' +JJJJ ?\>m0M='+l> /+l   d]@ŵLҦ]]L L}BBL]浍국絍뵍䵺L 鷎귭෍ᷩ췩緈JJJJx Lȿ L8ᷭ緍췩 緍i 8 `巬 췌`x (`(8`I`B` ``>J>J>VU)?`8'x0|&HhHh VY)'&Y)xꪽ)' `Hh`V0^*^*>&` aI꽌ɪVɭ&Y&&Y& 꽌ɪ\8`&&꽌ɪɖ'*&%&,E'зЮ꽌ɪФ`+*xS&x'8*3Ixix&& 8  '  & x)*++`FG8`0($ p,&"_]` L/  !"#$%&'()*+,-./0123456789:;<=>?  1#"""  (9"1 ( ,.(0# 2  /#0/#0 *?'#07#00/0/'#07#0:"4<*55/**5/*%5/)1/)1/)1/)'#0/#0*5/*75/**5/*:5//#0/#0'#07#0:::*::'#07#0EB H  @H !D)"E` @ $ C ` DQ &J80^݌Hh ü ü݌ ռ ռ ռA ļD ļ? ļAEDE?HJ>h Լ ռ ռ ռ`HJ>݌h Hh݌`HIHHHHhHH݌hHhHh݌H6 VDP (ED Z $0x8x D- ܸDD# H8`?E Vk *f???0xE Hh D#-EEE8` D ܸx D - ܸx8`-0ݩ?ʥD EEE`   vLDcpq` [` ~  LӜu`".Q`pNФbptťܥm2<(-Py0\|e<6e<g< JJJJj귍hI  aUL@ kU8  L  Q^R(jQ0l^l\  wUuW ԧ H h@ [_ /QSIRb_L`LLLL`ª`LQLYLeLXLeLee ўQH\(h0L& Ꝥ$`( R \ZLl8 ўR HH\`\Z[YS6`LxQɿu3'RͲʎRʎ]]]ɍuL͟ɍ}RLRɍg^H8 ^hZLɍR LͲɊRR% QLܤͲ Z@ -^ ş\[Z QY\[Z8`l6Lş_Ȍb_Ͳ] )Y h( ֭ͲLɍ [LLĦ__ ^ 9 LҦ3 9 a   0LjLY u< (_9 ˭ɠuɠK_9 ?LˆʎõĵL 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$3>L[dmx- `Ϡ@跻~!Wo*9~~~~ɬƬ~_ j ʪHɪH`Lc (L ܫ㵮赎 ɱ^_ J QL_Ls贩紎 DǴҵԵƴѵӵµȴ 7 ַ :ŵƴѵǴҵȴµ納贍﵎ٵ്ᵭⳍڵL^ѵ-I `  4 ò-յ!  8صٵ紭ﵝ 7L (0+BC  7L HH`LgL{0 HH` õL H hBL BH [ h`Lo õ ڬL B ڬ LʬH hB@ յյ [L (ȴ) ȴ 7L L ( L (ȴL{ƴѵ洩ƴǴҵ 7 ^* B0 HȱBh ӵԵ 8 L8 ݲ` ܫ  / / ED B / / ]ƴS0Jȴ ȴ)  紅D贅E B ƴ  / 0L Ν `HD٤DEEhiHLGh ` ŵBѵ-` ѵB-` ܫ XI볩쳢8 DH E𳈈췍Ȍ X0 · JLǵBȵC`,յp` 䯩 R-յյ`յ0` K R-յյ`ɵʵӵԵ` 4 K ( ѵҵLBȱBL8` DBHBH : ַ޵BȭߵBhhӵԵ RBܵmڵ޵ȱBݵm۵ߵ` 䯩LR˵̵ֵ׵`êĪLR E( 8` R` ELRŪƪ`췌 յյI뷭鷭귭ⵍ㵍跬ª 뷰` Lf ݵܵߵ޵ ^`8ܵ i B8` 4L ֵȱB׵ ܯ䵍൭嵍 ` DȑB׵Bֵ  ַ յյ`굎뵎쵬 뵎쵌``õĵBCõĵ`µµ`L õBĵCصص Qƴ0"Bƴ 󮜳` 0۰ϬBƴ8`i#`ЗLw!0>ﵭ` m ﳐ 7i볍 8 ЉLw`H h ݲL~ `浍국䵍뵩嵠Jm赍嵊mjnnn浈m浍浭m䵍䵐`"L ŵ8ŵH ~(`# d àz# u`TER!":22:100506H13376:D13:(7)::19:"END"GRD$;"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 FOR MORE PROBLEMS.":>"SEE YOU LAS 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 THID,":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$;", IF YS START BY SOLVING THE TORQUE":o" EQUATION:";:10060:::" Rc*Ty + Rd*Tx - Rb*W - Ra*Wb = 0"::1006013376:"REMEMBER THAT THE WEIGHT OF THE BEAM"::"IS M*g = 15*9.8 = 147 NT AND THE WEIGHT":"OF THE LOAD IS GIVEN AS 200 NT."::"AN FOR THE SUM OF"::"THE TORQUES.";:10060:v:" (Ty) + (Tx) - (W) - (Wb) = 0":::"DID YOU GET THE SIGNS RIGHT?"X13:Y142:9000:X75:9000:X138:9000:X195:9000::10060 13376:"#12 NOW WE HAVE EVERYTHING WE NEED."::" LET'CH TORQUE"::" IN TERMS OF 'R' AND 'F'.";:10060:X" (Wb) = Ra * Wb"::" (W) = Rb * W"::" (Ty) = Rc * Ty"::" (Tx) = Rd * Tx";bX26:Y134:9000:Y150:9000:Y166:9000:Y182:9000:10060,l13376:"NOW WRITE AN EQUATION M":O0" Rb = 7 * cos( ) = 5.36 M"::" Rc = 10 * cos( ) = 7.66 M"::" Rd = 10 * sin( ) = 6.43 M"::X127:Y118:9010:Y134:9010:Y150:9010:Y166:9010D"HOW DID YOU DO, ";Z$;"?";:100606N13376:"#10 WRITE AN EXPRESSION FOR EA^13376:"#9 YOU TRY TO CALCULATE THE RADIUS ARM"::" TO EACH LINE OF ACTION.";:100605:61,18116,18:47,2647,63:2:10:"Ra";:13:"Rb";:16:"Rc":6:5:"Rd":X73:Y26:9010&13376:"#9 HERE THEY ARE:"::" Ra = 5 * cos( ) = 3.83OTATION.";:10060::o"#8 NOW DRAW 'LINES OF ACTION' ON YOUR"::" SKETCH. SINCE Hx AND Hy ACT AT THE":" POINT OF ROTATION, WE DON'T NEED TO"::" WORRY ABOUT THEM.";:100606:118,44118,16:105,6441,64:94,3994,16:1:81,3181,16:"OK, ";Z$;"?";:10060S13376:"#6 YOU WRITE THE EQUATION FOR Fy."::10060:" T * sin( ) + Hy - Wb - W = 0":::X78:Y134:9020:" BE CAREFUL OF SIGN CONVENTIONS!"::10060:13376"#7 LET'S CHOOSE THE HINGE AS OUR POINT"::" OF R9:25:"Tx = T * cos( )"?X218:Y38:9020:X260:Y70:902013376:"YOU SHOULDN'T HAVE ANY TROUBLE HERE."::"FROM NOW ON, USE THE COMPONENTS OF T."::10060:"#5 Fx = 0.":X24:Y150:9030::" T * cos( ) - Hx = 0":X98:Y166:9020:"#3 ESTABLISH THE USUAL CONVENTIONS."::10060:"#4 RESOLVE INTO x AND y COMPONENTS."::" RESOLVE T YOURSELF AND";:27:100502:118,47118,64162,64:3:116,48120,48:160,63160,657:25:" ":X143:Y62:9020:5:19:"Ty = T * sin( )":6 h2:48,348,1810,18:94,4994,70:1:81,4181,56o r3:47,449,4:12,1712,19:93,6995,69:80,5582,55 |1:5:"Hy":4:1:"Hx":7:10:"Wb":10:15:"W" 13376:"IF YOU DIDN'T GET IT RIGHT, CORRECT"::"YOUR SKETCH NOW.";:10060::0 ROPE: ALPHA = 25 DEGREES":C @" WRITE DOWN THE DATA.";:10060 J13376:"#2 FREE-BODY DIAGRAM."::" MAKE YOUR OWN FREE-BODY SKETCH. THEN": T" CHECK BY PRESSING";:22:10050 ^62450:9100:1:118,64163,47:3:160,47162,49:7:25:"T"X15:45X,1345X,19::1:118,64251,13y "X72:Y22:9010:X210:9020:2:94,4994,64:90,6598,6598,7190,7190,65 ,34,12:13376:"#1 HERE'S THE PICTURE."::" BEAM: 15 KG, 10 M, THETA = 40 DEGREES": 6" LOAD: 200 NT, SUPPORTED AT 7 M"::"13376:3:"PROBLEM:"::"IN THE FOLLOWING PROBLEM, YOU ARE TO": "SOLVE FOR THE TENSION IN THE ROPE AND"::"THE x AND y COMPONENTS OF THE 'HINGE'": "FORCE.":::"WRITE THIS DOWN AND ";:21:10050 13376:5:20,12270,12:9100* 2:HANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL WORK ON A DETAILED ANALYSIS OF A": "BEAM IN STATIC EQUILIBRIUM. YOU SHOULD"::"HAVE ALREADY WATCHED THE PROGRAM:": " 'STATICS - METHOD'":::"IF YOU ARE READY, LET'S CONTINUE..."::10050C !d - STATICS - BEAM PROBLEMS.n:10000>x:16302,0V- TURN ON ROMPLUS+bD$(4)qD$;"PR#5"~R$(13)R$M$"":- THERE IS A HEREM$;"1A"13376:3:"HELLO!"::"WHAT'S YOUR NAME? ";Z$::M "T   H THE":" SUPPORT OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">UX115,6745,2151,15:#':100:8M'" ************************":q$'" 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 sin( )":9:25:"Tx = T * cos( )"HX218:Y38:9020:X260:Y70:902023:"*** NOW AGAIN: ";A$:13376:"#5 AND #6 HERE'S WHERE YOU WOULD SUM":" FORCES IN THE x-DIRECTION AND THE"::" y-DIRECTION.";:10060?:"#7 SELECT THE" THE TENSION VECTOR).";:10060v::"#3 ASSUME CONVENTIONS ARE ESTABLISHED."::10060::"#4 RESOLVE TENSION INTO COMPONENTS."::100602:118,47118,64162,64:3:116,48120,48:160,63160,65"7:25:" ":X143:Y62:9020:5:19:"Ty = T *WE WON'T DO THE WHOLE PROBLEM NOW."::"LET'S JUST LOOK AT HOW TO HANDLE THE":X"LOWER END WHERE THE CABLE IS CONNECTED."::10050b62450:9100:1:118,64163,47:3:160,47162,49:7:25:"T""l13376:"#2 HERE'S THE FREE-BODY (SHOWING ONLY"::AT A"::"SIMPLE EXAMPLE.";:10060E&13376:5:20,12270,12:9100x02:X15:45X,1345X,19::1:118,64251,13:X72:Y22:9010:X210:9020D34,10:13376:"#1 HERE'S OUR PICTURE--A BEAM OF LENGTH"::" L, SUPPORTED AS SHOWN."::ON"EPS 5,6, & 11 TO SOLVE FOR ANY":I" UNKNOWN QUANTITIES.";:10060::"THIS IS A LONG ONE! BEFORE WE GO ON TO"::"AN EXAMPLE, YOU MIGHT WANT TO REVIEW."::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A1250"::"OK, ";Z$;". LET'S LOOK ::" DICULAR DISTANCE FROM ORIGIN) TO"::" EACH LINE OF ACTION.":x">#10 CALCULATE EACH TORQUE = (r)(F).":">#11 SUM ALL TORQUES AND SET EQUAL TO"::" ZERO."::10050!13376:3:">#12 USE THE EQUATIONS DEVELOPED IN"::" ST IN THE y DIRECTION"::" AND SET EQUAL TO ZERO."::1005013376:3:">#7 SELECT A POINT OF ROTATION FOR"::" CALCULATION OF TORQUES.":">#8 DRAW A 'LINE OF ACTION' FOR ALL"::" FORCES.":J">#9 CALCULATE THE RADIUS ARM (PERPEN-"" COUNTER-CLOCKWISE TORQUES ARE ALL"::" POSITIVE.":">#4 RESOLVE ALL FORCES INTO x AND y"::" COMPONENTS (p AND n ON INCLINES).":">#5 SUM ALL FORCES IN THE x DIRECTION"::" AND SET EQUAL TO ZERO.":=">#6 SUM ALL FORCESE-BODY DIAGRAM. SHOW ONLY"::" THE FORCES ACTING ON THE OBJECT.":" (TENSION, GRAVITY, HINGE, FRICTION"::" SUPPORT REACTION, ETC.)";:1006013376:3:">#3 ESTABLISH CONVENTIONS: FORCES"::" POINTING UP OR TO THE RIGHT AND":@SIS. THERE"::"ARE QUITE A FEW STEPS AND IT'S EASY TO": h"GET LOST. BE SURE THAT YOU WORK"::" CAREFULLY AND ORDERLY! ";:10060:: r"METHOD - COPY IT DOWN!"::">#1 START WITH A DRAWING. SHOW ALL THE"::" FORCES.":H|">#2 DRAW A FRE FORCE OF A HINGE"::" W - WEIGHT = (M)(g)":w J" Fn - NORMAL FORCE"::" Ff - FRICTION FORCE = (u)(Fn)":: T"YOU SHOULD ALREADY BE FAMILIAR WITH"::"THESE. WRITE DOWN THE NOTATIONS."::100509 ^13376:3:"NOW ON TO THE METHOD OF ANALY0"::= "X30:Y92:9030:Y110:9030:Y126:9030:X44:9000 ,"GET THEM RIGHT, ";Z$;"? OF COURSE YOU"::"DID! LET'S LOOK AT SOME OF THE": 1"FORCES YOU'LL WORK WITH.";:10060 613376:3:"TYPES OF FORCES:"::" T - TENSION":1 @" H -::"CALCULATOR READY, LET'S GO."::10050 13376:3:"YOU SHOULD ALREADY KNOW THE CONDITIONS"::"FOR EQUILIBRIUM. WRITE THEM ON YOUR": "PAPER, THEN ";:13:10050: :"HERE THEY ARE:": " Fx = 0"::" Fy = 0"::" = HANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL DESCRIBE A METHOD FOR ANALYZING": "THE FORCES AND TORQUES WHICH ACT ON A"::"BODY IN STATIC EQUILIBRIUM. IT'S A": "LITTLE INVOLVED, SO PAY CLOSE ATTENTION!":* "IF YOU HAVE YOUR PAPER, PENCIL, AND"d- STATICS - METHOD&n:100006x:16302,0N- TURN ON ROMPLUS+ZD$(4)iD$;"PR#5"vR$(13)~R$M$"":- THERE IS A HEREM$;"1A"13376:3:"HELLO, FRIEND!"::"WHAT'S YOUR NAME? ";Z$::L "T     @ '3<JRXdq~ ,,$$36..$$%%..">'$$$..&%%666$$$--:?6.-= $,,-5>?<I9?'$$,$$$-557?$$$M1$";A$: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$;", YOU SHOULD PROBABLY WORK":G0l"SOME EXTRA PME: ";A::c.&"REMEMBER THAT X & Y MOTIONS DO NOT"::"AFFECT EACH OTHER, BUT THE FLIGHT TIME":.0"IS THE SAME. SO, T= 1.3 SECONDS."::10050.::3:"THEREFORE--"::" Y = V(0)Y*T - 1/2*G*T^2":/D:"SUBSTITUTING--"::" Y = 9.192*1. 10/7.714":3-5::"T = 1.3 SECONDS":::10050-:4:"HOW DO WE FIND THE ELEVATION WHERE THE"::"ARROW HITS THE CLIFF? ";:10050::-"HINT: HOW MUCH TIME DOES THE ARROW HAVE"::"TO GO UP AND DOWN?": ."INPUT THE ARROW'S FLIGHT TI0)Y--":a," V(0)X = 12*COS(50) = 7.714 M/S"::" V(0)Y = 12*SIN(50) = 9.192 M/S"::10050,::"SINCE WE KNOW THE RANGE (10 METERS) AND"::"AND THE HORIZONTAL VELOCITY (7.714 M/S)," -"WE CAN FIND THE FLIGHT TIME--"::" T = RANGE/V(X) =9,20;+T11400::22:" TIME MARKERS SHOW THE MOTION."]+Y501:7:203Y,11Y2:|+Y03:7:203Y,11Y2:+T11000::24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11620,::2:"HERE'S THE SOLUTION:":::" FIND V(0)X & V(24:100507*T::9000:1:X3035:14,36X::7:5,36*^22:" PRESS TO SHOOT THE ARROW.";A$: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$;"!"::"KEEP GOING...":22:10050*(:5:"SINCE T = 4 SECONDS, WE FINDSE: Y = V(0)Y*T - 1/2*G*T^2":::10050&::"REMEMBER THAT Y = -78.4 METERS, SINCE"::"THE FINAL POSITION IS BELOW THE START."::&" INPUT THE FLIGHT TIME: ";A::A41540':3:(7):"SORRY, ";Z$;".":::" 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$:"THEN,":" 0 = 173.2*T - 1/2*9.8*T^2":."AND":" -173.2*T = -1/2*9.8*T^2"::100508:"FINALLY,"::6::"T 73.21010"YES!";I9::"V(0)Y = V(0)*SIN(60) = 173.2 M/S":::"NOW USE THE VALUE OF V(0)Y AND THE"::"EQUATIONS OF MOTION FOR GRAVITATIONAL":"ACCELERATION TO FIND THE TIME OF FLIGHT.":/:"REMEMBER THAT THE SHELL HITS THE GROU:"INPUT YOUR ANSWER FOR V(0)X: ";A::A100970A"GOOD!";x10::"V(0)X = V(0)*COS(60) = 100 M/S":::10050:3:"REMEMBER THAT THIS IS THE VALUE OF THE"::"CONSTANT HORIZONTAL VELOCITY.":: "INPUT YOUR VALUE FOR V(0)Y: ";A::A1CANNON SHELL IS FIRED WITH A MUZZLE":}"VELOCITY OF 200 M/S AT AN ANGLE OF"::"60 DEGREES ABOVE HORIZONTAL. FIND THE":"RANGE OF THE SHELL. ";:10050::"COPY THE PROBLEM. THEN TRY TO FIND THE"::"X & Y COMPONENTS OF V(0).":3 THE GROUND IS X(0) = 0, Y(0) = 0.":|z"> THE STARTING TIME IS T(0) = 0."::"> THE ANGLE (THETA) IS MEASURED FROM ":" THE X-AXIS.":::"COPY THESE CONVENTIONS IF YOU NEED TO."::"THEN, ";:10050&:3:"NOW HERE'S A PROBLEM:":::"A METHOD FOR REFERENCE: ";:10050\:2:"THE CONVENTIONS WE USE ARE THE SAME AS"::"THOSE ESTABLISHED IN PREVIOUS PROGRAMS:"::f"> UP AND RIGHT ARE POSITIVE."::"> DOWN AND LEFT ARE NEGATIVE.":'p"> THE STARTING POSITION (EVEN IF ABOVE"::" E: V(X) = CONSTANT V(0)X!"::" V(0)Y = V(0)*SIN(THETA)">" NOTE: V(Y) WILL CHANGE!"::" 4. APPLY EQUATIONS OF MOTION IN THE"H" Y DIRECTION TO FIND THE TIME OF":" FLIGHT (T)":R" 5. FIND RANGE = V(X) * T"::"COPY THE ONE! ";:10050T:2:"METHOD:"::" 1. ESTABLISH CONVENTIONS": " 2. DETERMINE KNOWN & UNKNOWN VALUES"::" 3. FIND THE X & Y COMPONENTS OF THE"*" INITIAL VELOCITY--"::" V(0)X = V(0)*COS(THETA)">4" NOTE (THETA) OF":h"THE PROJECTILE. YOU ARE TO SOLVE FOR THE":"RANGE (MAX. HORIZONTAL DISTANCE)."::"OF COURSE, THE PROBLEM CAN BE CHANGED"::"AROUND (GIVEN RANGE AND ANGLE, FIND THE": "INITIAL VELOCITY). BUT, LET'S START WITH":"THE EASY NT TO WATCH AGAIN, ";Z$;"?"::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1370:3:"NOW LET'S LOOK AT A METHOD FOR SOLVING"::"PROJECTILE PROBLEMS. ";:10050::"IN THE BASIC PROBLEM, YOU ARE GIVEN THE"::"INITIAL VELOCITY AND ANGL IN THE X DIRECTION AND BY THE"::"ACCELERATED MOTION IN THE Y DIRECTION."::"THEREFORE, THE TIME IT TAKES THE PRO-"::"JECTILE TO GO UP AND DOWN IS EXACTLY":"THE AMOUNT OF TIME THE PROJECTILE HAS"::"TO MOVE SIDEWAYS."::P"DO YOU WA11Y15:T11000:TU11:10,36203Y:7:21,3611Y2:1:203Y,11Y2:24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1620::3:"YOU SHOULD SEE THAT THE PROJECTILE"::"POSITION IS DETERMINED BY THE CONSTANT":O"VELOCITY13:203Y,11Y2:Pb24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1370l::9000:1:5,36:22:" HERE IT IS IN SLOW MOTION.":T11500:vY501:T11000:T11:10,36203Y:7:4,2011Y2:1:203Y,11Y2:7:21,3613:5,36:T11000:&Y50.5Q13:203Y,11Y2:0:203Y,11Y2:_&YO5.5013:203Y,11Y2:0:203Y,11Y2::13:35,36DT11000::22:" TIME MARKERS SHOW THE MOTION."NY501:13:203Y,11Y2:XY05:1000:Y50.567:4,11Y2:0:4,11Y2:DY05.5m7:4,11Y2:0:4,11Y2::7:4,36T11000::22:" TIME MARKERS SHOW THE ACCELERATION."7:Y05:4,11Y2::24:10050:22:" NOW COMBINE THE MOTIONS." |11:5,10:T11000:( Y010.1T 11:53Y,10:0:53Y,10::11:35,10 T11000::22:" TIME MARKERS SHOW CONSTANT VELOCITY. " 11:Y010:53Y,10::24:10050 :22:" HERE'S A VERTICAL ACCELERATION."7:4,36:T1NSIDER THE PROJECTILE MOTION AS TWO":m ^"SEPERATE MOTIONS WHICH LAST FOR THE"::"SAME AMOUNT OF TIME.": h:"LET'S SEE HOW THE MOTIONS COMBINE."::"WATCH CAREFULLY, ";Z$;"!":::10050 r::9000:22:" HERE'S A CONSTANT HORIZONTAL VELOCITY."ICAL - ACCELERATION DUE TO"= ," GRAVITY"::H 610050 @:3:"THESE TWO MOTIONS OCCUR AT THE SAME"::"TIME, AND THE RESULT IS PARABOLIC": J"PROJECTILE MOTION. ";:10050::& T"THE TRICK IN SOLVING PROBLEMS IS TO"::"CO SOLVE PROBLEMS."::m "PROJECTILE MOTION CAN BE CONSIDERED AS"::"TWO SIMULTANEOUS MOTIONS WHICH ARE": :"INDEPENDENT": 12:13:"EXECPT FOR THE TIME FACTOR.":::"THE MOTIONS ARE:": "" 1. HORIZONTAL - CONSTANT VELOCITY"::" 2. VERTFFICULTY WITH":_ "THEM--SO PAY CLOSE ATTENTION.":::"YOU SHOULD HAVE ALREADY WATCHED:": :"KINEMATICS & GRAVITATIONAL ACCELERATION":::"WHEN YOU'RE READY, ";:10050 :3:"THE BASIC IDEA IS SIMPLE, BUT YOU MUST"::"UNDERSTAND IT WELL TO)d- PROJECTILE MOTION%n10000[:3:"HELLO!"::"PLEASE TELL ME YOUR NAME: ";Z$8:"THANK YOU, ";Z$;". THIS PROGRAM"::"IS DEVOTED TO THE SOLUTION OF PROJECTILE" "PROBLEMS. ALTHOUGH THEY'RE REALLY NOT "::"TOO HARD, STUDENTS HAVE DI                  0:::(7):"B'"";A$:+ 4000*OMEGA'":::" 0MEGA' = 60000/5250 = "`(#X2,Y2X2,Y2X2,Y2X2,Y2X2,Y2:n':100:8'" ************************":$'" CONSERVATION OF ANGULAR MOMENTUM".':" ************************" 8'255:A11007)::20:"END"2D11500:::(4);"RUN MENU"8?F@:21:P$]J2:140,40140,120wTL1P4:VP56.28P3^XA(V):YB(V)h3:9000:1:140,80X,Y:0:9000:140,80X,Yr2:140,50140,110:V|L:14!" 1250*OMEGA' 5 + 4000*5 =":qH" (50*5^2)*OMEGA' + 4000*OMEGA'"::::" 100000 + 20000 = 5250*OMEGA'"::R:"FINALLY, OMEGA' = 22.86 RAD/SEC":::10050p:12:"WELL, ";Z$;", THAT'S IT FOR NOW."::"BYE, BYE!":20:10050z:D13:(. WHAT WILL BE"::"THE NEW VELOCITY IF THE CORD IS REDUCED":*"TO 5 METERS?":::"WORK IT OUT, THEN CHECK YOUR ANSWER."::100504:2:"SOLUTION:"::" I1*OMEGA1 + I2*OMEGA2 =":>" I1'*OMEGA1' + I2'*OMEGA2'"::::" (50*20^2)*E RIGHT! SIMPLY ASSIGN A NEGATIVE"::"VELOCITY TO ONE OF THE WHEELS."::10050~ ::"ONE MORE PROBLEM: ";:10050:5:"A 50 KG MASS IS ATTACHED BY A 20 METER"::"CORD TO A CENTRAL HUB (I = 4000 KG-M^2).":> "BOTH ROTATE AT 5 RAD/SEC6.5 RAD/SEC"::10050q:3:"NOTICE THAT THE FINAL ANGULAR VELOCITY"::"IS BETWEEN THE ORIGINAL VALUES."::"WHAT WOULD YOU HAVE DONE IF THE TWO"::"WHEELS HAD BEEN ROTATING IN OPPOSITE":"DIRECTIONS? ";:10050:O:"YOU'R I1'*OMEGA1' + I2'*OMEGA2'":::"WHERE OMEGA1'=OMEGA2' (SINCE THE TWO":y"WHEELS ARE JOINED). ";:10050::"THEN SUBSTITUTE--"::".05*20 + .15*2 = .05*OMEGA' + .15*OMEGA'":" 1 + .3 = .20*OMEGA'"::"AND, OMEGA' = ERE":="BOTH ROTATING IN THE SAME DIRECTION?"::10050::"THIS PROBLEM IS LIKE THE 1-DIMENSIONAL"::"INELASTIC COLLISION PROBLEMS YOU'VE":"DONE BEFORE. ";:10050:2:"SOLUTION:"::" I1*OMEGA1 + I2*OMEGA2 =":O" AXIS."::10050Sv:2:"PROBLEM:"::"A WHEEL (I = .05 KG-M^2) ROTATES AT 20":"RAD/SEC. IT IS SUDDENLY JOINED TO A"::"SECOND WHEEL (I = .15 KG-M^2) WHICH IS":"ROTATING AT 2 RAD/SEC. WHAT IS THE FINAL":"SPEED OF THE TWO WHEELS IF THEY WNGULAR VELOCITIES CONFUSED!"::10050::X"NOW HERE'S SOMETHING YOU MIGHT NOT KNOW.":"THE MOMENT OF INERTIA OF TWO BODIES":b"(WITH THE SAME CENTER) MAY BE ADDED.":::"THIS ALLOWS US TO WORK 2-BODY PROBLEMS":l"IF WE KEEP THEM ON A SINGLE= 3 M/S"::10050j::"THIS IS VERY IMPORTANT! THE ANGULAR"::"VELOCITY DECREASES TO 1/4, WHILE THE":D"TANGENTIAL VELOCITY ONLY DECREASES TO"::"1/2 (WHEN THE RADIUS IS HALVED).":::10050)N:3:"PLEASE, ";Z$;", DON'T GET TANGENTIAL"::"& AGULAR VELOCITY DECREASES FROM"::"3.0 TO .75 RAD/SEC. ";:10050:3:"WHAT HAPPENS TO THE TANGENTIAL"::"VELOCITY? ";:10050:&"THE ORIGINAL VELOCITY IS 6 M/S. THE"::"FINAL VELOCITY = R * OMEGA'":0" = 4 * .75 = 6 M/S / 2 M = 3 RAD/SEC"::" OMEGA' = ?"::10050::"APPLY THE CONSERVATION LAW--"::" L1 = L1'"::" I1 * OMEGA1 = I1' * OMEGA1'":" 20 * 3 = 80 * OMEGA'"::" OMEGA' = .75 RAD/SEC"::10050F:"SO THE ANR A POINT MASS)"::" I = 5 * 2^2 = 20 KG-M^2":_" I' = 5 * 4^2 = 80 KG-M^2"::10050:"THE MOMENT OF INERTIA INCREASES WHEN THE":"RADIUS INCREASES."::"NOW FIND THE ANGULAR VELOCITY--"::" OMEGA = V/R"::100506:3:" OMEGA IF THE RADIUS IS LENGTHENED"::"TO 4 METERS?":::"COPY THE PROBLEM AND ";:10050:::"NOTICE THAT WE HAVE ONLY ONE BODY, SO:"::" L1 = L1'":::10050:3:"SOLUTION:"::"FIND THE MOMENT OF INERTIA--":2" I = M*R^2 (FOL RADIUS":8000L24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE? ";A:A1370::2:"HERE'S THE PROBLEM:"::"A 5 KG MASS TRAVELS AT 6 M/S IN A":"CIRCULAR PATH OF RADIUS 2 METERS. "::"WHAT ARE THE INITIAL AND FINAL ANGULAR":T"VELOCITIES16:"LET'S LOOK AT AN EXAMPLE:":::10050w r:P150:P210:P3.5:P44:P50:P$" ROTATING POINT MASS":8000 |P1100:P220:P3.1:P42:P5.3:P$" THE RADIUS IS LENGTHENED":8000P150:P210:P3.6:P46:P$" RETURN TO THE ORIGINA THE EVENT WHERE ANGULAR":E J"MOMENTUM = (I * OMEGA).":::10050 T:5:"NOTICE THAT CHANGING EITHER THE MASS OR"::"THE RADIUS OF A ROTATING BODY WILL ": ^"CHANGE THE MOMENT OF INERTIA, AND WE"::"EXPECT THE ANGULAR VELOCITY TO CHANGE.", hTION OF ANGULAR"::"MOMENTUM:": ," L1 + L2 = L1' + L2'"::" I1*OMEGA1 + I2*OMEGA2 ="::" I1'*OMEGA1' + I2'*OMEGA2'":: 6"WITH NO EXTERNAL TORQUES, ANGULAR "::"MOMENTUM BEFORE AN EVENT = ANGULAR": @"MOMENTUM AFTER VELOCITY"::" V = (R * OMEGA)": " I = MOMENT OF INERTIA"::" 1/2*M*R^2 - DISK"::" M*R^2 - POINT MASS": " M = MASS"::" R = RADIUS":: "WRITE THESE DOWN FOR REFERENCE.":::10050! ":3:"NOW THE LAW OF CONSERVAERVATION OF "::"ANGULAR MOMENTUM. WE'LL BE LOOKING AT": "SIMPLE 1-DIMENSIONAL PROBLEMS INVOLVING"::"ONE OR TWO BODIES.":: "LET'S LOOK AT THE VARIABLES INVOLVED:"::10050' :4:" OMEGA = ANGULAR VELOCITY (RAD/SEC)"::" V = TANGENTIIAL#)d- CONSERVATION OF ANGULAR MOMENTUM4n10000KxA(V)140P1(V)aB(V)80P2(V):4:"HELLO! WHAT'S YOUR NAME? ";Z$:7"WELL, ";Z$;", THIS PROGRAM WILL"::"HELP YOU LEARN TO SOLVE BASIC PROBLEMS":: "WHICH INVOLVE THE CONS             MOMENTUM"42.':" ************************"S28'255:A11000:::(7):h2B'"";A$:A19X,13.75X1 O1!23:"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A18400U1 !1(#" P1 + P2 = P1' + P2'"::" M1*V1 + M2*V2 = M1*V1' + M2*V2'"::1':100:81'" ************************": 2$'" CONSERVATION OF LINEAR:10,11154X:10,1116X:10,1115X:x0D 15,1612:15,169:23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A18200~0N 0 ::21:10:"2-DIMENSIONAL RECOIL":15:19,130 T1800::(7);(7)0 X115.51 1:19X,13:2:19,131.60X:12:110:10,111:12:10,11306/ T1800::(7):(7)/& X014:1:10,11X:10,11X1:12:10,1130X:T140::0:10,11X:10,11X1:10,1130X:/0 (7):21:15:"COLLISION"(0: X06.25:12:10,11154X:1:10,1116X:10,1115X:T135::0(7)P.^X010.25:1:10,11(10.9X):10,11(11.9X):13:10,11(122.7X).h0:10,11(10.9X):10,11(11.9X):10,11(122.7X):X.r23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A18000.|.. ::21:15:"COLLISION"/ 1:10,"::"SHOULD HOLD THE RIFLE FIRMLY!":::^-"THAT'S IT FOR NOW. SEE YOU LATER!":23:10050~-:D13:(7)::20:"END"-"D11500:::(4);"RUN MENU"-?-@::21:17:"RECOIL"-J1:10,1110:10,1111:13:10,1112.TT1800::(7): THE RECOIL WHEN A 95 KG MAN HOLDS":,"THE RIFLE FIRMLY AGAINST HIS SHOULDER"::"(SO THE RECOIL MASS IS 100 KG).":::10050,:4:"ANSWERS:"::" 1. V2' = -.90 M/S":," 2. V2' = -.045 M/S"::*-"THE DIFFERENCE EXPLAINS WHY A SHOOTERTINUE: ";A:A11160p+:4:"WELL, ";Z$;", THAT'S ABOUT IT.":::"HERE'S ONE LAST PROBLEM FOR PRACTICE:"::+"A 5 KG RIFLE FIRES A 15 GM BULLET WITH"::"A MUZZLE VELOCITY OF 300 M/S. FIND THE":%,"RECOIL VELOCITY OF THE RIFLE. ALSO,"::"FIND THETA = ARCTAN (4.267/-6.8)"::8::"THETA = 32.1 DEGREES"::*"WHERE THETA IS THE ANGLE ABOVE THE"::"NEGATIVE X-AXIS. ";:10050*8400:*:6:"IF YOU WOULD LIKE TO SEE THE LAST "::"PROBLEM AGAIN, TYPE 1. TYPE 2 TO":+"CON:24:10050d)x:3:"NOW WE MUST FIND THE TOTAL V3. WE USE"::"TRIG & THE PYTHAGOREAN THEOREM:":)10050::" V3 = SQRT";(91);" V3(X)^2 + V3(Y)^2 ]"::" V3 = SQRT";(91);" -6.8^2 + 4.267^2 ]":)3::"V3 = 8.03 M/S":::"AND,":D*" ::10050Y(P:4:"SUBSTITUTING:"::" 8*0 + 15*0 = 8*(-8) + 15*V3(Y)'"::10050(Z:"FINALLY,"::" 0 + 0 = -64 + 15*V3(Y)'":(d"AND,"::" V3(Y)' = 64/15 = 4.267 M/S":::10050 )n:"SO M3'S Y COMPONENT IS POSITIVE." 15*V3(X)' = -102":U'("AND,":" V3(X)' = -6.8 M/S"::10050::'2"SO THE X COMPONENT OF M3'S VELOCITY IS"::"IN THE NEGATIVE DIRECTION."::'<"NOW FOR THE Y DIRECTION:"::10050 (F:" M2*V2 + M3*V3(Y) = M2*V2' + M3*V3(Y)'"NOW LET'S LOOK AT THE X DIRECTION:"::10050g&:" M1*V1 + M3*V3(X) = M1*V1' + M3*V3(X)'"::10050:& "SUBSTITUTING:"::" 17*0 + 15*0 = 17*6 + 15*V3(X)'":&"OR,":" 0 + 0 = 102 + 15*V3(X)'"::10050':4:"THEN,":" "BOTH PARTS--IT WILL HAVE MOMENTUM COM-"::"PONENTS IN BOTH DIRECTIONS & WILL FLY":q%"OFF AT AN ANGLE."::%"FIND THE MASS OF PART #3."::"ENTER YOUR ANSWER WITHOUT UNITS: ";A%:3:"MASS OF #3:"::" M3 = 40 - 17 - 8 = 15 KG"::,&":"COPY THE PROBLEM, MAKE A SKETCH, AND"::10050=$8400:$:3:"YOU SHOULD NOTICE THAT PART #1 HAS ONLY"::"AN X COMPONENT WHILE PART #2 HAS ONLY A":$"Y COMPONENT. SINCE THE INITIAL TOTAL"::"MOMENTUM IS ZERO, PART #3 MUST BALANCE":U%IN THE POSITIVE X"::"DIRECTION AT 6 M/S. THE SECOND PART":#"(M = 8 KG) MOVES IN THE NEGATIVE Y"::"DIRECTION AT 8 M/S. WHAT IS THE VELOCITY"#"(MAGNITUDE & DIRECTION) OF THE THIRD"::"PART IF THE SHELL HAD A TOTAL MASS OF":1$"40 KG?":22SMALL. WE USUALLY"7"t"NEGLECT THIS EFFECT."::10050"~20:"GETTING TIRED, ";Z$;"? WELL, WE'RE"::"ALMOST DONE!":24:10050":3:"HERE'S A GOOD 2-DIMENSIONAL PROBLEM:"::"A SHELL EXPLODES INTO 3 PARTS. ONE PART":=#"(M = 17 KG) FLIES OFF :3:"SINCE THE SHELL HAS AN UPWARD MOMENTUM,"::"THERE MUST BE A BALANCING MOMENTUM DOWN."!`"BUT IT IS THE GUN AND THE EARTH WHICH"::"RECOIL IN THE Y DIRECTION. AND BECAUSE":"j"THE EARTH IS SO MASSIVE, IT'S RECOIL"::"VELOCITY IS VERY, VERY ." 0 + 0 = 1040 + 600*V2(X)'"::"OR,"::" 600*V2(X)' = -1040":} 8"AND,":8::"V2(X)' = -1.73 M/S"::: B"SO THE GUN RECOILS BACKWARDS WITH A"::"VELOCITY OF -1.73 M/S. WHAT ABOUT THE": L"Y DIRECTION? ";:10050]!VYOU SHOULD VERIFY THESE VALUES."::10050::|"NOW LET'S CONSIDER ONLY THE X-AXIS:"::" M1*V1(X) + M2*V2(X) =":" M1*V1(X)' + M2*V2(X)'"::10050$:3:"SUBSTITUTING:"::" 4*0 + 600*0 = 4*260 + 600*V2(X)'"::10050:OCITY OF THE GUN?":24:"COPY THE PROBLEM AND ";:10050:3:"SINCE THE SHELL'S VELOCITY IS AT AN"::"ANGLE, THE FIRST STEP IS TO FIND THE":"X & Y COMPONENTS:"::" V1(X) = V1*COS(30) = 260 M/S":-" V1(Y) = V1*SIN(30) = 150 M/S"::" M/S":::10050::n"NOW LET'S DO A 2-DIMENSIONAL PROBLEM:"::"A 600 KG GUN IS MOUNTED ON WHEELS. IT":"FIRES A 4 KG SHELL WITH A MUZZLE"::"VELOCITY OF 300 M/S AT AN ANGLE OF 30":"DEGREES ABOVE HORIZONTAL. WHAT IS THE"::"RECOIL VELOCOIL VELOCITY OF THE GUN?"::"ENTER YOUR ANSWER WITHOUT UNITS: ";A:A1.5960]::(7)"SORRY!"::" .015*0 + 6*0 = .015*600 + 6*V2'"::" 0 + 0 = 9 + 6*V2'"::10050:3:"THE CORRECT ANSWER IS:"::10::"V2' = -1.5400*V2'":" 0 + 0 = 100 + 400*V2'"::" 400*V2' = -100"::13::"V2' = -.25 M/S"::::10050:4:"LET'S CONTINUE:"::"A 6 KG RIFLE FIRES A 15 GM BULLET AT A":O"MUZZLE VELOCITY OF 600 M/S. WHAT IS "::"THE REULDN'T"::"THE SIGN BE NEGATIVE?"6f::10050:900p"EXCELLENT, ";Z$;"! THE RECOIL VELOCITY"::"IS -.25 M/S, ASSUMING YOU SAID THE MAN":z"HAD A POSITIVE VELOCITY.":::10050:920 :4:"SOLUTION:"::9000::" 50*0 + 400*0 = 50*2 + SO, YOUR NUMERICAL VALUE IS NOT"::"CORRECT."74850>"YOU SEEM TO UNDERSTAND THAT THE"::"VELOCITY OF THE BOAT SHOULD BE":H"NEGATIVE, BUT YOU HAVE THE WRONG VALUE."R::"TRY AGAIN.":::10050:720"\"YOU GOT THE RIGHT VALUE, BUT SHO4A.25880KNN1:(7);"SORRY, ";Z$;", YOUR ANSWER IS WRONG.":XN3900g A.25860tA0830 "SINCE THE INITIAL SYSTEM MOMENTUM IS"::"ZERO, AND THE MAN HAS A (+) VELOCITY,"::"THE BOAT SHOULD HAVE A (-) VELOCITY.":.*"AL":`"400 KG BOAT WITH A VELOCITY OF 2 M/S."::"WHAT IS THE RECOIL VELOCITY OF THE BOAT?":"RECOGNIZE THAT THE TOTAL MOMENTUM IS"::"INITIALLY ZERO, AND THAT YOU MAY WORK":"IN ONE DIMENSION.":20:"ENTER YOUR ANSWER WITHOUT UNITS: ";A::AL":^"MOMENTUM OF THE SYSTEM WAS ZERO, THE"::"'POSITIVE' MOMENTUM OF THE 15 KG MASS":"MUST BE BALANCED BY THE 'NEGATIVE'"::"MOMENTUM OF THE 5 KG MASS.) ";:10050:N0:4:"NOW YOU TRY ONE:"::"A 50 KG MAN DIVES HORIZONTALLY FROM A+ 5*V2'":D"OR,":" 0 + 0 = 45 + 5*V2'":21:10050x:3:"FINALLY,"::" 5*V2' = -45":"AND,":15::"V2' = -9 M/S":::10050::"YOU SHOULD RECOGNIZE THAT THE 5 KG MASS"::"MOVES TO THE LEFT. (SINCE THE INITICE THAT WE HAVE ASSUMED THE INITIAL"::"VELOCITY IS ZERO (SINCE IT ISN'T GIVEN),"v"AND THE 15 KG MASS HAS A (+) VELOCITY.":23:10050:3:"NOW THE CONSERVATION LAW:"::9000::10050:: "SUBSTITUTE THE DATA:"::" 15*0 + 5*0 = 15*3 THE RECOIL PROBLEM YOU HAVE":PD"SEEN?--OF COURSE YOU DO!":22:10050:8000N::5:"HERE'S HOW TO SOLVE THE PROBLEM:":X"DATA:"::" M1 = 15 KG V1 = 0 M/S V1' = +3 M/S":b" M2 = 5 KG V2 = 0 M/S V2' = ?"::10050::Rl"NOTI5 KG MASS ARE PLACED":s&"ON A FRICTIONLESS SURFACE. AN EXPLOSION"::"DRIVES THEM APART, AND THE 15 KG MASS":0"MOVES TO THE RIGHT AT 3 M/S. WHAT IS"::"THE VELOCITY OF THE 5 KG MASS?":::"COPY THE PROBLEM. DO YOU REALIZE THAT"::"THIS ISWRITE DOWN THE STEPS!":Q"ENTER 1 TO REVIEW, 2 TO CONTINUE? ";A:::A1370"NOW WE'RE READY TO SOLVE A SIMPLE"::"PROBLEM. WE'LL BEGIN WITH 1 DIMENSIONAL":"MOTION (NO Y COMPONENT). ";:10050:3:"PROBLEM:"::"A 5 KG MASS & A 1":3" P(X)=M*V(X) P(Y)=M*V(Y)":24:10050:3:" C. APPLY THE CONSERVATION LAW"::" P1(X)+P2(X)=P1(X)'+P2(X)'":" P1(Y)+P2(Y)=P1(Y)'+P2(Y)'":::" D. SOLVE FOR UNKNOWNS (USUALLY A":" VELOCITY).":::"BE SURE TO GN CONVENTION"::" IS NEEDED (UP & RIGHT ARE POSITIVE,":o11:" WHILE DOWN & LEFT ARE NEGATIVE.)"::"3. STEPS TO SOLUTION:"::" A. RESOLVE VELOCITIES INTO X & Y":" COMPONENTS - V(X) & V(Y)."::" B. FIND MOMENTUM COMPONENTS -M AFTER"::" THE CHANGE.":T9000::" WHERE (') MEANS AFTER THE CHANGE.":23:"WRITE DOWN THESE IDEAS AND ";:10050:3:"2. MOMENTUM IS A VECTOR. THEREFORE, WE"::" MUST WORK WITH X & Y COMPONENTS":<" SEPERATELY. ALSO, A SIET'S GET DOWN TO BUSINESS!":23:10050 r:2:"HERE'S WHAT YOU NEED TO KNOW:":::"1. IN THE ABSENCE OF EXTERNAL FORCES,": |" MOMENTUM IS CONSERVED. THEREFORE, THE":" TOTAL MOMENTUM BEFORE SOME CHANGE IS":" EQUAL TO THE TOTAL MOMENTUN BOTH SITUATIONS, MOMENTUM WILL BE"::"CONSERVED--THAT'S THE BASIS FOR SOLVING":m T"PROBLEMS."::10050 ^12:"IN THIS PROGRAM, WE'LL ONLY LOOK AT "::"RECOIL TYPE PROBLEMS. YOU SHOULD ALSO":& h"WATCH THE PROGRAMS ON COLLISIONS.":::"NOW, LBLEMS INVOLVE 2 OR "::"EVEN 3 DIMENSIONS. ";:10050 ,:6:"DON'T GET NERVOUS, ";Z$;". WE"::"WILL WORK ONLY IN 1 & 2 DIMENSIONS!" 613:"NOW LET'S LOOK AT AN EXAMPLE OF EACH"::"TYPE OF SITUATION.":20:10050 @8000:8200S J::3:"I TWO OR MORE BODIES"::" ARE BLOWN APART.": " 2. COLLISIONS, WHERE TWO OR MORE "::" BODIES RUN INTO EACH OTHER."::10050 ::"SIMPLER PROBLEMS INVOLVE ONLY ONE"::"DIMENSION--MOTION ALONG THE X-AXIS.":; ""MORE COMPLEX PROMENTUM LAW'"::d "IF YOU KNOW WHAT THE LAW SAYS, AND IF"::"YOU HAVE YOUR PAPER, PENCIL AND": "CALCULATOR HANDY, LET'S GO!":24:10050 :3:"MOST MOMENTUM PROBLEMS IN BASIC PHYSICS"::"FALL INTO TWO CATEGORIES:":4 " 1. RECOIL, WHEREi*(d- CONSERVATION OF LINEAR MOMENTUM3n10000]:3:"HI! WHAT'S YOUR NAME? ";Z$::"WELCOME, ";Z$;". IN THE NEXT FEW"::"PROGRAMS WE'LL BE LOOKING AT PROBLEMS": "WHICH CAN BE SOLVED MOST EASILY USING:":::" 'THE CONSERVATION OF MO                              :D$;"RUN DYNAMICS 1"5 :D$;"RUN DYNAMICS 2"P :D$;"RUN DYNAMICS 3"i :D$;"RUN ENERGY 1" :D$;"RUN ENERGY 2" :D$;"RUN ENERGY 3" :D$;"RUN MOMENTUM 1" :D$;"RUN MOMENTUM 2" & D$;"CATALOG" DJ ^D400,410,420,430,440,450,460,470,480,490,500,510,520,530,540,800,550e :D$;"RUN PROJECTILE"~ :D$;"RUN CIRCLE 1" :D$;"RUN CIRCLE 2" :D$;"RUN STATICS 1" :D$;"RUN STATICS 2" :D$;"RUN STATICS 3" :D$;"RUN STATICS 4""- "13. CONSERVATION OF ENERGY - NON-CON"X "14. CONSERVATION OF LINEAR MOMENTUM" "15. CONSERVATION OF ANGULAR MOMENTUM" "16. CATALOG" "17. EXIT THIS PROGRAM" ,:"TYPE IN THE NUMBER AND PRESS ";D JD110 YD1710" 6. STATICS - LADDERS"= d" 7. STATICS - INCLINED PLANES"Z n" 8. DYNAMICS - METHOD"| x" 9. DYNAMICS - TRANSLATION" "10. DYNAMICS TRANSLATION & ROTATION" "11. CONSERVATION OF ENERGY - METHOD" "12. CONSERVATION OF ENERGY - CON. FORCE :D$(4)A::"WHICH PROGRAM WOULD YOU LIKE TO VIEW ?"Gd(" 1. PROJECTILE MOTION"2" 2. CIRCULAR MOTION - CONSTANT SPEED"<" 3. CIRCULAR MOTION - CHANGING SPEED"F" 4. STATICS - METHOD"P" 5. STATICS - BEAM PROBLEMS" Z   ">">>""""">""6""">">*":/$'" CONS. OF ENERGY: NON-CONS. FORCES"Y.':" ************************"x8'255:A11000:::(7):B'"";A$:L'32:"";A$:6">H THE":" SUPPORT OF NATIONAL SC>">66PR#0":(zD11500:::D$;"RUN MENU".'#V(#1:X,YX10,YX10,Y8X,Y8X,Y:z#2:X,YX10,YX10,Y8X,Y8:#1:X,67X80,67X80,27X,67:T$3:X,YX10,Y5X16,Y5X6,Y10X,Y:':100:8'" ************************ (u)(M1)(g)(d)":hH"WHERE h1 = 1, h2 = .7, AND d = .7 ":::"THUS,"::" V = 1.91 M/S"::10060R13376:10:"THAT'S THE END OF THIS PROGRAM."::"IT'S BEEN NICE WORKING WITH YOU."\20:10050f13376:D13:(7)::20:"END"pD$;"'VE REALLY GOT THE IDEA.":20:10050:850V 3:"ANSWER:"::" INITIAL ENERGY--":*" Ug1 = (M1)(g)(h1)":" Ug2 = (M2)(g)(h2)":4" FINAL ENERGY--"::" K1 = 1/2(M1)(V^2)">" K2 = 1/2(M2)(V^2)":" Ug1 = (M1)(g)(h1)":" Ef =S SUSPENDED 0.7 METERS ABOVE THE":"FLOOR. WHEN RELEASED, BOTH MASSES ACCEL-":"ERATE. FIND THE IMPACT VELOCITY OF THE": "3 KG MASS ON THE FLOOR.":::"SOLVE THIS ONE COMPLETELY."::" ANSWER? ";A:13376:A1.91800)10:"EXCELLENT! YOUE"::"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"::"MASg)(.866)(d)"::10060T13376:"#5 SUBSTITUTE THE DATA:"::" (6)(9.8)(1.5) =":" 1/2(6)(V^2) + (.1)(6)(9.8)(.866)(3)"::10060:"FINALLY, WE FIND:"::" V = 4.93 M/S";:10060934,0:13376:10:"IF YOU WOULD LIKE TO REVIEW THES060X::"SO THE FINAL ENERGY IS:"::" 1/2(M)(V^2) + (u)(M)(g)(.866)(d)"::1006013376:"NOTICE THAT d IS THE LENGTH OF THE"::"INCLINE.";:10060::"#4 EQUATING INITIAL AND FINAL,"::" (M)(g)(h) = ":" 1/2(M)(V^2) + (u)(M)(S":::"#3 IN THE FINAL CONDITION, THERE IS":l" K = 1/2(M)(V^2) AND Ef = (u)(Fn)(d),"::" BUT NO Ug OR Us.";:10060v13376:"WHAT IS THE NORMAL FORCE ON THE MASS"::"WHEN IT IS ON THE INCLINE?";:10060:" Fn = M * g * cos(30)";:10800:iN::"#2 IN THE INITIAL CONDITION, THERE IS"::" ONLY Ug = (M)(g)(h). WHAT IS h?"::10060X3:99,26112,26:109,27109,66:110,27110,66:7:15:"h"/b13376:"THE HEIGHT = RAMP LENGHT * SIN(30)"::" h = (3)(0.5) = 1.5 METER29:"FINAL":8:7:"30";5:1:"u = .1":6:22:"V = ?"l&158,55170,49:159,53160,55:160,52162,56034,12:13376:"STUDY THE PROBLEM A MINUTE.";:10060:::"#1 SELECT A REFERENCE HEIGHT.";:10060D5:99,67149,67:267,67277,67:A1OF THE MASS AT":d"THE BOTTOM OF THE INCLINE IF THE CO-"::"EFFICIENT OF FRICTION IS 0.1?"::"COPY THE PROBLEM AND ";:22:1005013376:X13:9200:X181:920093,26:261,26X93:Y25:9300:X179:Y67:93001:5:"INITIAL";:METERS":::"THAT'S NOT SO HARD!"::1005071337634,0:13376:3:"READY FOR ANOTHER PROBLEM, ";Z$;"?"::"HERE YOU ARE:":::"A 6 KG MASS IS RELEASED FROM THE TOP OF":"A 30 DEGREE INCLINE WHICH IS 3 METERS"::"LONG. WHAT IS THE SPEED ::" 1/2(1000)(.06^2) = (.2)(.1)(9.8)(d)"::10060:K::::::20" WHAT'S THE ANSWER? ";A:13376:A9.1845015:"GREAT,";Z$;". SO FAR, SO GOOD!":::10050:460+12:" d = 1/2(1000)(.06^2)/(.2)(.1)(9.8)":::" d = 9.18 *(d)"::10060b:"#4 EQUATING INITIAL AND FINAL,"::" 1/2(K)(x^2) = (u)(Fn)(d)"::1006013376:"SINCE THE NORMAL FORCE (Fn) IS THE"::"WEIGHT OF THE MASS,":" 1/2(K)(x^2) = (u)(M)(g)(d)"::10060:7"#5 SUBSTITUTING VALUES,""#2 IN THE INITIAL CONDITION,"::" > THERE IS NO K, Ug (h=0), OR Ef,": h" BUT THERE IS Us = (1/2)(K)(x^2).":::10050 r13376:"#3 IN THE FINAL CONDITION,"::" > THERE IS NO K (V=0), Ug, OR Us,":|" BUT THERE IS Ef = (u)*(Fn)2:A,Y6A,Y2:K ;X223:9000:X154:9100:AX8X44:A,Y6A2,Y2:~ @1:5:"INITIAL";:26:"FINAL":6:26:"d = ?" J34,10:13376:10:"* HERE IT IS! *";:10060::"#1 SELECT A REFERENCE HEIGHT.";:10060 T5:36:"h=0":A1800::15L ^ESSED 6 CM AND RELEASED, DETERMINE":i "HOW FAR THE MASS WILL TRAVEL BEFORE"::"COMING TO REST.":: ""COPY THE PROBLEM, THEN WE'LL SOLVE IT."::10050 ,13376::16302,0:5:10,38100,38:140,38235,38 6X27:Y37:9000:X24:9100:AX8X10050p 13376:3:"HERE'S A STRAIGHT-FORWARD PROBLEM WHICH"::"INVOLVES ENERGY 'LOSS' DUE TO FRICTION.":: "A 100 GM MASS IS PLACED IN FRONT OF A"::"SPRING (K = 1000 NT/M) ON A HORIZONTAL":% "TABLE (u = 0.2). IF THE SPRING IS COM-"::"PR;Z$::Z "IN THIS PROGRAM, WE LOOK AT SOME"::"APPLICATIONS OF THE CONSERVATION OF": "ENERGY LAW WHICH DEAL WITH BOTH"::"CONSERVATIVE & NON-CONSERVATIVE FORCES.":: "IF YOU'VE ALREADY WATCHED THE 'METHOD'"::"PROGRAM, YOU'RE READY."::4d - CONSERVATION OF ENERGY: NON-CONS. FORCESAn:10000Qx:16302,0i- TURN ON ROMPLUS+uD$(4)D$;"PR#5"R$(13)R$M$"":- THERE IS A HEREM$;"1A" 13376:3:"HELLO!"::"NAME, PLEASE? "     ^1000:::(7):%B'"";A$:>L'32:"";A$:6">H THE":" SUPPORT OF NATIONAL SC>">66">">>""""">""6""">">#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 DEALS WITH NON-":Rz"CONSERVATIVE FORCES.":::"BYE!":::10050x13376:D13:(7)::20:"END"D$;"PR#0":D11500:::D$;"RUN MENU"'#(#5:X,YX10,YX10,Y8X,Y8X,Y:#1:X,YX4,Y:X2,Y2X2,Y2::)(.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/S":::10050p13376:9:"THAT'S THE END. YOU SHOULD ALSO SEE"::"THE PLINED AT AN":k4"ANGLE OF 30 DEGREES, HOW FAST IS THE"::"MARBLE GOING WHEN IT GETS TO THE TOP?":::>"SOLVE THE PROBLEM AND ";:23:10050H13376:3:"HERE'S THE ANSWER:"::"> INITIALLY, ALL ENERGY IS IN THE SPRING"0R" Us = 1/2(800HAT'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 KG MASS DROPS OUT, AND:"::" (3)(9.8)(0.7) = 1/2(9)(V^2)"::1006013376:"FINALLY,"::" V^2 = (2)(3)(9.8)(.7)/9":"AND,":" V = 2.14 M/S":" :"T:" 1/2(M1)(V^2)+(M1)(g)(h1)+1/2(M2)(V^2)":910060:"NOTICE THAT h1 IS THE SAME IN BOTH"::"PLACES, AND THE VELOCITIES OF THE":"MASSES ARE EQUAL.";:1006013376:"#5 NOW SUBSTITUTE THE DATA:"::" (6)(9.8)(1) + (3)(9.8)(0.7) =":5H THE":" SUPPORT OF NATIONAL SC>">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$;"RUN MENU"'#(#5:X,YX10,YX10,Y8X,Y8X,Y:Z#X,84X,94X8,94X8,84:Y84922:X,YX8,Y"::10050:810a "CORRECT! WORK IS EQUIVALENT TO THE"::"AMOUNT OF ENERGY GAINED."::10060:*34,0:13376:10:"THAT'S IT FOR THIS PROGRAM, ";Z$;".":4"YOU SHOULD NOW WATCH THE CONSERVATION"::"OF ENERGY PROGRAMS WHICH DEAL WITH":G>"COTING 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 * X. HEIGHT),"::" SO K = 0."::"AND...";:10060g13376:"> THE MASS HAS GRAVITATIONAL ENERGY."::"> THERE IS NO STORED ELASTIC ENERGY."::"> THERE ARE NO FRICTIONAL LOSSES."::"#4 SO EQUATING THE INITIAL & FINAL"::" ENERGIES...";:.";:10060Zv13376:"> THE SPRING IS COMPRESSED, SO THERE"::" IS ELASTIC ENERGY.":"> THERE ARE NO FRICTIONAL LOSSES."::1006013376:"#3 IN THE FINAL CONDITION:"::"> NOTHING IS MOVING (THE MASS IS":1" MOMENTARILY AT REST AT MAT YOU WISH."::10050xX13376:11:33:"h = 0":15:1:"WE'VE CHOSEN THE GROUND AS OUR REFERENCE":"HEIGHT.";:10060b13376:"#2 NOW WE NOTE THAT INITIALLY,"::"> NOTHING IS MOVING, SO K = 0.": l"> THE MASS IS AT h = 0, SO Ug = 0."::"AND..X72:9050:X162:9050:162,82170,84x:::" HERE IT IS.";:10060:::" THE PROBLEM IS TO FIND (h)."D3:178,22190,22:1:185,23185,82:7:29:"h = ?":22:10060N13376:"#1 SELECT A REFERENCE HEIGHT. CHOOSE"::" ANY HEIGHRING WHICH IS THEN RELEASED.":m:"HOW HIGH DOES THE BLOCK GO?":20:"COPY THE PROBLEM AND ";:22:1005013376:3:35,83215,83:0:72,8380,83:162,83170,83:X71:Y82:9000:X161:Y22:9000&2:2:"INITIAL";:31:"FINAL":14:34,14%02: THE ENERGIES"::"ASSOCIATED WITH EACH BODY IF MORE THAN":f"ONE OBJECT IS INVOLVED!"::1005013376:3:"HOW ABOUT A SAMPLE PROBLEM?":::"A SPRING (K = 1200 NT/M) IS COMPRESSED":"10 CM. A 0.4 KG BLOCK IS PLACED ON TOP"::"OF THE SP76:3:"METHOD:"::"1. ESTABLISH A REFERENCE HEIGHT (h = 0).":"2. DETERMINE THE INITIAL K, Ug, Us, Ef.":"3. DETERMINE THE FINAL K, Ug, Us, Ef."::"4. EQUATE #2 WITH #3.":"5. SOLVE FOR THE UNKNOWNS."::10050::>"BE SURE THAT YOU FIND OF ANY ACTION. ADD UP"::"ALL THE ENERGIES AT THE START, AND SET":"THEM EQUAL TO ALL THE ENERGIES AT THE"::"END. IT'S REALLY EASY, ";Z$;"!"::10050::"NOW WE'LL LIST THE METHOD IN DETAIL."::"BE SURE TO COPY IT DOWN."::10050k133z#Ҡ ҠΠŠŠϠ Š+ Š ӠӠ ӠӠҠ ! Ӡ"٠٠٠ՠ ͠, ͠ àΠӠ Ϡ  ӠӠ      Ġ٠  ŪϠؠĠȠ   Ӡ   3.46 M":^0" Rc = 8 * sin(55) = 6.55 M"::"YOU NEED TO BE VERY CAREFUL HERE!"::10060:13376:"#10 WRITE AN EXPRESSION FOR EACH TORQUE:":10060D" (Wl) = (Ra)(Wl)"::" (W) = (Rb)(W)"::" (Fw) = (Rc)(Fw)" NX27:Y134:900PLEASE.";A$]13376:"#9 YOU FIND THE RADIUS ARM TO EACH"::" LINE OF ACTION.";:100605:53,6984,69:49,6549,10:10:9:"Ra";:12:"Rb":6:5:"Rc"13376:"HERE THEY ARE:": &" Ra = 4 * cos(55) = 2.29 M"::" Rb = 6 * cos(55) =13376:"#8 YOU SHOULD DRAW 'LINES OF ACTION'"::" ON YOUR SKETCH. SINCE Fn AND Ff ACT":" AT THE POINT OF ROTATION, WE DON'T"::" NEED TO WORRY ABOUT THEM."::100606:76,6476,70:86,4886,70:93,947,9 :"*** AGAIN, "::10060:\"#6 NOW WRITE AN EQUATION FOR THE SUM"::" IN THE y-DIRECTION.";:1006013376:"HERE IT IS:"::" Fn - Wl - W = 0"::10060:"#7 WE'LL CHOOSE THE BOTTOM OF THE"::" LADDER AS OUR POINT OF ROTATION."::10060_ NEED TO RESOLVE FORCES"::" INTO x AND y COMPONENTS!";:10060::"#5 YOU WRITE THE EQUATION FOR THE SUM"::" OF THE FORCES IN THE x-DIRECTION:"::1006013376:"HERE IT IS:"::" Ff - Fw = 0": "REMEMBER THAT Ff = (u)(Fn)AT THERE IS NO FRICTION ON"::"THE WALL (THIS WILL ALWAYS BE TRUE FOR"::"THIS COURSE).";:1006013376:"#3 ESTABLISH THE USUAL SIGN CONVENTIONS.":" UP, RIGHT, AND COUNTER-CLOCKWISE":" ARE POSITIVE.";:10060E13376:"#4 THERE IS NO:5:11:"Fn"U h2:76,3976,51:86,2586,44:50,7050,91:32,6949,69:100,9118,9 r3:75,5077,50:85,4387,43:49,7151,71:47,6847,70:102,8102,10 |34,13:13376:"IF YOU DIDN'T GET IT RIGHT, CORRECT"::"YOUR SKETCH NOW.":b"NOTICE THCTION: ON THE FLOOR ONLY"::G 6" COPY THIS DATA AND ";:23:10050 @13376:"#2 FREE-BODY DIAGRAM."::" MAKE YOUR OWN SKETCH. THEN": J" CHECK BY PRESSING";:22:10050 T62450:9100 ^19:2:"Fw":14:6:"W":11:8:"Wl":3:9:"Ff"100:85,1087,1087,1185,11:5:30,70101,70101,1:2:86,1286,23:83,1489,14:88,1890,20q X60:Y67:9010 "34,12:13376:" #1 HERE'S THE PICTURE:"::" LADDER: 20 KG, 8 M, THETA = 55 DEGREES": ," MAN: 800 NT, 6 M UP THE LADDER"::" FRI13376:6:"PROBLEM:"::"IN THE FOLLOWING PROBLEM, YOU ARE TO": "FIND THE MINIMUM COEFFICIENT OF FRICTION":"WHICH WILL PREVENT THE LADDER FROM"::"SLIPPING, I.E. HOLD IT IN EQUILIBRIUM.": ::"WRITE THIS DOWN AND ";:21:10050] 13376:9KS, ";Z$;"! IN THIS PROGRAM"::"WE'LL WORK ON A DETAILED ANALYSIS OF A": "LADDER IN STATIC EQUILIBRIUM. YOU SHOULD":"HAVE ALREADY WATCHED THE PROGRAM:": " 'STATICS - METHOD'":::"IF YOU ARE READY, LET'S CONTINUE..."::10050@ ɴd- STATICS - LADDERS'n:100007x:16302,0O- TURN ON ROMPLUS+[D$(4)jD$;"PR#5"wR$(13)R$M$"":- THERE IS A HEREM$;"1A"13376:3:"HI THERE!"::"WHAT IS YOUR NAME? ";Z$::J "THAN        ">"">">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'" ************************":#$'"::"SYSTEM IN EQUILIBRIUM?"::J ,"COPY THE PROBLEM AND ";:22:10050u 613376:2:191,18111,1861,90191,90 @X87:Y49:9100:111,17108,13 JX149:Y17:9200 T3:148,13108,1383,47:X70:Y87:9010 ^1:22:"M2":4:26:"u2":5:16::"DEGREE INCLINE WHERE THE COEFFICIENT": "OF FRICTION IS 0.2. IT IS CONNECTED BY"::"A LIGHT CORD AND PULLEY TO AN 8 KG": "MASS RESTING ON A HORIZONTAL TABLE."::"WHAT IS THE MINIMUM COEFFICIENT OF":! ""FRICTION ON THE TABLE TO HOLD THE 13376:6:"IN SOME WAYS, THIS TYPE OF PROBLEM IS"::"EASY BECAUSE WE WILL NOT NEED TO WORK": "WITH TORQUES. ON THE OTHER HAND, WE"::"WILL WORK WITH MORE THAN ONE OBJECT.":21:10050* 13376:3:"PROBLEM:"::"A 10 KG MASS IS PLACED ON A 60": "THANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL WORK ON A DETAILED ANALYSIS OF AN": "INCLINED PLANE IN EQUILIBRIUM. YOU"::"SHOULD HAVE ALREADY WATCHED:": " 'STATICS - METHOD'":::"IF YOU ARE READY, LET'S CONTINUE..."::10050_ "d- STATICS - INCLINED PLANES/n:10000?x:16302,0W- TURN ON ROMPLUS+cD$(4)rD$;"PR#5"R$(13)R$M$"":- THERE IS A HERE.M$;"1A"13376:3:"HI THERE!"::"WHAT IS YOUR NAME? ";Z$::T     OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">700:8/'" ************************":T$'" STATICS - LADDERS"~.':" ************************"8'255:A11000:::(7):B'"";A$:L'31:"";A$:6">H THE":" SUPPORT 1500:::D$;"RUN MENU"'#(#3:X,Y5:X1,Y6:X2,Y6:X3,Y1X3,Y6:X4,Y:X4,Y5:X5,Y:X5,Y5:X6,Y1:X6,Y6:2#3:X1,YX3,Y:X4,Y1X4,Y5:X1,Y6X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:#3:50,6999,9:49,6898,8:':1OULD THE MAN BE":H "ABLE TO GO BEFORE THE LADDER SLIPPED?"::10050*13376:5:"THE ANSWER IS: 4.71 M":9:"THAT'S IT FOR NOW, ";Z$;".":4:"SEE YOU A LITTLE LATER!":20:10050>13376:D13:(7)::19:"END"HD$;"PR#0":RD176:8:"IF YOU WANT TO REVIEW THIS PROBLEM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A1240 13376:6:"HERE'S A PRACTICE PROBLEM FOR YOU."::"IF THE COEFFICIENT OF FRICTION IN THE":"LAST PROBLEM HAD BEEN ONLY 0.40, HOW"::"HIGH UP THE LADDER W800 = 996 NT"::"OK?";:10060p:"NOW FOR THE EASIEST PART--FIND THE"::"COEFFICIENT OF FRICTION.";:1006013376:"BY DEFINITION, Ff = (u)(Fn)"::"SO, u = Ff/Fn":" u = 0.49":::"THAT'S IT! YOU'RE DONE!!";:10060Y34,0:133"THE FORCE OF FRICTION, Ff.";:10060_13376:"OF COURSE, Ff = Fw = 491.1 NT"::10060::"NOW USE THE RESULTS OF STEP #6 TO FIND"::"THE NORMAL FORCE, Fn.";:1006013376:"FROM STEP #6,"::" Fn - Wl - W = 0":" Fn = 196 + (2.29)(196) -":Z" (3.46)(800) = 0":::"DON'T GET LAZY, ";Z$;"!":"YOU FIND Fw, THEN CHECK.";:1006013376:"THE ANSWER IS Fw = 491.1 NT."::"HOW DID YOU DO?";:10060::%"NOW USE THE RESULTS OF STEP #5 TO FIND"::RE READY TO SOLVE! REMEMBER"::" THAT Wl = M*g = (20)(9.8) = 196 NT,"v:" AND THE WEIGHT OF THE MAN IS 800 NT.":" SO, USING THE TORQUE EQUATION:":" (Rc)(Fw) - (Ra)(Wl) - (Rb)(W) = 0"::1006013376:"SO,"::" (6.55)(Fw) - 0:Y150:9000:Y166:9000:10060X13376:"#11 NOW WRITE THE EQUATION FOR THE SUM"::" OF THE TORQUES. BE CAREFUL OF SIGNS!"::10060:b" (Fw) - (Wl) - (W) = 0"::X33:Y158:9000:X96:9000:X159:9000:10060Il13376:"#12 NOW WE'#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$;"PR#0":v4D11500:::D$;"RUN MENU"|'#2#3:X1,YX3,Y:X4,Y1X4,Y5:X1,Y6X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:#1:X,YX6,Y8X16,Y2X10,Y6X,Y:$S AN ADJUSTABLE"::"ANGLE. TO WHAT MAXIMUM ANGLE CAN THE":q"INCLINE BE RAISED BEFORE THE MASS"::"SLIPS?"21:"SOLVE THIS ONE YOURSELF.";:10060 13376:4:"THE ANSWER IS 22 DEGREES.":10:"THAT'S THE END OF THIS PROGRAM.":"SEE YOU58":::"THAT'S IT. YOUR'RE DONE!";:1006034,0:13376:8:"IF YOU WANT TO REVIEW THIS PROBLEM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A126013376:5:"HERE'S ONE LAST PROBLEM:"::"A 25 KG MASS RESTS ON AN INCLINE":<"(u = 0.4) WHICH HA2, MAKING"::"SUBSTITUTIONS:";:10060^:" T - Ff2 = 0"::" T - u2*M2*g = 0"::"NOTE THE SUBSTITUTION FOR Fn2.";:1006013376:"NOW SOLVE FOR u2:";:10060:::" 75.1 - (u2)(8)(9.8) = 0":+:" u2 = 75.1/78.4 = .9NOWN EXECPT FOR THE"::"TENSION. YOU SOLVE IT!";:1006013376:"HERE IT IS:"::"(10)(9.8)(.866) - T -"::" (.2)(10)(9.8)(.500) = 0"::" T = 84.868 - 9.8 = 75.1 NT":::"IT'S EASY!";:10060&13376:"NOW REWRITE EQUATION ORQUES! ALL YOU NEED TO DO IS SOLVE"::"THE EQUATIONS FROM STEP #5!";:10060v13376:"REWRITE THE FIRST EQUATION, MAKING"::"SUBSTITUTIONS:";:10060:" Fp - T - Ff1 = 0"::" M1*g*sin(60) - T - u1*M1*g*cos(60) = 0":8:"EVERYTHING IS K WRITE THESE EQUATIONS NOW.";:10060cN13376:"HERE THEY ARE:"::" 1. Fp - T - Ff1 = 0":X" 2. T - Ff2 = 0":::"OK, ";Z$;"?";:10060b13376:"HERE'S THE NEAT PART: WE DON'T NEED TO"::"WORRY ABOUT THE y-DIRECTION OR THE":Ml"T"KNOW HOW TO DO THIS!";:10060r5::"FROM NOW ON WE'LL WORK ONLY WITH THE"::"COMPONENTS Fp AND Fn1.";:10060:13376:"#5 SUM OF FORCES IN THE x-DIRECTION:"::" THE POSITIVE x-DIRECTION IS DOWN":&D" THE RAMP AND LEFT ON THE TABLE."::":"::" Fn1 = Fg1 * cos(60) = M1*g*cos(60)":s" Fp = Fg1 * sin(60) = M1*g*sin(60)":::"ALSO NOTE THAT:":" Fn2 = Fg2 = M2*g":10060&13376:"RESOLVING THE FORCE OF GRAVITY INTO Fp"::"AND Fn IS VERY IMPORTANT. BE SURE  YOU":0S" ON THE INCLINE, WE MUST RESOLVE THE"::" Fg INTO p AND n COMPONENTS."::" YOU TRY IT ON YOUR OWN.";:1006064,4452,60:72,4382,49:51,5855,60:80,4982,47:6:9:"Fp":13:7:"Fn1":X72:Y56:9010.13376:"RESOLVED VECTORS THAT Ff = u * Fn"::10060u13376:"#3 ESTABLISH CONVENTIONS."::" WE'LL TAKE MOTION DOWN THE PLANE AND":" TO THE LEFT ON THE TABLE AS POSITIVE."::1006013376:"#4 RESOLVE FORCES."::" ON THE TABLE, FORCES ARE RESOLVED.":E NORMAL"::"FORCE IS THE WEIGHT OF THE MASS."::1006013376:"THE REACTIVE FORCES, Fr, ARE THE"::"SUPPORT PROVIDED BY THE INCLINE AND"::"TABLE. THEY ARE NEEDED TO BE COMPLETE,"::"BUT WE WON'T WORK WITH THEM.";:10060::"ALSO, REMEMBER73,2675,28:76,2978,31=5:67,4867,69:3:66,6868,68v12:3:"T":13:4:"Ff1":5:4:"Fr1":10:9:"Fg1"34,12:13376:"HERE IT IS. NOTE THAT THE FORCE OF"::"FRICTION OPPOSES THE DIRECTION OF":8"DESIRED MOTION. AND, FOR M2, TH42:92003:154,38168,38:180,38194,38:174,33174,20:174,43174,56:156,37156,39:192,37192,39:173,21175,21:173,55175,552:24:"Fr2":9:22:"Fn2 = Fg2":5:30:"Ff2":5:21:"T"62,3752,31:69,3475,26:72,3778,29:52,3354,31:"u1":6:9:"M1"g h34,14:13376:"#1 HERE'S OUR PICTURE. MAKE YOUR"::" OWN SKETCH.";:10060:: r"#2 FREE-BODY DIAGRAM."::" MAKE A FREE-BODY SKETCH FOR BOTH": |" BODIES. THEN CHECK BY ";:26:1005062450:X75:Y39:9100:X169:Y:21:"HERE'S X.":24:10050CX120:3:74,87173X,54X]5:61,90X61,903XxX4Ē1:21,110181,30D150:D 0:74,87173X,54X:5:59,14863,148:57,14665,146 3:X70:Y150:9100:X78:9600:X86:90006*:21:8000:8100:8200:570M::3:"THUS, THE DIRECTION OF X IS UP."::"WHAT IS THE DIRECTION OF X?"::"IF YOU SAID DOWN, YOU'RE RIGHT!"::" X = - X":::"LET'S LOOK AT IT.":24:10050"::8000:8100:8200b3:74,87194X,75X.l5:61,8861,883X=vD150:DV0:74,87194X,75X\}5:59,3063,30:57,3265,323:X70:Y31:9000:X78:9600:X86:9100:21:"TYPE 1 TO WATCH AGAIN.":"TYPE 2 TO CONTINUE. ";TT2710:ST BE A VECTOR IN THE +Z":v&"OR -Z DIRECTION. REMEMBER, THE CROSS"::"PRODUCT ANSWER MUST BE PERPENDICULAR TO":0"THE PLANE OF & . WATCH!":24:10050::16304,0:21:"ROTATE TO WITH YOUR RIGHT HAND.":24:10050DX12055)"::" X = 12 * 11 * 0.819"::x" X = 108.1 FT-LBS":::"THAT'S NOT SO BAD, IS IT? ";:10050:3:"NOW LET'S SEE HOW TO FIND THE DIRECTION"::"OF X."::"SINCE & LIE IN THE X-Y PLANE,"::"THE ANSWER MU480I:CHRS$(7):"OPPS! THAT'S INCORRECT. TRY AGAIN.":24:10050:390::"GOOD! NOW CALCULATE X."::"ENTER YOUR ANSWER WITHOUT UNITS: ";T::T107T110520(7):"ERROR! HOW DID YOU MESS UP?":*" X = 12 * 11 * SIN(>.":24:10050\:8200:21:"HERE'S . BOTH VECTORS ARE IN THE":"X-Y PLANE."::10050:X160:Y71:9500:21:"HERE'S THE ANGLE BETWEEN & .":24:10050::3:"WHAT IS THE ANGLE THETA?"::"ENTER THE NUMBER OF DEGREES? ";TT55ĺ::" = 12 FEET, 20 DEGREES N OF E":" = 11 POUNDS, 15 DEGREES E OF N":::"WRITE DOWN THE PROBLEM. THEN WE'LL"::"LOOK AT IT.":24:10050::8000:21:"HERE'S OUR COORDINATE SYSTEM.":24:10050:8100:21:"HERE'S VECTOR TO":l r" THROUGH THE SMALLEST ANGLE. YOUR"::" THUMB SHOWS THE RESULT.":: |"THIS RULE IS FOR X. TWIST TO"::" FOR X. MAKE A NOTE OF THIS!":24:10050,:3:"HERE'S A PROBLEM. FIND X IF:":B * SIN(THETA)":::"WHERE THETA IS THE SMALLEST ANGLE": T"BETWEEN & .":::"THE HARD PART IS FINDING THE DIRECTION": ^"OF THE RESULTING VECTOR. WE USE THE"::"RIGHT HAND RULE.":24:10050 h:3:"RIGHT HAND RULE:"::" USE YOUR RIG"NO"320 6:3:"NO? THEN YOU HAD BETTER WATCH THE"::"PROGRAM: 'DOT PRODUCTS - M/D FORM'."::"LOAD THAT PROGRAM BEFORE YOU GO ON.": ;:10050:1000 @:3:"GREAT! THE MATHEMATICS OF CROSS "::"PRODUCTS ARE EASY:"::: J" X = A * N DOT PRODUCTS, YOU KNOW":w "THE NOTATION WE USE & YOU KNOW THAT THE"::"CROSS PRODUCT OF TWO VECTORS PRODUCES": "A THIRD VECTOR AT RIGHT ANGLES TO THE"::"PLANE FORMED BY THE FIRST TWO.": ""RIGHT, ";Z$;:"? (TYPE YES OR NO): ";T$ ,T$, " 'DOT PRODUCTS - M/D FORM'": :"DON'T START THIS PROGRAM WITHOUT SEEING"::"THE OTHER ONE FIRST!":24:10050 :3:"WHAT'S YOUR NAME? ";Z$:::"THANK YOU, ";Z$;"." ::"LET'S GET STARTED. SINCE YOU'VE DONE"::"THE PROGRAM O!d-CROSS PRODUCTS - M/D FORM,n10000k:4:"HELLO!"::"THIS PROGRAM IS DESIGNED TO HELP YOU":"FIND THE CROSS PRODUCT OF TWO VECTORS"::"EXPRESSED IN MAGNITUDE/DIRECTION FORM.": "YOU SHOULD HAVE ALREADY WATCHED THE"::"PROGRAM:":          ">H THE":" SUPPORT OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">" KNOW THAT A DOT PRODUCT"::"CANNOT GIVE A VECTOR ANSWER. TRY AGAIN."::320|:"YES! A DOT PRODUCT OF A UNIT VECTOR"::"WITH ITSELF IS ALWAYS ONE."::"4. WHAT IS THE RESULT OF X?":" (1) <-I> (2) <-J> (3) "::"ANSW) J" (1) 0 (2) 1 (3) ":H T:"ANSWER 1,2 OR 3";::A[ ^A360,380,370 h:"INCORRECT. THE DOT PRODUCT USES THE"::"COSINE TERM. SINCE THE ANGLE BETWEEN A"::"VETOR AND ITSELF IS 0, THE COSINE IS 1."::320Mr:"SORRY. YOU MUST"2. ANOTHER NAME FOR THE CROSS PRODUCT"::"IS THE VECTOR PRODUCT.":d :"TRUE OR FALSE";::B${ ":(B$,1)"T"310 ,:"INCORRECT. TRY IT AGAIN."::270 6:"VERY GOOD, ";Z$;". LET'S CONTINUE.": @:"3. WHAT IS THE RESULT OF .?":":6 :"1. THE DOT PRODUCT . IS A SCALAR.":T :"TRUE OR FALSE";::B$k :(B$,1)"T"260 :"WRONG. YOU SHOULD KNOW A DOT PRODUCT"::"ALWAYS GIVES A SCALAR. TRY AGAIN."::220 :"EXCELLENT! DOT PRODUCTS YIELD SCALARS.":F :X::2,529:28,6:27,6:26,5:6,28:7,28o 13:3,3519:3,712:15,3:14,4:13,5:14,6:15,7:16,34:17,34 "THIS IS THE AXIS SYSTEM WE'LL USE.":"COPY IT DOWN. NOTE <-J> IS TOWARDS YOU.": 10050 ::"NOW WE NEED TO CHECK YOUR KNOWLEDGE.:: " MEANS THE VECTOR A"::" MEANS THE UNIT VECTOR I"::" . MEANS THE DOT PRODUCT OF A&B"::" X MEANS THE CROSS PRODUCT OF A&B": 10050 :1:7,3019:33,3516:16,2034:33,3520:4,17:5,17* 12:X928:X,38@ d100003n"HELLO! WHAT'S YOUR NAME? ";Z$:x"WELL, ";Z$;", THIS PROGRAM IS"::"DESIGNED TO HELP YOU FIND THE"::"DOT & CROSS PRODUCT OF TWO VECTORS.": "LET'S WORK WITH VECTORS IN UNIT VECTOR"::"FORM ONLY. WE'LL USE THESE NOTATIONS:"        X2,Y3:X3,Y2:X4,Y1:X4,Y5:X3,Y4:X1,Y2:X,Y1:K':100:8u'" *************************":$'" CROSS PRODUCTS - M/D FORM"::" *************************".'255:D11000:::(7):B'"";A$: :X4,Y1:X4,Y5:XT$X,Y6X,Y4:X1,Y3X3,Y3:X4,Y4X4,Y6:X2,Y2X2,Y:$X,Y6X4,Y6:X4,Y5:X3,Y5X,Y1:X,Y1:X,YX4,Y:%X,Y1X,Y5:X1,Y6X3,Y6:X4,Y5X4,Y1:X1,YX3,Y:X1,Y3X5,Y3:=%X,Y5:X1,Y4:181:Y50:9100:P(#X,YX,Y5:X1,Y6X3,Y6:X4,Y5X4,Y:X,Y3X4,Y3:#X,YX,Y6X3,Y6:X4,Y5X4,Y4:X3,Y3X,Y3:X4,Y2X4,Y1:X3,YX,Y:#X,Y:X,Y6:X4,Y6:X4,Y:X,Y1:X,Y5:X1,Y2:X1,Y4:X2,Y3:X3,Y2:X3,Y4:(4);"RUN MENU"?L@1:61,361,156:10,90229,90:21,110181,30J3:X224:Y102:9200:X49:Y10:9400:X170:Y27:9300:2:65,89200,76196,74196,79200,763:X206:Y81:9000: 2:65,89176,51169,51171,55176,51 3:XTHE DIRECTION IS NORTH! **":24:10050:6:"THAT'S ALL FOR NOW. SEE YOUR INSTRUCTOR"::"IF YOU'RE STILL HAVING TROUBLE."::"TYPE 1 TO REIVEW THIS PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A1320:D13:(7)::20:"END"D11500::"X = 62 * 15 * SIN(40) = 393 NT-M"::"WHAT IS THE DIRECTION OF X?"::"ENTER N, S, E, W, UP, OR DOWN: ";T$:T$"N"950(7):"INCORRECT. SINCE & ARE IN THE"::"X-Z PLANE, X IS EITHER +Y OR -Y.":&:" ** :T\"WHAT IS THE ANGLE BETWEEN & ?"::"ENTER THE ANGLE IN DEGREES: ";T:bfT25890p(7):"SLEEPING? TRY AGAIN."::860z:"GOOD, ";Z$;".":24:10050:3:"NOW SOLVE THE PROBLEM. ENTER YOUR"::"ANSWER WITHOUT UNITS. ";T/:"TYPE 1 TO WATCH AGAIN.":"TYPE 2 TO CONTINUE. ";TC4T2840L>740H::3:"THUS, THE DIRECTION OF X IS DOWN.":::"HERE'S ONE FINAL PROBLEM. FIND X:":R" = 62 NT, 25 DEGREES ABOVE EAST"::" = 15 M, DUE EAST":" -12 - 4 + 3"::255:840:"HERE'S ONE LAST DOT PRODUCT. SOLVE IT!":::"FIND . IF = 2 + 3 - 1":::" AND = 3 + 1 + 2"::"WHEN YOU'VE SOLVED THE PROBLEM, ":10050::"ANSWERJ> + 4":" -3 + 2 - 2":" -12 + 8 - 8":z"X = 6(0) - 4 + 4<-J>":" -3<-K> + 2(0) - 2":" -12 + 8<-I> - 8(0)":-"X = -2 - 8 - 4": = -10 - 16 - 1":|R:"TO SEE THE COMPLETE SOLUTION, INPUT 1."::"IF YOU WERE CORRECT, INPUT 2 (1 OR 2)?";A\A2910f:"SOLUTION:"::" = 2 - 1 - 4":" = 3 - 2 + "::3^p"X = 6 - 4X IF = 2 - 1 - 4"::" AND = 3 - 2 + 2"::>"WHEN YOU'VE SOLVED THE PROBLEM, ":10050H::"ANSWER: X 3 + 1":" = 4 + 6 - 5"::D13000::3". = 8 + 12 - 10":" - 12 - 18 + 15":" + 4 + 6 - 5": ". = 8 - 18 - 5 = -15 (A SCALAR)":::255&*"DO> + 15<-I> - 5(0)":e"X = 6 - 15 - 4":" -10 - 12 - 12":"X = -9 - 14 - 24":::255:"THAT'S IT!"::10050= :"NOW WE'LL SOLVE THE SAME PROBLEM":"FOR .:"::" = 2 -> + 1"::D11200::3"X = 8 - 12 + 4":" +12 - 18 + 6":" -10 + 15 - 5":"X = 8(0) - 12 + 4<-J>":" +12<-K> - 18(0) + 6":" -10 = 2 - 3 + 1"::" = 4 + 6 - 5":"WRITE IT DOWN & FIND X ."::10050:"HERE WE GO!"::" = 4 + 6 - 5":" = 2 - 3, , "::"TERMS AS A VECTOR."::= = = 1"::"*CROSS PRODUCTS - USE RIGHT-HAND RULE:"::" = 0 = <-K> = "::" = = 0 = <-I>"::" = <-J> = = 0"::D"YOU MUST BE ABLE TO DETERMINE THE DOT"::" EA + EB + EC"::" FA + FB + FC":b"DO NOT REVERSE ANY I,J,K TERMS! ":"IS NOT NECESSARILY EQUAL TO .":l10050v:"STEP 3 - EVALUATE TERMS:"::"*DOT PRODUCTS - ALL TERMS EQUAL ZERO"::"EXCEPT: M 2, TOP 1 BY BOTTOM 3, TOP 2 BY"::"BOTTOM 1, TOP 2 BY BOTTOM 2, ETC.":N"THE SEQUENCE IS: S1R1, S1R2, S1R3":" S2R1, S2R2, S2R3":" S3R1, S3R2, S3R3":CX"FOR THIS PROBLEM:"::" DA + DB + DC"::"BOVE R."::j&"EXAMPLE: = D + E + F":" TIMES = A + B + C":0"BE CAREFUL! REVERSING GIVES ERRORS.":::::10050JD:"STEP 2 - CROSS MULTIPLYING"::"MULTIPLY TOP 1 BY BOTTOM 1, TOP 1 BY"::"BOTTOBINE LIKE TERMS"::"ASSUME 2 VECTORS OF THE FORM:"::" = A + B + C"::" = D + E + F":::10050 :"STEP 1 - SET UP VECTORS:"::"TO FIND TIMES , WRITE R ABOVE S."::"FOR TIMES , WRITE S A. IF YOU MISSED"::"SOME, STUDY YOUR TEXT BEFORE CONTINUING."K:10050:"NOW WE'RE READY TO BEGIN MULTIPLYING."::"THERE ARE FOUR STEPS TO THE PROCESS:":" 1. SET UP VECTORS"::" 2. CROSS MULTIPLY"::" 3. EVALUATE TERMS"::" 4. COMER 1,2OR 3";::A:&A430,440,450:"NO. SINCE YOU HAVE & , THE"::"ANSWER MUST BE EITHER OR <-J>."::390:"WRONG. DID YOU USE THE RIGHT HAND RULE?"::390>:"YES! IF YOU GOT ALL THE ABOVE RIGHT,"::"YOU'RE READY TO GO ON "2. DENSITY":7000."3. VOLUME":7000M "4. KINETIC ENERGY":7000m*"5. ELECTRIC CHARGE":70004"6. ANGULAR VELOCITY (CROSS PROD.-RADIUS":" AND LINEAR VELOCITY)":7010>"7. SPEED":7000H"8. DISPLACEMENT":7010R"9. TIME": (Y/N)?";B$:(B$,1)"Y"580&9000W:"NOT SO GOOD. I THINK YOU NEED TO STUDY!"mD12000::9000sz2" VECTORS AND SCALARS, QUIZ 2":"****************************************":M0:"1. ACCELEREATION":7010ORCE & TIME)":7010Gj"9. MOMENTUM (PROD. OF MASS & VELOCITY)":7010zt"10.POWER (DOT PROD.-FORCE & VELOCITY)":7000~D1300::7:Z$;", YOU MISSED ";M;".":M1ī1220"VERY GOOD! DO ANOTHER.":"DO YOU WANT TO CONTINUE0:"1. AREA":70002$"2. VELOCITY":7010G."3. TIME":7000|8"4. WORK (DOT PROD.-FORCE & DISPLACEM'T)":7000B"5. TEMPERATURE":7000L"6. TORQUE (CROSS PROD.-RADIUS & FORCE)":7010V"7. MASS":7000`"8. IMPULSE (PRODUCT OF FS FOR YOU TO USE.":5 X:"CHOOSE QUIZ 1, 2, OR 3."S b:"WHICH DO YOU WANT?";X] g630s lX1000,2000,3000 v:"INPUT ERROR!"::600  2 " VECTORS AND SCALARS, QUIZ 1": "****************************************":MIF YOU CAN":2 "IDENTIFY VECTORS AND SCALARS."d &9:"FOR EACH QUANTITY LISTED, INDICATE IF": 0"IT IS A VECTOR OR SCALAR BY CHOOSING ": :"EITHER 'V' OR 'S'.":::"WHEN YOU ARE READY, PRESS : ";A$ D N:::"THERE ARE 3 QUIZE1)"F"460( "YOU HIT THE WRONG KEY."= D11500::390t :8:(7):"SORRY, ";Z$;", YOU NEED TO REVIEW.": "STUDY YOUR TEXT SOME MORE!": :10050:9000 "VERY GOOD, ";Z$;"!" D11500:  5 "HERE'S A CHANCE TO SEE $ ^"COME BACK WHEN YOU'RE READY."1 h:10050; r9000m |:"GREAT! NOW HERE'S ANOTHER."::D11000: :::"TRUE OR FALSE? A DOT PRODUCT OF TWO" :"VECTORS IS A SCALAR.": :"ANSWER T OR F";::B$:: (B$,1)"T"490 (B$,GNITUDE & DIRECTION. *":8 :"ENTER T OR F";::B$:M "(B$,1)"F"380b ,(B$,1)"T"330 6"YOU HIT THE WRONG KEY.":::(7) @D11200::260 J:8:(7):"YOU'RE NOT READY FOR THIS QUIZ!": T"STUDY THE DEFINITIONS IN YOUR TEXT.":qd- VECTOR/SCALAR QUIZ&n10000L:5:"HI! WHAT'S YOUR NAME? ";Z$S8"WELL, ";Z$;", I'M HERE TO HELP YOU.":"SO, LET'S GET STARTED!"::D12000:13"IS THIS STATEMENT TRUE OR FALSE?": ::"* A SCALAR HAS MA      **********"*#.'255:D11000:::(7):?#B'"";A$:S PROGRAM."::"TYPE 2 TO CONTINUE: ";A:A1160P":T13:(7)::20:"END"s"D11500:::(4);"RUN MENU"y"?"'"':100:8"'" ************************": #$'" DOT AND CROSS PRODUCTS - UNIT VECTORS"::" **************":" = -2 + 5 - 1"w!::"ANSWERS:"::"

. = -25"::"

X = -7 + 0 + 14":!::"SEE YOUR INSTRUCTOR IF YOU'RE STILL"::"HAVING TROUBLE. THAT'S ALL FOR NOW! "!::100500":10:"TYPE 1 TO REVIEW THI6 + 9 - 3":" 2 + 3 - 1":" 4 + 6 - 2": ". = 6(1) + 3(1) -2(1) "::255:930"!:"FINALLY, HERE'S 2 FOR YOU."::"FIND

. AND

X IF:"::"

= 4 - 3 + 2: . = 7 (A SCALAR)"::s:"TO SEE THE COMPLETE SOLUTION, INPUT 1"::"TO CONTINUE, INPUT 2 (1 OR 2)?";AA2990:"HERE'S THE SOLUTION:"::" = 3 + 1 + 2":" = 2 + 3 - 1"::D11200::3g ". = 02000:Z:) VOLTAGE=V, INITIALLY 01 V0l H=VISCOSITY COEFFICIENT BETWEEN LATEX SPHERE AND AIRz H.000018 G=ACCELERATION DUE TO GRAVITY G9.8 Q=COULOMBS PER ELECTRONIC CHARGE UNIT Q1.61019 D=DENSITY OF LATEX S******$  *PROGRAM MAY BE COPIED*B  *FOR EDUCATIONAL USE &*` *MAY NOT BE SOLD.COPY-*~ ! *RIGHT 1981 CUE,INC * " *********************** Y$(20) : N(4),M(4),R(4):FLAG0 (5):" MILLIKAN OIL DROP EXPERIMENT" Z SOFTSWAP *4  * *R  * 333 MAIN STREET *p  * REDWOOD CITY *  * CA 94063 *  *(415)364-5600 EX 4401*  * *  * *  ***************** EDWARDS& ***********************D * *b  * SAN MATEO COUNTY *  * OFFICE OF EDUCATION *  * & *  * COMPUTER-USING *  * EDUCATORS *  * *  *F  CHARGE JUL 20, 1981-  USER ENTER VOLTAGE AND MUST DETERMINE CHARGE ON STOPPED DROP   APPLE II W/APPLESOFT  AUTHOR UNKNOWN  REVISED BY CHRIS      膢 ȱ " <67 L  8ij v vo>";A$: "V"ī7013+dMM1:(7):25:"ERROR!"::3e:](#::"THIS IS THE END FOR NOW. BYE!"m2#D12000:<#X13:(7):::21:"END"F#D11500:::(4);"RUN MENU"P#':100:8'" ************************":<$'" VEC "DO YOU WANT TO CONTINUE (Y/N)?";B$:(B$,1)"Y"580A 9000q :"DON'T YOU THINK YOU SHOULD STUDY MORE?" D12000::9000X" V OR S?";C$Y(C$,1)"S"ī7003ZMM1:(7):25:"ERROR!"::[:b" V OR S?";C$ c(C$,1)00T0 "8. ELECTRIC FIELD INTENSITY (FORCE":" DIVIDED BY ELECTRIC CHARGE)":7010k: "9. ENERGY":7000D "10.ANGULAR VELOCITY":7010N D1300:X b 7:Z$;", YOU HAD ";M;" WRONG ANSWERS.":l M1ī3220v "TERRIFIC. DO ANOTHER.":7":" ELECT & MAG FIELD INTENSITIES":7010D "3. VELOCITY":7010 "4. ANGULAR MOMUNTUM (MOMENT OF INERTIA":" TIMES ANGULAR VELOCITY VECTOR)":7010 "5. DISPLACEMENT":7010 "6. MASS":7000& "7. WORK (DOT PROD.-FORCE & DISAPLM'T)":70000/:"YOU CAN DO BETTER. READ YOUR TEXT!"ED12000::9000K R 2} " VECTORS AND SCALARS, QUIZ 3": "****************************************": M0 :"1. DISTANCE":7000+ "2. INTENSITY OF E/M WAVES (CROSS PROD.-7000U\"10.ANGULAR MOMENTUM (CROSS PROD.-RADIUS":" AND LINEAR MOMENTUM)":7010dfD1300:jpz7:Z$;", YOU GOOFED ON ";M;".":M1ī2220"EXCELLENT. DO ANOTHER.":"DO YOU WANT TO CONTINUE (Y/N)?";B$:(B$,1)"Y"5809 ABOUT ME.": 2"AS YOU CAN SEE, THERE IS NO SQUARE ROOT KEY ON MY KEYBOARD. SO, IN ORDER TO FINDSQUARE ROOTS, YOU MUST TYPE:" <(37)3:15);"SQR(-----)." >(37)3:"WHERE '-----' IS THE NUMBER YOU WANT TO FIND THE ROOT OF." A1000::2SAME ";& 'A$:A$""Ģ((37)):18:39G ((A$)11Ģ(37)2:958:38 )((37)3):"HI ";A$;",":"I'M GLAD YOU CAME TO WORK WITH ME TODAY.":1000::2 *"TODAY'S LESSON IS ON THE DISTANCE BETWEEN 2 POINTS. BUT FIRST YOU MUST LEARN A LITTLE****"  *PROGRAM MAY BE COPIED*@  *FOR EDUCATIONAL USE &*^ *MAY NOT BE SOLD.COPY-*| ! *RIGHT 1981 CUE,INC * " *********************** #A$"ABCDEFGHIJKLMNOPQRSTUVWXYZ" %7000::6000::6 &"HELLO, I'M APPLE."::"WHAT IS YOU N SOFTSWAP *2  * *P  * 333 MAIN STREET *n  * REDWOOD CITY *  * CA 94063 *  * (415) 363-5472 *  * *  * *  *******************DWARDS$ ***********************B * *`  * SAN MATEO COUNTY *~  * OFFICE OF EDUCATION *  * & *  * COMPUTER-USING *  * EDUCATORS *  * *  * ! DISTANCE JUL 21,1981. M PROGRAM HELPS USER UNDERl -STAND AND CALCULATE THE DISTANCE BETWEEN POINTS. APPLE II W/APPLESOFT PIONEER HIGH SCHOOL REVISIONS BY CHRIS E      ?:370DI4000470EN::2:"GOODBYE":(4);"RUN MENU":TZ1(I$):Z$(I$,Z,1):(Z$)48(Z$)57ī518CZ:I(I$):520W"WHAT???":510bJ14t N0JI0540{J505:"CHARGE ON DROP "I" IS:"!.01(1021K2V.5)" * 10 ^-19 COULOMBS."&04705I3000580:N0N04(I$):Z$(I$,Z,1):(Z$)47(Z$)58ĂZ:480BZ$"."ĂZ:480ZZ$"-"Z1ĂZ:480n"WHAT???":460~V(I$):IV(I)1000495445I2000565"CALCULATION FOR WHICH DROP";I$:I$""Ģ((37)):27:515/(N(I)4)4420*R(I)R(.99.02(1))BM(I)4P1RRRD3IITI14310(I1));.1(107(K1VN(I)M(I)G)(K3R(I)).5);I34,8:"V=";I$:I$""Ģ((37)):470I$"I"FLAG1:315:FLAG0:445.Z1N02;25);N03;35)N04C" --- --- --- ---"UN00Ģ7:414" VELOCITY":"(METERS/SEC)":"(* 10 ^-6)" GENERATE RANDOM CHARGE,RADIUS, AND MASSFLAG1ıI14N(I)(33(6.283(1))(2((1))).5)-c"OR END THE PROGRAM (TYPE IN 4000).":Sh"PRESS TO CONTINUE =>";aj16368,0xmI$:(I$)13365r:24:" => ENTER I FOR INSTRUCTIONS <="s35,22t2:" NO ELECTRIC FIELD"|"DROP # ..."5);N01;15);{J"TYPE IN VOLTAGE BETWEEN -1000 AND 1000 IN ORDER TO MAKE THE VELOCITY PRINTED OUT AS CLOSE TO ZERO AS POSSIBLE,":T"REQUEST CALCULATION OF CHARGE FOR ANY STOPPED DROP (TYPE IN 2000),":^"REQUEST A NEW BATCH OF DROPS (TYPE IN 3000),":K2GD9MK36P1H%N00,4[""DO YOU NEED INSTRUCTIONS ? (Y OR N) =>";i'16368,0|,I$:I$""3001I$"N"3706I$"Y"295;::2:" ***** INSTRUCTIONS *****"@::"AFTER EACH QUESTION MARK (V=?), YOU MAY:"::PHERE D10007 R=AVERAGE RADIUS OF LATEX SPHEREE R5107_ D9=PLATE SEPARATIONj D9.02 P1=VALUE OF PI P13.141592 M=AVERAGE MASS OF LATEX SPHERE M4P1RRRD3  K1,K2, AND K3 ARE USEFUL CONSTANTS K1QD9(Z1$)13(Z1$)2710101(Z1$)27ė:5407wFLAG0:Z11(Z$):Z1$(Z$,Z1,1):(Z1$)48(Z1$)572020Z1Z1$"-"Z112005FLAG1:p(5)z"========================================":18)"DISTANCE";", THE DISTANCE IS 13."-1000::2x"IT'S BEEN FUN WORKING WITH YOU TODAY. LET'S DO IT AGAIN SOMETIME.":::&15);"GOOD-BYE, ";A$0(4);"RUN MENU":24:"PRESS TO GO ON - TO END";16368,0Z1$:ER IS WRONG, ";A$;".":"TRY AGAIN.":2A1:464g:2:"STILL WRONG, ";A$;".":"TRY ONCE MORE.":tA2:464z"I'M SORRY, ";A$;".":"THE ANSWER IS 13.""YOU NEED TO BE MORE CAREFUL ON THESE."510"THAT'S CORRECT, ";A$4600"CORRECT, ";A$;", THE DISTANCE IS 5."?1000::2v"WHAT IS THE DISTANCE FROM (10,3) TO (-5,9)?"Z$:Z$""Ģ(37):4652000:FLAG1Ģ(37):958:465R(Z$):R13500A1487A2494%:2:"THAT ANSW450I"THAT'S NOT RIGHT, ";A$;".":"YOU SHOULD HAVE DONE AS FOLLOWS:"Os5);"7 - 4 = 3......(3)^2 = 9"y5);"(-2)-2 = -4......(-4)^2 = 16"12);"9 + 16 = 2512);"SQR(25) = 5.":"SO THE DISTANCE IS 5."OORDINATES AND SQUARE."Ie3);"3. ADD 2 ANSWERS AND FIND SQUARE ROOT."Oh^m1000::2"SEE IF YOU CAN FIND THE DISTANCE BETWEENTHE POINTS (4,2) AND (7,-2)."Z$:Z$""Ģ(37):4102000:FLAG1Ģ(37):958:410T(Z$):T5===================="RIT3);"SQR((X(2)-X(1))^2 + (Y(2)-Y(1))^2)"QY:Z"========================================"\:^"IN WORDS, THE FORMULA SAYS TO:"_`3);"1. SUBTRACT X COORDINATES AND SQUARE."c3);"2. SUBTRACT Y-COORDINATES OF THE FIRST POINT AND X(2) AND Y(2) ARE THE SECOND POINTS."yE:"PUSH WHEN YOU ARE SURE OF THIS";F1005GH(2)J"WATCH CAREFULLY, ";A$;".":"I AM NOW GOING TO SHOW YOU THE FORMULA:"O:Q"====================R(100) = 10.""'310E,"EXCELLENT, ";A$;", 10 IS CORRECT."K1Z61000::2;"THE LAST PROBLEM IS AN EXAMPLE OF HOW TOUSE THE 'DISTANCE FORMULA' TO FIND THE DISTANCE BETWEEN TWO POINTS."G@:"REMEMBER X(1) AND Y(1) ARE THE C1Ģ(37):958:250!W(Z$):/ W10300{"NOPE! ";W;" IS NOT THE CORRECT ANSWER.":"WATCH THE FOLLOWING STEPS:"6);"9 - 3 = 6 , (6)^2 = 36"6);"2 - 8 = -8, (-8)^2 = 64"10);"36 + 64 = 100" 13);"SQ"16 + 9 = 25",20);"SQR(25) = 5."2;240l"FINE, ";A$;", 5 IS THE CORRECT":"ANSWER."r1000::2"HERE'S ANOTHER FOR YOU TO TRY:"5);"SQR((9-3)^2 + (2-10)^2) = ";Z$:Z$""Ģ(37):2502000:FLAG."+12);"SQR((4)^2 + (3)^2) = ?"FZ$:Z$""Ģ(37):207j2000:FLAG1Ģ(37):958:207xX(Z$):X5230A$;", THE ANSWER IS 5.":"YOU SHOULD HAVE DONE AS FOLLOWS:"12);"(4)^2 = 16 AND (3)^2=9" 18);SORRY, ";A$;", THE ANSWER IS 9.")200D:"REMEMBER, ";A$;",""WHEN YOU SQUARE A NEGATIVE NUMBER YOU GET A POSITIVE ANSWER.":"THE ANSWER IS 9."200"THAT'S RIGHT, ";A$;", (-3)^2 = 9."1000::2"TRY THIS ONE, ";A$;"3:130Z(Z$):!Z7150D"SORRY, ";A$;", SQR(49) = 7."M160o"GOOD, ";A$;", SQR(49) = 7."~1000::4"WHAT DOES (-3)^2 =";Z$:Z$""Ģ(37):1652000:FLAG1ė:4:165Y(Z$):Y9180Y9190 "L '^' MEANS RAISE TO A POWER AND 'X' IS THE POWER.":Vx"HERE ARE SOME EXAMPLES:"^}:u15);"SQR(9) = 3":17)"(9)^2 = 81"1000::3"NOW, ";A$;", WHAT DOES SQR(49) =";Z$:Z$""Ģ(37):958:1302000FLAG1ė:P"ANOTHER THING WHICH YOU WILL NOTICE ABOUT ME IS THE WAY I DO EXPONENTS.":Z"IF YOU WISH TO RAISE A NUMBER OR EXPRES-SION TO A POWER, YOU USE THE SHIFT KEY AND THE 'N' KEY AS FOLLOWS:"_:d15);"(---)^X"i:7n"WHERE THE SYMBOT$"TITLE"V$"XXX"# +J04J30@I124GIx(" ";2I<1000F1:24P255U500:90VI1JWBUZZ(49200)XI:KEY(49152):KEY1281500YZ24 dI113,1517:17,1415:13,1819:19,2215:13,1822:27,2413:13,1524:24,271515,1727:27,2418:16);"MATH SERIES":N03000:N::15,1727:27,2418:16);"MATH SERIES":N03000:N::A::8"========================================">L21)"BY"Rk15)"HOWARD JENSEN"q12)"PIONEER HIGH SCHOOL"N03000:NX::2:12,2911:11,2029:29,1220:20,1112:7:0,3937:18,1314:14,1713N BUT NO Ug":5" (h = 0), Us, OR Ef.";:10060m13376:"SO FINAL ENERGY = K1 + Ug1 + K2"::10050::"#4 SETTING INITIAL EQUAL TO FINAL,"::" (Ug1 + Ug2)i = (K1 + Ug1 + K2)f"::10050,13376:"THUS, (M1)(g)(h1) + (M2)(g)(h2) = ":ASS HAS A DIFFERENT Ug,":Cv" BUT NO K, Us, OR Ef.";:10060z13376:"SO INITIAL ENERGY = Ug1 + Ug2";:10060::"#3 IN THE FINAL POSITION,"::" > THE 6 KG MASS HAS K AND Ug, BUT": " NO Us OR Ef."::" > THE 3 KG MASS HAS K,T IS. NOW LET'S SOLVE:";:10060PN:"#1 SELECT A REFERENCE HEIGHT.";:10060:X81,70137,70:10:14:"h = 0":A11000::15b:"#2 IN THE INITIAL CONDITION,"::" > THE 6 KG MASS HAS Ug, BUT NO K,":l" Us, OR Ef."::" > THE 3 KG M140:9200:X187:Y29:9300:X217:Y70:9300x&3:26,2489,2489,31:90,2490,31:198,24221,24221,61:222,24222,610234,60234,69:235,60235,69:233,68236,68:232,67237,67:9:36:"V = ?":7:4:"INITIAL";:24:"FINAL":13376D"HERE ISPENDED 0.7 METERS ABOVE THE"y"FLOOR. WHEN RELEASED, BOTH MASSES ACCEL-":"ERATE. FIND THE IMPACT VELOCITY OF THE":"3 KG MASS ON THE FLOOR.":::"COPY THE PROBLEM AND ";:22:10050,13376:X8:9200:34,11:X15:Y29:9300:X85:Y40:9300:X%18:"EXCELLENT, ";Z$;"!";:10060j34,0:13376:3:"HERE'S AN EXAMPLE OF A TWO-BODY PROBLEM:"::"A 6 KG MASS RESTS ON A FRICTIONLESS"::"TABLE 1 METER HIGH. THE MASS IS CON-":"NECTED BY A LIGHT CORD AND PULLEY TO A"::"3 KG MASS SUE THAT THIS IS THE VELOCITY OF THE"::"CENTER OF MASS. WHAT IS THE VELOCITY":"OF THE TIP? ";:" ANSWER? ";A:13376:A8.86480"SORRY. SINCE THE TIP IS AT TWICE THE"::"RADIUS OF THE C of M, IT HAS TWICE THE"::"VELOCITY.";:10060:4900]:"#5 SUBSTITUTE VALUES: M = 7 KG, AND"::" h = -1 METER (HALF THE LENGTH BELOW":" THE REFERENCE HEIGHT).";:1006013376:"SOLVING:"::" 0 = (1/2)(7)(V^2) + (7)(9.8)(-1)":"AND,":" V = 4.43 M/S";:10060O:"NOT"#3 STUDY THE FINAL SITUATION."::" THERE IS K = 1/2 * M * V^2, AND":" Ug = M * g * h, BUT Us = Ef = 0."::" SO FINAL = 1/2 * M * V^2 + M * g * h"::1005013376:"#4 SET INITIAL = FINAL"::" 0 = 1/2 * M * V^2 + M * g * h"::100511200:d ^16:"#2 STUDY THE INITIAL SITUATION. YOU"::" SHOULD SEE THAT K = 0 (NO MOTION),": h" Ug = 0 (h = 0), AND Us = Ef = 0."::" SO, ALL ENERGY (Initial) = 0."::10050 r3:196,40214,40:211,11211,39 w4:29:"h"J|13376:185,74185,76:186,73186,77X @5:1:"INITIAL";:35:"FINAL":10:30:"V = ?":34,11k E13376:10050 J13376:"HERE'S THE SOLUTION:"::"#1 SELECT A REFERENCE HEIGHT.";:10060 T3:10,1035,10:114,10182,10:196,10214,10:2:33:"h = 0":A THE SPEED OF THE TIP"::"WHEN IT PASSES THE VERTICAL POSITION?"::n "COPY THE PROBLEM AND ";:22:10050 ":16302,0:X39:Y10:9100:X187:9100 ,2:42,7110,7110,1342,13:186,11186,71192,71192,11 61:77,10:189,40:3:184,75193,75::"THIS PROBLEM ILLUSTRATES THE NEED TO"::"USE THE CENTER OF MASS FOR EXTENDED": "BODIES.":::"A 7 KG BEAM, 2 METERS LONG, IS SUSPENDED" "VERTICALLY FROM ONE END. IT IS PUSHED"::"TO A HORIZONTAL POSITION AND THEN"E :"RELEASED. WHAT ISZ$::Y "IN THIS PROGRAM, WE LOOK AT SOME"::"APPLICATIONS OF THE CONSERVATION OF": "ENERGY LAW WHICH DEAL ONLY WITH"::"CONSERVATIVE FORCES.":: "IF YOU'VE ALREADY WATCHED THE 'METHOD'"::"PROGRAM, YOU'RE READY."::10050S 13376:3'/d - CONSERVATION OF ENERGY: CONS. FORCES<n:10000Lx:16302,0d- TURN ON ROMPLUS+pD$(4)D$;"PR#5"R$(13)R$M$"":- THERE IS A HEREM$;"1A" 13376:3:"HELLO!"::"WHAT'S YOUR NAME? ";          50F&::"OK, ";Z$;". LET'S LOOK AT A"::"SIMPLE EXAMPLE.";:10060013376:X100:Y30:9100:X161:Y24:9200:X82:Y60:9200:2:200,25120,25120,45100,45:100,30100,75200,75D1:87,5287,30:101,20161,208N2:18:"T2 M2":3:1RTIA":t" ROTATES, THE TENSION IN THE CABLE"::" IS NOT THE SAME ON EACH SIDE OF THE"::" PULLEY.";:1006013376:8:Z$;", THIS IS A LONG ONE. YOU"::"MIGHT WANT TO LOOK AT IT AGAIN.":::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A12 THE EQUATIONS DEVELOPED IN"::" STEPS #5, 6, & 8 FOR ANY UNKNOWNS."::10060::"SPECIAL NOTES:"::"> IF A PROBLEM INVOLVES TRANSLATION AND":" ROTATION, YOU'LL NEED a = (r)( ).":X225:Y110:9020:::"> IF A PULLEY WITH A MOMENT OF INEUAL TO (M)(ay)"::10060::t">#7 CALCULATE THE TORQUES ACTING ON ANY"::" BODY WHICH ROTATES.";:10060:::">#8 FOR EACH ROTATING BODY, SUM THE"::" TORQUES AND SET EQUAL TO (I)( )":X232:Y166:9020::10060U13376:3:">#9 SOLVE":{" x-DIRECTION AND SET EQUAL TO (M)(ax)"::"(SINCE IT'S ACCELERATING, F IS NOT 0.)":X182:Y134:9030::1006013376:3:">#6 FOR EACH BODY WHICH MOVES (i.e."::" TRANSLATES), SUM THE FORCES IN THE":" y-DIRECTION AND SET EQ IF THE BODIES GO IN DIFFERENT"::" DIRECTIONS.";:1006013376:3:">#4 RESOLVE ALL FORCES INTO x AND y"::" COMPONENTS (p AND n ON INCLINES)."::10060::">#5 FOR EACH BODY WHICH MOVES (i.e."::" TRANSLATES), SUM THE FORCES IN THE."::10060m13376:3:">#3 ESTABLISH CONVENTIONS: CHOOSE A"::" CONVENTION SO THAT THE DIRECTION":" OF MOTION IS POSITIVE. WHEN MASSES"::" ARE CONNECTED BY A CABLE, CHOOSE":<" THE DIRECTION OF CABLE MOVEMENT,"::" EVENORK":" h"ORDERLY.";:10060:: r"COPY DOWN THIS METHOD:":::">#1 START WITH A DRAWING. SHOW ALL THE"::" FORCES.";:10060: |">#2 DRAW A FREE-BODY DIAGRAM FOR EACH"::" OBJECT. SHOW ONLY THE FORCES": " ACTING ON THE OBJECTS110:9030:Y126:9030:X44:9000:X106:9020[ O:"GOOD! NOW REMEMBER THAT a AND ARE": T"RELATED: a = (r)( ).";:X211:Y159:9020:X120:Y175:9020:10060 ^13376:4:"NOW FOR THE METHOD. BE CAREFUL, THERE"::"ARE SEVERAL STEPS. YOU NEED TO W:3:"YOU SHOULD ALREADY KNOW THE BASIC"::"RELATIONS FOR BODIES WHICH ARE NOT IN": ;"EQUILIBRIUM. REMEMBER THEM?";:10060: @:"HERE THEY ARE:": E" Fx = (M)(ax)"::" Fy = (M)(ay)"::" = (I)( )"::, JX30:Y94:9030:Y Fx - FORCE IN THE x-DIRECTION"::" ax - ACCELERATION IN THE x-DIRECTION": " I - MOMENT OF INERTIA"::" - ANGULAR ACCELERATION": " - TORQUE":: "X28:Y94:9020:Y111:9000 ,18:"COPY THIS NOTATION.";:10060U 613376 ";Z$;"! IN THIS PROGRAM"::"WE'LL DESCRIBE A METHOD FOR ANALYZING": "THE FORCES AND TORQUES WHICH ACT ON "::"ACCELERATING BODIES. PAY CLOSE"::"ATTENTION.":: "IF YOU'RE READY, LET'S GO..."::10050 13376:3:"NOTATION:"::N " d- DYNAMICS - METHOD (n:100008x:16302,0P- TURN ON ROMPLUS+\D$(4)kD$;"PR#5"xR$(13)R$M$"":- THERE IS A HEREM$;"1A"13376:4:"HELLO!"::"WHAT'S YOUR NAME? ";Z$::F "THANKS,         * d::D$(4)),D$;"CATALOG"D(4)KD$;"BLOAD INTBASIC.X"U4096]:o D$;"CATALOG"}D$;"INT"D00:3:148,13108,1383,47:X70:Y87:9010a @1:22:"M2":4:26:"u2":5:16:"u1":6:9:"M1" J34,14:13376:"#1 HERE'S OUR PICTURE. MAKE YOUR"::" OWN SKETCH.";:10060::# T" NOTICE THAT WE HAVE AN 'IDEAL'"::" PULLEY--WE CAN IGNORE I "MASS RESTING ON A HORIZONTAL TABLE"::"(u = 0.1). WHAT IS THE ACCELERATION OF":z "THE MASS DOWN THE PLANE?":: ""COPY THE PROBLEM AND ";:22:10050 ,13376:2:191,18111,1861,90191,90:X87:Y49:9100:111,17108,13* 6X149:Y17:92PER, PENCIL AND"::"CALCULATOR READY, LET' GO..."::10050 13376:3:"PROBLEM:"::"A 15 KG MASS IS PLACED ON A 55"::"DEGREE INCLINE WHERE THE COEFFICIENT": "OF FRICTION IS 0.2. IT IS CONNECTED BY"::"A LIGHT CORD AND PULLEY TO A 12 KG":V"THANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL WORK ON A DETAILED ANALYSIS OF A": "DYNAMICS PROBLEM WHICH INVOLVES ONLY"::"TRANSLATIONAL MOTIONS. YOU SHOULD HAVE": "ALREADY WATCHED:"::" 'DYNAMICS - METHOD'"::; "IF YOU HAVE YOUR PAld- DYNAMICS - TRANSLATION,n:10000<x:16302,0T- TURN ON ROMPLUS+`D$(4)oD$;"PR#5"|R$(13)R$M$"":- THERE IS A HEREM$;"1A"13376:2:"HI THERE!"::"WHAT IS YOUR NAME? ";Z$::O            !!THE":" SUPPORT OF NATIONAL SC>">66">"">">>>"6>>>>">>">>"A8X,Y:':100:8?'" ************************":d$'" 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:XEQUATIONS FOR EACH OBJECT AND SOLVE.":k"WE'LL SEE HOW TO DO THIS IN FOLLOWING"::"PROGRAMS.";:1006034,0:13376:6:"IF YOU WANT TO REVIEW THIS METHOD,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A1250&12:"THAT'S IT FOR NOW, ";Z$;"."::"SE13376:"#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: UNKNOWN.";:10060vv13376:" THEREFORE, WITH FOUR UNKNOWNS, WE'LL"::" NEED FOUR EQUATIONS. WE'LL GET ONE":" EQUATION FOR EACH BODY & a = (r)( ).":X253:Y127:9020::10060:"#2 DRAW A FREE-BODY DIAGRAM FOR EACH"::" OBJECT.";:2:"I":4:14:"r":4:30:"u":6:11:"T1":9:11:"M1"X34,11:13376:"#1 HERE'S THE PICTURE. THE PROBLEM IS"::" TO FIND THE ACCELERATION OF THE":b" MASSES. ALSO NOTE THAT THE ANGULAR"::" ACCELERATION AND TWO TENSIONS ARE":l" '#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"EVIEW THE SOLUTION, TYPE 1.":E"TYPE 2 TO CONTINUE: ";A:A125013376:8:"TRY THIS ONE YOURSELF:"::"IN THE PREVIOUS PROBLEM, WHAT WOULD THE": "ACCELERATION OF M1 BE IF THE CORD WERE"::"SUDDENLY CUT?"::)"SOLVE THIS ONE COMPLETE::" 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 R"::"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 TWOGULAR ACCELERATION,"Ul"WE'RE READY TO SOLVE THIS PROBLEM RIGHT"::"NOW.";:10060v13376:"REWRITE THE FIRST EQUATION, MAKING"::"SUBSTITUTIONS:";:10060:" Fp - T - Ff1 = (M1)(ax)"::"OR, M1*g*sin(55) - T -":=" u1:" 1. Fp - T - Ff1 = (M1)(ax)":X" 2. T - Ff2 = (M2)(ax)":::"NOTICE THAT WE'VE USED ax FOR BOTH"::"MASSES SINCE THEIR ACCELERATIONS ARE"::"THE SAME.";:10060b13376:"SINCE THERE IS NO ACCELERATION IN THE"::"y-DIRECTION AND NO ANITH THE"::"COMPONENTS Fp AND Fn1."::10060:13376:"#5 SUM OF FORCES IN THE x-DIRECTION:"::" THE POSITIVE x-DIRECTION IS DOWN":D" THE RAMP AND LEFT ON THE TABLE."::" WRITE THESE EQUATIONS NOW.";:10060%N13376:"HERE THEY ARE:":Fp":13:7:"Fn1":X72:Y56:9010n13376:"RESOLVED VECTORS:"::" Fn1 = Fg1 * cos(55) = M1*g*cos(55)":" Fp = Fg1 * sin(55) = M1*g*sin(55)":::"ALSO NOTE THAT:":" Fn2 = Fg2 = M2*g":10060-&13376:"FROM NOW ON WE'LL WORK ONLY W4 RESOLVE FORCES."::" ON THE TABLE, FORCES ARE RESOLVED.":" ON THE INCLINE, WE MUST RESOLVE THE"::" Fg INTO p AND n COMPONENTS."::" YOU TRY IT ON YOUR OWN.";:10060!64,4452,60:72,4382,49:51,5855,60:80,4982,47:6:9:"::"ALSO, REMEMBER THAT Ff = u * Fn"::1006013376:"#3 ESTABLISH CONVENTIONS."::" WE'LL TAKE MOTION DOWN THE PLANE AND":" TO THE LEFT ON THE TABLE AS POSITIVE,":" SINCE THAT'S THE WAY THEY'LL MOVE."::10060?13376:"#MOTION. AND, FOR M2, THE NORMAL"::"FORCE IS THE WEIGHT OF THE MASS."::1006013376:"THE REACTIVE FORCES, Fr, ARE THE"::"SUPPORT PROVIDED BY THE INCLINE AND"::"TABLE. THEY ARE NEEDED TO BE COMPLETE,"::"BUT WE WON'T WORK WITH THEM.";:100602,3778,29:52,3354,31:74,2476,26:76,2978,31T5:67,4867,70:3:66,6968,6912:3:"T":13:4:"Ff1":5:4:"Fr1":10:9:"Fg1"34,12:13376:"HERE IT IS. NOTE THAT THE FORCE OF"::"FRICTION OPPOSES THE DIRECTION OF":O"DESIRED 75:Y39:9100:X169:Y42:92003:152,38168,38:180,38193,38:174,33174,20:174,43174,56:154,37154,39:191,37191,39:173,21175,21:173,55175,552:24:"Fr2":9:22:"Fn2 = Fg2":5:30:"Ff2":5:21:"T"062,3752,31:69,3476,24:72T IN THIS"::" PROBLEM.";:10060~ ^13376:"THEREFORE, THE TENSION IN THE CORD IS"::"THE SAME EVERYWHERE.";:10060:: r"#2 FREE-BODY DIAGRAM."::" MAKE A FREE-BODY SKETCH FOR BOTH": |" BODIES. THEN CHECK BY ";:26:1005062450:XFOR 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 MASSES AND THE": "ANGULAR ACCELERATION OF THE WHEEL IF:"::"M1 = 10 KG, M2 = 8 KG, I = .$::S "THANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL WORK ON THE SOLUTION OF A": "DYNAMICS PROBLEM WHICH INVOLVES BOTH"::"TRANSLATIONS AND ROTATIONS. YOU SHOULD": "HAVE ALREADY WATCHED:"::" 'DYNAMICS - METHOD'"::D "IF YOU HAV*d- DYNAMICS - TRANSLATION & ROTATION7n:10000Gx:16302,0_- TURN ON ROMPLUS+kD$(4)zD$;"PR#5"R$(13)R$M$"":- THERE IS A HEREM$;"1A" 13376:2:"HI THERE!"::"WHAT IS YOUR NAME? ";Z! ! ! ! !!!!!!!!!""" " " " " ">">6>>"">"">">">6>>">ARANSLATION"5.':" ************************"U8':255:A11000:::(7):jB'"";A$:L'31:"";A$:6">H THE":" SUPPORT OF NATIONAL SC>">66">""ERNAL WORK IS"::"DONE ON OR BY THE SYSTEM, IT MUST BE":X"ADDED IN.)";:10060::"COPY THE BASIC EQUATION AND ";:29:1005013376:4:"THE BASIC EQUATION SUGGESTS A METHOD"::"FOR SOLVING PROBLEMS. SIMPLY ANALYZE":E"THE END-POINTSAT FRICTIONAL ENERGY IS ENERGY"::"LOST TO HEAT DUE TO A NON-CONSERVATIVE ":e ^"FORCE.";:10060 h13376:4:"SO WE HAVE:"::" K + Ug + Us + Ef (Before) =": r" K + Ug + Us + Ef (After)"::10050:::|"(DON'T FORGET THAT IF EXTRE OTHER FORMS, BUT THESE":@ 6"WILL DO FOR NOW.";:10060:: @"KINETIC - (K) = 1/2 * M * V^2"::"GRAVITATIONAL - (Ug) = M * g * h": J"ELASTIC - (Us) = 1/2 * K * x^2"::"FRICTIONAL - (Ef) = u * Fn * d"::10060:O T"NOTE TH THE AMOUNT OF ENERGY IN"::"A SYSTEM BEFORE AN EVENT MUST BE THE":q "SAME AS THE AMOUNT AFTER AN EVENT.": ":"SO,"::"ALL ENERGY (Before) = ALL ENERGY (After)"::10060 ,13376:3:"HERE ARE THE ENERGIES YOU'LL NEED. OF"::"COURSE THERE AYOU HAVE YOUR PAPER, PENCIL AND"::"CALCULATOR READY, LET'S GO!";:10060Y :16302,0 13376:3:"YOU KNOW THAT THE LAW SAYS:":::" ENERGY CANNOT BE CREATED OR DESTROYED,": " BUT IT CAN BE CHANGED IN FORM."::10060:E "THIS MEANS THAT";Z$::X "THANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL LOOK AT A SIMPLE METHOD FOR": "SOLVING PROBLEMS USING THE LAW OF"::"CONSERVATION OF ENERGY. YOU'LL SEE THAT": "IT'S MUCH SIMPLER THAN USING DYNAMICS"::"AND KINEMATICS."I ::"IF )d - CONSERVATION OF ENERGY - METHOD6n:10000Fx:16302,0^- TURN ON ROMPLUS+jD$(4)yD$;"PR#5"R$(13)R$M$"":- THERE IS A HEREM$;"1A" 13376:3:"HELLO, FRIEND!"::"WHAT'S YOUR NAME? """""""     ONAL SC>">66">"">">6>>"">"">">">6>>">A ************************":M$'" 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 '" .84 M/S/S":9:Z$;", THAT'S ALL FOR NOW.":y"SEE YOUR INSTRUCTOR IF YOU WANT MORE"::"PROBLEMS TO PRACTICE WITH.":"BYE, BYE!":20:1005013376:D13:(7)::20:"END"D$;"PR#0":D11500:::D$;"RUN MENU"'#X(#3:X,YS PROBLEM THAT":"THE CORD WERE NOT ATTACHED TO M2, BUT"::"WRAPPED AROUND THE WHEEL. THEN THE"::"WHEEL WOULD TURN AS M1 FALLS. FIND":"THE ACCELERATION IN THIS CASE.":::"PRESS FOR THE ANSWER.";A$+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 PREVIOU32: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)( )":X2