' +JJJJ ?x\>0='+l>  / X  n e% "+鷎귌췠ᷠ 뷎67Lɯ <= <=` `庠宮    LҦ ³LɆLLҦ  µ)LɥLLѤLҦ :;N:NNH hN ;պ Ժ;L,&Z[\p2Q"(0 P3Q ɿi pLLRͲɊR ^׮p Ѷ8RP˵B̵CpBZ\ @ յյ\BIR -^ ➭\ (p\I Z𐐪L NLA巬 췌`x (`(8`rm`Ժsmaպ`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)*++`,G8`0($ p,&"H hH @D@ Jh  DLꢠ ~LLE ҭ !"#$%&'(')*+,-./0123456789:;<=>?  / X % "-+LŠ͠ˠҠԠĭӠɮԠĠΠŠ֠ӠӠФbptťܥm2<(-Py0\|mRǿx8ɍL)?GGLiѼټ ` `l8mµLTHIHHHHhHH݌hHhHh݌H6 ˆʎõĵL õ ĵµȌ aµ`` L̦µ_bJLLzH hB@L  ȟ QlXJ̥KlV  ȟ QlV eօ3LɠL4L³ &RL &QL d L4 Ne)n `ꭜiɴ LE f`L . tQLѤ LҦLL¼` OPu d L Ne)noon Rɍ`\õL$D@8 D@LBR \L RNLҦHv 3h`0h8` LQ õ`A@`8. ŵH fh  F 濭0 \  ȟ Q ^\lZl^=vH şh`fhjõĵ@OAP`u@`@&`QR`F Ls  @DAF@u`8` %@ @A@`@`@A`-µ µ L8@AWc@8@-@HAȑ@hHȑ@ȑ@hHȑ@Ȋ@ch8&ȑ@Hȑ@Ah@LHȑ@ȑ@ htphso`hMhL`LծHIDOLOASAVRUCHAIDELETLOCUNLOCCLOSREAEXEWRITPOSITIOOPEAPPENRENAMCATALOMONOMOPRINMAXFILEFINBSAVBLOABRUVERIFٯTLOATLISDUMDISATSAVDATŭApppp p p p p`" t""#x"p0p@p@@@p@!y q q p@p y@@y@y@p!y@p q@p@   DOS LANGUAGE ABSENRANGDISK PROTECTEEND DATFILE NOT FOUNVOLUME NUMBEI/FULL DISFILE LOCKESYNTAOUT OF BUFFERFILE TYPFILE TOO BIINDIRECT COMMANDAVID-DOS II.2 DISK ABSEN ERROR u`X`*2@MPYdjxϠ@跻~!b]*9~~e)rsrs`_ r j ʪHɪH`Lc (L ܫ㵮赎 ɱ^_J QL_Ls贩紎 DǴҵԵƴѵӵµȴ 7 ַ :ŵƴѵǴҵȴµ納贍﵎ٵ്ᵭⳍڵL^ѵ-I `  4 ò-յ! f 8صٵ紭ﵝ 7L (0BC  7LL{Lg0& JJ yõLõ ŤLµ%J y B B Ť L  Jn Ͱ,0 # ̵µ аL (ȴ) ȴ 7L L ( L (ȴL{ƴѵ洩ƴǴҵ e ^* B0 HȱBh ӵԵ 8 `L8 ݲ` ܫ f ; @D B ! , , ƴ0Lȴ ȴ)  紅D贅@ B ɴ  J$ , 0 Ν Ƀ`HD8H@ຐ@hDh h` ŵBѵ-`, ѵB8` Ꝣ u %ﵭ` fm ﳐ 7i볍 8 `ЉLw`H fh ݲL`~ `浍국䵍뵩嵠Jm赍嵊mjnnn浈m浍浭m䵍䵐`"L ŵ8ŵH P(` dƭ֍뷌ᷩz L*# ׳ܳ 䳩鳩곩i$,V6 LLټ%i[ci,,,ټ躥̺غi躭i3مLABCGحɠ іL³ƭȭ֍ 떙L³֠LXZdflpVVxT4=媗߷iȿ@JX ocu!*3RUN INSTRUCTIONS.A"dA$:A$" "ıeA$(1)į:HI:">RUN TEACHEN$,I))32)::N$M$2c0,140:"Thank you, "N$".";cN$^c2:0,175:"Press space bar"hc3000c1:0,65279,65:0,66279,66c0,131279,131c0,130279,130c0,70cN$", this program"c"lets you study the force needed"!d$(8)A$(127))N$""1120>bB(N$)12A$"A"A$"Z"1060TbLX((N$)),Y:A$;gbVN$N$A$:1060{b`X((N$)1),80bj(N$,1)btX((N$)1),80b~(N$)1N$"":1060bN$(N$,(N$)1):1060c(N$)1M$(N$,1):I2(N$):M$M$((($0000,4:18,4ab640Fal">BLOAD ^START FRAMES,A$7000":$7000Wav216,0:1000va216,0:(222)6ģHI:::a620a1:1:2000:2:2000:3a:2a7,80a3aN$""a49168,0a$A$a.A$(13)N$""1170b8(A@` "Start '`d(21)/`n::`xX(12)J`">BRUNGAP"V`$9F00o`HI(115)(116)256`,">BLOAD]CHOP,A$A500"`6">BLOAD]WAGO,A$800"`J">BLOAD]BLOCK,A$AE00"`$A500,1:10,1`$0800,2:13,2`$AE00,3:18,3a    ` "Hello 2`d43698,190:217,0@`n">HIDOS"L`s5:" "X`xS24576a`S,0w`103,1:104,S256`">RUN START"6">-START" S24576S,0103,1:104,S256">RUNSTART""F = 0"uRO168,123:"y"+uUO160,130CuVO"w = -mg cos(a)"VuXO169,133:"y"du_O160,140su`O"N = N"ubO169,143:"y"uiO160,150ujO"f = 0"umO169,153:"y"utOK2:21000u~O0,110vO"The net force in each of the directions"s in the"*t O"x and y directions are"7tO10,120FtO"F = F"XtO17,123:"x"etO10,130}tO"w = -mg sin(a)"t O20,133:"x"t#O10,140t$O"N = 0"t&O18,143:"x"t(O10,150t)O"f = -uN"t.O17,153:"x"tLO160,120 uQOsN"plane and the y axis perpendicular "@sN"to the plane."OsNK0:21000VsN3gsN65,3060,20xsN75,5085,70sN66,3061,20sN76,5086,70sN52,7sN"y"sN155,30sN"x"sNK1:21000sO0,95 tO"The components of the forceriction"2rN"between the block sliding over the"JrN"incline plane is"XrN100,145hrN"f = uN"srNK4ırN0,160rN"where u is the coefficient of friction."rNK1rN21000rN0,110rN"Let's make the x axis parallel to the"+ne."qbNK1:21000:0,110DqcN"The weight w acts downward with"SqfN"a force"`qkN90,131qqmN"w = m g"}qpN0,150qqN"where m is the mass and g = 9.8 m/sec ."quN266,146:"2"qzNK1:21000qN0,110rN"The magnitude of the force of f on the block on an"*pMN"incline plane."2p[NK1=p\N21000np]N0,110:"The force N is the normal force"p^N"exerted by the incline on the block."p`N"This normal force is a constant force"qaN"that keeps the block from penetrating":"the incli/N141,27:"f" o4N75,5075,751o5N76,5076,75No6N72,7175,7576,7579,71_o7N72,69:"w"po9N75,5065,30o:N76,5066,30oN66,12:"N"oBN0,110oCN"This is a free body diagram showing"pHN"all the forcesnFFF3FFF24050%nFFF140405nFFF14045Mn235,YY:"+":4100Wn4100on235,YY:"-":4100yn4100n235,YY:"right"n4100nn n N1n!N0,188n"N"Assistance"n%N0,12279,12n&N0,13279,13n)N1n*N199ocline plane."m 0,145Fm "The program will then tell you if you"om "used too much force or too little"}m "force."m! mmAAAA57.2958mFCMM9.8(AA)mFDUUMM9.8(AA)mF1FCFDmF2F11.05mF3F1.95mF104045ficient of friction "Cl " between the block and the incline"el "m - the mass of the block."pl 0,90l "Given these values you are then to"l "estimate the minimum value of the"l "force F necessary the push the block" m "up an in275,WW15:3kII(F110))kANII10=k225,WW:(AN)CkIk Qk 3kk 0,0:"Instructions"k 1:0,18279,18:0,19279,19:1k 0,30k "In this program you are asked to input"k "a - the angle of the incline"l "u - the coef15jMA1$"0"A1$"9"1115(jQ10257j["_":L1]j`A$A$A1$:LL1:A1$;:L1:1020dj0yj18,115276,160j3j:900j1:2j0,176:"One moment"j:">RUNEND"j only answer for last runjWWYY15 k0:210,WW)"."FD07iLL1:L1:(A$,1):LA$(A$,L):1060AiA$""Ki$1020Xi8A1$(A)i:A1$"?"į10:2:20000:1:10:1025i;E11085i<A1$"A"į10:1240:10i=A1$"C"1200i@A1$"Q"1220iBLFDXX1025jLA1$"."FD0FD1:1155h10:2:20000:1:10(h955.hFhX2,Y5XNN,Y15PhX,YlhA$"":L0:FD0:49168,0vhL1hL1:"_":L1hAhA13LN(A$):"_":X,Y:0:X2,Y5XNN,Y15:X,Y:hA8A127L01080h"_" i(A$,1A$),gFF2000Ē0:X2,Y5XNN,Y15:9396g4000RUN"R$(A)kDI19:(N$(I),4)"Inst"ĺ:">RUN"R$(I)kN:kXA1(A$):((A1NA1)(A$"0"N9F1))650kbA1A:A(A$)llF1A10NURN accepts"?jEĖ4:"1-"N", arrows select; RETURN accepts"GjA1Vj5W(A1)jjP3::N$(A):j8W(N1):29:" "B$B$;:A2ĺA2;j"_"B$;jI140:(49152)127I50jj(49152)127540j" "B$;kI125:(49152)127I0W(N1):V:Z1$;:40:Z2$;:`iT11W(N1):Z1$"______________________________________"Z2$B$ki^A1Nih5W(A1):P(A9):A". "N$(A)iri|8W(N1):13:"Select Segment: "; j10W(N1):4:Eĺ"1-"N" or "M$"J"M1$" "M$"K"M1$" select; RET$):N$(A)"Return to Program":R$(A)"#":AA1:11PP11ehN$(N$,2):(N$,1)"0"(N$,1)"9"280oh"NA1h,W2:N7W1h61:" ______________________________________"h@Z1$"____________";::" Teacher's Menu ";::"__________"Z2$;iJV31190=g$2F0,201,251,176,6,201,225,144,2,233,32,76,240,253Qg54,$2F0:1002[gAA1eg240vgN$(A),R$(A)gP1(21.5((N$(A))3)2):P1PPP1g190g216,0:(222)42ĦgN$:N$""290g(N$,1)"#"2907hN$(N$,2):S(Nf "Teacher Menu 2fd::(21):EfnN$(20),R$(20)jfx(64435)234(64448)234E1fB$(8):M$(27)(15):M1$(24)(14)fZ1$"|":Z2$Z1$:EZ1$M$"Z"M1$:Z2$M$"_"M1$fP12g(64435)234Ĺ49162,0:">PR#3":(17):110aL279,198L18:N$#aV:+a`486ajW12BatL114Xa~L2020,191:N$`a:ha16saW12aL114a295L20,0:N$a:a0a1aI1(28672)1aIaA$:A$" "1260aaIa ::NUN T `d(21)`n1:0*`$7000:$7000^`(8181)0ĺ">BLOAD]BLOCK,A$AE00":$1FF4,$AE00f`3`">BLOAD^END FRAMES,A$7000"`6:`0::3`1`N$`W12`L110`0,L1818:N$`$:`.32`8W12aBL  2į0,95279,191)x RK0į0,176279,1916x!RK3į1Cx$RK9į2Ix&RKK0,95279,191wx RK0į0,176279,191x!RK3į1x&Rx0uKinimum force to overcome"CwO"the friction and move the block up"]wO"the incline plane."lwOK9:21000vwO3000wOK4:21000wOwR1:0,176w R"Press space bar"wRA$wRA$" "21020wR21010wR0wRK1į0,110279,191xRKvO"is zero. Therefore"(vO42,140FvO"F - mg sin(a) - uN = 0"SvO60,155nvO"- mg cos(a) + N = 0"}vOK1:21000vO0,95vO"Solving for F gives"vO20,115vO"F = mg sin(a) + umg cos(a)"vOK0:21000vO0,140wO"F is the m????III --???-M ->;NII5?--5??wI->ۛmI 5??---??II III5?-5??- 5;-M)>7----5????wI 575I-----5????757mI???7----?5*MI)>?---??II --???-M ->;.>. --????.mI->;.MI)>?---??II -----5????wII)>-?-?-7575II --?----?--m)M۷IIRI757m????7----757JI ?7-5?.o III:????.----JJ:7->IIII5555555II --???-M ->;.MI)>;.MI)>;.MI)>;.MI)>?---??IIII 5?7-575757575*-->??VII- --???-M ->;.MI)>-?-?-?----5m)>7m)>*---->????Nm)>*---->????Nm)>7m)>߷II II5*--5???7m57---???Nm5---???N)>VI))m 5;?7mm5??- 55m?m-->m)-?;VII I)>--???- m-;.M5;.Mm??---??w)>VIII)?7-5?.RI 575757575w w575757575IRRIm)Mw--ad0[5EZg|?j6s1nCfNg-` 1 n  > l  Q x 5 b r $ O | )I'''0I575757575*>Vmm)>7m)>߷JI      Ӱ hIiHӅхhЅhIi҄f8Ъх` e8e҅ԥ k 0"H h gLy  k  g  (Ӱ eЅԥeP,pLZL^2^% TgTPNTRPE)L`&  38`` N  N `LNhh`Ϧ  ݩ`8`    `(LN&E=Q&&&` <=H <=hL ^ LHhH8H4 ⎺ -Ȏ`L   Ll0 LW  L> PQ`L:ۍРֳנ嬠ȍ&E%Q&Hh&̽h` ( 퍷` mLHHH)  9L L ) 8d* &'0H   hhhh`eȱe騥,LL] `'83̳,̰'$< H -m)?III?-5m)???N 5I-I??757--7??VII?I?m)7--;.M?III??wI75III?m)>w-;.M?IIwI?m)>w-->m)?IIR:R M :u -I>-M???7---?5VI :RI:?wI:?RI -I7-???7---?.??m???m5III*--w--*--޷II:m)w)m) m?.JI57R6I:75wIw75I Imm??w)??mmIINR)>--;.Iu ?*--RIRIJI I?m)>7m)>7m)?II->.>.?I I?m)>ad(:GQ`n6Lfy*C^u4Ef 9Qr0HYq &<Rh}(=O]nvI7575w:V m57mR-I7m?    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