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[1.1] Express in SI and US Customary units the speeds of a) satellite, b) racing car, e) Fighter plane, d) baseball, e)automobile. “piston. a) satellite = 20,000 mi /h 30000 mi, Safofe, Ih gases [hr Im 36005 = [1.2 cont] 6) Newtonian mechanies is basically exact For ordinary engineering speeds, c) A particle becomes infinitely massive as i+ approaches the speed of light. [43 | Evaluate the quality of data. dista fe, dom em Biyum/s b) race cor H 80 mi/h 1 Soft, In au Fe/s Var BW0S === F-mãa — m=E X-comp? m=55-3.akg 5 10 + : =— «<3ak SR Mr qua 3 Z-comp* m TT 3kg “There are errors in the data. The mass shauldi be the same for all three projections. tm E Pe, .308 Mm. q947 my, Is e ===> O plane Z 800 mi/h 800 mi, Sao % , lh (gays Var Im 3od0S Mi3%e , .3049m 3.b Ms Is We ===> d) baseball * 100 mi /h 100mi, Sato t, lh (jyç7ft)s Lhe Emi Jus —=== Mb 7 fe, 3048 mo Is Ve º) piston = 8,000 rev /min SOS rev, dAsirokes, Sin IR, Umin Ymin Vrey Vstroke Iâin GO se =4].l7 fels = 44,WM/s AboT Se | + 3648 m y = |aIM/s Is = [12] ayrind “o of mass a parti de changes iÊ ve O3xio! MA, )Ixine MAs, di) Puga! MG, à) o change = Imo-ml, jp Mo (axintja va me ção [ip Sg m mol! ES) eennoimo: elo « Lme= 1000000 mol, jo0 = 0% Mo ú E estos mim GxindE «D0005S ms So = Ie =. 00006 mol ; jo9 =. 005% Ms TEN wi) (1- e oii fatia ES = |.00504 mo %o= Ima = 1.00504 mol | j30 = 504% Ma [INT Find % Comp. FÊ, m=3 kg Fema x- Comp: Fy=3(25)=7,5N 4 Comp: Fy = 3(3.125) N 680] Assuming that +he calibration of the scale was conducted where if was assembled, the difference in weight of she meteorite will be proportional to the difference in force due to gravity between the earth and the meteorite, | E =ma- q,=q at elevation of assembl mesa a at Mt Mckinley ! Since ga 1.67 m 1.67 (1000) = ]l51lb mm 56!=blbin=55fe=1.83yd =]. bom = [bl mm nova Length: meter Mass: kilogram [km , 1Obmm lin (8 mi =0.baly aà setond Force: Newton “Tm “25 4mm Taim Sato eau [19 | MIN:m=]000kgmã/sa [Lib | Find the density in a) slug/P2, 6) 9 lemê, MNem is derived, a) 103 Istug im? y quj slug/ps Im asia 33166 => [110 T N.m/s = kg-mê/s? b) 163 a, 10009, Im | 9/mê Nem/s is derived, Vmê kg Tiooem3 * =" [LIT] Find 4he speed a) in Mi/s, BEM nin. a) cia mis lh 100yd, ó und A is faster than B. = RA 100 = 0,09M= 9,4% b) Does À or B run farther ? By what To? Runner À: IOkm, Im ay tn 1,609km toa14 mi > bmi, so Runner À runs Fanthen than 8 a = 0.0350=3. 5% [41% T Fina consumption in km/L. dom, Li im Pr assi MMA “gal” mi 3785L [1.40 | Find the relations between a)mi Zand km b)m2 and yd?, c)inZand yd? 3 Im$=(1. 609 km = 2599 km? db) Im? = (1094 yd)3 = |. 308 yd3 o) | ydê = Cb in)ã = ab inã [Ia | Find mass cf earth in a) bip yri/mi b)N-s2/m. a)MLORx Io! Ibrsê [hip day? | Iyá Saga [e boo ie (e goos (aus. 35 days Imi =a.b3x]0? kip-yra/mi b) 9.080 bes? UYSN, 3.281 Be 1RE Hb Im =5954 x10%! N-sY/m 1a | Find cost per miê, acre, m?, and t2, Á= OA miX va mi)= Vo mi? Cost=839,000/ Xe mi? =8513 000 /mi& 4519000 Imã ama Imit “dO aere dE 8514000, Imid Ami O (Am EE 2Sia000, Im? | Loniã Im (Gago pj? =80,90/m? =80,094/qe2 [133] ayFind po earth in slug/£e, = 408 y 10%? slu Lua arao A apo = 1.067 xt0! Slug /pe3 b)Find p in kg/m3. p E E stug Um? = 10.67 slug | 14.59 kg ( (3.28 882 s.soxjoê a, m ES T' isthe force exerted| an the rope by the weight. Wºi8 the gravitational attrachon exented on the earth by the weight. EA earth NOTE: Force notarion in magnitude and From the x-axis and negative angles are clockwise, For example, ta4F SN boo P=5N, O-bo? x direction — Positive angles are counterclockwise [3.7 ] Deseribe and illustrate recetion forces, a) FEIA List magnitude and direction of REC. R= IDON,0=4Y5º E= |aoN, 6=75º t= 90N, 6=135º D=5N, B=-135º | 3.9] Deseribe the characteristics of R and É. a) Rand B are collinear, DEisa Zero vector, o) Rand É are perpendicular, f Gravitational foree exerted on the earth by the bask. canth b) [=== 7 Floor pushes up en the legs, 3 Pao Gravitationa| pullon [|| the earth by the 7 desk. canto fo) Desk pushes up [23 ] Classify +he force systems. a) coplanar and concurrent b) coplanar and collincar e) concurrent and nonplanar d) collinear and coplanar [à4 | Desenibe a) divection line, b) magnitude, O sense For each force, É: 0) 30º clock wise from y-axis, bi SON; €) from O to À B: 0958 clokwise from y-axis, b) I80N, e e) from O to B a) 30º counterclockwise from graxis b)IOON, O from O +o E on the book, == [3.8 | How can either team win the tug of war? ve md Team À Team B Ta Se BForces = massX acceleration The forces on the ropa are equal but opposite in direction. Team À has a greater friction force that allows the team members to pull back on the rope, EMI Deseribe a) direction line, b)mag. , e) sense. [2:4 | a) What forces qet on the apple? R:0) along x-axis , b)3N, eYfromoO to À ES a) along y-axis; b)HN, c)fromo toB €: 0)53.13º countereloekwise from x-axis, b)5N 5 6) from O to C Weight Wap is the force exented by the carth on the apple. Nris the force exonted by te % L Nr table on the apple. (cyrtinueo) [29 com] Gywhat ane the reaction forces? || AU cont] Law of cosines: s00n Rt= 500% +5008- A(s0BP cos b0! AMA Na is the forca exented bo u) q ; a apple on the a (AIRES Law of sines: sin6 sinkd pe We is the force exerted A SOON E sob R E by theapple on the eanth, R=500N, B= bo! O What forces qet on the table? ii) Na Nyisthe force axented by Hhe apple R Law of cosines: RA A A nd OO EN RIAA oc -a (ain cos” Prue asented by +he earth on the table ER] Law of sines: sinB . sindo” N mg Ne is the fonce exented by the 100 e ij floor on the Table, Reaa3d kN, Omau.b? dyWhat are the neaction forces? vd am Ladiof cosiade: Nm | Nr is the force exented by the ERA qm RE= 4% ia cos 45! table on Hhe apple. We is the R Law of sines: Sinô . sinds qe force exerted by the table on 1 R rr Ane cant. Ny is the force R= 47 40,8 Nç Floon Na exerted by the table on the Floor. a [212] 0) Derive formula for magnitude of R, 2:10) ayDerive o Rand À. [2:10] ayDerive formulas for R an [3 q ip Re -a0n cos OA R a sl Am Rm (io- be cost) og/B —Õ z S V) Denive formula For GQ, o Aa] £ cosines: Rã=A24B2- QAB cos (1 Sb sind. sinô E Law of cosines [is cos (130-8) i = To-tesssa *hé R=(AR+Ba +2AB cos 0)4 al sin O lawofsines: RB Oeste So) sintrds) sinb Em desin (Ee e) O Evaluate for B=0",90º, Ig0º, R 6=0º R=(b- besst)a E aimé O= sin! a R=A kN, = 0º b) Check answers with B= 0,90, 180º, (lo-beos0)4) => — B=48 R=(jo-b costs) a a (Sinas O q Gesin Lt o cos) 6=0 R=(hi+Bi,9AB)2-A+B a a [Dea A) SER d= Sim (Esino)=0" Re Sb kN, Deca B=85 Re(id-locos tro) A desintf Reg UN, O Sivi RO! (Oro cos R8) 6-0" R=(pisBa +aNB cos (QU) 2 = (pas pad b= sin! (R Sindd) = sin*(B) 6-0 Re(AMma -ang)b SAB [2:13 | yDerive Formula for magnitude of R. TE 48, = sino! (E sin tes) = e A Re(3 18 - ala costino-6)8 SE cep oqdo e pirstemetsmBit O o E SO a (y 3 [AM] by Find magnitude and direction of resultant forces by Inigonometry. | Derive fsemula for G. U) Law cf cosines: Ri=4oo?+ 308 -aldoo(300) cos ad sin (5-8) a SD. Rs Law of sines : sin O = Sin lão” Ibrb cos 0) a l 300 R E R= Lot bs, 6=35.3º d-sir( x) = A A 5 [246] 6) Find magnitude and dinection of É by Arigonometny. R=(AS+B2-aAB cos los) Re(10%+ |53-a (10/18) cos 1058 B=usta, Sn& - Smlos R x=2878º [41] 6) Find magnitudes of RÉ by trigonometry. 9 A Sinos! sinyS (sin 30” 3 Nas Es EA A 8 F F A=193F B=141F RES t)Find magnitudes of RE, BO by trigonometry. e SnisS Snias sin 30 1000 AR Rd AP=I93a lb BP=a73a [833] 6) Find magnitude and direction of F by trigenometry, > AlDoN) adoN io * = Q008 4 LO a (G00K 00) cos so! PR Sina a=4178º F boo — B=-(185-5) F 2N,6 3.a” Loo) E E pot [8.84] Find R by trigaonom etry. (Ryt= 924 sa 1 R'=9.43 kips R ja 5 = val 6 . D=+an ($)-3s G=ar 8 RE (RA 442 a(RXY) cos O RE= 438 +48 a(43) cos AY R=b.1% kips Sins sinag” o Pla q “Lo B=90-30-at-1.7º = 14,3º R-blykips, 6=14.3º | 2:25] Show that the resultant is O. B = 80-0-b0'=b0º EE R= (Fispê-a(r/P costo)á=F R-ER Em Find nesultant forces exented on A and B. do C ed lb Er a(s 4 cos 45) 6 yo lb Ba == (48) Sb sinys d=tas do Ra Rn=20.blb; O="29.5" = Pago Rede ei - a (40) costas G=-(A0+0); $=21.5º E Sinb sinjas: do Re Res134b ,B=-llas* [SR] Find R by polygon construction. [Unit=100N A=220N Rem3n,0ess" EE Find R by polygon construction, pi [Unit=2.5]b D-$ib R= |0lb,O-laa” [230] Find É by polygon construetion. [23] Find R by polygon construction. VUnit=25N Th, D=80N “az E=100N É> 99N,8=4.,5 rad [932] Find R by polygon construction. tes 4x, HUnit=100b'%4 eso T B=10DIb Na A=3b0 Ib Rest, O=0.2yrad |Unit=0.2 Kia» C-aup Dea, A ado” E=05 wp R=1.5 ips, 6=% 4 [Unit =5 kN 4D=2kN EUA A=5EN ÇA “ F=13.6kN, B=|.0 rod lg IH cont; 2.29 AL Bs [E To E T] MicrosoR Excol Spresdshest orientaton, 2 magnitude [degrees [x pr |y projection IA 4 20] 376] 1397] s|8 10] 130] 43] 7.68] c E 60] 3.00] 5.20] D' 8] 225] 5.66] -5.66] [HE j E 5 E TR TOS] EX 3] E 230 Jal B Tc o=EE Tl] MirsoR Excel Spresáshest [orientation, 2 magnitude [degrees [x projection ly proj 3 02] 80] 0.000] 0.200] 418 10] 45] 0.707] 0707] 5 je E E 1.000] 1732 so Ex) 2700000] 2000) 7 05] 240] 9.250] EI TA 5063] 357] U208] B=-52.)3'+180º =jaig” ES a) Develop spreadaheet to find resultant of concwrrent, coplanan forces. [ Aa | Reduce vector equations to simplest form, a) a(3R-8)-E +ulB-4Ê)=0 bR-a8 -U +bE -ayt-0 bR 44B -ast=o 9 P=alucReatrst) (R+s0)-148] E=iaRragB+al-aR-st-4eÊ F=qR-aut rar 0) 138 +78 -b(4B -38)=0 I3R+7B -ayB + 18Ã=o aR-nB=0 | 243 | Find magnitudes and directions of a) b.0R, b)- so a F=i6R F-bl5=30, BF=-s0R F=|-si(9=a5, O=aio +) Use spreadshect to solve for 43/-33, 2% = D [TE 7 Eca Spresdshest orioniaiom 2 [magnitude radians |x projection |y projection 3JA E) 0.00] 50.00] 0.00] O E 052/2556] — 00) sic 40] 236] 28.28] 26.28] ejo 80] 3.67] 58.26] 40.00] 169] Chi 000/ —=10909] 5 EEE TED ZH5 S7 232 JaL BB Tc TDTE T Excel Spreadshest jorientabon. 2 magnitude (radians |x projection |y projection 3 380] 0.52] 31177] 180,00] 418 100] 1.57] 0.00] 100.00] sjc 180] 3,14] =180.00] 0.00] 81D 200] 419] 100.00] 173.21] TIE 140] 5.76] 121.24] Ei O 15737] 024] 15307 67 233 Jal B E ESSE T| MisiosoftExcoi Spresdo Torientation, 2 tudo lradians — |xprojocton |y projection 3 Ja 5 0.79) 3.54] 3.54] ao | 1.08] E) E sje 2| Am 9.00] -2.00] sjo E 2.36] -5.68] 5.66) é 550] 4.24] EF 7 TE To. TRT Refen to problem Solutians q.3/-33 for poly gon Construction, Note: tan Ta = tan “= tan (Ta em) The computer will return values Fon orientation between “Ta and Ta, “The actual orientation is either the returned value or the value +, B=L35+m 4.49 rad) “a Jar B [ c | E 1 licrosoR Excel Spraadshee! 2 magnitude joriantation, radians |x projection projection E ESrogaEar ParaNea [24] yYes. Two conturnent forces with equal sjc [EBS"COSICS) |=B5*SIN( N T ste e es, magnitudes but opposite senses have a zero E EBT'COS(CT) resultant, [ESORTIDS SEN [EATANCENDS] [ESUMDID) b) No. It is not possible, e) Yes. Polygon construction yeilas an equilatera | triangle. d) Yes. Às long as the polygon construction yeilôs a complete triangle, the resultart of the vestors is zero, [845 | show how Eand B can combine to form R with magnitudes (units) 7, 6) 1,95, dY any magnitude between land 7. a) so A=Dumts Ez4 units R=R+B Re3ry=Junits b) R=-3+4= lunit [o] TB Res Ei * R'=3arjã R=S units Re(32442 a(afa) cos 6) é R=(as-auy cos 6)! 0Ose<18oº 10 [3:56] a) Find R+B+E, R'= [1008 + 1068 -a(iosçios) cas (useg-s6)] é R'= 158.47 units Sin - Sin (45'+v0) Io8 158.7 A= alo-(37.5%t150º) = aa. 5º Re (ISSUT +00? a (158508) ens da. 5) a RE 6.54 units SMT . Sinaas! Sl] To.S4 B=- (98-137.8-30")=-aa.s Ra 1.54 unito, B=-2a.5º D Find AE-L. g4=37.5º F=ws-sas = laysº A-B=€= 00 units 150º =220 180º R'= (100%+ 1008 -a (1ob)100) cos 15º) É= [21,15 units Sud SniS 4 sas lodo 1a11S B= 3b0- aI0-30-5A5'= [97.5º Relloot+ 1a1.752-a(IoNKIal:75) cos 67,4) & Re lay.S units SNT. SinbLS qugo loo tay.s S=uS“+ 5as'+u79º = 45.4 Re tas units, de lyS 4º Pa c) Find DB such Mot R+BrTt+B-0. DB=-Rfrom (a), therefore, Bis R$ from Ra B=1654 units, 6 = aa. 5º +80 =157,5º d) Find D such that R+B-2-B=0, B=Rifrom (6) D= lau.Sunits, B= [45.40 [27] ay Find(R+ BT. [EM A= : R'=R+g 100, R'=VASBS =293 4, N By =210+ tan!(8) =a10+ tan” [E =3334º R= (223.424 lao?-a (aa 3ukiao) cos (astral) a R=aMEN Sina Sin Tb iao am? Sa= W0'-3.5'-ab.b*= 3014º R=amn.$N,8= 301 «=31,5º BD Find (BO +R. A=IDON R- Bro R'=(ams jaor- a(aooliad) cos 45) ta R=143.0 sind Sn4S a=80-34 2. aus ja apos ado 43 Gb ao janyo R=R+R Reliys% oi a(yaindcas la6:49)% R=aI.8N FERE UR = Bl? sind. sinlab4 qua loo al. Be= 365- 3h -3 17º =301.9º R=all.8N, Op= 3019º O Find R+B+2. Same steps as part (a), Mina Er R-HR R'=V adoro =433.L4N A = tanci(OO) = ab .sh* AETemE 8-9 Re(IDrbcos 0) A = 3.1, kN =sindo? = tan 18.43 aptas (& ei ss B=180º R=(lb+ locos 20º) !4= à kN =+an"! [Sin 180 o rias (a = [2.53 cont] +) Evaluate for B= 0,90", RG. 0=0 R=(otbsino)á= 3, IbakN cosd = var! ED, . o ir Sis (= aliro B=90* Relio+k sin IG)a= 4 kN === € Ans d= sina cos g0º ) -o ViDFbSindo) = [2:52] ay Denive formula for the magnitude of R, B=190º R= (10+b sin 1202 = 3.jba kN Pam cos(NO+B) Ry= sin(io+m)+ 3 Rj -ense Ry= 3-sin& R=(k-cosai+ (3- sina) a R= (costa —LosinB+ inte)! CosBA sina 6=| Re (lo- losine)fe === + Ane b) Derive Soc mula fon 4] e fi Ras p= sin (831) ! a ginol| cos0 ! 2 be sin me SE) SEvaluate for B=0º,90º, 120º. 8-0" R=(lo-losin 0) = 3.164 kN = ud cosO x $ ei Ds 8-9 R=(io- Lsindo)A= ak = " "* cosI0! | $ mM (e MIO = smãs; S=IR8 R=(lo-L sinto )É= 3,Iba kN 6b= sie cos ISO! ) =|843 ViO-G sinito [3:53] ayDerive formula for the magnitude of R. Ra= costR0+6) Ry= sintista-3 Ry=- cos O RM -SinO-3 R=(Censt+ (-sino-3y)a R= ( cos + sirAB + Lo sina +) 0058 + smiB =| R=(lorbsina)é bDenive formula for 6, VN, AS fan (4) e Mi de sin (188) $ da, Et = sin! [ cos Ito! ne mo e) [54] Find magnitudes of R,8. att R= kN Rzhs+By= cos losº +B Ry- Agr By = A sinos! vo a: Ryr loosb Ry= | sinto A cos 105º + B= costo” Asin 108º = Sin LO? A=087 kN B=0.732kN ESA magnitudes of RR. Rx Azi Ba= Nos 05 =R Ry= Ay+B, = Asingys-10 =0 A=AU.N (bo R=101b [2.Sb | Find Gand magnitude of E. BA Roz lodO= IDO cos O + Bos 45] Ry=0=-1000 sn6+ Bainys B=WM b 6=90º A=1000Ib [2.STT Find Sand magnitude of R. R=304N Ry=30 cos B=I0+A cos bo Ry =30 sinB = A sinko! ANE) A=a37a kN, Q=43aº Bal0kN 13 [9.58 | Fina magnitude and dipection ok R. Ry= “IO cos 45" +ISensbO” Ry=0424 Ry=-1O sin - 15 sinto" Rg ="30.0% R=0 JE H(-as.00R = ab. Ol B=ato! + tan t(Matyo n)= alva? R=20.0b kips, O=am.a” TaslO kips TosIS lips EE Find É. Ry= 100 cos 45! + 900 cos 38 150 costs 100 cns36| =%.3 Ry=-100 sinS* + 00 sin BO! + 150 sinlol! 4 OO sin 30º = aú.a =Vasirasna? = au. a tam! (ARA 4a 3) = U8.5º = 234.8 Ib, o-ê sº [260] Fa RE. [3:42] Prove that R,B are perpendicular to each othen. As=hcoga Aç=À sina Br=BeosA B,=B sing Ac By + AyBç=0 ABeosacos A + ABsins SinÃ=0| cosa cosa + sina sinÃ=0 cos («-8)=0 “tos 8=0 6=cos!O or 8= 90 470º - RB are perpendicular [263] Find 6. Fy=FcosO F=Feosô Fr=FeosO F= (Rr nar Fra) am (3 Facas) =/3F cosO I=1 eso Ge cos!) = 547º [24] Find R= =(RuRpRz R=R+B- (ABr, At ThtBo) Re (75, SER E 4 CM 2.65 | a) Find dinection cosines of R. 4 Rx=O= 300+100 cos45*-B cosa A qolb Ry=O= -40D+H00 Sin4s! +A ae -Bsingo” E) ai] A=9537b B=b73o lb B Hoblb db! | Find 4 and 4 projections of FyFa. A mazme=m mea LE R a) Fm) 2 km Li ) as ar Rós Ç As Them soe arê TA = (Ra + (Ra)? -aR(M cosa) = (Rê+ RW - Ricos e) t= R(1.95- cose) Fazkm(MO km& Ri(las-cos6) ARR(Ias-coso) sing | SsinB R RU.a5- cosa s= sim! ( in & aces) (Las-cose) ag [o (aê) a: am * Fay mê ARa(1a5- tos) R=R+F « =(Ar+ By, Ay + By, AetBa) =(lo-4, =1+5, -3+7) = (2,4,4) R=(a2s pasa) a cos O,=Ry =4. R bo cosB,=Ry=4 =4 PRE 3 cosdp=Re=U-4 R ba b) Find angles that R* forms with axes [CITA Br=cos'! (14)=70.5º BC, =cos!(M)=42,aº | Allo | a) Find É. =Br="t.8,= -%0 N “08 Ban Bea 1; cos CT de= 3 By 0 N cos By = a O .B,=0 de TOS CE tos Bu =. e = ; Cuz -Nl,SO N cos de, = é 2 ste IN c030cz =Co =7 .Cr=2,37 N lodo 4% ? EA [3] Fina R. R=R+Bet R= 100(cos0't + cos 904 + coso É) = 100€ É = avo (cost t+ cos AD'Z + cos 128 ))= -go0% E = Wpolcosalt + eos 383 + cos bl R) =18.b$ +%0% R=(1000)4 (-aoof)+ (122.62 +08) R= o0t tI3t.64. -laof N A ft [3:15] Find R. RR R= 5(cosqd tteos 180% + cosdok) = -54 B=4U(cos las t+eos Ns'Z+cos lactãe) = at tasas -ak T= 3(cos SUNT +eos SUA reos 54 ado) =113t +73 4+1 713% R=Cspi+Cat ratag -af) Ut eiaagrat) R=-0At- 04407 -0.a48R kips É= lo(cos 3h 4 +cos 24 Yreos Mk 8) =-5t-101$-101k R=l92-3 86) HUI4Z 14 4) St 070 1.018) f Sl -aLaf ib [279] Find FR. R= Ret R= 350(cos Ta t +cos Ta + cos Ye 8) =1154 +303,1 & Bs avo (css O? +cos Th4 +cos Ta É) = got T= uol costa t.+cos th 4 + cos Wo) = 200 t +ataS4 - 34,4 D= 300(c082k + cos "3 3 +00s%%) =-ala.lt +1504, -150%. Re (1753+ 303,18) +(200%)+(200% taga 24. -3464) +l-ala nt +1503 -150R) R=1879t +00718 4 193.3 N [3% | Find R. RR [380] à) Develop spreadsheet to Find nesultant oF concurrent, non- coplanar forces, B= aolcos Ta? + cos 3 facas MO) == AAA GAMA Re bo(cos U5ºt+ cosas tens 135%) E =4ayt-Magà E E E qu E—? Be 4 (cosa + eos 4sº4 + cos 45º) É ê =a%34 +atat Tre q T= V(cos IWL + cos TS +cos GU) Ser = bI3t+3.012.+3.4aà TESE TE B= la(cos [35º +cos [397 +cos IB) - E = =-Q4IL-2U9 E 7 Ea R= (4 ayt- 4a fa gap ta 238) , E Em ms +(b43% +a.077 +32) +R4GA-BAIZ) Gidi : — - Reauat-2:684 rack ii BVevi£y by Solving Prob. 74, 415, 4.76. [ATT] Find É, E E O E R=R+B+] Esc CS SN SEA Re 4 cost +cos Tag +eos 4 A)= agatraçaa] | Eee a lola o B= blcos PAL teos Ta f +e0s Ns 8) ENE ER es E po = 4 ay? +3.004 +3.00k To RR, ê= 3(cos Ya t+ens a 4 +eosrR)=-3R O E [Sosa] Sra] aa] 151] JONES] 22 Relass tras) adt rag +38)+638) || ax R=-iuit+39+agat kN AH ETTA : E DS Sa 38] F; E E ri 5 c 3/ 5474] Sá74] 5474) 1732] 1732) 1732] RE a 5/8 al 120] 45 120] É Ze] = R+Br 71 D Dl d| [O d| [) d| Re lblcosTAt+ cos 4 +cos 2%) THE DR E = 13864 lo 9.80 cont. 27% AT aa LeTo Elcel Spread! Angles in à E) 48] Tá ES 5 EEEZ) [ETrTeTH Test Profaciions Y 5 B Microsoft x EK) d| E Ex 0] BEE] 26] z E: 283] 342] EE 201] EX 0) miojo/o)>| BEREBSREEE CE] ERES [2:81] ayDevelop spreadahoct to Find resultant! of concurnent, non- coplanar fores. AT B [= 0 = E) T EN 1 Microsol Excsl Sprendahesi Z Angles Tm Radians magnuda % d * 7 B E b TA E 5 SORTE SOS HOT |CACOS(ONSO [FACOS(ORES) | TE EE EE Projeciione Fi x z [=EETCOSICA [=384 TSESTOSES [=SBS'COS(C5) [|=SES"COS(DS) [=saS"COS(ES) [=598:COS(C6) |=SBECOSIDO) [= é 'COS(CT) [=SBT-COS(DT) [=SB7"COS(ET) E E Co: E IE b) Verify by solving Prob. 2.17-2.74. 277 AT B IcID[IEIJFISIA 1 Microsol Excel Spreadshesi 2 Ângles Tn Radians. Projections [mimos o ToTo (xi yIz ATA 4/0765[ 1571 [0765] 262] — 0/2828 SJB 8! 2.356] 1,047] 1047] -4.243] 3] 3] Efe spin 1srn[aia] oo a) AR o o dq o o q 5) SLE o õ o [O o] [O E O 4358] TOO] Biz] 0865] Taiá 3/ 282) 238 at Ss [IcIS[LELFICcIs 1 Microsoh Excel Spresdshost [ET Angles In Radians Tojecions 3 magnitude | 0, | 0 | O x y z S/A 16/1097] 36554772] a[588]— O) sIB 20] 1571] 2358] 2356] 0] 14.142] TE TAZ) 8] c 10] 2094] 2356] 2350] | 7,071] -7.071] ama 0] o! E O] o] o o] SJE O 0 o O 9] o] 5] 7 ATU6S] TASE] 2503] 2113] PESAR 279 AT 3 IcIDILETETETA 1 Microsof Excel Spreadshest z Angies in Radiana Projecione 3 magnitude [ O | 0 | 6 x Y z A[ A 350/ 1871] 1097/0529] 0[ TE[S0M 15 200] O] 1.571] 1,571] 200] o O] s| c 400] 1.047] 0.785] 2.618] 200] 282.8] -345.47 T|)D 300] 2358] 1047] 2098]-21213] 150] -150] SLE 0] E 0 0] 0] O] s 0483] 1.28] 04: 1.87] 167.87[ 607.84] -T03: [3.1] Draw free- body diagrams. a)Fig.P3,a WFig.P3,3 y Fa | F Fa "Nele ro Nha x E 3 le 7 % 32N 3b0 lb o) Fig-34 d) Fig. P35 Vkip A ABC is a night isoceles triangle . Theme fore B= 15º, e) Fig. 3.10 o & Vo ae. di qd clio w Ww kN Prq Paia 3 500N 9) Fig Pais 8 ah 4 AL ANse tor + wW Ta W Fig P3.17 Bo fl bor Ty E: W C(-015,1.0,0) D(6,-1.5,6) B(1.a5,0,0) Fs A(0,0,-2.0) | 3-6 cont] cy Ts +he ball im eguil. inta)? Explain 3.9 cont, d) The ball is in equilibrium honizontally because +he forces in the y-dinection e zero, The forces in the equal zero; therefore +he ball is not in equilibrium ventically. No, The resultant forez is not equal to zaro. - direction do not Equilibrium equation 5F,=0=-Tsinf +450sino T= 450smn8 sin b To maximize T, set 4E =0. dT, -450 086 6 de sin 6 ual cos6=0 — B=90º b) Find Tomas Fon (= 170º, Ib6º, 150. [3.7 | Find the charge on the spheres, BF4=0= 100+Tainte.1Qº Fe=o.bêa dynes T= |00.4 dynes Fe=ee que! fo=eã vê nê e=YFer& =Ylo. U$9)(RI) = 20.4 esu Tm sm FBDY4| AT ARNS | É 1 o a o 6=cos Bb. IR vo 8 co] SO] qui RA TTçÕ: a a Divide system into à parts isolating 8,C. 100 dynes FeD:€ Equilibrium equations % ZF,=0= -Fe+tTeos 88º ISN Equilibrium equations LFy=-T eos 30 +F=0 16, = T sin 30 -15-0 T=30N F=4598N [3% | 6) Devive à formula fon T. Fen: B E FBo 4| é ' Iodib Ra tos! s e + 15N 30N E Equilibrium equations A ad ] LF,=0= -T sin6+30 sin bo! Equil ibrjum equation: XFy=0=|b0-aTSnA 2F,20= Ty cos 8 - 30c55 be? -15 TRÃGO ao Taz 30 sinbO* = 30 cos bo!+Is Sin Sind Elvis smB cosB b) Find T fon 6=10",a0,45º, What can be SinB - 30enbl 6 concluded ? cos8 3Ocosbl +15 Conelusion: A langen pulling force vesults from a amaller angle. 6=409º =. N EM Wi sinf W Sina Show that T, = “= ane VE O qinlaso sintas) [39] a) Find Oto mayximize T. F8D ni Ta, T F8D Bla 1 $| 450N A 8 4 Eis W T BF=D= T sina -Ty ng (Eq lo 2 Fy=0= Testa +Ticosp-W (Eq) 19 3.11 cont, From Eg | Ta= LSina sn8g Substitute into Eg.4 W=T sina cosB*Teasa sinâ Wsin8 = T sinacosB+ Tosa ginf Wing =T, sin (= +8) T= Wsin8 Sin (a +8) Substitute into Eq. | Ta sinB= W sin Ena sin (248) RE W sina Equilibrium equations LE=% T- YA 70 2F,= Yes T+ 4a +500=0 =174.6N T=4SI.8N Fede Tas 45.8N GR 4! Equilibnium equations EF =45(Y) - Ya T=0 BR, = 4) = Ag +W=0 Tas bbbaN W=571.5N [313] Find 4. F8D TF the tension UR Gana ia + is equal on [De both sides of e 1 / 7 q * +he cable, the w * anlesane equal, | 3.13 cont] This problem con be solved with geometry alone, Le+L,=50 Alternatively, (LotL)= Le cos O +L cos0= go 40/0058. +*. cos B=0.8; Lo SMO-L,sno = 10 | 8=3681%, (LoL) sinbeig , += Lo cos6 Lotli= so. Solve using a computer equation Solver, =3687º 1,=33.33m Li=lbkim [34] Find e. paro TE FED T CON Equilibnium equations RF,=-Teos6+ 400 ess 30'=0 EFp= TsinB +a00 sin30º-400=0 T= 200 co830º | 900-a00 cin 30º cos ainB +ang=-SinB q00-200 sin 30º cosa âDO cos 36 6=11.% F8D:B q vA 8 T— IN Equilibrium equations Lty=-Tacos bl +T=0 ER, = TaSinbO-1=0 STIN TRSLISSN Ta=|.158N Equilibrium equations Fy= 1.155 cos bD' + Ta cos 45º —T, =O z Fy 1.155 sinbOº +Ts Sin45º -|=0 Ta= 2838 N Ty=asMN 20 
