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# solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 11

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• 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 11, Solution 2.( )232 mx t t= − −( )23 2 2 m/sdxv t tdt= = − −26 2 m/sdva tdt= = −(a) Time at a = 0.00 6 2 0t= − =013t = 0 0.333 st =(b) Corresponding position and velocity.3 21 12 2.741 m3 3x⎛ ⎞ ⎛ ⎞= − − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠2.74 mx = −21 13 2 2 3.666 m/s3 3v⎛ ⎞ ⎛ ⎞= − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠3.67 m/sv =
• 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 11, Solution 3.Position: 4 35 4 3 2 ftx t t t= − + −Velocity: 3 220 12 3 ft/sdxv t tdt= = − +Acceleration: 2 260 24 ft/sdva t tdt= = −When 2 s,t =( )( ) ( )( ) ( )( )4 35 2 4 2 3 2 2x = − − − 52 ftx =( )( ) ( )( )3 220 2 12 2 3v = − + 115 ft/sv =( )( ) ( )( )260 2 24 2a = − 2192 ft/sa =
• 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 11, Solution 4.Position: 4 3 26 8 14 10 16 in.x t t t t= + − − +Velocity: 3 224 24 28 10 in./sdxv t t tdt= = + − −Acceleration: 2 272 48 28 in./sdva t tdt= = + −When 3 s,t =( )( ) ( )( ) ( )( ) ( )( )4 3 26 3 8 3 14 3 10 3 16x = + − − + 562 in.x = !( )( ) ( )( ) ( )( )3 224 3 24 3 28 3 10v = + − − 770 in./sv = !( )( ) ( )( )272 3 48 3 28a = + − 2764 in./sa = !
• 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 11, Solution 5.Position: 500sin mmx kt=Velocity: 500 cos mm/sdxv k ktdt= =Acceleration: 2 2500 sin mm /sdva k ktdt= = −When 0.05 s, and 10 rad/st k= =( )( )10 0.05 0.5 radkt = =( )500sin 0.5x = 240 mmx = !( )( ) ( )500 10 cos 0.5v = 4390 mm/sv = !( )( ) ( )2500 10 sin 0.5a = − 3 224.0 10 mm/sa = − × !
• 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 11, Solution 6.Position: ( )21 250sin mmx k t k t= −Where 21 21 rad/s and 0.5 rad/sk k= =Let 2 21 2 0.5 radk t k t t tθ = − = −( )2221 rad/s and 1 rad/sd dtdt dtθ θ= − = −Position: 50sin mmx θ=Velocity: 50cos mm/sdx dvdt dtθθ= =Acceleration:dvadt=222250cos 50sin mm/sd dadtdtθ θθ θ⎛ ⎞= − ⎜ ⎟⎝ ⎠When 0,v = either cos 0θ =or 1 0 1 sdt tdtθ= − = =Over 0 2 s, values of cos are:t θ≤ ≤( )st 0 0.5 1.0 1.5 2.0( )radθ 0 0.375 0.5 0.375 0cosθ 1.0 0.931 0.878 0.981 1.0No solutions cos 0 in this range.θ =For 1 s,t = ( )( )21 0.5 1 0.5 radθ = − =( )50sin 0.5x = 24.0 mmx =( )( ) ( )( )50cos 0.5 1 50sin 0.5 0a = − − 243.9 mm/sa = −
• 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 11, Solution 7.Given: 3 26 9 5x t t t= − + +Differentiate twice. 23 12 9dxv t tdt= = − +6 12dva tdt= = −(a) When velocity is zero. 0v =( )( )23 12 9 3 1 3 0t t t t− + = − − =1 s and 3 st t= =(b) Position at t = 5 s.( ) ( )( ) ( )( )3 25 5 6 5 + 9 5 + 5x = − 5 25 ftx =Acceleration at t = 5 s.( )( )5 6 5 12a = − 25 18 ft/sa =Position at t = 0.0 5 ftx =Over 0 ≤ t < 1 s x is increasing.Over 1 s < t < 3 s x is decreasing.Over 3 s < t ≤ 5 s x is increasing.Position at t = 1 s.( ) ( )( ) ( )( )3 21 1 6 1 9 1 5 9 ftx = − + + =Position at t = 3 s.( ) ( )( ) ( )( )3 23 3 6 3 9 3 5 5 ftx = − + + =Distance traveled.At t = 1 s 1 1 0 9 5 4 ftd x x= − = − =At t = 3 s 3 1 3 1 4 5 9 8 ftd d x x= + − = + − =At t = 5 s 5 3 5 3 8 25 5 28 ftd d x x= + − = + − =5 28 ftd =
• 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 11, Solution 8.( )322 ftx t t= − −( )22 3 2 ft/sdxv t tdt= = − −(a) Positions at v = 0.( )2 22 3 2 3 14 12 0t t t t− − = − + − =214 (14) (4)( 3)( 12)(2)( 3)t− ± − − −=−1 21.1315 s and 3.535 st t= =1At 1.1315 s,t = 1 1.935 ftx = 1 1.935 ftx =2At 3.535 s,t = 2 8.879 ftx = 2 8.879 ftx =(b) Total distance traveled.0At 0,t t= = 0 8 ftx =4At 4 s,t t= = 4 8 ftx =Distances traveled.10 to :t 1 1.935 8 6.065 ftd = − =1 2to :t t 2 8.879 1.935 6.944 ftd = − =2 4to :t t 3 8 8.879 0.879 ftd = − =Adding, 1 2 3d d d d= + + 13.89 ftd =
• 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 11, Solution 9.0.23 ta e−=0 0v tdv a dt=∫ ∫0.2 0.20030 30.2tt t tv e dt e− −− = =−∫( ) ( )0.2 0.215 1 15 1t tv e e− −= − − = −At t = 0.5 s, ( )0.115 1v e−= − 1.427 ft/sv =0 0x tdx v dt=∫ ∫( )0.2 0.20010 15 1 150.2tt t tx e dt t e− −⎛ ⎞− = − = +⎜ ⎟⎝ ⎠∫( )0.215 5 5tx t e−= + −At 0.5 s,t = ( )0.115 0.5 5 5x e−= + − 0.363 ftx =
• 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 11, Solution 10.Given: 20 05.4sin ft/s , 1.8 ft/s, 0, 3 rad/sa kt v x k= − = = =0 0 0 05.45.4 sin costt tv v adt kt dt ktk− = = − =∫ ∫( )5.41.8 cos 1 1.8cos 1.83v kt kt− = − = −Velocity: 1.8cos ft/sv kt=0 0 0 01.81.8 cos sintt tx x vdt kt dt ktk− = = =∫ ∫( )1.80 sin 0 0.6sin3x kt kt− = − =Position: 0.6sin ftx kt=When 0.5 s,t = ( )( )3 0.5 1.5 radkt = =1.8cos1.5 0.1273 ft/sv = = 0.1273 ft/sv =0.6sin1.5 0.5985 ftx = = 0.598 ftx =
• 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 11, Solution 11.Given: 23.24sin 4.32cos ft/s , 3 rad/sa kt kt k= − − =0 00.48 ft, 1.08 ft/sx v= =( ) ( )0 0 0 00 03.24 sin 4.32 cos3.24 4.321.08 cos sin3.24 4.32cos 1 sin 03 31.08cos 1.08 1.44sint t tt tv v a dt kt dt kt dtv kt ktk kkt ktkt kt− = = − −− = −= − − −= − −∫ ∫ ∫Velocity: 1.08cos 1.44sin ft/sv kt kt= −( ) ( )0 0 0 00 01.08 cos 1.44 sin1.08 1.440.48 sin cos1.08 1.44sin 0 cos 13 30.36sin 0.48cos 0.48t t tt tx x vdt kt dt kt dtx kt ktk kkt ktkt kt− = = −− = += − + −= + −∫ ∫ ∫Position: 0.36sin 0.48cos ftx kt kt= +When 0.5 s,t = ( )( )3 0.5 1.5 radkt = =1.08cos1.5 1.44sin1.5 1.360 ft/sv = − = − 1.360 ft/sv = − !0.36sin1.5 0.48cos1.5 0.393 ftx = + = 0.393 ftx = !
• 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 11, Solution 12.Given: 2mm/s where is a constant.a kt k=At 0,t = 400 mm/s; at 1 s, 370 mm/s, 500 mmv t v x= = = =2400 0 012v t tdv a dt kt dt kt= = =∫ ∫ ∫2 21 1400 or 4002 2v kt v kt− = = +At 1 s,t = ( )2 31400 1 370, 60 mm/s2v k k= + = = −Thus 2400 30 mm/sv t= −At 7 s,t = ( )( )27 400 30 7v = − 7 1070 mm/sv = −2 2 2When 0, 400 30 0. Then 13.333 s , 3.651 sv t t t= − = = =For 0 3.651 s,t≤ ≤ 0 and is increasing.v x>For 3.651 s,t > 0 and is decreasing.v x<( )2500 1 1400 30x t tdx vdt t dt= = −∫ ∫ ∫( )3 31500 400 10 400 10 390tx t t t t− = − = − −Position: 3400 10 110 mmx t t= − +At 0,t =0 110 mmx x= =At 3.651 s,t = ( )( ) ( )( )3max 400 3.651 10 3.651 110 1083.7 mmx x= = − + =At 7 s,t = ( )( ) ( )( )37 400 7 10 7 110x x= = − + 7 520 mmx = −Distances traveled:Over 0 3.651 s,t≤ ≤1 max 0 973.7 mmd x x= − =Over 3.651 7 s,t≤ ≤2 max 7 1603.7 mmd x x= − =Total distance traveled: 1 2 2577.4 mmd d d= + = 2580 mmd =
• 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 11, Solution 13.Determine velocity. 0.15 2 20.15v t tdv adt dt−= =∫ ∫ ∫( ) ( )( )0.15 0.15 0.15 2v t− − = −0.15 0.45 m/sv t= −At 5 s,t = ( )( )5 0.15 5 0.45v = − 5 0.300 m/sv =When 0,v = 0.15 0.45 0 3.00 st t− = =For 0 3.00 s,t≤ ≤ 0, is decreasing.v x≤For 3.00 5 s,t≤ ≤ 0, is increasing.v x≥Determine position. ( )10 0 00.15 0.45x t tdx v dt t dt−= = −∫ ∫ ∫( ) ( )2 2010 0.075 0.45 0.075 0.45tx t t t t− − = − = −20.075 0.45 10 mx t t= − −At 5 s,t = ( )( ) ( )( )25 0.075 5 0.45 5 10 10.375 mx = − − = −5 10.38 mx = −At 0,t =0 10 m (given)x = −At 3.00 s,t = ( )( ) ( )( )23 min 0.075 3.00 0.45 3.00 10 10.675 mmx x= = − − = −Distances traveled:Over 0 3.00 s,t≤ ≤ 1 0 min 0.675 md x x= − =Over 3.00 s 5 s,t< <2 5 min 0.300 md x x= − =Total distance traveled: 1 2 0.975 md d d= + = 0.975 md =
• 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 11, Solution 14.Given: 29 3a t= −Separate variables and integrate.( )20 09 3 9v tdv a dt t dt= = − =∫ ∫ ∫30 9v t t− = − ( )29v t t= −(a) When v is zero. 2(9 ) 0t t− =0 and 3 s (2 roots)t t= = 3 st =(b) Position and velocity at 4 s.t =( )35 0 09x t tdx v dt t t dt= = −∫ ∫ ∫2 49 152 4x t t− = −2 49 152 4x t t= + −At 4 s,t = ( ) ( )2 449 15 4 42 4x⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠4 13 mx =( )( )24 4 9 4v = − 4 28 m/sv = −(c) Distance traveled.Over 0 3 s,t< < v is positive, so x is increasing.Over 3 s 4 s,t< ≤ v is negative, so x is decreasing.At 3 s,t = ( ) ( )2 439 15 3 3 25.25 m2 4x⎛ ⎞ ⎛ ⎞= + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠At 3 st = 3 3 0 25.25 5 20.25 md x x= − = − =At 4 st = 4 3 4 3 20.25 13 25.25 32.5 md d x x= + − = + − = 4 32.5 md =
• 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 11, Solution 15.Given: 2dva ktdt= =Separate variables dv = kt2dtIntegrate using v = –10 m/s when t = 0 and v = 10 m/s when t = 2 s.10 2 210 0dv kt dt−=∫ ∫10 310013tv kt−=[ ] ( )31(10) ( 10) 2 03k ⎡ ⎤− − = −⎢ ⎥⎣ ⎦(a) Solving for k,( )( )3 208k = 47.5 m/sk =(b) Equations of motion.Using upper limit of v at t,( )3 31001 110 7.53 3tvv kt v t−⎛ ⎞= + = ⎜ ⎟⎝ ⎠310 2.5 m/sv t= − +Then,310 2.5dxv tdt= = − +Separate variables and integrate using x = 0 when t = 2 s.( )310 2.5dx t dt= − +( )30 210 2.5x tdx t dt= − +∫ ∫420 10 0.625tx t t⎡ ⎤− = − +⎣ ⎦( )( ) ( )( )4410 0.0625 10 2 0.625 2t t ⎡ ⎤⎡ ⎤= − + − − +⎢ ⎥⎣ ⎦ ⎣ ⎦[ ]410 0.625 10t t= − + − −410 10 0.625 mx t t= − +
• 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 11, Solution 16.Note that is a given function of .a x ( )40 160 160 0.25a x x= − = −( ) Note that is maximum when 0, or 0.25 ma v a x= =( )Use 160 0.25 with the limitsvdv adx x dx= = −max0.3 m/s when 0.4 m and when 0.25 mv x v v x= = = =( )max 0.250.3 0.4160 0.25vvdv x dx= −∫ ∫( ) ( )0.252 22 2max0.40.25 0.150.3160 160 0 1.82 2 2 2xv ⎡ ⎤− −⎢ ⎥− = − = − − =⎢ ⎥⎣ ⎦2 2 2max 3.69 m /sv = max 1.921 m/sv =( ) Note that is maximum or minimum when 0.b x v =( )Use 160 0.25 with the limitsvdv adx x= = −0.3 m/s when 0.4 m, and 0 when mv x v x x= = = =( )00.3 0.4160 0.25mxvdv x dx= −∫ ∫( ) ( )( ) ( )( )2 22 20.40.3 0.250 160 80 0.25 80 0.152 2mxmxx−− = − = − − + −( )20.25 0.02306 0.25 0.1519 mm mx x− = − = ±0.0981 m and 0.402 mmx =
• 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 11, Solution 17.is a function of :a x ( ) 2100 0.25 m/sa x= −( )Use 100 0.25 with limitsvdv a dx x dx= = − 0 when 0.2 mv x= =( )0 0.2100 0.25v xvdv x dx= −∫ ∫( )( )220.21 10 100 0.252 2xv x− = − −( )250 0.25 0.125x= − − +So( ) ( )2 220.25 100 0.25 or 0.5 1 400 0.25v x v x= − − = ± − −Use( )2or0.5 1 400 0.25dx dxdx vdt dtv x= = =± − −Integrate:( )0 0.2 20.5 1 400 0.25t x dxdtx= ±− −∫ ∫Let ( )20 0.25 ; when 0.2 = 1 and 20u x x u du dx= − = = −So 1 11 211 1sin sin10 10 210 1uu dut u uuπ− − = = = −  −∫m m mSolve for .u 1sin 102u tπ−= m( )sin 10 cos 10 cos102u t t tπ = = ± =  m( )cos 10 20 0.25u t x= = −continued
• 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.Solve for and .x v10.25 cos1020x t= −1sin102v t=Evaluate at 0.2 s.t =( )( )( )10.25 cos 10 0.220x = − 0.271 mx =( )( )( )1sin 10 0.22v = 0.455 m/sv =
• 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 11, Solution 18.Note that is a given function ofa xUse ( ) ( )2 3600 1 600 600vdv adx x kx dx x kx dx= = + = +Using the limits 7.5 ft/s when 0,v x= =and 15 ft/s when 0.45 ft,v x= =( )15 0.45 37.5 0600 600vdv x kx dx= +∫ ∫15 0.4522 407.5600 6002 2 4vx kx⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦( ) ( )( )( ) ( ) ( )2 22 415 7.5300 0.45 150 0.452 2k− = +84.375 60.75 6.1509k= +Solving for ,k 23.84 ftk −=
• 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 11, Solution 19.Note that is a given function of .a xUse ( )3800 3200vdv adx x x dx= = +Using the limit 10 ft/s when 0,v x= =( )310 0800 3200v xvdv x x dx= +∫ ∫( )222 410400 8002 2vx x− = +2 4 2 21600 800 100 Letv x x u x= + + =Then ( )( )2 21 21600 800 100 1600 ,v u u u u u u= + + = − −1 2where and are the roots ofu u 21600 800 100 0u u+ + =Solving the quadratic equation,( ) ( )( )( )( )( )21,2800 800 4 1600 100 800 00.25 02 1600 3200u− ± − − ±= = = − ±21 2 0.25 ftu u= = −So ( ) ( )222 2 2 2 21600 0.25 1600 0.5 ft /sv u x= + = +Taking square roots, ( )2 240 0.5 ft/sv x= ± +
• 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.Use( )2 2or40 0.5dx dxdx vdt dtv x= = = ±+2 240 Use limit 0 when 00.5dxdt x tx= ± = =+12 20 0140 tan0.5 0.50.5t x dx xdtx−= ± = ±+∫ ∫( ) ( )1 140 2.0tan 2 or tan 2 20t x x t− −= ± = ±( ) ( )2 tan 20 or 0.5tan 20x t x t= ± = ±( ) ( ) ( )2 20.5 sec 20 20 10 sec 20dxv t tdt = = ± = ± At 0, 10 ft/s, which agrees with the given data if the minus sign is rejected.t v= = ±Thus, ( ) ( )210 sec 20 ft/s, and 0.5tan 20 ftv t x t= =At 0.05 s,t = 20 1.0 radt =( )221010sec 1.0cos 1.0v = = 34.3 ft/sv =( )0.5tan 1.0x = 0.779 ftx =
• 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 11, Solution 20.Note that is a given function of .a x 2712 28 12 m/s3a x x = − = −  7Use 12 with the limits3vdv adx x dx = = −  8 m/s when 0.v x= =228 08 07 12 7123 2 2 3xvv x vvdv x dx x    = − = −         ∫ ∫2 22 28 12 7 72 2 2 3 3vx    − = − −         2 2 22 2 7 7 7 48 12 123 3 3 3v x x      = + − − = − −             27 4123 3v x = ± − −  Reject minus sign to get 8 m/s at 0.v x= =(a) Maximum value of .x max0 whenv x x= =2 27 4 7 112 0 or3 3 3 9x x   − − = − =      max max7 1 8 22 m and m 2 m3 3 3 3x x x− = ± = = =Now observe that the particle starts at 0 with 0 and reaches 2 m. At 2 m, 0 andx v x x v= > = = =20, so that becomes negative and decreases. Thus, 2 m is never reached.3a v x x< =max 2 mx = !
• 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.(b) Velocity when total distance traveled is 3 m.The particle will have traveled total distance 3 md = when max maxd x x x− = − or 3 2 2 x− = −or 1 m.x =Using27 4123 3v x = − − −  , which applies when x is decreasing, we get27 412 1 203 3v = − − − = −  4.47 m/sv = − !
• 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.Chapter 11, Solution 21.Note that is a function of .a x ( )1 xa k e−= −( )Use 1 with the limits 9 m/s when 3 m, and 0 when 0.xvdv adx k e dx v x v x−= = − = = − = =( )0 09 31 xvdv k e dx−−= −∫ ∫( )02 0392xvk x e−−⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠( )2390 0 1 3 16.08552k e k⎡ ⎤− = + − − − = −⎣ ⎦(a) 2.5178k = 22.52 m/sk =( ) ( )Use 1 2.5178 1 with the limit 0 when 0.x xvdv adx k e dx e dx v x− −= = − = − = =( )0 02.5178 1v x xvdv e dx−= −∫ ∫( ) ( )202.5178 2.5178 12xx xvx e x e− −= + = + −( ) ( )1/225.0356 1 2.2440 1x xv x e v x e− −= + − = ± + −(b) Letting 2 m,x = −( )1/ 222.2440 2 1 4.70 m/sv e= ± − + − = ±Since begins at 2 m and ends at 0, 0.x x x v= − = >Reject the minus sign.4.70 m/sv =
• 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 11, Solution 22.0.000576.8 xdva v edx−= =0.000570 06.8v x xvdv e dx−=∫ ∫20.0005706.802 0.00057xxve−− =−( )0.0005711930 1 xe−= −When 30 m/s.v =( )( )20.000573011930 12xe−= −0.000571 0.03772xe−− =0.000570.96228xe−=0.00057 ln(0.96228) 0.03845x− = = −67.5 mx =
• 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 11, Solution 23.Given: 0.4dva v vdx= = −or 0.4dvdx= −Separate variables and integrate using 75 mm/s when 0.v x= =75 00.4 75 0.4v xdv v x= − − = −∫ ∫(a) Distance traveled when 0v =0 75 0.4x− = − 187.5 mmx =(b) Time to reduce velocity to 1% of initial value.(0.01)(75) 0.75v = =0.752.5ln75t = − 11.51 st =
• 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 11, Solution 24.Given:dva v kvdx= = − 2Separate variables and integrate using 9 m/s when 0.v x= =9 0v xdvk dxv= −∫ ∫ln9vkx= −Calculate using 7 m/s when 13 m.k v x= =( )( ) 3 17ln 13 19.332 10 m9k k − −= − = ×Solve for .x1ln 51.728 ln9 9v vxk= − = −(a) Distance when 3 m/s.v =351.728 ln9x⎛ ⎞= − ⎜ ⎟⎝ ⎠56.8 mx =(b) Distance when 0.v =( )51.728 ln 0x = − x = ∞
• 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 11, Solution 25.0 0, 0, 25 ft/svdv adx k vdx x v= = − = =1/21dx v dvk= −0 003/21 23vx vx vvdx vdv vk k= − = −∫ ∫( ) ( )3/23/2 3/2 3/2 3/20 02 2 2or 25 1253 3 3x x v v x v vk k k⎡ ⎤ ⎡ ⎤− = − = − = −⎢ ⎥ ⎣ ⎦⎣ ⎦Noting that 6 ft when 12 ft/s,x v= =3/2 32 55.626 125 12 or 9.27 ft/s3kk k⎡ ⎤= − = =⎣ ⎦Then,( )( )( )3/2 3/22125 0.071916 1253 9.27x v v⎡ ⎤= − = −⎣ ⎦3/2125 13.905v x= −( ) Whena 8 ft,x = ( )( ) ( )3/23/2125 13.905 8 13.759 ft/sv = − =5.74 ft/sv =( )b dv adt k vdt= = −1/ 21 dvdtk v= −( )01/21/2 1/201 22vvt v v vk k⎡ ⎤= − ⋅ = −⎣ ⎦At rest, 0v =( )( )1/21/20 2 2529.27vtk= = 1.079 st =
• 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.Chapter 11, Solution 29.x as a function of v.0.000571154xve−= −20.000571154x ve− ⎛ ⎞= − ⎜ ⎟⎝ ⎠20.00057 ln 1154vx⎡ ⎤⎛ ⎞− = −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦21754.4 ln 1154vx⎡ ⎤⎛ ⎞= − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦(1)a as a function of x.( )2 0.0005723716 1v e−= −( )( )20.000511858 0.000572xdv d va v edx dx−⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠20.000576.75906 6.75906 1154x va e−⎡ ⎤⎛ ⎞= = −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦(2)(a) v = 20 m/s.From (1), x = 29.843 x = 29.8 mFrom (2), a = 6.64506 a = 6.65 m/s2(b) v = 40 m/s.From (1), x = 122.54 x = 122.5 mFrom (2), a = 6.30306 a = 6.30 m/s2
• 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 11, Solution 30.( )0.3Given: 7.5 1 0.04 with units km and km/hv x= −(a) Distance at 1 hr.t =0.3Using , we get7.5(1 0.04 )dx dxdx vdt dtv x= = =−Integrating, using 0t = when 0,x =( ) ( ) ( )( ){ }0.700.30 0 01 1 1or [ ] 1 0.047.5 7.5 0.7 0.041 0.04xt x tdxdt t x−= = ⋅ −−∫ ∫( ){ }0.74.7619 1 1 0.04t x= − − (1)Solving for ,x ( ){ }1/0.725 1 1 0.210x t= − −When 1 h,t = ( )( ){ }1/0.725 1 1 0.210 1x ⎡ ⎤= − −⎣ ⎦ 7.15 kmx =(b) Acceleration when 0.t =0.7 0.7(7.5)(0.3)( 0.04)(1 0.04 ) 0.0900(1 0.04 )dvx xdx− −= − − = − −When 0t = and 0,x = 17.5 km/h, 0.0900 hdvvdx−= −2(7.5)( 0.0900) 0.675 km/hdva vdx= = − = −22(0.675)(1000)m/s(3600)= − 6 252.1 10 m/sa −= − ×(c) Time to run 6 km.Using 6 kmx = in equation (1),( )( ){ }0.74.7619 1 1 0.04 6 0.8323 ht ⎡ ⎤= − − =⎣ ⎦49.9 mint =
• 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 11, Solution 31.The acceleration is given by22dv gRv adr r= = −Then,22gR drv dvr= −Integrating, using the conditions esc0 at , andv r v v= = ∞ = at r R=esc0 22v Rdrv dv gRr∞= −∫ ∫esc02 21 12 v Rv gRr∞⎛ ⎞= ⎜ ⎟⎝ ⎠2 2esc1 10 02v gRR⎛ ⎞− = −⎜ ⎟⎝ ⎠esc 2v gR=6 2Now, 3960 mi 20.909 10 ft and 32.2 ft/s .R g= = × =Then, ( )( )( )6esc 2 32.2 20.909 10v = × 3esc 36.7 10 ft/sv = ×
• 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 11, Solution 32.The acceleration is given by6220.9 1032.21 ya×−=⎡ ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦6220.9 1032.21 ydyvdv ady×−= =⎡ ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦20 maxIntegrate, using the conditions at 0 and 0 at . Also, use 32.2 ft/s andv v y v y y g= = = = =620.9 10 ft.R = ×( ) ( )00 22 20 01v yRdy dyv dv g gRR y∞ ∞= − = −++∫ ∫ ∫max002 201 12yvv gRR y⎛ ⎞= ⎜ ⎟+⎝ ⎠( )2 2 2max0 0 max maxmax max1 1 10 22gRyv gR v R y gRyR y R R y⎡ ⎤− = − = − + =⎢ ⎥+ +⎣ ⎦maxSolving for ,y20max 202RvygR v=−Using the given numerical data,( )( )( )6 2 6 20 0max 9 26 20020.9 10 20.9 101.34596 102 32.2 20.9 10v vyvv× ×= =× −× −0( ) 2400 ft/s,a v =( )( )( ) ( )26max 2920.9 10 24001.34596 10 2400y×=× −3max 89.8 10 fty = ×0( ) 4000 ft/s,b v =( )( )( ) ( )26max 2920.9 10 40001.34596 10 4000y×=× −3max 251 10 fty = ×0( ) 40000 ft/s,c v =( )( )( ) ( )26max 2920.9 10 40000negative1.34596 10 40000y×= =× −Negative value indicates that 0v is greater than the escape velocity.maxy = ∞
• 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 11, Solution 33.( )( ) Given: sin na v v tω ϕ′= +At 0,t = 00 sin or sinvv v vvϕ ϕ′= = =′ (1)Let x be maximum at 1t t= when 0.v =Then, ( ) ( )1 1sin 0 and cos 1n nt tω ϕ ω ϕ+ = + = ± (2)Using ordxv dx v dtdt= =Integrating, ( )cos nnvx C tω ϕω′= − +At 0,t = 0 0cos or cosn nv vx x C C xϕ ϕω ω′ ′= = − = +Then, ( )0 cos cos nn nv vx x tϕ ω ϕω ω′ ′= + − + (3)max 0 1cos using cos 1nnv vx x tϕ ω ϕω ω′ ′= + + + = −Solving for cos ,ϕ( )max 0cos 1nx xvωϕ−= −′max 0With 2 ,x x= 0cos 1nxvωϕ = −′ (4)Using2 22 2 0 0sin cos 1, or 1 1nv xv vωϕ ϕ   + = + − =   ′ ′   Solving for givesv′( )2 2 20 00(5)2nnv xvxωω+′ =
• 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.( ) Acceleration:b ( )cosn ndva v tdtω ω ϕ′= = +2Let be maximum at when 0.v t t a= =Then, ( )2cos 0ntω ϕ+ =From equation (3), the corresponding value of x is( )00 0 02 2 2 20 0 00 0 20 0cos 1 23 122 2 2nn n nnn n nv v x vx x x xvv x vx xx xωϕω ω ωωω ω ω′ ′ ′ = + = + − = − ′ += − = −( )002032nvxxω −  
• 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 11, Solution 34.0( ) 1 sindx ta v vdt Tπ⎡ ⎤= = −⎢ ⎥⎣ ⎦0Integrating, using 0 when 0,x x t= = =00 0 01 sinx t t tdx v dt v dtTπ⎡ ⎤= = −⎢ ⎥⎣ ⎦∫ ∫ ∫0000costx v T tx v tTππ⎡ ⎤= +⎢ ⎥⎣ ⎦0 00 cosv T t v Tx v tTππ π= + − (1)When 3 ,t T= ( )0 00 023 cos 3 3v T v Tx v T v TTππ π⎛ ⎞= + − = −⎜ ⎟⎝ ⎠02.36x v T=0cosdv v tadt T Tπ π= = −When 3 ,t T= 0cos3vaTππ= − 0vaTπ=( ) Using equation (1) with ,b t T=0 01 0 02cos 1v T v Tx v T v Tππ π π⎛ ⎞= + − = −⎜ ⎟⎝ ⎠Average velocity is1 0ave 021x x xv vt T πΔ − ⎛ ⎞= = = −⎜ ⎟Δ ⎝ ⎠ave 00.363v v=
• 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 11, Solution 35.10 km/h 2.7778 m/s= 100 km/h 27.7778 m/s=(a) Acceleration during start test.dvadt=8.2 27.77780 2.7778adt vdt=∫ ∫8.2 27.7778 2.7778a = − 23.05 m/sa =(b) Deceleration during braking.dva vdx= =44 00 27.7778a dx v dv= =∫ ∫( ) ( )044 2027.777812a x v=( )2144 27.77782a = −28.77 m/sa = − deceleration 28.77 m/sa= − =
• 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 11, Solution 36.10 km/h 2.7778 m/s= 100 km/h 27.7778 m/s=(a) Distance traveled during start test.dvadt=00t vva dt dv=∫ ∫0at v v= − 0v vat−=227.7778 2.77783.04878 m/s8.2a−= =0 2.7778 3.04878v v at t= + = +)8.20 02.7778 3.04878tx v dv t dt= = +∫ ∫( )( ) ( )( )22.7778 8.2 1.52439 8.2= + 125.3 mx =(b) Elapsed time for braking test.dva vdx=00x vvadx vdv=∫ ∫2 202 2v vax = −( ) ( )( )( )2 2 201 10 27.77782 2 44a v vx= − = −28.7682 m/s= −dvadt=00t vva dt dv=∫ ∫0at v v= −0 0 27.77788.7682v vta− −= =−3.17 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 11, Solution 37.Constant acceleration. 0 00, 0A Av v x x= = = =0v v at at= + = (1)2 20 01 12 2x x v t at at= + + = (2)At point ,B 2700 ft and 30 sBx x t= = =(a) Solving (2) for a,( )( )( )2 22 2700230xat= = 26 ft/sa =(b) Then, ( )( )6 30Bv at= = 180 ft/sBv =
• 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 11, Solution 38.Constant acceleration. 0 0x =0v v at= + (1)20 012x x v t at= + + (2)Solving (1) for a,0v vat−= (3)Then, ( ) ( )200 0 0 0 01 1 12 2 2v vx x v t t x v v t v v tt−= + + = + + = +At 6 s,t = 0 61and 540 ft2v v x= =( )0 0 0 001 1 540540 6 4.5 or 120 ft/s2 2 4.5160 ft/s2v v v vv v⎛ ⎞= + = = =⎜ ⎟⎝ ⎠= =Then, from (3), 2 260 120 60ft/s 10 ft/s6 6a−= = − = −Substituting into (1) and (2), 120 10v t= −( ) 210 120 102x t t= + −At stopping, 0 or 120 10 0 12 ss sv t t= − = =( )( ) ( )( )210 120 12 10 12 720 ft2x = + − =( ) Additional time for stopping 12 s 6 sa = − 6 stΔ =( ) Additional distance for stopping 720 ft 540 ftb = − 180 ftdΔ =
• 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 11, Solution 39.20 01( ) During the acceleration phase2a x x v t at= + +0 0Using 0, and 0, and solving for givesx v a= =22xat=Noting that 130 m when 25 s,x t= =( )( )( )22 13025a = 0.416 m/sa =(b) Final velocity is reached at 25 s.t =( )( )0 0 0.416 25fv v at= + = + 10.40 m/sfv =(c) The remaining distance for the constant speed phase is400 130 270 mxΔ = − =For constant velocity,27025.96 s10.40xtvΔΔ = = =Total time for run: 25 25.96t = + 51.0 st =
• 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 11, Solution 40.Constant acceleration. Choose 0t = at end of powered flight.Then, 21 27.5 m 9.81 m/sy a g= = − = −(a) When y reaches the ground, 0 and 16 s.fy t= =2 21 1 1 11 12 2fy y v t at y v t gt= + + = + −( )( )221 11 2 210 27.5 9.81 1676.76 m/s16fy y gtvt− + − += = =1 76.8 m/sv =(b) When the rocket reaches its maximum altitude max,y0v =( ) ( )2 2 21 1 1 12 2v v a y y v g y y= + − = − −2 2112v vy yg−= −( )( )( )2max0 76.7627.52 9.81y−= − max 328 my =
• 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 11, Solution 41.Place origin at 0.Motion of auto. ( ) ( ) 20 00, 0, 0.75 m/sA A Ax v a= = =( ) ( ) ( )2 20 01 10 0 0.752 2A A A Ax x v t a t t⎛ ⎞= + + = + + ⎜ ⎟⎝ ⎠20.375 mAx t=Motion of bus. ( ) ( )0 0?, 6 m/s, 0B B Bx v a= = − =( ) ( ) ( )0 0 06 mB B B Bx x v t x t= − = −At 20 , 0.Bt s x= =( ) ( )( )00 6 20Bx= − ( )0120 mBx =Hence, 120 6Bx t= −When the vehicles pass each other, .B Ax x=2120 6 0.375t t− =20.375 6 120 0t t+ − =( )( )( )( )( )26 (6) 4 0.375 1202 0.375t− ± − −=6 14.69711.596 s and 27.6 s0.75t− ±= = −Reject the negative root. 11.60 st =Corresponding values of xA and xB.( )( )20.375 11.596 50.4 mAx = =( )( )120 6 11.596 50.4 mBx = − = 50.4 mx =
• 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 11, Solution 42.Place the origin at A when t = 0.Motion of A: ( ) ( ) 20 00, 15 km/h = 4.1667 m/s, 0.6 m/sA A Ax v a= = =( )04.1667 0.6A A Av v a t t= + = +( ) ( ) 2 20 014.1667 0.32A A A Ax x v t a t t t= + + = +Motion of B: ( ) ( ) 20 025 m, 23 km/h = 6.3889 m/s, 0.4 m/sB B Bx v a= = = −( )06.3889 0.4B B Bv v a t t= + = −( ) ( ) 2 20 0125 6.3889 0.22B B B Bx x v t a t t t= + + = + −(a) When and where A overtakes B. A Bx x=2 24.1667 0.3 25 6.3889 0.2t t t t+ = + −20.5 2.2222 25 0t t− − =( )( )( )( )( )22.2222 2.2222 4 0.5 252 0.5t± − −=2.2222 7.4120 9.6343 s and 5.19 st = ± = −Reject the negative root. . 9.63 st =( )( ) ( )( )24.1667 9.6343 0.3 9.6343 68.0 mAx = + =( )( ) ( )( )225 6.3889 9.6343 0.2 9.6343 68.0 mBx = + − =moves 68.0 mAmoves 43.0 mB(b) Corresponding speeds.( )( )4.1667 0.6 9.6343 9.947 m/sAv = + = 35.8 km/hAv =( )( )6.3889 0.4 9.6343 2.535 m/sBv = − = 9.13 km/hBv =
• 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 11, Solution 43.Constant acceleration ( )1 2anda a for horses 1 and 2.Let 0x = and 0t = when the horses are at point A.Then, 2012x v t at= +Solving for ,a( )022 x v tat−=Using 1200 ftx = and the initial velocities and elapsed times for each horse,( )( )( )21 11 2 212 1200 20.4 61.50.028872 ft/s61.5x v tat⎡ ⎤−− ⎣ ⎦= = = −( )( )( )22 22 2 222 1200 21 62.00.053070 ft/s62.0x v tat⎡ ⎤−− ⎣ ⎦= = = −1 2Calculating ,x x− ( ) ( ) 21 2 1 2 1 212x x v v t a a t− = − + −( ) ( ) ( ) 21 22120.4 21 0.028872 0.05307020.6 0.012099x x t tt t⎡ ⎤− = − + − − −⎣ ⎦= − +At point B, 21 2 0 0.6 0.012099 0B Bx x t t− = − + =(a)0.649.59 s0.012099Bt = =Calculating Bx using data for either horse,Horse 1: ( )( ) ( )( )2120.4 49.59 0.028872 49.592Bx = + − 976 ftBx =Horse 2: ( )( ) ( )( )2121 49.59 0.05307 49.59 976 ft2Bx = + − =When horse 1 crosses the finish line at 61.5 s,t =(b) ( )( ) ( )( )21 2 0.6 61.5 0.012099 61.5x x− = − + 8.86 ftxΔ =
• 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 11, Solution 44.Choose x positive upward. Constant acceleration a g= −Rocket launch data: Rocket :A 00, , 0x v v t= = =Rocket :B 00, , 4 sBx v v t t= = = =Velocities: Rocket :A 0Av v gt= −Rocket :B ( )0B Bv v g t t= − −Positions: 201Rocket :2AA x v t gt= −( ) ( )201Rocket : ,2B B B BB x v t t g t t t t= − − − ≥For simultaneous explosions at 240 ft when ,A B Ex x t t= = =( ) ( )22 2 20 0 0 01 1 1 12 2 2 2E E E B E B E B E E B Bv t gt v t t g t t v t v t gt gt t gt− = − − − = − − + −0Solving for ,v 02BEgtv gt= − (1)Then, when ,Et t= 21,2 2BA E E Egtx gt t gt = − −  or 2 20AE B Ext t tg− − =Solving for ,Et( )( )( ) ( ) ( )( )( )( )22 4 1 2 240232.24 1 4 46.35 s2 2AxB B gEt tt± + ± += = =
• 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.(a) From equation (1), ( )( )( )( )032.2 432.2 6.3482v = − 0 140.0 ft/sv =At time ,Et 0A Ev v gt= − ( )0B E Bv v g t t= − −(b) ( )( )32.2 4B A Bv v gt− = = / 128.8 ft/sB Av =
• 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 11, Solution 45.(a) Acceleration of A.( ) ( )0 0, 168 km/h 46.67 m/sA A A Av v a t v= + = =At 8 s,t = 228 km/h 63.33 m/sAv = =( )0 63.33 46.678A AAv vat− −= = 22.08 m/sAa =(b) ( ) ( ) 20 012A A A Ax x v t a t= + + ( ) ( ) 20 012B B B Bx x v t a t= + +( ) ( ) ( ) ( ) ( ) 20 0 0 012A B A B A B A Bx x x x v v t a a t⎡ ⎤− = − + − + −⎣ ⎦When 0,t = ( ) ( )0 038 mA Bx x− = and ( ) ( )0 00B Av v− =When 8 s,t = 0A Bx x− =Hence, ( )( )210 38 8 , or 1.18752A B A Ba a a a= + − − = −1.1875 2.08 1.1875B Aa a= + = + 23.27 m/sBa =
• 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 11, Solution 46.(a) Acceleration of A.( ) ( ) ( ) 20 0 01and2A A A A A A Av v a t x x v t a t= + = + =Using ( ) ( )0 00 and 0 givesA Av x= =21and2A A A Av a t x a t= =When cars pass at 1, 90 mAt t x= =( )( )21 12 902 180andAA AA A Axt v a ta a a= = = =For 0 5 s,t≤ ≤ ( )096 km/h 26.667 m/sB Bv v= = − = −For 5 s,t > ( ) ( ) ( )015 26.667 56B B B Av v a t a t= + − = − + −When vehicles pass, A Bv v= −( )1 1126.667 56A Aa t a t= − −1 17 5 16026.667 or 7 56 6A AAa t a ta− = − =Using 1180 7 180 160gives 5A AAta aa= − =Let1,Aua= 27 180 5 160u u− =or 2160 7 180 5 0u u− + =continued
• 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.Solving the quadratic equation,( )( ) ( )( )( )( )( )7 180 49 180 4 160 5 93.915 74.9672 160 3200.0592125 and 0.52776u± − ±= ==21285.2 m/s and 3.590 m/sAau= =The corresponding values for 1t are1 1180 1800.794 s, and 7.08 s285.2 3.590t t= = = =Reject 0.794 s since it is less than 5 s.Thus,23.59 m/sAa =(b) Time of passing. 1 7.08 st t= =(c) Distance d.( ) ( )0 00 5 s, 26.667B B Bt x x v t d t≤ ≤ = − = −At 5 s,t = ( )( )22.667 5 133.33Bx d d= − = −For 5 s,t > ( ) ( ) ( )201133.33 5 52B B Bx d v t a t= − + − + −( ) ( )21 3.59133.33 26.667 5 52 6Bx d t t = − − − + −  1When 7.08 s,t t= = 90B Ax x= =( )( )( )( )( )( )23.59 2.0890 133.33 26.667 2.082 6d= − − +90 133.33 55.47 1.29d = + + − 278 md =
• 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 11, Solution 47.For 0,t > ( ) ( ) ( )2 2 20 01 10 0 6.5 or 3.252 2A A A A Ax x v t a t t x t= + + = + + =For 2 s,t > ( ) ( ) ( ) ( ) ( )( )2 20 01 12 2 0 0 11.7 22 2B B B Bx x v t a t t= + − + − = + + −or ( )2 25.85 2 5.85 23.4 23.4Bx t t t= − = − +For ,A Bx x= 2 23.25 5.85 23.4 23.4,t t t= − +or 22.60 23.4 23.4 0t t− + =Solving the quadratic equation, 1.1459 and 7.8541 st t= =Reject the smaller value since it is less than 5 s.( )a 7.85 st =( )( )23.25 7.8541A Bx x= = 200 ftx =( )b ( ) ( )( )00 6.5 7.8541A A Av v a t= + = + 51.1 ft/sAv =( ) ( ) ( )( )02 0 11.7 7.8541 2B B Bv v a t= + − = + − 68.5 ft/sBv =
• 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 11, Solution 48.Let x be the position relative to point P.Then, ( ) ( )0 00 and 0.62 mi 3273.6 ftA Bx x= = =Also, ( ) ( )0 068 mi/h 99.73 ft/s and 39 mi/h 57.2 ft/sA Bv v= = = − = −(a) Uniform accelerations.( ) ( )( ) ( )0 0220 021or2A A AA A A A Ax x v tx x v t a t at⎡ ⎤− −⎣ ⎦= + + =( )( )( )222 3273.6 0 99.73 400.895 ft/s40Aa⎡ ⎤− −⎣ ⎦= = − 20.895 ft/sAa =( ) ( )( ) ( )0 0220 021or2B B BB B B B Bx x v tx x v a t at⎡ ⎤− −⎣ ⎦= + + =( )( )( )222 0 3273.6 57.2 420.988 ft/s42Ba⎡ ⎤− − −⎣ ⎦= = − 20.988 ft/sBa =(b) When vehicles pass each other .A Bx x=( ) ( ) ( ) ( )2 20 0 0 01 12 2A A A B B Bx v t a t x v t a t+ + = + +( ) ( )2 21 10 99.73 0.895 3273.6 57.2 0.9882 2t t t t+ + − = − + −20.0465 156.93 3273.6 0t t− − + =Solving the quadratic equation, 20.7 st = and 3390 s−Reject the negative value. Then, 20.7 st =(c) Speed of B.( ) ( )( )057.2 0.988 20.7 77.7 ft/sB B Bv v a t= + = − + − = −77.7 ft/sBv =
• 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 11, Solution 49.Let x be positive downward for all blocks and for point D.1 m/sAv =Constraint of cable supporting A: ( ) constantA A Bx x x+ − =( )( )2 0 or 2 2 1 2 m/sA B B Av v v v− = = = =Constraint of cable supporting B: 2 constantB Cx x+ =( )( )2 0 or 2 2 2 4 m/sC B C Bv v v v+ = = − = − = −(a) 4 m/sC =v(b) / 2 1B A B Av v v= − = − / 1 m/sB A =v(c) constant, 0D C D Cx x v v+ = + =4 m/sD Cv v= − =/ 4 1D A D Av v v= − = − / 3 m/sD A =v
• 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 11, Solution 50.Let x be positive downward for all blocks.Constraint of cable supporting A: ( ) constantA A Bx x x+ − =2 0 or 2 and 2A B B A B Av v v v a a− = = =Constraint of cable supporting B: 2 constantB Cx x+ =2 0, or 2 , and 2 4B C C B C B Av v v v a a a+ = = − = − = −Since Cv and Ca are down, Av and Aa are up, i.e. negative.( ) ( )220 02A A A A Av v a x x⎡ ⎤− = −⎣ ⎦( )( )( )( )( )2 222000.2 0( ) 0.04 m/s2 0.52A AAA Av va ax x− −= = = −⎡ ⎤ −−⎣ ⎦20.04 m/sAa =4C Aa a= − 20.16 m/sCa =( )( ) 2( ) 2 2 0.04 0.08 m/sB Ab a a= = − = −( )( )0.08 2 0.16 m/sB Bv a tΔ = = − = − 0.16 m/sBvΔ =( )( )221 10.08 2 0.16 m2 2B Bx a tΔ = = − = − 0.16 mBxΔ =
• 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 11, Solution 51.Let xA, xB, xC, and xD be the displacements of blocks A, B, C, and D relative to the upper supports, increasingdownward.Constraint of cable AB: constantA Bx x+ =0A Bv v+ = B Av v= −Constraint of cable BED: 2 constantB Dx x+ =1 12 0 or2 2B D D B Av v v v v+ = = − =Constraint of cable BCD: ( ) ( ) constantC B C Dx x x x− + − =12 0 or 2 02C B D C A Av v v v v v− − = + − =(a) Velocity of block A.12 (2)(4)2A Cv v= − = − 8 ft/sAv = − 8 ft/sAv =(b) Velocity of block D.14 ft/s2D Av v= = − 4 ft/sDv =
• 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 11, Solution 52.Let xA, xB, xC, xD, and xE be the displacements of blocks A, B, C, and D and cable point E relative to the uppersupports, increasing downward.Constraint of cable AB: constantA Bx x+ =0A Bv v+ = B Av v= −0A Ba a+ = B Aa a= −Constraint of cable BED: 2 constantB Dx x+ =1 12 02 2B D D B Av v v v v+ = = − =1 12 02 2B D D A Aa a a a a+ = = − =Constraint of cable BCD: ( ) ( ) constantC B C Dx x x x− + − =2 0 2 0C B D C Av v v v v− − = + =12 0 2 02C B D C Aa a a a a− − = + =14C Aa a= −Since block C moves downward, vC and aC are positive.Then, vA and aA are negative, i.e. upward.Also, vD and aD are negative.Relative motion: /12A D A D Av v v v= − =/12A D A D Aa a a a= − =
• 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) Acceleration of block C./ 2/2 (2)(8)2 3.2 ft/s5A DA A Dva at= = = =23.2 ft/sAa = −210.8 ft/s4C Aa a= − = 20.8 ft/sCa =Constraint of cable portion BE: constantB Ex x+ =0B Ev v+ = 0B Ea a+ =(b) Acceleration of point E.23.2 ft/sE B Aa a a= − = = − 23.2 ft/sEa =
• 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 11, Solution 53.Let x be position relative to the right supports, increasing to the left.Constraint of entire cable: ( )2 constantA B B Ax x x x+ + − =2 0 2B A A Bv v v v+ = = −Constraint of point C of cable: 2 constantA Cx x+ =2 0 2A C C Av v v v+ = = −(a) Velocity of collar A.( )( )2 2 300 600 mm/sA Bv v= − = − = − 600 mm/sAv =(b) Velocity of point C of cable.( )( )2 2 600 1200 mm/sC Av v= − = − − = 1200 mm/sCv =(c) Velocity of point C relative to collar B./ 1200 300 900 mm/sC B C Bv v v= − = − = / 900 mm/sC Bv =
• 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 11, Solution 54.Let x be position relative to the right supports, increasing to the left.Constraint of entire cable: ( )2 constant,A B B Ax x x x+ + − =1 12 0, or , and2 2B A B A B Av v v v a a+ = = − = −(a) Accelerations of A and B./ /1 22 3B A B A A A A B Av v v v v v v= − = − − = −( )2610 406.67 mm/s3Av = − = −( )( ) 20A0406.67 0, or 50.8 mm/s8A AA A Av vv v a t at− − −− = = = = −250.8 mm/sAa =( )1 150.82 2B Aa a= − = − − 225.4 mm/sBa =(b) Velocity and change in position of B after 6 s.( ) ( )( )00 25.4 6B B Bv v a t= + = + 152.5 mm/sBv =( ) ( ) ( )( )220 01 125.4 62 2B B B Bx x v t a t− = + = 458 mmBxΔ =
• 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 11, Solution 55.Let x be position relative to left anchor. At the right anchor, .x d=Constraint of cable: ( ) ( )2 constantB B A Ax x x d x+ − + − =2 22 3 0 or and3 3B A A B A Bv v v v a a− = = =Constraint of point D of cable: ( ) constantA Dd x d x− + − =0 or andA D D A D Av v v v a a+ = = − = −(a) Accelerations of A and B.( ) ( ) ( )0 026 in./s 6 4 in./s3B Av v= = =( ) ( )220 02A A A A Av v a x x⎡ ⎤− = −⎣ ⎦( )( )( ) ( )( )( )2 2 222002.4 40.512 in./s2 102A AAA Av vax x− −= = = −⎡ ⎤−⎣ ⎦20.512 in./sAa =( ) 23 30.512 0.768 in./s2 2B Aa a= = = − 20.768 in./sBa =(b) Acceleration of point D. ( )0.512D Aa a= − = − − 20.512 in./sDa =(c) Velocity of block B after 4 s.( ) ( )( )06 0.768 4B B Bv v a t= + = + − 2.93 in./sBv =Change in position of block B.( ) ( ) ( )( ) ( )( )220 01 16 4 0.768 42 2B B B Bx x v t a t− = + = + − 17.86 in.BxΔ =
• 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 11, Solution 56.Let x be position relative to left anchor. At right anchor .x d=Constraint of entire cable: ( ) ( )2 constantB B A Ax x x d x+ − + − = 2 3 0B Av v− =(a) Velocity of A: ( )2 2123 3A Bv v= = 8.00 in./sAv =Constraint of point C of cable: constantB B Cx x x+ − = 2 0B Cv v− =(b) Velocity of C: ( )2 2 12C Bv v= = 24 in./sCv =Constraint of point D of cable: constantA Cd x d x− + − = 0,A Dv v+ =(c) Velocity of D: 8.00 in./sD Av v= − = − 8.00 in./sDv =(d) Relative velocity. / 24 8C A C Av v v= − = − / 16.00 in./sC Av =
• 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 11, Solution 57.Let x be position relative to the anchor, positive to the right.Constraint of cable: ( ) ( )3 constantB C B C Ax x x x x− + − + − =4 2 3 0 4 2 3 0C B A C B Av v v a a a− − = − − = (1, 2)When 0,t = ( )050 mm/s and 100 mm/sB av v= − =(a) ( ) ( ) ( )( ) ( )( )001 12 3 2 50 3 1004 4C B Av v v⎡ ⎤ ⎡ ⎤= + = − +⎣ ⎦⎣ ⎦ ( )050 mm/sCv =Constraint of point D: ( ) ( ) ( ) constantD A C A C B Bx x x x x x x− + − + − − =2 2 2 0D C A Bv v v v+ − − =(b) ( ) ( ) ( ) ( )( ) ( )( ) ( )( )0 0 02 2 2 2 100 + 2 50 2 50D A B Cv v v v= + − = − − ( )00Dv =( ) ( ) 20 012C C C Cx x v t a t− = +(c)( ) ( ) ( )( )( )0 0 22 22 2 40 50 230 mm/s2C C CCx x v tat⎡ ⎤− − ⎡ ⎤−⎣ ⎦ ⎣ ⎦= = = −230 mm/sCa =Solving (2) for aA( ) ( )( ) ( )( ) 21 14 2 4 30 2 0 40 mm/s3 3A C Ba a a ⎡ ⎤= − = − − = −⎣ ⎦240 mm/sAa =
• 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.Chapter 11, Solution 58.Let x be position relative to the anchor, positive to the right.Constraint of cable: ( ) ( )3 constantB C B C Ax x x x x− + − + − =4 2 3 0 and 4 2 3 0C B A C B Av v v a a a− − = − − =(a) Accelerations of B and C.At 2 s,t = 420 mm/s and 30 mm/sA Bv v= = −( ) ( )( ) ( )( )1 12 3 2 30 3 420 300 mm/s4 4C B Av v v ⎡ ⎤= + = − + =⎣ ⎦( )00Cv =( )0C C Cv v a t= +( )0 300 02C CCv vat− −= = 2150 mm/sCa =( ) ( )( ) ( )( ) 21 14 3 4 150 3 270 105 mm/s2 2B C Aa a a ⎡ ⎤= − = − = −⎣ ⎦2105 mm/sBa =(b) Initial velocities of A and B.( )0A A Av v a t= − ( ) ( )( )0420 270 2 120 mm/sA A Av v a t= − = − = −( )0120 mm/sAv =( )0B B Bv v a t= − ( ) ( )( )030 105 2B B Bv v a t= − = − − − ( )0180 mm/sBv =Constraint of point E: ( ) ( )2 constantC A E Ax x x x− + − =3 2 0E A Cv v v− + =(c) ( ) ( ) ( ) ( )( ) ( )( )0 0 03 2 3 120 2 0 360 mm/sE A Cv v v= − = − − = −( )0360 mm/sEv =
• 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 11, Solution 59.Define positions as positive downward from a fixed level.Constraint of cable. ( ) ( ) ( )2 constantB A C A C Bx x x x x x− + − + − =3 2 constantC B Ax x x− − =3 2 0C B Av v v− − =3 2 0C B Aa a a− − =Motion of block C.( ) ( )20 00, 3.6 in./s , 18 in./s, 0A A B B Bv a v v a= = − = = =( ) ( ) ( )0 0012 6 in./s3C B Av v v = + = ( ) ( )( ) 21 12 0 2 3.6 2.4 in./s3 3C B Aa a a  = + = + − = − ( )06 1.2C C Cv v a t t= + = −( ) ( ) 2 20 016 0.62C C C Cx x v t a t t t− = + = −(a) Time at vC = 0.0 6 2.4t= − 2.5 st =(b) Corresponding position of block C.( ) ( )( ) ( )( )2016 2.5 2.4 2.52C Cx x − = + −  ( )07.5 in.C Cx x− =
• 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 11, Solution 60.Define positions as positive downward from a fixed level.Constraint of cable: ( ) ( ) ( )2 constantB A C A C Bx x x x x x− + − + − =3 2 constantC B Ax x x− − =3 2 0C B Av v v− − =3 2 0C B Aa a a− − =Motion of block C.( )00,Av = 22.5 in./s ,Aa t= − ( )00,Bv = 215 in./sBa =( ) ( ) ( )0 0012 03C B Av v v⎡ ⎤= + =⎣ ⎦( ) 21 12 (15 5 ) in./s3 3C B Aa a a t= + = −( ) 00tC C Cv v a dt= + ∫( )210 15 2.5 in./s3t t= + −( ) ( )2 3017.5 0.83333 in.3C Cx x t t− = −( ) Time at 0Ca v =( )210 15 2.5 03t t+ − = 0 and 6 st t= = 6 st =(b) Corresponding position of block C.( ) ( )( ) ( )( )2 3010 7.5 6 0.83333 63C Cx x ⎡ ⎤− = + −⎢ ⎥⎣ ⎦( )030 in.C Cx x− =
• 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 11, Solution 61.Let x be position relative to the support taken positive if downward.Constraint of cable connecting blocks A, B, and C:2 2 constant, 2 2 0A B C A B Cx x x v v v+ + = + + =2 2 0A B Ca a a+ + = (1)Constraint of cable supporting block D:( ) ( ) constant, 2 0D A D B D A Bx x x x v v v− + − = − − =2 0D B Aa a a− − = (2)Given: / 120 or 120C B C B C Ba a a a a= − = − = − (3)Given: / 220 or 220D A D A D Aa a a a a= − = = + (4)Substituting (3) and (4) into (1) and (2),( )2 2 120 0 or 2 3 120A B B A Ba a a a a+ + − = + = (5)( )2 220 0 or 440A A B A Ba a a a a+ − − = − = − (6)Solving (5) and (6) simultaneously,2 2240 mm/s and 200 mm/sA Ba a= − =From (3) and (4), 2 280 mm/s and 20 mm/sC Da a= = −(a) Velocity of C after 6 s.( ) ( )( )00 80 6C C Cv v a t= + = + 480 mm/sCv =(b) Change in position of D after 10 s.( ) ( ) ( )( )220 01 10 20 10 1000 mm2 2D D D Dx x v t a t− = + = + − = −1.000 mDxΔ =
• 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 11, Solution 62.Let x be position relative to the support taken positive if downward.Constraint of cable connecting blocks A, B, and C:2 2 constant,A B Cx x x+ + = 2 2 0,A B Cv v v+ + = 2 2 0A B Ca a a+ + =( ) ( ) ( ) ( ) ( ) ( )0 0 0 00 00, ,A B C A B Cv v v x x x= = = = = ( ) ( )/ /0 00, 0B A B Ax v= =( ) ( ) ( )2 2/ / / / /0 02B A B A P A B A B Av v a x x⎡ ⎤− = −⎣ ⎦( )2/ /0 2 0B A B A B Av a x x− = − −( ) ( )2 2/ 2/4010 mm/s2 2 160 80B AB AB Avax x= = =− −( ) ( ) 2 2/ / / / /0 01 10 02 2B A B A B A B A B Ax x v t a t a t= + + = + +( ) ( )/2/ /2 2 2 160 80, or 4 s10B A B AB A B Ax x xt ta a− −= = = =( ) ( ) 20 012A A A Ax x v t a t− = +(a)( ) ( ) ( )( )0 02 22 2 80 04A A AAx x v tat⎡ ⎤− − −⎣ ⎦= = 210 mm/sAa =/ 10 10B A B Aa a a= + = + 220 mm/sBa =( ) ( )( ) ( )( )2 2 2 20 2 10 60 mm/sC B Aa a a ⎡ ⎤= − + = − + = −⎣ ⎦( )( )00300 05 s60C CC C CCv vv v a t ta− − −= + = = =−Constraint of cable supporting block D:( ) ( ) constant, 2 0D A D B D A Bx x x x v v v− + − = − − =( ) ( )1 12 0, 10 20 15 mm/s2 2D A B D A Ba a a a a a− − = = + = + =(b) ( ) ( ) ( )( )220 01 10 15 52 2D D D Dx x v t a t− = + = + 187.5 mmDxΔ =
• 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 11, Solution 63.curvea t−1 212 m/s, 8 m/sA A= − =(a) curvev t−6 4 m/sv = −( )0 6 1 4 12v v A= − = − − − 8 m/s=10 4 m/sv = −(b) 14 10 2 4 8v v A= + = − + 14 4 m/sv =3 416 m, 4 mA A= = −5 616 m, 4 mA A= − = −7 4 mA =(a) curvex t−0 0x =4 0 3 16 mx x A= + =6 4 4 12 mx x A= + =10 6 5 4 mx x A= + = −12 10 6 8 mx x A= + = −(b) 14 12 7x x A= + 14 4 mx = −Distance traveled:0 4 s,t≤ ≤ 1 16 0 16 md = − =4 s 12 s,t≤ ≤ 2 8 16 24 md = − − =12 s 14 s,t≤ ≤ ( )3 4 8 4 md = − − − =Total distance traveled: 16 24 4d = + + 44 md =
• 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 11, Solution 64.(a) Construction of the curves.curvea t−1 212 m/s, 8 m/sA A= − =curvev t−0 8 m/sv =( )6 0 1 8 12 4 m/sv v A= + = + − = −10 6 4 m/sv v= = −14 10 2 4 8 4 m/sv v A= + = − + =3 416 m, 4 mA A= = −5 616 m, 4 mA A= − = −7 4 mA =curvex t−0 0x =4 0 3 16 mx x A= + =6 4 4 12 mx x A= + =10 6 5 4 mx x A= + = −12 10 6 8 mx x A= + = −14 12 7 4 mx x A= + = −(b) Time for 8 m.x >From the x t− diagram, this is time interval 1 2to .t tOver 0 6 s,t< < 8 2dxv tdt= = −continued
• 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.Integrating, using limits 0x = when 0t = and 8 mx = when1t t=8 2 21 10 08 or 8 8tx t t t t = − = − or 21 18 8 0t t− + =Solving the quadratic equation,( ) ( )( )( )( )( )218 8 4 1 84 2.828 1.172 s and 6.828 s2 1t± −= = ± =The larger root is out of range, thus 1 1.172 st =Over 6 10,t< < ( )12 4 6 36 4x t t= − − = −Setting 8,x = 2 28 36 4 or 7 st t= − =Required time interval: ( )2 1 5.83 st t− =
• 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.Chapter 11, Solution 65.The a–t curve is just the slope of the v–t curve.0 10 s,t< < 0a =10 s < 18 s,t < 218 61.5 ft/s18 10a−= =−18 s < 30 s,t < 218 183 ft/s30 18a− −= = −−30 s < 40 st < 0a =Points on the x–t curve may be calculated using areas of the v–tcurve.1 (10)(6) 60 ftA = =21(6 18)(18 10) 96 ft2A = + − =31(18)(24 18) 54 ft2A = − =41( 18)(30 24) 54 ft2A = − − = −5 ( 18)(40 30) 180 ftA = − − = −0 48 ftx = −01 0 1 12 ftx x A= + =81 10 2 108 ftx x A= + =24 18 3 162 ftx x A= + =30 24 4 108 ftx x A= + =40 30 5 72 ftx x A= + = −continued
• 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.(a) Maximum value of x.Maximum value of x occursWhen 0,v = i.e. 24 s.t =max 162 ftx =(b) Time s when 108 ft.x =From the x–t curve,18 s and 30 st t= =
• 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 11, Solution 66.Data from problem 11.65: 0 48 ftx = −The a–t curve is just the slope of the v–t curve.0 10 s,t< < 0a = !10 s < 18 s,t < 218 61.5 ft/s18 10a−= =−!18 s < 30 s,t < 218 183 ft/s30 18a− −= = −−!30 s < 40 s,t < 0a = !Points on the x–t curve may be calculated using areas of the v–tcurve. !1 (10)(6) 60 ftA = =21(6 18)(18 10) 96 ft2A = + − =31(18)(24 18) 54 ft2A = − =41( 18)(30 24) 54 ft2A = − − = −5 ( 18)(40 30) 180 ftA = − − = −0 48 ftx = − !0 01 1 12 ftx x A= + = !1018 2 108 ftx x A= + = !24x = 18 3x A+ = 162 ft !30x = 24 4x A+ = 108 ft !40 30 5 72 ftx x A= + = − !continued
• 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.(a) Total distance traveled during 0 30 st≤ ≤ .For 0 24 st≤ ≤ 1 24 0 210 ftd x x= − =For 24 s 30 st≤ ≤ 2 30 24 54 ftd x x= − =Total distance. 1 2d d d= + 264 ftd = !(b) Values of t for which 0.x =In the range 0 10 st≤ ≤0 0 48 6x x v t t= + = − +Set 0.x = 148 6 0t− + = 1 8 st = !In the range 30 s 40 s,t< <30 30 ( 30)x x v t= + −108 ( 18)( 30)t= + − −648 18t= −Set 0.x = 2648 18 0t− = 2 36 st = !
• 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 11, Solution 67.Sketch v t− curve as shown. Label areas 1 2, ,A A and 3A( )( )1 3 20 60 in.A = =1 12 in./sv at tΔ = =( ) 22 1 11in.2A v t t= Δ =( )( ) ( )3 1 1 120 2 20 in.A v t t t= Δ − = −Distance traveled: 12 ft 144 in.xΔ = =( )21 1 1total area, 144 60 2 20x t t tΔ = = + + −or 21 140 84 0t t− + =( )( )( )( )( )2140 40 4 1 842.224 s and 37.8 s2 1t± −= =Reject the larger root. 1 2.224 st =12 4.45 in./sv tΔ = =max 3 3 4.45v v= + Δ = + max 7.45 in./sv =
• 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 11, Solution 68.Let x be the altitude. Then v is negative for decent and a is positivefor deceleration.Sketch the v t− and x t− curves using times 1 2,t t and 3t asshown.Use constant slopes in the v t− curve for the constantacceleration stages.Areas of v t− curve:( )1 1 11180 44 112 ft2A t t= − + = −2 244A t= −( )3 3 3144 222A t t= − = −Changes in position: 1 1800 1900 100 ftxΔ = − = −2 100 1800 1700 ftxΔ = − = −3 0 100 100 ftxΔ = − = −Using i ix AΔ = gives 11000.893 s112t−= =−2170038.64 s44t−= =−31004.55 s22t−= =−(a) Total time: 1 2 3 44.1 st t t+ + =(b) Initial acceleration.( ) ( )44 1800.893vat− − −Δ= =Δ2152.3 ft/sa =
• 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 11, Solution 69.Sketch the v t− curveData: 0 64 km/h 17.778 m/sv = =32 4.8 km 4.8 10 mx = = ×1 32 km/hr 8.889 m/sv = =3 31 4.8 10 800 4.0 10 mx = × − = ×2 450 st =(a) Time 1t to travel first 4 km.( ) ( )31 1 0 1 1 11 14.0 10 17.778 8.8892 2x A v v t t= × = = + = + 1 300 st =(b) Velocity 2.v( )( ) ( )( )2 1 2 1 2 2 1 1 21 1800 450 3002 2x x A v v t t v v− = = = + − = + −2 1 10.667 mv v+ =2 10.667 8.889v = − 2 1.778 m/sv =(c) Final deceleration.22 1122 11.778 8.8890.0474 m/s450 300v vat t− −= = = −− −212 0.0474 m/sa =
• 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.Chapter 11, Solution 70.10 2010 min 20 s 0.1722 h60 3600= + =Sketch the v t− curve602535abctatata===( )( ) ( )1 1 11 1 1 160 60 25 60 1800 312.52 2a bA t t t ta a= − − = − −But 1 5 miA =1160 2112.5 5ta− = (1)( )2 1 1135 0.1722 35 6.0278 35 612.5cA t t ta= − − = − −But 2 8 5 3 miA = − =1135 612.5 3.0278ta+ = (2)11Solving equations (1) and (2) for and ,ta31 85.45 10 h 5.13 mint −= × =6 2160.23 10 h /mia−= ×( )( )( )33 2216.616 10 528016.616 10 mi/h3600a×= × = 26.77 ft/sa =
• 77. COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 71.Sketch the curve as showna t−Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,0 120 ft/s, 6 ft/sv v= = −( )1 12 16140 6 172A t1A t t= −= − − = −1 0 1v v A A2= + +1 16 20 6 17t t= − −(a) 1 0.6087 st = 1 0.609 st =2 1.4 st =2 1 0.7913 st t− =( )( )1 3 6 1.4 8.4 ft/sA A+ = − = −( )( )2 17 0.6087 10.348 ft/sA = − = −2 0 1 3 2 20 8.4 10.348v v A A A= + + + = − − 2 1.252 ft/sv =(b) ( )2 0 0 2 1 3 13 2 2 by moment-area methodx x v t A A x A x= + + + +( )0 2 1 3 2 2 2 11 102 3v t A A t A t t⎛ ⎞ ⎛= + + + + −⎜ ⎟ ⎜⎝ ⎠ ⎝⎞⎟⎠( )( ) ( ) ( ) ( )1 0.60870 20 1.4 8.4 1.4 10.348 1.42 3⎛ ⎞ ⎛ ⎞= + − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠2 9.73 ftx =)Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.
• 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 11, Solution 72.Note that1 5280mile 660 ft8 8= =Sketch v t− curve for first 660 ft.Runner A: 1 24 s, 25 4 21 st t= = − =( )( ) ( )1 max max14 22A AA v v= =( )2 max21 AA v=1 25280 ft= 660 ft8A A x+ = ∆ =( ) ( )max max23 660 or 28.696 ft/sA Av v= =Runner B: 1 25 s, 25.2 5 20.2 st t= = − =( )( ) ( )1 max max15 2.52B BA v v= =( )2 max20.2 BA v=1 2 660 ftA A x+ = ∆ =( ) ( )max max22.7 660 or 29.075 ft/sB Bv v= =Sketch v t− curve for second 660 ft. 3 30.3v a t t∆ = =23 max 3 3 3 max 31660 or 0.15 660 02A v t vt t v t= − ∆ = − + =( ) ( )( )( )( )( )( )22max max3 max max4 0.15 6603.3333 3962 0.15v vt v v± −  = = ± −  Runner A:( )max 28.696,Av = ( )3 164.57 s and 26.736 sAt =Reject the larger root. Then total time (a) 25 26.736 51.736 sAt = + =51.7 sAt =
• 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.Runner B: ( )max 29.075,Bv = ( )3 167.58 s and 26.257 sBt =Reject the larger root. Then total time 25.2 26.257 51.457 sBt = + =51.5 sBt =Velocity of A at 51.457 s:t =( )( )1 28.696 0.3 51.457 25 20.759 ft/sv = − − =Velocity of A at 51.736 s:t =( )( )2 28.696 0.3 51.736 25 20.675 ft/sv = − − =Over 51.457 s 51.736 s, runner covers a distancet A x≤ ≤ ∆(b) ( ) ( )( )ave120.759 20.675 51.736 51.4572x v t∆ = ∆ = + − 5.78 ftx∆ =
• 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.Chapter 11, Solution 73.Sketch the v t− curves.At 12 min 720 s,t = =( )( )truckbusbus19.44 720 14000 m14000 1200 15200 marea under curvexxx v t= == + == −( )( ) ( )( )1 11120 27.78 720 27.78 152002t t− + − =1 225.8 st =(a) When bus truck,x x= areas under the v t− curves are equal.( )( ) ( )1 2 1 2127.78 120 27.78 19.442t t t t− + − =With 1 225.8 s,t = 2 576 st =( )( )truck 19.44 576 11200 mx = = truck 11.20 kmx =(b) 0bus127.78 0120 225.8 120v vat− −= =− −2bus 0.262 m/sa =
• 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 11, Solution 74.( )032 km/h 8.889 m/s24 km/h 6.667 m/sABvv= == =Sketch the v t− curves.( )( )( )( ) ( )( )( )12 //1 20106.667 45 300 m1 12.222 45 452 250 22.5A BA BA AB BAA vvx x A Ax x A= == += += + += +( )/ / 20B A B Ax x A= −( )b /0 60 50 22.5 A Bv= − − / 0.444 m/sA Bv =/ 6.667 0.444 7.111 m/sA B A Bv v v= + = + =(a)( )0 7.111 8.88945A AAv vat− −= = 20.0395 m/sAa = −
• 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 11, Solution 75.( )( )0022 mi/h 32.267 ft/s13 mi/h 19.067 ft/sABvv= == =Sketch the v t− curves.Slope of v t− curve for car A.( )( )211213.20.14 ft/s13.294.29 s0.14113.2 94.29 622.3 m2attA= − = −= == =( )( )101 20B BA Ax x Ax x A A= += + +( ) ( )/ 2 20 0, or 0B A B A B Ax x x x x A d A= − = − − = −2d A= 622 md =
• 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 11, Solution 76.Construct the a t− curves for the elevator and the ball.Limit on 1A is 24 ft/s. Using 1 4A t=2 24 24 6 st t= =Motion of elevator.For 10 6 s,t≤ ≤ ( ) ( )0 00 0E Ex v= =Moment of 1A about 1:t t= 211 14 22tt t=( ) ( ) 2 21 1 10 02 2E E Ex x v t t t= + + =Motion of ball. At 2,t = ( ) ( )0 040 ft 64 ft/sB Bx v= =For 1 2 s,t > ( )2 132.2 2 ft/sA t= − −Moment of 2A about 2 :t t= ( ) ( )211 1232.2 2 16.1 22tt t−⎛ ⎞− − = − −⎜ ⎟⎝ ⎠( ) ( ) ( ) ( )( ) ( )21 10 021 12 16.1 240 64 2 16.1 2B B Bx x v t tt t= + − − −= + − − −When ball hits elevator, B Ex x=( ) ( )2 21 1 121 140 64 2 16.1 2 2 or18.1 128.4 152.4 0t t tt t+ − − − =− + =Solving the quadratic equation, 1 1.507 s and 5.59 st =The smaller root is out of range, hence 1 5.59 st =Since this is less than 6 s, the solution is within range.
• 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 11, Solution 77.Let x be the position of the front end of the car relative to the front end of the truck.Letdxvdt= anddvadt= .The motion of the car relative to the truck occurs in 3 phases, lasting t1, t2, and t3 seconds, respectively.Phase 1, acceleration. 21 2 m/sa =Phase 2, constant speed. 2 90 km/h 54 km/hv = −36 km/h = 10 m/s=Phase 3, deceleration. 23 8 m/sa = −Time of phase 1. 2110 10 05 s2vta− −= = =Time of phase 3. 2320 0 101.25 s8vta− −= = =Sketch the a t− curve.Areas: 1 1 2 10 m/sA t v= =3 3 10 m/sA t v= = −Initial and final positions.0 30 16 46 mx = − − = −30 5 35 mfx = + =Initial velocity. 0 0v =
• 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.Final time. 1 2 3ft t t t= + +0 0f f i ix x v t A t= + + ∑1 112ft t t= −25 1.25 2.5t= + + −23.75 t= +2 310.625 s2t t= =( )( ) ( )( )235 46 0 10 3.75 10 0.625t= − + + + + −249.754.975 s10t = =1 2 3 11.225 sft t t t= + + =Total time. 11.23 sft =1 2 9.975 st t+ =
• 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 11, Solution 78.Let x be the position of the front end of the car relative to the front end of the truck.Letdxvdt= anddvadt= .The motion of the car relative to the truck occurs in two phases, lasting t1 and t2 seconds,respectively.Phase 1, acceleration. 21 2 m/sa =Phase 2, deceleration. 22 8 m/sa = −Sketch the a–t curve.Areas: 1 12A t=2 28A t= −Initial and final positions0 30 16 46 mx = − − = −30 5 35 mfx = + =Initial and final velocities.0 0fv v= =0 1 2fv v A A= + +1 20 0 2 8t t= + −1 24t t=0 0f f i ix x v t A t= + + ∑1 2 1 2132t t t t= + =2 212t t=
• 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.( )( ) ( )2 2 2 2135 46 0 2 4 3 82t t t t = − + + + −   2281 20 t=2 2.0125 st =1 8.05 st =1 2 10.0625 s.ft t t= + =Maximum relative velocity.( )( )1 1 2 8.05 16.10 m/smv a t= = =60.0 km/hmv =Maximum velocity relative to ground.max 54 60.0Tv v v= + = +max 112.0 km/hv = !
• 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 11, Solution 79.Sketch acceleration curve.Let jerkdajdt= =Then, ( )maxa j t= Δ( ) ( )( )1 max max2122A a t a tj t= Δ = Δ= Δ0 1 21 22 10 0fv v A AA AA A= + −= + −=( ) ( )( ) ( )( ) ( ) ( )( )( )0 1 23 3 33 34 30 3 20.360.49322 2 1.5x v t A t A tj t j t j txtjΔ = Δ + Δ − Δ= + Δ − Δ = ΔΔΔ = = =(a) Shortest time: ( )( )4 4 0.4932 1.973 stΔ = =(b) Maximum velocity: ( )2max 0 1 0v v A j t= + = + Δ( )( )21.5 0.4932 0.365 m/s= =Average velocity: ave0.360.1825 m/s4 1.973xvtΔ= = =Δ
• 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 11, Solution 80.Sketch the a t− curve.From the jerk limit, ( )1 maxj t aΔ = or ( ) max11.255 s.0.25atjΔ = = =( )( )115 1.25 3.125 m/s2A = =( )( )max 1 22 max 122max32 km/hr 8.889 m/s 22 8.889 2 3.125 2.639 m/s2.6392.111 s1.25v A AA v AAta= = = += − = − =Δ = = =Total distance is 5 km 5000 m.= Use moment-area formula.( ) ( )( )0 0 1 2 1 2 1 2 1 2max 1 21 12 22 20 0 2f f ffx x v t A A t t t A A t tv t t t⎛ ⎞ ⎛ ⎞= + + + − Δ − Δ − + Δ + Δ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= + + − Δ − Δ(a) ( )( )1 2max50002 2 5 2.111 10 2.111 562.5 575 s8.889ffxt t tv= Δ + Δ + = + + = + + =9.58 minft =(b) ave50008.70 m/s575ffxvt= = = ave 31.3 km/hv =
• 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 11, Solution 81.Indicate areas 1 2andA A on the a t− curve.( )110.6 0.1 m/s2 3TA T= =( )21 20.6 0.2 m/s2 3TA T= =By moment-area formula,( )( )( )0 1 22 2 2 22 27 49 97 8 15 140 090 90 90 640 6 240 sx v t A T A TT T T TT⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= + + = == =(a) 15.49 sT =max 0 1 2 0 0.1 0.2 0.3v v A A T T T= + + = + + =(b) max 4.65 m/sv =Indicate area 3 4andA A on the a t− curve.( )( )1 3410.1 0.6 0.052 610.45 0.03752 6TA T A TTA T= = == =(c) 0 1 3 4 0.1875v v A A A T= + + + = 2.90 m/sv =By moment-area formula,( ) ( ) ( )0 1 3 422 2 12 2 9 3 6 3 650 0.1 0.05 0.0375 0.03541718 9 18T T T T Tx v A A AT T TT T T T⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + ⋅ + ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞= + + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠( )( )20.035417 15.49= 8.50 mx =
• 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 11, Solution 82.Divide the area of the a t− curve into the four areas1 2 3 4, , and .A A A A( )( )( )( )( )( )( )( )123423 0.2 0.4 m/s35 0.2 1 m/s15 2.5 0.1 0.375 m/s212.5 0.1 0.125 m/s2AAAA= == == + == =(a) 0Velocities: 0v =0.2 0 1 2v v A A= + + 0.2 1.400 m/sv =0.3 0.2 3v v A= + 0.3 1.775 m/sv =0.4 0.3 4v v A= + 0.4 1.900 m/sv =Sketch the v t− curve and divide its area into 5 6 7, , andA A A asshown.0.3 0.4 0.40.3 or 0.3x t tdx x vdt x vdt= − = = −∫ ∫ ∫At 0.3 s,t = ( )( )0.3 50.3 1.775 0.1x A= − −(b) With ( )( )520.125 0.1 0.00833 m3A = = 0.3 0.1142 mx =At 0.2 s,t = ( )0.2 5 6 70.3x A A A= − + −With ( )( )5 620.5 0.2 0.06667 m3A A+ = =and ( )( )7 1.400 0.2 0.28 mA = =0.2 0.3 0.06667 0.28x = − − 0.2 0.0467mx = −
• 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 11, Solution 83.Approximate the a t− curve by a series of rectangles of height ,ia each with its centroid at .it t= When equalwidths of 0.25 stΔ = are used, the values of andi it a are those shown in the first two columns of the tablebelow.At 2 s,t = ( )20 00 iv v adt v a t= + ≈ + Σ Δ∫( )( )0 iv a t≈ + Σ Δ(a) ( )( )00 7.650 0.25v≈ − 0 1.913 ft/sv =Using moment-area formula,( ) ( )( )( )( )( )20 0 0 000 022i i i ii ix x v t a t t dt x v t a t tx v t a t t= + + − ≈ + + Σ − Δ≈ + + Σ − Δ∫(b) ( )( ) ( )( )0 1.913 2 11.955 0.25≈ + − 0.836 ftx =it ia 2 it− ( )2i ia t−( )s ( )2ft/s ( )s ( )ft/s0.125 3.215− 1.875 6.028−0.375 1.915− 1.625 3.112−0.625 1.125− 1.375 1.547−0.875 0.675− 1.125 0.759−1.125 0.390− 0.875 0.341−1.375 0.205− 0.625 0.128−1.625 0.095− 0.375 0.036−1.875 0.030− 0.125 0.004−Σ ( )27.650 ft/s− ( )11.955 ft/s−
• 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 11, Solution 84.Approximate the a t− curve by a series of rectangles of height ,ia each with its centroid at .it t= When equalwidths of 2 stΔ = are used, the values of andi it a are those shown in the first two columns of table below.(a) At 8 s,t = ( )88 0 00 iv v adt a t= + ≈ + Σ Δ∫( )( )ia t= Σ ΔSince 8 s,t = only the first four values in the second column are summed:217.58 13.41 10.14 7.74 48.87 ft/siaΣ = + + + =( )( )8 48.87 2v = 8 97.7 ft/sv =(b) At 20 s,t = ( ) ( )( )2020 020 0 20o ix v t a t dt a t t= + − = + Σ − Δ∫( )( )990.1 2= 20 1980 ftx =it ia 20 it− ( )20i ia t−( )s ( )2ft/s ( )s ( )ft/s1 17.58 19 334.03 13.41 17 228.05 10.14 15 152.17 7.74 13 100.69 6.18 11 68.011 5.13 9 46.213 4.26 7 29.815 3.69 5 18.517 3.30 3 9.919 3.00 1 3.0Σ ( )990.1 ft/s
• 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 11, Solution 85.The given curve is approximated by a series of uniformly accelerated motions.For uniformly accelerated motion,( )2 22 2 2 12 1 2 12 or2v vv v a x x xa−− = − Δ =( )2 1 2 1v v a t t− = − or 2 1v vta−Δ =For the regions shown above,(a) ( ) 3.19 st t= Σ Δ =(b) Assuming 0 0,x = ( )0 62.6 mx x x= + Σ Δ =Region ( )1 m/sv ( )2 m/sv ( )2m/sa ( )mxΔ ( )stΔ1 32 30 3− 20.67 0.6672 30 25 8− 17.19 0.6253 25 20 11.5− 9.78 0.4354 20 10 13− 11.54 0.7695 10 0 14.5− 3.45 0.690Σ 62.63 3.186
• 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 11, Solution 86.Usedva vdx= noting thatdvdx= slope of the given curve.Slope is calculated by drawing a tangent line at the required point, and using two points on this line todetermine and .x vΔ Δ Then, .dv vdx xΔ=Δ(a) When 0.25,x =1.4 m/sv = from the curve1m/s and 0.25m from the tangent linev xΔ = Δ =( )( )114 s 1.4 42.5dvadx−= = = 25.6 m/sa =(b) When 2.0 m/s,v = 0.5mx = from the curve.1 m/s and 0.6m from the tangent line.v xΔ = Δ =( )( )111.667s , 2 1.6670.6dvadx−= = = 23.33 m/sa =
• 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 11, Solution 87.The a t− curve for uniformly accelerated motion is shown. The area of the rectangle is.A at=Its centroid lies at1.2t t=By moment-area formula,( ) ( )0 0 0 012x x v A t t x v t at t⎛ ⎞= + + − = + + ⎜ ⎟⎝ ⎠20 012x v t at= + +
• 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.Chapter 11, Solution 88.From the curve,a t− ( )( )1 2 6 12 m/sA = − = −( )( )2 2 2 4 m/sA = =Over 6 s 10 s,t< < 4 m/sv = −0 1 0 0, or 4 12, or 8 m/sv v A v v= + − = − =By moment-area formula,12 0 0 moment of shaded area about 12sx x v t t= + + =( )( ) ( )( ) ( )( )12 0 8 12 12 12 3 4 12 11x = + + − − + − 12 8 mx = −
• 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 11, Solution 89.(a) 0.2s.T =( )( )1224 0.2 3.2 ft/s3A = − = −( )( )2 1124 0.224 4.8A tt= − −= − +010 90 3.2 24 4.8fv v At= + Σ= − − +1 3.8167 st =2 86.80 ft/sA = −1 3.6167 st T− =By moment-area formula, 1 0 0 1 moment of areax x v t= + +( )( ) ( ) ( ) ( )13 3.61670 90 3.8167 3.2 0.2 3.6167 86.808 2x⎡⎛ ⎞ ⎛ ⎞= + + − + + −⎜ ⎟ ⎜ ⎟⎢⎝ ⎠ ⎝ ⎠⎣1 174.7 ftx =(b) 0.8 s.T =( )( )( )( )12 1 10 1 1224 0.8 12.8 ft/s,324 0.8 24 19.2or 0 90 12.8 24 19.2, 4.0167 sfAA t tv v A t t= − = −= − − = − += + Σ = − − + =1 23.2167s 77.2 ft/st T A− = = −By moment-area formula,( )( ) ( ) ( ) ( )13 3.21670 90 4.0167 12.8 0.8 3.2167 77.28 2x⎡ ⎤ ⎛ ⎞= + + − + + − ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠1 192.3 ftx =
• 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.Chapter 11, Solution 90.Data from Prob. 650 048 ft, 6 ft/sx v= − =The a – t curve is just the slope of the v – t curve.0 10 s,t< < 0a = !10 s < < 18 s,t18 61.5 ft/s18 10a−= =−!18 s 30 s,t< <18 183 ft/s30 18a− −= = −−!30 s < < 40 st 0a = !0 0 i ix x v t A t= + + ∑(a) Position when t = 20 s.( )( )1 18 10 1.5 12 ft/sA = − =1 20 14 6st = − =( )( )2 2 3 6 ft/sA = − = −2 20 19 1 st = − =( )( ) ( )( ) ( )( )20 48 6 20 12 6 6 1x = − + + + −20 138 ftx = !(b) Maximum value of position coordinate.x is maximum where 0.v =From velocity diagram, 24 smt =( )( )1 18 10 1.5 12 ft/sA = − =( )1 24 14 10 st = − =( )( )2 24 18 3 18 ft/sA = − − = −( )2 24 21 3 st = − =( )( ) ( )( ) ( )( )48 6 24 12 10 18 3mx = − + + + −162 ftmx = !
• 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 11, Solution 91.( )21= +x t ( ) 24 1−= +y t( )2 1= = +&xv x t ( ) 38 1−= = − +&yv y t2= =&x xa v ( ) 424 1−= = +&y ya v tSolve for (t + 1)2from expression for x. (t + 1)2= xSubstitute into expression for y.4yx=Then, 4xy =This is the equation of a rectangular hyperbola.(a) t = 0. 2 m/s, 8 m/sx yv v= = −( ) ( )2 22 8 8.25 m/sv = + − =1 8tan 76.02θ −  −= = − °  8.25 m/s=v 76.0°( ) ( )2 22 2 212 m/s , 24 m/s2 24 24.1 m/s24tan 85.22x ya aaθ −= == + = = = °  224.1 m/s=a 85.2°(b)1s.2t = 3 m/s,xv = 2.37 m/s= −yv( )2 23 (2.37) 3.82 m/sv = + =1 2.37tan 38.33θ − − = = − °  3.82 m/s=v 38.3°
• 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.2 m/s,xa = 24.74 m/sya =2 2 22 4.74 5.15 m/sa = + =1 4.74tan 67.22θ −  = = °  25.15 m/s=a 67.2°
• 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.Chapter 11, Solution 92.Let ( )2 3 29 18 9 18u t t t t t t= − + = − +Then,2223 18 18, and 6 18du d ut t tdt dt= − + = −6 0.8 mx u= − 4 0.6 my u= − +0.8dx dudt dt= − 0.6dy dudx dt= +0.60.75 constant0.8= = − = − =dydtdxdtdydxSincedydxdoes not change, the path is straight.(a) At 2 s,t =226, and 6.= − = −du d udt dt( )( ) ( )( )0.8 6 4.8 m/s, 0.6 6 3.6 m/s= = − − = = = − = −x ydx dyv vdt dt( )( ) ( )( )22 220.8 6 4.8 m/s , 0.6 6 3.6 m/s= = − − = = − = −x yd xa adt6.0 m/s=v 36.9 ,° 26.0 m/s=a 36.9°(b) At 3 s,t =229, and 0du d udt dt= − =( )( ) ( )( )0.8 9 7.2 m/s, 0.6 9 5.4 m/sx yv v= − − = = − = −0, 0x ya a= =9.0 m/s=v 36.9 ,° 0=a(c) At 4 s,t =226, and 6du d udt dt= − =( )( ) ( )( )0.8 6 4.8 m/s, 0.6 6 3.6 m/s= − − = = − = −x yv v( )( ) ( )( )2 20.8 6 4.8 m/s , 0.6 6 3.6 m/sx ya a= − = − = =6.0 m/s=v 36.9 ,° 26.0 m/s=a 36.9°
• 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 11, Solution 93.Substitute the given expressions for x and y into the given equation of the ellipse, and note that the equation issatisfied.( )( ) ( )( ) ( )22 2 22 22 2 22 216cos 16cos 4 9sin4 3 4 2 cos 3 2 cos4cos 4cos 1 3sin 4 4cos cos12 cos 2 cost tx y tt tt t t t tt tπ π ππ ππ π π π ππ π− ++ = +− −− + + − += = =− −Calculate x& and y& by differentiation.( )( )( )( ) ( )( )( )( )( )( )2 22 24cos 2 sin4 sin 6 sin2 cos 2 cos 2 cos3sin sin 3 2cos 13 cos2 cos 2 cos 2 cost tt txt t tt t ttyt t tπ π ππ π π ππ π ππ π π π ππ ππ π π−− −= − =− − −−= − =− − −&&(a) When 0 s,t = 0 and 3 ,x y π= =& & 9.42 m/s=v(b) When1s,3t =( )( )322126 43, 032x yππ−= = − − =−& & 7.26 m/s=v(c) When 1 s,t =( )( )23 30 and ,3x yππ−= = = −& & 3.14 m/s=v
• 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.Chapter 11, Solution 94.Sketch the path of the particle, i.e. plot of y versus x.Using 6 sin , and 6 3cosx t t y t= − = − obtain the values in the table below. Plot as shown.(a) Differentiate with respect to t to obtain velocity components.( ) ( )( )2 22 2 2 226 3cos and 3sin6 3cos 9sin 45 36cos m/s36sin 0 0, , and 2 in the range 0 2 .x yx ydx dyv t v tdt dxv v v t t td vt t tdtπ π π= = − = == + = − + = −= = = ≤ ≤When 0 or 2 ,t π= 2cos 1, and is minimum.t v=When ,t π= 2cos 1, and is maximum.t v= −( ) ( )22min45 36 9 m/s ,v = − = min 3 m/sv =( )t s ( )x m ( )y m0 0 32π6.42 6π 18.85 932π31.27 62π 37.70 3
• 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.( ) ( )2max45 36 81 m/s ,v = + = max 9 m/sv =(b) 0, 0, 3 m, 3 m/s, 0x yt x y v v= = = = =0t =( )3 m=r jtan 0yxvvθ = = 0θ =2 s, 12 m, = 3 m, 3 m/s, 0x yt x y v vπ π= = = =2 st π=( ) ( )12 m + 3 mπ=r i jtanyxvvθ = 0θ =s, 6 m, = 9 m, 9 m/s, 0x yt x y v vπ π= = = =st π=( ) ( )6 m + 9 mπ=r i jtanyxvvθ = 0θ = °
• 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 11, Solution 95.Given: ( ) ( )cos sin sin cosA t t t A t t t= + + −r i j( ) ( )( ) ( )( ) ( )sin sin cos cos cos sincos sincos sin sin cosdA t t t t A t t t tdtA t t A t tdA t t t A t t tdt= = − + + + − += += = − + +rv i ji jva i j(a) When r and a are perpendicular, 0⋅ =r a( ) ( ) ( ) ( )cos sin sin cos cos sin sin cos 0A t t t t t t A t t t t t t⎡ ⎤ ⎡ ⎤+ + − ⋅ − + + =⎣ ⎦ ⎣ ⎦i j i j( )( ) ( )( )2cos sin cos sin sin cos sin cos 0A t t t t t t t t t t t t⎡ ⎤+ − + − + =⎣ ⎦( ) ( )2 2 2 2 2 2cos sin sin cos 0t t t t t t− + − =21 0t− = 1 st =(b) When r and a are parallel, 0× =r a( ) ( ) ( ) ( )cos sin sin cos cos sin sin cos 0A t t t t t t A t t t t t t⎡ ⎤ ⎡ ⎤+ + − × − + + =⎣ ⎦ ⎣ ⎦i j i j( )( ) ( )( )2cos sin sin cos sin cos cos sin 0A t t t t t t t t t t t t⎡ ⎤+ + − − − =⎣ ⎦k( ) ( )2 2 2 2 2 2sin cos sin cos sin cos sin cos cos sin sin cos 0t t t t t t t t t t t t t t t t t t+ + + − − − + =2 0t = 0t =
• 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 11, Solution 96.Given: ( )/2130 1 20 cos21te ttππ−⎡ ⎤= − +⎢ ⎥+⎣ ⎦r i jDifferentiating to obtain v and a,( )/2 /22130 20 cos2 2 sin 221t tde t e tdt tπ πππ π π− −⎛ ⎞= = + − −⎜ ⎟⎝ ⎠+rv j( )/2230 120 cos2 2sin 221te t ttππ π π−⎡ ⎤⎛ ⎞= − +⎜ ⎟⎢ ⎥⎝ ⎠+ ⎣ ⎦i j( )( )/2 /232 130 20 cos2 2sin 2 sin 2 4 cos22 21t tde t t e t tdt tπ πππ π π π π π π− −⎡ ⎤⎛ ⎞= = − − − + + − +⎜ ⎟⎢ ⎥⎝ ⎠+ ⎣ ⎦va i j( )( )2 /236010 4sin 2 7.5cos21te t ttππ π π−−= − −+i j(a) At 0,t = ( )130 1 20 11⎛ ⎞= − +⎜ ⎟⎝ ⎠r i j 20 in.=r( )1 130 20 1 01 2π⎡ ⎤⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦v i j 43.4 in./s=v 46.3°( )( )26010 1 0 7.51π= − − −a i j 2743 in./s=a 85.4°(b) At 1.5 s,t = 0.25130 1 20 cos32.5e ππ−⎛ ⎞= − +⎜ ⎟⎝ ⎠r i j( ) ( )18 in. 1.8956 in.= + −i j 18.10 in.=r 6.0°( )0.75230 120 cos3 022.5e ππ π− ⎛ ⎞= − +⎜ ⎟⎝ ⎠v i j( ) ( )4.80 in./s 2.9778 in./s= +i j 5.65 in./s=v 31.8°( )( )2 0.7536010 0 7.5cos32.5e ππ π−= − + −a i j( ) ( )2 23.84 in./s 70.1582 in./s= − +i j 270.3 in./s=a 86.9°
• 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 11, Solution 97.Given: ( ) ( )cos sinn nRt t ct Rt tω ω= + +r i j kDifferentiating to obtain v and a.( ) ( )cos sin sin cosn n n n n ndR t t t c R t t tdtω ω ω ω ω ω= = − + + +rv i j k( ) ( )( ) ( )2 22 2sin sin cos cos cos sin2 sin cos 2 cos sinn n n n n n n n n n n nn n n n n n n ndR t t t t R t t t tdtR t t t t t tω ω ω ω ω ω ω ω ω ω ω ωω ω ω ω ω ω ω ω= = − − − + + −⎡ ⎤= − − + −⎣ ⎦va i ki kMagnitudes of v and a.( ) ( ) ( )2 2 2 22 222 2 2 2 2 22 2 2 2 2cos sin sin coscos 2 sin cos sinsin 2 sin cos cosx y zn n n n n nn n n n n nn n n n n nv v v vR t t t c R t t tR t t t t t t cR t t t t t tω ω ω ω ω ωω ω ω ω ω ωω ω ω ω ω ω= + +⎡ ⎤ ⎡ ⎤= − + + +⎣ ⎦ ⎣ ⎦⎡ ⎤= − + +⎣ ⎦⎡ ⎤+ + +⎣ ⎦( )2 2 2 21 nR t cω= + + ( )2 2 2 21 nv R t cω= + +( ) ( )2 2 2 22 22 2 22 2 2 3 4 2 2 2 23 4 2 22 sin cos 2 cos sin4 sin 4 sin cos cos 4 cos4 sin cos sinx y zn n n n n n n nn n n n n n n n nn n n n na a a aR t t t t t tR t t t t t t tt t t t tω ω ω ω ω ω ω ωω ω ω ω ω ω ω ω ωω ω ω ω ω= + +⎡ ⎤= − − + −⎢ ⎥⎣ ⎦⎡= + + +⎣⎤− + ⎦( )2 2 4 24 n nR tω ω= + 2 24n na R tω ω= +
• 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 11, Solution 98.Given: ( ) ( ) ( )2cos 1 sin from whichAt t A t Bt t= + + +r i j k2cos , 1, sinx At t y A t z Bt t= = + =22cos sin 1x z yt t tAt Bt A = = = −  2 2 2 22 2 2cos sin 1 1 orx z x yt t tAt At A B       + = ⇒ + = = +              Then,2 2 21y x zA A B     − = +          2 2 21y x zA A B     − − =          !For 3 and 1,A B= = ( ) ( ) ( )23 cos 3 1 sint t t t t= + + +r i j kDifferentiating to obtain v and a.( ) ( )( )( )( )32223 cos sin 3 sin cos113 2sin cos 3 2cos sin1d tt t t t t tdt tdt t t t t tdtt= = − + + ++= = − − + + −+rv i j kva i j k(a) At 0,t = ( ) ( ) ( )3 1 0 0 0= − + +v i j k 3 ft/sv = !And ( ) ( ) ( )3 0 3 1 2 0= − + + −a i j hThen, ( ) ( )2 223 2 13a = + = 23.61 ft/sa = !(b) If and are perpendicular, 0⋅ =r v r v( ) ( ) ( ) ( )( )2233 cos 3 cos sin 3 1 sin sin cos 01tt t t t t t t t t t tt  − + + + + =    + or ( ) ( ) ( )2 2 2 29 cos 9 sin cos 9 sin sin cos 0t t t t t t t t t t t− + + + =continued
• 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.With 0,t ≠ 2 29cos 8 sin cos 9 sin 0t t t t t− + + =210 8 sin cos 8cos 0t t t t− + =or 7 2cos2 2 sin 2 0t t t+ − =The smallest root is 2 7.631 st = 3.82 st = !The next root is 4.38 st =
• 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 11, Solution 99.(a) At the landing point, tan30y x= − °Horizontal motion: ( )0 00xx x v t v t= + =Vertical motion: ( ) 2 20 01 12 2yy y v t gt gt= + − = −from which 2 02 2 tan 30 2 tan 30y x v ttg g g° °= − = =Rejecting the 0t = solution gives( )( )0 2 25 tan302 tan309.81vtg°°= = 2.94 st =(b) Landing distance:( )( )0 25 2.94cos30 cos30 cos30x v td = = =° ° °84.9 md =(c) Vertical distance: tan30h x y= ° +or 201tan302h v t gt= ° −Differentiating and setting equal to zero,0tan30tan30 0 or odh vv gt tdt g°= ° − = =Then,( )( )20 0 0maxtan 30 tan 30 1 tan 302v v vh gg g° ° ⎛ ⎞°= − ⎜ ⎟⎝ ⎠( ) ( )( )( )2 22 20 25 tan30tan 302 2 9.81vg°°= = max 10.62 mh =
• 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 11, Solution 100.Horizontal motion: ( )0 000, orxxx x v t v t tv= + = =Vertical motion: ( )22 20 0 0 2001 1or2 2 2ygxy y v t gt y gt y yv= + − = − = −At ground level, 0,y = so that20 202gxyv=At 50 m,x =( )( )( )( )20 29.81 5013.625 m2 30y = =0 13 0.625 mh y= − =At 53 m,x =( )( )( )( )20 29.81 5315.31 m2 30y = =0 13 2.31 mh y= − =Range to avoid: 0.625 m 2.31 mh< <
• 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 11, Solution 101.Horizontal motion. 0 0xv v x v t= =Vertical motion. 212y h gt= −Eliminate t.220 02x gxt y hv v= = −Solve for v0.( )202=−gxvh yData: h = 3ft, g = 32.2 ft/s2(a) To strike corner C. 15 ft, 0x d y= = =( )( )( )( )2032.2 152 3 0=−v 0 34.7 ft/sv =To strike point B. 15 ft, 1 ftx y= =( )( )( )( )2032.2 152 3 1=−v 0 42.6 ft/sv =To strike point D. 15 1 14 ft, 0x y= − = =( )( )( )( )2032.2 142 3 0=−v 0 32.4 ft/sv =(b) Range to strike corner BCD. 032.4 ft/s < 42.6 ft/sv <
• 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.Chapter 11, Solution 102.Place origin of coordinates at point A.Horizontal motion: ( )090 mi/h 132 ft/s= =xv( )0 00 132 ft= + = +xx x v t tAt point B where 6.5 s,Bt =( )( )132 6.5 858 ft= =Bx(a) Distance AB.From geometry858cos 10d =°871 ftd =Vertical motion: ( ) 20 012= + −yy y v t gtAt point B( )( )21tan 10 0 32.2 6.52− ° = + −Bx h(b) Initial height. 529 fth =
• 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 11, Solution 103.Data: 20 25 ft/s, 90 55 35 , 32.2 ft/sv gα= = ° − ° = ° =Horizontal motion. ( )0 cos=x v tαVertical motion. ( ) 201sin2= + −y h v t gtαEliminate t.0 cosxtv α=2 20tan2 cosgxy h xvαα2= + −Solve for h.22 20tan2 cosgxh y xvαα= − +To hit point B. 20 ft, 0x y= =( )( )( )( )2232.2 200 20 tan35 1.352 ft2 25cos35= − ° + =°hTo hit point C. 24 ft, 0x y= =( )( )( )( )2232.2 240 24 tan35 5.31 ft2 25cos35= − ° + =°hRange of values of h. 1.352 ft < 5.31 fth <
• 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 11, Solution 104.Place the origin at A. Let β be the direction of the discharge velocity measured counterclockwise from thex-axisHorizontal motion. ( ) ( )0 00cos cos= =xv v x v tβ βSolve for t.0 cosxtv β=Vertical motion. ( ) 00sin=yv v β( ) 201sin2= −y v t gtβ22 20tan2 cosgxxvββ= −Geometry. At points B and C tany x α=Hence,22 20tan tan2 cosgxx xvα ββ= −Solve for x. ( )2 202 costan tan= −vxgββ αTo water point B. 090 90 40 50β φ= °− = °− ° = °( )( )( )2 22 24 cos 50tan50 tan10 15.01 ft32.2°= ° − ° =Bx15.01 ftBd =To water point C. 090 90 40 130= ° + = ° + ° = °β φ( )( )( )2 22 24 cos 130tan130 tan10 20.2 ft32.2°= ° − ° = −Cx20.2 ftC Cd x= − =
• 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.Chapter 11, Solution 105.0 0 013 m/s, 33 , 0, 0.6 mv x yα= = ° = =Vertical motion: 0 sinyv v gtα= −( ) 20 01sin2y y v t gtα= + −At maximum height, 0 sin0 oryvv tgα= =(a)13sin330.7217 s9.81t°= =( )( ) ( )( )2max10.6 13sin33 0.7217 9.81 0.72172y = + ° − max 3.16 my =1.8 m 3.16 m 3.7 m< < yesHorizontal motion: ( ) 00 00cos orcosx xx x v t tvαα−= + =At 15.2 m,x =15.2 01.3941 s13cos33t−= =°(b) Corresponding value of :y ( )( ) ( )( )210.6 13sin33 1.3941 9.81 1.39412y = + ° −0.937 my =
• 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 11, Solution 136.Velocities:/ 1 m/sA B A B= − =v v vAccelerations:2/ 0.25 m/sA B A B= − =a a a(a)( )( )2 22222/10019610.25100 96A AAAABBBAAA Bv vavvavvaρρ= =−= =−= − =250 625 0A Av v− + =25Av = ± 25 m/sAv = !(b) 25 1 24Bv = − = 24 m/sBv = !
• 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 11, Solution 137.22max, 0,n t nva a v aρρ= = =( )( ) ( )( )( )2 2 2max 25 3 25 3 9.81 735.35 m /sv g= = =max 27.125 m/sv = max 97.6 km/hv = !
• 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 11, Solution 138.( ) ( )( ) ( )( )( )( )( )2 22,0.660.097066.80.09706 0.09706 60 5.8235 mmc cc cn nA AA Ac A c B cn nA Bc nB AA c n BB Av va av a aaaρ ρρ ρρρρ ρ   = =      = =     = = =  = = =2 11.65 mmB Bd ρ= = !
• 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 11, Solution 139.Initial speed. 0 72 km/h 20 m/sv = =Tangential acceleration. 21.25 m/sta = −(a) Total acceleration at 0.t =( )2220 201.14286 m/s350nvaρ= = =( ) ( )2 22 21.25 1.14286t na a a= + = − + 21.694 m/sa = !(b) Total acceleration at 4 s.t =( )( )0 20 1.25 4 15 m/stv v a t= + = + − =( )222150.6426 m/s350nvaρ= = =( ) ( )2 22 21.25 0.6426t na a a= + = − + 21.406 m/sa = !
• 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.Chapter 11, Solution 140.Length of run. 130 metersL Dπ π= = (1)Radius of circle.165m2Dρ = =Tangential acceleration of starting portion of run.( )( )1 4 4 m/sm t t tv a t a a= = = (2)( )( )221 11 14 8 m2 2t t ts a t a a= = = (3)Constant speed portion of run. mv v=( )1 1ms s v t t= + − (4)Substituting (1), (2) and (3) into (4)( )130 8 4 54 4t ta aπ = + −Solving for .ta 21301.9635 m/s8 200taπ= =+From (2) ( )( )4 1.9635 7.854 m/smv = =Normal acceleration during constant speed portion of run.( )2227.8540.9490 m/s65mnvaρ= = =Maximum total acceleration.( ) ( )2 22 21.9635 0.9490t na a a= + = + 22.18 m/sa = !
• 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 11, Solution 141.For uniformly decelerated motion: 0 tv v a t= +At 9 s,t = ( ) 20 150 9 , or 16.667 ft/st ta a= − = −Total acceleration: 2 2 2t na a a= +( ) ( )1/21/2 2 22 2 2130 16.667 128.93 ft/sn ta a a   = − = − − =    Normal acceleration:21 5, where diameter ft2 12nva ρρ= = =( )2 2 25128.93 53.72 ft /s , 7.329 ft/s12nv a vρ = = = =  Time: 0 7.329 15016.667tv vta− −= =−8.56 st = !
• 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 11, Solution 142.Speeds: 0 10 65 mi/h 95.33 ft/sv v= = =Distance: ( )450 300 1006.86 ft2sπ= + =Tangential component of acceleration: 2 21 0 2 tv v a s= +( )( )( )22 221 0 95.33 04.5133 ft/s2 2 1006.86tv vas+−= = =At point B, 2 20 2B t Bv v a s= + where ( )450 706.86 ft2Bsπ= =( )( )( )2 2 20 2 4.5133 706.86 6380.5 ft /sBv = + =(a) 79.88 ft/sBv = 54.5 mi/hBv = !At 15 s,t = ( )( )0 0 4.5133 15 67.70 ft/stv v a t= + = + =Since ,Bv v< the car is still on the curve. 450 ftρ =Normal component of acceleration:( )22267.7010.185 ft/s450nvaρ= = =(b) Magnitude of total acceleration: ( ) ( )2 22 24.5133 10.185t na a a= + = + 211.14 ft/sa = !
• 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 11, Solution 143.(a) 420 km/hA =v , 520 km/hB =v 60°/B A B A= +v v v or ( )/B A B A B A= − = + −v v v v vSketch the vector addition as shown.( ) ( ) ( )( )( )2 2 2/2 22 cos60420 520 2 420 520 cos60B A A B A Bv v v v v= + − °= + − °or / 477.9 km/hB Av =sin sin60520 477.9α °= or 70.4α = °/ 478 km/hB A =v 70.4° !(b) 26 m/sAa = ( ) 22 m/sB ta = 60°520 km/h 144.44 m/sBv = =( )( )222144.44104.32 m/s200BB nvρ= = =a 30°( ) ( )/B A B A B B At n= − = + −a a a a a a[2= ] [60 104.32° + ] [30 6° − ]( ) ( )2 cos60 sin 60 104.32 cos30 sin30 6= − ° + ° + − ° − ° −i j i j i( ) ( )2 297.34 m/s 50.43 m/s= − −i j2/ 109.6 m/sB Aa = 27.4° !
• 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 11, Solution 144.(a) 180 km/h 50 m/sA = =v 30 , 162 km/h 45 m/sB° = =v 45°( ) ( )/ 45 cos45 sin 45 50 cos120 sin120B A B A= − = ° − ° − ° + °v v v i j i j56.82 75.12 94.2 m/s= − =i j 52.9°/ 339 km/hB Av = 52.9° !(b) ( ) 28 m/sA t=a ( ) 260 , 3 m/sB t° =a 45°( )( )222506.25 m/s400AA nAvρ= = =a 30°( )( )222456.75 m/s300BB nBvρ= = =a 45°( ) ( ) ( ) ( )/B A B A B B A At n t n= − = + − −a a a a a a a( ) ( )3 cos45 sin 45 6.75 cos45 sin 45= ° − ° + ° + °i j i j( ) ( )8 cos60 sin60 6.25 cos30 sin30− ° − ° − − ° − °i j i j( ) ( )2 28.31 m/s 12.07 m/s= +i jor 2/ 15.18 m/sB A =a 56.8° !
• 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 11, Solution 145.(a) As water leaves nozzle.8 m/sv =2sin55 9.81 sin55 8.04 m/sna g= ° = ° =2nvaρ=( )2288.04nvaρ = = 7.96 mρ = !(b) At maximum height of stream.( )08 sin55 6.55 m/sxv v= = ° =29.81 m/sna g= =2nvaρ=( )226.559.81nvaρ = = 4.38 mρ = !
• 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.Chapter 11, Solution 146.Horizontal motion. 0 0cos cosxv v x v tα α= =Vertical motion. 0 sinyv v gtα= −20 01sin2y y v t gtα= + −Eliminate t.20 2 20tan2 cosgxy y xvαα= + − (1)Solving (1) for 0v and applying result at point B( )( )( )( )( )( )220 2 209.81 62 tan cos 2 1.5 6 tan3 0.97 cos 3gxvy x yα α= =+ − + ° − °(a) Magnitude of initial velocity. 0 14.48 m/s=v !(b) Minimum radius of curvature of trajectory.2 2 2cosnnv v va ga gρρ θ= = = = (2)where θ is the slope angle of the trajectory.The minimum value of ρ occurs at the highest point of the trajectory where cos 1=θand 0 cos= =xv v v αThen( )2 22 20min14.48 cos 3cos9.81°= =vgαρmin 21.3 m=ρ !
• 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 11, Solution 147.(a) At point A, 0 120 ft/s 120 ft/sv v= = =v 60°A g=a 232.2 ft/s=( )2sin30 AA nAva gρ= ° =( )22120sin30 32.2sin30AAvgρ = =° °894 ftAρ = !(b) At the point where velocity is parallel to incline,0 sin30 120 sin30 60 ft/sxv v= ° = ° =tan30 60tan30 34.64 ft/sy xv v= ° = ° =( ) ( )2 260 34.64 69.282 ft/sv = + =2sin 60 BnBva gρ= ° =( )2269.282sin60 32.2sin 60BBvgρ = =° °172.1 ftBρ = !
• 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.Chapter 11, Solution 148.Compute x- and y-components of velocity and acceleration.( )22cos 1 3 sin, ,2 cos 2 cost tx xt tπ π ππ π− −= =− −&( )( )( )22 36 sin sin3 cos2 cos 2 cost ttxt tπ π π ππ ππ π−= +− −&&( )( )21.5 2cos 11.5sin, ,2 cos 2 costty yt tπ πππ π−= =− −&( )( )( )( )22 33 2cos 1 sin3 sin2 cos 2 cost ttyt tπ π π ππ ππ π−−= −− −&&(a) 0,t = 21, 0, 0, 1.5 , 3 ,x y x y xπ π= = = = = −& & &&1.5 ,v y π= =&& 23 ,na x π= − =&&( )2221.53nvaπρπ= = 0.75 ftρ = !(b)1,3t =23 2 20, , , 0, ,2 3 3x y x y yπ π= = = − = = −& & &&2,3v xπ= − = −&22,3na yπ= − =&&2 224 33.2nvaπρπ= = 1.155 ftρ = !(c) 1,t =21, 0, 0, , ,2 3x y x y xπ π= − = = = − =& & &&,2v yπ= − =&2,3na xπ= =&&2 2234nvaπρπ= = ⋅ 0.75 ftρ = !
• 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 11, Solution 149.Given:( )324m6tx t−= +( )231m6 4tty−= −Differentiating twice( )( )( )222 2 6 m/s2xv x−= = + =&( )242 m/s2tx t−= +&( )21m/s2 2tty−= −&24 2 2 m/sx t t= − + = −&& 21m/s2y t= −&&At 2 s.t =( ) ( )22 11.5 m/s2 2yv y= = − =&2 2 0xa x= = − =&&212 1.5 m/s2ya y= = − =&&(a) Acceleration. ( )21.5 m/s=a j !(b) Radius of curvature of path.1.5tan6yxvvθ = =14.036θ = °2 2 2 2 26 1.5x yv v v= + = +2 238.25 m /s=cos 1.5 cos14.036na a θ= = °21.45522 m/s=2nvaρ=238.251.45522nvaρ = = 26.3 mρ = !
• 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 11, Solution 150.x Av v=At point B ( )B Axv v=( )cos cosB AxBv vvθ θ= =cos ABvvθ =cos cosn Ba a gθ θ= =ABvgv=2 2B B BBn Av v va gvρ = =3BBAvgvρ = !
• 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.Chapter 11, Solution 151.Let θ be the slope angle of the trajectory at an arbitrary point C.Then, ( )2 2cos , orcosC CC CnCv va ggθ ρρ θ= = =But, the horizontal component of velocity is constant, ( ) ( )C A xxv v=where ( ) ( )0 cos cosA C Cx xv v v vα θ= =Then, 0 cos cosCv vα θ=or 0coscosCv vαθ=so that2 2 200 31 cos coscos cos cosCvvg gα αρθ θ θ = =  (a) Since 0, ,v α and g are constants, Cρ is a minimum at point B where cosθ is a maximum or 0.θ =Then,2 20mincosQ.E.D.Bvgαρ ρ= = !(b)2 2031 coscosCvgαρθ =    or min3Q.E.D.cosCρρθ= !
• 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 11, Solution 152.Let θ be the slope angle of the trajectory at an arbitrary point C.Then, ( )2 2cos orcosC CC CnCv va ggθ ρρ θ= = =But the horizontal component of velocity is constant, ( ) ( )C A xxv v=( ) ( ) ( ) ( )0 0 cosA Cx x xv v x v t v tα= = =0or (1)cosxtv α=where ( ) ( )0 0cos and cosA Cx xv v v vα θ= =Then, 0 cos cosCv vα θ=so that30 cosCCvgvρα= (2)The vertical motion is uniformly accelerated( ) ( )0 00sincosC y ygxv v gt vvαα= − = − (3)( ) ( ) ( )22 2 220 0 0 002 220 2 4 20 0But cos sincos2 tan1cosC x yxv v v v v gvgx g xvv vα αααα = + = + −   = − +   or3/22 23 30 2 4 20 02 tan1cosCgx g xv vv vαα = − +   (4)Finally, substituting (4) into (2) gives3/22 2 202 4 20 02 tan1cos cosv gx g xg v vαρα α = − +   !
• 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.Chapter 11, Solution 153.Given: ( ) ( )cos sinn nRt t ct Rt tω ω= + +r j kiDifferentiating to obtain v and a,( ) ( )( ) ( )( ) ( )2 22 2cos sin sin cossin sin cos cos cos sin2 sin cos 2 cos sinn n n n n nn n n n n n n n n n n nn n n n n n n ndR t t t c R t t tdtdR t t t t R t t t tdtR t t t t t tω ω ω ω ω ωω ω ω ω ω ω ω ω ω ω ω ωω ω ω ω ω ω ω ω= = − + + += = − − − + + − = − − + − rv i j kva i ki kMagnitudes of v and a.( ) ( ) ( )( )2 2 2 22 222 2 2 2 2 22 2 2 2 22 2 2 2cos sin sin coscos 2 sin cos sinsin 2 sin cos cos1x y zn n n n n nn n n n n nn n n n n nnv v v vR t t t c R t t tR t t t t t t cR t t t t t tR t cω ω ω ω ω ωω ω ω ω ω ωω ω ω ω ω ωω= + +   = − + + +    = − + +  + + + = + + ( )( ) ( )2 2 2 22 2 2 22 22 2 22 2 2 3 4 2 22 2 3 4 2 22or 12 sin cos 2 cos sin4 sin 4 sin cos cos4 cos 4 sin cos sin4nx y zn n n n n n n nn n n n n n nn n n n n n nv R t ca a a aR t t t t t tR t t t t t tt t t t t tRωω ω ω ω ω ω ω ωω ω ω ω ω ω ωω ω ω ω ω ω ω= + += + + = − − + −  = + ++ − + = ( )2 4 2 2 2or 4n n n nt a R tω ω ω ω+ = +Tangential component of acceleration:( )2 21/22 2 2 21tndv R n tadtR t cωω= = + + At 2 2 20, , 2 , 0n tt v R c a R aω= = + = =Normal component of acceleration: 2 22n t na a a Rω= − =2But nvaρ=or2nvaρ =2 22 nR cRρω+= !
• 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 11, Solution 154.With 3 and 1,A B= = the position vector is( ) ( ) ( )23 cos 3 1 sinr t t t t t= + + +i j kDifferentiating to obtain v and a,( ) ( )( )( )222233 cos sin sin cos1113 sin sin cos 31cos cos sin3 2sind tt t t t t tdt ttt ttdt t t tdt tt t t tt t  = = − + + + +     + −  +  = = − − − ++    + + −= − +rv i j kva i jk( )( )( )3/223cos 2cos sin1t t t tt+ + −+i j kMagnitude of 2.v( ) ( )22 22 2 2 2299 cos sin sin cos1x y ztv v v v t t t t t tt= + + = − + + ++Differentiating,( )( )( )( )( )22182 18 cos sin 2sin cos12 sin cos 2cos sindv tv t t t t t tdt tt t t t t t= − − − +++ + −2When 0, 3 2 , 9, 2 0dvt v vdt= = + = =a j k2 2 23 2 13a = + =Tangential acceleration: 0tdvadt= =Normal acceleration: 2 2 213 or 13n t na a a a= − = =But2 29or13nnv vaaρρ= = = 2.50 ftρ = !
• 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.Chapter 11, Solution 155.For the sun, 2274 m/s ,g =and ( )9 91 11.39 10 0.695 10 m2 2R D = = × = ×  Given that22ngRar= and that for a circular orbit2nvar=Eliminating na and solving for r,22gRrv=For the planet Earth, 6 3107 10 m/h 29.72 10 m/sv = × = ×( )( )( )2992274 0.695 10Then, 149.8 10 m29.72r×= = × 149.8 Gmr = !
• 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 11, Solution 156.For the sun, 2274 m/sg =and ( )9 91 11.39 10 0.695 10 m2 2R D = = × = ×  Given that22ngRar= and that for a circular orbit:2nvar=Eliminating na and solving for r,22gRrv=For the planet Saturn, 6 334.7 10 m/h 9.639 10 m/sv = × = ×Then,( )( )( )29122274 0.695 101.425 10 m9.639r×= = × 1425 Gmr = !
• 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 11, Solution 157.From Problems 11.155 and 11.156,22ngRar=For a circular orbit,2nvar=Eliminating na and solving for v,gv Rr=For Venus, 229.20 ft/sg =63761mi 19.858 10 ft.R = = ×63761 100 3861mi 20.386 10 ftr = + = = ×Then, 6 3629.2019.858 10 23.766 10 ft/s20.386 10v = × = ××16200 mi/hv = !
• 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 11, Solution 158.From Problems 11.155 and 11.156,22ngRar=For a circular orbit,2nvar=Eliminating na and solving for v,gv Rr=For Mars, 212.24 ft/sg =62070 mi 10.930 10 ftR = = ×32070 100 2170 mi 11.458 10 ftr = + = = ×Then, 6 3612.2410.930 10 11.297 10 ft/s11.458 10v = × = ××7700 mi/hv = !
• 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 11, Solution 159.From Problems 11.155 and 11.156,22ngRar=For a circular orbit,2nvar=Eliminating na and solving for v,gv Rr=For Jupiter, 275.35 ft/sg =644432 mi 234.60 10 ftR = = ×644432 100 44532 mi 235.13 10 ftr = + = = ×( )6 3675.35Then, 234.60 10 132.8 10 ft/s235.13 10v = × = ××90600 mi/hv = !
• 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 11, Solution 160.Radius of Earth ( )( ) 63960 mi 5280 ft/mi 20.908 10 ftR = = ×Radius of orbit ( )( ) 63960 10900 5280 78.4608 10 ftr = + = ×Normal acceleration22ngRar= and2nvar=Thus,2 22v gRr r= or22 gRvr=( )( )262 6 2 2632.2 20.908 10179.40 10 ft /s78.4608 10v×= = ××313.3941 10 ft/sv = ×Time T for one orbit. 2vT rπ=( )6332 78.4608 102= 36.806 10 s13.3941 10rTvππ ×= = ××10.22 hT = !
• 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.Chapter 11, Solution 161.Normal acceleration.22ngRar= and2 2nv varρ= =Solve for v2.22ngRv rar= =Data: 29.81 m/s ,g = 66370 km = 6.370 10 mR = ×3 6384 10 km = 384 10 mr = × ×( )( )262 6 2 269.81 6.370 101.0366 10 m /s384 10v×= = ××= 1.018 m/sv 3670 km/hv = !
• 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 11, Solution 162.From Problems 155 through 156,22ngRar=For a circular orbit,2nvar=Eliminating na and solving for v,gv Rr=For one orbit the distance traveled is 2 ;rπ hence, the time is2 rtvπ=or3 21 22 rtRgπ=For satellites A and B,3 2 3 21 2 1 22 2andA BA Br rt tRg Rgπ π= =Let number of orbits of .n B= For the next alignment,( )3 23 211 or11B BA BA ABAn t rn t ntn t rrn r ++ = = =    = −  Data: 36370 km 6.370 10 mR = = ×36370 190 6560 km 6.560 10 mAr = + = = ×36370 320 6690 km 6.690 10 mBr = + = = ×Then,3/2331 6.690 101 0.02987 or 33.4756.560 10nn ×= − = =  × continued
• 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.Time for orbit of satellite B is( )( )( )3 2631 262 6.690 105.449 10 s 1.5137 h6.370 10 9.81Btπ ×= = × =×Time for next alignment is( )( )33.475 1.5137Bnt = 50.7 hBnt = !
• 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 11, Solution 163.Differentiate the expressions for r and θ with respect to time.2 320.80.8 0.80.8 0.81 2 6 82 12 2412 480.5 sin30.4 sin3 1.5 cos30.32 sin3 1.2 cos3tt tt tr t t tr t tr te te t e te t e tθ πθ π π πθ π π π−− −− −= + − += − += − +== − += −&&&&&&0.8 2 0.81.2 cos3 4.5 sin3t te t e tπ π π π− −− −At 0.5 s,t = 21.5 ft, 2.00 ft/s, 12 ft/s ,r r r= = =& &&0.820.67032, sin 3 1, cos 3 00.33516 rad, 0.26812 rad/s, 29.56 rad/ste t tπ πθ θ θ−= = − == − = =& &&(a) Velocity of the collar.rr r θθ= +v e e&& ( ) ( )2.00 ft/s 0.402 ft/sr θ= +v e e !2 ft/s, 0.402 ft/srv vθ= = !(b) Acceleration of the collar.( ) ( )22r r rr r r r a aθ θ θθ θ θ= − + + = +a e e e e& && &&& &( )( )212 1.5 0.26812ra = − 211.89 ft/sra = !( )( ) ( )( )( )1.5 29.56 2 2 0.26812aθ = + 245.41 ft/saθ = !( ) ( )2 211.89 ft/s 45.41 ft/sr θ= +a e e !(c) Acceleration of the collar relative to the rod.( )212 ft/sr rr =e e&& !
• 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.Chapter 11, Solution 164.Differentiate the expressions for r and θ with respect to time.( ) ( )22 310 10 20mm, mm/s, mm/s6 6 6r r rt t t= = − =+ + +& &&24sin rad, 4cos rad/s 4 sin rad/st t tθ π θ π θ π ππ= = =& &&At 1s,t = 210 10 20mm; mm/s, mm/s7 49 343r r r= = − =& &&0, 4 rad/s, 0θ θ θ= = − =& &&(a) Velocity of the collar.0.204 mm/s, 5.71 mm/srv r v rθ θ= = = = −&&( ) ( )0.204 mm/s 5.71 mm/sB r θ= −v e e !(b) Acceleration of the collar.( )( ) ( ) ( )22 2220 104 22.8 mm/s343 710 102 0 2 4 1.633 mm/s7 49ra r ra r rθθθ θ = − = − − = −     = + = + − − =      &&&&& &&( ) ( )2 222.8 mm/s 1.633 mm/sB r θ= − +a e e !(c) Acceleration of the collar relative to the rod./20343B OA r rr= =a e e&& ( )2/ 0.0583 mm/sB OA r=a e !
• 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 11, Solution 165.Given ( )2 cos /2r B At B= /2At Bθ =Differentiating twice( )sin /2r A At B= −& /2A Bθ =&( ) ( )2/2 cos /2r A B At B= −&& 0θ =&&Components and magnitude of velocity.( )sin /2 sinrv r A At B A θ= = − = −&( ) ( )2 cos /2 /2 cosv r B At B A B Aθ θ θ = = = &(a) 2 2 2 2 2 2sin cosrv v v A A Aθ θ θ= + = + = v A= !Components and magnitude of acceleration.( ) ( ) ( ) [ ]22 2/2 cos /2 2 cos /2 /2ra r r A B At B At B A Bθ  = − = − +  &&&( )2/ cosA B θ= −( ) ( )2 0 (2) sin /2 /2a r r A At B A Bθ θ θ  = + = + − && &&2/ sinA B θ= −( ) ( )2 2 4 2 2 4 2 2/ cos / sinra a a A B A Bθ θ θ= + = += 2/A B 2/a A B= !From the figure a is perpendicular to vThus, 2/na a A B= =2nvaρ=2nvaρ =(b)( )22/ABA Bρ = = Bρ = !Since ρ is constant, the path is a circle of radius B.
• 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.Chapter 11, Solution 166.Differentiate the expressions for r andθ with respect to time.( )22 cos ,sin ,cos,,0r b tr b tr b ttππ ππ πθ πθ πθ= += −= −===&&&&&&(a) At 2 s,t = sin 0, cos 1t tπ π= =23 , 0, , 2 rad, rad/sr b r r bπ θ π θ π= = = − = =&& &&0 , 3 ,rv r v r bθ θ π= = = =&& 3 b θπ=v e( )2 2 2 23 4ra r r b b bθ π π π= − = − − = −&&&2 0,a r rθ θ θ= + =&& && 24 rbπ= −a e(b) Values of θ for which v is maximum.( )( )( )22 2 2 2 2 22 2 2 22 2sin2 cossin 2 cossin 4 4cos cos5 4cosrrv r b tv r b tv v v b t tb t t tb tθθπ πθ π ππ π ππ π π ππ π= = −= = − +⎡ ⎤= + = + +⎢ ⎥⎣ ⎦⎡ ⎤= + + +⎣ ⎦= +&&2v is maximum when cos 1 or 0, 2 , 4 , 6 , etct tπ π π π π= =But , hencetθ π= 2 , 0,1, 2,N Nθ π= = K
• 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 11, Solution 167.Differentiate the expressions for r andθ with respect to time.( )1 22 2 2 26 1 4 , 6 1 4 24 1 4r t t r t t t−= + = + + +&( ) ( )1 2 3 22 3 272 1 4 96 1 4 ,r t t t t− −= + − +&&( )12arctan 2 2 1 4 ,t tθ θ−= = +&( )2216 1 4t tθ−= − +&&(a) At 0,t = 0, 6 ft/s, 0r r r= = =& &&0, 2 rad/s, 0θ θ θ= = =& &&6 ft/s, 0,rv r v rθ θ= = = =&& ( )6 ft/s r=v e20,ra r rθ= − =&&& 22 24 ft/s ,a r rθ θ θ= + =&& && ( )224 ft/s θ=a e(b) At 0.5 s,t = 23 2 ft, 9 2 ft/s, 15 2 ft/sr r r= = =& &&2rad, 1rad/s, 2 rad/s4πθ θ θ= = = −& &&12.73 ft/s, 4.243 ft/srv r v rθ θ= = = =&&( ) ( )12.73 ft/s 4.24 ft/sr θ= +v e e( )22 215 2 3 2 1 16.97 ft/sra r rθ= − = − =&&&( ) ( )( )( ) 22 3 2 2 2 9 2 1 16.97 ft/sa r rθ θ θ= + = − + =&& &&( ) ( )2 216.97ft/s 16.97ft/sr θ= +a e e
• 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 11, Solution 168.Change to rectangular coordinates. cos and sinx yr rθ θ= =Equation of the path:3 3 3sin cosrry x y xr rθ θ= = =− −−from which 3 or 3.y x y x− = = +Also, 23 3 1tan 1 1y xx x x tθ+= = = + = +from which 2 23 and 3 1x t y t= = +Differentiating, 6 , 6x yv x t v y t= = = =& &6, 6x ya x a y= = = =&& &&(a) Magnitudes: 2 2x yv v v= + 6 2 ft/sv t=2 2x ya a a= + 26 2 ft/sa =(b) 3y x= + is the equation of a straight line.Hence, ρ = ∞
• 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.Chapter 11, Solution 169.Sketch the directions of the vectors v and .θecosv vθ θ θ= ⋅ = −v eBut v rθ θ= &Hence, cosr vθ θ= −&But from geometry,cosbrθ=2cos orcos cosb bv vθ θθθ θ= − = −& &Speed is the absolute value of v.2cosbvθθ=&
• 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 11, Solution 170.From geometry,cosbrθ=Differentiating with respect to time, 2sincosbrθθθ=&&Transverse component of acceleration222 sin2cos cosb ba r rθθ θθθ θθ θ= + = +&& &&& && (1)Sketch the directions of the vectors a and .θecosa aθ θ θ= ⋅ = −a e (2)Matching from (1) and (2) and solving for a,( )22 3222 sincos cos2tancosb babθ θθθ θθ θθθ= − −= − +&& &&& &Since magnitude of a is sought, 22| | 2tancosba θ θθθ= +&& &
• 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.Chapter 11, Solution 171.Sketch the geometry.( )180 180θ β α+ ° − + = °α β θ= −( )sin 180 sinr dβ α=° −sinsindrβα=Sketch the velocity vectors.( )cos 90v vθ θ α= ⋅ = ° −v esinv α=Butsinor sin ,sindv r vθβθ α θα= =& &or 2sinsindvβθα= &( )2sinsindvβθβ θ=−&
• 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 11, Solution 172.Looking at d and β as polar coordinates with 0,d =&2 2, 02 0,ddv d d v da d d a d d dβββ ωβ β β ω= = = == + = = − = −&&& &&&& & &Geometry analysis: 3r d= for angles shown.(a) Velocity analysis: Sketch the directions of v, and .r θe ecos120r rv r dω= = ⋅ = °v e&12r dω= −&cos30v r dθ θθ ω= = ⋅ = °v e&32cos303ddr dωωθ°= =& 12θ ω=&(b) Acceleration analysis: Sketch the directions of a, and .r θe e23cos1502r ra a a dω= ⋅ = ° = −e2 232r r dθ ω− = −&&&22 2 23 3 132 2 2r d r d dω θ ω ω⎛ ⎞= − + = − + ⎜ ⎟⎝ ⎠&&&234r dω= −&&2 21cos12022a d da r rθ θθω ωθ θ= ⋅ = ° = −= +a e&& &&( ) ( )21 1 1 1 12 22 2 23a r d dr dθθ θ ω ω ω⎡ ⎤⎛ ⎞⎛ ⎞= − = − − −⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦&& && 0θ =&&
• 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.Chapter 11, Solution 173.Rate of change of .θ 348.0 47.0 1.0 17.453 10 radθ −Δ = ° − ° = ° = ×0.5 stΔ =3317.453 1034.907 10 rad/s0.5tθθ−−Δ ×≈ = = ×Δ&Let r be a polar coordinate with origin at A.34 km 4 10 mb = = ×334 105.921 10 mcos cos47.5brθ×= = = ×°( )( )3 35.921 10 34.907 10 206.68 m/sv rθ θ −= = × × =&From geometry,206.68cos cos47.5vv θθ= =°306 m/sv =Alternate solution. tanx b θ=22seccosbv x bθθθθ= = =&&&( )( )3 324 10 34.907 10306 m/scos 47.5v−× ×= =°
• 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 11, Solution 174.Changes in values over the interval13600 12600 1000 ftrΔ = − =328.3 31.2 2.9 5.0615 10 radθ −Δ = ° − ° = − ° = − ×2t sΔ =Rates of change.1000500 ft/s2rrtΔ= = =Δ&335.0615 102.5307 10 rad/s2tθθ−−Δ − ×= = = − ×Δ&Mean values.12600 1360013100 ft2r+= =31.2 28.329.752θ° + °= = °Velocity components.500 ft/srv r= =&( )( )313100 2.5307 10 331.53 ft/sv rθ θ −= = − × = −&( ) ( )2 22 2500 331.53 600 ft/srv v vθ= + = + − =409 mi/hv =cos sinx rv v vθθ θ= −( )500cos29.75 331.53 sin 29.75 598.61 ft/s= ° − − ° =sin cosy rv v vθθ θ= +( )( )500sin 29.75 331.53 cos29.75 39.73 ft/s= ° + − ° = −39.73tan 0.06636598.61yxvvα−= = = 3.80α = °
• 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 11, Solution 175.2 21/2 1/2,r be r beθ θθθ= = &&( ) ( )2 221/2 1/222 2 2 1/2 2 2,1rrv r be v r bev v v beθ θθθθθθ θ θθ θ= = = == + = +& & &&&( )2 1/21 2 21v be θθ θ= + &
• 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.Chapter 11, Solution 176.2 32,b br r θθ θ= = − &&3 22,rb bv r v rθθ θ θθ θ= = − = =& & &&( )2 3 22 2 2 2 2 2 26 4 644rb b bv v vθ θ θ θ θθ θ θ= + = + = +& & &( )1 2234bv θ θθ= + &
• 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 11, Solution 177.( )( ) ( )2 2 22 22 2 221/2 1/2 1/2 22 22 1/2 2 2 1/21/2 1/2 2 1/2 2, ,2 2 2rr be r be r bea r r be bea r r be be beθ θ θθ θθ θ θθθθ θθ θ θθθ θθ θ θθ θ θθ θθθ θ θ θθ θ θθ⎡ ⎤= = = + +⎢ ⎥⎣ ⎦⎡ ⎤ ⎡ ⎤= − = + + − = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤= + = + = +⎣ ⎦& & & &&& &&& & & && & & &&&&&& & && & && &&But and 0θ ω θ= =& &&( ) ( )( ) ( )2 2221/2 1/2 222 2 2 1/2 4 2 4and 24rra be a bea a a beθ θθθθθω θωθ θ ω= == + = +( )2 1/21/2 2 24a be θθ θ ω= +
• 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 11, Solution 178.22 3 3 42 2 6, ,b b b br r rθ θ θθ θ θ θ= = − = − +& && && &&( )2 2 2 2 2 23 4 2 42 62 6rb b b ba r rθ θ θ θ θθ θ θ θθ θ θ θ= − = − + − = − + −& & & & && & &&&( ) ( )2 22 3 322 2 4b b ba r rθ θ θ θ θ θθ θθ θ θ⎛ ⎞= + = = − = −⎜ ⎟⎝ ⎠&& & && & && &&But and 0θ ω θ= =& &&( )2 2 24 346 andrb ba aθθ ω ωθ θ= − = −( )2 22 2 2 2 4 2 28 61636 12rb ba a aθ θ θ ω ωθ θ= + = − + +( )22 4 2836 4bθ θ ωθ= + +( )1 22 4 2436 4ba θ θ ωθ= + +
• 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 11, Solution 179.Sketch the geometry.Law of cosines: 2 2 22 cosr d h dh ϕ= + −Differentiating with respect to time and noting that d and h are constant,2 2 sinrr dh ϕϕ=& &sindhrrϕϕ=& &Law of sines:sin sinr dϕ θ=so that sin Q.E.Dr h θϕ=& &
• 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 11, Solution 180.Given: , ,1 1A CtR Bt zt tθ= = =+ +Differentiating with respect to time,( )( )( ) ( )( ) ( )2 2 23 31, ,1 1 12 2, 0,1 1C t CtA CR B zt t tA CR zt tθθ+ −= − = = =+ + += = = −+ +&& &&&&& &(a) 0.t = , 0, 0R A zθ= = =, ,2 , 0, 2R A B z CR A z Cθθ= − = == = = −&&& &&&&& &&, ,R zv R A v R AB v z Cθ θ= = − = = = =&& &2 2 2 2 2 2 2 2R zv v v v A A B Cθ= + + = + + 2 2 2 2v A A B C= + +2 2 2 2 2 2 2 42 4 4R Ra R R A AB a A A B A Bθ= − = − = − +&&&2 0 2a R R ABθ θ θ= + = −&& && 2 24a A Bθ =2za z c= = −&& 2 24za C=2 2 2 2 2 2 4 24 4R za a a a A A B Cθ= + + = + + 2 2 4 24 4a A A B C= + +(b) .t = ∞ 0, , , 0, , 0,R z C R B zθ θ= = ∞ = = = =&& &0, 0, 0R zθ= = =&&&& &&0, 0, 0,r zv R v R v zθ θ= = = = = =&& & 0v =2 20, 0, 0,r za R R a R R a zθθ θ θ= − = = − = = =& && &&& &&0a =
• 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 11, Solution 182.From problem 11.97, the position vector is ( ) ( )cos sin .n nRt t ct Rt tω ω= + +r i j kDifferentiating to obtain v and a,( ) ( )( ) ( )2 2cos sin sin cossin sin cos cos cos sinn n n n n nn n n n n n n n n n n ndR t t t c R t t tdtdR t t t t R t t t tdtω ω ω ω ω ωω ω ω ω ω ω ω ω ω ω ω ω= = − + + += = − − − + + −rv i j kva i k( ) ( )2 22 sin cos 2 cos sinn n n n n n n nR t t t t t tω ω ω ω ω ω ω ω = − − + − i k( ) ( ) ( )x y z y z z y z x x z x y y xx y zv v v v a v a v a v a v a v aa a a× = = − + − + −i j kv a i j k( ) ( )( )( )( )( )2 2 22 222 cos sin sin cos 2 sin coscos sin 2 cos sin2 sin cosn n n n n n n n n n nn n n n n n nn n n ncR t t t R t t t t t tR t t t t t tcR t t tω ω ω ω ω ω ω ω ω ω ωω ω ω ω ω ω ωω ω ω ω  = − + + − −  − − − + − − − ijk( ) ( ) ( )2 2 22cos sin 2 2sin cosn n n n n n n n n ncR t t t R t cR t t tω ω ω ω ω ω ω ω ω ω= − − + + +i j k( ) ( )1/222 2 2 2 2 4 2 2 2| | 4 2n n n nc R R tω ω τ ω ω × = + + +  v aThe binormal unit vector be is given by| |b×=×v aev aLet α be the angle between the y-axis and the binormal.( ) ( )( ) ( )21 22 2 2 2 2 4 2 2 22cos| |4 2n nbn n n nR tv ac R t R tω ωαω ω ω ω+× ⋅= ⋅ = =×  + + + v a je j( ) ( )1 22 2 2 2 2Let 2 , 4 ,n n n nA R t B cR tω ω ω ω= + = + 2 2C A B= + so that cos asACα =shown in the sketch. The angle that the osculating plane makes with they-axis is the angle .β( )( )2 21 22 22tan4nnR tAB c tωβω+= =+( )( )2 211 22 22tan4nnR tc tωβω−+=+!
• 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 11, Solution 183.For 3 and 1,A B= = ( ) ( ) ( )23 cos 3 1 sint t t t t= + + +r i j kDifferentiating to obtain v and a.( ) ( )( )( )( )23/223 cos sin 3 sin cos113 2sin cos 3 2cot sin1d tt t t t t tdt tdt t t t t tdt t= = − + + ++= = − − + + −+rv i j kva i j k(a) At 0,t = ( ) ( ) ( )3 1 0 0 0 3= − + + =v i j k i( )3(0) 3(1) 2 0 3 2= − + + − = +a i j k j k3 0 0 6 90 3 2× = = − +i j kv a j k2 2| | 6 9 10.817× = + =v a0.55470 0.83205| |b×= = − +×v ae j kv a2cos 0, cos 0.55470, cos 0.83205x yθ θ θ= = − =z90 , 123.7 , 33.7x yθ θ θ= ° = ° = ° !(b) At s,2tπ= 4.71239 2.53069v = − + +i j k6 0.46464 1.5708= − + −a i j k4.71239 2.53069 16 0.46464 1.57084.43985 13.4022 12.9946× = −−= − − +i j kv ai j kcontinued
• 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.( ) ( ) ( )1/22 2 2| | 4.43985 13.4022 12.9946 19.1883 × = + + =  v a0.23138 0.69846 0.67721| |b×= = − − +×v ae i j kv acos 0.23138, cos 0.69846, cos 0.67721x y zθ θ θ= − = − =103.4 , 134.3 , 47.4x y zθ θ θ= ° = ° = ° !
• 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 11, Solution 184.Given: 2 20 9 9ft/s , 36 ft, 144 ft, 27 ft/sa kt x x v= = = =2 30 0 013t tv v adt kt dt kt− = = =∫ ∫Velocity: 3013v v kt= +40 00112tx x vdt v t kt− = = +∫Position: 4 40 0 01 13612 12x x v t kt v t kt= + + = + +When 9 s,t = 144 ft and 27 ft/sx v= =( ) ( )40136 9 9 14412v k+ + =or 09 546.75 108v k+ = (1)( )3019 273v k+ =0 243 27v k+ = (2)Solving equations (1) and (2) simultaneously yields:40 7 ft/s and 0.082305 ft/sv k= =Then, 436 7 0.00686 ftx t t= + +37 0.0274 ft/sv t= +
• 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 11, Solution 185.(a) Determination of k.From( ),0.6 1dv dvdv adt dta kv= = =−Integrating, using the condition 0v = when 0,t =( )( ) ( )0 0 0 01 1or ln 1 ln 10.6 1 0.6 0.6vtt v dvdt t kv t kvkv k k = = − − = − − −∫ ∫ (1)Using 20 s when 6 mm/s,t v= = ( )120 ln 1 60.6kk= − −Solving by trial, 0.1328 s/mk =(b) Position when 7.5 m/s.v =From ,vdv a dx=( )0.6 1vdv vdvdxa kv= =−Integrating, using the condition 6 mx = when 0,v =( )6 00.6 1x v vdvdxkv=−∫ ∫( )001 1 1 16 1 ln 10.6 1 0.6vvx dv v kvk kv k k   − = − + = − − −   −   ∫( )1 16 ln 10.6x v kvk k = − + −  
• 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.Using 7.5 m/sv = and the determined value of k:( )( )( )( )( )1 16 7.5 ln 1 0.1328 7.50.6 0.1328 0.1328x = − + −  434 mx =(c) Maximum velocity occurs when a = 0. max1 10.1328vk= = max 7.53 m/sv =
• 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 11, Solution 186.Constant acceleration. 0 25 mi/h 36.667 ft/sv = =65 mi/h 95.333 ft/sfv = =0 0 and 0.1 mi 528 ftfx x= = =( )2 20 02f fv v a x x= + −(a) Acceleration.( ) ( )2 2 2 20 2095.333 36.6677.3333 ft/s2 528 02ffv vax x− −= = =−−27.33 ft/sa =(b) Time to reach 65 mph. 0f fv v at= +0 95.333 36.6677.3333ffv vta− −= = 8.00 sft =
• 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 11, Solution 187.Let x be position relative to the fixed supports, taken positive if downward.Constraint of cable on left: 2 3 constantA Bx x+ =2 22 3 0, or , and3 3A B B A B Av v v v a a+ = = − = −Constraint of cable on right: 2 constantB Cx x+ =1 1 12 0, or , and2 3 3B C C B A C Av v v v v a a+ = = − = =Block C moves downward; hence, block A also moves downward.(a) Accelerations.( )( ) 200456 0or 38.0 mm/s12A AA A A Av vv v a T at− −= + = = =238.0 mm/sA =a( ) 22 238.0 25.3 mm/s3 3B Aa a⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠225.3 mm/sB =a( ) 21 138.0 12.67 mm/s3 3C Aa a⎛ ⎞= = =⎜ ⎟⎝ ⎠212.67 mm/sC =a(b) Velocity and change in position of B after 8 s.( ) ( )( )00 25.3 8 203 mm/sB B Bv v a t= + = + − = −203 mm/sB =v( ) ( ) ( )( )220 01 10 25.3 8 811 mm2 2B B B Bx x v t a t− = + = + − = −811 mmBxΔ =
• 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 11, Solution 188.(a) Construction of the curves.Construct the a t− curve. slope of curvea v t= −0 10 s:t< < 10 s,t∆ = 0v∆ = 0vat∆= =∆10 s 26 s:t< < 16 s,t∆ = 80 m/sv∆ = − 25 m/svat∆= = −∆26 s 41 s:t< < 15 s,t∆ = 0v∆ = 0vat∆= =∆41 s 46 s:t< < 5 s,t∆ = 15 m/sv∆ = 23 m/svat∆= =∆46 s 50 s:t< < 4 s,t∆ = 0v∆ = 0vat∆= =∆Construct the curve.x t− area of curve.x v t∆ = −x is maximum or minimum where 0.v =For 10 s 26 s,t≤ ≤ ( )60 5 10v t= − −0v = when 60 5 50 0 or 22 st t− + = =Also 0 540 mx = −0 to 10 s ( )( )10 60 600 mx∆ = = 10 540 600 60 mx = − + =10 s to 22 s ( )( )112 60 360 m2x∆ = = 22 60 360 420 mx = + =22 s to 26 s ( )( )14 20 40 m2x∆ = − = − 26 420 40 380 mx = − =
• 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.26 s to 41 s ( )( )15 20 300 mx∆ = − = − 41 380 300 80 mx = − =41 s to 46 s ( )20 55 62.5 m2x− − ∆ = = −  46 80 62.5 17.5 mx = − =46 s to 50 s ( )( )4 5 20 mx∆ = − = − 50 17.5 20 2.5 mx = − = −(b) Total distance traveled.( )1 22 00 22 s, 420 540 960 mt d x x≤ ≤ = − = − − =2 50 2222 s 50 s, 2.5 420 422.5t d x x≤ ≤ = − = − − =Total: 1 2 1382.5 md d d= + = 1383 md =(c) Times when 0.x =For 0 10 s,t≤ ≤ 540 60 mx t= − +At 0,x = 540 60 0t− + = 9 st =For 46 s 50,t≤ ≤ ( )17.5 5 46 mx t= − −At 0,x = ( )17.5 5 46 0 46 3.5t t− − = − = 49.5 st =
• 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 11, Solution 189.( ) ( )0 0100 km/h 27.778 m/s 25 km/h 6.944 m/sA Bv v= = = =Sketch acceleration curve for car B over 0 5 s.t< <Using moment-area formula at 5 s.t =( ) ( ) ( )( )( )( )( )025 2.570 6.944 5 12.52.822 m/sB B o BBBx x v t aaa− = += +=Determine when reaches 100 km/h.B( ) ( ) 2027.778 6.944 2.8227.38 sB BfBBv v Att= += +=( )( )2 2.822 7.38 20.83 m/sA = =Then, ( ) ( ) 20 02BB B B Btx x v t A= + + by moment-area formulaand ( ) ( )0 0A A A Bx x v t= +Subtracting, ( ) ( ) ( ) ( ) 20 0 0 02BB A B A B A Btx x x x v v t A⎡ ⎤− = − + − +⎣ ⎦Then, ( )( ) ( )7.38120 6.944 27.778 7.38 20.832B Ax x⎛ ⎞− = + − + ⎜ ⎟⎝ ⎠/Car is ahead of car . 43.1 mB AB A x =
• 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 11, Solution 190.(a) Vertical motion: 0 1.5 m,y = ( )00yv =( ) ( )020 021or2yy yy y v t gt tg−= + − =At point B,( )02or By hy h tg−= =When 788 mm 0.788 m,h = =( )( )2 1.5 0.7880.3810 s9.81Bt−= =When 1068 mm 1.068 m,h = =( )( )2 1.5 1.0680.2968 s9.81Bt−= =Horizontal motion: ( )0 000, ,xx v v= =0 0or BBx xx v t vt t= = =With 12.2 m,Bx = 012.2we get 32.02 m/s0.3810v = =012.2and 41.11 m/s0.2968v = =032.02 m/s 41.11 m/sv≤ ≤ or 0115.3 km/h 148.0 km/hv ≤≤(b) Vertical motion: ( )0y yv v gt gt= − = −Horizontal motion: 0xv v=( )( ) 0tany BBx Bvdy gtdx v vα = − = − =For 0.788 m,h =( )( )9.81 0.3810tan 0.11673,32.02α = = 6.66α = °For 1.068 m,h =( )( )9.81 0.2968tan 0.07082,41.11α = = 4.05α = °
• 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 11, Solution 191.The horizontal and vertical components of velocity are00sin15cos15xyv vv v gt= °= ° −At point B,00sin15tan12cos15xyv vv v gt°= = − °° −or 0 0sin15 cos15 tan12 tan12v v gt° + ° ° = °00.46413 tan12v gt= °02.1836vtg=Vertical motion:( )20 02220 0201cos15212.1836cos15 2.183620.27486y y v t gtv vgg gvg− = ° −⎛ ⎞= ° − ⎜ ⎟⎝ ⎠= −( ) ( )( )20 02 283.638 3.638 32.2 01278.10 ft / sv g y y⎛ ⎞= − − = − − −⎜ ⎟⎝ ⎠=0 8.84 ft /sv =
• 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.Chapter 11, Solution 192.First determine the velocity Cv of the coal at the point where the coalimpacts on the belt.Horizontal motion: ( ) ( ) 01.8cos50C Cx xv v⎡ ⎤= = − °⎣ ⎦1.1570 m/s= −Vertical motion: ( ) ( ) ( )22002C Cy yv v g y y⎡ ⎤= − −⎣ ⎦( ) ( )( )( )( )22 21.8sin50 2 9.81 1.531.331 m /s5.5974 m/sC yv= ° − −== −( ) ( )2 22 2 25.5974tan 4.8379, 78.321.157032.669 m /sC C Cx yv v vβ β−= = = °−= + =5.7156 m/s, 5.7156 m/sC Cv = =v 78.32°or ( ) ( )1.1570 m/s 5.5974 m/sC = − + −v i jVelocity of the belt: ( )cos10 sin10B Bv= − ° + °v i jRelative velocity: ( )/C B C B C B= − = + −v v v v v(a) /C Bv is vertical. ( )/ 0C B xv =( ) ( )/ 1.1570 cos10 0, 1.175 m/sC B B Bxv v v= − − − ° = =1.175 m/sB =v 10°(b) /C Bv is minimum. Sketch the vector addition as shown.2 2 2/ 2 cos88.32B C B C B Cv v v v v= + − °Set the derivative with respect to Bv equal to zero.2 2 cos88.32 0B Cv v− ° =cos88.32 0.1676 m/sB Cv v= ° = 0.1676 m/sB =v 10°
• 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 11, Solution 193.Given: ( ) ( ) 200, 0.8 in./sAA A tdvv adt= = =Then, ( ) ( )00.8A A A tv v a t t= + =(a) 0,t = ( )20, 0AA A nvv aρ= = =( )A A ta a= 20.800 in./sAa =(b) 2 s,t = ( )( )0 0.8 2 1.6 in./sAv = + =( )( )2221.60.731 in./s3.5AA nvaρ= = =( ) ( ) ( ) ( )1/2 1/22 2 2 20.8 0.731A A At na a a⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦21.084 in./sAa =
• 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.Chapter 11, Solution 194.(a) At point A. Aa g= 29.81m/s=Sketch tangential and normal components of acceleration at A.( ) cos50A na g= °( )( )2229.81cos50AAA nvaρ = =°0.634 mAρ =(b) At point B, 1 meter below point A.Horizontal motion: ( ) ( ) 2cos50 1.286 m/sB Ax xv v= = ° =Vertical motion: ( ) ( ) ( )2 22B A y B Ay yv v a y y= + −( ) ( )( )( )22 22cos40 2 9.81 121.97 m /s= ° + − −=( ) 4.687 m/sB yv =( )( )4.687tan , or 74.61.286B yB xvvθ θ= = = °cos74.6Ba g= °( )( ) ( )2 22cos74.6B Bx yBBB nv vva gρ+= =°( )21.286 21.979.81cos74.6+=°9.07 mBρ =