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Solution Manual For Machines and Mechanisms: Applied Kinematic Analysis, 4th Edition

Solution Manual For Machines and Mechanisms: Applied Kinematic Analysis, 4th Edition offers the best solutions to textbook problems, helping you prepare for exams and assignments.

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    CHAPTER . Il net oy a le TT, SE eT TU E has handle Fg pa wing ww | \ ends (lower ansle fixed) = dr —| nas, FL Shan, (oven fed) ‘ - n=4, y=4, =| fot pede) oN % = (gest Yoek freed Tug - lL a ) We “—eylinder | seat ~ & 8 Q _ G SUE C reed ) : LI 1-S nd, yey Bey \-lp , ned, ged, Fel, EE _ | ID os 7 ’ [SY po = TH ) =~ 5 J = ow < oF (er hamdie £0 \ower a : uppes anel\e Died) WEES (uve rand flones Lo ied) Nel, y=, Fe ES] IPR [EI we 151 J Toy ~ ER Rr Ged). \\ AN . Coed) Nu < \ AN Net A 1 < + StudyXY ® ANIA

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    andl - \-\0 © n=5% FLY TNR R i i F= sey 1 ATE = S| EE | clamp @ oi Fel 45588 = PEs Ee a Fo ew el i ; ‘ _7 ZN NIA * & Shove) 7 hE ne 1 IC C Fel F=2 \irear achviotec Cargo Yoo ~~ | y AR \ i WT ~ ~ i MICIOWaVe oven i LZ Lr) carrier (revere Leama. fixed) Ln or n=\0, y= 13, F=\ leon n=l 0 o . § 3:1 (@ IA : cn & « 7m Q | ydroshe gm | [WY —— StudyXY /

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    _ os 1-17 1-18 Na a 4 platform <7 x ull Lr AG % ? Wdrashe YN . \f) 3 grey : 7 \ in (5) 5 Show) ) gee n= A y / Ha, n=, 1, y=\2 F223 (rock Srame. ved) Ed! bok] [20 ro | SSN " q hydeawhe — + ¥4 *& oy\wders =e [> 2 (ruck frame red) ‘ & A. . = ; n=9 c- ’ ey hrder C SF At a Neb, ye, Fe |-2\ applante. \- B= > Pass Set 2 | ap e n=7 : * Je=8 { n=4 F=\ HIN _ 1 »H ' , 7 Ro Spang 1s nok on mgd bedy (2) (2 F= awe h "ned Be n= 5) F= \ C $10 ’ Fe ~+ Studyxy

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    BN \-2.5 \-20 Lwks =H Jowts=4 (ews n=b 1 shdwe [5 321 WF = 34-D- 20x 1 F=1 121] Links= 84 Jomis 4 GQ pms) wore DoF: 3D -20)= | y | : n FE ea ee £ -2% = nts = a ~ eT Gey ak WE 3(aD-2ld =1/ sms [129 | Liked lows = reno Jr he A] bess oe, sangy [3B bikend Seo ood) i WF =3(4-H - 204) = \y WF = B(3-N-2(u)= 1g % : 1-3) Links = 9 downts = 4 \-272 L o - wks o ny ws =o do ? Te ded DOF: a-\-2(d) =| WoF= 3 (6-0-2 = 1 . Links= ( dows-1 Lnks= 4 Jombos A C (5pns,2 ny) [CRT noe = 3(b-N- 2-1 WE = 34D = 28) =1 1-235 Links= 9 Joa = © Links= A Joints = 4 (9 pne) [&S ons, \ sliding) ME= 3D -2(H= 2y WF: 3A) 2-1 V-3T1 Lnks=T joints=R Likes Yowdss . wh 2 shai) = HAC wy Toad DF: 30) 2 (= 20 BOF=3rd ~20D = 1 8 k= Joes 1 Like= Lb Joys= 7 (Sms, 2 shin) (bo pues, 1 Shaver) NOR = 34) - 2D = WF= 3L-N-2(N= 1 y =) Lnke= 1D doves 13 NEA Lwks= bb Aowls= 1 CO (0 pins, 3 Siding) (Lb pws, \ sliding) ME- 30D - 203 = Ly RE 4 + StudyXy

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    Vo Links =1D Jone = 122 199] Links=9 Sowts=1} i * "Rpm, 30000 ne “Qa prs, 28d) 0 DoF = 200-N-20) = 3 MF- 26-0 - 200 = 2y 1-95 | Links= bo Joints 1 Links =H Nowe =4 (pins, 26hdng) (1 pe _ DoF: 3-0-2) | DOE = (a=) EXC my i Poh id “=a Links nN Riot tS “48 | Linke =B dots = V0 Eo Figher Jomis=\ (gee pas, Hshdin . NN _ OF 3@D-2(90 = PE= 3LoN-2 Lv Vy \- Ld s le Jove =H \ = ats = 7 . | at hunks =4 Se | shan ne 3 Ces, shad) A ore 30-2d ly WE = 30) - 20) 91) $=15, 121, p= 12,924, 82H, 1=12, p=1,9:=5 S=sve . &=s1de 1.914 £ 14ury 43124 12+ 5: Oo 15.518 (ye) bl 17 (ed = crank - rocker = crank. - roekee 1-53) 6-35 1212, p= 8,q=4 [541 $23) =, 5.12, 55 ' s= sue © S=sde By 4 8 +4 Pars Is < 12 (no) Los Av (ye Farle rocker = crank rocker + StudyXY

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    © Answers to the Chapter 1 Case Study Questions: Case 1-1. 1. As link A rotates clockwise, 90°, slide C will move to the left. 2. As link A rotates clockwise, 90°, the ball trapped in slide C will drop down the lower, left chute. 3. As link A continues another 90°, clockwise, link A will be oriented straight down and slide C will return to the original position shown in the figure. 4. This device allows one feeder bowl to distribute balls to two separate stations. 5. The chamfers on slide C allow relief as a ball drops into the empty slot as slide C moves under chute D. 6. This device would be useful in that only one feeder bowl will need to be filled and monitored. Case 1-2. 1. As handle A is rotated counterclockwise, flapper C rotates clockwise. 2. As flapper C is raised, the water in the tank is allowed to flow through the opening. 3. Abuoyancy force will offset the water pressure, keeping the flapper in the upper position. 4. As the water level lowers to the level of the flapper, the flapper will lower with the water line. 5. Item D floats on top of the water. As the water level lowers, float D also lowers. 6. As item D rotates counterclockwise, item F is also rotates counterclockwise. 7. Item F is a valve that controls water flow, In the upper position, it fills the tank. In the lower position, water flow is cut off. C 8. These mechanisms allow a rapid flow of water from the tank, slowly refill the tank, then shut-off the water flow. 9. The water pressure in residential areas does produce the required flow rate for a water closet. Case 1-3. 1. In the shown configuration, the water level in bucket B, is raising, as the flow into the bucket exceeds the amount that is leaking from the holes at the bottom of the bucket. 2. Inthe shown configuration, the water level in bucket B, is lowering, as the water is leaking from the holes at the bottom of the bucket. 3. If bucket B, were forced upward, rocker arm C would rotate clockwise. 4. If bucket B; were forced upward, rocker arm R would rotate counterclockwise. 5. Rocker arm R controls a directional valve, channeling the water flow to either the upper pipe or the lower pipe. 6. As water drains from one bucket, making it lighter, and fills the other, making it heavier, the weight shift cases the rocker C to rotate and reverse the direction of the water flow. The process repeats itself and rotates rocker C back to the original position. The continual motion is oscillation of rocker C. 7. As rocker C rotates, channel S moves between left and right positions. This allows the steel rod, which is constantly moving, to be coiled onto a reel placed on the left side or a reel on placed on the right side. 8. Since water is abundant and a common cooling medium in most foundries, water flow can ® be used to drive some machinery.

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    Ge Answers to the Chapter 2 Case Study Questions: i Case 2-1. 1. Ashandle A is rotated, moving threaded rod B to the left, grip C also moves to the left and slightly upward. Notice that links E and F are pivoting in the middle, thus grip C is constrained to a swinging motion. 2. As handle A is rotated, moving threaded rod B to the left, grip D moves to the right and slightly downward. Since links E and F are pivoting in the middle, grip D will have motion opposing grip C. 3. The purpose of this mechanism is to serve as a machining clamp for the workpiece. 4. The spring, G, pulling on link D would cause is to return to an upward and rightward position. 5. The purpose of spring G is, ultimately, to keep a positive contact between the threaded rod and link C. 6. Links E and F have a peculiar configuration to avoid interference with the workpiece, throughout the range of motion of the clamp. 7. Such a device could be called a machining clamp. 8. Since link C is moving in a swinging motion, the rounded end on the threaded rod, assures a consistent point contact with link C. C

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    Co lompes 31 3-2 — eB: Ys» B= T3RYy Az JOB) -66Y = 11.3 my € to l= E> x= DA, 0: 180- (60+ 3% = 307 cos bl = Las R= ve dD tend: En Beso $0 ag g > tan p= 2 180 5 oop ep Ee R= + dY - 12 N p Y ~ 213 p N eg . Wodenose = 2S ie B 8 18a: (0d = Wy 25: {0s OFF 515k lms 84 Os Fw ge 5 Buy : biel O) tonB> 8 = B= I Ref + (8 =q.4 in Hl tan 35°F 30 x= N2uny Selle ee Ca a. Hew B= Fw x= Sumy 25%) ) «(WD $s: q4 4 Qos 39° = Yoo EN y= 6.4 wy 31 tan g- &= B= By <n by law of cosines or 0 pe TREN O 1 . 2s . £=11% wm J 3-1\ (BRR L Sana PL Cosp = ADIGE Tt Ts ny yo AH 46 (6 2(} =8 72 Bh | An - Ne Se 1D oe “nf Tah =22.0y A hse 35( 2a 0 13-6 o \aw 0% (osmes NX, wp SeA=T134 py 2 0 00-2 ot 2 = 45 Tm 25 Z Pe al Sn = Yq y= TH SEE EN tan B= 2305 0-38.70 © = Ey =38.0/ 1” or z of @¥+ 130)" =3g.4)in AN cosy B+ G8 Lo _ Fon GRA + StudyXY | Wf _. Y 2364 30+ [724 20] 5in (3 FEEGE—_G—_— 7

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    a 0 - : aT tind =i §=3e1° 320) c J BE - 384 R i B oO (aN - (soy B BN HEED) ' 3b =H Tremsuring + - ° : y= See 204 Tse (pet-) 1 R248 fg. EEE = bb + (ID si (0d): 323, fm [3-2 =f °C o J wa R A * . : R-433%/864 R=2\2.13 757 323) \ EA C c B £ B E R 7 D B R A _ 222). Jay | reqs Tl R=221.20 avy EH) — A R_ R B AT R=] (1% (135- 2000190 = 24 ALS = c Bzsiv! (13s 10) = 12.071° ) R=24.% /18.07° C R=110.4 0297Y :

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    3-271 B «B=3 221) R= 1190 502s (i)cofd ~ =212.13 @& > R polos Re 7h fos O° ars 3 B=150 qo Lf hb PAY) ry N / i qn [RSE Eyes Ts = 4.33 PUES tm |o-arldnentloag 0 Ren BY Ha | R-4 RIN 28888 « ———— S— smn 329) Ry= LRaChE S19 ws30 +1.5 eotbDd Ry? “RTS -100 & 20 oh sald LE ae Ry HLS SRS + 195i 0 155m ky = 3.84 Ry =D T5ssd0¢ mo $n 20-40sn10= =} A R=|(-0h+(asd 9.85 5 S a (510 * ah : ba (8): 01 Fr * UEOREE x R= q £9 gL6T \ R=) 5 bs 34s. 134 15-31 R, 2010 5m4D ~30-5beos 30 [3-32 N C -20m 90 = -5D.0S © |Ry=4D-b0eosUD - 0-505 3D + 2Deos SO = -9¢.1 Rx. R= [(- ¥ ¢ T 5 J = SES Fe B=tan WIG oer 291°! a R= 00 GTOTY T8074 ea. 3-33] A 3-34 J=A>B B K K=B~A L J K= 8.074 /b8.20° EE J=5.581 M30 + Studyxy

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    To NR EX | I @ A. | J=A~T Measuring: B K REVAYREAWES K=B=A ’ fg | Mewes: sip K=5581 3a) : =n E-} = iH ER 3=AiT \5=2()(B) cos 3D } (=B A Bl 3 = 8.01 i Ea ogc BOI 18-19) % | Measuring: K | Bon JR enn sy A=IR0NME 8 T=98.0M GEE C 53 eaten | 2) K={ 184152 (1n)(15)e0s30 5 I3-(4543) = 105 a. = 8.01 y 3 K <0 (SI +015 3) - on ; B oreo SB) 5 Ie NIRTE EY . [8130-6 = 682k Bs eg an) K=8.01 fobaw 01a = Le T5987 Azaw py, Th 105 ra ad) 1 Ges © O\ 70s SOE 8 “Hef E-2 (DB) eosl05 ! Ww TU F221 J 45 = 5.987 20 Er Bes" (gm sn 1) 312% 156° A ew (5 sm) . H5-8= RI B=1%0- or 1) or @ ¥=5.561 TY T=2un & —+ Studyxy

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    3-43 0320-9) J=CA~B @ EXE Re fisBv 150-2 (fideo A CE) ses i fe © B51 (50 sind) ¢ . pen 2 =45° aE 226.094 \I4g i 2nd eX a 3-49 K= B>A>C 3-46 J=CpA-B+D : 3 B B %* Cc K J I! m y c RALMNY : < | K © RASUMNG K=206.009 14D T2154 Tel 288 CBM] EosbrA-C 348] 3-Co pr BD A C Dp 2 D B C A Measunag : E J K=44824 NR9° Measuring: T=101.619 18.28 34] K=B-D+A+C+E [3-3D] A=D+B=C~E i AsB= CDE K D E D ENDS E G Measuring: A K=107.869 \Zu4% | ¢ —+ Studyxy

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    3-51] A>cuE=DsR T-CoA~B Oo A+E=B CHD Ty - (15 coslD) + (250s)- (Sed) D o =2558 . Jy 15 arid (R543) ~(15 sm) 2 == + ) v — T= sd (510) i SE al Beton" (SWass)= 14 hn. . AB T= 24.10 \JF 1 [353] K=B=A>C 3-4[T= Co A=B+D | Ke =(I5c0s3)- (12 500543) (15 00840) [T= (£100 eos2d)s (5D) (5 cos id) k =-299% +(-u4d eee 10) = -100.2 Ky (55w30)- (12 5549) ~(-T54nld) | 3, = (po sin 2B + (0) (TSsin 40) = Sb Ween + (-40 $1010) = 44.8 [] ~ i 3 - * > bo = - I - 3 =" 7 Ng CE a fu = 26.10 BN NE he) Sit? Re Botan ($Y): NAY O:tan (4% 9) = 24.) K=2L.10 nab T=100.4 z41°K ie 3-86{1= Cs AR» DE Y= (2150s 40) (~40 oe70) + (5D) l= Lusi 8) ( a= iy bound B= (IB) tdy LD 030) R= (550 - Lapsed + (5 - (29930) = ~9635 A = i CAs Jy= 0+ (F40)- (60 coniO)+ (-S0s1 3D - (ds) = 44.82 N Ke - (200s) = 3.89 B I G oo | Kz (0 + (HRY 44 ]2 RR ESET (1 \ Bota! (424): 847 [IAT — = rs . Btn (318g) = 18.7% = 44.8 \8T C i 1 T=10.68 Tay + Studyxy

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    ISNK=B>D>A>C»E 35B| AsBe(=D»E G Kx=(60sn40)- (5D cos DR) -(-30) ess + (-20 $1n50)= 4.34 £=\4.20 oy Ky = (~bDews0)~ (-50 an30)r (40)-0 E=11%2—> B - + (2050) = -48.)| C 59353 Fre E ET Kc) Xe (QS) + (sy [_ = SER X me Yo RNC 107.84 . 5 mi Bata (BLS i I= 107.8 \Ze5° I P-ATA+B=I>D-E~F ASBoTeD- SEwF + I - +E-B= Fo, & E B CAF F A c |. \ “B ) D Measuring : E D=38.\2 65° F=23%9 730 measuring Be8.8\ ash Cn asy 361 A+R +0 =DwE 5-2 psBeC-D= EF assume. diveckioms C7 Es fer @ssume directions DN, FA hon CLs: dui 0) (Deets) O+ (20coniS)+ Copshl = 1530+ = (qSeoskd) ~(Feos™® veer: C= (19.95 Fees 30) eon IS 0 10+ (20524S)* Coin bD=155m30 +0 Eye (ps0) (Den) C=-192 = (TS sm LD) - (Fm c Shshivhng © E=11.9 D- (Eao- ba on 4s @ N BUR Meovs C2 w7t E-Ms2—~ [PE et be an.iz F=2380/20 1=33.\) [EEE

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    -b A2T3>CoD= » EF C a dweenons BR , C7 To 0s3)- (-Beos49)- (Coos 4 >) +0 = - (-45 con 60) + (psa 30) wp, B= Gems 89.96)fss 45 —0 BE | (L0sn30)- (Bemid)-(Lants) + (4) Bw |= (455i) (3Peos30) a B= (-T1.0- CsmiSybmis —@ HEE | Solving © @simethancausly EH 8-3.2 , Cx ty po | P= ask C1 BY | * e ©

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