solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
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solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 15 Document Transcript

  • 1. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 1.Angular coordinate: ( )238 6 2 radianst tθ = − −Angular velocity: ( )224 12 2 rad/sdt tdtθω = = − −Angular acceleration: 248 12 rad/sdtdtωα = = −(a) When the angular acceleration is zero.48 12 0t − = 0.250 st =(b) Angular coordinate and angular velocity at t = 0.250 s.( )( ) ( )( )3 28 0.250 6 0.250 2θ = − − 18.25 radiansθ = −( )( ) ( )( )224 0.250 12 0.250 2ω = − − 22.5 rad/sω =
  • 2. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 2.30.5 cos4te tπθ π−=( )3 30.5 3 cos4 4 sin 4t tde t e tdtπ πθω π π π π− −= = − −( )( )2 3 2 3 2 3 2 32 3 2 30.5 9 cos4 12 sin 4 12 sin 4 16 cos40.5 24 sin 4 7 cos4t t t tt tde t e t e t e tdte t e tπ π π ππ πωα π π π π π π π ππ π π π− − − −− −= = + + −= −(a) 0,t = ( )0.5θ = 0.500 radθ =( )( )0.5 3 4.71ω π= − = − 4.71rad/sω = −( )( )20.5 7 34.5α π= − = − 234.5 rad/sα = −( ) 0.125 s, cos4 cos 0, sin 4 sin 12 2b t t tπ ππ π= = = = =30.30786te π−=( )( )( )0.5 0.30786 0 0θ = = 0θ =( )( )( )0.5 0.30786 4 1.93437ω π= − = − 1.934 rad/sω = −( )( )( )20.5 0.30786 24 36.461α π= = 236.5 rad/sα =
  • 3. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 3.7 /60 sin 4te tπθ θ π−=7 /6 7 /607sin 4 4 cos46t tde t e tdtπ πθ πω θ π π π− − = = − +  2 2 27 /6 7 /6 7 /6 2 7 /607 /6 2 2049 28 28sin 4 cos4 cos4 16 sin 436 6 649 2816 sin 4 cos436 3t t t ttde t e t e t e tdte t tπ π π ππω π π πα θ π π π π πθ π π π π− − − −− = = − − −     = − − +    ( )a 0 0.4 rad,θ = 0.125 st =7 (0.125)/60.63245, 4 , sin 1, cos 02 2 2e tπ π π ππ−= = = =( )( )( )0.4 0.63245 1 0.25298 radiansθ = = 0.253 radθ =( )( ) ( )70.4 0.63245 1 0.92722 rad/s6πω = − = −  0.927 rad/sω = −( )( ) ( )2 2490.4 0.63245 16 1 36.551rad/s36α π = − − = −  236.6 rad/sα = −( )b ,t = ∞ 7 /60te π−= 0θ =0ω =0α =
  • 4. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 4.Angular coordinate: 1800 rev = 3600 radiansπθ =Initial angular velocity: 0 6000 rpm = 200 rad/sω π=Angular acceleration: constantd ddt dω ωα ωθ= = =d dα θ ω ω=000d dθωα θ ω ω=∫ ∫2012αθ ω= −( )( )( )2220 20017.4533 rad/s2 2 3600πωαθ π= − = − = −(a) Time required to coast to rest.0 tω ω α= +0 0 20017.4533tω ω πα− −= =−36.0 st =(b) Time to execute the first 900 revolutions.900 rev = 1800 radiansθ π=2012t tθ ω α= +( ) 211800 200 17.45332t tπ π= −272 648 0t t− + =( ) ( )( )272 72 4 6482t± −= 10.54 st =
  • 5. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 5.( )( )1 0 12400 22400 rpm 80 rad/s, 0, 4 s60tπω π ω= = = = =( )a 211 080, 20 rad/s4t ttω πω ω α α α π= + = = = =( )( )2 21 01 1 1600 20 4 160 rad 80 rev2 2 2t tπθ ω α π ππ= + = + = = =1 80 revθ =(b) 1 2 2 180 rad/s, 0, 40 st tω π ω= = − =( ) 22 12 1 2 12 10 80, 2 rad/s40t tt tω ω πω ω α α π− −= + − = = = −−( ) ( ) ( )( ) ( )( )2 22 1 1 2 1 2 11 180 40 2 402 2t t t tθ θ ω α π π− = − + − = + −16001600 radians 800 rev2πππ= = = 2 1 800 revθ θ− =
  • 6. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 6.Angular acceleration: 0.230 t dedtωα −= =Angular velocity: 0 0tdtω ω α= + ∫0.200 30t te dt−= + ∫0.20300.2tte− =  −( )0.2150 1 t dedtθ−= − =When t = 0.5 s, ( )( )( )0.2 0.5150 1 eω −= −14.27 rad/sω =Angular coordinate: 0 0tdtθ θ ω= + ∫( )0.200 150 1t te dt−= + −∫0.201501500.2ttt e− = −  −( )0.2150 750 1 tt e−= − −When t = 0.5 s, ( )( ) ( )( )( )0.2 0.5150 0.5 750 1 eθ −= − −3.63 radiansθ =
  • 7. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 7.( ) 0.5 , 0.5 0.5da d ddωα ω ω ω ω θθ= − = − = −030 0Integrating, 0.5 30 0.5d dθω θ θ= − − = −∫ ∫6060 radians 9.55 revθπ= = =29.55 revθ =( ) 0.5 2d db dtdtω ωωω= − = −Integrating,00 302t ddtωω= −∫ ∫02 ln30t = − = ∞ t = ∞( )c ( )( )0.02 30 0.6 rad/sω = =0.60 302t ddtωω= −∫ ∫0.62 ln 2 ln 5030t = − = 7.82 st =
  • 8. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 8.dk k d k ddωα θ ω θ ω ω θ θθ= − = − = −Integrating,0 612 0d k dω ω θ θ= −∫ ∫2 212 60 02 2k − = − −   ( )a222129 s6k −= = 29.00 sk −=312 0d k dωω ω θ θ= −∫ ∫2 2 212 39 02 2 2ω  − = − −   ( )b ( )( )22 2 2 212 9 3 63 rad /sω = − = 7.94 rad/sω =
  • 9. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 9.( ) ( ) ( )/ 5 in. 31.2 in. 12 in.A O = + +r i j k( ) ( )/ 5 in. 15.6 in.B O = +r i j( ) ( ) ( )2 2 25 31.2 12 33.8 in.OAl = + + =Angular velocity.( )/6.765 31.2 1233.8A OOAlω= = + +r i j kωωωω( ) ( ) ( )1.0 rad/s 6.24 rad/s 2.4 rad/s= + +i j kωωωωVelocity of point B./B B O= ×v rωωωω1.0 6.24 2.4 37.44 12 15.65 15.6 0B = = − + −i j kv i j k( ) ( ) ( )37.4 in./s 12.00 in./s 15.60 in./sB = − + −v i j kAcceleration of point B.B B= ×a vωωωω1.0 6.24 2.4 126.1 74.26 245.637.4 12 15.60B = = − − +− −i j ka i j k( ) ( ) ( )2 2 2126.1 in./s 74.3 in./s 246 in./sB = − − +a i j k
  • 10. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 10.( ) ( ) ( )/ 5 in. 31.2 in. 12 in.A O = + +r i j k( ) ( )/ 5 in. 15.6 in.B O = +r i j( ) ( ) ( )2 2 25 31.2 12 33.8 in.OAl = + + =Angular velocity.( )/3.385 31.2 1233.8A OOAlω= = + +r i j kωωωω( ) ( ) ( )0.5 rad/s 3.12 rad/s 1.2 rad/s= + +i j kωωωωVelocity of point B./B B O= ×v rωωωω0.5 3.12 1.2 18.72 6 7.805 15.6 0B = = − + −i j kv i j k( ) ( ) ( )18.72 in./s 6.00 in./s 7.80 in./sB = − + −v i j kAngular Acceleration.( )/5.075 31.2 1233.8A OOAlα −= = + +r i j kαααα( ) ( ) ( )2 2 20.75 rad/s 4.68 rad/s 1.8 rad/s= − − −i j kααααAcceleration of point B./B B O B= × + ×a r vαααα ωωωω0.75 4.68 1.8 0.5 3.12 1.25 15.6 0 18.72 6 7.8B = − − − +− −i j k i j ka28.08 9 11.7 31.536 18.564 61.406= − + − − +i j k i j k( ) ( ) ( )2 2 23.46 in./s 27.6 in./s 73.1in./sB = − − +a i j k
  • 11. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 11.( ) ( ) ( ) ( ) ( ) ( )/ 500 mm 225 mm 300 mm 0.5 m 0.225 m 0.3 mB A = − + = − +r i j k i j k2 2 20.5 0.225 0.3 0.625 mABl = + + =Angular velocity vector.( )/100.5 0.225 0.30.625B AABlω= = − +r i j kωωωω( ) ( ) ( )8 rad/s 3.6 rad/s 4.8 rad/s= − +i j k( ) ( )/ 300 mm 0.3 mE B = − = −r k kVelocity of E./ 8 3.6 4.8 1.08 2.40 0 0.3E E B= × = − = +−i j kv r i jωωωω( ) ( )1.080 m/s 2.40 m/sE = +v i jAcceleration of E.8 3.6 4.8 11.52 5.184 23.0881.08 2.4 0E E= × = − = − + +i j ka v i j kωωωω( ) ( ) ( )2 2 211.52 m/s 5.18 m/s 23.1m/sE = − + +a i j k
  • 12. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 12.( ) ( ) ( ) ( ) ( ) ( )/ 500 mm 225 mm 300 mm 0.5 m 0.225 m 0.3 mB A = − + = − +r i j k i j k2 2 20.5 0.225 0.3 0.625 mABl = + + =Angular velocity vector.( )/100.5 0.225 0.30.625B AABlω= = − +r i j kωωωω( ) ( ) ( )8 rad/s 3.6 rad/s 4.8 rad/s= − +i j kAngular acceleration vector.( )/200.5 0.225 0.30.625B AABlα −= = − +r i j kαααα( ) ( ) ( )2 2 216 rad/s 7.2 rad/s 9.6 rad/s= − + −i j k.Velocity of C /C C B= ×v rωωωω( ) ( )/ 500 mm 0.5 mC B = − = −r i i8 3.6 4.8 2.4 1.80.5 0 0C = − = − −−i j kv j k( ) ( )2.40 m/s 1.800 m/sC = − −v j k.Acceleration of C /C C B C= × + ×a r vαααα ωωωω16 7.2 9.6 8 3.6 4.80.5 0 0 0 2.4 1.8= − − + −− − −i j k i j k4.8 3.6 18 14.4 19.2= + + + −j k i j k18 19.2 15.6= + −i j k( ) ( ) ( )2 2 218.00 m/s 19.20 m/s 15.60 m/sC = + −a i j k
  • 13. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 13.( ) ( ) ( )/ 200 mm + 120 mm 90 mmA D = − +r i j k( ) ( ) ( )2 2 2200 120 90 250 mmDAd = + + =0.8 + 0.48 + 0.36A/DDADAd= = −ri j kλλλλ( )( ) ( ) ( ) ( )75 0.8 + 0.48 + 0.36 60 rad/s + 36 rad/s + 27 rad/sDAω= = − = −i j k i j kωωωω λλλλ0DAddtω= =αααα λλλλ( ) ( )200 mm = 0.2 mB/A =r i iVelocity of corner B./ 60 36 27 5.4 7.20.2 0 0B B A= × = − = −i j kv r j kωωωω( ) ( )5.40 m/s 7.20 m/sB = −v j kAcceleration of corner B./B B A B= × + ×a r vαααα ωωωω0 60 36 27 405 432 3240 5.4 7.2= + − = − − +−i j ki j k( ) ( ) ( )2 2 2405 m/s 432 m/s 324 m/sB = − − +a i j k
  • 14. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 14.( ) ( ) ( )/ 200 mm + 120 mm 90 mmA D = − +r i j k( ) ( ) ( )2 2 2200 120 90 250 mmADd = + + =0.8 + 0.48 + 0.36A/DDAADd= = −ri j kλλλλ( )( ) ( ) ( ) ( )75 0.8 + 0.48 + 0.36 60 rad/s + 36 rad/s + 27 rad/sDAω= = − = −i j k i j kωωωω λλλλ( )( )600 0.8 + 0.48 + 0.36DAddtω= = − − i j kαααα λλλλ( ) ( ) ( )2 2 2480 rad/s 288 rad/s 216 rad/s= − −i j k( ) ( )200 mm = 0.2 mB/A =r i iVelocity of corner B./ 60 36 27 5.4 7.20.2 0 0B B A= × = − = −i j kv r j kωωωω( ) ( )5.40 m/s 7.20 m/sB = −v j kAcceleration of corner B./B B A B= × + ×a r vαααα ωωωω480 288 216 60 36 270.2 0 0 0 5.4 7.2= − − + −−i j k i j k43.2 57.6 405 432 324= − − − −j k i j k( ) ( ) ( )2 2 2405 m/s 389 m/s 266 m/sB = − − −a i j k
  • 15. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 15.993,000,000 mi 491.04 10 ft= ×6365.24 days 31.557 10 s, 1rev 2 radπ= × =Angular velocity.962199.11 10 rad/s31.557 10πω −= = ××Velocity of the earth.( )( )9 9 3491.04 10 199.11 10 97.77 10 ft/sv rω −= = × × = ×366.7 10 mi/hv = ×Acceleration of the earth.( )( )22 3 9 3 297.77 10 199.11 10 19.47 10 ft/sa rω − −= = × × = ×3 219.47 10 ft/sa −= ×
  • 16. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 16.323 h 56min 23.933 h 86.16 10 s, 1rev 2 radπ= = × =63272.925 10 rad/s86.16 10πω −= = ××66370 km 6.37 10 mR = = ×, cos sinR Rω φ φ= = +j r i jωωωω( )( ) ( )6 6cos72.925 10 6.37 10 cos 464.53cos m/sRω φφ φ−= × = −= − × × = −v r kk kωωωω( ) ( )2 3 2cos cos 33.876 10 cos m/sp R Rω ω φ ω φ φ−= × = × − = − = − ×a v j i i iωωωω( ) .a Equator ( )0 cos 1.000φ ϕ= ° =( )465 m/s= −v k 465 m/sv =( )3 233.9 10 m/s−= − ×a i 20.0339 m/sa =( ) .b Philadelphia ( )40 cos 0.76604φ φ= ° =( )( ) ( )464.52 0.76604 356 m/s= − = −v k k 356 m/sv =( )( )333.876 10 0.76604−= − ×a i( )3 20.273 10 m/s−= − × i 20.0259 m/sa =( ) .c North Pole ( )90 cos 0φ φ= ° =0v =0a =
  • 17. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 17.300 mm/sB Av v= = 120 mmBr =( ) 180 mm/sB Ata a= =( )a300, 2.5 rad/s120BB BBvv rrω ω= = = = 2.50 rad/s=ωωωω( )( ) 2180, 1.5 rad/s120B tB BtBaa rrα α= = = = 21.500 rad/s=αααα( )b ( ) ( )( )22 2120 2.5 750 mm/sB Bna r ω= = =( ) ( ) ( ) ( )2 2 2 2 2180 750 771mm/sB B Bt na a a= + = + =750tan , 76.5180β β= = ° 2771mm/sB =a 76.5°
  • 18. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 18.4 rad/sBω = , 120 mmBr =( ) ( )( )22 2120 4 1920 mm/sB B Bna r ω= = =22400 mm/sBa =( ) ( )22 2 2 22400 1920 1440 mm/sB B Bt na a a= − = − = ±( )( ) 21440, 12 rad/s120B tB BtBaa rrα α±= = = = ±212.00 rad/s or
  • 19. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 19.Let Bv and Ba be the belt speed and acceleration. These are given as 212 ft/s and 96 ft/s .B Bv a= =These are also the speed and tangential acceleration of periphery of each pulley provided no slipping occurs.(a) Angular velocity and angular acceleration of each pulley.Pulley A. 8 in. 0.6667 ftAr = =1218 rad/s0.6667A BAA Av vr rω = = = = 18 rad/sA =ωωωω296144 rad/s0.6667A BAA Aa ar rα = = = = 2144 rad/sA =ααααPulley C. 5 in. 0.41667 ftCr = =1228.8 rad/s0.41667C BCC Cv vr rω = = = = 28.8 rad/sC =ωωωω296230.4 rad/s0.41667C BCC Ca ar rα = = = = 2230 rad/sC =αααα(b) Acceleration of point P on pulley C. 5 in. 0.41667 ftCρ = =( ) 296 ft/sP Bta a= =( )( )22 2212345.6 ft/s0.41667P BP nC Cv vaρ ρ= = = =( ) ( ) ( ) ( )2 2 2 2 296 345.6 358.7 ft/sP P Pt na a a= + = + =96tan 15.52345.6β β= = °2359 ft/sP =a 15.52°
  • 20. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 20.1rev radians, 8 in. 0.6667 ft, 5 in. 0.41667 ft2A Cr rπ= = = = =22 2500120 0.002 ,120 0.002 60000d d dddω ω ω ω ωα ω ω θθ ω ω= − = = =− −Integrating and applying initial condition 0 at 0ω θ= = and noting that θ π= radians at the final state,( )220 00500250 ln 6000060000ddωω πω ωω θ πω= − − = =−∫ ∫( )22 60000250 ln 60000 ln 60000 250 ln60000ωω π− − − − = − = 2/2506000060000e πω −−=2 /250 2 260000 1 749.26 rad /se πω − = − = 27.373 rad/sω =( )( )2120 0.002 120 0.002 749.26 118.50 rad/sα ω= − = − =(a) Tangential velocity and acceleration of point B on the belt.( )( )0.6667 27.373 18.249 ft/sB A Av v r ω= = = =( )( ) 20.6667 118.50 79.0 ft/sB A Aa a r α= = = =279.0 ft/sBa =(b) Acceleration of point P on pulley C. 5 in. 0.41667 ftCρ = =18.249 ft/sP Bv v= =( )2 2218.249799.3 ft/s0.41667BP nCvρ= = =a( ) 279.0 ft/sP Bta= =a( ) ( )2 2 2799.3 79.0 803.2 ft/sPa = + =79tan , 5.64799.3β β= = °2803 ft/sP =a 5.64°
  • 21. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 21.Left pulley.Inner radius r1 = 50 mmOuter radius r2 = 100 mm0.6 m/s = 600 mm/sAv =126006 rad/s100Avrω = = =Speed of intermediate belt.( )( )1 1 1 50 6 300 mm/sv rω= = =Right pulley.Inner radius r3 = 50 mmOuter radius r4 = 100 mm1243003 rad/s100vrω = = =(a) Velocity of C.( )( )3 2 50 3 150 mm/sCv r ω= = =0.1500 m/sC =v(b) Acceleration of point B.( )( )22 24 2 100 3 900 mm/sBa r ω= = =20.900 m/sB =a
  • 22. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 22.Left pulley.Inner radius r1 = 50 mmOuter radius r2 = 100 mm0.6 m/s = 600 mm/sAv =( ) 1.8 m/s = 1800 mm/sA ta = −126006 rad/s100Avrω = = =( ) 212180018 rad/s100A tarα = = =Intermediate belt.( )( )1 1 1 50 6 300 mm/sv rω= = =( ) ( )( ) 21 1 1 50 18 900 mm/sta rα= = =Right pulley.Inner radius r3 = 50 mmOuter radius r4 = 100 mm1243003 rad/s100vrω = = =( )1 2249009 rad/s100tarα = = =(a) Velocity and acceleration of point C.( )( )3 2 50 3 150 mm/sCv r ω= = =0.150 m/sC =v( ) ( )( )3 2 50 9 450 mm/sC ta rα= = =0.450 m/sC =a
  • 23. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) Acceleration of point B.( ) ( )( )22 24 2 100 3 900 mm/sB na r ω= = =( ) 20.900 m/sB n=a( ) ( )( ) 24 2 100 9 900 mm/sB ta r α= = =( ) 20.900 m/sB t=a21.273 m/sB =a 45°
  • 24. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 23.(a) Let point C be the point of contact between the shaft and the ring.1C Av rω=12 2C ABv rr rωω = =12ABrrωω =21( ) : A Ab On shaft A a rω=21A Arω=a22 12 22: AB BrOn ring B a r rrωω = =   2 212ABrrω=a
  • 25. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 24.(a) Let point C be the point of contact between the shaft and the ring.( )( )1 0.5 25 12.5 in./sC Av rω= = =212.55.0 rad/s2.5CBvrω = = = 5.00 rad/sBω =( ) On shaft :b A ( )( )221 0.5 25A Aa rω= =2312.5 in./s ,= 226.0 ft/sA =aOn ring :B ( )( )222 2.5 5.0B Ba r ω= =262.5 in./s ,= 25.21ft/sB =a( ) At a point on the outside of the ring,c 3 3.5 in.r r= =( )( )22 23.5 5.0 87.5 in./sBa rω= = = 27.29 ft/sa =
  • 26. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 25.( )a( )( )600 2600 rpm 20 rad/s.60Aπω π= = =Let points A, B, and C lie at the axles of gears A, B, and C, respectively.Let D be the contact point between gears A and B.( )( )/ 2 20 40 in./sD D A Av r ω π π= = =/40 6010 rad/s 10 300 rpm4 2DBD Bvrπω π ππ= = = = ⋅ =300 rpmB =ωωωωLet E be the contact point between gears B and C.( )( )/ 2 10 20 in./sE E B Bv r ω π π= = =( )/20 603.333 rad/s 3.333 100 rpm6 2ECE Cvrπω π ππ= = = = =100 rpmC =ωωωω(b) Accelerations at point E.( )222/20On gear : 1973.9 in./s2EBE BvB arπ= = =21974 in./sB =a( )222/20On gear : 658 in./s6ECE CvC arπ= = =2658 in./sC =a
  • 27. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 26.( ) At timea 2 s,t =( )( )600 2600 rpm 20 rad/s60Aπω π= = =2, 10 rad/sAA A Attωω α α π= = =Let D be the contact point between gears A and B.( ) ( )( ) 2/ 2 10 20 in./sD D A Ata r α π π= = =( ) 2/205 rad/s4D tBD Barπα π= = = 215.71rad/sB =ααααLet E be the contact point between gears B and C.( ) ( )( ) 2/ 2 5 10 in./sE E B Bta r α π π= = =( ) 2/101.6667 rad/s6E tCE Carπα π= = = 25.24 rad/sC =αααα( ) Atb 0.5 s.t = ( )( )For gear , 5 0.5 2.5 rad/sB BB tω α π π= = =( ) ( )( )22 2/ 2 2.5 123.37 in./sE E B Bnr ω π= = =a( ) 210 in./sE tπ=a 231.416 in./s=( ) ( ) ( ) ( )2 2 2 2 2123.37 31.416 127.3 in./sE E En ta a a= + = + =31.416tan , 14.29123.37β β= = ° 2127.3 in./sE =a 14.29°
  • 28. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )( )For gear , 1.6667 0.5 0.83333 rad/sC CC tω α π π= = =( ) ( )( )22 2/ 6 0.83333 41.123 in./sE E C Cna r ω π= = =( ) 231.416 in./sE ta =( ) ( ) ( ) ( )2 2 2 2 241.123 31.416 51.75 in./sE E En ta a a= + = + =31.416tan 37.4 ,41.123β = = ° 251.8 in./sE =a 37.4°
  • 29. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 27.( ) For the pulley,a ( )1 1, 0.3 0.15 m2 2Ar d r= = =( )10.2 0.1 m2Br = =( )//,A A B B A B A B A BA BA Bv r v r v v v r rvr rω ω ωω= = = − = −=−At 0,t =00.816 rad/s0.15 0.1ω = =−At 0.25 s,t = 10.48 rad/s0.15 0.1ω = =−21 2 8 1632 rad/s0.25tω ωα− −= = = − 232 rad/s=A Ar α=a ( )( ) 20.15 32 4.80 m/s= =B Br α=a ( )( ) 20.1 32 3.2 m/s= =2/ 1.6 m/sA B A B= − =a a a 2/ 1.600 m/sA B =a( )b ( ) ( ) 2/ / /012A B A B A Bt t= +x v a( )( ) ( )( )210.8 0.25 1.6 0.25 0.15 m2= − + = −/ 150.0 mmA B =x
  • 30. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 28.For the pulley, ( )1 1, 0.3 0.15 m2 2Ar d r= = =( )10.2 0.1m2Br = =( )/,A A B B A B A Bv r v r v r rω ω ω= = = −/A BA Bvr rω =−At 0,θ =00.918 rad/s0.15 0.1ω = =−At1rev radians,2θ π= =0.459 rad/s0.15 0.1ω = =−d dd ddt dω ωα ω ω ω α θθ= = =( )( )2 29 218 09 1838.675 rad/s2 2d dπω ω α θ α π α= − = = −∫ ∫238.7 rad/sα =( ) 2 20.15 38.675 5.8012 m/s or 5.8012 m/sA Ar α= = − = −a( ) 2 20.1 38.675 3.8675 m/s or 3.8675 m/sB Br α= = − = −a( )a 2 2/ 1.9337 m/s or 1.9337 m/sA B A B= − = −a a a2/ 1.934 m/sA B =a( )b ( ) ( ) 2/ / /012A B A B A Bt t= +x v a( )( ) ( )( )210.9 0.3 1.9337 0.3 0.1830 m2= + − =/ 183.0 mmA B =x
  • 31. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 29.(a) Motion of pulley.( ) ( )0 08 in./sE A= =v v ( ) 210 in./sE At= =a aFixed axis rotation.( )( )00 008 in./s2 rad/s4 in.EE AAvv rrω ω= = = =( )( ) 2210 in./s2.5 rad/s4 in.E tE AtAaa rrα α= = = =Since α is constant,0 2 2.5t tω ω α= + = +2 20 010 2 1.252t t t tθ θ ω α= + + = + +For t = 3s, ( )( )2 2.5 3 9.5 rad/sω = + =( )( ) ( )( )20 2 3 1.25 3 17.25 radθ = + + =In revolutions,17.252θπ= 2.75 revθ =(b) Motion of load B. t = 3s( )( )6 9.5B Bv r ω= = 57.0 in./sB =v( )( )6 17.25B By r θ∆ = = 103.5 in.By∆ =(c) Acceleration of point D. t = 00 2 rad/sω ω= = 22.5 rad/s=αααα( ) ( )( ) 26 2.5 15 in./sD Dtr α= = =a( ) ( )( )22 26 2 24 in./sD Dnr ω= = =a228.3 in./sD =a 32.0°
  • 32. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 30.22.4 rad/sα = 0 0ω =Use equations for constant angular acceleration.0 2.4t tω ω α= + =2 20 011.22t t tθ θ ω α= + + =At t = 4s,( )( )2.4 4 9.6 rad/sω = =( )21.2 4 19.2 radθ = =(a) Load A. at t = 4s, 4 in.Ar =( )( )4 in. 9.6 rad/sA Av r ω= = 38.4 in./sA =v( )( )4 in. 19.2 radA Ay r θ= = 76.8 in.A =y(b) Load B. at t = 4s, 6 in.Br =( )( )6 in. 9.6 rad/sB Bv r ω= = 57.6 in./sB =v( )( )6 in. 19.2 radB By r θ= = 115.2 in.B =y
  • 33. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 31.When contact is made, 240 rpm 8 rad/sAω π= =Let C be the contact point between the two gears.( )( ) 20.15 8 1.2 m/sC A Av r ω π π= = =1.26 rad/s0.2CBBvrπω π= = =8 rad/sA tω π α= =( )6 2 rad/sB tω π α= = −Subtracting, ( )( ) 22 2 rad/sπ α α π= =( )a 23.14 rad/sα =( )b8 88 stπ πα π= = = 8.00 st =
  • 34. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 32.( )0240 rpm 8 rad/sAω π= = ( ) 118A A tω π α= −3 32 2 21 1 11 1 1 0.1542 2 2 0.2AB B A ABrt t trθ π α α α   = = = =     ( )3210.28 59.574 radians0.15A tα π = =  ( )31 1 110.150.4218750.2B B A At t tω α α α = = =  Let Cv be the velocity at the contact point.( )( )1 10.15 8 1.2 0.15C A A A Av r t tω π α π α= = − = −and ( )( )1 10.2 0.421875 0.084375C B B A Av r t tω α α= = =Equating the two expressions for Cv ,1 1 11.2 0.15 0.084375 or 16.0850 rad/sA A At t tπ α α α− = =Then,211159.5743.7037 s16.0850AAtttαα= = =( )a 216.08504.3429 rad/s3.7037Aα = = 24.34 rad/sAα =( )320.154.3429 1.83218 rad/s0.2Bα = =  21.832 rad/sBα =( ) From above,b 1 3.70 st =
  • 35. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 33.Motion of disk B. ( )0500 rpm 52.360 rad/sBω = =Assume that the angular acceleration of disk B is constant.( )0B B Btω ω α= +At 60 s,t = 0Bω =( ) 20 0 52.3600.87266 rad/s60B BBtω ωα− −= = = −( )( )3 52.360 0.87266 157.08 2.618 in./sB B Bv r t tω= = − = −Motion of disk A. ( )00,Aω = 23 rad/sAα =( )03A A At tω ω α= + =( )( )2.5 3 7.5 in./sA A Av r t tω= = =If disks are not to slip, A Bv v=7.5 157.08 2.618t t= −(a) 15.52 st =(b) ( )( )3 15.52 46.6 rad/sAω = =( )( )52.360 0.87266 15.52 38.8 rad/sBω = − =445 rpmA =ωωωω , 371 rpmB =ωωωω
  • 36. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 34.Wheel B. ( )0300 rpm 31.416 rad/sBω = =At 12s,t = 75 rpm =7.854 rad/sBω =Angular acceleration.( )0 7.854 31.4161.9635 rad/s12B BBtω ωα− −= = = −Velocity at contact point with disk A at 12 s:t =( )( )3 7.854 23.562 in./sB B Bv r ω= = =Wheel A. ( )0300 rpm = 31.416 rad/sAω =Assume that slipping ends when 12 s.t =Then, 23.562 in./sAv =23.5629.4248 rad/s2.5AAAvrω = = =9.4248 rad/sAω = 9.4248 rad/s= −( ) 20A9.4248 31.4163.4034 rad/s12A Atω ωα− − −= = = −(a)23.40 rad/sA =αααα21.963 rad/sB =αααα(b) Time when ωA is zero.( )00A A A tω ω α= + =( )0 0 31.4163.4034A AAtω ωα− −= =−9.23 st =
  • 37. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 35.Let one layer of tape be wound and let v be the tape speed.2 andv t r r bπ∆ = ∆ =2 2r bv bt rωπ π∆= =∆For the reel:1 1d d v dv dvdt dt r r dt dt rω    = = +      2 22a v dr a v br dt rr rωπ= − = −2102barωπ = − =  202bωπ=a
  • 38. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 36.Let one layer of paper be unrolled.2 andv t r r bπ∆ = ∆ = −2r bv drt r dtπ∆ −= =∆21 10d d v dv d v drvdt dt r r dt dt r dtrωα   = = = + = −      22 32 2v bv bvrr rπ π−  = − =    232bvrπ=αααα
  • 39. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 37.Velocity analysis.150 mm/sB =v 15°A A=v v/ 500B A ω=v 50°Plane motion Translation with Rotation about .B B= +/A B A B= +v v vDraw velocity vector diagram.180 50 75 55φ = ° − ° − ° = °Law of sines./sin75 sin sin50A B A Bv v vφ= =° °( )a /sin75 150 sin75189.14 mm/ssin50 sin50BA Bvv° °= = =° °/ 189.140.378 rad/s500A BABvlω = = =0.378 rad/s=ωωωω( )bsin 150 sin55160.4 mm/ssin50 sin50BAvvφ °= = =° °160.4 mm/sA =v
  • 40. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 38.Velocity analysis.225 mm/sA =vB Bv=v 15°/ /B A B Av=v 30°Plane motion Translation with Rotation about .A A= +/B A B A= +v v vDraw velocity vector diagram.180 60 75 45φ = ° − ° − ° = °Law of sines./sin75 sin60 sinB A B Av v vφ= =° °( )a /sin75 225sin75307.36 mm/ssin sin 45AB Avvφ° °= = =°/ 307.360.615 rad/s500B AABvlω = = =0.615 rad/s=ωωωω( )bsin 60 225sin60276 mm/ssin sin 45ABvvφ° °= = =°276 mm/sB =v 15°
  • 41. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 39.Geometry.12sin , 36.8720β β= = °12tan , 67.385θ θ= = °Velocity analysis.4.2 ft/sA =v/2012B A AB ABrω ω= =v βB Bv=v θPlane motion Translation with Rotation about .A A= +/B A B A= +v v vDraw velocity vector diagram.( )180 90 59.49φ θ β= ° − − ° − = °Law of sines.( )/sin sin 90 sinB A B Av v vθ β φ= =° −/sin 4.2sin67.384.5 ft/ssin sin59.49AB Avvθφ°= = =°( )a4.52.7 rad/s20/12ABω = =2.70 rad/sAB =ωωωω( )bcos 4.2cos36.873.90 ft/ssin sin59.49ABvvβφ°= = =°3.90 ft/sB =v 67.4°
  • 42. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 40.Geometry.12sin , 36.8720β β= = °12tan , 67.385θ θ= = °Velocity analysis.4.2 rad/sAB =ωωωω( )/ /204.2 7.0 ft/s12B A B A ABr ω = = =  v βB Bv=v θA Av=vPlane motion Translation with Rotation about .A A= +/B A B A= +v v vDraw velocity vector diagram.( )180 90 59.49φ θ β= ° − − ° − = °Law of sines.( )/sin sin 90 sinB AA Bvv vφ β θ= =° −( )a/ sin 7sin59.496.53 ft/ssin sin 67.38B AAvvφθ°= = =°6.53 ft/sA =v( )b/ cos 7cos36.876.07 ft/ssin sin67.38B ABvvβθ°= = =°6.07 ft/sB =v 67.4°
  • 43. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 41.In units of m/s, ( )/ / 0.6 0.6 0.6 0.6B A B Aω ω ω ω= × = × + = − +v k r k i j i j/ / 1.2 1.2C A C Aω ω ω= × = × =v k r k i j/A B A= +v v vB( ) ( )7.4 7 0.6 0.6B Ay xv v ω ω− + = − − +i j i j i jComponents. ( ): 7.4 0.6A xv ω− = −i (1)( ): 7 0.6B yv ω= − +j (2)/A C A= +v v vC( ) ( )1.4 7 1.2C A xyv v ω− + = − +i j i j jComponents. ( ): 1.4 A xv− =i (3)( ): 7 1.2C yv ω= − +j (4)From (3), ( ) 1.4 m/sA xv = −( ) From (1),a( )7.4 1.410 rad/s,0.6ω− − −= =−10.00 rad/s=ωωωωFrom (2), ( ) ( )( )7 0.6 10 1m/sB yv = − + = −( )b ( ) ( )7.40 m/s 1.000 m/sB = − −v i j
  • 44. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 42.In units of m/s, ( )/ / 0.6 0.6 0.6 0.6B A B Aω ω ω ω= × = × + = − +v k r k i j i j/ / 1.2 1.2C A C Aω ω ω= × = × =v k r k i j/B A B A= +v v v( ) ( )7.4 7 0.6 0.6B Ay xv v ω ω− + = − − +i j i j i jComponents. ( ): 7.4 0.6A xv ω− = −i (1)( ): 7 0.6B yv ω= − +j (2)/C A C A= +v v v( ) ( )1.4 7 1.2C A xyv v ω− + = − +i i j jComponents. ( ): 1.4 A xv− =i (3)( ): 7 1.2C yv ω= − +j (4)From (3), ( ) 1.4 m/s, 1.4 7A Axv = − = − −v i jFrom (1),( )( )7.4 1.410 rad/s, 10.00 rad/s0.6ω− − −= = =−kωωωω(a) ( )/ / 10 0.6O A O A A O A A= + = + × = + ×v v v v r v k iωωωω1.4 7 6 1.4 1= − − + = − −i j j i j( ) ( )1.400 m/s 1.000 m/sO = − −v i j(b) ( )0 O x y= + × +v i jωωωω( )0 1.4 1 10 1.4 1 10 10x y x y= − − + × + = − − + −i j k i j i j j jComponents. : 0 1.4 10 ,y= − −i 0.14 my = − 140.0 mmy = −: 0 1 10 ,x= − +j 0.1 mx = 100.0 mmx =
  • 45. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 43.In units of mm/s, ( )/ / 125 75 75 125B A B Aω ω ω ω= × = × + = − +v k r k i j i j( )/ / 50 150 150 50C A C Aω ω ω ω= × = × + = − +v k r k i j i j/B A B A= +v v v( ) ( )75 100 75 125B x yv v ω ω− = + − +i j i j i jAComponents. ( ): 100 75B xv ω= −i (1)( ): 75 125A yv ω− = +j (2)/C A C A= +v v v( ) ( )400 100 150 50C A yyv v ω ω+ = + − +i j i j i jComponents. : 400 100 150ω= −i (3)( ) ( ): 125C A yyv v ω= +j (4)(a) From (3), 2 rad/sω = − ( )2 rad/s= − kωωωω(b) From (2), ( ) ( )75 125 75 125 2 175 mm/sA yv ω= − − = − − − =( ) ( )100.0 mm/s 175.0 mm/sA = +v i j
  • 46. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 44.In units of mm/s, ( )/ / 125 75 75 125B A B Aω ω ω ω= × = × + = − +v k r k i j i j( )/ / 50 150 150 50C A C Aω ω ω ω= × = × + = − +v k r k i j i j/B A B A= +v v v( ) ( )75 100 75 125B x yv v ω ω− = + − +i j i j i jAComponents. ( ): 100 75B xv ω= −i (1)( ): 75 125A yv ω− = +j (2)/C A C A= +v v v( ) ( )400 100 150 50C A yyv v ω ω+ = + − +i j i j i jComponents. : 400 100 150ω= −i (3)( ) ( ): 125C A yyv v ω= +j (4)From (3), ( )2 rad/sω = − kFrom (2), ( ) ( )75 125 75 125 2 175 mm/sA yv ω= − − = − − − =( ) ( )100 mm/s 175 mm/sA = +v i jFind the point with zero velocity. Call it D. 0D =v( ) ( )/ or 0 100 175 2A D A x y= + = + + × +v v v i j k i jD0 100 175 2 2 0x y= + + − =i j j iComponents. : 0 100 2 , 50 mmy y= − =i: 0 175 2 , 87 mmx x= + = −jRadius of locus.200100 mm2vrω= = =Circle of 100.0 mm radius centered at 87.5 mm, 50.0 mmx y= − =
  • 47. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 45.7Slope angle of rod. tan 0.7, 3510θ θ= = = °1012.2066 in. 20 7.7934 in.cosAC CB ACθ= = = − =Velocity analysis.25 in./sA =v , C Cv=v θ/C A ABACω=v θ/C A C A= +v v vDraw corresponding vector diagram./ sin 25sin35 14.34 in./sC A Av v θ= = ° =( )a/ 14.341.175 rad/s12.2066C AABvACω = = =1.175 rad/sAB =ωωωωcos 25cos 20.479 in./sC Av v θ θ= = =( )( )/ 7.7934 1.175B C ABv CBω= =9.1551in./s=/ /has same direction as .B C C Av v/B C B C= +v v vDraw corresponding vector diagram./ 9.1551tan , 24.0920.479B CCvvφ φ= = = °( )b20.47922.4 in./s 1.869 ft/scos cos24.09CBvvφ= = = =°59.1φ θ+ = °1.869 ft/sB =v 59.1°
  • 48. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 46.Instantaneous geometry. Law of sines:sin sin12010 15φ °=10sin sin120 0.5773515φ = ° =35.264φ = °Velocity analysis.1.2 ft/sA =v 14.4 in./s=/ 10B A ABω=v 60°B Bv=v φ/B B A Av v= +vUse the triangle construction to perform the vector addition.60 24.736β φ= ° − = °90 125.264γ φ= ° + = °Law of sines./sin sin30 sinB A B Av v vγ β= =°/sin 14.4 sin125.26428.10 in./ssin sin 24.736AB Avvγβ°= = =°(a)28.1010ABω = 2.81 rad/sAB =ωωωω(b)sin30 14.4 sin3017.21 in./ssin sin 24.736ABvvβ° °= = =1.434 ft/sB =v 35.3°
  • 49. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 47.Label the contact point between gears A and B as 1, the center of gear Bas 2, and the contact point between gears B and C as 3.Gear A: 1 80 Av ω= (1)Arm AB: 2 120 ABv ω= (2)Gear B: 1 2 40 Bv v ω= − (3)3 2 80 Bv v ω= + (4)Gear C: 3 200 Cv ω= (5)Data: 0, 5 rad/sA Cω ω= =From (1), 1 0,v =From (5), ( )( )3 200 5 1000 mm/sv = =From (3), 2 40 0Bv ω− = (6)From (4), 2 80 1000Bv ω+ = (7)Solving (6) and (7) simultaneously,(a)10008.333 rad/s120Bω = = 8.33 rad/sB =ωωωω2 (40)(8.333) 333.33 mm/sv = =(b) From (2),333.332.78 rad/s120ABω = = 2.78 rad/sAB =ωωωω
  • 50. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 48.Label the contact point between gears A and B as 1, the center ofgear B as 2, and the contact point between gears B and C as 3.Gear A: 1 80 Av ω= (1)Arm AB: 2 120 ABv ω= (2)Gear B: 1 2 40 Bv v ω= − (3)3 2 80 Bv v ω= + (4)Gear C: 3 200 Cv ω= (5)Data: 20 rad/s, 0B Cω ω= =From (5), 3 0.v =From (4), ( )( )2 80 80 20Bv ω= − = − 1600 mm/s=From (3), ( )( )1 1600 40 20 2400v = − − = − 2400 mm/s=From (1), 1 /80 30 rad/sA vω = = −(a) 30.0 rad/sA =ωωωωFrom (2), 1600 120 ABω=160013.33 rad/s120ABω = − = −(b) 13.33 rad/sAB =ωωωω
  • 51. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 49.Data: 3600 rpm 376.99 rad/s, 0A Bω ω= = =11.25 in.2A Ar d= =diameter of ball 0.5 in.d = =Velocity of point on inner race in contact with a ball.(1.25)(376.99) 471.24 in./sA A Av r ω= = =Consider a ball with its center at point C./A B A Bv v v= +0A Cv dω= +471.240.5ACvdω = =942.48 rad/s=/C B C Bv v v= +10 (0.25)(942.48) 235.62 in./s2dω= + = =(a) 236 in./sCv =(b) Angular velocity of ball.942.48 rad/sCω = 9000 rpmCω =(c) Distance traveled by center of ball in 1 minute.(235.62)(60) 14137.2 in.C Cl v t= = =Circumference of circle: 2 2 (1.25 0.25)rπ π= +9.4248 in.=Number of circles completed in 1 minute:14137.22 9.4248lnrπ= = 1500n =
  • 52. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 50.Contact point 1 between gears A and B.Contact point 2 between gears B and C.Gear B: 6 rad/sBω =1 (6 rad/s)(10 in.) 60 in./sv = = (1)2 (6 rad/s)(5 in.) 30 in./sv = = (2)Arm ABC: ABC ABCω=ωωωω15A ABCω=v 15C ABCω=vGear A: 3 rad/sAω =1 (5)(3)Av v= + 15 ABCω= 15+ (3)Matching expressions (1) and (3) for 1,v(a) : 60 15 15ABCω= + 3.00 rad/sABC =ωωωωGear C: C Cω ω=2 10C Cv v ω= + 15= 10 Cω+ (4)Matching expressions (2) and (4) for 2,v(b) : 30 15 10 Cω= + 1.500 rad/sC =ωωωω
  • 53. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 51.Let a be the radius of the central gear A, and let b be the radius of theplanetary gears B, C, and D. The radius of the outer gear E is 2 .a b+Label the contact point between gears A and B as 1, the center of gear Bas 2, and the contact point between gears B and E as 3.1Gear : AA v aω= (1)( )2Spider: Sv a b ω= + (2)2 1Gear : BB v v bω= + (3)3 2 Bv v bω= + (4)( )3Gear : 2 EE v a b ω= + (5)( )2From (4) and (5), 2B Ev b a bω ω+ = + (6)2 1From (1) and (3), B Av b v aω ω− = = (7)( )2 22Solving for and ,2E ABa b av vω ωω + + =( )22E ABa b abω ωω + − =From (2),( )( )222E AS Sa b ava b a bω ωω ω + + = =+ +Data: 60 mm, 60 mm, 2 180 mm, 120 mma b a b a b= = + = + =( )a( )( )180 601.5 0.52 60E AB E Aω ωω ω ω−= = −( )( ) ( )( )1.5 120 0.5 150 105 rpm= − =105.0 rpmB =ωωωω( )b( )( )180 600.75 0.252 120E AS E Aω ωω ω ω+= = +( )( ) ( )( )0.75 120 0.25 150 127.5 rpm= + =127.5 rpmS =ωωωω
  • 54. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 52.Let a be the radius of the central gear A, and let b be the radius of theplanetary gears B, C, and D. The radius of the outer gear E is 2 .a b+Label the contact point between gears A and B as 1, the center of gear Bas 2, and the contact point between gears B and E as 3.1Gear : AA v aω= (1)( )2Spider: Sv a b ω= + (2)2 1Gear : BB v v bω= + (3)3 2 Bv v bω= + (4)( )3Gear : 2 EE v a b ω= + (5)( )2From (4) and (5), 2B Ev b a bω ω+ = + (6)2 1From (1) and (3), B Av b v aω ω− = = (7)2Solving for and ,Bv ω( )222E Aa b avω ω + + =( )22E ABa b abω ωω + − =From (2),( )( )222E AS Sa b ava b a bω ωω ω + + = =+ +1Data: 0,5E S Aω ω ω= =
  • 55. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )a( )( )( )2 015 2AS Aa b aa bωω ω + + = =+( ) ( )1 1 1,2 5 52 1 baaa b= =+ +2 1 5ba + =  1.500ba=( )b( )( )( )202 2 2 1.5 3E A A A ABa b a ab bω ω ω ω ωω + − = = − = − = −13B Aω=ωωωω
  • 56. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 53.Label the contact point between gears A and B as 1 and that between gears B and C as 2.Rod ABC: 75 rpm 2.5 rad/sABCω π= =0Av =(12)(2.5 ) 30 rad/sBv π π= =(12 7)(2.5 ) 47.5 rad/sCv π π= + =Gear A: 10, 0, 0A Av vω = = =Gear B: 1 4 0B Bv v ω= − =30 4 0Bπ ω− =7.5 rad/sBω π=2 4B Bv v ω= +30 30 60 in./sπ π π= + =Gear C: 2 3C Cv v ω= −60 47.5 3 Cπ π ω= −4.1667 rad/sCω π= −Summary: 225 rpmB =ωωωω125.0 rpmC =ωωωω
  • 57. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 54.Label the contact point between gears A and B as 1 and that between gears B and C as 2.Rod ABC: 0, 80 rpm 8 /3 rad/sC ABCv ω π= = =7 (7)(8 /3) 56 /3 in./sB ABCv ω π π= = =(7 12) (19)(8 /3) 152 /3 in./sA ABCv ω π π= + = =Gear C: 20, 0, 0C Cv vω = = =Gear B: 2 4 56 /3 4 0B B Bv v ω π ω= − = − =14 /3 rad/sBω π=1 4 56 /3 56 /3B Bv v ω π π= + = +112 /3 in./sπ=Gear A: 1 8A Av v ω= −112 /3 152 /3 8 Aπ π ω= −5 /3 rad/sAω π= −Summary: 50.0 rpmA =ωωωω140.0 rpmB =ωωωω
  • 58. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 55.Geometry.( ) ( )sin sinOA ABθ β=( )sin 10 sin30sin , 1.79160OAABθβ β°= = = °Shaft and eccentric disk. (Rotation about O), 900 rpm 30 rad/sOAω = = π( ) ( )( )10 30 300 mm/sA OAOA ω= = π = πvRod AB. (Plane motion = Translation with A + Rotation about A.)[/B A B A Bv= +v v v ] [ Av= ] /60 A Bv° +  ]βDraw velocity vector diagram. 90 88.21β° − = °180 60 88.21 31.79φ = ° − ° − ° = °Law of sines.( )sin sin 90B Av vφ β=° −( )( )300 sin 31.79sinsin 90 sin 88.21497 mm/sABvvφβπ °= =° − °=497 mm/sB =v
  • 59. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 56.Geometry.( ) ( ) ( )sin 180 sinOA ABθ β° − =( ) ( )sin 180 10 sin60sin , 3.10160OAABθβ β° − °= = = °Shaft and eccentric disk. (Rotation about O) 900 rpm 30 rad/sOAω = = π( ) ( )( )10 30 300 mm/sA OAOA ω= = π = πvRod AB. (Plane motion = Translation with A + Rotation about A.)[/B A B A Bv= +v v v ] [300= π ] /30 B Av° +  ]βDraw velocity vector diagram. 90 93.10β° + = °180 30 93.10 56.90φ = ° − ° − ° = °Law of sines.( )sin sin 90B Av vφ β=° +( )( )300 sin 56.90sinsin 90 sin 93.10791 mm/sABvvφβπ °= =° + °=791 mm/sB =v
  • 60. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 57.Disk ABar :BD15 rad/sAω = , 2.8 in.AB =Rotation about a fixed axis.( ) ( )( )2.8 15 42 in./sB Av AB ω= = =( )a 0 .θ = ° 42 in./sB =v2.8sin , 16.26010β β= = °/D B D B= +v v vDv [42= ] /D Bv+  β /4243.75 in./scosD Bvβ= =/ 43.75,10D BDBvDBω = = 4.38 rad/sDB =ωωωωtan ,D Bv v β= 12.25 in./sD =v( ) 90 .b θ = ° 42 in./sB =v5.6sin , 34.0610β β= = °Bar :BD /D B D B= +v v vDv [42= ] /D Bv+  β Components: : / 0D Bv =: D Bv v=0DB =ωωωω42.0 in./sD =v
  • 61. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )c 180 .θ = ° 42 in./sBv =2.8sin , 16.2610β β= = °Bar :BD /D B D B= +v v vDv [42= ] /D Bv+  β /4243.75cosD Bvβ= =/ 43.7510D BDBvDBω = =4.38 rad/sDB =ωωωωtanD Bv v β= 12.25 in./sD =v
  • 62. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 58.2.8 in., 10 in.A BDr l= =From geometry, sin sinBD A Al r rβ θ= + (1)is tangent to the circular path of ,B Bvthus B A Ar ω=v θFor rod BD /B D BD BDl ω=v β/ /0B D B D B D= + = +v v v v/B D B=v v( ) For matching directiona or 180θ β θ β= = ° +For ,β θ= sin sin so that sin sinBD A Al r rβ θ β β= = +2.8sin , 22.9 ,10 2.8ABD Arl rβ β= = = °− −22.9θ = °For 180 , sin sin ,θ β θ β= ° + = −sin sinBD A Al r rβ β= −2.8sin , 12.610 2.8ABD Arl rβ β= = = °+ +192.6θ = °( ) For matching magnitudesb /D B Bv v=( )( )2.8 20, 5.6 rad/s10A ABD BD A A BDBDrl rlωω ω ω= = = =For 22.9 ,θ = ° 5.60 rad/sBD =ωωωωFor 192.6 ,θ = ° 5.60 rad/sBD =ωωωω
  • 63. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 59.Rod BE: 0Ev =/ (192)(4) 768 mm/sB B E BEr ω= = =vRod ABD: D Dv=v/D B D B= +v v vDv 768= 360 ADω+ 60°Draw diagram for vector addition.(a) 768 360 sin30ADω= °4.2667 rad/sADω =4.27 rad/sAD =ωωωω(b) 768 tan30Dv= °1330 mm/sDv =1.330 m/sD =v(c) / 240A B ADω=v 60 1024 mm/s° = 60°/ 768A B A B= + =v v v 1024+ 60 1557 mm/s° = 34.7°1.557 m/sA =v 34.7°
  • 64. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 60.Rod BE: 0E =v B Bv=vRod ABD: 1.6 m/sD =v AD ADω=ωωωω/B D B D= +v v vBv 1.6= 0.360 ADω+ 60°Draw diagram for vector addition.(a)1.60.360sin60ADω =°5.13 rad/sAD =ωωωω(b) 1.6tan30 0.92376Bv = ° =0.924 m/sB =v(c) / 0.240A B ADω=v 60 1.2317 m/s° = 60°/ 0.92376A B B A= + =v v v 1.2317+ 60°[1.5396= ] [1.0667+ ] 1.873= 34.7°1.873 m/sA =v 34.7°
  • 65. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 61.1000 rpmABω =( )( )1000 2104.72 rad/s60π= =( )a 0 . . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r( )( )/ 3 104.72 314.16 in./sB B A ABv ω= = =v.Rod BD (Plane motion Translation with Rotation about )B B= +/D B D Bv= +v vDv [314.16= ] /D Bv+  ]/0, 314.16 in./sD D Bv v= =P D=v v 0P =v314.168BBDvlω = = 39.3 rad/sBD =ωωωω( )b 90 . . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r( )( )/ 3 104.72 314.16 in./sB B A ABr ω= = =v.Rod BD (Plane motion Translation with Rotation about .)B B= +/D B D B= +v v vDv 314.16= /D Bv+  ]β/ 0, 314.16 in./sD B Dv v= =/,D BBDvlω = 0BD =ωωωω314.16 in./sP D= =v v 26.2 ft/sP =v
  • 66. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 62.( )( )1000 21000 rpm 104.72 rad/s60ABωπ= = =60 , . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r 30°( )( )/ 3 104.72 314.16 in./sB B A ABω= = =v r 60°Rod BD. (Plane motion Translation with Rotation about .)B B= +Geometry. sin sinl rβ θ=3sin sin sin60818.95rlβ θβ= = °= °/D B D B= +v v v[ Dv ] [314.16= ] /60 D Bv° +  ]β Draw velocity vector diagram.( )180 30 90 78.95φ β= ° − ° − ° − = °Law of sines.( )/sin sin30 sin 90D BD Bvv vφ β= =° ° −sin 314.16 sin78.95326 in./scos cos18.95BDvvφβ°= = =°P Dv v= 27.2 ft/sP =v/sin30 314.16sin30166.08 in./scos cos18.95BD Bvvβ° °= = =°/ 166.088D BBDvlω = = 20.8 rad/sBD =ωωωω
  • 67. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 63.Bar AB. (Rotation about A)( ) ( ) ( )/ 4 0.25 1.00 m/sB AB B A= × = − × − = −v r k j iωωωωBar ED. (Rotation about E)( )/ 0.075 0.15 0.15 0.075D DE D E DE DE DEω ω ω ω= × = × − − = −v k r k i j i jBar BD. (Translation with B + Rotation about B.)/ / 0.2 0.2D B BD D B BD BDω ω ω= × = × =v k r k i j/D B D B= +v v v0.15 0.075 1.00 0.2DE DE BDω ω ω− = − +i j i jComponents:: 0.15 1.00, 6.6667 rad/sDE DEω ω= − = −i 6.67 rad/sDE =ωωωω: 0.075 0.2DE BDω ω− =j( )( )0.075 6.66670.2BDω− −= 2.50 rad/sBD =ωωωω
  • 68. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 64.Bar AB. (Rotation about A.) ( ) 0.18B AB ABAB ω ω= =v 30°Bar DE. (Rotation about E.) ( ) 0.18D DE DEDE ω ω= =vBar BGD. (Plane motion = Translation with B + Rotation about .B )[/D B D B Dv= +v v v ] [ Bv= ]30° /D Bv+  ]30°Draw the velocity vector diagram.Equilateral triangle./ 0.18D B B ABv v ω= =/ 0.180.18D B ABBD ABBDvlωω ω= = =/ /10.092G B GB BD D B ABv l vω ω= = =/G B G B= +v v vDraw vector diagram.
  • 69. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Law of cosines( ) ( ) ( )( )2 220.18 0.09 2 0.18 0.09 cos60G AB AB AB ABv ω ω ω ω= + − °2 20.0243G ABv ω=( )( )6.415 6.416 2.5AB Gvω = =16.04 rad/sAB =ωωωω16.04 rad/sBD =ωωωω0.18D B ABv v ω= =0.180.18D ABDE ABDEvlωω ω= = =16.04 rad/sDE =ωωωω
  • 70. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 65.Bar AB. (Rotation about A.) 25 rad/sAB =ωωωω( )( )/ 8 25 200 in./sB B A ABr ω= = =vBar ED. (Rotation about E.)D Dv=v 8D DEv ω=Plate BDHF. (Translation with B + Rotation about .B )[/D B D B Dv= +v v v ] [ Bv= ] /D Bv+  ]30°Draw velocity vector diagram./200230.94 in./scos30 cos30BD Bvv = = =° °/ 230.9414.4338 rad/s16B DBDHFBDvlω = = =(a) 14.43 rad/sBDHF =ωωωω( )( )/ 8 14.4338 115.47 in./sF B BDHFv BFω= = =/ 115.47 m/sF B =v 30°[/ 200 in./sF B F B= + =v v v ] [115.47 in./s+ ]30°(b) [142.265 in./sF =v ] [100 in./s+ ] 173.9 in./sF =v 54.9°
  • 71. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 66.Rod DE. (Rotation about E.) 35 rad/sDEω =( )( )/ 8 35 280 in./sD D E DEr ω= = =vRod AB. (Rotation about A.)B Bv=v 8B ABv ω=Plate BDHF. (Translation with D + Rotation about .D )[/B D B D Bv= +v v v ] [ Dv= ] /B Dv+  ]30°Draw velocity vector diagram.//280560 in./ssin30 sin3056035 rad/s16DD BD BBDHFDBvvvlω= = =° °= = =(a) 35.0 rad/sBDHF =ωωωωPoint of zero velocity lies above point D./2808 in.35DC DBDHFvyω= = =(b) 8 in. above point D.
  • 72. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 67.(15 rad/s)AB = − kωωωω BD BDω= kωωωω DE DEω=ω k( )/ 0.2mB A = −r j ( ) ( )/ 0.6 m 0.25 mD B = − −r i j ( )/ 0.2 mD E = −r i( ) ( ) ( )/ 0 15 0.25 3 m/sB A AB B A= + × = + − × − = −v v ω r k j i( ) ( )/ / 6 0.2 0.25 0.6D B BD D B BD BD BDω ω ω= × = × − − = −v r k i j i jωωωω/ 3 0.25 0.6D B D B BD BDω ω= + = − + −v v v i i j (1)( ) ( )/ /0 0.2 0.2D E D E DE D E DE DEω ω= + = + × = × − = −v v v ω r k i j (2)Equate the expressions (1) and (2) for vD and resolve into components.: 3 0.25 0BDω− + =i 12 rad/sBDω =: 0.6 0.2BD DEω ω− = −j 36 rad/sDEω =( )a Angular velocity of rod BD. 12 rad/sBD =ωωωω(b) Velocity of the midpoint M of rod BD.( ) ( )/ /10.3 m 0.125 m2M B D B= = − −r r i j( )/ / 3 12 0.3 0.125M B M B B BD M B= + = + × = − + × − −v v v v ω r i k i j( ) ( )1.5 m/s 3.6 m/s= − −i j3.90 m/sM =v 67.4°
  • 73. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 68.Bar AB. ( ) ( )/ 0.300 m + 0.125 m ,B A =r i j ( )3 rad/sAB = kωωωω0A =v/ /0B A B A AB B A= + = + ×v v v rωωωω( )3 0.3 + 0.125 0.375 + 0.9= × = −k i j i jBar BD. ( )/ 0.325 mD B = −r j BD BDω= kωωωω( )0.375 + 0.9 + 0.325D B D/B BDω= + = − × −v v v i j k j0.375 + 0.9 + 0.325 BDω= − i j iBar DE. ( ) ( )/ 0.150 m 0.200 m ,E D = − +r i j DE DEω= kωωωω/ /E D E D D DE E D= + = + ×v v v v rωωωω( )0.150 0.200 0D DEω= + × − + =v k i j0.375 0.9 0.325 0.15 0.2 0BD DE DEω ω ω− + + − − =i j i j iComponents:j: 0.9 0.15 0DEω− = 6 rad/sDEω =i: 0.375 0.325 0.2 0BD DEω ω− + − =( )( )0.325 0.375 0.2 6BDω = +4.85 rad/sBDω = 4.85 rad/sBD =ωωωω6.00 rad/sDE =ωωωω
  • 74. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 69.80 km/h 22.222 m/sA = =v 0C =v560 mm 280 mm 0.28 m2dd r= = = =22.22279.364 rad/s0.28Avrω = = =/ / /B A D A E Av v v rω= = =( )( )0.28 79.364 22.222 m/s= =[/ 22.222 m/sB A B A= + =v v v ] [22.222 m/s+ ]44.4 m/sB =v[/ 22.222 m/sD A D A= + =v v v ] [22.222 m/s+ ]30°42.9 m/sD =v 15.0°[/ 22.222 m/sE A E A= + =v v v ] [22.222 m/s+ ]31.4 m/sE =v 45.0°
  • 75. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 70.(a) 0β = .Wheel AD 0, 45 in./sC D= =v v4511.25 rad/s4DADvCDω = = =( ) ( ) 4 2.5 1.5 in.CA CD DA= − = − =( ) ( )( )1.5 11.25 16.875 in./sA ADv CA ω= = =Rod AB. /B A B A= +v v v[ Bv ] [16.875= ] /B Av+  ]ϕ 16.88 in./sB =v/ 0B Av = 0ABω =(b) 90β = ° .Wheel AD 0, 11.25 rad/sC ADω= =v2.5tan , 32.0054DADCγ γ= = = °4.7170 in.cosDCCAγ= =( ) ( )( )4.7170 11.25 53.066 in./sA ADv CA ω= = =[53.066 in./sA =v ]32.005°Rod AB. B Bv=v4sin , 18.66312.5φ φ= = °Plane motion = Translation with A + Rotation about A.
  • 76. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.[/B A B A Bv= +v v v ] [ Av= ] [r + vB/A ]φDraw velocity vector diagram.( )180 90δ γ φ= ° − − ° +90 32.005 18.663 39.332= ° − ° − ° = °Law of sines.( )/sin sin sin 90B AB Avv vδ γ φ= =° +( )( )53.066 sin 39.332sinsin 90 sin108.663ABvvδφ°= =° + °35.5 in./s= 35.5 in./sB =v( )/sin (53.066)sin32.005sin 90 sin108.663AB Avvγφ°= =° + °29.686 in./s=/ 29.6862.37 rad/s12.5B AABvABω = = = 2.37 rad/sAB =ωωωω
  • 77. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 71.0 180 km/h = 50 m/s=v( )( )180 2180 rpm = 18.85 rad/s60πω = =Top View0v zω=0 502.65 m18.85vzω= = =Instantaneous axis is parallel to the y axis and passes through the point0x =2.65 mz =
  • 78. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 72.1.5 1.0 1rad/s1.5 3E DEDv vlω− −= = =131.03 mDCEvlω= = =(a) 1.5 2 3 0.5 mACl = + − = lies 0.500 m to the right ofC A(b) ( )10.5 0.1667 m/s3A ACv l ω = = =  A 0.1667 m/s=v
  • 79. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 73.Contact points:1 between gears A and B.2 between gears B and C.Arm ABC: 4 rad/sABCω =( )( )15 4 60 in./sAv = =( )( )15 4 60 in./sCv = =Gear B: 8 rad/sBω =( )( )1 10 8 80 in./sv = =( )( )2 5 8 40 in./sv = =Gear A:1 80 605 5AAv vω− −= =4 rad/sAω =6015 in.4AAAvω= = =l
  • 80. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Gear C:2 60 4010 10CCv vω− −= =2 rad/sCω =6030 in.2CCCvω= = =l(a) Instantaneous centers.Gear A: 15 in. left of AGear C: 30 in. left of C(b) Angular velocities.4.00 rad/sA =ωωωω2.00 rad/sC =ωωωω
  • 81. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 74.Since the drum rolls without sliding, the point of contact C with the fixed surface is the instantaneous center.Let point A be the center of the cylinder and point B the point where the cord breaks contact with thecylinder.0 6 in./sC B Dv v v= = =(a) Angular velocity of cylinder/B B Cv r ω=/66 rad/s1BB Cvrω = = =6.00 rad/s=ωωωω(b) Velocity of point A. /A A Cv r ω=( )( )5 6 30= =30.0 in./sA =v(c) Rate of winding of cord. Since ,A Bv v> the cord is wound up at rate of30 6 24 in./s.A Bv v− = − =Winding rate 24.0 in./s.=
  • 82. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 75.12 1.535 rad/s0.3O AAOv vlω− −= = =(a) 31.542.86 10 m35ACAvlω−= = = ×42.86 mm=lies 42.9 mm below .C A(b) 0.6 0.04286 0.64286 mCBl = + =( )( )0.64286 35 22.5 m/sB CBv l ω= = =22.5 m/sB =v(c) 0.3 m, 0.3 0.04286 0.34286 mOD COl l= = + =( ) ( )2 20.3 0.34286 0.45558 mCDl = + =( )( )0.45558 35 15.95 m/sD CDv l ω= = =0.3tan , 41.20.34286ODCOllθ θ= = = °15.95 m/sD =v 41.2°
  • 83. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 76.12 1.545 rad/s0.3O AAOv vlω− += = =(a) 31.533.33 10 m45AACvlω−= = = ×33.33 mm=lies 33.3 mm above .C A(b) ( )0.3 0.3 0.0333 0.56667 mCBl = + − =( )( )0.56667 45 25.5 m/sB CBv l ω= = =25.5 m/sB =v(c) 0.3 m, 0.3 0.03333 0.26667 mOE OCl l= = − =( ) ( )2 20.3 0.26667 0.4014 mCEl = + =( )( )0.4014 45 18.06 m/sE CEv l ω= = =0.3tan , 48.40.26667OEOCllθ θ= = = °18.06 m/sE =v 48.4°
  • 84. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 77.(b) Velocity of point B.(a) Location of instantaneous axis.8 in./sE D= =v v3 in./sA =v/E A E A= + ×v v rωωωω/E A E Av v r ω= +8 3 3ω= +1.6667 rad/sω = −/C A C A= + ×v v rωωωω0 3 1.6667y= −1.800 in. above point .y A=/B A B A= + ×v v rωωωω( )( )3 3 3 3 1.6667 2Bv ω= − = − = −2.00 in./sB =v(c) Since ,D E Av v v= > the paper unwinds.Rate of unwinding: 8 3 5 in./sE Av v− = − =
  • 85. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 78.10 in./sD =v , 8 in./sB =v10 84 rad/s4.5D Bv vBDω+ += = =102.5 in.4DvCDω= = =3.0 2.5 0.5 in.CA = − =(a) C lies 0.500 in. to the right of A.(b) ( )( )0.5 0.5 4 2 in./sAv ω= = =2.00 in./sA =v(c) 12 in./sD A− =v vCord DE is unwrapped at 12.00 in./s.6 in./sB A− =v vCord BF is unwrapped at 6.00 in./s.
  • 86. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 79.Rod AD.( )( )/ 0.192 4 0.768 m/sB B E BEr ω= = =v(a) Instantaneous center C is located by noting that CD is perpendicularto vD and CB is perpendicular to vB/ 0.360 sin30 0.180 mB Cr = ° =/0.7684.26670.180BADB Cvrω = = =4.27 rad/sAD =ωωωω "(b) Velocity of D. / 0.360 cos30 0.31177 mD Cr = ° =( )( )/ 0.31177 4.2667D D Cv r ω= =1.330 m/sD =v "(c) Velocity of A.0.240cos30 0.20785 mAEl = ° =0.600sin30 0.300 mCEl = ° =0.20785tan0.300β = 34.7β = °( ) ( )2 20.20785 0.300 0.36497 mCAl = + =( )( )0.36497 4.2667 1.557 m/sA CA ADv l ω= = =1.557 m/sA =v 34.7° "
  • 87. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 80.15 rad/sABω =( ) ( )( )0.200 15 3 m/sB ABv AB ω= = =B Bv=v D Dv=vLocate the instantaneous center (point C) of bar BD by noting thatvelocity directions at points B and D are known. Draw BC perendicular toBv and DC perpendicular to .Dv( )a312 rad/s0.25BBDvBCω = = = 12.00 rad/sBD =ωωωω(b) Locate point M, the midpoint of rod BD. Draw CM.( ) ( )2 20.6 0.25 0.65 mBD = + =0.25tan 22.62 90 67.380.6β β β= = ° ° − = °( )10.325 m2CM DM MB BD= = = =( ) ( )( )0.325 12 3.9 m/sMv CM ω= = = 3.90 m/sM =v 67.4°
  • 88. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 81.Bar DC. (rotation about D)( )( )( ) 18 10C CDv CDω= =180 in./s=180 in./sC =v 30°Bar AB. (rotation about A)B Bv=v 30°Locate the instantaneous center (point I) of bar BC by noting that velocity directions at two points are known.Extend lines AB and CD to intersect at I. For the given configuration, point I coincides with D.10 in., 10 3 in.IC IB= =18018 rad/s10CBCvICω = = =( ) ( )( )10 3 18 311.77 in./sB BCv IB ω= = =(a)311.7731.177 rad/s10BABvABω = = = 31.2 rad/sAB =ωωωω(b) 18.00 rad/sBC =ωωωω(c) Locate point M, the midpoint of bar BC.Triangle ICM is an equilateral triangle. 10 in.IM =( ) ( )( )10 18 180 in./sM BCv IM ω= = = 15.00 ft/sM =v 30°
  • 89. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 82.Bar AB. (rotation about A)B Bv=v 60°Bar CD. (rotation about D)C Cv=vBar BC. Locate its instantaneous center (point I) by noting that velocity directions at two points are known.Extend lines AB and CD to intersect at I. For the given configuration, point I coincides with D.Locate point M, the midpoint of bar BC. From geometry, triangle ICM is an equilateral triangle.10 in., 10 3 in.IM AB CD IB= = = =( )( )7.8 129.36 rad/s10MBCvIMω = = =(a) ( ) ( )( )10 3 9.36 162.12 in./sB BCv IB ω= = =162.1216.21 rad/s10BABvABω = = = 16.21 rad/sAB =ωωωω(b) 9.36 rad/sBC =ωωωω(c) ( ) ( )( )10 9.36 93.6 in./sC BCv IC ω= = =93.69.36 rad/s10CCDvDCω = = = 9.36 rad/sCD =ωωωω
  • 90. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 83.800 mm/sB =v A Av=vLocate the instantaneous center of rod ABD by noting that velocity directions at points A and B are known.Draw AC perpendicular to Av and BC perpendicular to Bv .(a)8003.0792 rad/s300cos30BABDvBCω = = =°3.08 rad/sABD =ωωωω( ) ( )2 2600cos30 300sin30 540.83 mmCDl = ° + ° =300sin30tan 16.10 90 73.9600cos30γ γ γ°= = ° ° − = °°(b) ( )( ) 3540.83 3.0792 1.665 10 mm/sD CD ABDv l ω= = = ×1.665 m/sD =v 73.9°
  • 91. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 84.40 ,θ = ° 0.6 m/sB =v , A Av=vLocate the instantaneous center (point C) by noting that velocity directionsat points A and B are known. Draw AC perpendicular to Av and BCperpendicular to .Bv( )sin 2sin 40 1.28557 mBC AB θ= = ° =(a)0.60.46672 rad/s1.28557BABDvBCω = = =0.467 rad/sABD =ωωωω/ / /D C B C D B= +r r r[1.28557 m= ] [2 m+ ]40°2.9930 m= 30.79°30.79β = °(b) ( )( )/ 2.9930 0.46672D D C ABDv r ω= =1.397 m/s=1.397D =v β90 59.2β° − = °1.397 m/sD =v 59.2°
  • 92. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 85.2 2320 240 400 mmDE = + =240tan 0.75320β = = 36.87β = °( )( ) ( )( )400 15 6000 mm/s 6 m/sD DEv DE ω= = = =6 m/sD =v β B Bv=vLocate point C, the instantaneous of bar DBF, by drawing BC perpendicular to vB and DC perpendicularto vD.From the figure:540720 mmtanACβ= =( )720 320 100 300 mmBC AC AB= − = − + =Since triangles FCB and BDK are similar,300100b DKBC BK= =( )( )300 300900 mm.100b = =( )a Distance b. 0.900 mb =2 2300 400 500 mm 0.5 mCD = + = =61.2 rad/s5DBDFvCDω = = =( )( )0.900 1.2 10.8 m/sF BDFv bω= = =( )b Velocity of point F. 10.80 m/sF =v
  • 93. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 86.Locate the instantaneous center I of rotation of bar ABD as theintersection of line AI perpendicular to Av and line BI perpendicular toBv Triangle IAB is equilateral.300 mmIA IBl l= =(a)900 mm/s3 rad/s300 mmAABDIAvlω = = =3.00 rad/sABD =ωωωω(b) By the law of cosines, ( )600cos30 mmIDl = °( )( )600cos30 3 1559 mm/sD ID ABDv l ω= = ° =1.559 m/sD =v
  • 94. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 87.A Av=v 45 ,° 7.5 ft/sB =vLocate the instantaneous center (point C) of rod AB by noting that velocitydirections at points A and B are known. Draw AC perpendicular to Av andBC perpendicular to .BvLet 24 in. 2 ftl AB= = =Law of sines for triangle ABC.2.8284 ftsin75 sin60 sin 45b a l= = =° ° °2.4495 ft, 2.73205 fta b= =7.52.7452 rad/s2.73205Bvbω = = =(a) ( )( )2.4495 2.7452 6.724 ft/sAv aω= = =6.72 ft/sA =v 45.0°(b) 2.75 rad/s=ωωωω(c) Let M be the midpoint of AB.Law of cosines for triangle CMB.( ) ( ) ( )( )( )22 22 22 cos602 22.73205 1 2 2.73205 1 cos602.3942 ftl lm b bm = + − °  = + − °=Law of sines.2sin sin60 1sin60, sin , 21.22.3942lmββ β° °= = = °( )( )2.3942 2.7452 6.573 ft/s,Mv mω= = =6.57 ft/sM =v 21.2°
  • 95. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 88.Bar DE. ( )( )24 8 192 in./sEDDv eω= = =192 in./sD =vBar AB. 8ABB ABv aω ω= =8B ABω=v 30°Locate the instantaneous center (point C) of bar BD by noting that velocitydirections at points B and D are known. Draw BC perpendicular to Bv andDC perpendicular to .DvLet 24 in.l BD= =Law of sines for triangle CBD.2448 in.sin120 sin30 sin30 sin30b d l= = = =° ° ° °41.569 in., 24 in.b d= =(a)1928 rad/s24DBDvdω = = = 8.00 rad/sBD =ωωωω(b) ( )( )41.569 8 332.55 in./sB BDv bω= = =332.5541.6 rad/s8BABvaω = = = 41.6 rad/sAB =ωωωω(c) Law of cosines for triangle CMD.( ) ( )( )( )22 2222 cos1202 224 12 2 24 12 cos12031.749 in.l lm d dm = + − °  = + − °=Law of sines.( )212 sin120sin sin120, sin , 19.131.749lmββ β°°= = = °Velocity of M. ( )( )31.749 8 253.99 in./sM BDv mω= = =21.2 ft/sM =v 19.1°
  • 96. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 89.( ) ,B ABF B Bv AB vω= =v 75°200 mm/sD =vLocate the instantaneous center (point C) of bar DBE by noting that the velocity directions at points B and Dare known. Draw BC perpendicular to Bv and DC perpendicular to .DvLaw of sines for triangle .sin150 sin15 sin15CD BC BDBCD = =° ° °180sin150180 mm 347.73 mmsin15BC BD CD°= = = =°2000.57515 rad/s347.73DDBEvCDω = = =( ) ( )( )180 0.57515 103.528 mm/sB DBEv BC ω= = =103.5280.57515 rad/s180BABFvABω = = =( ) ( )( )300 0.57515 172.546 mm/sF ABFv AF ω= = =Law of cosines for triangle DCE. ( ) ( ) ( ) ( )( )2 2 22 cos15CE CD DE CD DE= + − °( ) ( )( )( )2 2 2347.73 300 2 347.73 300 cos15 , 96.889 mmCE CE= + − ° =
  • 97. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.sin15 300 sin15EH DE= ° = °300sin15cos 36.796.889EHCEβ β°= = = °(a) ( ) ( )( )96.889 0.57515 55.7 mm/s,E BCDv CE ω= = =55.7 mm/sE =v 36.7°(b) 172.5 mm/sF =v 75.0°
  • 98. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 90.3 rad/sDE =ωωωω( ) ( )( )160 3480 mm/sD DEv DE ω= ==is perpendicular to .D DEvB Bv=vLocate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D areknown. Draw BC perpendicular to Bv and DC perpendicular to .Dv( )120 mm, cos30 120cos30BD DK BD= = ° = °120 cos30cos , 49.495 , 180 30 100.505160DKEDβ β φ β°= = = ° = ° − ° − = °Law of sines for triangle BCD.sin30 sin sinCD BC BDφ β= =°( )( )sin30 120sin3078.911 mmsin sinsin 120sin155.177 mmsin sinBDCDBDBCβ βφ φβ β° °= = == = =
  • 99. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Law of cosines for triangle ABC. ( ) ( ) ( ) ( )( )2 2 22 cos150AC BC AB AB BC= + − °( ) ( )( )( )2 2 2155.177 200 2 155.177 200 cos150 , 343.27 mmAC AC= + − ° =Law of sines.sin sin150 200 sin150sin , 16.9343.27AB ACγγ γ° °= = = °(a)4806.0828 rad/s78.911DABDvCDω = = =6.08 rad/sABD =ωωωω(b) ( ) ( )( )343.27 6.0828 2088 mm/sA ABDv AC ω= = =2.09 m/sA =v 73.1°
  • 100. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 91.AB = 20 in.Instantaneous centers:at I for BC.at J for BD.Geometry( )1211 8.25 in.16IC = =  ( )1221 15.75 in.16JD = =  ( )1120 13.75 in.16AI = =  20 in. 13.75 in. = 6.25 in.BI AB AI= − = −6.25 in.BJ BI= =Member BC. 33 in./sC =v334 rad/s8.25CBCvICω = = =( ) ( )( )6.25 in. 4 rad/s 25 in./sB BCv BI ω= = =Member BD.25 in./s4 rad/s6.25 in.BBDvBJω = = =(a) ( ) ( )( )= = 15.75 in. 4 rad/sD BDJD ωv 63.0 in./sD =vMember AB.(b)25 in./s20 in.BABvABω = = 1.250 rad/sAB =ωωωω
  • 101. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 92.6 in./sA =v , B Bv=v 30°Locate the instantaneous center (point C) of rod AB by noting that velocity directions at points A and B areknown. Draw AC perpendicular to Av and BC perpendicular to Bv .Triangle ACB. Law of sines.sin 30 sin 30 sin120AC CB AB= =° ° °10sin 305.7735 in.sin120AC CB°= = =°(a)61.03923 rad/s5.7735AABvACω = = =1.039 rad/sAB =ωωωω( ) ( )( )5.7735 1.03923 6 in./sB ABCB ω= = =v 30°D Dv=vLocate the instantaneous center (point I) of rod BD by noting that velocity directions at points B and Dare known. Draw BI perpendicular to Bv and DI perpendicular to Dv .Triangle BID. Law of sines.sin120 sin 30 sin 30BI DI BD= =° ° °10 sin12017.3205 in.sin 30BI°= =°, 10 in.DI =
  • 102. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.60.34641 rad/s17.3205BBDvBIω = = =0.346 rad/sBD =ωωωω(b) ( ) ( )( )10 0.34641 3.46 in./sD BDv DI ω= = =3.46 in./sD =v
  • 103. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 93.Method 1Assume Dv has the direction indicated by the angle β as shown. DrawCDI perpendicular to .Dv Then, point C is the instantaneous center of rodAD and point I is the instantaneous center of rod BD..Geometry ( ) ( )2 20.27 0.36 0.45 mAD = + =( ) ( )2 20.18 0.135 0.225 mBD = + =0.27 0.18sin 0.6, sin 0.80.45 0.225θ φ= = = =( ) ( )0.45 0.45 0.225 0.225sin sin 90 cos sin sin 90 cosc dθ β β φ β β= = = =° + ° −0.45sincoscθβ=0.225sincosdφβ=0.36 0.27tan 0.135 0.18tana bβ β= − = +.Kinematics 0.4 m/s, 1 m/sA Bv v= =,A DADv va cω = = B DBDv vb dω = =A BDcv dvva b= =0.45sin cos 0.40.6cos 0.225sin 1ABcva b b bdvθ ββ φ= = ⋅ ⋅ =( )0.36 0.27tan 0.6 0.135 0.18tanβ β− = +0.279 0.378tan , tan 0.73809, 36.43β β β= = = °( )( )0.36 0.27 0.73809 0.1607 ma = − =( )( )0.135 0.18 0.73809 0.2679 mb = + =( )( ) ( )( )0.45 0.6 0.225 0.80.3356 m 0.2237 mcos cosc dβ β= = = =
  • 104. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )a0.42.4891 rad/s0.1607AADvaω = = = 2.49 rad/sAD =ωωωω( )b13.733 rad/s0.2679BBDvbω = = = 3.73 rad/sBD =ωωωω(c) ( )( )0.3356 2.4891 0.835 m/s,D ADv cω= = = 0.835 m/sD =v 53.6°Method 2Consider the motion using a frame of reference that is translating withcollar A. For motion relative to this frame.0.4 m/sA =v1 m/sB =v/ /0, 1.4 m/sA A B A= =v v0.27tan , 36.870.36θ θ= = °0.360.45 mcosADθ= =/ 0.45D A ADω=v θLocate the instantaneous center (point C) for the relative motion of barBD by noting that the relative velocity directions at points B and D areknown. Draw BC perpendicular to B A/v and DC perpendicular to .D A/v( )0.180.3 msincos 0.135 0.375 mCDBC CDθθ= == + =/ 1.43.73 rad/s0.375B ABDvCBω = = =( ) ( )( )/ 0.3 3.73 1.120 m/sD A BDv CD ω= = =( )a/ 1.1200.45D AADvADω = = 2.49 rad/sAD =ωωωω( )b 3.73 rad/sBD =ωωωω( )c [/ 0.4 m/sD A D A= + =v v v  [1.120+ ]θ0.835 m/s= 53.6° 0.835 m/sD =v 53.6°
  • 105. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 94.( )( )0.2 0.2 4.5 0.9 m/sB ABω= = =vLet point C be the instantaneous center of bar BD. Define angle β andlengths a and b as shown.0.9BBDvb bω = =0.9D BDaabω= =v β0.7 m/sE =vLet point I be the instantaneous center of bar DE. Define lengths c and das shown.0.9DDEv ac bcω = =( )( )0.9 0.20.2 0.7E DEa dv dbcω−= − = = (1)0.25 0.15 0.25, , 0.25tan , 0.15tancos cos 0.15aa c b dcβ ββ β= = = = =Substituting into (1) ( )0.25 0.2 0.15tan0.9 0.7 or 1.2 0.9tan 0.7tan0.15 0.25tanββ ββ− = − =  1.2tan 0.75 36.87 cos 0.81.6β β β= = = ° =0.250.3125 m, 0.1875 m, 0.1875 m, 0.1125 m0.8a b c d= = = = =( )a0.94.8 rad/s0.1875BDω = = 4.80 rad/sBD =ωωωω( )b( )( )( )( )0.9 0.31258 rad/s0.1875 0.1875DEω = = 8.00 rad/sDE =ωωωω( )c ( )( )0.3125 4.8 1.5 m/sDv = = 1.5 m/sD =v 53.1°
  • 106. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 95.5 rad/sABCω = , vA = vAB Bv=v , E Ev=vLocate point I, the instantaneous center of rod ABD by drawing IAperpendicular to vA and IB perpendicular to vB.9tan 26.56518φ φ= = °1820.125 in.cosIDlφ= =( )( )5 20.125D ABD IDv lω= =100.6 in./sD =v φLocate point J, the instantaneous center of rod DE by drawing JDperpendicular to vD and JE perpendicular to vE.1820.125 in.cosJDlφ= =100.65 rad/s20.125DDEJDvlω = = =(a) 5.00 rad/sDE =ωωωω9 cos 27 in.JE JDl l φ= + =( )( )27 5E JE DEv l ω= =(b) 135 in./sE =v
  • 107. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 96.12 in./sA =vB Bv=vPoint C is the instantaneous center of bar AB.1220cos30BABvACω = =°0.69282 rad/s=10 in.CD =( ) ( )( )10 0.69282 6.9282 in./sD ABv CD ω= = =6.9282 in./sD =v 30°E Ev=v 30°Point I is the instantaneous center of bar DE.20cos30DI = °( )a6.92820.4 rad/s20cos30DDEvDIω = = =°0.400 rad/sDE =ωωωω( )b ( ) ( )( )20sin30 0.4 4 in./sE DEv EI ω= = ° = 0.333 ft/sE =v 30°
  • 108. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 97.Let points A, B, and C move to , , andA B C′ ′ ′ as shown.Since the instantaneous center always lies on the fixed lower rack, the space centrode is the lower rack.: lower rackspace centrodeSince the point of contact of the gear with the lower rack is always a point on the circumference of the gear,the body centrode is the circumference of the gear.: circumference of gearbody centrode A
  • 109. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 98.Draw x and y axes as shown with origin at the intersection of the twoslots. These axes are fixed in space.A Av=v , B Bv=vLocate the space centrode (point C) by noting that velocity directions atpoints A and B are known. Draw AC perpendicular to Av and BCperpendicular to .BvThe coordinates of point C are sinCx l β= − and cosCy l β=( )22 2 2300 mmC Cx y l+ = =The space centrode is a quarter circle of 300 mm radius centered at O.Redraw the figure, but use axes x and y that move with the body. Placeorigin at A.( )( )( )2coscos 1 cos22sincos sin sin 22CCx AClly ACllββ βββ β β== = +== =( )2 222 2 2150 1502 2C C C Cl lx y x y   − + = = − + =      The body centrode is a semi circle of 150 mm radius centeredmidway between A and B.
  • 110. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 99.80 km/h 22.222 m/sAv = =0C =v1560 mm, 0.280 mm 0.28 m2d r d= = = =Point C is the instantaneous center.22.22279.364 rad/s0.28Avrω = = =2 0.56 mCB r= =( ) ( )( )0.56 79.364 44.4 m/sBv CB ω= = =44.4 m/sB =v( )130 152γ = ° = °( )( )2 cos15 2 0.28 cos15 0.54092 mCD r= ° = ° =( ) ( )( )0.54092 79.364 42.9 m/s,Dv CD ω= = =42.9 m/sD =v 15.0°2 0.28 2 0.39598 mCE r= = =( ) ( )( )0.39598 79.364 31.4 m/s,Ev CE ω= = =31.4 m/sE =v 45.0°
  • 111. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 100.900 rpm 30 rad/sOAω π= =( ) ( )( )10 30 300 mm/sA OAOA ω π π= = =vA Av=v 60 ,° B Av=vLocate the instantaneous center (point C of bar BD by noting thatvelocity directions at point B and A are known. Draw BC perpendicular toBv and AC perpendicular to .Av( )sin30 10sin30sin , 1.79160OAABβ β° °= = = °( ) ( )cos30 cos 10cos30 160cosOB OA AB β β= ° + = ° +168.582 mm=168.58210 184.662 mmcos30 cos30OBAC OA= − = − =° °( )tan30 97.377 mmBC OB= ° =A BABv vAC BCω = =( )( )97.377 300497 mm/s,184.662B ABCv vACπ = = =  497 mm/sB =v
  • 112. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 101.Bar AB.4 rad/sABω =( )( )0.25 4B AB ABv l ω= =1 m/sB =vBar DE.75tan 26.565150φ φ= = °0.15= 0.167705 mcosEDlφ=0.167705D ED ED EDv l ω ω= =0.167705D EDω=v φBar BD. Locate point I, the instantaneous center of bar BD by drawing IBperpendicular to vB and ID perpendicular to vD.0.20.4 mtanIBlφ= =0.40.44721 mcosIDlφ= =12.5 rad/s0.4BBDIBvlω = = =2.50 rad/sBD =ωωωω1.11803 m/sD ID BDv l ω= =6.67 rad/sDEDEDvlω = = 6.67 rad/sED =ωωωω
  • 113. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 102.(a) 0β = .Wheel AD 0, 45 in./sC D= =v v4511.25 rad/s4DADvCDω = = =( ) ( ) 4 2.5 1.5 in.CA CD DA= − = − =( ) ( )( )1.5 11.25 16.875 in./sA ADv CA ω= = =Rod AB. B Bv=vSince vA and vB are parallel, the instantaneous center of rod AB lies at infinity.B A=v v 16.88 in./sB =v0ABω =(b) 90β = ° .Wheel AD 0, 11.25 rad/sC ADω= =v2.5tan , 32.0054DADCγ γ= = = °4.7170 in.cosDCCAγ= =( ) ( )( )4.7170 11.25 53.066 in./sA ADv CA ω= = =[53.066 in./sA =v ]32.005°Rod AB. B Bv=v4sin , 18.66312.5φ φ= = °
  • 114. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Locate point I, the instantaneous center of bar AB, by drawing IA perpendicular to vA and IBperpendicular to vB. For triangle ABI,( )180 90 39.332δ γ φ= ° − − + ° = °( )12.5sin sin 90 sinIA IBγ φ δ= =+ °12.5sin108.66322.345 in.sin32.005IA°= =°12.5sin39.33214.9486 in.sin32.005IB°= =°53.0662.3748 rad/s22.345AABvIAω = = =( ) ( )( )14.9486 2.3748 35.5 in./sB ABv IB ω= = =35.5 in./sB =v2.37 rad/sΑΒ =ωωωω
  • 115. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 103.Velocity analysis. 3 rad/sAB =ωωωω CD CDω=ωωωω , BDE BDω=ωωωω( ) ( )( )12 3 36 in./sB ABAB ω= = =v [, 9D CDω=v ]/D B D B= +v v v9 CDω [36= ] [7.5 BDω+ ]Components: : 7.5 0 0BD BDω ω− = =: 9 36 4 rad/sCD CDω ω= =Acceleration analysis. 0,AB CD CDα= =αααα αααα , BCD BDα=αααα( )B ABAB α= a ( ) 2ABAB ω +  ( )( )12 0=   ( )( )212 3+ 2108 in./s= ( )D CDCD α= a ( ) 2CDCD ω +   [9 CDα= ( )( )29 4+ [9 CDα =] 2144 in./s+ ( ) ( )/ /D B D B B Dt n= + +a a a a( ) ( )/D B BDtBD α= a 7.5 BDα =( ) ( ) 2/D B BDnBD ω= a ( )( )27.5 0 = 0=( ) ( )/ /D B D B D Bt n= + +a a a a[9 CDα ] [144+ ] [108= ] [7.5 BDα+ ] [ ]0+Components: : 9 0 0CD CDα α= =2: 144 108 7.5 4.8 rad/sBD BDα α= + =( )a ( )( )9 0D = a ] [144+ ] 2144 in./s= 212.00 ft/sD =a( )b ( ) ( )/E D ADtDE α= a ( )( )7.5 4.8 = 236 in./s =( ) ( ) 2/E D ADnDE ω= a ( )( )20.75 0 =  0 =( ) ( )/ /E D E D D Et n= + +a a a a[144= ] [36+ ] [0+ ] 2180 in./s= 215.00 ft/sE =a
  • 116. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 104.Velocity analysis. 3 rad/sAB =ωωωω CD CDω=ωωωω , BDE BDω=ωωωω( ) ( )( )12 3 36 in./sB ABAB ω= = =v [, 9D CDω=v ]/D B D B= +v v v9 CDω [36= ] [7.5 BDω+ ]Components: : 7.5 0 0BD BDω ω− = =: 9 36 4 rad/sCD CDω ω= =.Acceleration analysis 0,AB CD CDα= =αααα αααα , BCD BDα=αααα( )B ABAB α= a ( ) 2ABAB ω +  ( )( )12 0 =  ( )( )212 3+ 2108 in./s=( )D CDCD α= a ( ) 2CDCD ω +  [9 CDα = ] ( )( )29 4+ [9 CDα =] 2144 in./s+ ( ) ( )/ /D B D B B Dt n= + +a a a a( ) ( )/D B BDtBD α= a 7.5 BDα =( ) ( ) 2/D B BDnBD ω= a ( )( )27.5 0 = 0=( ) ( )/ /D B D B D Bt n= + +a a a a[9 CDα ] [144+ ] [108= ] [7.5 BDα+ ] [ ]0+Components: : 9 0 0CD CDα α= =2: 144 108 7.5 33.6 rad/sBD BDα α= − + =
  • 117. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )a ( )( )9 0D = a [144+ ] 2144 in./s= 212.00 ft/sD =a( )b ( ) ( )/E D ADtDE α= a ( )( )7.5 33.6= 2252 in./s =( ) ( ) 2/E D ADnDE ω= a ( )( )27.5 0 0  = =   ( ) ( )/ /E D E D E Dt n= + +a a a a[144= ] [252+ ] 20 396 in./s+ = 233.0 ft/sE =a
  • 118. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 105.0,ABC ABCω α= = =ωωωω αααα , B Ba=a 260 , 7.2 m/sD° =a/ 0.45 mB D =r /30 , 0.9 mA D° =r 30°( ) ( )/ /B D B D D Bt n= + +a a a a[ Ba ] [60 7.2° = ] [0.45α+ ] ( )( )260 0.45 0° +30 °Components.( ): cos60 7.2 0.45cos60 0Ba α° = − ° + (1)( ): sin60 0.45sin60 0.45B Ba aα α° = ° =Substituting into (1), ( ) ( )0.45cos60 7.2 0.45cos60α α° = − °( )a 27.216 rad/s0.9cos60α = =°216.00 rad/sABC =αααα( ) From (2),b ( )( ) 20.45 16 7.2 m/sBa = = 27.20 m/sB =a 60°( )c ( ) ( )/ /A D A D A Dt n= + +a a a a[7.2= ] [0.9α+ ] ( )( )260 3 0° +30 °[7.2= ] [14.4+ ] [ ]60 0° + 212.47 m/sA =a
  • 119. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 106.20, 12 rad/sABC ABCω= = =ωωωω αααα ,Collar B slides along a straight rod. ( )B Ba=a 60°Collar D slides along a straight rod. D Da=a/ 0.45 mB D =r /30 0.9 mA D° =r 30°( ) ( )/ /B D B D B Dt n= + +a a a a[ Ba ] [60 Da° = ] ( )( )0.45 12+  ] ( )( )260 1.5 0° + 30 °Components.: cos 60 5.4cos60 0B Da a° = + ° + (1)2: sin60 5.4sin60 , 5.4 m/s ,B Ba a− ° = ° = −( ) From (1),a 2cos60 5.4cos60 10.8cos60 5.4 m/sD Ba a= ° − ° = − ° = −25.40 m/sD =a( )b 25.40 m/sB =a 60°( )c ( ) ( )/ /A D A D A Dt n= + +a a a a[5.4= ] ( )( )0.9 12+  ( )( )260 3 0° + 30 °[5.4= ] [10.8+ ] [ ]60 0° + 29.35 m/sA =a
  • 120. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 107.0ω =20.8 rad/sα =20.9 ft/sC =aAcceleration of A./A C A C= +a a a20.9ft/s=  ] [( )AC α+ ]20.9 ft/s=  ( )( )210.8 ft 0.8 rad/s+27.74 ft/s=27.74 ft/sA =aAcceleration of B./B C B C= +a a a20.9 ft/s=  ] [( )BC α+ ]20.9 ft/s=  ] [(1.8)(0.8)+ ] 20.54 ft/s=20.54 ft/sB =a
  • 121. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 108.Geometry. Let φ be the angle between rod ADE and the vertical.4tan7.5φ = 28.072φ = °( )2 2 10.5( ) (7.5) (4) 8.5 in. ( ) = 8.5 11.9 in.7.5AD AB= + = =Velocity analysis. A Av=v D Dv=vLocate point I, the instantaneous center of rod ADB, bydrawing IA perpendicular to vA and ID perpendicular to vD.0AABvIAω = =( ) 0D ABv ID ω= =0DDEvEDω = =Acceleration analysis. 2 20.8 ft/s 9.6 in./sA = =a0,ABω = 0DEω =Rod DE. (Rotation about E) DE DEα=αααα( ) ( )D D t D n= +a a a [6 DEα= ] 26 DEω+  [] 6 DEα= ] (1)Rod ADB.29.6 in./sA =a AB ABα=αααα(Plane motion = Translation with A + Rotation about A)( ) ( )/D A D/A D At n= + +a a a a[9.6= [] 8.5 ABα+ ] 28.5 ABφ ω+  ]φ[9.6= [] 8.5 ABα+ ] 0φ + (2)Match expressions (1) and (2) for Da and resolve into components.( ): 0 9.6 8.5 cos ABφ α= − + 21.2799 rad/sABα =
  • 122. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(a) Angular acceleration of rod ADB. 21.280 rad/sAB =αααα(b) Acceleration of point B. ( ) ( )/ /B A B A B At na= + +a a a[9.6B =a [] 11.9 ABα+ ] 2[11.9 ABφ ω+ ]φ[9.6= ] [15.2319+ ] 0φ +[3.84= ] [7.1678+ ]28.1316 in./s= 61.8° 20.678 ft/sA =a 61.8°
  • 123. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 109.Geometry. Let point C be the center of the center cylinder, B its contact point withthe crate, and D its contact point with the ground. Let r be the radius ofthe cylinder. 100 mm.r =Velocity analysis. Since the contacts at B and D are rolling contacts without slipping,200 mm/sBv = and 0.D =v Point D is the instantaneous center ofrotation.200 mm/s1 rad/s2 200 mmBvrω = = =Point C moves on a horizontal line.Acceleration analysis. [C Ca=a ]/ /( ) ( ) [D C D C t D C n Ca= + + =a a a a ] [rα+ 2] [rω+ ]Component : Ca rα0 −==== (1)/ /( ) ( ) [B C B C t B C n Ca= + + =a a a a ] [rα+ 2] [rω+ ]Component :2400 mm/s Ca rα= + (2)Solving (1) and (2) simultaneously,2200 mm/sCa =2200 mm/srα =(a)2200 mm/s100 mmα = 22.00 rad/sα =/ /( ) ( ) [A C A C t A C n Ca= + + =a a a a ] [rα+ 2] [rω+ ]2[200 mm/s= 2] [200 mm/s+ 2] + [(100 mm) (1 rad/s) ]2[100 mm/s= 2] [200 mm/s+ ](b) 223.6 mm/sAa =20.224 m/sA =a 63.4°
  • 124. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 110.Let point C be the contact point.( )0, 0,C C tω= = =v a ωωωω , α=αααα/ 0O C O C bω= + = +v v v bω=Point O moves parallel to x-axis.( ) ( )/ /C O C O C Ot n= + +a a a a( )C ta ( )C na+  [ Oa= ] [bα+ ] 2bω+ ( )0 C na+  [ Oa= ] [bα+ ] 2bω+ From x-component, ,O Oa b bα α= − =aAcceleration of the point with coordinates ( ),x y[O xα= +a a ] [yα+ ] 2xω+ 2yω + [bα= ] [xα+ ] [yα+ ] 2xω+ 2yω+ Set 0=a and resolve into components.( ) 2: 0b y xα ω+ + = (1)2: 0x yα ω− = (2)( )2 4From (2), , From (1), 0y yx b yω ωαα α= + + =4 42 22 2/,1 1b b yy xω ωα αω α ωα−= − = =+ +Data: 212 in., 2 rad/sb ω= =2, 3 rad/sα =2 4 42 24 16 25, 13 9 9ω ω ωα α α= = + =( )912 4.32 in.,25y = − = − 5.76 in., 4.32 in.x y= − = −
  • 125. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 111..Velocity analysis 1.5 in. 10 rad/sr ω= =Point C is the instantaneous center of the wheel.( )( )1.5 10 15 in./sA rω= = =v.Acceleration analysis 230 rad/sα =Point moves on a circle of radius .A ρ 6 1.5 7.5 in.R rρ = + = + =Since the wheel does not slip, C Ca=a( ) ( )/ /C A C A C At na= + +a a a[ Ca ] ( )A ta= 2Avρ + [rα+] 2rω+ ( )A ta=  ]( )2157.5+( )( )1.5 30 + ( )( )21.5 10 + ( )A ta= [30+ ] [45+ ] [150+ ]Components. ( ) ( ): 45 0 45 in./sA At ta a− + = =: 30 150 120 in./sC Ca a= − + =(a) Acceleration of point A.245 in./sA= a 230 in./s + 254.1 in./sA =a 33.7°( )b Acceleration of point B. ( ) ( )/ /B A B A B At n= + +a a a a[45B =a ] [30+ ] [rα+ ] 2rω+ [45= ] [30+ ] ( )( )1.5 30+  ( )( )21.5 10+ 2105 in./s= 275 in./s+ 2129.0 in./sB =a 35.5°( )c Acceleration of point C. C Ca=a 2120 in./sC =a
  • 126. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 112.Velocity analysis.Let point G be the center of the shaft and point C be the point of contact with therails. Point C is the instantaneous center of the wheel and shaft since that pointdoes not slip on the rails.24, 0.8 rad/s30GGvrrω ω= = = =v.Acceleration analysisSince the shaft does not slip on the rails, C Ca=a 20°Also,210 mm/sG= a 20 °( ) ( )/ /C G C G C Gt n= + +a a a a[ Ca ] 220 10 mm/s° =  [20 30α° + ] 220 30ω° +  20 °Components 220 : 10 30 0.33333 rad/sα α° = − =(a) Acceleration of point A.( ) ( )/ /A G A G A Gt n= + +a a a a[10= ] [20 360α° + ] 2360ω+ [9.3969= ] [3.4202+ ] [120+ ] [230.4+ ][129.3969= ] [233.4202+ ] 2267 mm/sA =a 61.0°(b) Acceleration of point B.( ) ( )/ /B G B G B Gt n= + +a a a a[10= ] [20 360α° + ] 2360ω+ [9.3969= ] [3.4202+ ] [120+ ] [230.4+ ][110.6031= ] [226.9798+ ] 2252 mm/sB =a 64.0°
  • 127. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 113..Velocity analysis 160 mm/sD A= =v vInstantaneous center is at point .B ( ) ( ), 160 100 60Av AB ω ω= = −4 rad/sω =.Acceleration analysis [B Ba=a ] for no slipping α=αααα2600 mm/sA= a ( )A na + [G Ga=a ]( ) ( )/ /B A B A B At n= + +a a a a[ Ba ] [600= ] ( )A na+  ( )100 60 α + − ( ) 2100 60 ω + − Components 2: 0 600 40 15 rad/sα α= − + =( ) ( )/ /B G B G B Gt n= + +a a a a[ Ba ] [ Ga= ] [100α+ ] 2100ω+ Components 2: 0 100 100 1500 mm/sG Ga aα α= − + = =( )( )2 2: 100 4 1600 mm/sBa = = 21600 mm/sB =a
  • 128. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( )/ /A G A G A Gt n= + +a a a a[1500= ] [60α+ ] 260ω+ [1500= ] [900+ ] [960+ ]2600 mm/s= 2960 mm/s +  21132 mm/sA =a 58.0°( ) ( )/ /C G C G C Gt n= + +a a a a[1500= ] [100α+ ] 2100ω+ [1500= ] [1500+ ] [1600+ ]23100 mm/s= 21500 mm/s +  23440 mm/sC =a 25.8°
  • 129. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 114..Velocity analysis 160 mm/sD B= =v vInstantaneous center is at point .A ( ) ( ), 160 100 60Bv AB ω ω= = −4 rad/sω =.Acceleration analysis [A Aa=a ] for no slipping. α=αααα2600 mm/sB= a ( )B na + [G Ga=a ]( ) ( )/ /A B A B A Bt n= + +a a a a[ Aa ] [600= ] ( )B na+  ( )100 60 α + − ( ) 2100 60 ω + − Components 2: 0 600 40 15 rad/sα α= − + =( ) ( )/ /A G A G A Gt n= + +a a a aAa [ Ga= ] [60α+ ] 260ω+ Components 2: 0 60 60 900 mm/sG Ga aα α= − = =( )( )22 2: 60 60 4 960 mm/sAa ω= = = 2960 mm/sA =a
  • 130. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( )/ /B G B G B Gt n= + +a a a a[900= ] [100α+ ] 2100ω+ [900= ] [1500+ ] [1600+ ]2600 mm/s= 21600 mm/s +  21709 mm/sB =a 69.4°( ) ( )/ /C G C G C Gt n= + +a a a a[900= ] [100α+ ] 2100ω+ [900= ] [1500+ ] [1600+ ]2700 mm/s= 21500 mm/s + 21655 mm/sC =a 65.0°
  • 131. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 115.Velocity analysis. Point A is the instantaneous center of rotation of the cylinder. 12 in./s.C Ev v= =( )( )122 2 rad/s2 2 3CCvv rrω ω= = = =.Acceleration analysis ( )( )22 23 2 12 in./srω = =[A Aa=a ] ( )C C ta= a ( )C na +  [ Ea= ( )C na+ ( )/C A C A ta=a a ( )/C A ta+ [ Ea ] ( )C na+  [ Aa= ] [2rα+ ] 22rω+ [19.2 ] ( )C na+  [ Aa= ] [6α+ ] [24+ ] (1)From (1), Components 2: 19.2 6 , 3.2 rad/sα α= =( ) ( )/ /G A G A G At n= + +a a a a[ Ga ] [ Aa= ] [rα+ ] 2rω+ [ Ga ] [ Aa= ] 29.6 in./s+ 212 in./s + From which 2 2 29.6 in./s and 12 in./s 9.6 in./sG A Ga a= = =aFrom (1), Components ( ) 2: 24 12 24 12 in./sC Ana a= − + = − + =Then 219.2 in./sC= a 212 in./s + 222.6 in./sC =a 58.0°( ) ( )/ /D G D G D Gt n= + +a a a a[9.6= ] [rα+ ] 2rω+ [9.6= ] [9.6+ ] [12+ ][ 221.6 in./s= ] 29.6 in./s+ 223.6 in./sD =a 66.0°
  • 132. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 116.Velocity analysis./A B A B= +v v v[9 in./s ] [6 in./s= ] [(5 in.) ω+ ]Components9 6 5ω= − +3 rad/sω =.Acceleration analysisPoint A moves on a path parallel to the belt. The path is assumed to be straight.236 in./sA =a 30°Since the drum rolls without slipping on the belt, the component of acceleration of point B on thedrum parallel to the belt is the same as the belt acceleration. Since the belt moves at constantvelocity, this component of acceleration is zero. ThusB Ba=a 60°Let the angular acceleration of the drum be α ./ /( ) ( )B A B A t B A n= + +a a a a[ Ba ] [36= ] [rα+ 2] [rω+ ]Components : 0 36 5α= −7.2 rad/sα =
  • 133. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies..Acceleration of point D/ /( ) ( )D A D A t D A n= + +a a a a[ Aa= 30 ] [rα° + 260 ] [rω° + ]30°[36= 30 ] [(5)(7.2)° + 260 ] [(5)(3)° + 30 ]°Components: 30 :° 236 45 9 in./s− + =Components: 60 :° 236 in./s2 2 29 36 37.1 in./sDa = + =36tan 76.09β β= = °30 46.0β − °= °237.1 in./sD =a 46.0°
  • 134. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 117..Geometry Define angles and as shown.θ β0.25cos 0.4 0.25cosβ θ= − (1)0.25sin 0.25sin 0.2 0.25 sinxβ θ θ= + = + (2)Dividing (1) by (2) gives0.2 0.25 sintan0.4 0.25 cosθβθ+=−(3)Squaring (1) and (2), adding, and rearranging gives0.2 cos 0.1 sin 0.2θ θ− = (4)Let sin .u θ= 2cos 1 uθ = −20.2 1 0.2 0.1u u− = +Squaring both sides,( )2 20.04 1 0.04 0.04 0.01u u u− = + +20.05 0.04 0u u+ =0 or 0.8u u= = −Reject the negative root. sin 0 0θ θ= =From (3),0.2tan0.4 0.25β =−53.13β = °.Velocity analysis 1.2 m/sD = =v
  • 135. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.0.25B ABω=v/D B D B= +v v v[1.2 ] [0.25 ABω= ] [0.25 BDω+ ]βComponents: : 0.25 sin 0, 0BD BDω β ω= =: 1.2 0.25 4.8 rad/sAB ABω ω= =.Acceleration analysis 0, 0,A D AB ABα= = =a a αααα BD BDα=,,,, αααα( ) ( )/ /B A B A B At n= + +a a a a[0 0.25 ABα= + ] 20.25 ABω+  [0.25 ABα + ] [5.76+ ]( ) ( )/ /D B D B D Bt n= + +a a a a[0 0.25 ABα= ] [5.76+ ] [0.25 BDα+ β 20.25 BDω+  β [0.25 ABα= ] [5.76+ ] [0.25 BDα+ ] 0β +( )a 2: 0 0 5.76 0.25 sin , 28.800 rad/sBD BDα β α= + + = −228.8 rad/sBD =αααα( )b ( )( ): 0 0.25 0.25 28.800 cos ,ABα β= + −217.28 rad/sAB =αααα
  • 136. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 118..Geometry Define angles and as shown.θ β0.25cos 0.4 0.25cosβ θ= − (1)0.25sin 0.25sinxβ θ= + (2)With 0,x = equation (2) gives .β θ=From (1), 0.25 cos 0.4 0.25 cosβ β= −0.4cos0.5β = 36.87β = °36.87θ = °.Velocity analysis 0.6 m/sD = =v[0.25B ABω=v ]θ/D B D B= +v v v[0.6 ] [0.25 ABω= ] [0.25 BDθ ω+ ]βComponents: : 0.6 0.25 cos 0.25 cosAB BDω θ ω β= +: 0 0.25 sin 0.25 sinAB BDω θ ω β= −1.5 rad/sABω = , 1.5 rad/sBDω =
  • 137. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies..Acceleration analysis 0, 0,A D AB ABα α= = =a a , BD BDα α=( ) ( )/ /B A B A B At n= + +a a a a[0 0.25 ABα= + ] 20.25 ABθ ω+  θ [0.25 ABα= ] [0.5625θ + ]θ( ) ( )/ /D B D B D Bt n= + +a a a a[0 0.25 ABα= ] [0.5625θ + ] [0.25 BDθ α+ ] 20.25 BDβ ω+  ]β[0.25 ABα= ] [0.5625θ + ] [0.25 BDθ α+ ] [0.5625β + ]β: 0.25 cos 0.5625sin 0.25 cos 0.5625sin 0AB BDα θ θ α β β+ + − =: 0.25 sin 0.5625cos 0.25 sin 0.5625cos 0AB BDα θ θ α β β− − − =Solving simultaneously, 2 23 rad/s , 3 rad/sAB BDα α= = −( )a 23.00 rad/sBD =αααα( )b 23.00 rad/sAB =αααα
  • 138. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 119.Law of sines.sin sin60, 16.7792 6ββ°= = °.Velocity analysis 900 rpm 30 rad/sABω π= =2 60 in./sB ABω π= =v 60°D Dv=v BD BDω=ωωωω/ 6D B BDω=v β/D B D B= +v v v[ Dv ] [60π= ] [60 6 BDω° + ]βComponents : 0 60 cos60 6 cosBDπ ω β= ° −60 cos6016.4065 rad/s6cosBDπωβ°= =.Acceleration analysis 0ABα =( )( )22 22 2 30 17765.3 in./sB ABω π= = =a 30°D Da=a BD BDα=αααα[/ 6D B ABα=a ] 26 BDβ ω+  β [6 BDα= ] [1615.04β + ]β( )/ Resolve into components.D B D B= +a a a: 0 17765.3cos30 6 cos 1615.04sinBDα β β= − ° + +22597.0 rad/sBDα =( )( ): 17765.3sin30 6 2597.0 sin 1615.04cosDa β β= ° − +25931 in./s= P D=a a 2494 ft/sP =a
  • 139. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 120.Law of sines.sin sin120, 16.7792 6ββ°= = °.Velocity analysis 900 rpm 30 rad/sABω π= =2 60 in./sB ABω π= =v 60°D Dv=v BD BDω=ωωωω/ 6D B BDω=v β/D B D B= +v v v[ Dv ] [60π= ] [60 6 BDω° + ]βComponents : 0 60 cos60 6 cosBπ ω β= − ° +60 cos6016.4065 rad/s6cosBDπωβ°= =.Acceleration analysis 0ABα =( )( )22 22 2 30 17765.3 in./sB ABω π= = =a 30°D Da=a BD BDα=αααα[/ 6D B BDα=a ] 26 BDβ ω+  β [6 BDα= ] [1615.04β + ]β/ Resolve into components.D B D B= +a a a: 0 17765.3cos30 6 cos 1615.04cosBDα β β= − ° + +22597.0 rad/sBDα =( )( ): 17765.3sin30 6 2597.0 sin 1615.04cosDa β β= ° + −211835 in./s= P D=a a 2986 ft/sP =a
  • 140. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 121..Geometry and velocity analysis 0θ =( ) ( )( )60 16 960 mm/sBv AB ω= = =960 mm/sB =v , D v=vInstantaneous center of bar lies at .BD C120sin 0.6, cos 0.8200β β= = =36.9 , 200cos 160 mmCBβ β= ° = =9606 rad/s160BBDvCBω = = =.Acceleration analysis 0ABα =[60B ABα=a ] 260 ABω+  ( )( )20 60 16 = +  215360 mm/s =Point moves on a straight line.D D Da=a( ) [/ 120D B BDtα=a ] [160 BDα+ ]( ) 2/ 160D B BDnω= a 2120 BDω +  [5760 = ] [4320+ ]( ) ( )/ / . Resolve into components.D B D B D Bt n= + +a a a a2: 0 0 160 4320 27 rad/sBD BDα α= + − =( )a ( )( ) 2: 15360 120 27 5760 24360 mm/sDa = − − − = − 224.4 m/sD =a( ) [/ 60G B BDtα=a ] [80 BDα+ ] [1620= ] [2160+ ]( ) 2/ 80G B BDnω= a 260 BDω +  [2880 = ] [2160+ ]( )b ( ) ( )/ /G B G B G Bt n= + +a a a a[15360= ] [1620+ ] [2160+ ] [2880− ] [2160+ ]219860 mm/s= 219.86 m/sG =a
  • 141. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 122..Geometry and velocity analysis 90θ = °60sin 0.3, 17.458200β β= = = °and are parallel, thus the instantaneous center lies at .D B ∞v v0BDω =.Acceleration analysis 0, 16 rad/s, 0AB AB BDα ω ω= = =[60B ABα=a ] 260 ABω+  ( )( )20 60 16 = +  215360 mm/s =Point moves on a straight line.D D Da=a( ) [/ 60D B BDtα=a ] [200cos BDβα+ ( ) 2/ 60D B BDnω= a 2200cos BDβω +  0 =( ) ( )/ / . Resolve into components.D B D B D Bt n= + +a a a a215360: 0 15360 200cos , 80.508 rad/s200cosBD BDβα αβ= − + = =( )a ( )( ) 2: 0 60 0 60 80.508 4831 mm/sD BDa α= − + = − = −24.83 m/sD =a( ) [/ 30G B BDtα=a ] [100cos BDβα+ ] [2415= ] [7680+ ]( ) 2/ 30G B BDnω= a 2100cos BDβω +  0 =( )b ( ) ( )/ /G B B G B Gt n= + +a a a a[15360= ] [2415+ ] [7680+ ] 0+22415 mm/s= 27680 mm/s + 28.05 m/sG =a 72.5°
  • 142. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 123..Disk A 15 rad/sAω = , 0, 2.8 in.A ABα = =( ) ( )( )2.8 15 42.0 in./sB Av AB ω= = =( ) ( )( )22 22.8 15 630 in./sB Aa AB ω= = =( ) 0.a θ =42 in./sB =v , D Dv=v2.8sin 16.2610β β= = °Instantaneous center of bar BD lies at point C.424.375 rad/s10cosBBDvBCωβ= = =2630 in./sB =a , D Da a= , BD BDα=αααα( )/D B BDDB α= a ( ) 2BDDBβ ω +  β [10 BDα= ] [191.406β + ]β/ Resolve into components.D B D B= +a a a( ) 2: 0 0 10cos 53.594, 5.5826 rad/sBD BDβ α α= − + = −( )( ) 2: 630 10sin 5.5826 183.75 430.62 in./sDa β= − − =25.58 rad/sBD =αααα , 2431 in./sD =a
  • 143. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) 90 .b θ = °42 in./sB =v , D D=v v5.6sin 34.05610β β= = °Instantaneous center of bar BD lies at .∞0BDω =2630 in./sB =a , D Da=a( )/D B BDDB α= a ( ) 2BDDBβ ω +  10 BDβ α = β/ Resolve into components.D B D B= +a a a( ) 2: 0 630 10cos 76.042 rad/sBD BDβ α α= − − = −( )( ) 2: 0 0 10 76.042 sin 425.8 in./sD Da aβ= + − − =276.0 rad/sBD =αααα , 2426 in./sD =a( ) 180 .c θ = °42 in./sB =v D Dv=v2.8sin 16.2610β β= = °Instantaneous center of bar BD lies at point C.424.375 rad/scos 10cosBBDvBDωβ β= = =2630 in/sB =a D Da=a( )/D B BDBD α= a ( ) 2BDBDβ ω +  β [10 BDα= ] [191.406β + ]β/ Resolve into components.D B D B= +a a a2: 0 10 cos 53.594 5.5826 rad/sBD BDα β α= − + =( )( ) 2: 630 10 5.5826 sin 183.75 829.4 in./sDa β= + + =25.58 rad/sBD =αααα , 2829 in./sD =a
  • 144. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 124.Crank AB.0, 0, 1500 rpm 157.08 rad/sA A ABω= = = =v a , 0ABα =/ 0 [2B A B A ABω= + = +v v v 45 ] 314.16 in./s° = 45°( ) ( )/ /B A B A B At n= + +a a a a0 [2 ABα= + 45 ]° 2[2 ABω+ 45 ]°2[(2)(157.08)= 245 ] 49348 in./s° = 45°Rod BD. D Dv=v 45° BD BDω=ωωωω/D B B D= +v v vDv 45 [314.16°= 45 ] [7.5 BDω° + 45 ]°Components 45 : 0 314.16 7.5 41.888 rad/sBD BDω ω° = − =D Da=a 45°( ) ( )/ /D B D B D Bt n= + +a a a a[ Da 45 ] [49348° = 45 ] [7.5 BDα° + 245 ] [7.5 BDω° + 45 ]°Components 45 :° 2 249348 (7.5)(41.888) 62507 in./sDa = + =25210 ft/sD =a 45°
  • 145. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Rod BE.2sin , 15.466 , 45 29.5347.5γ γ β γ= = ° = ° − = °E Ev=v 45°Since Ev is parallel to ,Bv 0BEω =E Ea=a 45° ( ) 2/ 7.5 0B E BEna ω= =( ) ( )/E/B E Bt ta=a β ( )/E B B E t= +a a aDraw vector addition diagram.45 15.466γ β= ° − = °tanE Ba a γ=49348tan15.466= °213654 in./s=21138 ft/sE =a 45°
  • 146. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 125.Geometry. Triangle ABD is equilateral. ( ) 180 mmAD =Velocity analysis. 5 rad/sAB =ωωωω , B Bv=v 30 , D Dv° =vLocate point I, the instantaneous center of bar BD, by drawing IB perpendicular to Bv and IDperpendicular to .Dv Point I coincides with A.(0.18)(5) 0.9 m/sBv = =5 rad/s0.18BBDvω = =0.18 0.9 m/sD BDv ω= =5 rad/s0.18DDEvω = =Acceleration analysis. 0AB =ααααBar AB. (Rotation about A)( ) ( ) ( ) 20 0.18B B B ABt nω= + = +a a a 260 4.5 m/s° = 60°Bar BD. BDα( Plane motion Translation with Rotation aboutB B= + )( ) ( )/ /D B D B D Bt n= + +a a a a[4.5= 60 ] [0.18 BDα° + 230 ] [0.18 BDω° + 60 ]°[4.5= 60 ] [0.18 BDα° + 30 ] [4.5° + 60 ]° (1)Bar DE. (Rotation about E) DEα( ) ( ) [0.18D D D DEt nα= + =a a a 2] [0.18 DEω+ ][0.18 DEα= ] [4.5+ ] (2)
  • 147. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Match the expressions (1) and (2) for Da and resolve into components.: 4.5 2.25 0.18 cos30 2.25BDα= − + ° −2957.735 rad/s0.18cos30BDα = =°: ( )( )0.18 3.8971 0.18 57.735 sin30 3.8971DEα = − ° −257.735sin30 28.868 rad/sDEα = − ° = −(a) Angular acceleration of bar BD. 257.7 rad/sBD =αααα(b) Angular acceleration of bar DE. 228.9 rad/sDE =αααα
  • 148. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 126.60 120tan , 26.565 , 134.164 mm120 cosDEβ ββ= = ° = =.Velocity analysis 4 rad/sABω =( ) ( )( )200 4 800 mm/sB ABv AB ω= = =B Bv=v , D Dv=v βPoint C is the instantaneous center of bar BD.160320 mmtan tanBDBCβ β= = =357.77 mmcosBCCDβ= =8002.5 rad/s320BBDvBCω = = =( ) ( )( )357.77 2.5 894.425 mm/sD BDv CD ω= = =894.4256.6667 rad/s134.164DDEvDEω = = =.Acceleration analysis 0, 4 rad/sAB ABα ω= =( )B ABAB α= a ( ) 2ABAB ω+ ( )( )20 200 4= + 23200 mm/s =
  • 149. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )/D B BDBD α= a ( ) 2BDBD ω + [160 BDα= ] ( )( )2160 2.5+ [160 BDα= ] 21000 mm/s+ ( )D DEDE α= a ( ) 2DEDE ωβ +  β [134.164 DEα= ] [5961.5β + ]β/ Resolve into components.D B D B= +a a a: 134.164 cos 5961.5sin 0 1000DEα β β+ = −230.55 rad/sDEα = −( )( ): 134.164 30.55 sin 5961.5cos 3200 160 BDβ β α− − + = +( )a 224.8 rad/sBD =αααα( )b 230.6 rad/sDE =αααα
  • 150. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 127..Velocity analysis 19 rad/sABω =( ) ( )( )8 19 152 in./sB ABv AB ω= = =B Bv=v , D Dv=vInstantaneous center of bar BD lies at C.15219 rad/s8BBDvBCω = = =( ) ( )( )19.2 19 364.8 in./sD BDv CD ω= = =2364.824 rad/s15.2DDEvDEω = = =.Acceleration analysis 0.ABα =( ) 2B ABAB ω= a ( )( )28 19 = 2888 in./s =( )D DEDE α= a ( ) 2DEDE ω + [15.2 DEα= ] 28755.2 in./s+ ( ) [/ 19.2D B BDtα=a ] [8 DBα+ ]( ) 2/ 19.2D B BDnω= a 28 BDω +  ]26931.2 in./s= 22888 in./s + ( ) ( )/ / Resolve into components.D B D B D Bt na= + +a a a: 8755.2 0 8 6931.2BDα= + +( )a 2228 rad/sBD =αααα( )( ): 15.2 2888 19.2 228 2888DEα = − + −( )b 92 rad/sDEα = − 292.0 rad/sDE =αααα
  • 151. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 128..Velocity analysis 18 rad/sDEω =( ) ( )( )15.2 18 273.6 in./sD DEDE ω= = =vD Dv=v , B Bv=vPoint C is the instantaneous center of bar BD.273.614.25 rad/s19.2DBDvCDω = = =( ) ( )( )8 14.25 114 in./sB BDv CB ω= = =11414.25 rad/s8BABvABω = = =.Acceleration analysis 0DEα =( ) 2D DEDE ω= a ( )( )215.2 18 = 24924.8 in./s = ( )B ABAB α= a ( ) 2ABAB ω + [8 ABα= ] 21624.5 in./s+ ( ) [/ 19.2D B BDtα=a [8 BDα +( ) 2/ 19.2D B BDnω= a 28 BDω + 23898.8 in./s= 21624.5 in./s + ( ) ( )/ / Resolve into components.D B D B D Bt na= + +a a a
  • 152. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.: 0 1624.5 19.2 1624.5,BDα= − +2169.21875 rad/sBDα =( )( ): 4924.8 8 8 169.21875 3898.8ABα= + +240.96875 rad/sABα = −( )a ( )( )8 40.96875B = −a 21624.5 in./s + 2327.75 in./s= 21624.5 in./s + ,2138.1 ft/sB =a 78.6°( )b ( )/ /12G B G B B D B= + = +a a a a a( ) ( )1 12 2B D B B D= + − = +a a a a a327.75 4924.82− += 1624.52 + 22298.5 in./s= 2812.25 in./s+ 2203 ft/sG =a 19.5°
  • 153. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 129.Geometry. Lengths are expressed in meters./ 0.150 0.200 , 0.300 , 0.150 0.200B A D/B E/Dr += − = =r i j i r i j/ 0.150 0.200H B = −r i jVelocity analysis. 4 rad/sAB =ωωωω 4 rad/s= − kLink AB. 0A =v/ /0B A B A AB B A= + = + ×v v v rωωωω( ) ( ) ( ) ( )4 0.150 0.200 0.8 m/s 0.6 m/s= − × − = − −k i j i jCross BD. ( )/ / 0.300 0.3BD BD D B BD D B BD BDω ω ω= = × = × =k v r k i jωωωω ωωωω/ 0.8 0.6 0.3D B D B BDω= + = − − +v v v i j jLink DE. DE DEω= kωωωω( )/ / 0.150 0.200 0.2 0.15E D DE E D DE DE DEω ω ω= × = × + = − +v r k i j i jωωωω/ 0.8 0.6 0.3 0.2 0.15E D E D BD DE DEω ω ω= + = − − + − +v v v i j j i jSince point E is fixed, 0E =vComponents. : 0 0.8 0.2 4 rad/sDE DEω ω= − − = −i( )( ): 0 0.6 0.3 0.15 4BDω= − + + −j(a) 4BDω = 4 rad/sBD =ωωωωAcceleration analysis. 0, 4 rad/sAB ABω= =ααααLink AB. 0A =a( ) ( ) 2/ / /0 0B A B A B A AB B At na ω= + + = + −a a a r( )( ) ( ) ( )2 216 0.150 0.200 2.4 m/s 3.2 m/s+= − − = −i j i j
  • 154. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Cross BD. , 4 rad/sBD BD BDα ω= =kαααα( ) ( ) 2/ / / / /D B D B D B BD D B BD D Bt nω= + = × −a a a r rαααα( ) ( ) ( )20.300 4 0.300BDα= × −k i i0.3 4.8BDα= −j i/D B D B= +a a a7.2 3.2 0.3 BDα= − + +i j jLink DE. 4 rad/sDE DE DEα ω= = −kαααα( ) ( ) 2/ / / / /E D E D E D DE E D DE E Dt nω= + = × −a a a r rαααα( ) ( ) ( )20.150 0.200 4 0.150 0.200DE +α= × + −k i j i j0.15 0.2 2.4 3.2DE DEα α= − − −j i i j/E D E D= +a a a0.3 0.15 0.2 9.6BD DE DEα α α= + − −j j i iSince point E is fixed, 0E =aComponents.2: 0 0.2 9.6 48 rad/sDE DEα α= − − = −i( )( ): 0 0.3 0.15 48BDα= + −j(b) 224 rad/sBDα = 224.0 rad/sBD =αααα(c) Acceleration of point H.( ) ( )/ /H B H B H Bt n= + +a a a a2/ /B BD H B BD H Bω= + −a r rαααα( ) ( ) ( )22.4 3.2 24 0.150 0.200 4 0.150 0.200+= − + × − − −i j k i j i j10= j 210.00 m/sH =a
  • 155. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 130..Velocity analysis 4 rad/sDEω =( ) ( )( )8 4 32 in./sD DEv DE ω= = =D Dv=v , B Dv v= 45°Instantaneous center of bar BD lies at point B. 0Bv =324 rad/s8DBDvBDω = = =0BABvABω = =.Acceleration analysis 210 rad/sDEα = ( )22, 16 rad/sDEω =( )D DEDE α= a ( ) 2DEDE ω + ( )( )8 10=  ( )( )8 16 +  280 in./s= 2128 in./s + ( )/B D BDDB α= a ( ) 2BDDB ω + [8 BDα= ] ( )( )28 4+ [8 BDα= ] 2128 in./s+ ( )B ABAB α= a ( ) 245 ABAB ω + ° 45 °[8 ABα= ]45 0° +/ Resolve into components.B D D B= +a a a
  • 156. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.2: 8cos45 128 128, 45.255 rad/sAB ABα α− ° = + = −( )( ): 8sin 45 45.255 80 8 BDα° − = − −( )a 222.0 rad/sBD =αααα( )/G D BDDG α= a ( ) 2BDDG ω + ( )( )4 22=  ( )( )24 4+ 288 in./s= 264 in./s + ( )b 2/ 168 in./sG D G D= + = a a a 2192 in./s + 2255.1 in./s= 41.2°221.3 ft/sG =a 41.2°
  • 157. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 131.From the solution to Problem 15.68 the angular velocities are:3 rad/sAB =ωωωω , 4.85 rad/sBD =ωωωω , 6 rad/sDE =ωωωω .Bar AB. ( ) ( )/ 0.300 m 0.125 m , 0B A AB= + =r i j αααα0A =a( ) ( ) 2/ / / /0B A B A B A AB B A AB B At nω= + + = + × −a a a a r rαααα( ) ( )20 0 3 0.3 0.125 2.7 1.125= + − + = − −i j i jBar BD. ( )/ 0.325 m ,D B BD BDα= − =r j kαααα( ) ( ) ( ) ( ) ( )2/ / / 0.325 4.85 0.325D B D B D B BDt nα= + = × − − −a a a k j j0.325 7.6448BD +α= i j/ 2.7 6.5198 0.325D B D B BDα= + = − + +a a a i j iBar DE. ( ) ( )/ 0.150 m 0.200 m ,E D DE DEα= − + =r i j kαααα( ) ( ) 2/ / / / /E D E D E D DE E D DE E Dt nα ω= + = × −a a a k r r( ) ( ) ( )20.15 0.2 6 0.15 0.2DE +α= × − + − −k i j i j0.15 0.2 5.4 7.2DE DEα α= − − + −j i i j/E D E D= +a a a2.7 0.4448 0.325 0.15 0.2BD DE DEα α α= + + − −i j i j iSince point E is fixed, 0E =aComponents.: 0 0.6802 0.15 4.5347DE DEα α= − = −j( )( ): 0 2.7 0.325 0.2 4.534BDα= + − −i(a) 11.10BDα = − 211.10 rad/sBD =αααα(b) 24.53 rad/sDE =αααα
  • 158. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 132.3tan , 36.874β β= = °45 in.cosABβ= =45 in.cosDEβ= =( ) ( )( )5 1575 in./sB ABv AB ω= ==B Bv=v , D Dvβ =v βPoint C is the instantaneous center of bar BD.5 756.25 in. 12 rad/scos 6.25BBDvCBCBωβ= = = = =( ) ( )( )56.25 in. 6.25 12 75 in./scosD BDCD v CD ωβ= = = = =7515 rad/s5DDEvDEω = = =.Acceleration analysis 0ABα =( )B ABAB α= a ( ) 2ABABβ ω+  β ( )( )20 5 15= + 21125 in./sβ  =β( )/D B BDBD α= a ( ) 2BDBD ω+  [10 BDα= ( )( )210 12 +  [10 BDα= 21440 in./s+ 
  • 159. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )D DEDE α= a ( ) 2DEDEβ ω+  β [5 DEα= ] ( )( )25 15β + β [5 DEα= ] 21125 in./sβ +  β / Resolve into components.D B D B= +a a a( )a : 5 sin 1125cos 1125cos 1440DEα β β β+ = − −21080 rad/sDEα = − 21080 rad/sDE =αααα( )b ( )( )5 1080D = −a 21125 in./sβ +  β 25400 in./s= 21125 in./sβ  +  β 1125tan 11.775400γ γ= = °2 2 25400 1125 5516 in./sDa = + =2460 ft/s=90 64.9β γ° − + = °2460 ft/sD =a 64.9°
  • 160. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 133..Relative position vectors ( ) ( )/ /200 mm , 160 mmB A D B= − =r j r i( ) ( )/ 60 mm 120 mmD E = − −r i j.Velocity analysis ( )4 rad/sAB = −ω k( ) ( ) ( )/ 4 200 800 mm/sB AB B A= × = − × − = −v r k j iωωωω/ / 160 160D B BD D B BD BDω ω= × = × =v r k i jωωωω( )/ 60 120 120 60D DE D E DE DE DEω ω ω= + = × − − = −v r k i j i jωωωω/ Resolve into components.D B D B= +v v v:i 120 800 0 6.6667 rad/sDE DEω ω= − + = −:j 60 160 0 2.5 rad/sDE BD BDω ω ω− = + =.Acceleration analysis ( )0, 4 rad/sAB ABα ω= = − k( ) ( ) ( )22 2/ / 0 4 200 3200 mm/sB AB B A AB B Aω= × − = − − =a r r j jαααα( ) ( ) ( )22/ / / 160 2.5 160D B BD D B BD B D BDω α= × − = × −a r r k i iαααα( )2160 1000 mm/sBDα= −j i( ) ( ) ( )22/ / 60 120 6.6667 60 120D DE D E DE D E DEω α= × − = × − − − − −a r r k i j i jαααα( ) ( )2 2120 60 2666.7 mm/s 5333.3 mm/sDE DEα α= − + +i j i j/D B D B= +a a a Resolve into components.:i 2120 2666.7 0 1000 30.556 rad/sDE DEα α+ = − = −:j 260 5333.3 160 3200 24.792 rad/sDE BD BDα α α− + = + =( )a 224.8 rad/sBD =αααα( )b 230.6 rad/sDE =αααα
  • 161. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 134.Relative position vectors. ( ) ( )/ 4 in. 3 in.B A = −r i j( )/ 10 in.D B =r i ( ) ( )/ 4 in. 3 in.D E = − −r i jVelocity analysis. ( )15 rad/sAB =ω k( ) ( ) ( ) ( )/ 15 4 3 45 in./s 60 in./sB AB B A= × = × − = +v ω r k i j i j/ / 10 10D B BD D B BD BDω ω= × = × =v ω r k i j( )/ 4 3 3 4D DE D E DE DE DEω ω ω ω= × = × − − = −v k r k i j i j/D B B D= +v v v Resolve into components.:i 3 45 0DEω = + 15 rad/sDEω =:j 4 60 10DE ABω ω− = + 12 rad/sBDω = −Acceleration analysis. 0,AB =α ( )15 rad/sAB =ω k( ) ( ) ( ) ( )22 2 2/ / 0 15 4 3 900 in./s 675 in./sB AB B A AB B Aω= × − = − − = − +a α r r i j i j( ) ( ) ( )22/ / / 10 12 10D B BD D B BD D B BDω α= × − = × −a α r r k i i( )210 1440 in./sBDα= −j i( ) ( ) ( )22/ / 4 3 15 4 3D DE D E DE D E DErω α= × − = × − − − − −a α r k i j i j( ) ( )2 23 4 900 in./s 675 in./sDE DEα α= − + +i j i j/D B D B= +a a a Resolve into components.:i 3 900 900 1440DEα + = − − 21080 rad/sDEα = −(a) 21080 rad/sDE =αααα(b) ( )( ) ( )( ) ( ) ( )2 23 1080 4 1080 900 675 2340 in./s 4995 in./sD = − − − + + = − +a i j i j i j25516 in./sD =a 64.9° 2460 ft/sD =a 64.9°
  • 162. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 135.( ) At the instantaneous center ,a C 0C =v/ /A C A C A C= + × = ×v v r rωωωω ωωωω( ) 2/ /A A C A Cω× = × × = −v r rωωωω ωωωω ωωωω/ / /2 2orA AA C C A C Aω ω× ×= − = − =v vr r rωωωω ωωωω2AC Aω×− =vr rωωωω2AC Aω×= +vr rωωωω( )b / /A C A C A C= + × + ×a a r vαααα ωωωω( )( )22AC A CC A AC A Aαωαωωαω×= − × + × −= − × × + ×= + + ×va k v va k k v ω va v vωωωωωωωωωωωωSet 0.C =a A A Aαω= + ×a v vωωωω
  • 163. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 136.Geometry. Let φ be the angle ADB as shown. By the law of cosines for triangle ABD( ) ( )( )( ) ( )22 24 2 4 cos 180c b b b b θ= + − ° −( )2 217 8cosc b θ= + (1)17 8cosc b θ= +Law of sines: sin sinc bφ θ= (2)Also, cos 4 cosc b bφ θ= +Velocity analysis. Angular velocity of crank AB: =ω θ&Angular velocity of rod DE: =DEω φ&Differentiating (1),22 42 8 sin sinbcc b ccθθ θθ= − = −& && &Differentiating (2), cos sin cosc c bφφ φ θθ+ =& &&( )244 cos sin sin cosb bb b bc cθ φ θ θ θ θθ  + + − =     & & &( )2 2 3 22cos 4 sin4 cosb c bb bcθ θθ φ θ++ =& &( ) 217 8cos cos 4sin17 8cosbθ θ θθθ+ +=+&24 17cos 4cos17 8cosbθ θθθ+ +=+&( )( )4 cos 1 4cos17 8cosbθ θθθ+ +=+&1 4cos17 8cosθφ θθ+=+& &1 4cos17 8cosDEθωθ+=+ωωωω
  • 164. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 137.From horizontal distances,( )sin 1 sincos coscoscosl bl bblφ θφφ θθθφ θφ= +==& && &From vertical distances,( )cos cossin sincossin sincosy l by l bbl blφ θφϕ θθθφ θ θθϕ= −= − += − +&& && &[ ]sin tan cosbθ θ φ θ= −&(1)Now, ,Dy v θ ω= =&&From geometry, ( )sin 1 sinblφ θ= +( )( )( )1/22221/222 2cos 1 1 sin1 sinsintancos1 sinblbl bφ θθφφφθ = − +  += = − +  Substituting into (1),( )( )1/222 21 sin cossin1 sinDbv bl bθ θω θθ  + = −   − +    
  • 165. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 138.From horizontal distances, ( )sin 1 sinl bφ θ= +cos coscoscosl bblφφ θθθφ θφ==& && &Set constant.θ ω= =&( ) ( )222coscoscos sin cos sincoscos sin sincoscoscos sin cos sincos coscosblblblb bl lω θφφφ θθ θ φφωφφω θ φ θθφφφω θ φ ω θ θωφ φφ=− − −= = −   = ⋅ −  && &&&&&2 23cos sin sincoscosb bl lω θ φ θφφ = −  (1)Now, BDφ α=&&From geometry, ( )sin 1 sinblφ θ= +( )1/21/2 22 2 21cos 1 sin 1 sinl blφ φ θ  = − = − +    Substituting into (1),( )( ) ( )223/2 1/22 22 2 2 2cos 1 sin sin1 sin 1 sinBDb b lb lll b l bθ θω θαθ θ  + = −     − + − +        ( )( ) ( )3 223/2 1/22 22 2 2 2cos 1 sin sin1 sin 1 sinBDb bl b l bθ θ θα ωθ θ  + = −     − + − +        
  • 166. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 139.,A Ax r y rθ= =sinP Ax x r θ= −sinr rθ θ= −cosP Ay y r θ= −cosr r θ= −Axrθ =, 0,,cos 1 cosA AAP xvx v yrvtx vtrvt vx v r r rr rθθθ θθ= = == = = = − = −  && && &&1 cosxvtv vr = −  sin sinP yvt vy v r rr rθθ = = =   &&sinyvtv vr=
  • 167. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 140.tan cot AAb xux bθ θ= = =1cot uθ −=21uuθ = −+&&( )( )2 22211uu u uuuθ = −++& & &&&&But, andω θ α θ= =& &&, , 0A A A Ax x v vu u ub b b b= = = − = − =& && &&Then,( )2 2 2,1AAvAbx Abbvb xω = =++2 2AAbvb x=+ωωωω( )( )( ) ( )222 22 2 22 20 ,1A AAx vb b A Ax Abbx vb xα = − =  ++  ( )222 22 A AAbx vb x=+αααα
  • 168. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 141.( ) ( )1/2 1/22 2 2 2sin , cos AA Ab xb x b xθ θ= =+ +( )1/22 2cosB AAAAx l xlxxb xθ= −= −+( )1/22 2sinBAlby lb xθ= =+( ) ( ) ( )21/2 3/2 3/22 2 2 2 2 2A A A A AB A AA A Alx lx x x lb xx x xb x b x b x= − − = −+ + +& & && & &( )3/22 2A ABAlbx xyb x= −+&&But, ( ) ( ), ,A A B B B Bx yx v x v y v= − = =& & &( )( )23/22 2AB AxAlb vv vb x= −+( )( )3/22 2A AB yAlbx vvb x=+
  • 169. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 142.costan sinCb byθθ θ= =( )( ) ( )( )2 2sin sin cos cossin sinCCdy d b dv bdt dt dtθ θ θ θ θ θθ θ− −= = = −2 20sin sinCd v vdt b bθ θ θω= − = − =Angular acceleration.20 02 sin cos sind d d v vdt d dt b bω ω θ θ θ θαθ − − = = =       2302 sin cosvbθ θ =   αααα
  • 170. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 143.From geometry,sinryθ=2cossindy r ddt dtθ θθ= −0But, anddy dvdt dtθω= − =0 2cossinrvθωθ=20 sincosvrθωθ=Angular acceleration.d d d ddt d dt dω ω θ ωα ωθ θ= = =( )2 3 20 022cos sin sin sincoscosv vr rθ θ θ θθθ+  =    ( )2 32031 cos sincosvrθ θθ+ =   ( )22 301 cos tanvrθ θ = +  αααα
  • 171. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 144.Law of cosines for triangle ABE.( )2 2 22 cos 180u l b bl θ= + − ° −2 22 cosl b bl θ= + +coscosθφ+=l busintancosθφθ=+bl b( )( )( ) ( )( )( )22cos cos sin costan seccosθ θ θ θ θ θφ φφθ+ += =+& && l b b b bddt l b( ) ( )( )2 2 2 22cos cos cos sincosφ θ θ θ θφθ + + =+&&bl bl b( )22 2 2coscos2 cosb b lbl bu l b blθθθ θθ++= =+ +& &But, , , andθ ω φ ω= = = −& & &BD Ev uHence,( )2 2cos2 cosBDb b ll b blθωθ+=+ +ωωωωDifferentiate the expression for 2.u2 2 sinuu bl θθ= − &&( )1/22 2sin2 cosEblv ul b blθωθ= − =+ +&( )1/22 2sin2 cosEbll b blω θθ=+ +v 1 sintancosbl bθθ−   + 
  • 172. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 145.Law of cosines for triangle ABE.( )2 2 22 cos 180u l b bl θ= + − ° −2 22 cosl b bl θ= + +coscosl buθφ+=sintancosbl bθφθ=+( )( )( ) ( )( )( )22cos cos sin costan seccosl b b b bddt l bθ θ θ θ θ θφ φφθ+ += =+& &&( ) ( )( )2 2 2 22cos cos cos sincosbl bl bφ θ θ θ θφθ + + =+&&( )22 2 2coscos2 cosb b lbl bu l b blθθθ θθ++= =+ +& &( )2 2cos2 cosb b ll b blθφ θθ+=+ +&& &&( )( ) ( )( )( )2 2222 22 cos sin cos 2 sin2 cosl b bl bl b b l bll b blθ θ θ θθθ+ + − − + −++ +&( ) ( )( )2 2222 2 2 2sincos2 cos 2 cosbl l bb b ll b bl l b blθθθ θθ θ−+= −+ + + +&& &But, , 0, BDθ ω θ ω φ α= = = =& && &&&( )2 222 2sin2 cosBDbl l bl b blθωθ−=+ +αααα
  • 173. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 146.Define angles andθ φ as shown., tθ ω θ ω= =&Since the wheel rolls without slipping, the arc OC is equal to arc PC.( )r Rφ θ φ+ =rR rθφ =−r rR r R rθ ωφ = =− −&&r tR rωφ =−( )sin sinPx R r rφ θ= − −( ) ( )cos cosP Pxv x R r rφφ θθ= = − −& &&( ) ( )( )cos cosr t rR r r tR r R rω ωω ω  = − −  − −  ( ) cos cosP xr tv r tR rωω ω = − − ( )cos cosPy R R r rφ θ= − − −( ) ( )sin sinP Pyv y R r rφφ θθ= = − +& &&( ) ( )( )sin sinr t rR r r tR r R rω ωω ω  = − +  − −  ( ) sin sinP yr tv r tR rωω ω = + − 
  • 174. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 147.Define angles andθ φ as shown., , 0tθ ω θ ω θ= = =& &&Since the wheel rolls without slipping, the arc OC is equal to arc PC.( ) 2r R rφ θ θ θ+ = =0φ =φ θ ω= =& &0φ θ= =&& &&( )sin sin sin sin 0Px R r r r rφ θ θ θ= − − = − =The path is the -axis.y( )cos cos cos cosPy R R r r R r rφ θ θ θ= − − − = − −( )1 cosR θ= −sinPv y R θθ= = && ( )sinR tω ω=v j( )2 2cos sin cosa v R Rθθ θθ ω θ= = − =& &&&( )2cosR tω ω=a j
  • 175. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 148.Geometry.180 120 20 40APB = ° − ° − ° = °Law of sines.sin 40 sin 20 sin120b AP PB= =° ° °0.5321 0.15963 m1.3473 0.40419 mAP bBP b= == =Let P′ be the point on rod AD that coincides with P at the instant shown.Let u = u 60° be the velocity of P relative to rod AD.( )/P P P D AAP ω′ = + = v v v ] [30 u° + ]60°[1.5963 m/s= ] [30 u° + ]60°But collar P is attached to rod BP. Β Bω=ωωωω( ) 0.40419P B Bv BP ω ω= =0.40419P Bω=v 70°Equating the two vectors for vP gives the vector diagram shown.From the vector diagram,1.59680.40419cos40Bω =°(a) 5.16 rad/sBω =1.5968 tan 40 1.340 m/su = ° =(b) 1.340 m/sP/AD =v 60°
  • 176. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 149.Geometry.30θ = °250288.68 mmcos30AP = =°250tan30 144.34 mmBP = ° =Velocity analysis. 4 rad/sA =ωωωω , 5 rad/sB =ωωωωRod AE. Let P′ be the point on rod AE coinciding with the pin P.( ) ( )( )288.68 4 1154.70 mm/sP Av AP ω′ = = =1154.70 mm/sP′ =v θ/P/AE P AEv=v θP P P/AE′= +v v v[1154.70 mm/s= θ ] /P AEv+  ]θRod BD. Let P′′ be the point on rod BD coinciding with the pin P.( ) ( )( )144.34 5 721.69 mm/sP Bv BP ω′′ = = =721.69 mm/sP′′ =v/P/BD P BDv=vP P P/BD′′= +v v v[721.69 mm/s= ] /P BDv+  ]
  • 177. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Equate the two expressions for vP and resolve into components. ( )30θ = °: /1154.70 sin30 cos30 721.69 0P AEv− ° + = − +/ 166.67 mm/sP AEv = − / 166.67 mm/sP AE =v 30°:θ /1154.70 0 721.69 sin30 cos30P BDv+ = ° + °/ 916.67 mm/sP BDv = / 916.67 mm/sP BD =v[1154.70P =v ] [30 166.67° + ]30°[721.69 mm/s= ] [916.67 mm/s+ ]1167 mm/sP =v 51.8°
  • 178. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 150.Coordinates.( )0,0,, 0sincosA A AB BC A CP APx x r y rx y rx x yx x ey r eθθθ= + == == == += +Data: ( )024 in.Ax =10 in.7 in.re==Velocity analysis. AC ACω=ωωωω , BD BDω=ωωωω ,[/P A P A ACrω= + =v v v ] [ ACeω+ ]θ[P P BDx ω′ =v ] ( )cos BDe θ ω+  ]/ [ cosP F u β=v ] [ sinu β+ ]Use /P P P F′= +v v v and resolve into components.( ) ( ) ( ): cos cos cosAC BDr e e uθ ω θ ω β+ = + (1)( ): sin ACe θ ω = P BDx ω ( )sin uβ− (2)
  • 179. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(a) 0.θ = 24 in. 24 in. 20 rad/sA P ACx x ω= = =cos 7tan , 16.2624Pexθβ β= = = °Substituting into Eqs. (1) and (2),( )( ) ( )10 7 20 7 cos16.26BD uω+ = + ° (1)( )0 24 sin16.26BD uω= − ° (2)Solving simultaneously, 3.81 rad/s,BDω = 3.81 rad/sBDω =326.4 in./s,u = / 27.2 ft/sP F =v 16.26°( )b 90 .θ = ° ( )24 10 7 46.708 in.2Pxπ = + + =  0β =Substituting into Eqs. (1) and (2),( )( )10 20 u= (1)200 in./s = 16.67 ft/su =( )( )7 20 46.708 BDω= (2)2.9973 rad/s,BDω = 3.00 rad/sBD =ωωωω/ 16.67 ft/sP F =v
  • 180. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 151.Coordinates.( )0,0,, 0sincosA A AB BC A CP APx x r y rx y rx x yx x ey r eθθθ= + == == == += +Data: ( )024 in.Ax =10 in.7 in.re==Velocity analysis. AC ACω=ωωωω , BD BDω=ωωωω ,[/P A P A ACrω= + =v v v ] [ ACeω+ ]θ[P P BDx ω′ =v ] ( )cos BDe θ ω+  ]/ [ cosP F u β=v ] [ sinu β+ ]Use /P P P F′= +v v v and resolve into components.( ) ( ) ( ): cos cos cosAC BDr e e uθ ω θ ω β+ = + (1): ( )sin ACe θ ω = P BDx ω ( )sin uβ− (2)
  • 181. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )30 , 24 10 7sin30 32.736 in.6Pxπθ = ° = + + ° =  7cos30tan 10.49132.736β β°= = °Substituting into Eqs. (1) and (2)( )( ) ( ) ( )10 7cos30 20 7cos30 cos10.491BD uω+ ° = ° + ° (1)( )( ) ( )7sin30 20 32.736 sin10.491BD uω° = − ° (2)Solving simultaneously, 3.82 rad/sBDω =303.13 in./su = / 25.3 ft/sP F =v 10.49°
  • 182. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 152.10 rad/sEFω = , /E E F EFr ω= v ] ( )( )0.375 10 3.75 m/s.= =[/ 3.75 m/s.B E B E′ = + =v v v ] [0.5 DEω+ ]/B ED u=v[/ 3.75 m/s.B B B ED′= + =v v v ] [0.5 DEω+ ] [u+ ]( )B ABAB ω= v ( )0.37545 15cos45 ° =   ° [45 5.625 m/s° =] [5.625 m/s.+ ]Equate the two expressions for Bv and resolve into components.: 3.75 5.625, 1.875 m/s.u u+ = =: 0.5 5.625, 11.25 rad/sDE DEω ω= =(a) 11.25 rad/sDEω =(b) / 1.875 m/s.B DE =v
  • 183. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 153.15 rad/sEFω = /E E F EFr ω= v ( )( )0.375 15 5.625 m/s = =[/ 5.625 m/sB E B ED′ = + =v v v ] [0.5 DEω+ ] [5.625 m/s= ] [5 m/s+ ]/B ED u=v[/ 5.625 m/sB B B ED′= + =v v v ] [5 m/s+ ] [u+ ]( )B ABAB ω= v0.37545cos45ABω° =  °[45 0.375 ABω° =] [0.375 ABω+ ]Equate the two expressions for Bv and resolve into components.: 5.625 0.375 ABu ω− + = − (1):55 0.375 13.333 rad/s0.375AB ABω ω= = =(a) 13.33 rad/sABω =From (1), ( )( )5.625 0.375 13.333 0.625 m/su = − =(b) / 0.625 m/sB ED =v
  • 184. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 154.Bar AB. (Rotation about A) 8 rad/sABω =( )( )12 8 96 in./sB = =vRod EF. (Rotation about E ) 6 rad/s.EFω =( )( )12 6 72 in./sD′ = =vBar BD. Assume angular velocity is BDω .Plane motion = Translation with B + Rotation about B.[96D B D/B= + =v v v ] [24 BDω+ ] [12 BDω+ ] (1)Collar D. Sliding on rotating rod EF with relative velocity u .[72D D D/EF′= + =v v v ] [u+ ] (2)Matching the expressions (1) and (2) for v ,DComponents : 96 12 72BDω+ = − 14BDω = −(a) 14.00 rad/sBDω =Components : 24 BD uω = ( )( )24 14 336 in./su = − = −(b) = 28.0 ft/su
  • 185. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 155.Bar AB. (Rotation about A) 4 rad/sABω =( )( )1 ft 4 rad/s 4 ft/sB = =vBar BD. Angular velocity is BDω .Plane motion = Translation with B + Rotation about B.[4D B D/B= + =v v v ] [2 BDω+ ] [1 BDω+ ]Magnitude of : 20 ft/sD Dv =v( ) ( ) ( )2 2 224 2 20D BD BDv ω ω= + + =25 8 384 0BD BDω ω+ − =8 8810BDω− ±= Positive root 8 rad/sBDω =[4D =v ] ( )( )2 8+  ] ( )( )1 8+  ] [12= ] [16+ ] (1)Rod EF. (Rotation about E) Angular velocity = EFω( )1D EFω= v ]Collar D. Slides on rotating rod EF with relative velocity u .[1D D D/EF EFω= + =v v v ] [u+ ] (2)Matching the expressions (1) and (2) for ,Dv(a) Component : 12 1 EFω= 12.00 rad/sEFω =(b) Component : 16 = u 16 ft/su =
  • 186. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 156.Body ACD rotates about point A with angular velocity1.6 rad/sA =ωωωω ( )1.6 rad/s= kIf the rod EF were not moving relative to body ACD, the velocity of point H′ currently at point H, would be( )/ 1.6 600 0.96 mm/sH A H A′ = × = × =v r k i jωωωωThe velocity of point H relative to H′ is/ 300 mm/sH H′ =v ( )300 mm/s= i(a) Velocity of tip H.( ) ( )/ 960 mm/s + 300 mm/sH H H H′ ′= + =v v v j iH 1006 mm/s=v 72.6°Angular acceleration of body ACD: 0A =αααα( ) ( )22/ 0 1.6 600H A H/A A H Aω′ = × − = −a r r iαααα( )21536 mm/s= − iSince the sliding motion of H relative to H′ occurs at constant velocity,/ 0H H′ =aCoriolis acceleration. 2c A H/H ′= ×a vωωωω( )( ) ( ) ( )22 1.6 300 960 mm/sca = × =k i j(b) Acceleration of tip H./H H H H c′ ′= + +a a a a( ) ( )2 21536 mm/s + 0 + 960 mm/s= − i j21811 mm/sH =a 32.0°
  • 187. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 157.Body ACD rotates about point A with angular velocity1.6 rad/sA =ωωωω ( )1.6 rad/s= kIf the rod EF were not moving relative to body ACD, the velocity of point ,G′ currently at point G, would be( )1/ 1.6 600 400G A G A= × = × +v r k i jωωωω( ) ( )640 mm/s 960 mm/s= − +i jThe velocity of point G relative to G′ is/ 300 mm/sG G′ =v ( )300 mm/s= i(a) Velocity of tip G./ 640 960 300G G G G′ ′= + = − + +v v v i j i( ) ( )340 mm/s 960 mm/s= − +i j1018 mm/sG =v 70.5°Angular acceleration of body ACD is 0.A =αααα2/ /G A G A A G Aω′ = × −a r rαααα( ) ( )20 1.6 600 + 400= − i j( ) ( )2 21536 mm/s 1024 mm/s= − −i jSince the sliding motion of G relative to G′ occurs at constant velocity,/ 0G G′ =aCoriolis acceleration. /2c A G G′= ×a vωωωω( )( ) ( ) ( )22 1.6 300 960 mm/sca = × =k i j(b) Acceleration of tip G./G G G G c′ ′= + +a a a a1536 1024 0 960= − − + +i j j( ) ( )2 21536 mm/s 64 mm/s= − −i j21537 mm/sG =a 2.4°
  • 188. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 158.For each pin: /P P P F c′= + +a a a aAcceleration of the coinciding point P of the plate.′2For each pin, towards the center .P r Oω′ =aAcceleration of the pin relative to the plate.1 2 4For pins , , and ,P P P / 0P F =a3For pin ,P2/P Fur=a.cCoriolis acceleration aFor each pin 2ca uω= with ca in a direction obtained by rotating u through 90° in the sense of ,ωωωω i.e. .Then, 21 rω= a [2 uω + ] 21 2r uω ω= −a i j22 rω= a [2 uω + ] 22 2 u rω ω= −a i j23 rω= a2ur + [2 uω+]223 2ur urω ω = − + +   a i24 rω= a [2 uω + ] ( )24 2r uω ω= +a j
  • 189. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 159.For each pin: /P P P F c′= + +a a a aAcceleration of the coinciding point P of the plate.′2For each pin towards the center .P r Oω′ =aAcceleration of the pin relative to the plate.1 2 4For pins , , and ,P P P / 0P F =a3For pin ,P2/P Fur=a.cCoriolis acceleration aFor each pin 2ca uω= with ca in a direction obtained by rotating u through 90° in the sense of .ωωωωThen, 21 rω= a [2 uω + ] 21 2r uω ω= +a i j22 rω= a [2 uω + ] 22 2 u rω ω= − −a i j23 rω= a2ur + [2 uω+]223 2uu rrω ω = − −   a i24 rω= a [2 uω + ] ( )24 2r uω ω= −a j
  • 190. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 160.16 in. 1.3333 ft, 16 2 in. 1.3333 2 ftAP BP= = = =Given: 6 rad/sAE =ωωωω , /0, 8 ft/sAE P AE= =vαααα , / 0.P AE =aFind: /and .BD P BDaααααVelocity of coinciding point P′ on rod AE.( ) ( )( )1.3333 6 8 ft/sP AEAP ω′ = = =v = ( )8 ft/s iVelocity of P relative to rod AE. ( )/ 8 ft/sP AE =v jVelocity of point P. ( ) ( )/ 8 ft/s 8 ft/sP P P AE′= + = +v v v i jVelocity of coinciding point P′′ on rod BD.( )P BDBP ω′′ =v 45 1.3333 2 BDω° = 45 1.3333 1.3333BD BDω ω° = − +i jVelocity of P relative to rod BD. ( ) ( )/ cos45 sin 45P BD u u= ° + °v i jVelocity of point P. /P P P BD′′= +v v v( ) ( )1.3333 1.3333 cos45 sin 45P BD BD u uω ω= − + + ° + °v i j i jEquating the two expressions for Pv and resolving into components.( ): 8 1.3333 cos45BD uω= − + °i (1)( ): 8 1.3333 sin 45BD uω= + °j (2)Solving (1) and (2), ( ) ( )/0, 8 2 ft/s, 8 ft/s 8 ft/sBD P BDuω = = = +v i jAcceleration of coinciding point P′ on rod AE.( ) ( ) ( )( ) ( )22 20 1.3333 6 48 ft/sP AE AEAP APα ω′ = − = − = −a i j j j
  • 191. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Acceleration of P relative to rod AE. / 0P AE =aCoriolis acceleration. ( )( ) ( )2/2 2 6 8 96 ft/sAE P AE× = − × =v k j iωωωωAcceleration of point P.( ) ( )2 2/ /2 96 ft/s 48 ft/sP P P AE AE P AE′= + + × = −a a a v i jωωωωAcceleration of coinciding point P′′ on rod BD.2/ / 1.3333 1.3333 0P BD P B BD P B BD BDα ω α α′′ = × − = − + +a k r r i jAcceleration of P relative to rod BD. ( ) ( )/ cos45 sin 45P BD r ra a= ° + °a i jCoriolis acceleration. /2 0BD P BD× =vωωωωAcceleration of point P./ /2P P P AE BD P BD′′= + + ×a a a vωωωω( ) ( )1.3333 1.3333 cos45 sin 45BD BD r ra aα α= − + + ° + °i j i jEquating the two expressions for Pa and resolving into components.( ): 96 1.3333 cos45BD raα= − + °i (3)( ): 48 1.3333 sin 45BD raα− = + °j (4)Solving (3) and (4), 254 rad/s , 24 2 ft/sBD raα = − =(a) 254 rad/sBD =α(b) 2/ 33.9 ft/sP BD =a 45°
  • 192. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 161.16 in. 1.3333 ft, 16 2 in. 1.3333 2 ftAP BP= = = =Given: 6 rad/sBD =ωωωω , /0, 10.5 ft/sBD P BD= =vαααα /45 , 0.P BD° =aFind: /and .AE P AEα aVelocity of coinciding point P′ on rod AE.( ) 1.3333P AE AEAP ω ω′ = =v 1.3333 AEω= − iVelocity of P relative to rod AE. /P AE u=v jVelocity of point P. / 1.3333P P P AE AE uω′= + = − +v v v i jVelocity of coinciding point P′′ on rod BD.( )P BDBP ω′′ =v ( )( )45 1.3333 2 6° = ( ) ( )45 8 ft/s 8 ft/s° = − +i jVelocity of P relative to rod BD. ( ) ( )/ 5.25 2 ft/s 5.25 2 ft/sP BD = − −v i jVelocity of point P. ( ) ( )/ 15.4246 ft/s 0.57538 ft/sP P P BD′′= + = − +v v v i jEquating the two expressions for Pv and resolving into components.: 1.3333 15.4246 11.5685 rad/sAE AEω− = − =i ω: 0.57538u = −j / 0.57538 ft/sP AE =vAcceleration of coinciding point P′ on rod AE.( )( )22/ / 1.3333 1.3333 11.5685P AE P A AE P A AEω α′ = × − = − −a k r r i jαααα1.3333 178.440AEα= − −i j
  • 193. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Acceleration of P relative to rod AE. /P AE ra=a jCoriolis acceleration. ( )( )( ) ( )2/2 2 11.5685 0.57538 13.3126 ft/sAE P AEω × = − = −k v i iAcceleration of point P. / /2P P P AE AE P AE′= + + ×a a a vωωωω( )1.3333 13.3126 178.440P AE raα= − + + − −a i j i jAcceleration of coinciding point P′′ on rod BD.( ) ( ) ( ) ( )22 2 2/ / 0 6 1.3333 1.3333 48 ft/s 48 ft/sP BD P B BD B Dα ω′′ = × − = − + = − −a k r r i j i jAcceleration of P relative to rod BD. / 0P BD =aCoriolis acceleration. ( )( )( )/2 2 6 10.5 126BD P BDvω = = 45°( ) ( )2 2126cos45 ft/s 126sin 45 ft/s= ° − °i jAcceleration of point P. / /2P P P AE BD P BDvω′′= + +a a a( ) ( )2 248 48 126cos45 126sin 45 41.0955 ft/s 137.0955 ft/sP = − − + ° − ° = −a i j i j i jEquating the two expressions Pa and resolving into components.: 1.3333 13.3126 41.0955AEα− − =i 240.81 rad/sAEα = −: 178.440 137.0955ra − = −j 241.345 ft/sra =(a) 240.8 rad/sAE =αααα(b) 2/ 41.3 ft/sP AE =a
  • 194. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 162.Earth makes one revolution ( )2 radiansπ in 23.933 h = 86160 s.( )6272.926 10 rad/s86160π −= = ×Ω jSpeed of sled. 600 mi/hr 880 ft/su = =Velocity of sled relative to the Earth.( )/earth 880 sin cosφ φ= − +Pv i jCoriolis acceleration. /earth2c P= ×a vΩΩΩΩ( )( ) ( )62 72.926 10 880 sin cos0.12835sinc φ φφ− = × × − + =a j i jkAt latitude , 40 ,φ = °0.12835sin 40c = °a k20.0825 ft/s= k20.0825 ft/s westc =a
  • 195. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 163.Earth makes one revolution ( )2 radiansπ in 23.933 h (86160 s).( )6272.926 10 rad/s86160π −= = ×j jΩΩΩΩVelocity relative to the Earth at latitude angle .φ( )/earth 12.2 cos sinP φ φ= − −v i jCoriolis acceleration .ca( )( ) ( )( )/earth6322 72.926 10 12.2 cos sin1.7794 10 cosc Pφ φφ−−= × = × × − − = ×a vj i jkΩΩΩΩ(a) 0 , cos 1.000φ φ= ° = 3 21.779 10 m/s westc−= ×a(b) 40 , cos 0.76604φ φ= ° = 3 21.363 10 m/s westc−= ×a(c) 40 , cos 0.76604φ φ= − ° = 3 21.363 10 m/s westc−= ×a
  • 196. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 164.Geometry: ( ) ( )/ 16 3 in. 16 in.P A = +r i j.Motion of coinciding point P on rod AB′( ) ( ) ( ) ( )( ) ( )( ) ( )/22/ /2 25 16 3 16 80 in./s 80 3 in./s0 5 16 3 16400 3 in./s 400 in./sP AB P AP AB P A AB P Aω′′= × = × + = − += × − = − += − −v v k i j i ja r r i ji jωωωωααααMotion of collar P relative to rod AB. 6 ft/s 72 in./s=( ) ( )/ /72cos30 72sin30 36 3 in./s 36 in./s , 0P AB P AB= − ° − ° = − − =v i j i j aCoriolis acceleration. /2 AB P AB× vωωωω( )( ) ( ) ( ) ( )2 22 5 72cos30 72sin30 360 in./s 360 3 in./s× − ° − ° = −k i j i j(a) Motion of collar P./ 80 80 3 36 3 36P P P AB′= + = − + − −v v v i j i j( ) ( )142.35 in./s 102.56 in./s ,= − +i j 14.62 ft/sP =v 35.8°/ /2 400 3 400 0 360 360 3P P P AB AB P AB′= + + × = − − + + −a a a v i j i jωωωω( ) ( )2 2332.82 in./s 1023.54 in./s ,= − −i j 289.7 ft/sP =a 72.0°(b) Motion of point D./ 142.35 102.56 8D P D P PDω= + = − + +v v v i j j16D DEω= −v iEquating the two expressions for Dv and resolving into components,: 142.35 16 DEω− = −i 8.897 rad/sDEω =: 0 102.56 8 12.820 rad/sPD PDω ω= + + = −j( )( ) ( )16 8.897 142.35 in./sD = − = −v i i 11.86 ft/sD =v
  • 197. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( )( )2/28 8332.82 1023.54 8 1314.8216 16 16 1266.5D P P D P PD PDPDD DE DE DEα ωαα ω α= + = + × −= − − + −= × − = − −a a a a k i ii j j ia k j j i jEquating the two expressions for and resolving into components,Da: 1647.64 16 ,DEα− = −i 2102.98 rad/sDEα =2: 1023.54 8 1266.5, 30.375 rad/sPD PDα α− + = − = −j( )( ) ( ) ( )2 216 102.98 1266.5 1647.7 in./s 1266.5 in./sD = − − = − −a i j i j2173.2 ft/sD =a 37.5°
  • 198. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 165.Geometry. ( ) ( )/ 16 3 in. 16 in.P A = +r i jMotion of coinciding point P on rod AB.′( ) ( ) ( ) ( )/ 4 16 3 16 64 in./s 64 3 in./sP AB P A′ = × = − × + = −v r k i j i jωωωω( ) ( )22/ / 0 4 16 3 16P AB P A AB P Aω′ = × − = − +a r r i jαααα( ) ( )2 2256 3 in./s 256 in./s= − −i jMotion of collar P relative to rod AB. 9.6 ft/s 115.2 in./s=( ) ( )/ /115.2cos30 115.2sin30 57.6 3 in./s 57.6 in./s , 0P AB P AB= − ° − ° = − − =v i j i j aCoriolis acceleration. /2 AB P AB× vωωωω( )( ) ( ) ( ) ( )2 22 4 115.2cos30 115.2sin30 460.8 in./s 460.8 3 m/s− × − ° − ° = − +k i j i jMotion of collar P./ 64 64 3 57.6 3 57.6P P P AB′= + = − − −v v v i j i j( ) ( )35.7661in./s 168.451in./s= − −i j/ /2 256 3 256 0 460.8 460.8 3P P P AB AB P AB′= + + × = − − + − +a a a v i j i jωωωω( ) ( )2 2904.205 in./s 542.129 in./s= − +i j/ 35.7661 168.451 8D P D P PDω= + = − − +v v v i j j16D DEω= −v iEquating the two expressions for Dv and resolving into components,: 35.7661 16 ,DEω− = −i 2.2354 rad/sDEω =: 0 168.451 8 ,PDω= − +j 21.0564 rad/sPDω =( ) ( )2/ 8 8D P P D P PD PDα ω= + = + × −a a a a k i i904.205 542.129 8 3546.98PDα= − + + −i j j i( )216 16 16 79.952D DE DE DEα ω α= × − = − −a k j j i j
  • 199. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Equating the two expressions for Da and resolving into components,: 4451.18 16 ,DEα− = −i 2278.20 rad/sDEα =: 542.129 8 79.952,PDα+ = −j 277.76 rad/sPDα = −(a) Angular velocities. 21.1 rad/sPD =ωωωω , 2.24 rad/sDE =ωωωω(b) Angular accelerations. 277.8 rad/sPD =αααα , 2278 rad/sDE =αααα
  • 200. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 166.For each configuration ( ) ( )/ 275 mm 100 mmD A = +r i jAcceleration of coinciding point .D′ 2/ /D D A D Aα ω′ = × −a k r r( ) ( ) ( ) ( )2 2 20 2.4 275 100 1584 mm/s 576 mm/sD′ = − + = − −a i j i jAcceleration of point D relative to arm AB. / 0D AB =aLength .CD 2 275 100 125 mmCD = + =Velocity of point D relative to the arm AB.Case (a)( )/ 250 mm/sD AB =v iCase (b) ( ) ( ) ( )/25075 100 150 mm/s 200 mm/s125D AB = + = +v i j i jCoriolis acceleration. /2 D Bω ×k vCase (a) ( )( ) ( )22 2.4 250 1200 mm/s× =k i jCase (b) ( )( ) ( ) ( ) ( )2 22 2.4 150 200 960 mm/s 720 mm/s× + = − +k i j i jAcceleration of nozzle D. / /2D D D AB D AB′= + + ×a a a k vωωωω(a) ( ) ( )2 21584 576 1200 1584 mm/s 624 mm/sD = − − + = − +a i j j i j( ) ( )2 2 21584 624 1702 mm/sD = + =a624tan , 21.51584β β= = °21.702 m/sD =a 21.5°(b) ( ) ( )1584 576 960 720 2544 mm/s 144 mm/sD = − − − + = − +a i j i j i j( ) ( )2 2 22544 144 2548 mm/sD = + =a144tan , 3.242544β β= = °22.55 m/sD =a 3.24°
  • 201. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 167.For each configuration ( ) ( )/ 275 mm 100 mmD A = +r i j.Acceleration of coinciding point D′ 2/ /D D A D Aα ω′ = × −a k r r( ) ( ) ( ) ( )2 2 20 2.4 275 100 1584 mm/s 576 mm/sD′ = − + = − −a i j i j.Acceleration of point D relative to arm AB / 0D AB =aLength .CD 2 275 100 125 mmCD = + =Velocity of point D relative to the arm AB.Case (a) ( )/ 250 mm/sD AB = −v iCase (b) ( ) ( ) ( )/25075 100 150 mm/s 200 mm/s125D AB = + = − −v i j i jCoriolis acceleration. /2 D Bω ×k vCase (a) ( )( ) ( ) ( )22 2.4 250 1200 mm/s× − = −k i jCase (b) ( )( ) ( ) ( ) ( )2 22 2.4 150 200 960 mm/s 720 mm/s× − − = −k i j i jAcceleration of nozzle D. / /2D D D AB D AB′= + + ×a a a k vωωωω(a) ( ) ( )1584 576 1200 1584 mm/s 1776 mm/sD = − − − = − −a i j j i j( ) ( )2 2 21584 1776 2380 mm/sD = + =a1776tan , 48.3 ,1584β β= = ° 22.38 m/sD =a 48.3°(b) ( ) ( )2 21584 576 960 720 624 mm/s 1296 mm/sD = − − + − = − −a i j i j i j( ) ( )2 2 2624 1296 1438 mm/sD = + =a1296tan , 64.3 ,624β β= = ° 21.438 m/sD =a 64.3°
  • 202. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 168.Rod AB. (Rotation about A ) 15 rad/sAB =ωωωω , 0ABα =( )( )0.4 sin 45 15 4.2426 m/sE Ev ′ ′= ° =v( ) ( )( ) ( )0.4sin 45 0 0E Et ta ′ ′= ° =a( ) ( )( ) ( )2 20.4sin 45 15 63.64 m/sE En na ′ ′= ° =aCollar E. (Slides on rotating rod AB) 3 m/su = , 0u =&/ [4.2426E E E AB′= + =v v v ] [3+ ]Coriolis acceleration: ( )( )( )2 2 15 3 90 m/sABuω = =290 m/sc =a/E E E AB c′= + +a a a a[63.64= ] [u+ & ] [90+ ] [63.64= ] [90+ ]Rod DE. Angular velocity and acceleration: DEωωωω , DEααααD Dv=v , D Da=aPlane motion Translation with Rotation aboutE E.= +/D E D E= +v v v[ Dv ] [4.2426= ] [3+ ] [0.4 DEω+ 45 ]°Components : 0 3 0.4 sin 45 10.6066DE DEω ω= + ° = −( ) ( )/ / /D E D E E D E D Et n= + = + +a a a a a a[ Da ] [63.64= ] [90+ ] 0.4 DEα+ 245 ] [0.4 DEω° + 45 ]°Components : 0 63.64 0.4 sin 45 45sin 45DEα= + ° − °2112.5 rad/sDEα = −(a) Angular velocity of rod DE. 10.61 rad/sDE =ωωωω(b) Angular acceleration of rod DE. 2112.5 rad/sDE =αααα
  • 203. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 169.Rod AB. (Rotation about A ) 10 rad/sABω = , 0ABα =( )( )0.4 sin 45 10 2.8284 m/sE Ev ′ ′= ° =v( ) ( )( ) ( )0.4sin 45 0 0E Et ta ′ ′= ° =a( ) ( )( ) ( )2 20.4sin 45 10 28.284 m/sE En na ′ ′= ° =aCollar E. (Slides on rotating rod AB) 2 m/su = , 0u =&/ [2.8224E E E AB′= + =v v v ] [2+ ]Coriolis acceleration: ( )( )( ) 22 2 10 2 40 m/sABuω = =240 m/sc =a/E E E AB c′= + +a a a a[28.284= ] [u+ & ] [40+ ] [28.284= ] [40+ ]Rod DE. Angular velocity and acceleration: DEωωωω , DEααααD Dv=v , D Da=aPlane motion Translation with Rotation aboutE E.= +/D E D E= +v v v[ Dv ] [2.8284= ] [2+ ] [0.4 DEω+ 45 ]°Components : 0 2 0.4 sin 45 7.0711DE DEω ω= − + ° =Components : ( )( )2.8284 0.4 7.07011 cos45 4.8284D Dv v= + ° =( ) ( )/ / /D E D E E D E D Et n= + = + +a a a a a a[ Da ] [28.284= ] [40+ [] 0.4 DEα+ 245 ] [0.4 DEω° + 45 ]°
  • 204. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Components : 0 28.284 0.4 sin 45 20sin 45DEα= + ° − °50 rad/sDEα = −Components : ( )( )40 0.4 50 cos45 20cos45Da = + − ° + °40Dα =(a) Velocity of collar D. 4.83 m/sD =v(b) Acceleration of collar D. 240.0 m/sD =a
  • 205. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 170.6 rad/sω = , 0, 180 mm/s 0.18 m/s, 0u uα = = = =&( )222 2180200 mm, 162 mm/s 0.162 m/s200uρρ= = = =( )( )( )2 2 2 2 236 rad /s , 2 2 6 180 2160 mm/s 2.16 m/suω ω= = = =(a) Point A. /0, 0.18 m/sA A F= =r v2/0,A A Fuρ′ = =a a 20.162 m/s=Coriolis acceleration. 2 uω 22.16 m/s=[/ 2A A A F uω′= + +a a a ] 22.322 m/s= 22.32 m/sA =a(b) Point B. 0.2 2 mB =r /45 , 0.18 m/sB F° =v( )( )236 0.2 2B Bω′ = − = −a r 245 7.2 2 m/s° = 45°22/ 0.162 m/sB Fuρ= =aCoriolis acceleration. 22 2.16 m/suω =[/ 2B B B F uω′= + +a a a ] 29.522 m/s= 27.2 m/s + 211.94 m/sB =a 37.1°(c) Point C. 0.4 mC =r /, 0.18 m/sC F =v( )(236 0.4C Cω′ = − = −a r ) 214.4 m/s=22/ 0.162 m/sC Fuρ= =aCoriolis acceleration. 22 2.16 m/suω =[/ 2C C C F uω′= + +a a a ] 216.722 m/s= 216.72 m/sC =a
  • 206. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 171.( )( )20 2 220 rpm rad/s, 0, 90 radians60 3 2π π πω α θ= = = = = ° =.Uniform rotational motion 0 tθ θ ω= +0 2230.75 stππθ θω−= = =.Uniform motion along rod 0r r ut= +0/20 10 40 40in./s, in./s0.75 3 3P ABr rut− −= = = =vAcceleration of coinciding point P on the rod.′ ( )20 in.r =( )2 22 22 8020 in./s3 9P rπ πω′ = = =  a 287.730 in./s=.Acceleration of collar P relative to the rod / 0P AB =aCoriolis acceleration. ( ) 2/2 402 2 2 55.851 in./s3 3P AB uπω  × = = =    vωωωω.Acceleration of collar P / /2P P P AB P AB′= + + ×a a a vωωωω287.730 in./sP= a 255.851 in./s + 2104.0 in./sP =a 57.5° 2104.0 in./sPa =
  • 207. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 172.Given:75 mm/sB =v0B =a150 mmb =Definitions:ω=ωωωω , α=αααα902θφ = ° −2 cosAB b φ=Point B′ is point on rod AD that currentlycoincides with pin B.Velocity analysis /B B B AD′= +v v vB Bv=v( ) ( )2 cosB AB bω φ ω′ = =v/ /B AD B AD=v v φ[ Bv ( )] [ 2 cosb φ ω= /] [ B ADv+ ]φComponents.( ): cos 2 cos sin 0Bv bφ φ φ ω φ = + 2 sinBvb φ=ω (1)( ) /: 0 2 cos cosB ADb vφ ω φ= −/ 2sinBB ADvbωφ= =v φ (2)
  • 208. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Acceleration analysis. /B B B AD c′= + +a a a a0B =a[( )B AB α′ =a 2] ( )[AB ω+ ] [(2 cos )b φ α= 2] [(2 cos )b φ ω+ ]/ /[( )B AD B AD ta=a /] [( )B AD naφ + /] [( )B AD taφ =2/]B ADvφρ+ φ/[( )B AD ta=22]sinBvbφφ+ /[2c B ADvω=a22]sinBvbφφ= φComponents.2: 0 (2 cos ) sin (2 cos ) cosb bφ φ α φ φ ω φ= −2 22 20sin sinB Bv vb bφ φ+ − +2 cossinφα ωφ= (3)Data: 75 mm/s, 150 mm, 110Bv b θ= = = °90 55 35φ = ° − ° = °(a) From (1),( )( )750.43586 rad/s2 150 sin35ω = =°0.436 rad/s=ω(b) From (3), ( )2 2cos350.43586 0.271 rad/ssin35α°= =°20.271 rad/s=α
  • 209. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 173.Given:75 mm/sB =v0B =a150 mmb =Definitions:ω=ωωωω , α=αααα902θφ = ° −2 cosAB b φ=Point B′ is point on rod AD that currentlycoincides with pin B.Velocity analysis /B B B AD′= +v v vB Bv=v( ) ( )2 cosB AB bω φ ω′ = =v/ /B AD B AD=v v φ[ Bv ( )] [ 2 cosb φ ω= /] [ B ADv+ ]φComponents.( ): cos 2 cos sin 0Bv bφ φ φ ω φ = + 2 sinBvb φ=ω (1)( ) /: 0 2 cos cosB ADb vφ ω φ= −/ 2sinBB ADvbωφ= =v φ (2)
  • 210. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Acceleration analysis. /B B B AD c′= + +a a a a0B =a[( )B AB α′ =a 2] ( )[AB ω+ ] [(2 cos )b φ α= 2] [(2 cos )b φ ω+ ]/ /[( )B AD B AD ta=a /] [( )B AD naφ + /] [( )B AD taφ =2/]B ADvφρ+ φ/[( )B AD ta=22]sinBvbφφ+ /[2c B ADvω=a22]sinBvbφφ= φComponents.2: 0 (2 cos ) sin (2 cos ) cosb bφ φ α φ φ ω φ= −2 22 20sin sinB Bv vb bφ φ+ − +2 cossinφα ωφ= (3)Data: 75 mm/s, 150 mm, 90Bv b θ= = = °90 45 45φ = ° − ° = °(a) From (1),( )( )750.35355 rad/s2 150 sin 45ω = =°0.354 rad/s=ω(b) From (3), ( )2 2cos450.35355 0.1250 rad/ssin 45α°= =°20.1250 rad/s=α
  • 211. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 174.Coordinates.( )0,0,, 0sin , cosA A AB BC A CP A Px x r y rx y rx x yx x e y r eθθ θ= + == == == + = +Data: ( )024 in.Ax =10 in.r =7 in.e =0 24 in.Pxθ = =Velocity analysis.20 rad/sAC =ωωωω , BD BDω=ωωωω( )P ACr e ω= +v( )( )10 7 20= +340 in./s=[P P BDx ω′ =v ] [ BDeω+ ][/ cosP F u β=v ] [ sinu β+ ]7tan2416.260Pexββ= == °Use /P P P F′= +v v v and resolve into components.: 340 7 cosBD uω β= + (1): 0 24 sinBD uω β= − (2)Solving (1) and (2), 3.8080 rad/s, 326.4 in./sBD u= =ωωωω
  • 212. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Acceleration analysis. 0,AC BD BDα= =α αα αα αα α( )( )22 2/0 7 20 2800 in./sA P A ABrω= = = =a a2/ 2800 in./sP A P A= + =a a aP P BDx α′ = a [ Beα + [ 2P BDx ω +2BDeω + 24 BDα=  [7 BDα + ( )( )224 3.8080 +  ( )( )27 3.8080+  24 BDα=  [7 BDα + [ 2348.021in./s + [ 2101.506 in./s+ [/ cosP F u β=a & [ sinu β + & Coriolis acceleration.( )( )( ) [ 22 2 3.8080 326.4 2485.86 in./sBDuω = = ]βUse / 2P P P F BD uω′ = + + a a a ]β and resolve into components.: 0 7 348.021 cos 2485.86sinBD uα β β= − + +&or 7 cos 348.02BD uα β+ = −& (3): 2800 24 101.506 sin 2485.86cosBD uα β β= + − +&or 24 sin 312.07BD uα β− =& (4)Solving (3) and (4), 8.09 rad/s, 421.48 in./sBD uα = = −&( )a 28.09 rad/sBD =αααα !( )b 2/ 35.1 ft/sP F =a 16.26° !
  • 213. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 175.Coordinates.( )0,0,, 0sin , cosA A AB BC A CP A Px x r y rx y rx x yx x e y r eθθ θ= + == == == + = +Data: ( )024 in.Ax =10 in.r =7 in.e =( ) ( )90 24 10 7 46.708 in., 02Pxπθ β = ° = + + = =  Velocity analysis.20 rad/sAC =ωωωω , BD BDω=ωωωω/P A P A= +v v v[ ACrω= ] [ ACeω+ ]( )( )10 20=  ( )( )7 20+  [200 in./s= [140 in./s] + ]p BDP x ω′ = v 46.708 BDω =  , /P F u=vUse /P P P F′= +v v v and resolve into components.: 200 u= 200 in./s.u =: 140 46.708 BDω= 2.9973 rad/sBDω =Acceleration analysis. 0,AC BD BDα= =α αα αα αα α( )( )22 2/0, 7 20 2800 in./sA P A ABrω= = = =a a2/ 2800 in./sP A P A= + =a a aP P BDx α′ = a 2P BDx ω +  [46.708 BDα = ( )( )246.708 2.9973+  46.708 BDα= 2419.616 in./s + /P F u=a &
  • 214. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Coriolis acceleration ( )( )( ) 22 2 2.9973 200 1198.92 in./sBDuω = =Use / 2P P P F Buω′= + +a a a and resolve into components.2: 2800 419.616 , 2380.4 in./s ,u u− = − + = −& &: 0 46.708 1198.92BDα= + 225.7 rad/sBD = −αααα(a)225.7 rad/sBD =αααα !(b) 2/ 2380 in./sP F =a2/ 198.4 ft/sP F =a !
  • 215. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 176.Geometry.5tan , 26.56510β β= = °1011.1803 in.cosAElβ= =Velocity analysis. 12 rad/sAB =ω , BD BDω=ω( ) ( )( )5 12 60 in./sB ABAB ω= = =v( )E B BDBE ω′ = +v v β[60= ] [11.1803 BDω+ ]β[/E BD u=v ], 0Eβ =vUse /E E E BD′= +v v v and resolve into components.+ : 0 60sin 11.1803 , 2.400 rad/sBD BDβ β ω ω= − + =+ : 0 60cos , 53.666 m/su uβ β= − =[60E′ =v ] ( )( )11.1803 2.400+  53.7 in./sβ  = 63.4°Acceleration analysis.( ) ( )( )22 25 12 720 in./sB ABAB ω= = =a( )E B BDBE α′ = + a a ( ) 2BDBEβ ω +  β [720= ] [11.1803 BDα+ ] [64.399β + ]β[/E BD u=a & 0Eβ  = aCoriolis acceleration.( )( )( ) [2 2 2.400 53.666 257.60BDuω = = ]β
  • 216. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Use [/ 2E E E BD BD uω′= + +a a a ]β and resolve into components.+ : 0 720cos 11.1803 257.60BDβ β α= − + +234.56 rad/sBDα =+ 2: 0 720sin 64.399 , 257.59 in./su uβ β= − + − = −& &[720E′ =a ] ( )( )11.1803 34.56+  ] [64.399β + ]β[720= ] [386.39+ ] [64.399β + ]β2365 in./s= 18.4°Summary:(a) 2.40 rad/sBD =ωωωω , 234.6 rad/sBD =αααα(b) 53.7 in./sE′ =v 63.4 ,° 2365 in./sE′ =a 18.4°
  • 217. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 177..Geometry1818 in., 30 , 20.7846 in.cos30R ABθ= = ° = =°:Pin B B Bv=v , B Ba=aLet rod be a rotating frame of reference.AC 26 rad/s, 4 rad/sω α= =( ) [20.7846B AB ω ω′ = =v ]30°[20.7846B α′ =a ] 230 20.7847ω° +  30 °Motion of B relative to the frame.[/B AC u=v ] [/30 , B AC u° =a & 30 °.Velocity Analysis / , 6 rad/sB B B AC ω′= + =v v vBv [ 20.7846ω= ] [30 u° + ]30°( )( ): 0 20.7846 6 sin30 cos30u= − ° + °72.000 in./s,u =( )( ): 20.7846 6 cos30 72.000sin30Bv = ° + °144.000 in./sBv = 144.0 in./sB =v.Coriolis acceleration ( )( )( ) 22 2 6 72.000 864 in./suω = =  30 °.Acceleration analysis [/ 2B B B AC uω′= + +a a a ] 230 , 4 rad/sα° =[ Ba ] [20.7846α= ] 230 20.7847ω° +  ] [30 u° + & [30 1728° + ]30°[83.1384= ] [30 748.246° + [30 u° + & [30 864° + ]30°[947.14= ] [30 748.246° + ] [30 u° + & 30 °: 0 947.14sin30 748.246cos30 cos30u= − ° − ° + °&21295.51 in./su =&: 947.14cos30 748.246sin30 1295.51sin30Ba = ° − ° + °21094 in./s= 21094 in./sB =a
  • 218. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 178.Geometry.Law of cosines.( )( )( )2 2 21.25 2.50 2 1.25 2.50 cos301.54914 in.rr= + − °=Law of sines.sin sin 301.25 rβ °=23.794β = °Let disk S be a rotating frame of reference. Sω=ΩΩΩΩ , Sα=&ΩΩΩΩMotion of coinciding point P′ on the disk.1.54914P S Srω ω′ = =v β[2/ / 1.54914P S P O S P O Sα ω α′ = − × − =a k r r ] 21.54914 Sβ ω+  β Motion relative to the frame./P S u=v /P S uβ =a & βCoriolis acceleration. 2 Suω β[/ 1.54914P P P S Sω′= + =v v v ] [uβ + ]β/ 2P P P S Suω= + +a a a[1.54914 Sα= ] 21.54914 Sβ ω+  [uβ  +& [2 S uω + ]βMotion of disk D. (Rotation about B)( ) ( )( )1.25 8 10 in./sP DBP ω= = =v 30°( )P DBP α= a ( ) 260 SBP ω° +  ( )( )230 0 1.25 8° = + 30 °280 in./s= 30°
  • 219. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Equate the two expressions for Pv and resolve into components.( ): 1.54914 10cos 30Sβ ω β= ° +10cos53.7943.8130 rad/s1.54914Sω°= =3.81 rad/sS =ωωωω( ): 10sin 30 10sin53.794 8.0690 in./suβ β= ° + = ° =Equate the two expressions for Pa and resolve into components.( ): 1.54914 2 80sin 30S Suβ α ω β− = ° +( )( )( ) 280sin53.794 2 3.8130 8.069081.4 rad/s1.54914Sα° += =281.4 rad/sS =αααα
  • 220. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 179.Geometry.Law of cosines.( )( )( )2 2 21.25 2.50 2 1.25 2.50 cos 451.84203 in.rr= + − °=Law of sines.sin sin 451.25 rβ °=28.675β = °Let disk S be a rotating frame of reference. Sω=ΩΩΩΩ , Sα=&ΩΩΩΩMotion of coinciding point P′ on the disk.1.84203P S Srω ω′ = =v β[2/ / 1.84203P S P O S P O Sα ω α′ = − × − =a k r r ] 21.84203 Sβ ω+  β Motion relative to the frame./P S u=v /P S uβ =a & βCoriolis acceleration. 2 Suω β[/ 1.84203P P P S Sω′= + =v v v ] [uβ + ]β/ 2P P P S Suω= + +a a a[1.84203 Sα= ] 21.84203 Sβ ω+  [uβ  + & [2 S uω + ]βMotion of disk D. (Rotation about B)( ) ( )( )1.25 8 10 in./sP DBP ω= = =v 30°( )P DBP α= a ( ) 245 SBP ω ° +  ] ( )( )245 0 1.25 8° = + 45 °280 in./s= 45°
  • 221. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Equate the two expressions for Pv and resolve into components.( ): 1.84203 10cos 45Sβ ω β= ° +10cos73.6751.52595 rad/s1.84203Sω°= =1.526 rad/sS =ωωωω( ): 10sin 45 10sin73.675 9.5968 in./suβ β= ° + = ° =Equate the two expressions for Pa and resolve into components.( ): 1.84203 2 80sin 45S Suβ α ω β− = ° +( )( )( ) 280sin73.675 2 1.52595 9.596857.6 rad/s1.84203Sα° += =257.6 rad/sS =αααα
  • 222. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 180.Geometry. 30θ = °250288.68 mmcos30AP = =°250tan30 144.34 mmBP = ° =Velocity analysis. 4 rad/sA =ω , 5 rad/sB =ωSee Solution of Problem 15.149, which obtains/ 166.67 mm/sP AE =v 30°/ 916.67 mm/sP BD =vAcceleration analysis. 0, 0A B= =α αRod AE. Let P′ be the point on rod AE coinciding with the pin P.( ) ( ) 0P Ata AP α′ = =( ) ( ) ( )( )22 2288.68 4 4618.9 mm/sP Ana AP ω′ = = =24618.9 mm/sP =a 30°/ /P AE P AEa=a 30°Coriolis acceleration: ( ) /2c A P AEAEa vω=( )( )( ) 22 4 166.67 1333.3 mm/s= =( ) 21333.3 mm/sc AE=a 60°( )/P P P AE c AE′= + +a a a a= 24618.9 mm/s ] /30 P AEa° +  ] 230 1333.3 mm/s° + 60°
  • 223. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Rod BD. Let P′′ be the point on rod BD coinciding with the pin P.( ) ( ) 0P Bta BP α′′ = =( ) ( ) ( )( )22 2144.34 5 3608.5 mm/sP Bna BP ω′′ = = =( ) 23608.5 mm/sP′′ =a/ /P BD P BDa=aCoriolis acceleration. ( ) /2c B P BDBDa vω=( )( )( ) 22 5 916.67 9166.7 mm/s= =( ) 29166.7 mm/sc BD=a( )/P P P BD c BD′′= + +a a a a[3608.5= ] /P BDa+  ] 29166.7 mm/s+  ]Equate the two expressions for Pa and resolve into components./: 4618.9cos30 cos30 1333.3cos60P AEa− °+ °+ °0 0 9166.7= + −2 2/ /6735.7 mm/s 6735.7 mm/sP AE P AEa = − =a 30°/60 : 0 0 1333.3 3608.5sin60 sin 60 9166.7cos60P BDa° + + = °− °− °/ /3223.4 mm/s 3223.4 mm/sP BD P BDa = − =aUsing Pa from rod AE.4618.9P =a 30 6735.7°+ 30 1333.3°+ 60°29166.7 mm/s=  ] 26832.0 mm/s+  ]Check using Pa from rod BD.[3608.5P =a ] [3223.4+ ] [9166.7+ ]29166.7 mm/s=  ] 26831.9 mm/s+  ]211430 mm/sP =a 36.7° 211.43 m/sP =a 36.7°
  • 224. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 181.Position of A. ( ) ( ) ( )100 mm + 250 mm 300 mmA = − −r i j kVelocity of A. ( ) ( ) ( )10 mm/s + + 80 mm/sA A yv=v i j kA A= ×v rωωωωSince vA is perpendicular to rA, the scalar product0A A⋅ =r v( )( ) ( )( ) ( )( )100 10 250 300 80 0A A A yv⋅ = − + − =r v( ) 100 mm/sA yv =
  • 225. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 182.Radius of ball: 4.3 in. = 0.35833 ftAt the given instant, the origin is not moving.: 14.4 14.4 10.80.35833 0.35833 0A A x y zω ω ω= × − + =i j kv r i j kωωωω( )14.4 14.4 10.8 0.35833 0.35833 0.35833z z x yω ω ω ω− + = − + + −i j k i j k( ): 0.35833 14.4 40.186 rad/s: 0.35833 14.4 40.186 rad/s: 0.35833 10.8 30.140 rad/sz zz zx y x yω ωω ωω ω ω ω− = = −= − = −− = − =ijk: 28.8 21.60 0.71667 0D D x y zω ω ω= × + =i j kv r i kωωωω28.8 21.6 0.71667 0.71667z xω ω+ = − +i k i k: 0.71667 28.8 40.186 rad/s: 0.71667 21.6 30.140 rad/sz zx xω ωω ω− = = −= =ik30.140 0y xω ω= − =( ) .a Angular velocity ( ) ( )30.1 rad/s 40.2 rad/s= −i kωωωω( )b Velocity of point C.( )30.140 40.186 0.3583314.4 10.8C C= × = − ×= +v r i k ji kωωωω( ) ( )14.4 ft/s 10.8 ft/sC = +v i k
  • 226. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 183.At the given instant, the origin is not moving.: 10.8 14.4 14.40 0.35833 0.35833B B x y zω ω ω= × − + =i j kv r i j kωωωω( )10.8 14.4 14.4 0.35833 0.35833 0.35833y z x xω ω ω ω− + = − − +i j k i j k( ): 0.35833 10.8 30.140 rad/s: 0.35833 14.4 40.186 rad/s: 0.35833 14.4 40.186 rad/sy z y zx xx xω ω ω ωω ωω ω− = − =− = − == =ijk: 21.6 28.80 0.71667 0D D x y zω ω ω= × + =i j kv r i kωωωω21.6 28.8 0.71667 0.71667z xω ω+ = − +i k i k: 0.71667 21.6 30.140 rad/s: 0.71667 28.8 40.186 rad/sz zx xω ωω ω− = = −= =ik30.140 0y zω ω= + =( ) .a Angular velocity ( ) ( )40.2 rad/s 30.1 rad/s= −i kωωωω( )b Velocity of point C.( ) ( )40.186 30.140 0.3583310.8 14.4C C= × = − ×= +v r i k ji kωωωω( ) ( )10.8 ft/s 14.4 ft/sC = +v i k
  • 227. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 184..Total angular velocity 2 1ω= +j kωωωω ωωωω( ) ( )4 rad/s 5 rad/s= +j kωωωω.Angular acceleration1Frame is rotating with angular velocity .Oxyz ω=Ω k( )( )( )1 2 1 1 205 4 20Oxyzω ω ω ω ω= = + ×= + × + = −= − = −k j k ii i& &αααα ωωωω ωωωω ΩΩΩΩ ωωωωαααα( )220.0 rad/s= − iαααα
  • 228. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 185.( ) ( )1 450 rpm 15 rad/sπ= − = −i iωωωω( ) ( )2 3 rpm 0.100 rad/sπ= − = −j jωωωωLet the frame rotate with the motor housing.OxyzRate of rotation of frame :Oxyz ( )2 0.100 rad/sω π= = − jΩΩΩΩ.Angular acceleration( ) ( )1 2 1 2 1 2Oxyz= + = + + × +& & & &αααα ωωωω ωωωω ωωωω ωωωω ΩΩΩΩ ωωωω ωωωω( ) ( ) ( )0 0 0.100 15 0.100π π π= + + − × − −j i j21.500π= − k ( )214.80 rad/s= − kαααα
  • 229. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 186.Let be the spin of gear about the axle .s A ADω.Total angular velocity ( )1 sin cossω ω θ θ= + − +j i jωωωω (1)( ) ( )sin cos cos sinC L r L rθ θ θ θ= − − +r i jSince gear is fixed,B 0.C =v( )( ) ( )1sin cos 0 0sin cos cos sin 0C C s sL r L rω θ ω ω θθ θ θ θ= × = − + =− − +i j kv rωωωω( ) ( )sin cos sin cos sin coss sL r L rω θ θ θ ω θ θ θ + − − k( )1 sin cos 0L rω θ θ− − =k( ) ( )2 21sin cos sin coss sr r L rω θ θ ω ω θ θ+ = = −1 sin cossLrω ω θ θ = −  ( ) .a Angular velocity ( )1 1 sin cos sin cosLrω ω θ θ θ θ = + − − +  j i jωωωω1 sin cos sin sin cosL Lr rω θ θ θ θ θ    = − + +        i jωωωω( ) .b Angular acceleration1Frame is rotating with angular velocity .Oxyz ω= jΩΩΩΩOxyz= = + ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω1 10 sin cos sin cos sinL Lr rω ω θ θ θ θ θ      = + × − + +           j i j21 sin sin cosLrω θ θ θ = −  kαααα
  • 230. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 187.Let be the spin of gear about the axle .s A ADω.Total angular velocity ( )1 sin cossω θ θ= + − +j i jωωωω ωωωω( ) ( )sin cos cos sinC L r L rθ θ θ θ= − − +r i j2Since gear is rotating with angular velocity , on gear ,B Bω j( )2 / 2 sin cosC C B L rω ω θ θ= × = − −v j r kOn gear A ( )( ) ( )1sin cos 0sin cos cos sin 0C C s sL r L rω θ ω ω θθ θ θ θ= × = − +− − +i j kv rωωωω( ) ( )( )1sin cos sin cos sin cossin cosC s sL r L rL rω θ θ θ ω θ θ θω θ θ = + − − − −v kkEquating the two expressions for and solving for ,C sωv( )1 2 sin cossLrω ω ω θ θ = − −  ( ) .a Angular velocity ( ) ( )1 1 2 sin cos sin cosLrω ω ω θ θ θ θ = + − − − +  j i jωωωω( )1 2sin cos sin sin cos cos sin sin cosL L Lr r rω θ θ θ θ θ ω θ θ θ θ      = − + + + − − +            i j i jωωωω( ) .b Angular acceleration1Frame is rotating with angular velocity .Oxyz ω= jΩΩΩΩOxyz= = + ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω( )1 1 1 20 sin cosLrω ω ω ω θ θ = + × = − −  j kωωωω( )1 1 2 sin cosLrω ω ω θ θ = − −  kαααα
  • 231. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 188.Angular velocity. 1 2ω ω= +i kωωωω( ) ( )5 rad/s 4 rad/s= +i kωωωω( )a Angular acceleration.1Frame is rotating with angular velocity .Oxyz = iΩΩΩΩ ωωωωOxyz= = + ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω( )1 1 2 1 20 ω ω ω ω ω= + × + = −i i k j( )( )4 5 20= − = −j j ( )220.0 rad/s= − jαααα( ) 0.b Acceleration of point P.θ =( ) ( )( )( )60 mm 0.06 m5 4 0.06 0.2420 0.06 5 4 0.241.2 1.2 0.96 0.96 2.4PP PP P P= == × = + × == × + × = − × + + ×= + − = − +r i iv r i k i ja r v j i i k jk k i i kωωωωαααα ωωωω( ) ( )2 20.960 m/s 2.40 m/sP = − +a i k( ) 90 .c Acceleration of point P.θ = °( )( )( ) ( )0.06 m5 4 0.06 0.24 0.320 0.06 5 4 0.24 0.30 0 1.5 0.96 0 2.46PP PP P P== × = + × = − += × + ×= − × + + × − += + − − + = −r jv r i k j i ka r vj j i k i kj j jωωωωαααα ωωωω( )22.46 m/sP = −a j
  • 232. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 189.Angular velocity. 1 2ω ω= +i kωωωω( ) ( )5 rad/s 4 rad/s= +i kωωωωAngular acceleration.1Frame is rotating with angular velocity .Oxyz = iΩΩΩΩ ωωωωOxyz= = + ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω( )1 1 2 1 20 ω ω ω ω ω= + × + = −i i k j( )( )4 5 20= − = −j j ( )220.0 rad/s= − jαααα( )( )( )( )30 , 60 mm cos30 sin300.06 m cos30 sin30Pθ = ° = ° + °= ° + °r i ji j( ) ( ) ( )5 0 40.06cos30 0.06sin30 00.12 m/s 0.20785 m/s 0.15 m/sP P= × =° °= − + +i j kv ri j kωωωωAcceleration of point P.0 20 0 5 0 40.06cos30 0.06sin30 0 0.12 0.20785 0.151.03923 0.8314 1.23 1.03925P P P= × + ×= − +° ° −= − − +a r vi j k i j kk i j kαααα ωωωω( ) ( ) ( )2 2 20.831 m/s 1.230 m/s 2.08 m/sP = − − +a i j k
  • 233. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 190.1230Let rad/s rad/s180 610rad/s rad/s180 18ddtddtβ π πγ π π   = − = − = −         = − = − = −      j j ji i iωωωωωωωω( )a Angular velocity. 1 2= +ωωωω ωωωω ωωωω( ) ( )0.1745 rad/s 0.524 rad/s= − −i jωωωω( ) .b Angular acceleration1Frame is rotating with angular velocityOxyz =ΩΩΩΩ ωωωω( )1 1 1 2 1 20Oxyz= = + × = + × + = ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω ωωωω ωωωω ωωωω ωωωω ωωωω26 18 108π π π   = − × − = −      j i k ( )20.0914 rad/s= − kαααα( )c Velocity and acceleration of point P.For 90 and 30 ,β γ= ° = ° ( )( )12 ft sin30 cos30P = ° + °r j k018 60 12sin30 12cos305.4414 1.81380 1.04720P Cπ π   = × = − −      ° °= − + −i j kv ri j kωωωω( ) ( ) ( )5.44 ft/s 1.814 ft/s 1.047 ft/sP = − + −v i j k( )26 10.3923 0108 18 65.4414 1.81380 1.047200.54831 0 0.54831 0.18277 3.1657P P Pπ π π= × + ×   = − × + + − −      − −= + + − −a r vi j kk j ki j i j kαααα ωωωω( ) ( ) ( )2 2 21.097 ft/s 0.1828 ft/s 3.17 ft/sP = − −a i j k
  • 234. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 191.Contact points. ( ) ( ) ( ) ( )1 26 in. 9 in. 4 in. 3 in.= − + = +r i j r i jGear C. ( )15 rad/s,C = iωωωω( ) ( )1 1 15 6 9 135 in./sC= × = × − + =v r i i j kωωωωGear D. ( )30 rad/sD = iωωωω( ) ( )2 2 30 4 3 90 in./sD= × = × + =v r i i j kωωωωGears A and B. x y zω ω ω= + +i j kωωωω( )1 1 9 6 9 66 9 0x y z z z x yω ω ω ω ω ω ω= × = = − + + +−i j kv r i j kωωωωMatching expressions for 1,v : 9 0, : 6 0z zω ω− = =i j: 9 6 135x yω ω+ =k (1)( )2 2 3 4 3 44 3 0x y z z z x yω ω ω ω ω ω ω= × = = − + + −i j kv r i j kωωωωMatching expressions for 2.v : 3 0, : 4 0z zω ω− = =i j: 3 4 90x yω ω− =k (2)Solving (1) and (2) simultaneously, 20 rad/s, 7.5 rad/sx yω= = −ωωωω(a) Angular velocity. ( ) ( )20.0 rad/s 7.50 rad/s= −i jωωωωShaft HFG and thus the frame Fxyz rotates with angular velocity( )20 rad/s .xω= =i iΩΩΩΩ( )0 20 20 7.5Fxyz= = + × = + × −i i j& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω
  • 235. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) Angular acceleration. ( )2150.0 rad/s= − kαααα(c) Acceleration of tooth of gear B at point 2.2 2 20 0 150 20 7.5 04 3 0 0 0 90450 600 0 675 1800 0= × + ×= − + −= − + − − +a r vi j k i j ki j k i j kαααα ωωωω( ) ( )2 2225 in./s 2400 in./s= − −i j ( ) ( )2 22 18.75 ft/s 200 ft/s= − −a i j
  • 236. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 192.Results from the solution to Prob. 15.186.Position vector of contact point C: ( ) ( )sin cos cos sinC L r L rθ θ θ θ= − − +r i jVelocity of contact point C on gear A: 0C =vAngular velocity of gear A: 1 sin cos sin sin cosL Lr rω θ θ θ θ θ    = − + +        i jωωωωAngular acceleration of gear A: 21 sin sin cosLrα ω θ θ θ = −  kAcceleration of point C on gear A.C C C Cr= × + × = ×a r vαααα ωωωω αααα( ) ( )21 sin sin cos sin cos cos sinCLL r L rrω θ θ θ θ θ θ θ  = − × − − +    a k i j21 sin sin cos cos sin sin cosCL L Lrr r rω θ θ θ θ θ θ θ      = − + + −            a i j
  • 237. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 193.Since the cone rolls on the xz plane, 0 0A= =v vThe instantaneous axis of rotation lies along the x-axis.Total angular velocity. ω= iωωωω(a) Rate of spin. ( )cos sins sω β β= +i jωωωω1 spinω= +jωωωω ωωωω( ) ( ) 1cos sins sω ω β ω β ω= + +i i j jComponents.111 1s: 0 sinsincos: cossin tans sωω β ω ωβω β ωω ω ββ β= + = −= = − = −ji1sinsωωβ=(b) Angular velocity. 1tanωβ= − iωωωω(c) Angular acceleration. The vector ωωωω rotates about the y-axis at a rate ω1.1ω= jΩΩΩΩ11tanωωβ = × = × −  j iαααα ΩΩΩΩ ωωωω21tanωβ= kαααα
  • 238. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 194.Geometry.plane. Law of cosines.yz( )2 2 211.2 8 2 8 cos120d d= + − °By solving the quadratic equation 4.8 in.d =Let point be the midpoint of rod .H AB ( ) ( )/ 4.8 in. sin60 10.4 in.H D = ° −r j k( ) ( ) ( ) ( ) ( ) ( )/ /4 in. 4.8 in. sin60 10.4 in. , 4 in. 4.8 in. sin60 10.4 in.B D A D= + ° − = − + ° −r i j k r i j kLet be a unit vector normal to plane .ABDλλλλ sin30 cos30= ° + °j kλλλλThe projection of onto the normal is zero.Hv λλλλ/0 sin30 cos308.8 0, 00 4.8sin 60 10.4H H D x y z x xω ω ω ω ω ω° °⋅ = ⋅ × = = = =° −λ v λ r/Also, the projection of onto the normal is zero.B Av λλλλ( )/ /0 sin30 cos300 8 sin30 cos308 0 00 3B A B A y z z yz yω ω ω ωω ω° °⋅ = ⋅ × = = ° − °= =λ v λ rωωωωThen, 3y yω ω= +j kωωωω( ) ( ) ( )/ 0 34 4.8sin60 10.417.6 6.9282 4B B D y yy y yω ωω ω ω= × =° −= − + −i j kv ri j kωωωω( ) ( ) ( ) ( )2 2 22 2 2 2 2 217.6 6.9282 4 4B B B B yx y zv v v v ω= + + = + + =2373.76 16 0.20690 rad/sy yω ω= = −( )3 0.20690 0.35836 rad/szω = − = −
  • 239. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )a Angular velocity. 0.20690 0.35836= − −j kωωωω( ) ( )0.207 rad/s 0.358 rad/s= − −j kωωωω( ) .b Velocity of point A /A A D= ×v rωωωω0 0.20690 0.358364 4.8sin60 10.4A = − −− ° −i j kv( ) ( ) ( )3.64 in./s 1.433 in./s 0.828 in./sA = + −v i j k
  • 240. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 195.From the solution to Problem 15.194, ( ) ( )0.20690 rad/s 0.35836 rad/s= − −j kωωωω( ) ( ) ( ) ( ) ( )/ /4.8 in. sin 60 10.4 in. , 4 in. 4.8 in. sin60 10.4 in.H D B D= ° − = + ° −r j k r i j ksin30 cos30 , 4 in./sBv= ° + ° =j kλλλλNote that ωωωω is parallel to .λλλλThe projection of Ha onto the direction λλλλ is zero./ / 0H H D H H D⋅ = ⋅ × + ⋅ × = ⋅ × +a r v rλλλλ λλλλ αααα λλλλ ωωωω λλλλ αααα0 sin30 cos308.8 0 00 4.8sin 60 10.8x y z x xα α α α α° °= = =° −Also, the projection of /B Aa onto the direction λλλλ is zero./ / / / 0B A B A B A B A⋅ = ⋅ × + ⋅ × = ⋅ × +a r r rλλλλ λλλλ αααα λλλλ ωωωω λλλλ αααα( )0 sin30 cos300 8 sin30 cos30 0 38 0 0y z z y z yα α α α α α° °= ° − ° = =Velocity at B. /B B D= ×v rωωωω( ) ( ) ( )0 0.20690 0.35836 3.6414 in./s 1.4334 in./s 0.8276 in./s4 4.8sin60 10.4B = − − = − +° −i j kv i j kUnit vector tangent to the path of point B.BtBv=ve0.91036 0.35836 0.20690t = − +e i j k
  • 241. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Component of acceleration tangent to the path( ) / / 00.91036 0.35836 0.206900 3 19.3334 4.8sin60 10.4B t B t B D t B t B Dty y yaα α α= ⋅ = ⋅ × + ⋅ × = ⋅ × +−= = −° −e a e r e v e rαααα ωωωω ααααBut ( )B ta is given as 28 in./s , thus 219.333 8 in./syα− =( )2 20.41380 rad/s , 3 0.41380 0.71673 rad/sy zα α= − = − = −(a) Angular acceleration. ( ) ( )2 20.414 rad/s 0.717 rad/s= − −j kααααNormal component of acceleration. ( )B Bn= ×a vωωωω( )( ) ( ) ( )( ) ( ) ( ) ( )2 2 22 2 2 20 0.20690 0.358363.6414 1.4334 0.82760.6849 in./s 1.3049 in./s 0.7534 in./s0.6849 1.3049 0.7534 1.6551in./sB nB na= − −−= − − += + + =i j kai j kBut ( )( )( )22 249.67 in.1.6551B BB nB nv vaaρρ= = = =(b) Radius of curvature of path. 9.67 in.ρ =
  • 242. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 196.Geometry. ( ) ( )2 22 2 2 2 2 2/ / / : 325 195 100 , 240 mmAB A B A B A Bl x y z c c= + + = − + + − =( ) ( ) ( )/ 195 mm 240 mm 100 mmA B = − + −r i j kVelocity of collar B. ( )1m/sB = −v k ( )1000 mm/s= − kVelocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0.A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )195 240 100 195 240 100 1000Av− + − ⋅ = − + − ⋅ −i j k j i j k k240 100000 or 416.67 mm/sA Av v= = ( )0.417 m/sA =v j
  • 243. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 197.Geometry. ( ) ( )/2 22 2 2 2 2 2/ / : 325 195 156 , 208 mmA BAB A B A Bl x y z c c= + + = − + + − =( ) ( ) ( )/ 195 mm 208 mm 156 mmA B = − + −r i j kVelocity of collar B. ( ) ( )1.6 m/s 1600 mm/sB = =v k kVelocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0.A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )195 208 156 195 208 156 1600Av− + − ⋅ = − + − ⋅i j k j i j k k208 249600 or 1200 mm/sA Av v= − = − ( )1.200 m/sA = −v j
  • 244. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 198.Geometry. ( ) ( ) ( ) ( )/2 in. 16 in. 8 in. 2 in.B/A C B= − = + +r k r i j k( ) ( )/ 12 in. 8 in.C D = +r i j 2 212 8 208 in.CDl = + =Velocity at B. ( ) ( )0 24 2 48 in./sB B/Aω= × = × − = −v j r j k iVelocity of collar C./ 12 8208D CCDCDl+= =r i jλλλλ( )0.83205 0.55470C C CD Cv v= = +v i jλλλλ/ / /whereC B C B C B BC C B= + = ×v v v v rωωωωNoting that /C Bv is perpendicular to / ,C Br we get / / 0.C B C B⋅ =r vForming / ,C B C⋅r v we get ( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /C B C C B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )16 8 2 0.83205 + 0.55470 16 8 2 48Cv+ + ⋅ = + ⋅i j k i j i j k i++++ −−−−17.7504 768Cv = − 43.267 in./sCv = −( )( )43.267 0.83205 0.55470C = −v i j++++( ) ( )36.0 in./s 24.0 in./sC = − −v i j
  • 245. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 199.Geometry. ( ) ( ) ( ) ( )/2 in. 16 in. 8 in. 2 in.B/A C B= − = + +r k r i j k( ) ( )/ 12 in. 8 in.C D = +r i j 2 212 8 208 in.CDl = + =Velocity at B. ( ) ( )0 24 2 48 in./sB B/Aω= × = × − =v i r i k jVelocity of collar C./ 12 8208D CCDCDl+= =r i jλλλλ( )0.83205 0.55470C C CD Cv= = +v v i jλλλλ/ / /whereC B C B C B BC C B= + = ×v v v v rωωωωNoting that /C Bv is perpendicular to / ,C Br we get / / 0.C B C B⋅ =r vForming / ,C B C⋅r v we get ( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /C B C C B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )16 8 2 0.83205 + 0.55470 16 8 2 48Ev+ + ⋅ = + ⋅i j k i j i j k j++++17.7504 384Cv = 21.633 in./sCv =( )( )21.633 0.83205 0.55470C =v i j++++( ) ( )18.00 in./s 12.00 in./sC = +v i j
  • 246. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 200.Geometry. ( ) ( ), 4.5 in. 9 in.A D Cy= = =r j r i r k( ) ( ) ( ) ( )2 2/ 4.5 in. 9 in. 4.5 9 10.0623 in.D C D C CDl= − = − = + − =r r r i k( ) ( )( ) ( )//4 4.5 92 in. 4 in.9 9D CB Cc −= = = −r i kr i k( ) ( )/ 9 2 4 2 in. 5 in.B C B C= + = + − = +r r r k i k i k( ) ( ) ( )/ 2 in. in. 5 in.A B A B y= − = − + −r r r i j k( ) ( )2 22 2 2 2 2 2/ / : 15 2 5AB A B A Bl x y z y= + + = − + + −( ) ( ) ( )/14 in., 2 in. 14 in. 5 in.A By = = − + −r i j kVelocity of collar B./D CB BCDvl=rv( )( )( ) ( )2.5 4.5 91.11803 in./s 2.23607 in./s10.0623B−= = −i kv i kVelocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v ω rNoting that /A Bv is perpendicular to / ,A Br we get / / 0.A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )2 14 5 2 14 5 1.11803 2.23607Av− + − ⋅ = − + − ⋅ −i j k j i j k i k( )( ) ( )( )14 2 1.11803 5 2.23607 or 0.63888 in./sA Av v= − + − − =( )0.639 in./sA =v j
  • 247. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 201.Geometry. ( ) ( ), 4.5 in. 9 in.A D Cy= = =r j r i r k( ) ( ) ( ) ( )2 2/ 4.5 in. 9 in. 4.5 9 10.0623 in.D C D C CDl= − = − = + − =r r r i k( ) ( )( ) ( )//6 4.5 93 in. 6 in.9 9D CB Cc −= = = −r i kr i k( ) ( )/ 9 3 6 3 in. 3 in.B C B C= + = + − = +r r r k i k i k/ 3 3A B A B y= − = − + −r r r i j k2 2 2 2 2 2 2/ / : 15 3 3AB A B A Bl x y z y= + + = + +2( ) ( ) ( )/14.3875 in., 3 in. 14.3875 in. 3 in.A By = = − + −r i j kVelocity of Collar B./D CB BCDvl=rv( )( )( ) ( )2.5 4.5 91.11803 in./s 2.23607 in./s10.0623B−= = −i kv i kVelocity of Collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0.A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )3 14.3875 3 3 14.3875 3 1.11803 2.23607Av− + − ⋅ = − + − ⋅ −i j k j i j k i k( )( ) ( )( )14.3875 3 1.11803 3 2.23607 or 0.23313 in./sA Av v= − + − − =( )0.233 in./sA =v j
  • 248. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 202.Geometry. ( ) ( )/2 22 2 2 2 2 2/ / /: 0.5 0.24 0.4A BAB A B A B A Bl x y z y= + + = − + + −( ) ( ) ( )/ /0.18 m, 0.24 m 0.18 m 0.4 mA B A By = = − + −r i j k( ) ( ) ( ) ( )2 2/ 0.24 m 0.18 m , 0.24 0.18 0.3 mD C CDl= − = + − =r i jVelocity of collar B./D CB BCDrvl=v( )( )( ) ( )0.24 0.180.200 0.16 m/s 0.12 m/s0.3B−= = −i jv i jVelocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0.A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )0.24 0.18 0.4 0.24 0.18 0.4 0.16 0.12Av− + − ⋅ = − + − ⋅ −i j k j i j k i j( )( ) ( )( )0.18 0.24 0.16 0.18 0.12 or 0.333 m/sA Av v= − + − = −( )0.333 m/sA = −v j
  • 249. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 203.Geometry. ( )/22 2 2 2 2 2/ / /: 0.5 0 0.4A BAB A B A B A Bl x y z y= + + = + + −( ) ( )/ /0.3 m, 0.3 m 0.4 mA B A By = = −r j k( ) ( ) ( ) ( )2 2/ 0.24 m 0.18 m , 0.24 0.18 0.3 mD C CDl= − = + − =r i jVelocity of collar B./D CB BCDvl=rv( )( )( ) ( )0.24 0.180.200 0.16 m/s 0.12 m/s0.3B−= = −i jv i jVelocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0.A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )0.3 0.4 0.3 0.4 0.16 0.12Av− ⋅ = − ⋅ −j k j j k i j( )( )0.3 0.3 0.12 0.12 m/sA Av v= − = −( )0.1200 m/sA = −v j
  • 250. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 204.Angular velocity of shaft AC. 1AC ω= kωωωωLet 3ω j be the angular velocity of body D relative to shaft AD.Angular velocity of body D. 1 3D ω ω= +k jωωωωAngular velocity of shaft EG. ( )2 cos20 sin 20EG ω= ° − °k jωωωωLet 4ω i be the angular velocity of body D relative to shaft EG.Angular velocity of body D. ( )2 4cos20 sin 20D ω ω= ° − ° +k j iωωωωEquate the two expressions for Dω and resolve into components.4: 0 ω=i (1)3 2: sin 20ω ω= − °j (2)1 2: cos20ω ω= °k (3)From (3),12cos20ωω =°
  • 251. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 205.Angular velocity of shaft AC. 1AC ω= kωωωωLet 3ω i be the angular velocity of body D relative to shaft AD.Angular velocity of body D. 1 3D ω ω= +k iωωωωAngular velocity of shaft EG. ( )2 cos20 sin 20EG ω= ° − °k jωωωωLet 4ω λ be the angular velocity of body D relative to shaft EG.Where λ is a unit vector along the axis the clevis axle attached to shaft EG.cos20 sin 20= ° + °j kλλλλ4 4 4cos20 sin 20ω ω ω= ° + °j kλλλλAngular velocity of body D. 4D EG ω= +ωωωω ωωωω λλλλ( ) ( )4 2 4 2cos20 sin 20 sin 20 cos20D ω ω ω ω= ° − ° + ° + °j kωωωωEquate the two expressions for Dωωωω and resolve into components.3: 0ω =i (1)4 2: 0 cos20 sin 20ω ω= ° − °j (2)1 4 2: sin 20 cos20ω ω ω= ° + °k (3)From (2) and (3), 2 1cos20ω ω= °
  • 252. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 206.Geometry. ( ) ( ) ( )640 mm , 24 in. 10 in.C C Bx= + = +r i j r j k2 224 10 26 in.ABl = + =( ) ( )/ 8 in. 10 in.C B Cx= + −r i j kLength of rod BC. 2 2 2 2 242 8 10BC Cl x= = + +Solving for ,Cx 40 in.Cx =( ) ( ) ( )/ 40 in. 8 in. 10 in.C B = + −r i j kVelocity. ( ) ( ) ( )19.524 10 18 in./s 7.5 in./s26B = − − = − −v j k j kC Cv=v iAngular velocity of collar C. C Cω= iωωωωThe axle of the clevis at C is perpendicular to the x-axis and to the rod BC.A vector along this axle is /C B= ×p i r( ) ( ) ( )2 240 8 10 10 in. 8 in.10 8 12.806 in.p= × + − = += + =p i i j k j kLet λλλλ be a unit vector along the axle. 0.78087 0.62470p= = +pj kλλλλLet s sω=ωωωω λλλλ be the angular velocity of rod BC relative to collar C.0.78087 0.62470s s sω ω= +j kωωωωAngular velocity of rod BC. BC C s= +ωωωω ωωωω ωωωω0.78087 0.62470BC C s sω ω ω= + +i j kωωωω/C B BC C B= + ×v v rωωωω18 7.5 0.78087 0.6247040 8 10C C s sv ω ω ω= − − +−i j ki j k
  • 253. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Resolving into components,: 12.806C sv ω= −i (1): 0 18 10 24.988C sω ω= − + +j (2): 0 7.5 8 31.235C sω ω= − + −k (3)Solving the simultaneous equations (1), (2), and (3),1.4634 rad/s, 0.13470 rad/s, 1.725 in./sC s Cvω ω= = = −(a) Angular velocity of rod BC.( )( ) ( )( )1.4634 0.78087 0.13470 0.62470 0.13470BC = + +i j kωωωω( ) ( ) ( )1.463 rad/s 0.1052 rad/s 0.0841 rad/sBC = + +i j kωωωω(b) Velocity of collar C. ( )1.725 in./sC = −v i
  • 254. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 207.Geometry. Determine the position of collar A.( ) ( )/, 6 in. 2 in.6 2A A B B BA B Az x yz= = + = += − −r k r i j i jr k i jLength of rod AB: 2 2 2 2 211 6 2AB Al z= = − −Solving for ,Az 9 in.Az =( ) ( ) ( )/ 6 in. 2 in. 9 in.A B = − − +r i j kVelocity. ( ) ( )4.5 ft/s 54 in./s ,B A Av= − − =v j j v k====Angular velocity of collar B. B Bω= jωωωωThe axle of the clevis at B is perpendicular to both the y-axis and the rod AB.A vector along this axle is /A B= ×p j r( ) ( ) ( )/2 26 2 9 9 in. + 6 in.9 6 10.8167 in.A Br= × = × − − + == + =p j j i j k i kpLet λ be a unit vector along the axle. 0.83205 0.55470p= = +pλ i kLet s sω=ω λ be the angular velocity of rod AB relative to collar B.0.83205 0.55470s s sω ω= +i kωωωωAngular velocity of rod AB. AB B s= +ωωωω ωωωω ωωωω0.83205 0.55470AB s B sω ω ω= + +ω i j k/A B AB A B= + ×v v rωωωω54 0.83205 0.554706 2 9A s B sv ω ω ω= − +− −i j kk j
  • 255. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Resolving into components,: 0 0 9 1.1094B sω ω= + +i (1): 0 54 10.8167 sω= − −j (2): 0 6 1.6641A B sv ω ω= + −k (3)From (2), 4.9923 rad/ssω = −From (1), 0.61539 rad/sBω =(a) Angular velocity of rod AB.( )( ) ( )( )0.83205 4.9923 0.61539 0.55470 4.9923AB = − + + −i j kωωωω( ) ( ) ( )4.15 rad/s 0.615 rad/s 2.77 rad/sAB = − + −i j kωωωω(b) Velocity of collar A.From (3), ( )( ) ( )( )6 0.61539 1.6641 4.9923 12.00 in./sAv = − − =( )1.000 ft/sA =v k
  • 256. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 208.Geometry. ( ) ( )2 22 2 2 2 2 2/ / / : 325 195 100 , 240 mm.AB A B A B A Bl x y z c c= + + = − + + − =( ) ( ) ( )/ 195 mm 240 mm 100 mmA B = − + −r i j kVelocity of collar B. ( ) ( )1m/s 1000 mm/sB = − −v k k====Velocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0.A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )195 240 100 195 240 100 1000Av− + − ⋅ = − + − ⋅ −i j k j i j k k240 100000 or 416.67 mm/sA Av v= =Relative velocity. /A B A B= −v v v( ) ( ) ( ) ( )2 22 2 2 2/ /416.67 mm/s 1000 mm/s 416.67 1000 1083.33 mm /sA B A Bv= + = + =v j kAcceleration of collar B. 0B =aAcceleration of collar A. A Aa=a j/ / / /, whereA B A B A B AB A B AB A B= + = × + ×a a a a r vαααα ωωωωNoting that /AB A B× rαααα is perpendicular to / ,A Br we get / / 0A B AB A B⋅ × =r rααααWe note also that / / / / /A B AB A B A B A B A B⋅ × = ⋅ ×r v v rωωωω ωωωω( )2/ / /A B A B A Bv= − ⋅ = −v v
  • 257. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Then, ( ) ( )2 2/ / / /0A B A B A B A Bv v⋅ = − = −r aForming / ,A B A⋅r a we get ( )/ / / / / /A B A A B A A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r aor ( )2/ / /A B A A B B A Bv⋅ = ⋅ −r a r a (2)From (2), ( ) ( ) ( )2195 240 100 0 1083.33Aa− + − ⋅ = −i j k j( )2 2240 1083.33 4890 mm/sA Aa a= − = − ( )24.89 m/sA = −a j
  • 258. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 209.Geometry. ( ) ( )2 22 2 2 2 2 2/ / / : 325 195 156 , 208 mmAB A B A B A Bl x y z c c= + + = − + + − =( ) ( ) ( )/ 195 mm 208 mm 156 mmA B = − + −r i j kVelocity of collar B. ( ) ( )1.6 m/s 1600 mm/sB =v k k====Velocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )195 208 156 195 208 156 1600Av− + − ⋅ = − + − ⋅i j k j i j k k208 249600 or 1200 mm/sA Av v= − = −Relative velocity. /A B A B= −v v v( ) ( ) ( )2 2 2 2 2/ /1.2 m/s 1.6 m/s 1.2 1.6 4 m /sA B A Bv= − − = + =v j kAcceleration of collar B. 0B =aAcceleration of collar A. A Aa=a j/ / / /, whereA B A B A B AB A B AB A B= + = × + ×a a a a r vαααα ωωωωNoting that /AB A B× rαααα is perpendicular to / ,A Br we get / / 0A B AB A B⋅ × =r rααααWe note also that / / / / /A B AB A B A B A B A B⋅ × = ⋅ ×r v v rωωωω ωωωω( )2/ / /A B A B A Bv= − ⋅ = −v v
  • 259. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Then, ( ) ( )2 2/ / / /0A B A B A B A Bv v⋅ = − = −r aForming / ,A B A⋅r a we get ( )/ / / / / /A B A A B A A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r aor ( )2/ / /A B A A B B A Bv⋅ = ⋅ −r a r a (2)From (2), ( ) ( )0.195 0.208 0.156 0 4Aa− + − ⋅ = −i j k j0.208 4Aa = −( )219.23 m/sA = −a j
  • 260. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 210.Geometry. ( ) ( ) ( ) ( )/2 in. 16 in. 8 in. 2 in.B/A C B= − = + +r k r i j k( ) ( )/ 12 in. 8 in.C D = +r i j 2 212 8 208 in.CDl = + =Velocity at B. ( ) ( )0 24 2 48 in./sB B/Aω= × = × − = −v j r j k iVelocity of collar C./ 12 8208D CCDCDl+= =r i jλλλλ( )0.83205 0.55470C C CD Cv v= = +v i jλλλλ/ / /whereC B C B C B BC C B= + = ×v v v v rωωωωNoting that /C Bv is perpendicular to / ,C Br we get / / 0.C B C B⋅ =r vForming / ,C B C⋅r v we get ( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /C B C C B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )16 8 2 0.83205 + 0.55470 16 8 2 48Cv+ + ⋅ = + ⋅ −i j k i j i j k i++++17.7504 768Cv = − 43.267 in./sCv = −( )( )43.267 0.83205 0.55470C = −v i j++++(36.0 in./s) (24.0 in./s)= − −i jRelative velocity. /C B C B= −v v v/ 36 24 ( 48 ) (12 in./s) (24 in./s)C B = − − − − = −v i j i i j2 2 2 2 2/( ) (12) (24) 720 in. /sC Bv = + =Acceleration at B. 2/ /B AB B A D B Aω= × −a r rαααα2 20 (24) ( 2 ) (1152 in./s )B = − − =a k kAcceleration of collar C. (0.83205 0.55470 )C C CD Ca a= = +a i jλλλλ/ / /whereC B C B C/B BC C B BC C B= + = × + ×a a a a α r vωωωω
  • 261. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Noting that /BC C B×α r is perpendicular to / ,B Cr we get / /( ) 0.C B BC C B⋅ × =r α rWe also note that / / / /C B BC C B C B C B BC⋅ × = ⋅ ×r ω v v r ω2/ / /( )C B C B C B= − ⋅ = −v v vThen, 2 2/ / / /0 ( ) ( )C B C B C B C Bv v⋅ = − = −r aForming /C B C⋅r a we get/ / / / / /( )C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r aor 2/ / /( )C B C C B B C Bv⋅ = ⋅ −r a r a (2)From (2), (16 8 2 ) (0.83205 0.55470 ) (16 8 2 ) (1152 ) 720Ca+ + ⋅ + = + + ⋅ −i j k i j i j k k217.7504 1584 89.237 in./sC Ca a= =(89.237)(0.83205 0.55470 )C = +a i j2 2(74.3 in./s ) (49.5 in./s )C = +a i j
  • 262. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 211.Geometry. ( ) ( ) ( ) ( )/ /2 in. 16 in. 8 in. 2 in.B A C B= − = + +r k r i j k( ) ( ) 2 2/ 12 in. 8 in. 12 8 208 in.C D CDl= + = + =r i jVelocity at B. ( ) ( )0 / 24 2 48 in./sB B Aω= × = × − =v i r i k jVelocity of collar C./ 12 8208D CCDCDl+= =r i jλ( )0.83205 0.55470C C CD Cv v= = +v λ i j/ / /whereC B C B C B BC C B= + = ×v v v v ω rNoting that /C Bv is perpendicular to / ,C Br we get / / 0C B C B⋅ =r vForming /,C B C⋅r v we get ( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ = ⋅r v r v v r v r vor / /C B C C B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )16 8 2 0.83205 0.55470 16 8 2 48Cv+ + ⋅ + = + + ⋅i j k i j i j k j17.7504 384 21.633 in./sC Cv v= =( )( ) ( ) ( )21.633 0.83205 0.55470 18 in./s 12 in./sC = + = +v i j i jRelative velocity. /C B C B= −v v v( ) ( )/ 18 12 48 18 in./s 36 in./sC B = + − = −v i j j i j( ) ( ) ( )2 2 2 2 2/ 18 36 1620 in. /sC Bv = + =Acceleration at B.2/ 0 /B AB B A B Aω= × −a α r r( ) ( ) ( )2 20 24 2 1152 in./sB = − − =a k kAcceleration of collar C. ( )0.83205 0.55470C C CD Ca aλ= = +a i j/ / / /whereC C C B C B BC C B BC B C= + = × + ×a a a a r ω vααααNoting that /BC C B×α r is perpendicular to / ,C Br we get / / 0C B BC C B⋅ × =r α rWe also note that ( )2/ / / / / / /C B BC C B C B C B BC C B C B C Bv⋅ × = ⋅ × = − ⋅ = −r ω v v r ω v v
  • 263. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Then, ( ) ( )2 2/ / / /0C B C B C B C Bv v⋅ = − = −r aForming / we getC B C⋅r a( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r aor ( )2/ / /C B C B C B C Bv⋅ = ⋅ −r a r a (2)From (2) ( ) ( )16 8 2 0.83205 0.55470 C+ + ⋅ +i j k i j a( ) ( )16 8 2 1152 1620= + + ⋅ −i j k k217.7504 684 38.534 in./sC Ca a= =( )( )38.534 0.83205 0.55470C = +a i j( ) ( )2 232.1 in./s 21.4 in./sC = +a i j
  • 264. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 212.Geometry. ( ) ( ), 4.5 in. 9 in.A D Cy= = =r j r i r k( ) ( ) ( ) ( )2 2/ 4.5 in. 9 in. 4.5 9 10.0623 in.D C D C CDl= − = − = + − =r r r i k( ) ( )( ) ( )//4 4.5 92 in. 4 in.9 9D CB Cc −= = = −r i kr i k( ) ( )/ 9 2 4 2 in. 5 in.B C B C= + = + − = +r r r k i k i k/ 2 5A B A B y= − = − + −r r r i j k( ) ( )2 22 2 2 2 2 2/ / : 15 2 5AB A B A Bl x y z y= + + = − + + −( ) ( ) ( )/14 in., 2 in. 14 in. 5 in.A By r= = − + −i j kVelocity of collar B./D CB BCDvl=rv( )( )( ) ( )2.5 4.5 91.11803 in. 2.23607 in./s10.0623B−= = −i kv i kVelocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )2 14 5 2 14 5 1.11803 2.23607Av− + − ⋅ = − + − ⋅ −i j k j i j k i k( )( ) ( )( )14 2 1.11803 5 2.23607 or 0.63888 in./sA Av v= − + − − =Relative velocity. /A B A B= −v v v( ) ( ) ( )( ) ( ) ( ) ( ) ( )/2 2 2 2 2/0.63888 in./s 1.11803 in./s 2.23607 in./s0.63888 1.11803 2.23607 6.6582 in./sA BA Bv= − += + − + =v j i k
  • 265. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Acceleration of collar B. 0B =aAcceleration of collar A. A Aa=a j/ / / /, whereA B A B A B AB A B AB A B= + = × + ×a a a a r vαααα ωωωωNoting that /AB A B× rαααα is perpendicular to / ,A Br we get / / 0A B AB A B⋅ × =r α rWe note also that / / / / /A B AB A B A B A B A B⋅ × = ⋅ ×r v v rωωωω ωωωω( )2/ / /A B A B A Bv= − ⋅ = −v vThen, ( ) ( )2 2/ / / /0A B A B A B A Bv v⋅ = − = −r aForming / ,A B A⋅r a we get ( )/ / / / / /A B A A B A A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r aor ( )2/ / /A B A A B B A Bv⋅ = ⋅ −r a r a (2)From (2), ( ) ( )2 14 5 0 6.6582Aa− + − ⋅ = −i j k j14 6.6582Aa = − ( )20.476 in./sA = −a j
  • 266. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 213.Geometry. ( ) ( ), 4.5 in. 9 in.A D Cy= = =r j r i r k( ) ( ) ( ) ( )2 2/ 4.5 in. 9 in. 4.5 9 10.0623 in.D C D C CDl= − = − = + − =r r r i k( ) ( )( ) ( )//6 4.5 93 in. 6 in.9 9D CB Cc −= = = −r i kr i k( ) ( )/ 9 3 6 3 in. 3 in.B C B C= + = + − = +r r r k i k i k/ 3 3A B A B y= − = − + −r r r i j k2 2 2 2 2 2 2 2/ / : 15 3 3AB A B A Bl x y z y= + + = + +( ) ( ) ( )/14.3875 in. 3 in. 14.3875 in. 3 in.A By = = − + −r i j kVelocity of collar B./D CB BCDvl=rv( )( )( ) ( )2.5 4.5 91.11803 in./s 2.23607 in./s10.0623B−= = −i kv i kVelocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v rωωωωNoting that /A Bv is perpendicular to / ,A Br we get / / 0A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )3 14.3875 3 3 14.3875 3 1.11803 2.23607Av− + − ⋅ = − + − ⋅ −i j k j i j k i k( )( ) ( )( )14.3875 3 1.11803 3 2.23607 or 0.23313 in./sA Av v= − + − − =Relative velocity. /A B A B= −v v v( ) ( ) ( )/ 0.23313 in./s 1.11803 in./s 2.23607 in./sA B = − +v j i k( ) ( ) ( ) ( ) ( )2 2 2 2 2/ 0.23313 1.11803 2.23607 6.30435 in./sA Bv = + − + =Acceleration of collar B. 0B =aAcceleration of collar A. A Aa=a j/ / / /, whereA B A B A B AB A B AB A B= + = × + ×a a a a r vαααα ωωωω
  • 267. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Noting that /AB A B× rαααα is perpendicular to / ,A Br we get / / 0A B AB A B⋅ × =r rααααWe note also that / / / / /A B AB A B A B A B A B⋅ × = ⋅ ×r v v rωωωω ωωωω( )2/ / /A B A B A Bv= − ⋅ = −v vThen, ( ) ( )2 2/ / / /0A B A B A B A Bv v⋅ = − = −r aForming / ,A B A⋅r a we get ( )/ / / / / /A B A A B A A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r aor ( )2/ / /A B A A B B A Bv⋅ = ⋅ −r a r a (2)From (2), ( ) ( )3 14.3875 3 0 6.30435Aa− + − ⋅ = −i j k j14.3875 6.30435Aa = −( )20.438 in./sA = −a j
  • 268. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 214.Geometry. / / /20(24 in.) (10 in.)24B A D A B A= + =r i j r r( ) ( )/ / / 8 in. 8.3333 in.D C D A C A= − = +r r r i j2 224 10 26 in.ABl = + =Unit vector along AB./ 12 513 12B AABABl= = +rλ i jLet Oxyz be a frame of reference currently coinciding with OXYZ but rotating with angular velocity( )1 6 rad/sω= =Ω j j(a) Velocity of D. /D D D AB′= +v v v( ) ( )/ 6 8 8.3333 48 in./sD D C′ = × = × + = −v Ω r j i j k( ) ( )/12 578 72 in./s 30 in./s13 13D AB ABu = = + = +  v λ i j i j( ) ( ) ( )72.0 in./s 30.0 in./s 48 in./sD = + −v i j k(b) Acceleration of D. / /2D D D AB D AB′= + + ×a a a Ω v( )( ) ( )22 21 8 6 288 in./sD rω′ = − = − = −a i i i/ 0D AB =a( )( ) ( ) ( )2/2 2 6 72 30 864 in./sD F× = × + = −Ω v j i j k( ) ( )2 2288 in./s 864 in./sD = − −a i k
  • 269. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 215.Geometry. ( )( )/ 10 in. cos30 sin30D C = ° − °r i j( ) ( )5 3 in. 5 in.= −i jLet frame Cxyz, which at the instant shown coincides with CXYZ, rotate with angular velocity( )1 10 rad/s .ω= =Ω j jMotion of coinciding point D′ in the frame.( ) ( )/ 10 5 3 5 50 3 in./sD D C′ = × = × + = −v Ω r j i j k( ) ( )2 2 210 10cos30 500 3 in./sD r′ = −Ω = − ° = −a i iMotion of point D relative to the frame 4.5 ft/s 54 in./su = =( ) ( ) ( )/ sin30 cos30 27 in./s 27 3 in./sD F u= ° + ° = +v i j i j( )2/ cos30 sin30D Fuρ= ⋅ − ° + °a i j( )254cos30 sin3010= − ° + °i j( ) ( )2 2145.8 3 in./s 145.8 in./s= − +i j(a) Velocity of point D. /D D D F′= +v v v( ) ( ) ( )27 in./s 27 3 in./s 50 3 in./sD = + −v i j k( ) ( ) ( )2.25 ft/s 3.90 ft/s 7.22 ft/sD = + −v i j kCoriolis acceleration. /2 D F×Ω v( )( ) ( ) ( )2/2 2 10 27 27 3 540 in./sD F× = × + = −Ω v j i j k(b) Acceleration of point D. / /2D D D F D F′= + + ×a a a Ω v( ) ( ) ( )2 2 2645.8 3 in./s 145.8 in./s 540 in./sD = − + −a i j k( ) ( ) ( )2 293.2 ft/s 12.15 ft/s 45.0 ft/sD = + −a i j k
  • 270. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 216.With the origin at point A, ( ) ( ) ( )0.2 m 0.3 m 0.15 mD = + −r i j k( ) ( ) 2 2/ 0.2 m 0.15 m , 0.2 0.15 0.25 mC B BCl= − = + =r i kLet the frame Axyz rotate with angular velocity ( )1 10 rad/sω= =i iΩΩΩΩ(a) ( ) ( ) ( )10 0.2 0.3 0.15 1.5 m/s 3 m/sD D′ = × = × + − = +v r i i j k j kΩΩΩΩ( ) ( ) ( )/0.60.6 m/s, 0.2 0.15 0.48 m/s 0.36 m/s0.25D Fu = = − = −v i k i k( ) ( ) ( )/ 0.48 m/s 1.5 m/s 2.64 m/sD D D F′= + = + +v v v i j kVelocity of D. ( ) ( ) ( )0.480 m/s 1.500 m/s 2.64 m/sD = + +v i j k(b) ( ) ( ) ( )2 210 1.5 3 30 m/s 15 m/sD D′ ′= × = × + = − +a v i j k j kΩΩΩΩ/ 0D F =a( )( ) ( ) ( )2/2 2 10 0.48 0.36 7.2 m/sD F× = × − =v i i k jΩΩΩΩ( ) ( )2 2/ /2 22.8 m/s 15 m/sD D D F D F′= + + × = − +a a a v j kΩΩΩΩAcceleration of D. ( ) ( )2 222.8 m/s 15.00 m/sD = − +a j k
  • 271. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 217.Geometry. ( ) ( ) ( )0.36 m , 0.15 m 0.52 mB D= = +r j r i k( ) ( ) ( ) 2 2 2/ 0.15 m 0.36 m 0.52 m 0.15 0.36 0.52 0.65 mB D DBl= − + − = + + =r i j k( ) ( ) ( ) ( )10.075 m 0.18 m 0.26 m2C B D= + = + +r r r i j kLet frame Axyz rotate with angular velocity ( )1 5 rad/sω= =k kΩΩΩΩVelocity Analysis. 975 mm/s 0.975 m/su = =( ) ( ) ( )5 0.075 0.18 0.26 0.9 m/s 0.375 m/sC C′ = × = × + + = − +v r k i j k i jΩΩΩΩ( ) ( ) ( ) ( )/ /0.9750.15 0.36 0.52 0.225 m/s 0.54 m/s 0.78 m/s0.65C F D BDBul= = − + − = − + −v r i j k i j k/C C C F′= +v v v( ) ( ) ( )1.125 m/s 0.915 m/s 0.780 m/sC = − + −v i j kAcceleration Analysis. / 0C F =a( ) ( ) ( )2 25 0.9 0.375 1.875 m/s 4.5 m/sC C′ ′= × = × − + = − −a v k i j i jΩΩΩΩ( )( ) ( ) ( ) ( )2 2/2 2 5 0.225 0.54 0.78 5.4 m/s 2.25 m/sC F× = × − + − = − −v k i j k i jΩΩΩΩ/ /2C C C F C F′= + + ×a a a vΩΩΩΩ( ) ( )2 27.28 m/s 6.75 m/sC = − −a i j
  • 272. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 218.Geometry. With the origin at A, ( )6.75 in.B =r jLet frame Axyz rotate about the Y axis with constant angular velocity ( )1 9 rad/s .ω= =k kΩΩΩΩ Then, themotion relative to the frame consists of rotation about the x axis with constant angular velocity( )2 2 12 rad/s .ω= =i iωωωωMotion of coinciding point .B′( )9 6.75 60.75 in./sB B′ = × = × = −v r k j iΩΩΩΩB B B′ ′= × + ×a r vαααα ΩΩΩΩ( ) ( )20 9 60.75 546.75 in./s= + × − = −k i jMotion relative to the frame.( )/ 2 12 6.75 81 in./sB F B= × = × =v r i j kωωωω( )/ 2 2 /20 12 81 972 in./sB F B B F= × + ×= + × = −a r vi k jωωωωαααα(a) Velocity of point B. /B B B F′= +v v v( ) ( )60.8 in./s 81.0 in./sB = − +v i kCoriolis acceleration. /2 B F× vΩΩΩΩ( )( )/2 2 9 81 0B F× = × =v k kΩΩΩΩ(b) Acceleration of point B. / /2B B B F B F′= + + ×a a a vΩΩΩΩ( )21518.75 in./sB = −a j( )2126.6 ft/sB = −a j
  • 273. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 219.Geometry. With the origin at A, ( ) ( )6.75 in. 4.5 in.C = +r j kLet frame Axyz rotate about the Y axis with constant angular velocity ( )1 9 rad/s .ω= =k kΩΩΩΩ Then, the motionrelative to the frame consists of rotation about the x axis with constant angular velocity( )2 2 12 rad/s .ω= =i iωωωωMotion of coinciding pointC′ in the frame.( ) ( )9 6.75 4.5 60.75 in./sC C′ = × = × + = −v r k j k iΩΩΩΩC C C′ ′= × + ×a r vαααα ΩΩΩΩ( ) ( )20 9 60.75 546.75 in./s= + × − = −k i jMotion relative to the frame.( ) ( ) ( )( ) ( ) ( )/ 2/ 2 2 /2 212 6.75 4.5 54 in./s 81 in./s0 12 54 81 972 in./s 648 in./sC F CC F C C F= × = × + = − += × + ×= + × − + = − −v r i j k j ka r vi j k j kωωωωαααα ωωωω(a) Velocity of point C. /C C C F′= +v v v( ) ( ) ( )60.8 in./s 54.0 in./s 81.0 in./sC = − − +v i j kCoriolis acceleration. /2 C F× vΩΩΩΩ( )( ) ( ) ( )2/2 2 9 54 81 972 in./sC F× = × − + =v k j k iΩΩΩΩ(b) Acceleration of point C. / /2C C C F C F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 2972 in./s 1518.75 in./s 648 in./sC = − −a i j k( ) ( ) ( )2 2 281.0 ft/s 126.6 ft/s 54.0 ft/sC = − −a i j k
  • 274. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 220.Geometry. ( )( )0.36 m cos20 sin 20C = ° − °r i jLet frame Oxyz rotate about the Y axis with angular velocity 1ω= jΩΩΩΩ and angular acceleration 0.=&ΩΩΩΩThen, the motion relative to the frame consists of rotation with angular velocity 2ω=2 kωωωω and angularacceleration 2 0=αααα about the z axis.(a) ( )3 0.36cos 20 0.36sin 20 1.08cos 20C C′ = × = × ° − ° = − °v r j i j kΩΩΩΩ( )/ 2 4 0.36cos 20 0.36sin 201.44sin 20 1.44cos 20C F C= × = × ° − °= ° + °v r k i ji jωωωω/ 1.44sin 20 1.44cos 20 1.08cos 20C C C F′= + = ° + ° − °v v v i j k( ) ( ) ( )0.493 m/s 1.353 m/s 1.015 m/sC = + −v i j k(b) ( )3 1.08cos 20 3.24cos 20C C′ ′= × = × − ° = − °a v j k iΩΩΩΩ( )/ 2 / 4 1.44sin 20 1.44cos 205.76cos 20 5.76sin 20C F C F= × = × ° + °= − ° + °a v k i ji jωωωω( )( ) ( )/2 2 3 1.44sin 20 1.44cos 20 8.64sin 20C F× = × ° + ° = − °v j i j kΩΩΩΩ( )/ /23.24 5.76 cos 20 5.76sin 20 8.64sin 20C C C F C F′= + + ×= − + ° + ° − °a a a vi j kΩΩΩΩ( ) ( ) ( )2 2 28.46 m/s 1.970 m/s 2.96 m/sC = − + −a i j k
  • 275. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 221.Geometry. ( )( ) ( )0.36 m cos 20 sin 20 0.180 mD = ° − ° +r i j kLet frame Oxyz rotate about the Y axis with angular velocity 1ω= jΩΩΩΩ and angular acceleration 0.=&ΩΩΩΩThen, the motion relative to the frame consists of rotation with angular velocity 2 2ω= kωωωω and angularacceleration 2 0=αααα about the z axis.(a) ( )3 0.36cos20 0.36sin 20 0.180D D′ = × = × ° − ° +v r j i j kΩΩΩΩ0.540 1.08cos 20= − °i k( ))/ 2 4 0.36cos20 0.36sin 20 0.1801.44sin 20 1.44cos20D F D= × = × ° − ° += ° + °v r k i j ki jωωωω( )/ 0.540 1.44sin 20 1.44cos 20 1.08cos 20D D D F′= + = + ° + ° − °v v v i j k( ) ( ) ( )1.033 m/s 1.353 m/s 1.015 m/sD = + −v i j k(b) ( )3 0.540 1.08cos20D D′ ′= × = × − °a v j i kΩΩΩΩ3.24cos 20 1.62= − ° −i k( )/ 2 / 4 1.44sin 20 1.44cos20D F D F= × = × ° + °a v k i jωωωω5.76cos 20 5.76sin 20= − ° + °i j( )( ) ( )/2 2 3 1.44sin 20 1.44cos 20 8.64sin 20D F× = × ° + ° = − °v j i j kΩΩΩΩ( ) ( )/ /23.24 5.76 cos 20 5.76sin 20 1.62 8.64sin 20D D D F C F′= + + ×= − + ° + ° − + °a a a vi j kΩΩΩΩ( ) ( ) ( )2 2 28.46 m/s 1.970 m/s 4.58 m/sD = − + −a i j k
  • 276. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 222.With the origin at point A, ( ) ( ) ( )0.2 m 0.3 m 0.15 mD = + −r i j k( ) ( ) 2 2/ 0.2 m 0.15 m , 0.2 0.15 0.25 mC B BCl= − = + =r i kLet the frame Axyz rotate with angular velocity ( )1 10 rad/sω= =i iΩΩΩΩ and angular acceleration( )21 15 rad/s .ω= = −i i& &ΩΩΩΩ(a) ( ) ( ) ( )10 0.2 0.3 0.15 1.5 m/s 3 m/sD D′ = × = × + − = +v r i i j k j kΩΩΩΩ( ) ( ) ( )/0.60.6 m/s, 0.2 0.15 0.48 m/s 0.36 m/s0.25D Fu = = − = −v i k i k( ) ( ) ( )/ 0.48 m/s 1.5 m/s 2.64 m/sD D D F′= + = + +v v v i j kVelocity of D. ( ) ( ) ( )0.480 m/s 1.500 m/s 2.64 m/sD = + +v i j k(b) D D D′ ′= × + ×a r v&ΩΩΩΩ ΩΩΩΩ( ) ( )15 0.2 0.3 0.15 10 1.5 3= − × + − + × +i i j k i j k( ) ( )2 22.25 4.5 30 15 32.25 m/s 10.5 m/s= − − − + = − +j k j k j k2rel 3 m/sa =( ) ( ) ( )2 2rel30.2 0.15 2.4 m/s 1.8 m/s0.25= − = −a i k i k( )( ) ( ) ( )2/2 2 10 0.48 0.36 7.2 m/sD F× = × − =v i i k jΩΩΩΩ( ) ( ) ( )rel /2 2.4 m/s 25.05 m/s 8.70 m/sD D D F′= + + × = − +a a a Ω v i j kAcceleration of D. ( ) ( ) ( )2 2 22.40 m/s 25.1m/s 8.70 m/sD = − +a i j k
  • 277. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 223.Geometry. ( ) ( ) ( )0.36 m , 0.15 m 0.52 mB D= = +r j r i k( ) ( ) ( ) 2 2 2/ 0.15 m 0.36 m 0.52 m 0.15 0.36 0.52 0.65 mB D DBl= − + − = + + =r i j k( ) ( ) ( ) ( )10.075 m 0.18 m 0.26 m2C B D= + = + +r r r i j kLet frame Axyz rotate with angular velocity ( )1 5 rad/s .ω= =k kΩΩΩΩVelocity analysis. 2 2975 mm/s 0.975 m/su = =( ) ( ) ( )5 0.075 0.18 0.26 0.9 m/s 0.375 m/sC C′ = × = × + + = − +v r k i j k i jΩΩΩΩ( ) ( ) ( ) ( )/ /0.9750.15 0.36 0.52 0.225 m/s 0.54 m/s 0.78 m/s0.65C F D BDBul= = − + − = − + −v r i j k i j k/C C C F′= +v v v( ) ( ) ( )1.125 m/s 0.915 m/s 0.780 m/sC = − + −v i j kAcceleration analysis. 210 rad/s, 6.5 m/suα = = −k &( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )2 2/2 2 25 0.9 0.375 1.875 m/s 4.5 m/s6.510 0.075 0.18 0.26 0.15 0.36 0.520.651.8 m/s 0.75 m/s 1.5 m/s 3.6 m/s 5.2 m/sC CC F C′ ′= × = × − + = − −= × +−= × + + + − + −= − + + − +a v k i j i ja r uk i j k i j ki j i j k&ΩΩΩΩαααα( )( ) ( ) ( ) ( )2 2/2 2 5 0.225 0.54 0.78 5.4 m/s 2.25 m/sC F× = × − + − = − −v k i j k i jΩΩΩΩ/ /2C C C F C F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 27.58 m/s 9.60 m/s 5.20 m/sC = − − +a i j k
  • 278. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 224Geometry. With the origin at A, ( ) ( )6.75 in. 4.5 in.C = +r j kLet frame Axyz rotate about the y axis with angular velocity ( )1 9 rad/sω= =k kΩΩΩΩ and angular acceleration( )21 45 rad/s .α= = −k kαααα Then, the motion relative to the frame consists of rotation about the x axis withangular velocity ( )2 2 12 rad/sω= =i iωωωω and angular acceleration ( )2 2 60 rad/s .α= = −i iααααMotion of coinciding pointC′ in the frame.( ) ( )( ) ( ) ( )( ) ( )2 29 6.75 4.5 60.75 in./s45 6.75 4.5 9 60.75303.75 in./s 546.75 in./sC CC C C′′ ′= × = × + = −= × + ×= − × + + × −= −v r k j k ia r vk j k k ii jΩΩΩΩαααα ΩΩΩΩMotion relative to the frame.( ) ( ) ( )( ) ( ) ( )( ) ( )/ 2/ 2 2 /2 212 6.75 4.5 54 in./s 81 in./s60 6.75 4.5 12 54 81270 405 972 648702 in./s 1053 in./sC F CC F C C F= × = × + = − += × + ×= − × + + × − += + − − −= − −v r i j k j ka r vi j k i j kj k j kj kωωωωαααα ωωωω(a) Velocity of point C. /C C C F′= +v v v( ) ( ) ( )60.8 in./s 54.0 in./s 81.0 in./sC = − − +v i j kCoriolis acceleration. /2 C F× vΩΩΩΩ( )( ) ( ) ( )2/2 2 9 54 81 972 in./sC F× = × − + =v k j k iΩΩΩΩ(b) Acceleration of point C. / /2C C C F C F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 21275.75 in./s 1248.75 in./s 1053 in./sC = − −a i j k( ) ( ) ( )2 2 2106.3 ft/s 104.1ft/s 87.8 ft/sC = − −a i j k
  • 279. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 225.Let frame Oxyz rotate with angular velocity 1 .ω= jΩΩΩΩLet s s DAω=ωωωω λλλλ be the spin of gear A about the axle AD, where sin cosDA θ θ= − +λ i j is a unit vector alongthe axle.( ) ( )sin cos cos sinC L r L rθ θ θ θ= − − +r i jDue to motion of the frame, ( )1 cos sinC C r Lω θ θ′ = × = −v r kΩΩΩΩDue to spin ,sωωωω / /C F s C A srω= × =v r kωωωωThen, ( )/ 1 cos sinC C C F sr L rω θ θ ω′  = + = − + v v v kSince gear B is fixed, 0C =v1 sin cossLrω ω θ θ = −  (a) Angular velocity. s= +ωωωω ΩΩΩΩ ωωωω( )1 1 sin cos sin cosLrω ω θ θ θ θ = + − − +  j i jωωωω1 sin cos sin sin cosL Lr rω θ θ θ θ θ    = − + +        i jωωωω(b) Angular acceleration.1 10 sin cos sin cos sinOxyzL Lr rωω ω θ θ θ θ θ= = + ×      = + × − + +           j i j& &αααα ωωωω ΩΩΩΩ ωωωω21 sin sin cosLrω θ θ θ = −  kαααα
  • 280. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 226.Let frame Oxyz rotate with angular velocity 1 .ω= jΩΩΩΩLet s s DAω=ωωωω λλλλ be the spin of gear A about the axle AD, where sin cosDA θ θ= − +λ i j is a unit vector alongthe axle.( ) ( )sin cos cos sinC L r L rθ θ θ θ= − − +r i jDue to motion of the frame, ( )1 cos sinC C r Lω θ θ′ = × = −v r kΩΩΩΩDue to spin ,sωωωω / /C F s C A srω= × =v r kωωωωThen, ( )/ 1 cos sinC C C F sr L rω θ θ ω′  = + = − + v v v kSince gear B is rotating with angular velocity 2 ,ω j on gear B( )2 / 2 sin cosC C B L rω ω θ θ= × = − −v j r kEquating the two expressions for Cv and solving for ,sω( )1 2 sin cossLrω ω ω θ θ = − −  (a) Angular velocity. ( ) ( )1 1 2 sin cos sin cosLrω ω ω θ θ θ θ = + − − − +  j i jωωωω1 sin cos sin cos sinL Lr rω θ θ θ θ θ    = − + +        i jωωωω( )2 cos sin sin cosLrω θ θ θ θ + − − +  i j(b) Angular acceleration.Frame Oxyz is rotating with angular velocity 1 .ω= jΩΩΩΩ( )1 1 1 20 sin cosOxyzLrω ω ω ω θ θ= = + × = + × = − −  j k& &αααα ωωωω ωωωω ΩΩΩΩ ωωωωωωωω( )1 1 2 sin cosLrω ω ω θ θ = − −  kαααα
  • 281. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 227.Let frame Oxyz rotate with angular velocity ( )1 5 rad/s .ω= =i iΩΩΩΩ The motion relative to the frame is the spin( )2 4 rad/s .ω =k k(a) Angular acceleration.2Oxyz ω= = + × k& &αααα ωωωω ωωωω ΩΩΩΩ0 5 4 20= + × = −i k j ( )220.0 rad/s= − jαααα(b) 0. .Acceleration at point Pθ =( ) ( )( )( ) ( )( )( ) ( ) ( )( ) ( )1/ 21 12/ 2 2 /2/2 2/60 mm 0.06 m5 0.06 04 0.06 0.24 m/s00 4 0.24 0.96 m/s2 2 5 0.24 2.4 m/s0.96 m/s 2.4 m/sPP PP F PP P PP F P P Fc P FP P P F cvωωω ωω ω′′ ′′= == × = × == × = × == × + × == × + × = + × = −= × = × == + + = − +r i iv i r i iv k r k i ja i r i va k r k v k j ia i j ka a a a i k&&ΩΩΩΩ( ) ( )2 20.960 m/s 2.40 m/sP = − +a i k(c) 90 . .Acceleration at point Pθ = °( ) ( )( )( )( )( ) ( )( )( ) ( )1/ 221 12/ 2 2 ///60 mm 0.06 m5 0.06 0.3 m/s4 0.06 0.24 m/s0 5 0.3 1.5 m/s0 4 0.24 0.96 m/s2 2 5 0.24 0PP PP F PP P PP F p P Fc P FP P P F cωωω ωω ω′′ ′′= == × = × == × = × = −= × + × = + × = −= × + × = + × − = −= × = × − == + +r j jv i r i j kv k r k j ia i r i v i k ja k r k v k i ja v i ia a a a&&ΩΩΩΩ( )22.46 m/sP = −a j
  • 282. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 228.Let frame Oxyz rotate with angular velocity ( )1 5 rad/sω= =i iΩΩΩΩThe motion relative to the frame is the spin ( )2 4 rad/sω =k k( )( )30 , 60 mm cos30 sin30Pθ = ° = ° + °r i j( )( )0.06 m cos30 sin30P = ° + °r i j( ) ( )1 5 0.06cos30 0.06sin30 0.15 m/sP Pω′ = × = × ° + ° =v i r i i j k( )/ 2 4 0.06cos30 0.06sin30P F Pω= × = × ° + °v k r k i j( ) ( )0.12 m/s 0.20785 m/s= − +i j( )21 1 0 5 0.15 0.75 m/sP P Pω ω′ ′= × + × = + × = −a i r i v i k j&( )/ 2 2 / 0 4 0.12 0.20785P F P P Fω ω= × = × = + × − +a k r k v k i j&( ) ( )2 20.8314 m/s 0.48 m/s= − −i j( )( ) ( ) ( )2/2 2 5 0.12 0.20785 2.0785 m/sc P F= × = × − + =a v i i j kΩΩΩΩAcceleration at point P./P P P F c′= + +a a a a( ) ( ) ( )2 2 20.831m/s 1.230 m/s 2.08 m/sP = − − +a i j k
  • 283. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 229.Geometry. ( )( ) ( ) ( )/ 6 m sin30 cos30 3 m 3 3 mB A B= = ° + ° = +r r j k j kMethod 1Let the unextending portion of the boom AB be a rotating frame of reference.Its angular velocity is ( ) ( )2 1 0.40 rad/s 0.25 rad/s .ω ω= + = +i j i jΩΩΩΩIts angular acceleration is ( )21 2 1 2 0.10 rad/s .ω ω ω ω= × = − = −j i k kααααMotion of the coinciding point B′ in the frame.( ) ( ) ( ) ( ) ( )( ) ( ) ( )2 2 20.40 0.25 3 3 3 0.75 3 m/s 1.2 3 m/s 1.2 m/s0 0 0.10 0.40 0.25 00 3 3 3 0.75 3 1.2 3 1.20.30 0.30 0.48 1.15614 0.60 m/s 0.48 m/s 1.15614 m/sB BB B B′′= × = + × + = − += × + × = − +−= + − − = − −v r i j j k i j ki j k i j ka r vi i j k i j kΩΩΩΩαααα ΩΩΩΩMotion relative to the frame.( ) ( ) ( )//sin30 cos30 0.45 m/s sin30 0.45 m/s cos300B FB Fu= ° + ° = ° + °=v j k j kaVelocity of point B. /B B B F′= +v v v0.75 3 1.2 3 1.2 0.45sin30 0.45cos30B = − + + ° + °v i j k j k( ) ( ) ( )1.299 m/s 1.853 m/s 1.590 m/sB = − +v i j kCoriolis acceleration. /2 B F× vΩΩΩΩ( )( ) ( )( ) ( ) ( )/2 2 22 2 0.40 0.25 0.45sin30 0.45cos300.194856 m/s 0.31177 m/s 0.18 m/sB F× = + × ° + °= − +v i j j ki j kΩΩΩΩAcceleration of point B. / /2B B B F B F′= + + Ω ×a a a v( ) ( ) ( )0.60 0.194856 0.48 0.31177 1.15614 0.18B = + − + + − +a i j k( ) ( ) ( )2 2 20.795 m/s 0.792 m/s 0.976 m/sB = − −a i j k
  • 284. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Method 2Let frame ,Axyz which at the instant shown coincides with AXYZ, rotate with an angularvelocity ( )1 1 0.25 rad/s .ω= =j jΩΩΩΩ Then, the motion relative to this frame consists of turningthe boom relative to the cab and extending the boom.Motion of the coinciding point B′ in the frame.( ) ( )( ) ( )20.25 3 3 3 0.75 3 m/s0.25 0.75 3 0.1875 3 m/sB BB B′′ ′= = × + =×= × = × = −v j j k ira v j i kΩΩΩΩΩΩΩΩMotion of point B relative to the frame.Let the unextending portion of the boom be a rotating frame with constant angular velocity( )2 2 0.40 rad/s .ω= =i iΩΩΩΩ The motion relative to this frame is the extensional motion with speed u.( ) ( ) ( )( ) ( ) ( )22 220.40 3 3 3 1.2 3 m/s 1.2 m/s0.40 1.2 3 1.2 0.48 m/s 0.48 3 m/sB BB B′′′′ ′′= × = × + = − += × = × − + = − −v r i j k j ka v i j k j kΩΩΩΩΩΩΩΩ( ) ( ) ( )/boom/boomsin30 cos30 0.45 m/s sin30 0.45 m/s cos300BBu= ° + ° = ° + °=v j k j ka( )( ) ( )( ) ( )2 /boom2 22 2 0.40 0.45sin30 0.45cos300.31177 m/s 0.18 m/sB× = × ° + °= − +v i j kj kΩΩΩΩ/ /boom 1.2 3 1.2 0.45sin30 0.45cos30B F B B′′= + = − + + ° + °v v v j k j k( ) ( )1.85346 m/s 1.58971 m/s= − +j k( ) ( )/ /boom 2 /boom2 220.48 0.48 3 0 0.31177 0.18 0.79177 m/s 0.65138 m/sB F B B B′′= + + ×= − − + − + = − −a a a vj k j k j kΩΩΩΩVelocity of point B. /B B B F′= +v v v0.75 3 1.85346 1.58971B = − +v i j k( ) ( ) ( )1.299 m/s 1.853 m/s 1.590 m/sB = − +v i j kCoriolis acceleration. 1 /2 B F× vΩΩΩΩ( )( ) ( )1 /2 2 0.25 1.85346 1.58971 0.79486B F× = × − + =v j j k iΩΩΩΩAcceleration of point B. / 1 /2B B B F B F′= + + ×a a a vΩΩΩΩ0.1875 3 0.79177 0.65138 0.79486B = − − − +a k j k i( ) ( ) ( )2 2 20.795 m/s 0.792 m/s 0.976 m/sB = − −a i j k
  • 285. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 230.Geometry. ( )( ) ( )( )160 mm sin 45 cos45 0.16 m sin 45 cos45A = ° + ° = ° + °r j k j kLet frame ,Oxyz which coincides with the fixed frame OXYZ at the instant shown, be rotating about they axis with constant angular velocity ( )1 0.8 rad/s .ω= =j jΩΩΩΩ Then, the motion relative to the frame consistsof a rotation about the x axis with constant angular velocity( ) ( )2 300 rpm 10 rad/s .ddtθπ= − = − = −i i iωωωωMotion of the coinciding point A′ in the frame.( ) ( )( )20.8 0.16sin 45 0.16cos 45 0.128 m/s cos 450.8 0.128cos 45 0.1024 m/s cos 45A AA A′′ ′= × = × ° + ° = °= × = × ° = − °v r j j k ia v j i kΩΩΩΩΩΩΩΩMotion relative to the frame.( )( ) ( )( )( ) ( )/ 2/ 2 /2 2 2 210 0.16sin 45 0.16cos451.6 m/s cos45 1.6 m/s sin 4510 1.6 cos45 1.6 sin 4516 m/s sin 45 16 m/s cos45A F AA F A Fππ ππ π ππ π= × = − × ° + °= ° − °= × = − × ° − °= − ° − °v r i j kj ka v i j kj kωωωωωωωωVelocity of point A. /A A A F′= +v v v( ) ( ) ( )0.128 m/s cos45 1.6 m/s cos45 1.6 m/s sin 45A π π= ° + ° − °v i j k( ) ( ) ( )0.0905 m/s 3.55 m/s 3.55 m/sA = + −v i j kCoriolis acceleration. /2 A F× vΩΩΩΩ( )( ) ( ) ( )2/2 2 0.8 1.6 cos 45 1.6 sin 45 2.56 m/s sin 45A F π π π× = × ° − ° = − °v j j k iΩΩΩΩAcceleration of point A. / /2A A A F A F′= + + ×a a a vΩΩΩΩ( )2 22.56 sin 45 16 sin 45 16 cos 45 0.1024cos 45A π π π= − ° − ° − ° + °a i j k( ) ( ) ( )2 2 25.68 m/s 111.7 m/s 111.7 m/sA = − − −a i j k
  • 286. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 231.Geometry. ( )( ) ( ) ( )60 , 8 in. cos60 sin 60 4 in. 4 3 in.Aθ = ° = ° + ° = +r i j i jLet the frame Oxyz rotate about the Y axis with constant angular velocity ( )1 12 rad/s .ω= =j jΩΩΩΩ Then, themotion relative to the frame consists of rotation about the z axis with constant angular velocity( )2 2 8 rad/s .ω= =k kωωωωMotion of the coinciding point A′ in the frame.( ) ( )( ) ( )111212 4 4 3 48 in./s0 12 48 576 in./sA AA A Addtωωω′′ ′= × = × + = − = × + ×  = + × − = −v j r j i j ka j r j vj k iMotion relative to the frame.( ) ( ) ( )( ) ( ) ( )/ 22/ 2 /2 28 4 4 3 32 3 m/s 32 m/s0 8 32 3 32 256 in./s 256 3 in./sA F AA F A A Fddtωω= × = × + = − + = × + ×  = + × − + = − −v k r k i j i ja k r k vk i j i jωωωωVelocity of point A. /A A A F′= +v v v( ) ( ) ( )32 3 in./s 32 in./s 48 in./sA = − + −v i j k( ) ( ) ( )4.62 ft/s 2.67 ft/s 4.00 ft/sA = − + −v i j kCoriolis acceleration.( )( ) ( ) ( )2/2 2 12 32 3 32 768 3 in./sA F× = × − + =v j i j kΩΩΩΩAcceleration of point A. / /2A A A F A F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 2832 in./s 256 3 in./s 768 3 in./sA = − − +a i j k( ) ( ) ( )2 2 269.3 ft/s 37.0 ft/s 110.9 ft/sA = − − +a i j k
  • 287. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 232.Geometry. ( )( ) ( ) ( )60 , 8 in. cos60 sin 60 4 in. 4 3 in.Aθ = ° = ° + ° = +r i j i jLet the frame Oxyz rotate about the Y axis with angular velocity ( )1 12 rad/sω= =j jΩΩΩΩ and angularacceleration ( )2116 rad/s .ddtω = = −  j jαααα Then, the motion relative to the frame consists of rotation aboutthe z axis with constant angular velocity ( )2 2 8 rad/sω= =k kωωωω and angular acceleration( )222 10 rad/s .ddtω = =  k kααααMotion of the coinciding point A′ in the frame.( ) ( )( ) ( ) ( ) ( ) ( )1112 212 4 4 3 48 in./s16 4 4 3 12 48 64 in./s 576 in./sA AA A Addtωωω′′ ′= × = × + = − = × + ×  = − × + + × − = −v j r j i j ka j r j vj i j j k k iMotion relative to the frame.( ) ( ) ( )( ) ( )( ) ( )/ 22/ 2 /8 4 4 3 32 3 in./s 32 in./s10 4 4 3 8 32 3 3240 3 40 256 256 3 256 40 3 256 3 40A F AA F A A Fddtωωω= × = × + = − + = × + ×  = × + + × − += − + − − = − + − −v k r k i j i ja k r k vk i j k i ji j i j i jVelocity of point A. /A A A F′= +v v v( ) ( ) ( )32 3 in./s 32 in./s 48 in./sA = − + −v i j k( ) ( ) ( )4.62 ft/s 2.67 ft/s 4.00 ft/sA = − + −v i j kCoriolis acceleration.( )( ) ( ) ( )2/2 2 12 32 3 32 768 3 in./sA F× = × − + =v j i j kΩΩΩΩAcceleration of point A. / /2A A A F A F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )832 40 3 256 3 40 768 3 64A = − + − − + +a i j k( ) ( ) ( )2 2 275.1ft/s 33.6 ft/s 116.2 ft/sA = − − +a i j k
  • 288. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 233.. Let moving frame rotate about the axis with angular velocityFrame of reference Axyz Y( )1 0.15 rad/s .ω= =j jΩΩΩΩ.Geometry ( ) ( )/ 5cos20 5sin 20 4.6985 m 1.7101 mB A = − ° + ° = − +r i j i jPlace point O on Y axis at same level as point A.( ) ( ) ( )/ / / / 0.8 m 3.8985 m 1.7101 mB O B A A O B A= + = + = − +r r r r i i jMotion of corresponding point B′ in the frame.( ) ( ) ( )/ 0.15 3.8985 1.7101 0.58477 m/sB B O′ = × = × − + =v r j i j kΩΩΩΩ( ) ( ) ( )20.15 0.58477 0.087715 m/sB B′ ′= × = × =a v j k iΩΩΩΩ.Motion of point B relative to the frame ( )2 0.25 rad/sddtθ= − =k kωωωω( ) ( )/ 2 / 0.25 4.6985 1.7101B F B A= × = − × − +v r k i jωωωω( ) ( )0.42753 m/s 1.17462 m/s= − +i j( ) ( )/ 2 / 0.25 0.42753 1.17462B F B F= × = − × − +a v k i jωωωω( ) ( )2 20.29365 m/s 0.106881 m/s= +i j.Velocity of point B /B B B F′= +v v v( ) ( ) ( )0.428 m/s 1.175 m/s 0.585 m/sB = − + +v i j k.Coriolis acceleration ( )( ) ( )/2 2 0.15 0.42753 1.17462B F× = − × − +v j i jΩΩΩΩ( )20.128259 m/s= − k.Acceleration of point B / /2B B B F B F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 20.381 m/s 0.1069 m/s 0.1283 m/sB = + −a i j k
  • 289. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 234.Geometry. Dimensions in meters./ 0.1C B =r k / 0.070711 0.070711D C = − +r j k/ 0.070711 0.170711D B = − +r j kMethod 1Let the rigid body ABC be a rotating frame of reference.Its angular velocity and angular acceleration are( )1 8 rad/s ,ω= = = 0j j &ΩΩΩΩ ΩΩΩΩMotion of the coinciding point D′ in the frame.( )/ 8 0.070711 0.170711 1.36569D D B′ = × = × − + =v r j j k iΩΩΩΩ/ 0 8 0.136569 10.9255D D B D′ ′= × + × = + × = −a r v j i k&ΩΩΩΩ ΩΩΩΩMotion of point D relative to the frame.( )/ 2 / 4 0.070711 0.070711D ABC D Cω= × = × − +v i r i j k0.28284 0.28284= − −j k/ 2 / 2 /D ABC D C D ABCω ω= × + ×a i r i v&( )5 0.070711 0.070711= × − +i j k( )+ 4 0.28284 0.28284× − −i j k0.77781 1.48492= −j kVelocity of point D. /D D D ABC′= +v v v( ) ( ) ( )1.366 m/s 0.283 m/s 0.283 m/sD = − −v i j kCoriolis acceleration. /2 D ABC× vΩΩΩΩ( )( ) ( )/2 2 8 0.28284 0.28284 4.5255D ABC× = × − − = −v j j k iΩΩΩΩAcceleration of point D. / /2D D D ABC D ABC′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 24.53 m/s 0.778 m/s 12.41 m/sD = − + −a i j k
  • 290. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Method 2Bent rod ABC. (Rotation about the fixed y-axis)1 8ABC ω= =j jωωωω 1 0ABC ω= =j&αααα/ 8 0.1 0.8C ABC C B= × = × =v r j k iωωωω/ 0 8 0.8 6.4C ABC C B ABC C= × + × = + × = −a r v j i kαααα ωωωωRod CD. 1 2 8 4CD ω ω= + = +j i j iωωωω1 2 1 2 0 5 8 4 5 32CD ω ω ω ω= + + × = + + × = −j i j i i j i i k& &αααα/ / 4 8 00 0.070711 0.0707110.56569 0.28284 0.28284D C CD D C= × =−= − −i j kv ri j kωωωω/ / /5 0 4 8 00 0 070711 0 070711 0.56569 0 28284 0 282842.26276 0.35356 0.35356 2.26272 1.13136 5.65688D C CD D C CD C D= × + ×= −32 +− . . − . − .= − − − − + −a r vi j k i j ki j k i j kαααα ωωωωVelocity of point D. /D C D C= +v v v( ) ( ) ( )1.366 m/s 0.283 m/s 0.283 m/sD = − −v i j kAcceleration of point D. /D C D C= +a a a( ) ( ) ( )4.53 m/s 0.778 m/s 12.41 m/sD = − + −a i j k
  • 291. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 235.Geometry. ( ) ( ) ( ) ( )/ /7.5 in. 9 in. 18 in. , 9 in.A D A C= + − =r i j k r jLet frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the Y axiswith constant angular velocity ( )1 8rad/s .ω= =j jΩΩΩΩ Then, the motion relative to the frame consists of arotation of the disk AB about the bent axle CD with constant angular velocity ( )2 2 12 rad/s .ω= =k kωωωωMotion of the coinciding point A′ in the frame.( ) ( ) ( )/ 8 7.5 9 18 144 in./s 60 in./sA A D′ = × = × + − = − −v r j i j k i kΩΩΩΩ( ) ( ) ( )2 28 144 60 480 in./s 1152 in./sA A′ ′= × = × − − = − +a v j i k i kΩΩΩΩMotion of point A relative to the frame.( )/ 2 / 12 9 108 in./sA F A D= × = × = −v r k j iωωωω( ) ( )2/ 2 / 12 108 1296 in./sA F A F= × = × − = −a v k i jωωωωVelocity of point A. /A A A F′= +v v v( ) ( )144 60 108 252 in./s 60 in./sA = − − − = − −v i k i i k( ) ( )21.0 ft/s 5.00 ft/sA = − −v i kCoriolis acceleration. /2 A F× vΩΩΩΩ( )( ) ( ) ( )/2 2 8 108 1728 in./sA F× = × − =v j i kΩΩΩΩAcceleration of point A. / /2A A A F A F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 2480 1152 1296 1728 480 in./s 1296 in./s 2880 in./sA = − + − + = − − +a i k j k i j k( ) ( ) ( )2 2 240.0 ft/s 108.0 ft/s 240 ft/sA = − − +a i j k
  • 292. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 236.Geometry. ( ) ( ) ( ) ( )/ /7.5 in. 9 in. 18 in. , 9 in.B D B C= − − = −r i j k r jLet frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the Y axiswith constant angular velocity ( )1 8 rad/s .ω= =j jΩΩΩΩ Then, the motion relative to the frame consists of arotation of the disk AB about the bent axle CD with constant angular velocity ( )2 2 12 rad/s .ω= =k kωωωωMotion of the coinciding point B′ in the frame.( ) ( ) ( )( ) ( ) ( )/2 28 7.5 9 18 144 in./s 60 in./s8 144 60 480 in./s 1152 in./sB B DB B′′ ′= × = × − − = − −= × = × − − = − +v r j i j k i ka v j i k i kΩΩΩΩΩΩΩΩMotion of point B relative to the frame.( ) ( )( )/ 2 /2/ 2 /12 9 108 in./s12 108 1296 in./sB F B DB F B F= × = × − == × = × =v r k j ia v k i jωωωωωωωωVelocity of point B./B B B F′= +v v v( ) ( )144 60 108 36 in./s 60 in./sB = − − + = − −v i k i i k( ) ( )3.00 ft/s 5.00 ft/sB = − −v i kCoriolis acceleration./2 B F× vΩΩΩΩ( )( ) ( ) ( )2/2 2 8 108 1728 in./sB F× = × = −v j i kΩΩΩΩAcceleration of point B./ /2B B B F B F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 2480 1152 1296 1728 480 in./s 1296 in./s 576 in./sB = − + + − = − + −a i k j k i j k( ) ( ) ( )2 2 240.0 ft/s 108.0 ft/s 48.0 ft/sB = − + −a i j k
  • 293. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 237.Geometry. ( ) ( )120 mm 180 mmA = +r i kMethod 1Let the rigid body DCB be a rotating frame of reference.Its angular velocity is ( ) ( )1 2 1.2 rad/s 1.5 rad/s .ω ω= + = +i k i kΩΩΩΩIts angular acceleration is ( )21 2 1 2 1.8 rad/s .ω ω ω ω= × = − = −i k j jααααMotion of the coinciding point A′ in the frame.( ) ( )( )1.2 1.5 120 180216 180 36 mm/sA A′ = × = + × += − + = −v r i k i kj j jΩΩΩΩ( ) ( ) ( ) ( )( ) ( )2 21.8 120 180 1.2 1.5 36216 324 43.2 54 270 mm/s 172.8 mm/sA A A′ ′= × + ×= − × + + + × −= − − + = − +a r vj i k i k jk i k i i kαααα ΩΩΩΩMotion of point A relative to the frame.( )/ /60 mm/s , 0A F A Fu= = =v i i aVelocity of point A. /A A A F′= +v v v( ) ( )60.0 mm/s 36.0 mm/sA = −v i jCoriolis acceleration. /2 A F× vΩΩΩΩ( )( ) ( )2/2 2 1.2 1.5 60 180 mm/sA F× = + × =v i k i jΩΩΩΩAcceleration of point A. / /2A A A F A F′= + + ×a a a vΩΩΩΩ( ) ( ) ( )2 2 2270 mm/s 180.0 mm/s 172.8 mm/sA = − + +a i j kMethod 2Let frame Dxyz, which at instant shown coincides with DXYZ, rotate with an angular velocity1 1.2 rad/s.ω= =i iΩΩΩΩ Then the motion relative to the frame consists of the rotation of body DCB about the zaxis with angular velocity ( )2 1.5 rad/sω =k k plus the sliding motion ( )60 mmu= =u i i of the rod ABrelative to the body DCB.Motion of the coinciding point A′ in the frame.( ) ( )( ) ( )21.2 120 180 216 mm/s1.2 216 259.2 mm/sA AA A′′ ′= × = × + = −= × = × − = −v r i i k ja v i j kΩΩΩΩΩΩΩΩ
  • 294. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Motion of point A relative to the frame.( ) ( ) ( )( ) ( )( )( ) ( )( ) ( )/ 2/ 2 2 2 22 21.5 120 180 60 180 mm/s 60 mm/s20 1.5 180 0 2 1.5 60180 mm/s 270 mm/sA F AA F A Auu uωω ω ω= × + = × + + = += × + × × + + ×= + × + + ×= −v k r i k i k i j ia k r k k r i j ik j k ij i&ααααVelocity of point A. /A A A F′= +v v v216 180 60A = − + +v j j i( ) ( )60 mm/s 36 mm/sA = −v i jCoriolis acceleration. /2 A F× vΩΩΩΩ( )( ) ( ) ( )2/2 2 1.2 180 60 432 mm/sA F× = × + =v i j i kΩΩΩΩAcceleration of point A. / /2A A A F A F′= + + ×a a a vΩΩΩΩ259.2 180 270 432A = − + − +a k j i k( ) ( ) ( )2 2 2270 mm/s 180.0 mm/s 172.8 mm/sA = − + +a i j k
  • 295. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 238.Geometry. ( ) ( )120 mm 180 mmA = +r j kMethod 1Let the rigid body DCB be a rotating frame of reference.Its angular velocity is ( ) ( )1 2 1.2 rad/s 1.5 rad/s .ω ω= + = −i k i kΩΩΩΩIts angular acceleration is ( )21 2 1 2 1.8 rad/s .ω ω ω ω= × = − =i k j jααααMotion of the coinciding point A′ in the frame.( ) ( )( ) ( ) ( )1.2 1.5 120 180144 216 180 180 mm/s 216 mm/s 144 mm/sA A′ = × = − × += − + = − +v r i k j kk j i i j kΩΩΩΩ( ) ( )2 20 1.8 0 1.2 0 1.50 120 180 180 216 144324 324 442.8 259.2 442.8 mm/s 259.2 mm/sA A A′ ′= × + ×= + −− += − + − = −a r vi j k i j ki i j k j kαααα ΩΩΩΩMotion of point A relative to the frame.( )/ /60 mm/s , 0A F A Fu= = =v j j aVelocity of point A. /A A A F′= +v v v180 216 144 60A = − + +v i j k j( ) ( ) ( )180.0 mm/s 156.0 mm/s 144.0 mm/sA = − +v i j kCoriolis acceleration. /2 A F× vΩΩΩΩ( )( ) ( ) ( )2 2/2 2 1.2 1.5 60 180 mm/s 144 mm/sA F× = − × = +v i k j i kΩΩΩΩAcceleration of point A. / /2A A A F A F′= + + ×a a a vΩΩΩΩ442.8 259.2 180 144A = − − + +a j k i k( ) ( ) ( )2 2 2180 mm/s 443 mm/s 115.2 mm/sA = − −a i j k
  • 296. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Method 2Let frame Dxyz, which at instant shown coincides with DXYZ, rotate with an angular velocity1 1.2 rad/s.ω= =i iΩΩΩΩ Then the motion relative to the frame consists of the rotation of body DCBabout the z axis with angular velocity ( )2 1.5 rad/sω = −k k plus the sliding motion ( )60 mmu= =u i jof the rod AB relative to the body DCB.Motion of the coinciding point A′ in the frame.( ) ( ) ( )( ) ( ) ( )2 21.2 120 180 216 mm/s 144 mm/s1.2 216 144 172.8 mm/s 259.2 mm/sA AA A′′ ′= × = × + = − += × = × − + = − −v r i j k j ka v i j k j kΩΩΩΩΩΩΩΩMotion of point A relative to the frame.( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( )/ 2/ 2 2 2 22 21.5 120 180 60 180 mm/s 60 mm/s20 1.5 180 0 2 1.5 60270 180 180 mm/s 270 mm/sA F AA F A Auu uωω ω ω= × + = − × + + = += × + × × + + ×= + − × + + − ×= − + = −v k r j k j k j i ja k r k k r j k jk i k jj i i j&ααααVelocity of point A. /A A A F′= +v v v216 144 180 60A = − + − +v j k i j( ) ( ) ( )180.0 mm/s 156.0 mm/s 144.0 mm/sA = − +v i j kCoriolis acceleration. /2 A F× vΩΩΩΩ( )( ) ( ) ( )2/2 2 1.2 180 60 144 mm/sA F× = × + =v i i j kΩΩΩΩAcceleration of point A. / /2A A A F A F′= + + ×a a a vΩΩΩΩ172.8 259.2 180 270 144A = − − + − +a j k i j k( ) ( ) ( )2 2 2180.0 mm/s 443 mm/s 115.2 mm/sA = − −a i j k
  • 297. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 239.Geometry. ( ) ( )6 in. 9 in.A = −r i jLet frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the x axiswith constant angular velocity ( )1 6 rad/s .ω= =i iΩΩΩΩ Then, the motion relative to the frame is a spin about theaxle CD of angular velocity ( )2 8 rad/s .ω− = −j jMotion of the coinciding point A′ in the frame.( ) ( )( ) ( )26 6 9 54 in./s6 54 324 in./sA AA A′′ ′= × = × − = −= × = × − =v r i i j ka v i k jΩΩΩΩΩΩΩΩMotion of point A relative to the frame.( ) ( )( )/ 2/ 2 /8 6 9 48 in./s8 48 384 in./sA F AA F A F= − × = − × − == − × = − × = −v j r j i j ka j v j k iωωωωωωωωVelocity of point A./A A A F′= +v v v( )54 48 6 in./sA = − + = −v k k k( )0.500 ft/sA = −v kCoriolis acceleration./2 A F× vΩΩΩΩ( )( ) ( )2/2 2 6 48 576 in./sA F× = × = −v i k jΩΩΩΩAcceleration of point A./ /2A A A F A F′= + + ×a a a vΩΩΩΩ( ) ( )2 2324 384 576 384 in./s 252 in./sA = − − = − −a j i j i j( ) ( )2 232.0 ft/s 21.0 ft/sA = − −a i j
  • 298. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 240..Geometry ( ) ( )6 in. 14 in.B = −r i jLet frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the x axiswith constant angular velocity ( )1 6 rad/s .ω= =i iΩΩΩΩ Then, the motion relative to the frame is a spin about theaxle CD of angular velocity ( )2 8 rad/s .ω− = −j jMotion of the coinciding point B′ in the frame.( ) ( )( ) ( )26 6 14 84 in./s6 84 504 in./sB BB B′′ ′= × = × − = −= × = × − =v r i i j ka v i k jΩΩΩΩΩΩΩΩMotion of point B relative to the frame.( ) ( )( )/ 2/ 2 /8 6 14 48 in./s8 48 384 in./sB F BB F B F= − × = − × − == − × = − × = −v j r j i j ka j v j k iωωωωωωωωVelocity of point B./B B B F′= +v v v( ) ( ) ( )84 in./s 48 in./s 36 in./sB = − + = −v k k k( )3.00 ft/sB = −v kCoriolis acceleration./2 B F× vΩΩΩΩ( )( ) ( )2/2 2 6 48 576 in./sB F× = × = −v i k jΩΩΩΩAcceleration of point B./ /2B B B F B F′= + + ×a a a vΩΩΩΩ( ) ( )2 2504 384 576 384 in./s 72 in./sB = − − = − −a j i j i j( ) ( )2 232.0 ft/s 6.00 ft/sB = − −a i j
  • 299. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 241.Let the moving frame of reference be the unit less the pulleys and belt. It rotates about the Y axis with constantangular velocity ( )1.6 rad/s .ω =1= j jΩΩΩΩ The relative motion is that of the pulleys and belt with speed90 mm/s.u =(a) Acceleration at point A.( ) ( )100 mm 380 mmA = − +r i j( ) ( )1.6 100 380 160 mm/sA A′ = × = × − + =v r j i j kΩΩΩΩ( )21.6 160 256 mm/sA A′ ′= × = × =a v j k iΩΩΩΩ( )/ 90 mm/sA F u= =v k k( )2 22/90135 mm/s60A Fuρ   = − = − = −         a j j j( )( ) ( ) ( )2/2 2 1.6 90 288 mm/sA F× = × =v j k iΩΩΩΩ/ /2 256 135 288A A A F A F′= + + × = − +a a a v i j iΩΩΩΩ( ) ( )2 2544 mm/s 135.0 mm/sA = −a i j(b) Acceleration of point B.( ) ( ) ( )100 mm 200 mm 60 mmB = − + +r i j k( ) ( ) ( )1.6 100 200 60 96 mm/s 160 mm/sB B′ = × = × − + + = +v r j i j k i kΩΩΩΩ( ) ( ) ( )2 21.6 96 160 256 mm/s 153.6 mm/sB B′ ′= × = × + = −a v j i k i kΩΩΩΩ( )/ 90 mm/sB F u= − = −v j j/ 0B F =a( )( ) ( )/2 2 1.6 90 0B F× = × − =v j jΩΩΩΩ/ /2 256 153.6 0 0B B B F B F′= + + × = − + +a a a v i kΩΩΩΩ( ) ( )2 2256 mm/s 153.6 mm/sB = −a i k
  • 300. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 242.Let the moving frame of reference be the unit less the pulleys and belt. It rotates about the Y axis with constantangular velocity ( )1 1.6 rad/s .ω= =j jΩΩΩΩ The relative motion is that of the pulleys and belt with speed90 mm/s.u =(a) Acceleration at point C.( ) ( )100 mm 80 mmC = − +r i j( ) ( )1.6 100 80 160 mm/sC A′ = × = × − + =v r j i j kΩΩΩΩ( )21.6 160 256 mm/sC A′ ′= × = × =a v j k iΩΩΩΩ( )/ 90 mm/sC F u= − = −v k k( )2 22/90135 mm/s60C Fuρ   = = =         a j j j( )( ) ( ) ( )2/2 2 1.6 90 288 mm/sC F× = × − = −Ω v j k i/ /2 256 135 288C C C F C F′= + + × = + −a a a v i j iΩΩΩΩ( ) ( )2 232.0 mm/s 135.0 mm/sC = − +a i j(b) Acceleration at point D.( ) ( ) ( )100 mm 200 mm 60 mmD = − + −r i j k( ) ( ) ( )1.6 100 200 60 96 mm/s 160 mm/sD B′ = × = × − + − = − +v r j i j k i kΩΩΩΩ( ) ( ) ( )2 21.6 96 160 256 mm/s 153.6 mm/sD B′ ′= × = × − + = +a v j i k i kΩΩΩΩ( )/ 90 mm/sD F u= =v j j/ 0D F =a( )( ) ( )/2 2 1.6 90 0D F× = × − =v j jΩΩΩΩ/ /2 256 153.6 0 0D D D F D F′= + + × = + + +a a a v i kΩΩΩΩ( ) ( )2 2256 mm/s 153.6 mm/sD = +a i k
  • 301. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 243.( ) ( ) ( ) 2 2 2/ 16 in. 16 in. 8 in. 16 16 8 24 in.A E AEl= − + + = + + =r i j kAngular velocity.( )/1216 16 824A EAElω= = − + +r i j kωωωω( ) ( ) ( )8 rad/s 8 rad/s 4 rad/s= − + +i j kωωωω( ) ( )/ 16 in. 6 in.C E = − +r i jVelocity of C./ 8 8 4 24 64 8016 6 0C C E= × = − = − − +−i j kv r i j kωωωω( ) ( ) ( )24.0 in./s 64.0 in./s 80.0 in./sC = − − +v i j kAngular Acceleration. 0=ααααAcceleration of C./ 0 8 8 424 64 80C C E C= × + × = + −− −i j ka r vαααα ωωωω896 544 704= + +i j k( ) ( ) ( )2 2 2896 in./s 544 in./s 704 in./sC = + +a i j k
  • 302. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 244.( )( )02400 22400 rpm 80 rad/s, 1.00 in.60rπω π= = = =1 10, 10 stω = =(a) Before power is turned off, 0α =( )( )0 1.00 80 251.33 in./s,Cv rω π= = = 20.9 ft/sCv =2 22251.3363165 in./s1.00Cnvar= = =20, 63165 in./s ,t Ca r aα= = = 25260 ft/sCa =(b) Uniformly decelerated motion. 1 0 1tω ω α= +21 010 808 rad/s10tω ω πα π− −= = = −At 9 s, ( )( )0 80 8 9 8 rad/stω ω α π π π= + = − =( )( )1.00 8 25.133 in./s,Cv rω π= = = 2.09 ft/sCv =( )2 2225.133631.65 in./s1.00CC nCvar= = =( ) ( )( ) 21.00 8 25.133 in./sC Cta r α π= = − = −( )22 2 2 2631.65 25.133 632.15 in./sC n ta a a= + = + =252.7 ft/sCa =
  • 303. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 245.Arm . Fixed axis rotation.ACB/ /6 mm,A C A A C ABr r ω= =v ( )( )6 40= 240 mm/s=/ /36 mm,B C B B C ABr r ω= =v ( )( )36 40= 1440 mm/s=Disk B: Plane motion = Translation with B + Rotation about B./12 mm,B D B B Ar = = −v v v0 1440= 12 Bω+1440120 rad/s12Bω = =/E B E B= +v v v1440= ( )( )6 120+ 2880 mm/s=Disk A: Plane motion = Translation with A + Rotation about A./30 mm,A E A E Ar = = −v v v2880 240= 30 Aω+2880 240104 rad/s30Aω+= =Answers. (a) 104.0 rad/sA =ωωωω(b) 120.0 rad/sB =ωωωω
  • 304. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 246.Rod AB. (Rotation about A) ( ) 0.12B AB ABAB ω ω= =v 45°Rod DE. (Rotation about E) ( ) 0.15D DE DEED ω ω= =v 45°Rod BGD. Plane motion = Translation with B + Rotation about .B[/D B D B Dv= +v v v ] [45 Bv° = ]45° + [ /D Bv ]Draw velocity vector diagram./// //2 0.12 20.12 2 120.24 210.06 22Draw vector diagram.sin 45 .12 sin 45D B B ABD B ABBD ABBDG B BG BD D B ABG B G BG B ABv vvll vv vωωω ωω ωω= == = == = == += ° = °vv v v3.6,0.12sin 45 0.12sin 45GABvω = =° °42.4 rad/sAB =ωωωω( )12 42.4 ,2BDω =   30.0 rad/sBD =ωωωω0.12 5.0912 m/sD B ABv v ω= = =5.09120.15DDEEDvlω = = 33.9 rad/sDE =ωωωω
  • 305. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 247.Since the drum rolls without sliding, its instantaneous center lies at B.120 mm/sE D= =v v/ /,A A B D D Bv r v rω ω= =(a)/1203 rad/s100 60DD Bvrω = = =−3.00 rad/s=ωωωω(b) ( )( )60 3.00 180 mm/sAv = =180 mm/sA =vSince Av is to the right and Dv is to the left, cord is being unwound.180 120 300 mm/sA Ev v− = + =(c) Cord unwound per second = 300 mm
  • 306. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 248.16 in./sB =v5tan12EFDFγ = = D Dv=v γLocate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D areknown. Draw BC perpendicular to Bv and DC perpendicular to .Dv( ) ( )5tan 6.4 2.6667 in.,12CJ DJ γ = = =  12.8 2.6667 10.1333 in.CB JB CJ= − = − =(a)161.57895 rad/s10.1333BABDvCBω = = = 1.579 rad/sABD =ωωωω10.1333 7.2 17.3333 in.CK CB BK= + = + =3.6tan , 11.733 , 90 78.317.3333KACKβ β β= = = ° ° − = °17.333317.7032 in.cos cos 11.733CKACβ= = =°(b) ( ) ( )( )17.7032 1.57895 28.0 in./s,A ABDv AC ω= = = 28.0 in./sA =v 78.3°
  • 307. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 249..Velocity analysis 30 in./sD =v , 12 in./sC α=v , ω[/ or 12C D C D= +v v v ] [30= ] [6ω+ ]4212 30 6 , 7 rad/s6ω ω− = − = =.Acceleration analysis 236 in./sD =a , C Ca=a , α( ) ( )/ /C D C D C Dt n= + +a a a a[ Ca ] [36= ] [6α+ ] 26ω+ Components: 2: 0 36 6 6 rad/sα α= − =( )( )2 2: 6 7 294 in./sCa = = 224.5 ft/sC =a( ) ( )/ /A D A D A Dt n= + +a a a a[36= ] [6α+ ] 26ω+  [36= ] [36+ ] 294+  272 in./s= 2294 in./s + 225.2 ft/sA =a 76.2°( ) ( )/ /B D B D B Dt n= + +a a a a[36= ] [6α+ ] 26ω+ [36= ] [36+ ] [294+ ]2258 in./s= 236 in./s + 221.7 ft/sB =a 7.9°
  • 308. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 250..Velocity analysis 8 rad/sABω =( ) ( )( )12 8 96 in./sB ABv AB ω= = =B Bv=v , D Dv=vInstantaneous center of bar BD lies at .∞0, 96 in./sBD D Bω = = =v v968 rad/s12DDEvDEω = = =.Acceleration analysis ( )220, 64 rad/sAB ABα ω= =( )B ABAB α= a ] ( ) 2ABAB ω+  ( )( )0 12 64= + 2768 in./s =[/ 2D B BDlα=a ] [ BDlα+ ] 22 BDlω+ 2BDlω + [24 BDα= ] [12 BDα+ ] 0+ [24 BDα= ] [12 BDα+ ]( )D DEDE α= a ( ) 2DEDE ω+  [12 DEα= ] ( )( )12 64+  [12 DEα= ] 2768 rad/s+ / Resolve into components.D B D B= +a a a2: 768 768 24 64 rad/sBD BDα α= − + =[/ /1122C B D B BDα= =a a ] [6 BDα+ ]2768 in./s= 2384 in./s + [/ 768C B C B= + =a a a ] [768+ ] [384+ ]2384 in./sC =a
  • 309. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 251.0.5500 mm 0.5 m, 0.5tan30 ,cos30AB AP BP= = = ° =°8 rad/sAE =ωωωω , 3 rad/sBD =ωωωωLet P′ be the coinciding point on AE and 1u be the outward velocity of the collar along the rod AE.( )/P P P AE AEAP ω′ = + = v v v ] [ 1u+ ]Let P′′ be the coinciding point on BD and 2u be the outward speed along the slot in rod BD.( )/P P P BD BDBP ω′′ = + = v v v ] [ 230 u° + ]60°Equate the two expressions for Pv and resolve into components.: ( )( )1 20.53 cos30 cos60cos30u u = ° + ° ° or 1 21.5 0.5u u= + (1): ( )( ) ( ) 20.50.5tan30 8 3 sin30 sin60cos30u − ° = − ° + ° ° [ ]211.5 tan30 4tan30 1.66667 m/ssin60u = ° − ° = −°From (1), ( )( )1 1.5 0.5 1.66667 0.66667 m/su = + − =( )( )0.5tan30 8P = °v  [0.66667+ ] 2.3094 m/s=   [0.66667 m/s+ ]2 22.3094 0.66667 2.4037 m/sPv = − + =2.3094tan 73.90.66667β β= = °2.40 m/sP =v 73.9°
  • 310. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 252.Let the arm AB be a rotating frame of reference. 0.75 rad/s=ΩΩΩΩ ( )0.75 rad/s= − kLink 1: ( )1 1/2 in. , AB u= − =r i v ( )4 in./s= j( ) ( ) ( )221 1 0.75 2 1.125 in./s′ = −Ω = − − =a r i i2 221/48 in./s2ABuρ= = =a ( )28 in./s= i( )( ) ( ) ( )/2 2 0.75 4 6 in./sP AB× = − × =v k j iΩΩΩΩ( )21 1 1/ 1/2 15.125 in./sAB AB′= + + × =a a a v iΩΩΩΩ21 15.13 in./s=aLink 2: ( ) ( )2 2/8 in. 2 in. AB u= + =r i j v ( )4 in./s= i( ) ( ) ( ) ( )22 2 22 2 0.75 8 2 4.5 in./s 1.125 in./s′ = −Ω = − + = − −a r i j i j2/ 0AB =a( )( ) ( ) ( )22/2 2 0.75 4 6 in./sAB× = − × = −v k i jΩΩΩΩ( ) ( )2 2 2/ 2/2 224.5 1.125 6 4.5 in./s 7.125 in./sAB AB′= + + ×= − − − = − −a a a Ω vi j j i j( ) ( )2 2 22 4.5 7.125 8.43 in./sa = + =7.125tan , 57.74.5β β= = °22 8.43 in./s=a 57.7°
  • 311. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 253.Geometry. ( ) ( ) ( ) ( )/ /0.16 m , 0.5 m 0.4 m 0.16 mB C A B= − = − + −r k r i j kVelocity at B. ( ) ( )0 / 3 0.16 0.48 m/sB B Cω= × = × − =v j r j k iVelocity of collar A. A Av=v j/ / /, whereA B A B A B AB A B= + = ×v v v v ω rNoting that /A Bv is perpendicular to / ,A Br we get / / 0.A B A B⋅ =r vForming / ,A B A⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r vor / /A B A A B B⋅ = ⋅r v r v (1)From (1), ( ) ( ) ( ) ( )0.5 0.4 0.16 0.5 0.4 0.16 48Av− + − ⋅ = − + − ⋅i j k j i j k i0.4 0.24 or 0.6 m/sA Av v= − = − ( )0.600 m/sA = −v j
  • 312. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 15, Solution 254.Let frame Oxyz rotate with angular velocity ( )1 8 rad/s .ω= =i iΩΩΩΩNote that point B does not move.Geometry. ( )/ sin 1 cos , 120 mm, 60A B ρ θ ρ θ ρ θ= + − = = °r i j( ) ( )/ 103.923 mm 60 mmA B = +r i jMotion of coinciding point A′( ) ( )/ 8 103.923 60 480 mm/sA A B′ = × = × + =v r i i j kωωωω( )2/ 0 8 480 3840 mm/sA A B A′ ′= × + × = + × = −a r v i k jαααα ωωωωMotion relative to the frame.( )/ cos sin , 600 mm/sP F u uθ θ= + =v i j( ) ( )/ 300 mm/s 519.62 mm/sP F = +v i j( )2/ sin cosA Fuθ θρ= − +a i j( ) ( )2 2/ 2598.1 mm/s 1500 mm/sA F = − +a i jCoriolis acceleration. /2c P F= ×a vΩΩΩΩ( )( ) ( ) ( )22 8 300 519.62 8313.9 mm/sc = × + =a i i j k(a) Velocity of A. /A A A F′= +v v v( ) ( ) ( )300 mm/s 519.62 mm/s 480 mm/sA = + +v i j k( ) ( ) ( )0.300 m/s 0.520 m/s 0.480 m/sA = + +v i j k(b) Acceleration of A. /A A A F c′= + +a a a a( ) ( ) ( )2 2 22598.1 mm/s 2340 mm/s 8313.9 mm/sA = − − +a i j k( ) ( ) ( )2 2 22.60 m/s 2.34 m/s 8.31m/sA = − − +a i j k