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OOGgccqpxsssssssppxxx|<<~?>~||xxxxppp```@@@`?~|x|~|||||||p@@p|~><|xp`@>|x|xpx}yyyyyyyAA?~|xpp`ppx|~? ?? 33@``@@glloLL LlOgloloG @`@glolo@@gLLL GaamGGlg`GGlg`G  xL| x@@xMLLxMyyM MyxAxMx|Yq @lllLglllgG LlL`@GLOLo033366m~0>3>33~G<`G}rccb``acccq{{{{8008????~ |lL COoooog CgoooocCFFLLXXPpqa88`}}}}>????~ | ~FCCFL ____|GG|GGccp{{{{ ;1q``qp`aas{{{{8008 0y| |}}}}AAQg((h' $C@0j ed`y < I1@@q q s8EE9EE8> ` `$DC0ANPQN@'h(' WPG((H@stq q8AyEx:NPQ@GhHHG#$p# @ (*Y9AyEy: QQQO~C%%C~_DDD qAAAaq9`ppxxx|xxx|AsGN8088 CcˆʎõĵL õ ĵµ aµ`` L̦µ_bJLuLz`  ȟ QlXJ̥KlV  ȟ QlV eօ3L e3L &RL &QL d L4 Ne)n `@-eff L f`L . tQLѤ LҦL` OPu d L Ne)noon 8ɍ` ^f\õL ^NR  RΩLҦ)\Z ʽ LHv 3h`0h8` [L NС õ`A@` ŵL^L iõ`  \ 濭0 \  ȟ Q ^\lZl^?cqH şch`fhjõĵ@OAP`u@`@&`QR`E Ls  @DAE@u`8` %@ @A@`@`@A`Mµ ) LЦ`8@AWc@8@-@HAȑ@hHȑ@ȑ@hHȑ@Ȋ@ch8&ȑ@Hȑ@Ah@LHȑ@ȑ@ htphso`hMhL`9V8U897T6S67`INILOASAVRUCHAIDELETLOCUNLOCCLOSREAEXEWRITPOSITIOOPEAPPENRENAMCATALOMONOMOPRINMAXFILEFINBSAVBLOABRUVERIF!pppp p p p p`" t""#x"p0p@p@@@p@!y q q p@  LANGUAGE NOT AVAILABLRANGE ERROWRITE PROTECTEEND OF DATFILE NOT FOUNVOLUME MISMATCI/O ERRODISK FULFILE LOCKESYNTAX ERRONO BUFFERS AVAILABLFILE TYPE MISMATCPROGRAM TOO LARGNOT DIRECT COMMANč$3>L[dmx- ( t Ϡ@跻~!Wo*9~~~~ɬƬ~_ j ʪHɪH`Lc (L ܫ㵮赎 ɱ^_ J QL_Ls贩紎 DǴҵԵƴѵӵµȴ 7 ַ :ŵƴѵǴҵȴµ納贍﵎ٵ്ᵭⳍڵL^ѵ-I `  4 ò-յ!  8صٵ紭ﵝ 7L (0+BC  7L HH`LgL{0 HH` õL H hBL BH [ h`Lo õ ڬL B ڬ LʬH hB@ յյ [L (ȴ) ȴ 7L L ( L (ȴL{ƴѵ洩ƴǴҵ 7 ^* B0 HȱBh ӵԵ 8 L8 ݲ` ܫ  / / ED B / / ]ƴS0Jȴ ȴ)  紅D贅E B ƴ  / 0L Ν `HD٤DEEhiHLGh ` ŵBѵ-` ѵB-` ܫ XI볩쳢8 DH E𳈈췍Ȍ X0 ĪLR E( 8` R` ELRŪƪ`췌 յյI뷭鷭귭ⵍ㵍跬ª 뷰` Lf ݵܵߵ޵ ^`8ܵ i B8` 4L ֵȱB׵ ܯ䵍൭嵍 ` DȑB׵Bֵ  ַ յյ`굎뵎쵬 뵎쵌``õĵBCõĵ`µµ`L õBĵCصص Qƴ0"Bƴ 󮜳` 0۰ϬBƴ8`i#`ЗLw!0>ﵭ` m ﳐ 7i볍 8 ЉLw`H h ݲL~ `浍국䵍뵩嵠Jm赍嵊mjnnn浈m浍浭m䵍䵐`"L ŵ8ŵH ~(` d ֠z# u`SE: Y = V(0)Y*T - 1/2*G*T^2":::10050&::"REMEMBER THAT Y = -78.4 METERS, SINCE"::"THE FINAL POSITION IS BELOW THE START."::&" INPUT THE FLIGHT TIME: ";A::A41540':3:(7):"SORRY, ";Z$;".":::" Y = V(0)Y*T - 1/2*G*T^2"::"BECAUSE THE BALL IS KICKED HORIZONTALLY.":T%" ENTER V(0)Y: ";A::A01460b%"SURE!";%25::"V(0)Y = 0 M/S":::"SINCE THE BALL HAS NO INITIAL VERTICAL"::"VELOCITY. ";:10050*&:4:"NOW FIND THE TIME OF FLIGHT--":::"U$n::2:"NOW FOR THE SOLUTION. YOU SHOULD BE ABLE"$x"TO GET THIS ONE ON YOUR OWN.":::"START BY FINDING THE X & Y COMPONENTS":$"OF THE INITIAL VELOCITY."::" ENTER V(0)X: ";A::A121430$"OF COURSE!";.%25::"V(0)X = 12 M/S"::13X,11:#2Y05.5A#<9:162Y,11Y2:0:162Y,11Y2:a#F(7):9:26,36:T11200:#P22:" TIME MARKERS SHOW THE MOTION."#Z9:14,11:Y05:162Y,11Y2::T11000:#d24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A112805OCITY IS":]""12 METERS/SECOND?":::"COPY THE PROBLEM. THEN WE'LL LOOK AT IT."::10050"::9000:3:X515:12,36X::9:13,11" 22:" PRESS TO KICK THE BALL.";A$":T1800::(7)"X02:9:13X,11 #(T150:T:0:NNON TO SHOOT THAT"::"HIGH.":22:10050T!:3:"HOW ABOUT ANOTHER PROBLEM?"::!"A MAN KICKS A BALL HORIZONTALLY OFF THE"::"ROOF OF A BUILDING 78.4 METERS TALL.": ""HOW FAR FROM THE BASE OF THE BUILDING"::"DOES IT LAND IF THE INITIAL VEL (173.2)^2 - 2*9.8*Y"::" Y = (173.2)^2/(2)(9.8)":j "AND,":9::"Y = 1530.5 METERS"::: "IT'S A BIG CANNON!":::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A1910 1240)!:10:"EXCELLENT WORK, ";Z$;"!"::"IT'S A PRETTY BIG CAON HAS NO EFFECT ON HOW":i"HIGH IT GOES.":::"INPUT YOUR MAX. HEIGHT: ";A::A1530A15311230:4:(7):"SORRY. HERE'S THE CORRECT SOLUTION:":"AT MAXIMUM HEIGHT, V(Y) = 0. SO,"::" V(Y)^2 = V(0)Y^2 - 2*G*Y":< " 0^2 =EVIEW THIS":Yt"PROBLEM SOLUTION AGAIN, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A1910~::"AS A CONTINUATION OF THIS PROBLEM, FIND"::"THE MAXIMUM HEIGHT OF THE SHELL."::10050:3:"HINT: THE FACT THAT THE SHELL MOVES IN"::"THE X DIRECTI= 173.2/4.9 = 35.35 SEC":::kB"* NOW YOU CALCULATE THE RANGE."::" INPUT YOUR ANSWER (NO UNITS):";AL:5:A35351120V"GOOD!";`10::"RANGE = V(X)*T = 3535 METERS"::: j"THAT'S NOT SO BAD, IS IT?"::::"IF YOU WOULD LIKE TO RND,":"SO: Y = Y(0) = 0 METERS ";:10050y:3:"USE THIS EQUATION:"::" Y = V(0)Y*T - 1/2*G*T^2"::10050$:"THEN,":" 0 = 173.2*T - 1/2*9.8*T^2":."AND":" -173.2*T = -1/2*9.8*T^2"::100508:"FINALLY,"::6::"T 73.21010"YES!";I9::"V(0)Y = V(0)*SIN(60) = 173.2 M/S":::"NOW USE THE VALUE OF V(0)Y AND THE"::"EQUATIONS OF MOTION FOR GRAVITATIONAL":"ACCELERATION TO FIND THE TIME OF FLIGHT.":/:"REMEMBER THAT THE SHELL HITS THE GROU:"INPUT YOUR ANSWER FOR V(0)X: ";A::A100970A"GOOD!";x10::"V(0)X = V(0)*COS(60) = 100 M/S":::10050:3:"REMEMBER THAT THIS IS THE VALUE OF THE"::"CONSTANT HORIZONTAL VELOCITY.":: "INPUT YOUR VALUE FOR V(0)Y: ";A::A1CANNON SHELL IS FIRED WITH A MUZZLE":}"VELOCITY OF 200 M/S AT AN ANGLE OF"::"60 DEGREES ABOVE HORIZONTAL. FIND THE":"RANGE OF THE SHELL. ";:10050::"COPY THE PROBLEM. THEN TRY TO FIND THE"::"X & Y COMPONENTS OF V(0).":3 THE GROUND IS X(0) = 0, Y(0) = 0.":|z"> THE STARTING TIME IS T(0) = 0."::"> THE ANGLE (THETA) IS MEASURED FROM ":" THE X-AXIS.":::"COPY THESE CONVENTIONS IF YOU NEED TO."::"THEN, ";:10050&:3:"NOW HERE'S A PROBLEM:":::"A METHOD FOR REFERENCE: ";:10050\:2:"THE CONVENTIONS WE USE ARE THE SAME AS"::"THOSE ESTABLISHED IN PREVIOUS PROGRAMS:"::f"> UP AND RIGHT ARE POSITIVE."::"> DOWN AND LEFT ARE NEGATIVE.":'p"> THE STARTING POSITION (EVEN IF ABOVE"::" 7,13:<%780,75:888,75:996,75:10104,75:2112,75:J':100:8t'" ************************":$'" CIRCULAR MOTION - CONSTANT SPEED".':" ************************"8'255:D1900:::(7):B'"";A$:3.143.14.1:13069.6(T),8060(T)::>#1:0:T3.14z#712875.5(T),8365(T):712875.5(T),8365(T)$TT.2:T14.859240$9210$7186,41:T$1175,21:5185,21:3191,21:6198,21:$182,40151,13151,18:151,1315NSTANT SPEED.":`"YOU SHOULD NOW WATCH:":::3::"CIRCULAR MOTION - CHANGING SPEED":::"SEE YOU LATER, ";Z$;"!":24:10050:D13:(7)::20:"END"D11500:::(4);"RUN MENU"'#(#:3:128,80132,80:130,78130,82,2#T60162,60Nx21:"COPY THIS DIAGRAM FOR ROTATIONS":"WITH CONSTANT SPEED.":a10050:::10"IF YOU WOULD LIKE TO REVIEW, TYPE 1."::"TO CONTINUE, TYPE 2: ";A:A1290:7"THIS CONCLUDES OUR DISCUSSION OF"::"CIRCULAR MOTION WITH COIS NOT CONSTANT"::" - ANGULAR ACCELERATION IS ZERO"::" - TANGENTIAL ACCELERATION IS ZERO"::" - RADIAL ACCELERATION = V^2/R":::10050<9000:9200:9300:9500:2160,71:5169,71:4176,71:6184,71P5:9400 d2:182,41156,60156,55:156,***********":21$'" PROJECTILE MOTION"\1.':" ************************"{18'255:A11000:::(7):1B'"";A$:AROBLEMS FOR PRACTICE. LOOK"::"IN YOUR TEXT OR SEE YOUR INSTRUCTOR.":q0v"GOOD LUCK. BYE FOR NOW!":24:100500:D13:(7)::20:"END"0D11500:::(4);"RUN MENU"0?0(#12:2,3737:2,3738:0':100:81'" *************3 - 1/2*9.8*(1.3)^2"::I/N"SOLVING--"::7::"Y = 3.67 METERS":::/X"THAT ALL THERE IS TO IT! TYPE 1 TO"::"REVIEW, 2 TO CONTINUE: ";A:A11570/b:6:"THAT'S THE END OF THIS PROGRAM."::Z$;", YOU SHOULD PROBABLY WORK":G0l"SOME EXTRA PME: ";A::c.&"REMEMBER THAT X & Y MOTIONS DO NOT"::"AFFECT EACH OTHER, BUT THE FLIGHT TIME":.0"IS THE SAME. SO, T= 1.3 SECONDS."::10050.::3:"THEREFORE--"::" Y = V(0)Y*T - 1/2*G*T^2":/D:"SUBSTITUTING--"::" Y = 9.192*1. 10/7.714":3-5::"T = 1.3 SECONDS":::10050-:4:"HOW DO WE FIND THE ELEVATION WHERE THE"::"ARROW HITS THE CLIFF? ";:10050::-"HINT: HOW MUCH TIME DOES THE ARROW HAVE"::"TO GO UP AND DOWN?": ."INPUT THE ARROW'S FLIGHT TI0):L:֨ؠӨͯӢ::֨٠ΨͯӢ:0BV:::ŠŠנŠŠөĢ::ĠŠ̠٠ͯөt:ŠΠĠŠԠŭ::Ԡů֨ة9,20;+T11400::22:" TIME MARKERS SHOW THE MOTION."]+Y501:7:203Y,11Y2:|+Y03:7:203Y,11Y2:+T11000::24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11620,::2:"HERE'S THE SOLUTION:":::" FIND V(0)X & V(24:100507*T::9000:1:X3035:14,36X::7:5,36*^22:" PRESS TO SHOOT THE ARROW.";A$:T11200:::(7)*hY50.5*r7:203Y,11Y2:0:203Y,11Y2:*|YO3.5*7:203Y,11Y2:0:203Y,11Y2:+(7):7:2REES ABOVE THE"::"HORIZONTAL AND FIRED WITH AN INITIAL":)6"VELOCITY OF 12 M/S TOWARD A VERTICAL"::"CLIFF 10 METERS AWAY. HOW HIGH ABOVE":)@"THE GROUND DOES THE ARROW STRIKE THE"::"CLIFF?":: *J"COPY THE PROBLEM AND THEN WE'LL WATCH.": THE RANGE--"::" RANGE = V(X)*T":(" RANGE = 12 * 4"::8::"RANGE = 36 METERS"::::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A11240(":3:"LET'S DO ONE MORE PROBLEM, THIS TIME"::"IN REVERSE:"::;),"AN ARROW IS AIMED 50 DEG;'"OR,"::" -78.4 = 0*T - 1/2*9.8*T^2"::10050:b'"THEN,"::" T^2 = 78.4/4.9":'"AND,"::5::"T = 4 SECONDS":::10050:1550':10:"GREAT WORK, ";Z$;"!"::"KEEP GOING...":22:10050*(:56G]ZWQ4@4)4 4GQW[ZPG84CQ4R]ZPD,":M" Tx = T*cos(25) = .906*T"::" Ty = T*sin(25) = .423*T";:1006013376::"THEN:"::"(7.66)(.423*T) + (6.43)(.906*T) -":" (5.36)(200) - (3.83)(150) = 0":::"YOU SOLVE FOR THE VALUE OF 'T'."::10060>13376:Z$;", IF YH THE":" SUPPORT OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">UX115,6745,2151,15:#':100:8M'" ************************":q$'" STATICS - METHOD".':" ************************"8'255:D11000:::(7):B'"";A$:L'31:"";A$:6">5:X5,Y:X5,Y5:X6,Y1:X6,Y6:y2#3:X1,YX3,Y:X4,Y1X4,Y5:X1,Y6X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:<#3:X,Y2X,Y4:X1,Y1X5,Y5:X1,Y5X5,Y1:F#3:X5,Y1X5,YX,YX3,Y3X,Y6X5,Y6X5,Y5:#3:51,15121,61OU WANT TO REVIEW,"::"TYPE 1. IF YOU'RE READY TO END THIS":b>"PROGRAM, TYPE 2: ";A:A1250H13376:D13:(7)::19:"END"RD$;"PR#0":\D11500:::D$;"RUN MENU"'#%(#3:X,Y5:X1,Y6:X2,Y6:X3,Y1X3,Y6:X4,Y:X4,YNOWNS.";:10060:p13376:::"THAT'S IT FOR NOW. YOU SHOULD BE"::"ABLE TO RESOLVE FORCES, FIND RADIUS": "ARMS, AND CALCULATE TORQUES. OTHER"::"PROGRAMS SHOW HOW TO SOLVE COMPLETE":*"PROBLEMS.";:10060>434,0:13376:10:Z$;", IF Y)(Ty)";/X31:Y166:9000:Y182:9000:1006013376::"NOTICE THAT BOTH TORQUES ARE POSITIVE"::"(COUNTER-CLOCKWISE).";:10060::"#11 AND #12 HERE YOU WOULD SET THE SUM"::" OF ALL TORQUES EQUAL TO ZERO AND": " SOLVE FOR ANY UNK80:Y60:9010R13376:"#9 CALCULATE THE RADIUS ARM FOR EACH"::" FORCE:":" Ra = L * sin( )"::" Rb = L * cos( )";X127:Y118:9010:Y134:9010:10060:"#10 CALCULATE THE TORQUES:"::" a = (Ra)(Tx)"::" b = (Rb ROTATION POINT. WE'LL"::" CHOOSE THE HINGE POINT.";:10060{:"#8 DRAW LINES OF ACTION FOR THE FORCES."::100606:118,44118,10:105,6435,64:24:"*** NOW AGAIN: ";A$ 5:62,18116,18:47,2647,63:2:14:"Rb":7:5:"Ra":X*************",8'255:D1900:::(7):AB'"";A$:1:f&"GREAT!"::<&:"ANSWER TRUE OR FALSE:";B$::R&(B$,1)"T"9930n&"ERROR!":(7):GG1:&"GREAT!"::':100:8'" ************************":$'" CIRCULAR MOTION - CHANGING SPEED".':" ***********(T):712875.5(T),8365(T)9$TT.2:T14.859240C$9210S$7186,41:T$1175,21:5185,21:3191,21:6198,21:$182,40151,13151,18:151,13157,13:H&:"ANSWER TRUE OR FALSE:";B$::R&(B$,1)"F"9830\&"ERROR!":(7):GG0,78130,82>2#T3.143.14.1:13069.6(T),8060(T)::T#1:1:T3.14:N0#712875.6(T),8365(T):712875.6(T),8365(T)#TT(.1N):NN.5#T14.859150#9110#7186,41:#1:0:T3.14 #712875.5(T),8365PROGRAM,"::"PLEASE TYPE 1. IF YOU WANT TO CONTINUE,"::"TYPE 2: ";A:A11580 :10:"THAT'S ALL FOR NOW, ";Z$;".":::" BYE!":20:10050 :D13:(7)::20:"END" D11500:::(4);"RUN MENU"'# (#:3:128,80132,80:13OOD. BUT, YOU MIGHT":[ "WANT TO WATCH BOTH CIRCULAR MOTION"::"PROGRAMS AGAIN.":2790 ::"EXCELLENT, ";Z$;"! A PERFECT SCORE."::"IT LOOK LIKE YOU REALLY UNDERSTAND": "THE MATERIAL." 24:10054Pu~NEDNV=ɼȼӼ˼ϼE FIRST FEW SECONDS THE FAN"::"WAS SPEEDING UP? YOU HAD BETTER REVIEW": "BOTH PROGRAMS ON CIRCULAR MOTION. THEN"::"SEE YOUR INSTRUCTOR IF YOU'RE STILL": "HAVING TROUBLE.":2790 :"WELL, ";Z$;", YOU ONLY MISSED ONE."::"THAT'S PRETTY G2, MAKING"::"SUBSTITUTIONS:";:10060^:" T - Ff2 = 0"::" T - u2*M2*g = 0"::"NOTE THE SUBSTITUTION FOR Fn2.";:1006013376:"NOW SOLVE FOR u2:";:10060:::" 75.1 - (u2)(8)(9.8) = 0":+:" u2 = 75.1/78.4 = .9NOWN EXECPT FOR THE"::"TENSION. YOU SOLVE IT!";:1006013376:"HERE IT IS:"::"(10)(9.8)(.866) - T -"::" (.2)(10)(9.8)(.500) = 0"::" T = 84.868 - 9.8 = 75.1 NT":::"IT'S EASY!";:10060&13376:"NOW REWRITE EQUATION OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">700:8/'" ************************":T$'" STATICS - LADDERS"~.':" ************************"8'255:A11000:::(7):B'"";A$:L'31:"";A$:6">H THE":" SUPPORT 1500:::D$;"RUN MENU"'#(#3:X,Y5:X1,Y6:X2,Y6:X3,Y1X3,Y6:X4,Y:X4,Y5:X5,Y:X5,Y5:X6,Y1:X6,Y6:2#3:X1,YX3,Y:X4,Y1X4,Y5:X1,Y6X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:#3:50,6999,9:49,6898,8:':1fܔgeoLS}fddadd~l9Q^e8]}e)ŗeVč^76:8:"IF YOU WANT TO REVIEW THIS PROBLEM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A1240 13376:6:"HERE'S A PRACTICE PROBLEM FOR YOU."::"IF THE COEFFICIENT OF FRICTION IN THE":"LAST PROBLEM HAD BEEN ONLY 0.40, HOW"::"HIGH UP THE LADDER W800 = 996 NT"::"OK?";:10060p:"NOW FOR THE EASIEST PART--FIND THE"::"COEFFICIENT OF FRICTION.";:1006013376:"BY DEFINITION, Ff = (u)(Fn)"::"SO, u = Ff/Fn":" u = 0.49":::"THAT'S IT! YOU'RE DONE!!";:10060Y34,0:133"THE FORCE OF FRICTION, Ff.";:10060_13376:"OF COURSE, Ff = Fw = 491.1 NT"::10060::"NOW USE THE RESULTS OF STEP #6 TO FIND"::"THE NORMAL FORCE, Fn.";:1006013376:"FROM STEP #6,"::" Fn - Wl - W = 0":" Fn = 196 + ">"">">6>>"">"">">">6>>">A>" STATICS - BEAM PROBLEMS"M.':" ************************"l8'255:A11000:::(7):B'"";A$:L'31:"";A$:6">H THE":" SUPPORT OF NATIONAL SC>">666X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:_<#3:X,Y2X,Y4:X1,Y1X5,Y5:X1,Y5X5,Y1:F#3:X5,Y1X5,YX,YX3,Y3X,Y6X5,Y6X5,Y5:#3:51,15121,61115,6745,2151,15:':100:8'" ************************":#$'TER!":22:100506H13376:D13:(7)::19:"END"GRD$;"PR#0":h\D11500:::D$;"RUN MENU"n'#(#3:X,Y5:X1,Y6:X2,Y6:X3,Y1X3,Y6:X4,Y:X4,Y5:X5,Y:X5,Y5:X6,Y1:X6,Y6:'2#3:X1,YX3,Y:X4,Y1X4,Y5:X1,YW THIS PROBLEM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:34,0:A1240*34,0:13376:8:"THAT'S THE END OF THIS PROGRAM."::Z$;", YOU REALLY SHOULD WORK":4"A FEW PROBLEMS ON YOUR OWN. SEE YOUR"::"INSTRUCTOR FOU;&1T$&S PART EASY!";:10060::\"NOW YOU USE THE RESULTS OF STEP #6 TO"::"FIND Hy.";:10060 13376:"HERE IT IS:"::" T*sin(25) + Hy - 150 - 200 = 0":" Hy = 273.23 NT"::"THAT'S IT. WE'RE DONE!";:10060C ::"IF YOU WANT TO REVIEOU FOUND T = 181.5 NT"::"YOUR ALGEBRA IS EXCELLENT!";:10060x::"FROM STEP #5,"::" T * cos(25) - Hx = 0":"YOU SOLVE FOR Hx, THEN ";:24:1005013376:" (181.5)(.906) - Hx = 0"::" Hx = 164.4 NT":"ISN'T THI"::"UNKNOWNS. YOU SOLVE THEM!";:10060i34,0:13376:3:"SOLUTION:"::" (15)(9.8)(.819) - T -":" (.2)(15)(9.8)(.574) = (15)(ax)"::"AND,":" T - (.1)(12)(9.8) = (12)(ax)"::10060:)"THEN,"::" 103.51 - T = (15)(ax)"*M1*g*cos(55) = M1*ax"::"THERE ARE TWO UNKNOWNS.";:1006013376:"NOW REWRITE THE SECOND EQUATION MAKING"::"SUBSTITUTIONS:";:10060:" T - Ff2 = (M2)(ax)"::"OR, T - u2*M2*g = M2*ax":("SO NOW WE HAVE TWO EQUATIONS WITH TWOTHE":" SUPPORT OF NATIONAL SC>">66">"">">>>"6>>>>">>">>"A8X,Y:':100:8?'" ************************":d$'" DYNAMICS - METHOD".':" ************************"8'255:A11000:::(7):B'"";A$:L'31:"";A$:6">H 4,Y1X4,Y5:X1,Y6X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:u<#3:X,Y2X,Y4:X1,Y1X5,Y5:X1,Y5X5,Y1:F#3:X5,Y1X5,YX,YX3,Y3X,Y6X5,Y6X5,Y5:#3:A3.143.14.1:X11.6(A),Y10(A)::#X,YX10,YX10,Y8X,YE YOU IN THE NEXT PROGRAM.":20:10050L 13376:D13:(7)::20:"END"]D$;"PR#0":~ D11500:::D$;"RUN MENU"'#(#3:X,Y5:X1,Y6:X2,Y6:X3,Y1X3,Y6:X4,Y:X4,Y5:X5,Y:X5,Y5:X6,Y1:X6,Y6:=2#3:X1,YX3,Y:XĢġīĥķ^ ^ƳèķĬİģƲƯƠ\^ƴT7]h-F^ƭĽijİĶűŨ_aDZİİħߥI4 F^ư÷ĭĢƫܽ]]Ŵ13376:"#3 CONVENTIONS:"::" WE CHOOSE DOWN FOR M1, LEFT FOR M2,"::" AND COUNTER-CLOCKWISE FOR THE WHEEL":" AS POSITIVE SINCE THAT IS THE WAY"::" THEY'LL MOVE."::10060'13376:"THE REST IS PRETTY EASY. SIMPLY SET UP"::"144,24144,26:114,51116,51:214,59216,59:214,12216,121:31:"Fr":8:"T1":22:"T2":8:16:"T1":5:26:"T2";:35:"Ff":9:7:"Fg1";:28:"Fn = Fg2"13376:"SINCE THE WHEEL DOESN'T TRANSLATE,"::"WE'VE LEFT OFF Fg AND Fr.";:10060z10060>62450:X128:Y35:9100:X48:Y39:9200:X210:92001:53,4053,60:53,3153,16:209,35195,35:221,35233,35:131,25145,25:115,35115,525:215,11215,31:215,40215,60;3:52,5954,59:52,1754,17:196,34196,36:232,34232,36:A">H THE":" SUPPORT OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">"#5:X,YX10,YX10,Y8X,Y8X,Y:2':100:8\'" ************************":$'" STATICS - INCLINED PLANES".':" ************************"8'255:A11000:::(7):B'"";A$:L'31:"";A$:6 LATER ON, ";Z$;"!":20:10050D 13376:D13:(7)::19:"END"U*D$;"PR#0":v4D11500:::D$;" emf} (T3 (3 pqpqpqpqpqpqpqpqpqpq(3 pqpqpqpqpq( 4S AN ADJUSTABLE"::"ANGLE. TO WHAT MAXIMUM ANGLE CAN THE":q"INCLINE BE RAISED BEFORE THE MASS"::"SLIPS?"21:"SOLVE THIS ONE YOURSELF.";:10060 13376:4:"THE ANSWER IS 22 DEGREES.":10:"THAT'S THE END OF THIS PROGRAM.":"SEE YOU58":::"THAT'S IT. YOUR'RE DONE!";:1006034,0:13376:8:"IF YOU WANT TO REVIEW THIS PROBLEM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A126013376:5:"HERE'S ONE LAST PROBLEM:"::"A 25 KG MASS RESTS ON AN INCLINE":<"(u = 0.4) WHICH HAND_NַԣԽԱԠԠNNֵԺֲװNDTMx=VN֠ӧԽԲԠԤϮMMծפԺԣԠԷNNֻԱԤԣ԰ԣLLԵTING THE"::"MASS FROM h = 0 TO h = 1.53 METERS?":Y" ANSWER? ";A:13376:A6800 "SINCE WORK & ENERGY ARE EQUIVALENT, THE"::"AMOUNT OF WORK DONE IS THE AMOUNT OF":"ENERGY GAINED BY THE MASS:"::" W = Ug = 0.4 * 9.8 * 1.53 = 6 JOULESh":D12500:s" OR,":" 6 = 3.92 * h":" AND,":" h = 1.53 METERS":1005013376:"THAT IS EASY!"::"LET'S DO ONE FINAL PROBLEM FOR THIS"::"PROGRAM.";:10060413376:"HOW MUCH WORK IS DONE IN LIF10060o13376:25:"(K + Ug + Us + Ef) = (K + Ug + Us + Ef)"::"(0 + 0 + Us + 0 ) = (0 + Ug + 0 + 0 )":"OR,":" 1/2 * K * x^2 = M * g * h"::255:1005013376:"#5 NOW WE SUBSTITUTE VALUES:"::" 1/2 * 1200 * (.1)^2 = .4 * 9.8 * ONAL SC>">66">"">">6>>"">"">">">6>>">A ************************":M$'" DYNAMICS - TRANSLATION & ROTATION"w.':" ************************"8'255:A11000:::(7):B'"";A$:L'31:"";A$:6">H THE":" SUPPORT OF NATI5:X1,Y6:X2,Y6:X3,Y1X3,Y6:X4,Y:X4,Y5:X5,Y:X5,Y5:X6,Y1:X6,Y6:<#3:X,Y2X,Y4:X1,Y1X5,Y5:X1,Y5X5,Y1:#3:A3.143.14.1:X11.6(A),Y10(A)::#X,YX10,YX10,Y8X,Y8X,Y:':100:8 '" ٹp傚鏃4{䍁䀃E-h q`F"䀡mO党󂑃(LY q`#6党R!32hS PROBLEM THAT":"THE CORD WERE NOT ATTACHED TO M2, BUT"::"WRAPPED AROUND THE WHEEL. THEN THE"::"WHEEL WOULD TURN AS M1 FALLS. FIND":"THE ACCELERATION IN THIS CASE.":::"PRESS FOR THE ANSWER.";A$+13376:6:"THE ANSWER IS 7 YOUR ALGEBRA SKILLS ARE WEAK!"::1006034,0:13376:8:"THAT'S THE END OF THIS PROBLEM. IF YOU"::"WANT TO GO THROUGH IT AGAIN, TYPE 1.":"TYPE 2 TO CONTINUE: ";A:A125013376:5:"HERE'S ONE LAST PROBLEM:"::"SUPPOSE IN THE PREVIOU>">6>>"">"">">">6>>">ARANSLATION"5.':" ************************"U8':255:A11000:::(7):jB'"";A$:L'31:"";A$:6">H THE":" SUPPORT OF NATIONAL SC>">66">""'#X2#3:X1,YX3,Y:X4,Y1X4,Y5:X1,Y6X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:#1:X,YX6,Y8X16,Y2X10,Y6X,Y:#5:X,YX10,YX10,Y8X,Y8X,Y:':100:8'" ************************": $'" DYNAMICS - TLY, THEN ":"FOR THE ANSWER: ";A$} 13376:6:"THE ANSWER IS 6.9 M/S/S.":10:"THAT'S THE END OF THIS PROGRAM.":*"SEE YOU LATER, ";Z$;"!":20:10050413376:D13:(7)::20:"END">D$;"PR#0":HD11500:::D$;"RUN MENU"ôѴڸѴ.юlѴ۴Ӭ׬;FR _4,ʹǴѴҮ..ڴѴǴٸ¶ҶӴ,t,ڴҴ٥ѴҴѴдѶ..ʹ../еƵе::" T - 11.76 = (12)(ax)"::10060}13376:3:"ADDING THE EQUATIONS:"::" 91.75 = (27)(ax)"::"AND,":" ax = 3.4 M/S/S":::"ISN'T THAT SLICK!";:10060:::"SO WE'VE SOLVED THE PROBLEM. IF YOU"::"WANT TO R륤襮6ď6ʯ羯罯˯믲555ܣ55ٯ߮616X,0=ZBx`t{ZKX3?X59++X76X,0=X>477*VZBBBZ+74.=X,01+X76=X;75(4=,=4!VZBBZq96+/=*GqXZC9BIKKONB9IVAI@HHxQan{IHBZ= ;=44=6,YX!7-E"::"PROBLEMS, TYPE 1. TO CONTINUE, TYPE 2:";A:A1240q13376:3:"NOW HERE'S ONE TO DO ON YOUR OWN:"::"A 6 KG MASS RESTS ON A TABLE (u = 0.1)"::"ONE METER HIGH. THE MASS IS CONNECTED":#"BY A LIGHT CORD AND PULLEY TO A 3 KG"::"MAS^1000:::(7):%B'"";A$:>L'32:"";A$:6">H THiRȻȧȦȻN2^\^#2:X,70X70,70X70,30X,30X,70:1:X71,29X81,24:bT$5:X,YX10,YX10,Y8X,Y8X,Y:p':100:8'" ************************":$'" CONSERVATION OF ENERGY: CONS. FORCES".':" ************************"8'255:A1ROGRAM WHICH DEALSx\@(FGF%*2Zr *Kk   h fjrrrj  ijrrr д,$a)xiO* 슨!&쌓ӤҪ} 9hy:/횸8`*+ypypqyqy E4*+yp+pqypp )(.05^2)":::"> FINALLY, THERE IS K AND Ug.":u\" K = 1/2(.05)(V^2)"::" Ug = (.05)(9.8)(2*SIN 30)":f:"> EQUATING INITIAL AND FINAL,"::" V = 4.52 M/Sz::10050p13376:9:"THAT'S THE END. YOU SHOULD ALSO SEE"::"THE PLINED AT AN":k4"ANGLE OF 30 DEGREES, HOW FAST ISxŢ::ŠǠΠԠӠϠŠп::::ŠŠ숊>@$a q ndxuc`by~w0+b3eC HAT'S ALL THERE IS TO IT!";:10060[34,0:13376:3:"NOW SOLVE THIS ONE YOURSELF:":: "A SPRING (K = 800 NT/M) IS COMPRESSED"::"5 CM. IT IS THEN USED TO SHOOT A 50 GM":*"MARBLE UP A 2 METER LONG FRICTIONLESS"::"RAMP. IF THE RAMP IS INC" 1/2(6)(V^2)+(6)(9.8)(1)+1/2(3)(V^2)"::10060:"THE Ug OF THE 6 KkȥȬȧȩRRRX8d!RʮRRȾRRʩ꜁ e~veO@e~v (?*)>+ gf}`jl }ah} A `z }al {iemafjg|`  xdikm{ifl|`m~mdgka|am{Pyl{㌀gy w{bchiiohXOZԑikkonbz{mxx   x x bzbbzxplsrav`qpiqxsxpkqpav`qphvoqxezbXmH THE":" ?Q">66">"">">6>>"">"">">">6>>">A]2::':100:8>'" ************************":j$'" CONSERVATION OF ENERGY - METHOD".':" ************************"8'255:A11000:::(7):B'"";A$:L'32:"";A$:6">NSERVATIVE & NON-CONSERVATIVE FORCES.":::"SEE YOU LATER!":22:10050mH13376:D13:(7)::20:"END"~RD$;"PR#0":\D11500:::D$;"RUN MENU"'#(#5:X,YX10,YX10,Y8X,Y8X,Y:Z#X,84X,94X8,94X8,84:Y84922:X,YX8,YEB,XDJ/_@\F[FYJ/W-55-KF]JL[F@A/N[/9/B \!˯E]gDcEO*GZG_G, NG*(1"4G.*D0,!D*!#%0-2!D=F^^F -6!'0-+*D%0D\D)K7JD3,%0D-7IJ B^nyusz}`apqp}fqw`}{z{r`|q`|}fpduf`}r`|qg|qxx|upu`{`uxyugg{r40s 쎀ԤԠԱNDw֊fBڿجâددߪBBڹؼZP;POzbbbnoȿ`b cҠjatevp$! hnb cil`vjccgRJңФУибФнJJҤнвбвн`bvrag{d{zp{ avp|zz}g{vjwzavpgz|}r}trusqecu 0"Z4dxuuqbdxyaa|~r``zevzg`avp|z ev|\.3>G.4G1"5>KG1"5>G ." 0 + 0 = 1040 + 600*V2(X)'"::"OR,"::" 600*V2(X)' = -1040":} 8"AND,":8::"V2(X)' = -1.73 M/S"::: B"SO THE GUN RECOILS BACKWARDS WITH A"::"VELOCITY OF -1.73 M/S. 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Fp - T - Ff1 = (M1)(ax) (=p* (\Q\:N\A\T1NUTU^FFF^23(5?9\(4=(\+9[*9\)/98\\:3.\>3(4^FF^1=//9/\/52?9\(495.\=??909.=(532/jj򄘕Ht1Bګجرضعjj𑞔)4(@4( 22PLLDOUR!Gq!@OE!Go0/#;;01171;T'н۽D+!=0U Y0=&17 =;:NVNNVTUT <1T$;'= ="1T Y0jttvnT@Vvtttt5H8YV\8̸ָݸݶ""ߺߺɺʹ*ԘS ߺúߠЂ6Αppγ,i󚂚ΑǑӈ푊NJӈсƐ硼ĸ硼333Qnn}ow:glmmkm]pM{㇓Ꮞᎏᖄ捍ᖎᎏ蚍莇Ǽ ( 9ٳ&w'ӽؽٰټȼ۽پ&&Ӽ苇ƾH`@/.@9/52@/7.NB[ZQPPTQcBpbUWOWWVQOUSYTQOWP[QOWZYVROV[VVOUS[[SOWZ[QOWTYUYZYA::"ALSO, REMEMBER THAT Ff = u * Fn"::1006013376:"#3 ESTABLISH CONVENTIONS."::" WE'LL TAKE MOTION DOWN THE PLAbtvnTEUvtttTO THE Lmn|gf|`m!|ijdm!턊ǩ|`i|{|`miq|`A K$Meƶv|,@vikp'&R95Pqlrdb~sxyd{wV00̩ڪͨܨΨͨۦFƔ.//)/-//+*&>HUY>NX\^ITKX=[RO^XN1=[o1=\OX=IUX=%%&&>H]^PY25=5(735=5(7=>537(67713*47<;=9&??;=?&?? htp莺j^rꎦ莯j^rꎮk_scT|.|ڴǶ..ٺ$ʕ].f۾ھϾ&&ټټ瘕`jjݢR"TӠ<1}]%kgmcg`taa fk`}woapglbkv~c|J>hieFDDD&+ -!7JD0,!*D',!'/D&=DF_^VR^VUUPUeka׵ qq a +K*%){Ƶń'?ȷ'>񶦾&>𴧿&B~|Ȏ} L= ԧW.\KJMZ1?R^TZ?d}y  d|e |w <1h=rrit~xi㗴$K홙2]QCC~ `ɜ h0艋臎rȲr꜀腉 ȼ 铎xdE䔖̭Rt*JZM2xLƗ}Lyxpdy{HbD~Iy|qrx}rqz|\}$슒ɺnĿڶڽ@+**/*+))-, )#:HJWZT]U":"":Y8)-8S_8UYKKA:4KGECB&KJ$E$11&>>&@ACVAA$MJkui {di~i xdi ocijjeldk`z .$*/cj j~eox冗ᄜݣ qIQ$6T??S%=#:Q>?Q0Q54%08=45Q0`Itqv 7>7pmxtT:HYrڴƱƯƩ\\IJ"ᆞBBĹ˼QQ̯vW+,'DAYYcXicA* ܩܴмjx~mb`mxecb,$B,,$ a).+(*4(L5ϣͿ5`< P{FjضH8lB^ h ^½έ1Lx~]pY0/6%  d!JRʠȼRlʿǴƴѫϲ--؞E-||||||||||||brbqbpbwbvbuavay`x`{`z`32:Y126:9020:X106:Y142:9020:Xl:"SOLVE THESE SIMULTANEOUS EQUATIONS!"::10060v13376:"YOU SHOULD FIND THAT:"::" a = 3.63 M/S/S":"AND = 18.17 RAD/S/S":X42:Y126:9020::("IF NOT, CHECK YOUR ALGEBRA. GET HELP"::"IF" a = (r)( )":X204:Y143:9020:X92:Y159:9020::"NOW MAKE THE DATA SUBSTITUTIONS."::10060X13376:"1. (10)(9.8) - T1 = (10)(a)"::"2. T2 - (.3)(8)(9.8) = (8)(a)":"b"3. (.2)(T1) - (.2)(T2) = (.1)( )"::"4. a = (.2)( )":X2FOR THE UNKNOWNS."::" WITH FOUR UNKNOWNS, WE NEED FOUR":m5" EQUATIONS. NOW WE HAVE THEM."::10060:13376:"HERE THEY ARE:"::" (M1)(g) - T1 = (M1)(a)":D" T2 - (u)(M2)(g) = (M2)(a)"::" (r)(T1) - (r)(T2) = (I)( )":dNVE.":X40:Y150:9000:X151:9000::1006013376:"#8 SUM THE TORQUES."::" YOU WRITE THE EQUATION FOR TORQUES."::10060:&" HERE IT IS:"::" 1 - 2 = (I)( )":X47:Y158:9000:X82:9000:X148:9020::10060=013376:"#9 SOLVE UT NOTE THAT ax = ay, SO WE CAN DROP"::"THE x,y NOTATION.";:10060a2:128,25128,35115,3513376:"#7 CALCULATE THE TORQUES."::" 1 = (r)(T1)"::" 2 = (r)(T2)"::*X26:Y110:9000:Y126:9000:"NOTE: 1 IS POSITIVE, 2 IS NEGATILY M2 ACCELERATES IN THE x-":[" DIRECTION..."::" T2 - Ff = (M2)(ax)"::1006013376:"#6 SUM THE FORCES IN THE y-DIRECTION."::" ONLY M1 ACCELERATES IN THE y-":" DIRECTION..."::" Fg1 - T1 = (M1)(ay)"::10060:D"BUSE THE DIRECTION OF MOTION (DOWN,":i" COUNTER-CLOCKWISE, AND LEFT) AS"::" POSITIVE.";:1006013376:"#4 RESOLVE FORCES."::" THE FORCES ARE ALREADY RESOLVED!"::10060::13376:"#5 SUM THE FORCES IN THE x-DIRECTION."::" ON"T1":5:26:"T2";:35:"Ff":9:7:"Fg1";:28:"Fn = Fg2"13376:"SINCE THE WHEEL DOESN'T TRANSLATE,"::"WE'VE LEFT OFF Fg AND Fr."::"CORRECT YOUR DRAWINGS IF YOU NEED TO"::"DO SO.";:10060%13376:"#3 ESTABLISH CONVENTIONS."::" 1:53,4053,60:53,3153,16:209,35195,35:221,35233,35:131,25145,25:115,35115,523:52,5954,59:52,1754,17:196,34196,36:232,34232,36:144,24144,26:114,51116,51:214,59216,59:214,12216,12:1:31:"Fr":8:"T1":22:"T2":8:16:* T:" MAKE YOUR OWN SKETCH.";:10060} ^13376:"#2 FREE-BODY DIAGRAM."::" MAKE A FREE-BODY SKETCH FOR EACH OF": h" THE THREE OBJECTS.";:10060 r62450:X128:Y35:9100:X48:Y39:9200:X210:9200:5:215,11215,31:215,40215,60X| M2":3:12:"I":4:14:"r":4:30:"u":6:11:"T1":9:11:"M1" @34,11:13376:"#1 HERE'S THE PICTURE. NOTICE THAT"::" SINCE WE HAVE A REAL WHEEL, THE": J" TENSION IN THE CABLE IS NOT EQUAL"::" ON EACH SIDE OF THE WHEEL."::100601 KG-M^2,":/ "r = .2 M, AND u = 0.3."::X "COPY THE PROBLEM AND ";:22:10050 "13376:X100:Y30:9100:X161:Y24:9200:X82:Y60:9200 ,2:200,25120,25120,45100,45:100,30100,75200,75:1:87,5287,30:101,20161,20E 62:18:"T2 E YOUR PAPER, PENCIL AND"::"CALCULATOR READY, LET' GO..."::10050 13376:3:"PROBLEM:"::"FOR THE FOLLOWING DIAGRAM, FIND THE"::"ACCELERATION OF THE MASSEWYXhHTuN41fQ??90ttttnvnnvetitedtxtftitltxttitz&&ϕN&ϰƸҼ̿Y!tfootlqhlmwkfջݻڹ!6D"˸ոиɺҸ A2SpfdF둋z/&z∁zbzzcaaaaaaaaf alM蘉ggݙ-gqԸn߮$YBWA\Z["'IsxyyyyIA1Is$#&%'•-ѿѼϬ)vA.r6VIP" QGPrx|.XTMOU|t|.X|t|1X^a^FQ\(49,;^7-^?^B=*.0S/45:(S1d DLD `fufDAMDȍuwvrswg e d