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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 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 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 Chapter 1 Solution For Person Ftsec Slugs Solution Wmg Slug Slugs Wmg Slugs Sec
Solution 1.1 () 2 180 lb 801 N lb : 801 N 9.81 /s = == Wmg m 5.59 slugs 4.4482 N 81.6 kg = = m m Solution 1.2 () () () 2 2 1500 kg […]
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 Gsdvdt See Art Tan Tan Dsv Tan Tdtkstds Tanh Cos Skhe
s g Solution 2.38 2 g tv dv akv dt dv dt kv =− =− = (see Art. C.10): − =1 0 g thv k () () − = == = = 1 00 1tan g g gtan […]
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 […]
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 See Plotsa Mssolution Particle Displacement Displacement Particle Velocity Ftsec Elocityv Particle
Solution 2.1 0t, s –100 0 6 0t, s –100 200 2 0 a, m/s 6 2 2 20 100 50 See plots 40 100 0 : 40 100 0, 2.5 s A t 2.5 s, 20(2.5) 100(2.5) 50 75 […]
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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 () […]