Archives: Solution Manual
Management Chapter 18 Homework Page The Community And The Corporation Introduction Strong Relationship Benefits Both
Page 18-1 CHAPTER 18 THE COMMUNITY AND THE CORPORATION INTRODUCTION A strong relationship benefits both business and its community. Communities look to businesses for civic leadership and for help in coping with local problems, while businesses expect to be treated […]
Management Chapter 17 Homework Page Business And Its Suppliers Introduction Corporations Have Complex Relationships With
Page 17-1 CHAPTER 17 BUSINESS AND ITS SUPPLIERS INTRODUCTION Corporations have complex relationships with their suppliers, other firms that provide them with goods and services and in some cases manufacture their products. In today’s interconnected world, many firms are embedded […]
Management Chapter 16 Homework Page Managing Diverse Workforce Introduction The Workforce The United States More
Page 16-1 CHAPTER 16 MANAGING A DIVERSE WORKFORCE INTRODUCTION The workforce in the United States is more diverse than it has ever been, reflecting the entry of women into the workforce, immigration from other countries, the aging of the population, […]
Management Chapter 15 Homework Page Employees And The Corporation Introduction Employees And Employers Are Engaged
Page 15-1 CHAPTER 15 EMPLOYEES AND THE CORPORATION INTRODUCTION Employees and employers are engaged in a critical relationship affecting the corporation’s performance. There is a basic economic aspect to their association: Employees provide labor for the firm, and employers compensate […]
Management Chapter 14 Homework Page Consumer Protection Introduction Safeguarding Consumers While Continuing Supply Them With
Page 14-1 CHAPTER 14 CONSUMER PROTECTION INTRODUCTION Safeguarding consumers while continuing to supply them with the goods and services they want, at the prices they want, is a prime social responsibility of business. Many companies recognize that providing customers with […]
Management Chapter 12 Homework Page Regulating And Managing Technology Introduction The Rapid Advances Technology Just
Page 12-1 CHAPTER 12 REGULATING AND MANAGING TECHNOLOGY INTRODUCTION The rapid advances in technology in just the past few decades have radically changed how business operates. More information than ever before—about individuals, organizations, and governments—is located on servers or in […]
Management Chapter 11 Homework Page The Role Technology Introduction Technology Unmistakable Economic And Social Force
Page 11-1 CHAPTER 11 THE ROLE OF TECHNOLOGY INTRODUCTION Technology is an unmistakable economic and social force in both business and the world where we live. Global and local communications, business exchanges, the science that affects the quality of our […]
Management Chapter 10 Homework Page Managing For Sustainability Introduction Growing Public Concern About Sustainability Has
Page 10-1 CHAPTER 10 MANAGING FOR SUSTAINABILITY INTRODUCTION Growing public concern about sustainability has prompted political, corporate, and civil society leaders to become increasingly responsive to environmental issues. In the United States and other nations, government policymakers have moved toward […]
Mechanical Engineering Appendix B Solution Ooaoggg Dimensions Element Composed Three Parts And Let Massunit Face Area
Solution B.43 () () 22 11 62 11 0.160 0.080 Table D.4 12 3 54.6 10 kg m OO Im m ρ − =+ =× ⋅ () () () () () () 2 2 2 2 2 262 2 2 […]
Mechanical Engineering Appendix B Solution Zdx Bxmdxax Use Quartercircular Shell Element Shell Height The Bounding Curve
Solution B.21 ()() 0 32 2 0 23 3 0 zz zz IdIdmr xzdx xaxdx ρ == = = − 22 215m 23 so 15 2 15 22 a z a a bax a x […]
Mechanical Engineering Appendix B Solution Ydxds Cos Lds Dxcos Cos Xdx Cos Cos Cos Cos Cos
Solution B.1 cos 2 1cos cos 3 12 cos L xm LL mL β ρρβ βρ β − == = 22 1 sin 2 22 sin 2 32 22 11 sin sin L m LmL β ρβ […]
Mechanical Engineering Chapter 8 Solution For The Bar Combined Let The Equilibrium Position Shown And Choose
Solution 8.87 For the bar, 2 22 1313 12 10 75 Imm m =+ = Combined: 2 222 01 2 1 2 113113 2 5 75 50 75 Im m m m =+=+ Let θ […]
Mechanical Engineering Chapter 8 Solution Ggr Mgkxcx Roll With Slip The Xequation Reduces Rad Sec Solution
Solution 8.103 2 : x GG mg FmxkxcxFmx =−−+= == Roll with no slip: xr α =− The x-equation reduces to 2 2 10 G k mxcxkx r +++= () () 2 […]
Mechanical Engineering Chapter 8 Solution For For Rad For Rad For Rad Thus Prohibited Range Solution
Solution 8.47 2 10.15 1.5 X === Solution 8.48 , iB iB xx xxxx=+ =− () ()() 2 From text: B iB iB i ii BB cx kx m x x dt cx x kx x mx ckk c xx […]
Mechanical Engineering Chapter 8 Solution Solution Rad Secn Rad Cycle Hzn Sec Rad Solution Sin Tan
Solution 8.1 () 39.81 736 N m 736 4.20 lb in. m 175.13 N m lb 12 in. 4.20 50.4 lb ft in. ft W k k k == = == == Solution 8.2 () […]
Mechanical Engineering Chapter 7 Solution Zooz Yrorc Crelorc Let The Angular Velocity The Axes Xyz Sin
Solution 7.119 () τ Ω= + Relative to the xyz axes, O′ is fixed and C moves with speed () rel 2 cR π υτ = So () () rel rel 2 cR j j rr υπ ωτ =−=− 2 […]
Mechanical Engineering Chapter 7 Solution Case Rad Case From The Precessional Rate Cos Cos Rad From
Solution 7.105 ππ × Solution 7.106 I = moment of inertia about its longitudinal axis () 12 ma a a ′′ = += I0 = moment of inertia about transverse axis through () 22 1 0,82 12 ma a=+== ′′ […]
Mechanical Engineering Chapter 7 Ysolution Xyx Rad Sec Revminym Kgcdbabbbbzgnb Bxgcrzgragdxx From About Xyzzyxzz Mbryzxz Bxxx
y CD x 0, 1200 2 60 125.7 rad sec xy z ωω π ω == × = = Solution 7.74 N = 1200 rev/min From Eq. 7.23 , about G, 22 , 0 x yzz y xzz z MI […]
Mechanical Engineering Chapter 7 Solution Angular Velocity Xyz Axes Yrela Aba Relbarelxob Where Rpi Rel Thus
A Yy υ rel Solution 7.42 Angular velocity of x–y–z axes is 12 iJ ωω Ω=− + υυ υ υ ==+Ω× + () ()() () ()() ωω ω ωωω υωω ω ωω ωω Ω× Ω× = − + × − […]
Mechanical Engineering Chapter 7 Solution Zzyzyy Firstsequencex Yxemememxxzz Secondyzyy Sequencey Xememxemxx Final Positions Are Different Finite
Solution 7.1 Solution 7.2 Final positions essentially the same – the more so the smaller the angle. Infinitesimal angles add as proper vectors. y x x x y y y z z z EM xxx EM yyy z Second sequence […]
Mechanical Engineering Chapter 6 Solution Madaafa Nagma Nfa Noxxo Ysolution Gmay Madc Solution Constant Fmph Asf
Oy Solution 6.1 () == = AA Mmad 30 20 600 N x ma Solution 6.2 () () = = mad 6 66 2 C M a x = g 3g a y Gma 2″ 2″6″ 6 /b […]
Mechanical Engineering Chapter 6 Wheel Assume Roll Slip Shown Mgbb Xgxb Bbxb Aygyf Ggf Bbyffbf Fsb
BB () () () :11 MI Frmk αα == 2 :0 9 :g 10 x yB B GG f B ω F fB B y xGxB yGyf B FmaBN FmaBF m ma ωω =−= […]
Mechanical Engineering Chapter 6 Solution Yox For Dydu Gdtd Tttdmim Madx Doo Adx Madx Madx Dxdvd
Solution 6.130 222 =+ = 2 18 OO adx 233 22 madx m madx () =++=+ + = 2 3mg 2 dx 123 gggg 0mg mg dx d VdVVV dx =+ 83 T hus mg 32 Pdx ma dx dx […]
Mechanical Engineering Chapter 6 Solution Angular Impulse Negligible During Impact Sin Sin Mga During Rotation Cos
13″ 2 Solution 6.157 Angular impulse of mg is negligible during impact. 2 1 0:m sin 23 3sin 2 L υ ωθ = G mg A L HmL υθ ω Δ= = During rotation, ΔT + ΔVg = 0 () […]
Mechanical Engineering Chapter 6 Solution Mgmagigragopoaoo Aaaa Gog Rigmagr Mgmroopr Drmrmrma Sin Cos Sin Cos Ado
Solution 6.84 2 2 5 Imr= ααα =+ =−= = 25 :57 GOGO a aaa aar r r () ωθ αθθ =+ 7 5 a r () () ωω α θ ωω θ θ θ ωθθ ==+ =−+ :gsincos […]
Mechanical Engineering Chapter 6 Solution Rad Tcaabktm Llm Ldagl Glggalm Kvg Obo Brboad Instant Shown Bar
Solution 6.107 ! ! radians radians 22 2 2 2 1111 1 2mg sin 2mg sin 22 2 2 2 2 2 TOOACABD T LL kII I k ω θ φ ωωω θ φ ↑ ↑ ′ +=+++ + Kinematics […]
Mechanical Engineering Chapter 6 Solution Rad Rad Jmk Mgoy Rxgxxgx Mag Magyyygy Oot Solving Right Oxy
tR 100 lb-in. 2 0.323 lb-ft-sec = 2 2 20 3 0.323 0.3623 lb-ft-sec 32.2 12 O I =+ = () () 2 2 100 ; 0.3623 , 23.0 rad sec 12 3 ;23.05.75ftsec 12 20 ; 5.75 […]
Mechanical Engineering Chapter 6
Solution 6.62 , but 0 so 0 G MI I M α == = Hence no friction force and 0 s µ = θα == ;mgsin gsin xx A A r αθα αθ == = 2 ;mg […]
Mechanical Engineering Chapter 5 Solution Insec Constant Aba Sin Cbb Tan Sin Sin Seca Insec Rad
A O 16″ 6″ 4″ υ A β 2 υ A/B υ A δ = 90 + β – γ = 78.3° δ 16 16 − 12 tan 3 33.7 γ − = =° 10 AB υ = γ 3 […]
Mechanical Engineering Chapter 5 Solution Secb Mib Mixv Sec Vab Relav Angular Velocity Axes Rad Sec
B A x y 5 mi 9 mi υ B υ A rel AB Angular velocity of axes = () 704 0.01481 rad sec 9 5280 B vkkk ωρ − =−= =− × () rel velocity of relative to 5 […]
Mechanical Engineering Chapter 5 Solution From The Solution Prob Rad Abx Bab Aaabb Aabb Terms Ibb
A y B 0.2 m BA BA BABAAAB AB β Solution 5.113 From the solution to Prob. 5.82: β ω =° =44.4 , 4.37 rad s CW AB 2 aaa a r r αω =+ =+ × − (1) Terms […]
Mechanical Engineering Chapter 5 Solution Icr Sin Sin Sin Bcac Rad Ccw Down Bb Solution Radscoag
Solution 5.77 0.4 0.359 m BC AC BC () υ ωω υ ω === == = 0.5 , 1.394 rad s CCW 0.359 0.293 1.394 0.408 m s down B A BC AC υ A 45° 0.4 m […]
Mechanical Engineering Chapter 5 Solution Xblbc Sin Hbahcd Cbd For Vertical Motion Only Its Horizontal Coordinate
Solution 5.40 For vertical motion only of B, its horizontal coordinate remains constant so () () 2 1cos 2tan tan 1 1cos cos 2 dLx Lx x b Lx Lx c bb θ θθδ ++ == ++ () 1 where […]
Mechanical Engineering Chapter 5 Solution Rev Mint Rev Solution Solution Rad Rad Sin Cos Cos Sin
Solution 5.1 ()() () ωω α ωω ω ωαθθ α −− == = − =+ == − == 2 21 22 22 21 21 22 900 300 6000 rev min 660 2, 2 900 300 60 rev 2 6000 t […]
Mechanical Engineering Chapter 4 Solution Djmd Idjmd Rii Mid Vjmvi Mvm Vmvi Mvdk Mvdk Mvdkoiiih Fdkoo
Solution 4.1 () () () () () == =+ == + === 2 222 66 6 11 11 22 3 i i i i rij mm FFi rmm Tmv mvmv mv +++ == ++ […]
Mechanical Engineering Chapter 4 Solution Pav Sin Cos Solution Ftbcct Pavx Sec Fttpc Lbin Vvpb Lbinc
Solution 4.42 π 2 Solution 4.43 () () () () − ′= − ′== ′== 4 4 144 0.840 1b sec 0.349 50 0.455 32.2 ft 0.0760 0.0873 2.06 10 B Ccc mpAv m mvv T/2 x p B = […]
Mechanical Engineering Chapter 4 Solution Maxxxmgx Constant Sec Propulsion Time Sec Sin Odtt Sin When Secv
Solution 4.70 θ Σ= − = :sin 2(400) 24.8 lb constant mg T x 10° 32.2 Propulsion time 20 10sec t== xx x F ma T mg ma 2 0 000 S o( )sin( ) sin θ −= o […]
Mechanical Engineering Chapter 3 Solution Annyy Mgf Mnyfnn Vnn Nnn Totyn Out Ktotk Ttk Tt
Solution 3.276 22 0: mg : : ma m F µ = − 2 gvm += 2 m , 10.75 s aa=− yy nn t tk 4 2 ρ tt As in part (a), Ny = mg 2 a nd […]
Mechanical Engineering Chapter 3 Solution Axx Xxx Oaox Xdx Axa Xoo Xxoad Kxkxx For Max And
Solution 3.253 ()() 00 : 22 xx xx k x o a o Fmakxx mxa mm =−−=− =+ − = =+ =+ 00 So , kx m a kx axx 22 0 0 00 0 […]
Mechanical Engineering Chapter 3 Solution For Interval Lbv Tan Inlbkk Spring Compression Ink Lbin Cos Inlbe
Solution 3.136 () 0; 0.01294 34.64 17.24 0 1345 in./sec , 36.7 in./sec or 3.06 ft/se c TV V v vv v Δ +Δ +Δ = − + = = = = 2 g 2 2 e 0.01294 v = […]
Mechanical Engineering Chapter 3 Interesting Examine The Yvelocity The Extremes Gty Cdd For Cdvc Adx Coo
It is interesting to examine the y-velocity at the extremes: ()() () 2.04 0.8 2 9.81 4 9.81 1.060 3.31m/s y v =−=− d 0 0 0 0 2g g 2 ve d t d dv =− − C y […]
Mechanical Engineering Chapter 3 Solution The Speed The Plug Upon Impact Ftsec Angular Momentum About Preserved
Solution 3.182 The speed of the plug upon impact is 2g 2(32.2)(2) 11.35 ft/secvh== = Solution 3.183 2 12 1 1 00 0 01 1 44 2 0 00 2 (2 3 ) 0.4( 2 ) 1 (4 2t) (3 […]
Mechanical Engineering Chapter 3 Solution Lby Mgr Propulsive Forcefrx Tan Sin Sin Solution For Amp Max
Solution 3.92 60 50 110 lb propulsive force tan 0.06 3.43 mg 3600 lb F θ = ==° = F N x θ R F =+= () a 0 : 110 3600 sin3.43 […]
Mechanical Engineering Chapter 3 Solution Mgn Solution Mrynn Rys Mgonn Rss Solution Ftsec Ftsec Average Per
Solution 3.48 22 2 g3.13ms 0.981 N == = v ρ Solution 3.49 2 ; nn FmaNmr ω Σ == n T = O mg t mg y · θ = ω :mg 1 g :mg 1 Σ= − = […]
Mechanical Engineering Chapter 3 Solution Maxxkxxa Kinematics Knn Solution Cos Throughout Equilibrium Check Sin Fnf Motion
y N ()() 2 06 2 3.92 xx −=− − Solution 3.1 0 : mg 0, mg :mg xxk x Fma ma µ Σ= − = = Σ= − = y FN N 0 4.59 mx= […]
Mechanical Engineering Chapter 2 Solution Aax Abbn Mihra Mihrsecv Mihr Constant Aabv Sin Cos Sin Aba
Solution 2.193 A y x 2 20° 45° υ A , a A υ B Solution 2.194 υ 1 υ 2 60° 45° 45° t 1 t 2 y x cos 200 cos60 100 ft/sec sin g 200 sin60 32.2 […]
Mechanical Engineering Chapter 2 Solution From Prob Radsa Cos Sin Cos Sin Jabab Sin I Cos Cos I
Solution 2.161 From Prob. 2.160… / 112.1 , 40.2 49.8 , 19.75 15.43 m/s, 0.01446 rad/s ==° =° = ° =− = rm r φ θβ θ . == AB r r aae ()() () […]
Mechanical Engineering Chapter 2 Solution Msv Cos Cos Osculating Planeat Cos An Sin Msn Solution Sin
Solution 2.133 222 6 3 2 m/s, 6 3 2 7 m/s =−+ = ++ = = vijk v v p 22 78.59 m v pa ∴= = = 5.70 n Solution 2.134 () () 0 0 g 520 32.2 […]
Mechanical Engineering Chapter 2 Solution Sin Asinx Where Amp Radius Curvature Dyd Cos Sinx Dxdxd Set
Solution 2.102 T = 2L y x 3 m ω =sin y Ax, where 3mA=& ω =2 π T Radius of curvature 3 22 22 1dy dx + dydx ρ = dx dx […]
Mechanical Engineering Chapter 2 Solution Kmhr Msbc Landing Zone Solving Other Solution Has Negative Cos Mfy
Solution 2.66 y x A8° 37.5° Landing zone BC = 45 m C S B 60 m Solving… 2.92 s f t=, =41.0 ms (Other solution has negative s) 22 2 2 60 41.0 cos37.5 92.5 m 30 41.0 sin37.5 […]