o, - ss Alt): ma DSAt) = Recoliná (1): Í To (o/nkê-r)= ma b - Gra A (no 4) dt E To ME d integating from bao, 50 fo E= E, vim Edo fat £ Vw E 2 [- 66 tu lê A t My & = & = 0U Eta EM tbm? q 2.23 =ma&: nOxDêN 2,25 0"=(25 mb ta)a Substhuting az Ms da a voxmê- pas 47 =(25x10 )u BY AX dx Ja -f 25xwwdy o 2 dio gas” 25 xp” [Em goxio” 225 uv" & 205) (homo-a, o — AE fo Hoxi0? “5 HOMDÍ 2,45 VU; For = 2h0 fem/A =66.67mk 3 > x - 22X & 40xto o ' 45 - É Gaxio- AISO 5 S5ebe 1,333 = 1 5482x10!m x 1516 Am «q 12.25 £Tr=ma: [ ma P-hy ma —a = [D+ P — af (PR) [6] Feto Initial comditions: É=0, Pro. dy ac p(PA) de [E is =i far E [to (pdoe) -E dn Pote Bt P dt Phv = Pem (2) Sob stibuing dar Poder from (2) into (O): E aste Solviny (2) for Nr; t vs) v=E(-e (3) Meteireç n-=defátim (3): Pre & s AM e) “ a EO e Jdt = Res Rê ] 2-ft-Bn(i-eé ) - l2.2%| F-ma; moxIÊN = 2.25 vê= (25x toa Ja Subeh'iohng az de : moxio? - 225 = (5x1) dr dd 4 Vfas xo) dr Er "dE - = E (IX É f HOxID 2,25 é Foxit? é o timixio? [or (mo + p” 4 CCRVTIRERO | ELI = - CE BL 133,33+U, dL6o7 bn 133.37 -W For a = 240 fem/h = 667 mk - 1333346667. my 667 bu B B= 44667 ln ra gago 6 tos = HH. 667 (1,0986) t= 45.85 «q 12.26] Belutiors getueen cxcetercims Let A, be ocel. 0 pulley D (+ doem) ler à, be auelar B relatim to DO+ doca) Them: ja, 2-2! 2 = (ara) O) ash a & = (2-0) - 4 Ler The tessim ir Cord EC a, e Then Tension in cord AD is ET] Sine DX of neghgint meat. 3h 7 Block 8: E = 8! HEF=ma: Puis-T= lata) (3 5h P Metas Para) 1 ma Glaa) Bloko; cli= E Gg! Lei HTFema: Tu=fla-a) (tr 65 dh Aga Ego (2) ama (3), amd subtract (19): MPG A = a 2, +ã (nta)-Flaa) (ConTINU ED) 48  12,26 CoNTINVED The tstegoation yictas P= E +pmrga, pára, P= 4 a, a, =5 g Add now Eme (3)and( vo): Pero = E (rtp(ra dl) P=30 a, a -5 g Substifute fer O, and ao from (S)and (6) into The relations CO): (o (6; U=4,= 53 24-68! (1) Cecnta=farãa dar €) -2â-2,=P astra ta-fa a cfyto () (4) Motion df E relative fo À Tam Egs. (1)onet (B): o Ep = pal ste batém Cp E Es! = É (uz)= custe Vo Momo + gg É = O HEM MES) = 2578 th mp7 258 Ha, a db) Mafia FC relative do À Frera Es Jana CM: EL bat Ea t=o Veja = (ea), + Cop t = 0+0(h)=0, EyoO 2.27 derivalion 0€ Evo, (1) thrugh (6) and E). Motinn of B Comi tovni actelo rates) Ze: 06), E + 4 nat Male Ya = 104%, (,=2, te2s neto is) Gg = 6 Ft/s* KR) Substilte Ag = 6 int Bs (B) .P = 120 mb» FP = dp CATETE - = 37. 6) Teo 7 F= o 4 [A=[A | e Dee cuejeto =2745 154 =Í2 =277 b P=3707 g q 15 + 8,723-7 = 2,735 T= sm db Sime weight of pull, Dis negrgibl, HE FE ma: Tg =2T= A (Sh) Tags 319 -« 12.28 | We first check Jhab static eguil. ix not maintained: [ARCA RT Cmarmog = O2n(Fro)g< 69 Since Wo = 64 =/09 > 3,63, Oguil is pol maintained, MasjMag 4, Block A: =: Ma =tTag a fa PAM =ALMAI . Fast Da AZhcma, Tolma-ma, () r Wa Mg aa Block G: z6: Meo Tg LIM = [+ Fez MeNço o Z0kg NT Ectas, ATR=ma: T-02mg=ma (0) 27 Block B: HZF =, = Lg Meg 27 =mpas (3) 64, Tmkinemahes: à, (e) emsionin cord, Giten Daka! m,=5 ty, El): Tozog-sa a=07T-02 g (e): T-ozlidg za. & or T-n2g (6) 10): log-27.=i0ag Sobatidute jato (4): g-027=1(027-02g 4087-025) l2g = 0.857 Tebtg = 2t(a8imit) TE SENA 6) Substituto tor Tinto (5) (Dan (6): p= OBS) -029 = 0 HBSTERBIM) Da HE ME» A rar g-02(Eg) = 0.3143( Elmo) ag 308mp A ((E49)-02 20. M8206(BBIMÃO) aus lvolmpe ag AB =]-42T See solution 0 Pb o ze for he | 9) gre FreFcheck Fut fobia efuT ie cof maznieimed” Cah t EA (ntmg z atu(sstog =67 Sire Wa-mag - vg > 63, egurhis pot maintained. We shall aseuime thatall 3 blocks move amd use El, (03), 0) elerived in solofimm et Pos IZ.28. Q) Tension in cord, Given Dela! mp=Sbg, Mez Obs, mozÃo kg Ep(0): T-02(s)g =sap Ap=02T-A (5) BO) Tag =20m a sposr-034 (e) EG (3) Og -2T =D Ag ag 9-0.27 (3 Substituto into (0) g-02r= I02T-02g+005T- 0.25) lag =032sT Tl = SB GImh) T-Hin Substihde inh (624 q cc0s(ea)-d2g= Doing Cirpossidh This mem Mar our assempation Fra? block O moves nes wroi, Pisummg now fust C dos mk me, we how O, =0. «J Sutahtubing 2,5D and from(5/) and (6) into Ep CH): g-02T= 4 (0271-028) Hg=03T 0 T=lg =g(t81mb') 535 7 T=380N «q Substitotirs fo Tinto Ejs. (5) and (7): Ap = Oia] 02 0558 (Abas) de 5I3mf ya asc g-02(Ug)= O RecI(noIh) a asma] 4 We ao cchatihet Toligois less fhen (Fo) coo4fiog)on ga, The tube assumed hr die usar 2-0 4 as welias that oblninea for FT! T='36.0N 4 49  44 x o x LEfama: (Mrts)snid=- Ra «) 8 - The quelêradios of 8 may be retabres A = E ty + Eua HE tolowa that Fe Vector Mada Mem be reprecevded 23 shown: t5ib ma E] —— "64 gp = !E IB = a” % 47% a" ac Éo, HEG= nas 15=My= E ap sin3o (2) & Bhema 0= Pag Gapesar Rap = Apos (3) IR) Multiply (2) by SinH= 4 and add Poli): = 2804, ay=fo(ea Helda 22,40 fijo MALHA EI3O a (6) Fra (3): Cam = ALHO emtot= 1940 146º [Ent 00 &p SR x (T+) Sinês' = My (T+ Zog)sinasº = 200, q) Block B, The ateleratom 0F B pk fe cum of the acefematim of À amd of fhe acceleva bia of B relative fo As (ConTINVED) 12,3% CONTINVED TEC + 2942 15 2874 , =+ abre” o Mp2, = isa, a Mig 424 mass S2Toma, 15g-T= |5 ay sints* (2) O = 15 24g +15 Ap cos 28º : Peja = Ap tosEs" () Molhiply (2)by sin 25 amal add 151): Log amis Hs g sin Te '=(o pis sines Ja a, = 2EsmIS (qnh= 6398 me” à Po (a, 207/15 8in 2º as sto mea as" «a Free (0): To 15 (28)- 1506898) Sin25"= 106, 994 T=/06b N a Note: From (3) me hau Toja= 68 csis'= EOO met da = 580 gra Cop E IM — JE Za) ca O. 4 A fxT-ma: E mus, tos, Temis"= 204, (O Blax B The aueleratom at Bi Ge cum SÊ the accejeço fimo 08 Pl and of te aúeferatio o: & relativo iu Bi a, + A a “8 “3h . 2 SE VA t/Eh,-Ma,: 15g costs 7 = 154,8) Ce 2 6-ma (gsiniS=/sagy-15aprosis gp = ps mIsra, wu (2 Moltiply (2) by sm 75ard eta ba (1): 154 SinBSfeos2s* = (Z0+ I5 sio'25t aa A, = JosniScostst(o a) q 4 E aaa E Rea (BI) = 2, MB SA! E emb a - From (1): T= 22 mesp. 76 p Sm 25º TE76N Note: rm (3), me have “ops TB sines + 2. y52 Cos ps" Sp bM int zs* 5O 000000000 0000000000000000000000000000000000000%  mr o A BE coma. Te Sos Eine e; +*Hz ; =0: T Tens3"FentsT W =o Te tOS3O Teasis'= w () ma-by Add, Ege (Donate). Sine sin4tS=csys” we obtam and force +riangle, Law cf sines: 12.43 |(a) Before cDis cut, Ci See solution of Prob.iZb2 for Es. dingram 2.4 Taio Ssmiê vilibein). Te 035wW «4 E) After CD has been cut (cce lerated motion) AN "Ne JN EE=ma ao! Tr e” “ts Ter Wsinsb'= 0 Te (05364 sim30)= W(1+ 75 = T = Wen | & — ne Sobstitelimg gue data, W=Blb,C=5H,gesez th | AS ma, -076W a 13660 7; De =6(1+ =) wo re = 49H Te mio! [6] FZ. 45 - nes ate Tó! ira pissible) tra) E track when N=0: For Reid: Hc 274% (15)- fel =251,36 A W= 1585 ts fr OST E 15h, we must have « VEISES HA A sp / Fr the conditinve reeotinç wire BC, we multipl . . Eq (O) by cosa, Egito by sir as cend Sub facts HZtcma mg = age! Tec (SinNT tos 304 cat t Dada: = 382 fm/h = 282 m/s = 6H mf = WS cio sine?) - 8 = Moinho tom p= titêm « e Sin7S'= V Ge cos o te sin 30) Es f 1 Bimjor 4 Ce) e o With given date; D.ES93 T = 8 (peer = -0,5) mento do. dá d — me vê as + nam Be 6) = 8838 mk da T, ae = Ara Terena +ZEema,: W-Ne my for To=lbb: O-47056 4 20,007 (15) = 420.64 1 ? z b + ! , Ú , = W. mu ra , Jo. B3,33mk) e 7 210 Ab Ms Wo RES 5 4X GB mio) (75 4) Elin For 0<T, <tblb, me munt have N=BSISN-AStI NEZOLDEN N=782N4 «8 Sept << <T| At bottom of hill; é Gmbir ing the resolls cbitined, me conclude Fhal rarge Ra mamã of allouabk palpes je 26% df cre rasto - º *2.4] | See peevioo. page = - IZ 42 o. Betore AB is cot (Eguilhriom) EE re S es, a z -W- Y pe 1 w +12 N-W= me - ET N=W+ MES 735,75 V+453,67M E omew = 89.4 (6) Ate A8 das gera trata motion) N=negnto fo A,=5E=0 422 Ecmaco T Sa - Weszo'= O - Tp? Wes zo” — MA, To TOW ad 53  CoMan=tym (at) fama, a-E- =M(gre) Givei Data: Mp = 020, W=72 km/h = 20 ms G=L8im/s E =30m = o20[g61 +] A= 6806 «4 (zo 582] RENA oo [get — 2 = 01842 m/s TZ, 16 | Note: Bpparenf weignt”2F pilo? x mesored by the force Nº exerted on pit by his seat. ny means thaf N=0. É paint A: MAs 1 = de, too ma, io w= a E uatrod % vs 126.87 Ff Va =6ESmh ma Mt print Br j ” TE=ma: J New = g E At f z Esp-lto = 160 322 F0) LE L dp ssoib - W=i6o "= preso) 310 D =39, 244 Vo = 198.1 Há B f % = BSdmifh ad 12.47) We note lhak sine speed dt belt Te constant; = as lorg as package does mk slip- (e HZE-maç: W-N= ma, - Wen e Wa” E .E y Ne W(i- E) reto N=0303h4 (b) When slipping é impendino, we have F= RE MN=240N ard Sth a,=0 DAS Ma, “dr 3€ LZfoma-o; Wsing-F=o 075Smb-04N =O N=4875 stry cÚ Hera Wiocs- N = ê O 7Scoso—N' = 212 (o 32.2 UHE) Sobetilvimg fm N Frowl!) into 2) OPS caso -ABISsinê= DHWIZO msB = ESsint+ 0.5G627 Squarmg butt members, and seftmg cor bis! sn'B = E 2ssinp + 2. 4D13Ssinb Mass + Alssin D+ aBIas sd O dente = SinºB to qHZLsnô - 0,DBB97! =D Sind = COMNEL E QgLy au Z Emb= 0 1SES 8= 201º -4 or sinb= -9,56NB B=-Iug' Coxcrsresr) [2.48 ,P YEL = ma, | mi Prugeinô-g=- "5 D home conta wir Sprcipied Surface, W=m ta Wo neod R>0,0r 4 = Prmg sino -mvdo in L "Pp 5 r25(E £) c) Dada: m=ntsokg, aro 30/, Pe 15 n, E'=09)m Sushi fude into (): sino» + | ef - É] galo? ag Sn X 0, 40775 AUV<O<ITES 5% DO0C000C000100000001000000000000000000000000000006 e e e e º e e e º e e e e º e º º o e e e e e e o é e e e e e º º e e e º º º e o e º º e e e º s IL49] v=usmjn= cof e= 60H-24t = Sgt w Pre = [EE- na ge Rsin(B4d) -am Ma. + =0: tg ER FAN Gruen: P=200m, = 32040m/h = Cesta p= 5108" Bo, 809 A =7605* tizh=o Nico fentem (10) ten(s1o8t do) = ff eedi. noz SE + this Ne GE (O derrote mir prtang = 0u5 Divide (1) by (0: E At minimum speed M- 38 = (22.2 (58) M= 043 <a TZ P a Ge» 3 12.50] Minimum safr pres . -— v +Th= ma: MA, = E Rein(0-8) = 10 n(b-a)= me q) stZn-o Rex(9-+)- mg =0 Res(6-4 )= ma DB o . Dividle (1) by (0): tuo(8-6,) = ua lt 2 Fa EA Rein(e-d) mg [62) 3º HTE, Res (o-4) = mg =0 vs farto a) « ' Res (0-4) " (4) Hopiu ente Speed! o ng ETE-ma Dinide (Di (O tom (0-) = x € 12,51 CONTINVEDT We have Mheve fee Rsin(9+8,) ma (1 Reos (048) Mi (2) Divide O byl): tan(t+ 4) = vie ge tas (B-8) Revn(bed) = mo” Reos(g4++) — (+ McottnNuED) POSTE qasbia)-ng-o [Gia Polom, O-d = 5hBB=aMD= 217 Pra Pos (By) =mg CO) are qBi(zoo) tan Z7 7 = 1031 Dinde (1) ty (1): ' Í fan lDé ho ge v= 32.1mk Ve Ze kelh A v =43 FENESA) «a 2,51 [005 4 caled speed (F=0) 12.52 dotar mato o 75 Wemg 1/9 =36.87º 2 — ma mi g=/8 8 e E = 00 ft K=N s J x ao to TE cm Mg Sing = mn e cosa a E A ne +TE-na, conta) ZE O Given: W=/BOfm/h= 50 mjs, E=200m er a tino Ob = nor postos. espa)! 2570 Peos(g+an)-M=O j Edo, Reos(0+h) = W (2) Ch) bt magimor Sprod Tg + Diwde (1) by la): ui — . fan (Or) = Es T ma cm£ a o [A 1% ge fem (046%) = 32 2(000)fam (19% 86.87) = 32,2(400) tan 5487"= 18, 306 =13530 Bh, VIM A 55  Eg (1) which defives vertical distorce 3, traveled by electrm as th ns Plates: 12.61 | Porizenta! motion ( uni turn 12,6%] Kimematios te. L-te “Time between : É foto Ls 5 Pb plates: taste? 0=28 TT % E =6L-38 p= 4t fe L Time “beyond plates: 4 =6-6E É =s tel ftot Kechal gl +€) (0) When t=0: ferTrca) teto "zo 8-0 Betel e calado acceleratef motion k=o g=0 1 tra Fem: Er 3a ty dad” EV ama ev us ASTM Gm aê cólo for ape vm pt a, = 4B4248 =0 = ae 3 4 = sadios Ele 0) | 0) Men gos ts: - opa iso FE Dell as eec)iame ri au ã = nápi É . £=60)-s)= sh B-4D=udk CH PAHO E = 1 eve LL!) . 1 = 5 condi "má TE 4 26-6()=0 9 =4 rodk” = eve ft t.L eveL tr A mi Mot-tOsO-T(4)=-—32ft)s [2.62] See solotimof Peob.t2.61 fr deriatim ot | apo AMADA (MU) n4 30 HA? Kineties Dente by E ane Lº He new levgius of lake and tube Sine Sand all ser characters are to remain unthanged, we musr have 5-5" eVEL Lever “Mwra — md £) = for Sine = E we have Z'= £ = s=5 18 The lengta of the plates should bedovbled «4 =) EX(E y (2) vlhen t = 0º % Ema, =E te = E eltram is not fo sfrike positive plabe, re musthag 22290) ROTAS «q 4<£ Ezmaç=t (0) G=0 «a eve gd 3p.7 Tmdw Tê (b) When é = (s: - vV TT a g > E $2/£. Rec2 bo q Simeltest alloupble valve; A ey a E=5.18/b 4 £ ma TZ.63 12.65] kKinemabies, Sec the em presstoma obteiaca Fem solotim of Bob. 12,6]: ED VPL " ' Tnhe Solo Fion of Pub. 12,64 for 4, 8, À, dna É õ'= mA Whe = 38: B=2(3)z/brad a=3(3)-G)'-o B=4(a)= 12 rea E=6(3)-3(3)- Gts Eee) = th Duty rdp? a cE-agie (e = Ap apar Brabo Of!) +20) = 216 HE Kinefics E ma, == (es2) E-- dest «a F= -A68 lb «di 58 000C000000000000000000000000000000000002000000000%  12,66] Kinematies 4=2(Heos2Tb) 82 amb 4 cbr sinimk g=21 é&=- 69'cosank 2 =o E) nhen Eco: 4= 2(4)=4m $=0 t=0 g=2m7 mdf bo-Brntmh? é o 4,2 E-t6'=- em uam) =-247'mp? age tBsIAB=O CB) lhes E = 0,758: to 2(1 scam) 2(tro)=zm B=41(b75)=157 E ma trsinBÃo + um. 8 =27 E=-eTlas=o Ê =0 ,=E-t8=0-2 (2n)=- mms ag= 4842t6= 2lo)+2 (emor) = 46 mp” Kinefics Wheu E =0: Eme, (058 MT nk) E--HB4N formas = (05 taxo) Rs 0 «4 When E= 0,758: I2,67 ContINVED etics - Motion of Block “oo era / SA a) ZRemg: Tema, = mf-tá )= -me bo Temtê 4 dema,-m (26) ==20mb bo RE Ambê «q Note: From E (1), ve mobo fra E describes am Archimedian spirak 12,68] Sec sotitron ef Pob, 12.67 h obtair ae z 6) Thy: Tome 6 jal= tmbê Wit given do ta mz? bg e=09m b:0,06m À, = 8 radf we have: E) T=(RhgÃos m)( Brest) q Nose) eredh T=t40N A Ainemafies Dime he leni cirad CDE must remain Cons tent, we have Are CD + DE = constant bort-d=044-d Where 4, = kngth DE when dx RO t>-b80--b8 So . a õ-. Aectóo AgEtB+AEÔ co + 2-bÍ)O, 266, (ContTIN VED) Rem -(0s bits mm), Fo SIN at al= E (ty (0.08 mat * formas (OS (EM M/S), F=7ON A a alo cenas LE SEN « [izier 2.6 n [12,67] 12.69 Mb a, Das” (1) E-k . y ” Ei Fo -htom ty ouém) - tem a, fio fon Dc -Têm EO: a, = (ou m(o)+ 2 (18 msi radfo)=+432m/r04 (6) Fm Eg(1): E = 4,44 8" Eectampis(O mf izradfo)” EJHZR ma, Fama, Lo2%fus im) F= 86eN O a, =t8+ 2tô (5 IB CG) TRE-MA: E=-livomi! Not fhot the acceleratin à, of fe slider and ste relotive aceleration É wito cespret fo arm AB are not equal 53  (0) +, Epema: Substilule this velo for As va Ep (0): -Br=i-ad! &- «(6-8)» (irado) el t=-56t -ha = MA, Fa W= nõ = (0,3m ta radfs) (6) Form Eg.(3); A --É + = ea (2.3m) m=-Sumj Fm e Cysinee go do = core tantos =4BHIib<0r2m6 = Tm: Es(oztgk-si6mh) Gr+3é0n a =2(-2.40mN i20096) A, =-Si6 me: aq) fternete solofion: Ene nÔ=U, = sin Ê, and E="MS2N O a FILTO Where "12.71 CONTINUED Eua fo a,=8-n6 cn MES 36 (013 -0,08) + bt (0.5-0.8)= wo? % = ap =+B+226 (2) Ve = 300mh E ma, Since 1 ia decreasing: V=-s0mj «ad = 16 =(03mfizradi) 3) (6) Fm Eq (3): 2,=- I60(0,3)-84 a <-elmb” ad From om Ep (0,5 since É =b, =tonctunt: Apsnfsaiê= oszmd=z( 3.00 m/eX12 rede) = +3,60r/5 | Gp=-120nh «6 E de da [OA Tema: Ex(D2tyf-720m/e) FRA Thus! e-leionN «df ia te e ida 2.72 Equation E path: Ev cosó : get Bm as apta 229 =h utero fr= 40= 8% 4 Nos ô=A “pa = (1) Es cos Wf= 36 (0.25-0,09)= Ko gs . A 4% + AL = É 2H0 mA Hz E spsimojo =("405 DU esr Sine tisdeereesimgs = Lô = ágind + “costa B=20 ct uso E - Sotosd & cost cos*B VE fes = fintóscosp = E ae Rt abate” rs Ap = tio na os oro ge pe (009) 4 sine á=%0, é (iZue)yietds tusinP = % us Fi] ma, We nom have 8 hé=m CimigiatoPi nº . — be Es A(tto) = º— 4º amd LT LCP we have Estes) - ma, with Fe = 36 Nom o P=016, snd-ç ã E and 4 =-0.045m Thus arise 4,55 (3) We recall that a,=0-48 0) But fre eqratiom ps fr as = eb rocô €) bb tos b 4 o “ co = (a)sEtoma: -f(t-4)= mar beriotra Ér a) % — — biftoto) (36N.m) , vos g cse in (2 Ascág Goig (+ vovo) E = tos ars tbt (0:90) 4 gi: Aç=-/B0% - 810 c) & Sobsbilute : fra, fm (3) into E, 12,73 | Eqvatim of patn: é =tofcosão -/B0+-B10 = E -t57 Ep (RAT): EB =b where h=076 28% =4-4 (forodh 8 = É- ta Vo o e 0 E=-36t-8.10 (ems)” Pojaza” cost ã =é -(Lbe'nzo 20) &5) Thus: wu MR = -s6 4-8 ua 5 « dinda = “erenio)da 4=t tó es SE co 28) 5 cume! reu Jos a = cost8 «4 (ConTINVED) 60  12.83] Girevlor grbito See solution OP Pb, 12.80 Em derivada of Cn E En But fem Em. (12.30): GH- gq Rº=(822 Hp amto xs280 Fe 14.08 x10/8 Orbil throagh A: e 45 = 240 mi + 3P60mi = 430 Mi= 24,70X10 + through 8 bu = 22,230m) +3960mi =26,1900m) = 128,28x 101 vo Tem [reosxn é fleje é Vea — Vazzxe — 24, 900 fifa vo det! Vela O / 5 Obxio 19 pop th so £ Orbil Trenstrr orbit - temstro othi Cs (adm +4TA = 26 DO A+ 770 ex ALEN = 32,6H0 HA [O Conserva io af amp momen tum Ci ! Letwuren Rand 8. mm > mta mto =& mb) e s é . (p= (0m,= ale (8640 fh)= 5360 fik 4% om (Va), = 1000-5360 = 4730 Fih AM =4730 fts 4 (A) Sine forte T exerted by cord and weight W huve gero moment a bovky oia We write &a, (1219) in terms or comp! EM,=0 H = Rmy= constant Rm = Rem %e (E sin) 4; = (Cosinte) 16 Squarig both members: Lºcinto, apfsim as (1) and fhus Therefore : 4 T As ball deseribes circle 0 cacos Rs 1 = Cal a $2L-ma Tsing=m 4 w aa ut zé ? K Tsng=m A, qt DE spa (2 [56 cm o Mie sia *DTEco! Tesg-wW=0, qa Wo = mg Coib) — CotBj Subatide for Tinto (2) A E sim8 /ons6, (3) Similarty: = q E sin Ba /cosê, 6) Sobstitude for Wand v;" From (3 Jam Coin O: EPsint (26, sind /cosy)= Eisin BG E sit fas ) Esin?B tanô, = Coina, tan «di Ch) Mekias E,230%m., 8.030, 6,5 20.5 (CoPsin 3o'trnao'= (20) sim, tam & sima and, = 024957 Solrnp by trial end error: B=409" 0 «4 [2.6 da) Line no fone arte om fhe ball afror fhe E teru cut and ontif the tal) hi stop E, F=F,=0 «a Contil ad hits shop B, the bel will move in a Sha gat line mith the absolute velocity am) Z,=2,=0 é) Recadiny that G=É -s6" we have 1a + 2 . E=a+t6 =0s4(i)o E -(20nA) e É, E, ITA. E =+300n/5"00 4 E) Sine the fire exertme by shp Bor due ball ic directed fovacd O, we have congervafim af angular mpmentum abovf O: E=30ink E yyB Er = 15im, B cin Ho= my = mv (30 inh) U-6 in fe ao Um = Ein Co tt Isin «4 12,86/ 2) Transwerse Component of Velocih, Since the enhy hormtontal force excerted on colar 1s the forre E crentes 5 Spring, we have Fa =D ane the anpolar momento is Confersed aboul Fe Shaft: tom (Bs tam va = ÉS (Va, Sine lob= 4,6: = ta 6 = (24600) Cerodb) UG =0450nj «ad : &) Components of Accelerativa, *| mk horizonth/ force exerteot om coflar is Fexerkil by spo. For = 0 bm: F=fx= (SMm)f0,15m-0bm)= Tso N ma, +) = & bit-ma, : F=0,70N E, 050 N (0,3 kg) 4 A, 242,50 04 «ad HIf= ma, O=ma,, a,=0 = (E) Peceleratica Retative to Rod We fish not that, for £ =06m, m=tó, 6=3B = QugomA = 0,75rudf We recall thar a, a Accel, relstiw torrado = a + tê = E-+2 50040 bm(0Tscu=+2,8378 mb! Peel, relativo to cod=+ 2,84 m/s? 44 63  FIL .S7 a) component of veloute ao aebihay distante te 0:75m fem omd The fo exerted by the Pes é on coilar is no E (EMO TS A On the GHhw hard, we haveo à, = Em 8, Thus ste) -(oz0om z-28) OR But sine Ho, we have conservapim of angular mormentom Chou the shaft; 8 = “ê or, Since €, 2 0,15m and â= 2rod/s ; EBAH = 0.270. Sobstilutina for ê into Ep (1) and sevimy for 6, E= A (east) + (22 z 2as- ro s67 ep DER - dk da dt But o dk = dj de coa” e . os: a bg = (5-Ibébra » MOMO) do integeating in 4 from O to Ve ama in 4 From Om to D.6m, we have fes - Lo das megIe + ;t [ns tease tee Es asõe 05) 8:333/0.6)4 85 (0,15 Fe=mas 5= 227 o SUE ar L 2 . he shaft, Lhe. Sprimp is compressed ty foram E) Sn Sa F= mo 20528 (fre spo): - imbicotos = RAS -3+ DUBTS O. DILS ALELOS ar cecmllnç xl: e ) g NOS M=AIUI O M=H940b «6 * Fe - tudo (ae mf (b ) Valve 5£ 8 Abe: The — Sã indicales fhat Fit repulsive Sine ! [2.90 4-b Fa B=mas: jo=m 4g=0 8 De Am 4. o But ag=tôriio, Thos 5 a 2. ce) da rôrziô=o Era Bud g=06m 4 =U, VU T42 mk EA -0 (3) do Di? o 227 - Po O qa 000 atá Tim GR gui (nu) (e) Substihiog inh E9 f): ” do? Cem) É 2 (27432 mp0 cade) <o | Substitotina rom (2) and (3) into Co): . no 2 2 B=-1358 reta =-036 todjo Femi(Eo4 2) =ma(er or, recaliiny Ea (1): . 4 F-+ má «4 (BED) 12.88)... —. €) Ecs ” €H sinB Lo contê din tofbrcsdanto — I+2 tando (2) do” dy cosT8 CT Acabo Frrm Ep(12,87)' QU bt * (Ga) (u) Substiduting from (2) amd (3) into Ch): F= mk (rettand at) = 2mb (tanto! bptrostg Gespo o %o tosa or, Sine trend = es o nsinto SEE a aà amplo tico (Eme) and , recalling E. co: e InddeS < “as (R.e.D) Note! The + sign indicados that E it attractive 2.89 £ = des (1) a=t= 6828 (2) due -.2 si o = % Sin 28 EL) (3) E a Fam Ey (229): Fomáio( die +a) O) Subetituting From (Dart into (4): Note; The + siga indrcates that Fis altiactive 6% TOC00000C00000000000000000000000000000000000000006  12.91. 4=+, et 0 [2.95 . DO mel = Let? 6) For Voyager T, we shl have” . Tó, “58 veda. GM (1) gu. -be 1 E Mes E ab 'p dia 4 pe (3) Fa Voy agar2, weuse E (12,39' with E=0: dor e += GE (imzaso)= SE (142) From Eq (12.87): & mu Fem “(di tu) (w | But ATA É Setastluting É ta ÉSISO, += 1450) 4% = |£506M Sutetilotina Hom (2) ond (3) to Cs é ep (tsO ago (Aid 2 Femdil 4 Le lg pet) Dividiag (2) by (1), member by member: Ui 125%, = mf 8 md! z Vó Feqfiftoe = uisd) 7; [tgé or, monica Es (): . W = [hEt - (26,98 0) en Baerkm) E Fo BE) apegs 6 nb) TER Am (QED) = 21.04 Bm/s -« Notes The + cign indicates lhat Fix affractive. [296 a teftmoy Bz0i . (12,39): 12.92] Por a posaloite tro! me have from Eg (12,43): , “EM (t sr ( 7 cn r SEM a = Us =yégo O) . “ ao , 2, feftima &= 480º in same eg va tio: o EM (1-4) or Mo = (25970 úm(es, puro mo), + Sa (1-8 (2) E “2(66: 3x4" Pla s) A t k be 7 n em “ M= 190 xy aa di (and (2), member fo m r TZ Ly 26M (80) «a 95 | for Tethys, we hove from Ego(l2 4): 4 Ai = (6 or usa €) [12.97 | gdipticorbif . Me vm tre equation derived . Ar in previu problem: where Me masa of Soto, ' 1 = 264 e) Far Voyager 1, we use Eg (12,39) with Bo: 6 lã Ta Lc EM (148080) = EM (145) A with ds = 360 2 HOM dy A . ) e z = 4000 mi=21 120 xw'fe Bot = Ns + Subetioting CARA £, =3960my 4 240 mi — = a nm ut. (rti)6M 2 4200 m = ER lTáxio E “apa gt! ) frilem (2) em= q Ri Gee Fh Aeronsanoje prorgpero | Dica (3) by (0, member ty member: Substihule into (0: açuora vio) di (5) tr . oe “amas Ar - = + Ta = mé = (E Zbystpe sa 3en5 mo ras ongs! = 3R.tuas ato? Sobatitra fio 6 a sapo A Ion sto Asssuseg do” - i je - His = dai (Gera E genti 2150, E=L150 Aisne A=b% se zu nda as, 16, (30pxHO PER [2,9%] Ling Ey, (1) DE Prob, 12,92 for ench Sme Bit, 4, — Est E au.0B59x 16 Fte of the Voyager Spareenafts, we nave 2217610 a gra Zço Circular dir ae rr se ae rsrs Ee [taprniwel = as. 1450 1 z Vire = 22,/6x 10% a) snes Yo Do = RE ISDP NICHO, Vó NH ms mem o “I&) mereage rr spert at B; x A WU =% E =(26 à) [Pit 18,82 A Va o Cr US AS ISSO X 10 filo 208859 x 10" Fo E ne (EX Ohm = Osouxoftk, AVa= 709 As “4. : teASSSCCdECECIE Cd er acdcdnt Iate ec cata tttetesdeas 65  o Cireolamgebits - [12.109 CONTINVED] Elliphic trajectory of LEM (Pl A bpmR - º Eq dm) wilh kem ve me Ep CRI TEM Luso 4 pa amo | <a * - b err as ephie crash trajecto PE A fopraed), De 120 and tdo emos Balbi Ep (RO): do grGc mo caçiso TEM Censo (| MFB, b=Bo ad t=R: tr DEM ras O, q Comm=b- E) Mt point As [RR s Sp dr We have Dividing (4) by (3), member by member o Pago, ben 1. EM Sobstitutms 0 : casa = E (67) um + , mrGe de ca O E Hr puntB' Mehove Of 6a R, Sobra no) Det As %o VA fiaiiNa + GM Cesst or Casgo H -S” fo ara Fem (1): GM =% em GM — AgaÊ Dividing (e) by (8) member bay member: Tu o = gi ço O É Muay” az £— Feia] (5) Sobstilhme into (5) and ends homerater ces = and devominttor by bo, E (E WaR . (E ) We nofê fat csô, = cé . . j 2 2atr* Ure foton RO), Ac anRp'u RT Sobstr futmy é “ É” Ben . Given: 3 ROB=50) G,=/B/-50= 136, casógo- 0,64279 Bene = a RP EM += 860 Am, Rei7yo km, de= 106897 ando GM « EM . Substifvieg into O Biro dC aRpER O nRfº (6) lo jpcorm (7) Ryo Sobsty toing from (6 Jinto Cs): Pim (3) ' : , ()raemas= nose o.sezmy (ye gu -E api! é va cos 8 = É E. o Y aRP O aR Recolhng tr le Poa ao E : s; 98) = 1p0'- ha o ecnlliny fre value url erre e & (pos) Eg» we go “058 UnE E o. USS62UIm/s)= 1852 m/s Cos (ROB) = =" | Lu fromcan E MBSA- MEU 3 e (321 mis A piicn shows that S(A06) cleperds enty upon a and, Note sis size mes fhal LErii cost Pomard e tear 12.109] Cotcular prbit of command module (2. [10] See colubror DE Prod, 209 for derivalvon We use Eq (Musa) of Egs. (t)and (6). e Mi du = [ER ol Gin: R= 1740 bm to to =1740%km +150 em = 1890 Rn = /89%0n wi = to — eae — hr 086, wi (em: e te = ai s 62! — XJO Jo Bot GR Fam Eq (DB): Cm "= aci ano). feog, mk = Gretmto *Y6.3300%m)” Yen = 1604 Ms — PO = 152% mk as a = 398 x mA a É sora sa = 4/0735 Thus: Vere = di “ê, os = [pormeçase xo) Gl cxeiitulmo for L/R amd (luis faro) into Bp.( E) Let p= 1740 bm + 20 bn paso bm Ler Io m cosbg = LOBERITIIIIS o, 19673, 6, = 104 Substitotme fr sin tB), we Find 110755 +1 . er = A ADE = (BOB, = IBO-IOLS Aob= 75.5" «dl Mana dE ET 16223 Ph (CoNTINVED) ê * 68  But, from Em. (0:35), ai sn Thus 4 Ate O esco (60) = 0,8836446 M md SM. EM ' A” DBBBrAGI!. DBB3ESA Substituto into (4) od molhynby img rumera for amd dlecomineta by ABBA) ave recallra Hat Eg= 130": aber (talo) - f- à 8836 o, sete já = + (I-DBB3e)cos 135! = 0.92548 asso = Tha ! $-L = óleo) a Ela) = tycto AF B) Fromsolotim 2E rob, 12.100 and Eg (3) E = 390 -15 , 300+15 Second transfer grbit: 4 = 15 xtolem, 2,< tg = 3008/0Am &= 0.905 3 = 9x107f mm, €, = tg = 300% bbw 00-92. s,= O qu2 2" 30049 2. 113 From eg. (12,39); 123 |(a) teus Eq. (12.393; - [77 8a = 180% SU = tã0 Ê = SR +Cuso () + (tt cost) A: B=l80, t=4g Le Mc Mós tsbi 1 EM fi42) c) to bo At a Das, o o oc Dólares: Ze SE ( :) (o) Divite (1) by (1), member by member: or Ceostg a 4 — J4Ê bo fa Dividing (O) by (O, membar by member; t-LE = ty ttoE qa t= gire » cos E CR () = Rc 3 «a Za Mehr Actpva of, sine ap = Ha, Mofez This eguatioo may also be obieined by realiza a A from gesme try tal g = fa, where Ás ta los) À =08E% e, Cm £= semi-foced distance = Fltato) -« , EC oqane R Cogasa — DBBIG , 0.663 (3404200) = 39730 IZ.Hy AMilode 0E B =39750 mi-3960mic /30m. IZ 112 orbit passes farvogh fue center af fire q, Comparing Es: Care (2), we conclude tuto ar apogee or perigee ywe have «brtfisà «a É 2 (+ E) (2.£,9) br apomer or periger. the nosmal do fre | from Ep (1239), A4t- 26M b Ho A 44% = 2 %%, = b* 4 Eb & vara” &m See solution of Prob. 12,96 for derivatim of We mas alo weite thie equation às t M Thus, af each of fhes povis we have tora Ze E.=F = Elm det ” E Recalirg fo Eq (1246) Mot Also, A = é and from Ep, (12H47) that A Medos seed law: ve ter Za - 26M or L=na,: EM mm E o 4º o. 2 o From pu (ts): b - e DI DGM EM “a T= dio 5 Tp LO Lu ae gr. GTA 2 AITAbO We recoii from Prob. 12.97 tha AP EMb/a La lo. Z6M | a x = 28m (2) For ol planeis, Glihas fe same valve, since M represents the mass of the sum ond Gis à imiversal constant, Them toe z 3 Cm a (2z0) 4 69  | We recall Equltor 39) 12.117 CONTINVED += sa (142000) for O= o, 4=4, . 1. EM =2 q %” E fo am Te 2 Por B=/80º, tab: . , 1 EM (ico) or 4,-ÉD (3) 4 E em fã Recalling the definition ef “a given im Eq (124) and vsing (and (2)! Bor goz2L-Lcos3o'=2(h2)-bicosdo'= 4361m Thos! h5e/-L(48)t'=0 E=0, 5275 Pegjechon on horizontal floor (uniform motion) XSM+(Wpt=Lsinso+O, X=06Mm 2=8, +(Wytr D+ L 843 (0.52) = 047 fra” Radius ofeircle; à actos) (nan) Astigio «df Nofe: The drop travel in a ver bio! plane parade! to me ye plane. el - ALAN RE Int Ce lo ae zen dio) LA Bei aczês o to * em r em Tê Solviap for B: 4 - VERA) -« (B,E.D) 12.116] Goiag vp: Car io ataticeguiliprivea, Cinco Spredl 1 con portas HH SP wWp +ZE- o! Pwsind- o P- Wen da Wtanbz002W (1) nas Going ctomm : — Ss a Efoma “Prusia de ma But sind x fan b = 0,03 “pros W= ma é no Substituto fer P From (1): O.02W+ 0,D3W = ma as aos É = 005g=0,05(% 8! mt) 0,88 ma mad I24L7 fest velocity of drop = velocity of bucket p=" ZH =0: Teszoemy (1) U £EE= ma; Teinto'ema, (2) = Tão DividetDMD.o tom 30 = Su = PA 4 3 4 Thus; V*=8 8 tando Bot god sinta (hmsja3)=0, 6m Thus: vi=06(AB)tando'= 3398 W=h843 mf Assumiag Fhe buktef fo rotate eloekwice (when riewed from abote) And vans the axes Shown, we Final that the componentá af Fe initral velocih of he drop are (= 0, (4) = 0 (05) = 43h E Eme fat ot rop sus ast-pstto grega ds When drop str'kes fosr: 4=00 Yo tg t'=0 CCONTINÓED) E” [2.118] From solotim af Prob. Ih 157 we Sontlude Fhat a,=-Hbcos 8 a age Absinto a) Ez ma, = -4mbb; Jos & = ma, =-umbô, snô Gi Cie) — 0.25 = =/2 radfe ven Dako. ! qm = GM) 2 025, po lê pro tt, ê Hj Fec(EN faso «4 E (671 b)sirê 46 At (RE peer, -=- “ (GaSKiDUZV sine, (b) Project alog Bor E Gasb Eri copQc Q-ETbI2o Project alog perpendicular h BO: = PrOsna P=5 - sinto bNsing- 6Ttsing P-0 4 We note that Par, Tárt is becavie Folk are reter lar compo meris of the force Fexerted on pin B pmhile Gamd Pare obligue com = E, gecr Cemacic . The Same resulk conlé hum obtamed ba recallina from fhe solohim ot Fob. HIST fat He ameleratin à of pin DS x directed tomará A mi has me mag arado = bb = a (15 meta radh)'o BU tejor Thus the fuce Feperkim Bis also divected toward A and hos te Naide imaz bi é (864 ]= eu db LÊ folow “ar j E, 67 b)cosb Ga (67 4)sinô amd alo tah D=F-67lhy26 and P=O MODO ODODOOCDCD00O0CD000000D0 7000000 C0000000000004  CORE EC E TE IZ6TAr A; 4,23 9kbmi + 170 Mi = 41300 12.127] Black B PI Crer) = 2h 8064 xr vo mas 6%. ir E =32,6 x DF p Ha g>0%4 ” o , f = 3 dn MN = Mo Wa At, p= (28064 ao intao mt = Rm q “EO so(6t UBimp) - =8 - e) Lte emation decised in Prob. 12.96: . Fem m=rn655 N (9 pe se or de ALiema: E-Pomas ) 4 Fo 22 HO xs2 8)" Bracker A P= EM (2 Zracevxim * 2 (rraBgxnt)r CO : $ ss nixio es Bsexio!= gas? 2? = =10y t= tonHgx nt = 19,222 mi Max. altitude = t-P= 0322-3960 =15, 2600 4 ) E EE=mas -E UM, ) Om oh: (tor woxibEE) = 0. B no! Ah > ama er jm E Cada e G= 70065 KO Éh Te TODA A Subshitute for E Fm (2) into O): 4 o is = í x f % 2 MAD E = Ma Cy Mp *2Ag às = Fo Sobstioby for En fam!) avo fre Mp ant ma, and obterviny that Gg =p, Sine block dos nob sirde, log, +2(6)ag =I7658 ascág= o, 20264 2 =000mbi, «df From (2): Po esg-eto, cozeu) * =/2,84N P-izBino 4 0000000000000000000000000000000000000000000000006. 73  Dosem tech ASSUME Bom ELOCI AM WED Prever. Ler g = coprpjemenT of Pero RT Bl SVEpeS =30* Kiregpap Tres lose atso Shmas Progrim (24) q kimezies Glock 8 (pssume seram o E = EQ Wa E E Mag: My Ma tos O =— Mg Ap SS m= hgcas O - ds a, she de= [agia 7 e] S-30 aa, Dea mr iba, cast ER CREA AA 2< Geo a 3 “ns WEDHE br M FEas, ST E= ma, o MEDE mo 0 egos Va Ra o 2 A ea = fá dada + == pay ot Ma A COS O cepM, SME p= No Mas tos O gain O,+VAEO Sutistirere Mo From 6) vovo (B: NIDA e e55O capNlcos O tapsim O Jr Mg = Ação M [a 8 -Zgess «sftsin O Tras = dica LET BRtcre7 = [mo “Speos > si6] Pres mus mero A, [BRACET] > gh a aa suesprme M reomD) tri (O: [higes O - Sean o Deraceer] gm Em ngcos e [5 enceer] mm 5% + asno feznceer] 5 t KG] +Pa EF = tmado: “We DO LOM, AA E Ra ga O a 2) & Setur For aq: Wa cos O [beaceer) -s Vá - o ei A * o vw Rr e [Bene 7º < a + Wa são [etneie va seesrrore 4 reomD) nro): > In Eq emo +Hg DO = gfmg css O - Gê a, sir O) Ovas ex “ag: Aga” du(cos E te si S+a (= O caes 8) &g a JF Ag <O, THEM Wecse Doês No7T Move Am < From WITH RA 50, Us Hr ves Zap” gem O cor 8) < E ATO, METRER WELOE NOR (Loc MOVES: NO MO Try < SEL MEX7 AEE EO Prestar aro leurs 7% DO000000000000000000000000000000000000000000000004%  3 16" e Caem - no tor PROBLEM je,c1 + fue PROBLEM 12,C1 . ao te BASIC VERSION no[t FORTRAN VERSION . * “6 rent a ese (|O Ea E o Definition of Symbols e Definition of Syatols ” E E WA = weight of edge (pounds) c MA = meight of medos (Iba) Ea ME» weight of block (poumds) e WD = mesght of biprk (Ibs) 166 * MU = cosfficient of friction at all surface c MU = comfficiant of friction at al) surtaces ue + e 129 * tfccelerationa ara expressed in tt/9"2) e (hecelorations are mypresaed ín ft/sm2) az » BA » acceleration of Hedos , ponitáve to lhe right c A » acceleration Dt wadge A, positive to the right 149 + AGA = accuieration of block-uith respact to edge, c ABA = acceleration Df biGck with respect to wedge, 15 + directed dom along the top aurfara of the madgo je direction dom along tha top surfaco Of the wecge so + 8 « acceleration of gravity e 6 = acemleration of gravity 176 * c seg * TH º angla formed by medoe and horizontal (38 dagreme: c TH - angle forned by medor and horizontal (28 degrees) 199 * BRACKET = function of NU ahoxn in derivation E BRACKET = Function 6f MU shown in derivation ea + e BIS K= avaTerI)/ISA [a ag » e eo Ma= 3 we I2 TH = 20 aa ABA! tese Cr/mcarsPRINT E = 4 MATANCI VIDE. EBB FORM =0 401 STEP.81 e E BRACKET = ISINCKATHO «PAMUECOS(KATH) —MUNDASINCKATH) 5 URITE des? trsBRiávt) + MU aa amar E AA = CUMBACOSIKOTHIABRACKET — Misa) 7 MRITE Les Ms BRSANO + Fte rtrmmpr tt + MBESINCKOTHISBRECKET) 526 MRITE (ao? (2484) aa ABA = AMSLCOSLKATHO + MUASINCKETHO) + GrCBINIKATA! e = MUNDOS UKeTHD) my = 0. ES IF GA < 8 GnTO 380 STE = 9.81 38 PRINT UBING “44 t44 “or AA, ABA iD BRACKET = (SINÍK&TH) - PiEMUSDOSLKATH) — MUntDaSINCKETHIS 48 MET RA = ((HESCOSCKKTHIHBRACKE] -NUSHA) *CHAeBRGINCKATH) BRACKET) Me des PRINT PRINT" (Hedge does not move)"iha = 5 BA = AROCCOS(KATH? CPLBSENCKATH) + GSAGINAKATHI-HUSCOS (KATE) nas + IF (AA.L7-8.) GOTO 526 a78 *Detersina value pf Mi at smpllest 6.1 interval mich is BRITE (4,28) M9,00,0BA E “greater that the last valua of MI at ahich meiga soves. 29 FORMAT (9X4F6,9,10X,F6.3, 16%, Fá.3) SPP MI = CINTIMMIO) + 11/10 MU = MU + STEP Bo FORM = MU TDI STEP IF CMU.LT.1.3 GOTO 10 “a ABR = BrfSINIKATH) — MUSCDSCKATAD) e «eo IF ABA < 9 BOTO 444 220 MRITE (6! (fo SMA?) Piledom does not move)” “se PRINT USING “4.444 “PMI AA ARA MRITE (44747347) mo NEXT mag. e c aa PRINTS PRINT * tNo Motion)"; AÉ = 4 ABA = & e Determine value of MU at saallest 8.1 interval mich às “78 PRINT USING "Whieho “SMU,AA ABA z nrester than the last value of MU for which medgo moved. BU = CINTIMIMIG,5+1.)/10, E RUN STEP = 6.1 nu BA 28 ABA = GA(SINHKETH) — MUCOSLKATIO teres rerssg IE (ABA.LT-9.) GOTO 450 WRITE (4,40) MU, AA, ABA 5.8 28,49% 48 FORMAT (3XF6.3,20X.F6,3,0%,F6.3) testa 19.muz MO = MU + STEP “158 19.185 TF AMU.LT.1.) goTa as Bare 18.519 c E 17.846 “38 URITE (44! (7 LBKAN' “(No Motion)? arma 12.165 URITE (e, (7407) ess . a =. 1.887 ADA = 1.385 WRITE (4,50) MU,5A,ADA 9.638 SO FORMAT (3h,F&.3, 19X, F4.3, 40X, F6,9) 2.347 END (Hedge údas rot move) tu SA ABA 0.249 D.gs2 19.523 trema ftisene 2.908 0.080 7.784 9.485 s.20 a.94% o em Bea .. . . “ a 19.842 soe v.508 2.1s7 -s20 4.159 19.185 co no -eaa a.a94 15.519 9.480 ese gos «848 3.200 17.846 «eso 2.748 17.165 E) 2.Bão 16.475 «sro 1.807 15.977 .eeo 1.325 15.970 896 838 14.384 «186 7 +a,629 tsmage doca not nove) 7) mo 19.523 RE 7) 7.784 “ssa «Bop ad «som «oo ERC (No Motion) Rr) «898 «589 75  PROBLEM I2.€CR ComTinupreo OE RUN or Base Frog D = TANCITH + PHIDMK) = (CRROMEGA"2)/6) 6SINETHEO Theta e ” 12 Record range(s) of the value of theta for which O = & in ordor to procead, enter any digit. 3 O = TANCETH + PHTDMEI — ECREDMEGASE) 5) NSINCTHSIO Lower tanit of range O Upper limit of range 16 Motion impends for theta = 8.197 degrees. <] Do you hava another range of theta to investigate for a zero of 0 = TANLETH + PAIN) — ((REOMEGAND) /B)nSINETHA 1 VES enter 1, if NO enter O. 1 Lower limit of range 54 dpper limit of range 69 Motion inpends for thata = 52.124 dagroas. <q Do you have another range Of theta to investigate for a zero of 0 = TANCITH + PHITSKO = ((RUOMEGADD:/GHSINCTHHO. Tf VES enter 1, if NO enter 8. 1 Lomer limit 0f range 176 Upper ligit of conçe 188 Motion impede for theta = 176.357 degrees. aq Do you have another range of theta to ênvestiçate for a zero of UE TONLCTH + PHIIMO = CURADMEGA E) /BISEINCTIMO Ff YES enter 1, if NO enter 8. 5 % CONPNEUATON OF UM DE er ram Placa = TANCCTH — PHLIMO — LIRGDNEGARAD) /G)ASINETHECI Lower Liait of range = 80 Uppor 1init of range = 39 84.731 degross <q Do you have another range of theta ta investigate for à zero of O = TAMECTH — PHIHAK) = (CREOMEBAMS! /EHASINLTHEKO Motion inpends for theta = if VES entar 1, 17 NDenter 0. O O = TONHITH + PHIJMK) - ((RAOMEBAMAD)/2)ASINATHRIO asa Record rangeis) 61 valus of theta for mach G= & In order to proceed, enter any digit. 3 O = TANCCTA + PHDMO - CURSIMEGARED)/BHASINCTHSO Lower liait of range = & Upper limit of range = 2 B.197 degrema <] Do vou have another range of theta to investigate Tor a zero of Ee TANCITH + PAINSK) — CCRADMEGDAAB) /G)ASIRCTHEK) Motion impends for theta = If VES enter 1, if NO enter. & Lomer limit of ranga = 59 Upper iámit of range = 68 Hotson impends for tnota = 52.124 degrees < do you nave arpther range of teta to investigate Tor à zero of 2 TANCCTA + PHIISO — CCREDMEGARD) /BIASINCTHRK] If VES entar 1, if NO enter D. à Lower limit of range = 170 Upper limit of range = 1Bb 176.805 degrees Motion impende for cheta = Do you have another range of theta to investigate for a zero of Q= TANCITH + PADSEI - CORMDMEBADAS) /G)ASINCTHEO 1f YES enter 1, if MDenter 0. WE mave Foro Trar MOTO IMAPENOS AT: S=82" E E=S21º . e-8%7" a 6 =/m8º a From frsuer ce Fog 125% me Em TRAT FOR Mfs =D, COLAR US 18 E Quarg erora AT So, Cro po Eve" Õ (raso vácves Com GE cnscunTeo E Remine THE Suapary D = LOU Aguas bia fe =D = eps Sine TUE VaLvES Ena MEO SIDET PREL torno THE Besreo CAGE, iu puras fios or E rr er 84” pum = 4% “o IH PR ria ré ogoca? 4 SEE) SIP LO E sus <CO EO BBOVE SRCERAS zm zo) v8  o Fheogresm 12 €2 Block EesT SEMSE FoR roça (szé pos fita) Ad av= a, dé ; turem Lidos Do TRA dAs= vurefga o (ore danat Ta H «qr a o OEA? A For E >= RO, Erock serps ReLarne TO THE Gecr ro coptacamo DE FRiCIOM 15 a = Mg = 0184 tWy=ma meras! o E Fun o Me NG + =» E YZF= oa, macaé -M = Pa, WD? o mlgemo- E PG SAS BN = o Ra aD NAN EF= ma, s az -gere -Suquso- z < ; Ms as gere AE s 1º Mrradt= (9 são -gueaso + pv )ar ASg= 45/r «1 (NCEtptrT AT srers: Af=000/s, o-s+a8 4 CALemare AT Enem s77p Value of A vsme(D) A Lock pull LEAVE BEL? oi NE O, 157 e T ee reais arena Bo tm PROBLEM 12.C3 cjce PROBLEM 12.03 « 3 re BASIC VERSION + los FORTRAN VERSION . ag» ss eras eesas | os au eremasa cemaasenanes se» c “o Definition of Symbol e Definition of Syabols E c so TH = angie theta (degrees) c TH = angle lheta (degrees) o HU = coefficient of kinetic friction ç My = coefficient of kinetic friction 100 + V = volocity DF nait (Ft/9) r V = velocity of belt cft/ay ne R = Radius uf pulley (rot? e Ra radiys of pulley (feet) 126 + DT = Intervai ef tine for integration (seconds? E DT = interval Df tioe for integration (seconds) 130 " r E 146 K = GmTucDAIDA REAL HU, ey N 150 ma as My = 9.55 lg vas ves 10 R=1812 Ra 19.12. 189 DT = 68 DT = 9.8 190 iego + Angle at «bich package First slips with respect to belt: Ê fmgle tneto at which package first Slips with respect to belt; ais To = 9.01 idegrees) See Prod. 1a.hg, ma = 9.81 E e tiegrees Soo Prob. 12.48 e teTg G= ses c ma + K = 4 *ATANCI.I/180, 2 MILEN O Ta tuo aso DE = (BNSINHKATIS - GANCOSLKETHO + Myey-2/R)aDT | G=m22 E v=veDy c 206 DS = «+ DUaaDT N=6. 26 DTM = DS/R *incregent of angle theta in radians 240 LE (NLGE,9,) THEN 209 TH = TH + <DTHS/K DV = (GRSINCKATA) = GAMUACOS!KATH) + MUMVMAZ/RIADT E N= CGACOB(KaTD — Wr2zm) va vem ss vem DS + (4+Dw/e.1aDT aa DTH = D5/R 38 PRINT “For given incresent of toa: DT = p.89i seconds”iPRINT Le Increnent of argle in radians 350 PRINT “package leaves belt aheni” Tua TH + tDTNIZK 368 PRINT USING theta = 44.46 degreses;TH N = (GULOSUeTH) — vesz/R) ao PRINT 886 PRINT “orresponding. speed of packagest soro 338 MO PRINT USING * vm ag.ema rtrossy enDir GOTO esa E t = 339 WRITE (4,7 C/SXsBAND! A For given increment of time 1', For gives increent of tra: DT = 9.86) seconds à a, Package leaves belt vieni SEITE CAL ISKAVSS “Package Imavem Beit uihen 1 thata = 22.64 degrees “0 FORMAT (7420, Theta = ",F5.2,º degrees”) Eorresponding apesd cf package: WRITE (44 "(/,SH5ASH'S *Corresponding aposd of package :” EL 4.78 this WRITE tas59) y 3a FORMAT (Shy = 2,F5,0,) 16/87) END For given increment of time à DT = 8.4M second Package leaves belt ven : Theto = 23.8 degrees Correspanding speas of package + vo = 4,784 this 792  Fomiura L2C4 FRBe FLISHT OM S/MCECRAFT STARTS AT Re db 15 FReAUEL TO SURANcE Of ESTA, MLTITULE AT Tits de 15 REU E Da Doszanco Freods CENTER OF ERRTH 5 vr=R td pn to Luar of Pora 45 LL. EM fora): q GA tCea8 or, r= ss +cese) < umemelro ia sm: h=6% a pro fez ua): Cx E - a aq VIME OF RE del SEM? Lele: RES ah Sove dor dé: a(o on dé= 748 WE som DÊ To Fio TIME ECRPSED Friorm LRUNMNS NOTE: 4 45 EmTrRSO mi Pop Ano CONVERTED TO nfs 1 LAH CE THE Factos FEL ema u arena era PROBLEM 1é 6 - | o ROSLEM 18.Ck + FASIC VERSION + jo FORTRAM VERSTOM * nesareessa sai rea Correm € definition of Svatols E vefimtion nf Syabole c emqlo thata (degrees! E TH = angle theta (degrees) radius of the earth (ka) e RE = radius of tne earth Ckm) distance from center of earth to spacecraft (im) E & = distance from center of earth to spacecrart (km) altitude of cpareerart cko) c D = altitude of spacecraft (kn) altitute of pacecraft at launch (km) z DO = altitude of the spocecraft at launch (ka velocity of launen (km/h) e VB = velocity of apacecraft at launch (ka/h) distance from center of earth to apacecraft E RB - distance fria contar 6f earth to spacecrart at launch Ika) c a Jaunching cka) c Om > (mass ot parthiacconstant af gravitatiom c OM = (maus of carth)xiconstant of gravitationh = es Ipe (nrazera) c = 990 + IOasi2 (nstB/54kD) t N = counter E N = counter NBR = number of incresents in a printina interval c NBR = muaber 0f increments in a printima anterval c DATA REAL H+ Es N E DATA INPUT "Enter LAUNCH velocity VD expressad ih ken. “ova e INPUT "Enter largest valuo of theta tin deqraas) “sTHMAISPRINT MAITE (as! 4BL4AA)!) “Enter Launch velocity VP expressas in kaíh IF Vê <> 38759! BOTO 256 READ test VE PRINT USING 2PARABOLIO TRAJECTORVE vB - MH km/h" ;VB:GOTO Sao WRITE 19, (2X,AN7) “Enter lorgoet value of theta (in degrees) * IF V6 > S87SE! GOTO a1g READ (9,8) TRMAL PRINT USING "ELLIPTI ORBIT: v2 » 20924 ka/hº] VIIGOTO 22m WRITE (ea) Tr PRINT USING “HYPERBOLIC TRAJECTORY: vZ = SMA Ko/h"; va e 1F (v6.NE.38756) BOTO PO 16 FORMAT (BX,'PARABULEC YRATECTORV: VB a *,F6.8º kafh?) 28 IF (W0.67.98750) GDTO 4d WRITE (2,20) 48 ae FORMAT (2X, ELLIPTIC DRDIT: VO = ",F6.6 knfh') Soro 68 “8 WRITE (e,50) va SA FORMAT (2%, "HYPERBOLIC TRAJECTORY: VÊ» 'oF6dr kmfn'o Contras CoNDA ES