solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 19

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Transcript of solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 19

  • 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 19, Solution 2.Eq. 19.15: 2m m n m m nv x a x = =Given data 20.2 m/s 4 m/sm mv a= =: 0.2 m/sm m n m mv x x = = (1)2 2 2: 4 m/sm m n m ma x x = = (2)Divide Equ. (2) by Equ. (1):24 m/s20 rad/s0.2 m/sn = =Eq. (1): ( )0.2 m/s 20 rad/smx=0.01 mmx = 10 mmmx = !Frequency20 rad/s2 2nnf = = 3.18 Hznf = !

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 19, Solution 3.cycle 2 radsin , 6s cyclem n nx x t = = sin 12mx x t=12 cos12mx x t =&2144 sin12mx x t = &&12 4 ft/smx =40.1061 ft12mx= =1.273 in.mx = !( )2 2Max Acc. 144 0.1061 150.8 ft/s= = ! 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 19, Solution 4.Simple Harmonic Motion20 lb0.2222 in.90 lb/in.sWk = = =( )( )90 1241.699 rad/s 22032.2nkfm = = = = (a) Amplitude 0.222 in.s mx= = =0.222 in.mx = !41.699 rad/s6.63662f= =6.64 Hzf = !(b) ( )( )41.699 rad/s 0.2222 in. 9.2655 in./sm n mv x= = =9.27 in./smv = !( ) ( )22 241.699 rad/s 0.2222 in. 386.36 in./sm n ma x= = =232.197 ft/s=232.2 ft/sma = ! 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 19, Solution 5.Simple Harmonic Motion(a) ( )sinm nx x t = +( )( )29000 lb/ft70 lb 32.2 lb/snkm = =64.343 rad/s=20.90765 snn= =0.0977 sn = !110.240 Hznnf= =10.24 Hznf = !(b) At 0 0 00: 0, 10 ft/st x x v= = = =&( )( )0 0 sin 0 0m nx x = = + =( )( )0 0 cos 0m n n m nx v x x = = + =&Substituting 10 ft/s 64.343 rad/smx=or 0.1554 ft 1.865 in.mx = =1.865 in.mx = !( )( )220.15542 ft 64.343 rad/sm m na x = =2643.4 ft/s=2643 ft/sma = ! 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 19, Solution 6.In Simple Harmonic Motion(a) 2m m na x =Substituting ( )2 250 m/s 0.058 m n=or ( )22862.07 rad/sn =29.361 rad/sn =Now29.361 rad/s4.6729 Hz2 2nnf = = =Then( )( )1 cycle 1in Hz Hz1 min 60 s/min 60f = =So( ) 116060Hz 4.6729 Hz280.37 r/minHzf= =and 280 rpm(b) ( )( )0.058 m 29.361 rad/s 1.7029 m/sm m nv x = = =1.703 m/smv = 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 19, Solution 7.Simple Harmonic Motion(a) ( )sinm nt = +( )2 21.35 snn = =4.833 rad/s=( )cosm n nt = +&m m n =&m m m nv l l = =&Thus, mmnvl= (1)For a simple pendulumngl =Thus,( )22 29.81 m/s4.833 rad/sngl= =0.420 m=From (1)( )( )0.4 m/s0.42 m 4.833 rad/smmnvl= =0.197 rad=or 11.28711.29m = ! 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.(b) Nowta l= &&Hence, the maximum tangential acceleration occurs when && ismaximum.( )2sinm n nt = +&&2m m n =&&( ) 2t m nma l =or ( ) ( )( )( )20.42 m 0.197 rad 4.833 rad/st ma =21.9326 m/s=( ) 21.933 m/st ma = ! 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 19, Solution 8.Simple Harmonic Motion:260 lb/ft6.2161 rad/s50 lb32.2 ft/snkm = = =6.2161Hz2 2nnf = =(a) 2.4 in. 0.2 ftmx = = 0.2 ftmx = !0.989 Hznf = !( )( )220.2 ft 6.2161 rad/sm m na x = =27.728 ft/s=(b) f m sF ma mg= =or227.728 ft/s0.24032.2 ft/smsag = = = ! 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 19, Solution 9.6 lb/in. 72 lb/ft, 55 in./s, 4 lb.mk v W= = = =: 0kF ma kx mx x xm= = + =&& &&Thus: 2 72579.6 24.025 rad/s432.2km = = = = Eq. (19.15): m mv x =( )55 in./s 24.025 rad/smx=2.2845 in.mx = 2.28 in.mx = !( )( )2 2 22.2845 in. 579.6 rad /sm ma x = =21324.1 in./sma = 2110.3 ft/sma = ! 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 19, Solution 10.( )60cos 10 45sin 103x t t = + ( )60cos 10 45 sin10 cos cos10 sin3 3t t t = + 22.5sin10 21.02886cos10t t = + (1)Nowsin(10 ) sin10 cos cos10 sinm m mx t x t x t + = + (2)Comparing (1) and (2) gives22.5 cos , 21.02866 sinm mx x = =(a)2 20.2 s10nn = = = !(b) 2 2 2(22.5) (21.02866)mx = +30.8 mmmx = !(c)21.02866tan22.5 = 0.7516 rad 43.1 = = ! 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 19, Solution 11.At both 600 rpm and 1200 rpm, the maximum acceleration is just equal to g.(a) 600 rpm 62.832 rad/s = =Eq. (19.15):( )2262.832m m mga x x= =SI:( )329.812.4849 10 m62.832mx = = 2.48 mmmx =US:( )232.20.008156 ft62.832mx = = 0.0979 in.mx =(b) 1200 rpm 125.664 rad/s = =Eq. (19.15):( )22125.664m m mga x x= =SI:( )629.81621.2 10 m125.664mx = = 0.621 mmmx =US:( )232.20.002039 ft125.664mx = = 0.0245 in.mx = 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 19, Solution 12.Simple Harmonic Motion, thus( )sinm nx x t = +400 N/m16.903 rad/s1.4 kgnkm = = =Now (0) 0 sin(0 ) 0mx x = = + =Then(0) cos(0 0)m nx x = +&or 2.5 m/s ( )m(16.903 rad/s) 0.14790 mm mx x= =Then ( ) ( )0.14790 m sin 16.903 rad/sx t = (a)At ( )0.06 m: 0.06 m (0.14790 m)sin 16.903 rad/sx t = = or1 0.06 msin0.14790 m0.02471 s16.903 rad/s = =t0.0247 st = !(b) Now( )cosm n nx x t =&( )2sinm n nx x t = &&Then, for 0.024713 st =( )( ) ( )( )0.1479 m 16.903 rad/s cos 16.903 rad/s 0.024713 sx = &2.285 m/s= 2.29 m/sx =& !And( )( ) ( )( )20.1479 m 16.903 rad/s sin 16.903 rad/s 0.024713 sx = &&217.143 m/s= 217.14 m/sx =&& ! 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 19, Solution 13.Referring to the figure of Problem 19.12( )sinm nx x t = +( )cosm n nx x t = +&( )2sinm n nx x t = +&&Using the data from Problem 19.13: 0, 0.1479 m, 16.903 rad/sm nx = = =, , arex x x& &&And ( ) ( )0.14790 m sin 16.903 rad/sx t = So, at 0.9 s,t =( ) ( )( )0.1479 m sin 16.903 rad/s 0.9 sx = 0.0703 m= 70.3 mmx = !( )( ) ( )( )0.1479 m 16.903 rad/s cos 16.903 rad/s 0.9 sx = &2.19957 m/s= 2.20 m/sx =& !( )( ) ( )( )20.1479 m 16.903 rad/s sin 16.903 rad/s 0.9 sx = &&220.083 m/s= 220.1 m/sx =&& ! 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 19, Solution 14.(a)sin( )m nx x t = +( )( )29000 lb/ft70 lb32.2 lb/s = =nkm64.343 rad/s=20.9765 snn= =With the initial conditions: (0) 15 in. 1.25 ft, (0) 0x x= = =&( )1.25 ft sin 0mx = +(0) 0 cos(0 )2m nx x = = + =&1.25 ftmx =Then( ) (1.25 ft)sin 64.3432x t t = + (1.5) (1.25 ft)sin 64.343(1.5 s) 0.80137 ft2x = + = ( )(1.5) (1.25 ft)(64.343)cos 64.343 1.5 s 61.726 ft/s2x = + = &In 1.5 s, the block completes1.5 s15.361 cycles0.09765 s/cycle= 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.So, in one cycle, the block travels4(1.25 ft) 5 ft=Fifteen cycles take15(0.09765 s/cycle) 1.46477 s=Thus, the total distance traveled is15(5 ft) 1.25 ft (1.25 0.80137)ft 77.1 ft+ + =Total 77.1 ft= !(b)2(1.5) (1.25 ft)(64.343 rad/s) sin (64.343 rad/s)(1.5 s)2x = + &&23317.68 ft/s= 23320 ft/sx =&& ! 16. COSMOS: Complete Online Solutions Manual Organization SystemChapter 19, Solution 15.2210 lb0.31056 lb s /ft32.2 ft/sm = = With the given properties:250 lb/ft12.