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

  1. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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

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