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

1. 1. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 2.Eq. 19.15: 2m m n m m nv x a xω ω= =Given data 20.2 m/s 4 m/sm mv a= =: 0.2 m/sm m n m mv x xω ω= = (1)2 2 2: 4 m/sm m n m ma x xω ω= = (2)Divide Equ. (2) by Equ. (1):24 m/s20 rad/s0.2 m/snω = =Eq. (1): ( )0.2 m/s 20 rad/smx=0.01 mmx = 10 mmmx = !Frequency20 rad/s2 2nnfωπ π= = 3.18 Hznf = !
2. 2. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 3.cycle 2 radsin , 6s cyclem n nx x tπω ω= = ×sin 12mx x tπ=12 cos12mx x tπ π=&2144 sin12mx x tπ π= −&&12 4 ft/smxπ =40.1061 ft12mxπ= =1.273 in.mx = !( )2 2Max Acc. 144 0.1061 150.8 ft/sπ= = !
3. 3. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 4.Simple Harmonic Motion20 lb0.2222 in.90 lb/in.sWkδ = = =( )( )90 1241.699 rad/s 22032.2nkfmω π= = = =   (a) Amplitude 0.222 in.s mxδ= = =0.222 in.mx = !41.699 rad/s6.63662fπ= =6.64 Hzf = !(b) ( )( )41.699 rad/s 0.2222 in. 9.2655 in./sm n mv xω= = =9.27 in./smv = !( ) ( )22 241.699 rad/s 0.2222 in. 386.36 in./sm n ma xω= = =232.197 ft/s=232.2 ft/sma = !
4. 4. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 5.Simple Harmonic Motion(a) ( )sinm nx x tω φ= +( )( )29000 lb/ft70 lb 32.2 lb/snkmω = =64.343 rad/s=20.90765 snnπτω= =0.0977 snτ = !110.240 Hznnfτ= =10.24 Hznf = !(b) At 0 0 00: 0, 10 ft/st x x v= = = =&( )( )0 0 sin 0 0m nx x ω φ φ= = + ⇒ =( )( )0 0 cos 0m n n m nx v x xω ω φ ω= = + =&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 ω= =2643.4 ft/s=2643 ft/sma = !
5. 5. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 6.In Simple Harmonic Motion(a) 2m m na x ω=Substituting ( )2 250 m/s 0.058 m nω=or ( )22862.07 rad/snω =29.361 rad/snω =Now29.361 rad/s4.6729 Hz2 2nnfωπ π= = =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 ω= = =1.703 m/smv =
6. 6. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 7.Simple Harmonic Motion(a) ( )sinm ntθ θ ω φ= +( )2 21.35 snnπ πωτ= =4.833 rad/s=( )cosm n ntθ θ ω ω φ= +&m m nθ θ ω=&m m m nv l lθ θ ω= =&Thus, mmnvlθω= (1)For a simple pendulumnglω =Thus,( )22 29.81 m/s4.833 rad/snglω= =0.420 m=From (1)( )( )0.4 m/s0.42 m 4.833 rad/smmnvlθω= =0.197 rad=or 11.287°11.29mθ = °!
7. 7. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) Nowta lθ= &&Hence, the maximum tangential acceleration occurs when θ&& ismaximum.( )2sinm n ntθ θ ω ω φ= − +&&2m m nθ θ ω=&&( ) 2t m nma lθ ω=or ( ) ( )( )( )20.42 m 0.197 rad 4.833 rad/st ma =21.9326 m/s=( ) 21.933 m/st ma = !
8. 8. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 8.Simple Harmonic Motion:260 lb/ft6.2161 rad/s50 lb32.2 ft/snkmω = = =6.2161Hz2 2nnfωπ π= =(a) 2.4 in. 0.2 ftmx = = 0.2 ftmx = !0.989 Hznf = !( )( )220.2 ft 6.2161 rad/sm m na x ω= =27.728 ft/s=(b) f m sF ma mgµ= =or227.728 ft/s0.24032.2 ft/smsagµ = = = !
9. 9. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 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/s432.2kmω ω= = = =   Eq. (19.15): m mv x ω=( )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 ω= =21324.1 in./sma = 2110.3 ft/sma = !
10. 10. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 10.( )60cos 10 45sin 103x t tππ π = + −  ( )60cos 10 45 sin10 cos cos10 sin3 3t t tπ ππ π π = + −  22.5sin10 21.02886cos10t tπ π= + (1)Nowsin(10 ) sin10 cos cos10 sinm m mx t x t x tπ φ π φ π φ+ = + (2)Comparing (1) and (2) gives22.5 cos , 21.02866 sinm mx xφ φ= =(a)2 20.2 s10nnπ πτω π= = = !(b) 2 2 2(22.5) (21.02866)mx = +30.8 mmmx = !(c)21.02866tan22.5φ = 0.7516 rad 43.1φ = = °!
11. 11. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 11.At both 600 rpm and 1200 rpm, the maximum acceleration is just equal to g.(a) 600 rpm 62.832 rad/sω = =Eq. (19.15):( )2262.832m m mga x xω= =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/sω = =Eq. (19.15):( )22125.664m m mga x xω= =SI:( )629.81621.2 10 m125.664mx −= = × 0.621 mmmx =US:( )232.20.002039 ft125.664mx = = 0.0245 in.mx =
12. 12. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 12.Simple Harmonic Motion, thus( )sinm nx x tω φ= +400 N/m16.903 rad/s1.4 kgnkmω = = =Now (0) 0 sin(0 ) 0mx x φ φ= = + ⇒ =Then(0) cos(0 0)m nx x ω= +&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 tω ω=&( )2sinm n nx x tω ω= −&&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. 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 tω φ= +( )cosm n nx x tω ω φ= +&( )2sinm n nx x tω ω φ= − +&&Using the data from Problem 19.13: 0, 0.1479 m, 16.903 rad/sm nxφ ω= = =, , 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. 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 tω φ= +( )( )29000 lb/ft70 lb32.2 lb/sω = =nkm64.343 rad/s=20.9765 snnπτω= =With the initial conditions: (0) 15 in. 1.25 ft, (0) 0x x= = =&( )1.25 ft sin 0mx φ= +(0) 0 cos(0 )2m nx xπω φ φ= = + ⇒ =&1.25 ftmx =Then( ) (1.25 ft)sin 64.3432x t tπ = +  (1.5) (1.25 ft)sin 64.343(1.5 s) 0.80137 ft2xπ = + = −  ( )(1.5) (1.25 ft)(64.343)cos 64.343 1.5 s 61.726 ft/s2xπ = + = −  &In 1.5 s, the block completes1.5 s15.361 cycles0.09765 s/cycle=
15. 15. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.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)2xπ = − +  &&23317.68 ft/s= 23320 ft/sx =&& !
16. 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/ft12.6886 rad/s0.31056 lb s /ftnkmω = = =⋅From free fall of the collar( )0 2 2 1.5 ft 3 9.82853 ft/sv gh g g= = = =The free-fall time is thus:( )12 1.5 ft2 30.30523 sytg g g= = = =Now to simplify the analysis we measure the displacement from theposition of static displacement of the spring, under the weight of thecollar:Note that the static deflection is:10 lb0.2 ft50 lb/ftstWkδ = = =Then 0,mx kx+ = where x is measured positively up from theposition of static deflection. The solution is:( )sin ,m nx x tω φ= + with velocity( )cosm n nx x tω ω φ= +Now to determine and ,mx φ impose the conditions at impact andcount the time from there.Thus:At impact:0, 0.2 ft and 9.82853 ft/sstt x vδ= = = = − (down)or 0.2 ft sinmx φ=( )9.82853 ft/s 12.6886 rad/s cosmx φ− =Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.
17. 17. COSMOS: Complete Online Solutions Manual Organization SystemSolving for andmx φ0.800 ftmx = −0.25268 radφ = −So, from time of impact, the ‘time of flight’ is the time necessary for thecollar to come to rest on its downward motion. Thus, is the time such2tthat( )2 20 12.68862x t tπφ= ⇒ + =or 212.6886 0.252682tπ− =Hence, 2 0.14371 st =(a) Thus, the period of the motion is( ) ( )1 22 2 0.30523 s 0.14371 st tτ = + = +0.89788 s= 0.898 sτ =(b) After 0.4 seconds, the velocity is( ) ( )10.4 cos 0.4m n nx x tω ω φ⎡ ⎤= − +⎣ ⎦( )( ) ( )( )12.6886 rad/s 0.8 ft cos 12.6886 rad/s 0.4 0.30523 s⎡= − −⎣]0.25268 rad−5.91ft/s= − ( )0.4 5.91 ft/sv =Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.
18. 18. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 16.sin , cos ,m n m n n ngt tlθ θ ω θ θ ω ω ω= = =&9.812.8592 rad/s,1.2nω = =0.180:1.2m nt θ θ ω= = =&0.052462 radiansmθ∴ =At 1.5 s,t = ( )( )0.052462 sin 2.8592 1.5θ =(a) 0.047826θ = − radians 2.74= − ° !(b) ( )( ) ( )( )1.2 0.052462 2.8592 cos 2.8592 1.5v =74.0 mm/sv = !( )( ) ( )( )21.2 0.052462 2.8592 sin 2.8592 1.5a =2469 mm/sa = !
19. 19. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 17.(a)( )sinm nx x tω φ= +( )0 sin 0 0.75 ftmx x φ= + =( )0 0 cos 0 ,2m nx xπω φ φ= = + ⇒ =&0.75 ftmx∴ =When the collar just leaves the spring, its acceleration isg (downward) and 0.v =Now( )0.75 ft cos2n nx tπω ω = +  &( ) ( )0 0 0.75 ft cos ,2 2 2n n nx v t tπ π πω ω ω = = = + ⇒ + =  &And( ) 20.75 ft sin2n na g tπω ω = − = − +  or ( ) 20.75ft ng ω− = −232.2 ft/s6.5524 rad/s0.75 ftnω = =Then( )22210 lb, 6.5524 rad/s32.2 ft/sn nkk mmω ω= ⇒ = =13.333 lb/ft=13.33 lb/ftk = !(b) 6.5524 rad/snω =
20. 20. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.At 1.6 s:t =( ) ( )( )0.75 ft sin 6.5524 rad/s 1.6 s 0.36727 ft2xπ = + = −  0.367 ftx = − above equilibrium !( )( ) ( )( )0.75ft 6.5524 rad/s cos 6.5524 rad/s 1.6 s 4.2848ft/s2v xπ = = + =  &4.28 ft/sv = !( )( ) ( )( )2 20.75 ft 6.5524 rad/s sin 6.5524 rad/s 1.6 s 15.768 ft/s2a xπ = = − + =  &&215.77 ft/sa = !
21. 21. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 18.Determine the constant k of a single spring equivalent to the three springs shown.Springs 1 and 2:1 1 11 21 2, andP P Pk k kδ δ δ= + = +′Hence1 21 2k kkk k′ =+Where k′is the spring constant of a single spring equivalent of springs 1 and 2.Springs k′and 3 Deflection in each spring is the sameSo 1 2 1 2 3, and , ,P P P P k P k P kδ δ δ′= + = = =Now 3k k kδ δ δ′= +1 23 31 2k kk k k kk k′= + = ++or( )( )( ) ( )33.5 kN/m 2.1 kN/m2.8 kN/m 4.11 kN/m 4.11 10 N/m3.5 kN/m 2.1 kN/mk = + = = ×+(a)32 20.361 s4.11 10 N/m13.6 kgnkmπ πτ = = =×1 12.77 Hz0.3614 snnfτ= = =
22. 22. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) ( )sinm nx x tω φ= +And 44 mm 0.044 mmx = =( )( )2 2 2.77 Hz 17.384 rad/sn nfω π π= = =And( ) ( )0.044 m sin 17.4 rad/sx t φ = + ( )( ) ( )0.044 m 17.4 rad/s cos 17.4 rad/sx t φ = + &Then ( )( )max 0.044 m 17.4 rad/s 0.766 m/sv = =( )( ) ( )20.044 m 17.4 rad/s sin 17.4 rad/sx t φ = − + &&Then ( )( )2 2max 0.044 m 17.4 rad/s 13.3 m/sa = =max 0.765 m/sv =2max 13.31 m/sa =
23. 23. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 19.(a) First, calculate the spring constant( ) ( ) ( ) ( )24 kN/m 12 kN/m 12 kN/m 48 kN/mP δ δ δ δ= + + =48 kN/mk∴ =Then340 10 N/m30.984 rad/s50 kgnkmω×= = =20.20279 snnπτω= =14.9312 Hznnfτ= =0.203 snτ = !4.93 Hznf = !(b) Now( )sinm nx x tω φ= +And, since 0 0.060 mx =( ) ( )0.060 m sin 30.984 rad/sx t φ = + ( )( ) ( )0.060 m 30.984 rad/s cos 30.984 rad/sx t φ = + &( )( ) ( )20.060 m 30.984 rad/s sin 30.984 rad/sx t φ = − + &&Hence max 1.859 m/sv = !2max 57.6 m/sa = !
24. 24. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 20.Equivalent spring constant2 2 4k k k k′ = + =(Deflection of each spring is the same.)( )1226.8 s10 lb32.2 ft/sπτ = =nk2210 6.832.2k π ⇒ =   0.2651 lb/ftk =( )22246 lb32.2 ft/sπτ =nk( )( )224 0.2651 lb/ft 32.2 ft/s6 lbπ=2.633 s=2.63 snτ =
25. 25. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 21.Equivalent SpringsSeries: 1 21 2sk kkk k=+Parallel: 1 2pk k k= +2 2 2 2;s ps ps pk km mπ π π πτ τω ω= = = =( )( )( )2 221 21 21 2 1 21 252ττ  ++ = = = =       +psp sk k kk kk kk k kk k( )( ) 2 21 2 1 1 2 26.25 2k k k k k k= + +( ) ( )2 2 22 2 214.25 4.25 42k k kk−=m122.125 3.516kk= m1214 or4kk=
26. 26. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 22.For a static load P the total elongation of the spring is:1 2 3P P Pk k kδ = + +1 1 18 kN/m 12 kN/m 16 kN/mP = + +  6 4 3 1348 48δ+ + = =  P P483.6923 kN/m13Pkδ= = =33.6923 10 N/m( ) 12.153 rad/s25 kgkamω×= = =2 20.517 s12.153π πτω= = = 0.517 sτ = !12.1531.9342 Hz2 2fωπ π= = = 1.934 Hzf = !(b) For 30 mm = 0.03 mmx =( )( )0.03 m 12.153 rad/sm mv x ω= = 0.365 m/smv = !( )( )220.03 m 12.153 rad/sm ma x ω= = 24.43 m/sma = !
27. 27. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 23.For equilibriumEqual stretch12 N 16 N,A C Bk kδ δ δ= = =12 N ,A CT T kx= = + 16 NBT kx= +( ) ( ) ( )4.08 9.81 2 12 16mx kx kx= × − + − +&&3 0mx kx+ =&&( )30.0125 coskx tm=(a) Then ( )( )16 0.0125 1 0BT k= + − =1.280 kN/mk = !(b)( )( )2 2 23 12803941.76 rad /s40 9.81nkmω = = =30.688 rad/s 4.88 Hzn fω = ⇒ = !(c) ( )312 1280 0.0125 cosAkT tm= +3Minimum when cos 1ktm= = −( ) ( )min12 1280 0.0125 4AT = − = −Max Comp: 4 N !
28. 28. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 24.Springs in parallel, eq 1 22k k k= +12 16,000 2 2100.2nkmπ πω πτ+= = = =After removal12 280.25nkmπω π= = =Elimination: 2 12 16,000100kmπ+=2 1264kmπ =(a) 2 2 16,000100 64 ,mπ π∴ = × 45.0 kgm = !(b) 2 11264 , 14,222 N m45.0316kkπ = =1 14.22 kN/mk = !
29. 29. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 25.The equivalent spring constant for springs in series is:1 21 2ek kkk k=+1 2For andk k( )1 21 22 2π πτ = =+eA Ak k km m k k1For alonek12πτ′ =Akm(a)22 1 2k k kττ = + ′ 0.2 s0.166670.12 sττ= =′And with 2 3.5 kN/mk =( )( )213.5 kN/m 1.6667 3.5 kN/mk= +or 1 6.22 kN/mk =(b)12πτ′ =AkmAAWmg=( )( )21 122Akmτπ=And with 31 6.22 kN/m 6.22 10 N/mk = = ×( ) ( )( )2 3220.12 s 6.22 10 N/m2.2688 Ns /m 2.2688 kg2Amπ×= = =2.27 kgAm =
30. 30. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 26.The equivalent spring constant, accounting for the springs in series, iseq2kk k= +( )( ) ( )( )eq222 20.4 s3 /2100 lb 32.2 ft/s100 lb 32.2 ft/sk kπ πτ = = =It follows that 510.849 lb/ftk =511 lb/ftk = !After the changeseq20.4 s120/32.2kπτ = =Thus eq 919.528 lb/ftk =SinceeqAAkkk kk k= ++or( ) ( )( )eqeq510.849 919.528 510.8492 2 510.849 919.528Ak k kkk k− −= =− −2043.4 lb/ft=2040 lb/ftAk = !
31. 31. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 27.Initially21.6 sAkmπτ = =After the 14 lb weight is added to A,22.1 sA Bkm mπτ′ = =+(a) A BAm mmττ′ +=22.11.6A BAm mm+ =  ( )( )1.7227 A A Bm m m= + Note 214 lb32.2 ft/sBm =214 lb1.3838 1.383832.2 ft/sA Bm m = =   ( )2214 lb32.2 ft/s 1.3838 19.373 lb32.2 ft/sA AW m = = =  19.37 lbAW = !(b)2Akmπτ =( )( )( )222 Amk πτ=( )( )22219.373 lb32.2 ft/s21.6 sk π   =9.278 lb/ftk =9.28 lb/ftk = !
32. 32. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 28., 2 2 2s sd dx x F kx k xh h= = =( ) ( )eff: 2A AM M Fd mgx mx hΣ = Σ − = − &&2dk x d mg x mxhh − = −  &&2220kd gx xhmh + − =  &&222 222 2,kd g k d gh m h hmhω ω   = − = −     (1)Data: 0.3 m; 0.7 m; 20 kgd h m= = =(a) For 1 s:τ =2;πτω= 2 221 s = ; 4πω πω=Eq. (1):22 2 0.3 9.81420 0.7 0.7kπ = −  2912.4 N/mk = 2.91 kN/mk =(b) For Infinite:τ =2;πτω= 0ω =Eq. (1):22 2 0.3 9.81020 0.7 0.7kω = = −  763.0 N/mk = 763 N/mk =
33. 33. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 29.(a) The equation of motion is:( )( )( ) ( )( ): 0.75 0.75 2 1.2 m 1.2AM kθ θΣ − = &&1.125 1.44k mθ θ− = &&1.44 1.125 0m kθ θ+ =&&( )31.125 1.35 1001.44mθ θ×+ =&&1054.690mθ θ+ =&&Then 2 1054.69 2, and 4n nnsmπω τω= = =22 1054.69 1054.69 2or4 4m mπ π = =   427.45 kgm =427 kgm =(b)( )sinm ntθ θ ω φ= +At 0t =0.060.03636 rad1.65mmθ = =
35. 35. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 30.(a)( )( )( )( )6 2 2 2 3 43 348 30 10 lb/in. 144 in. /ft 2 10 ft4815 ftP EIkLδ−× ×= = =122880 lb/ft= 51.229 10 lb/ftk = ×(b)( )2122880 lb/ft51.360 rad/s1500 lb 32.2 ft/snkmω = = =or1 51.368.17 Hz2 2 2nnkfmωπ π π= = = =
36. 36. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 31.(a) , ,ePL AEP k PAE Lδ δ δ = = =   ( )2630.32 in.29 10 psi4129.57 10 lb/in.18 in.eAEkLπ    ×   = = = ×3129.6 10 lb/in.ek = × !(b)( )( )( )32129.57 10 lb/in.16 lb32.2 ft/s81.272 Hz2 2enkmfπ π×= = =81.3 Hznf = !
37. 37. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 32.stWkδ=Wmg=2 stWn Wstgk gmδωδ= = =12 2nnstgFωπ π δ= = !
38. 38. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 33.(a) ( )( )20.10 kg 9.81 m/s 0.981 Nmg − =122000.981 N44F mg x x = = ⇒ =   0.06015 m=0 60.1 mmx = !(b) At 0,x ( )1 12 20042 0.060152xdFxdx− − = =  8.1549 N/m=08.1549 N/mexdFkdx = =  8.1549 N/m0.10 kg1.4372 Hz2 2π π= = =enkmF1.437 HznF = !
39. 39. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 34.Using the Binomial Theorem we write1222 211 sin sin21 sin sin2mmθφθφ−  = −       −   2 211 sin sin2 2mθφ= + + ⋅ ⋅ ⋅ ⋅Neglecting terms of order higher than 2 and setting ( )2 1sin 1 cos2 ,2φ φ= − we have( )2 201 14 1 sin 1 cos22 2 2mnldgπ θτ φ φ  = + −    ∫2 2 201 14 1 sin sin cos24 2 4 2m mldgπ θ θφ φ = + −  ∫22 201 14 sin sin sin 24 2 8 2m mlgπθ θφ φ φ  = + −    214 sin 02 4 2 2mlgπ θ π  = + +    212 1 sin4 2mnlgθτ π = +  !
40. 40. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 35.For small oscillations: 0 2lgτ π=We want ( )01.01 1.01 2lgτ τ π= =Using the formula of Prob. 19.34, we write20 011 sin 1.014 2mθτ τ τ = + =  ( )2sin 4 1.01 1 0.04; sin 0.2; 11.542 2 2m m mθ θ θ= − = = = °23.1mθ = ° !
41. 41. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 36.Eq. (19.20)22k lgτ ππ =    For 0.8 m:l =20.8 m2 2 1.7943 s9.81 m/slgπ π= =Thus: ( )21.7943 skτπ=Using Table 19.1:(a) For small oscillations:20; 1mkθπ= =1.7943 sτ = !(b) For230 : 1.017mkθπ= ° =( )( )1.017 1.7943 sτ = 1.825 sτ = !(c) For290 : 1.180mkθπ= ° =( )( )1.180 1.7943 sτ = 2.12 sτ = !
42. 42. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 37.From Equation 19.20:22nk lgτ ππ  =      For 60 , 1.686m kθ = ° =( ) ( )( )( )3 s 2 1.686 232.2l=30.44484 0.1978832.2 6.744 32.2l l= = ⇒ =Thus 6.3718 ft 76.462 in.l = =76.5 in.l = !
43. 43. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 38.( ) ( )0.6 , 0.4A st B stA BF k l F k lθ δ θ δ   = + = +   (a) ( )eff:C CM MΣ = Σ( ) ( )0.6 0.6 0.1 0.4 0.4 0.1st st tA Blk l lmg lk l I lmaθ δ θ δ α   − + + − + = +    (1)Equilibrium position: 0θ =( ) ( )0.6 0.1 0.4 0st stA Blk lmg lkδ δ− + − = (2)Substitute (2) into (1)( ) ( )2 20.1 0.6 0.4 0tI lma l k l kα θ θ+ + + =or 20.1 0.52 0tI lma l kα θ+ + =But2, 0.1 , ,12tmlI a lα α θ= = = &&So ( )22 20.1 0.52 012mll m l kθ θ + + =  &&or 20.09333 0.52 0l kθ θ+ =&&25.5714 0l kθ θ+ =&&( )850 N/m5.5714 09 kgθ θ + =  &&2526.19 s 0θ θ−+ =&&
44. 44. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Thus 2526.19 s 22.939 rad/s, 3.6508 Hz2nn nfωωπ−= = = =3.65 Hznf =(b) ( ) ( )sin , cosm n m n nt tθ θ ω φ θ θ ω ω φ= + = +&m m nθ θ ω=&( ) ( )0.6A mmx l θ= &&( )( )( )0.0011 m/s 0.6 0.6 m 22.939 rad/smθ=0.0001332 rad 0.0076mθ = = °0.0076mθ = °
45. 45. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 39.( )eff:B BM MΣ = ΣDiskcos2 2 2ABL L Lmg kxr I m Iθ α α α  − = + +    (1)where where from statics2st stLx r mg k rθ δ δ= + =into (1) and assuming small angles ( )cos 1θ ≈2Lmg θ 2stkr krθ δ− −2Disk2ABLI m I α   = + +    so22Disk 04ABmLI I krθ θ + + + =   &&so2 222 22 2Disk Disk1 14 12 4 2nABkr krmL mLI I mL M rω = =+ + + +(2)Disk: 0.9 m7.5 kg0.25 m into (2)6 kg5 kN/mData LmrMk= = = = = 
46. 46. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )( )( )( ) ( )( ) ( )( )222 2 25000 N/m 0.25 m1 1 17.5 kg 0.9 m 7.5 kg 0.9 m 6 kg 0.25 m12 4 2nω =+ +2 2141.243 rad/snω =11.8846 rad/snω =(a) so2 211.8846nπ πτω= = 0.529 sτ = !(b) 0.020 mmx =( )0.02 11.8846 0.2377 m/sm m nv x ω= = =max 238 mm/sv = !
47. 47. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 40.At equilibrium, ( )spring tension 20 lb sin15= °( )020 lb sin15stretch50 lb/ftx°= =20sin15PMΣ = °( ) 045012x−412x  +    24121 20 4 20 416.667 0.3105592 32.2 12 32.2 12xx x x       = + − =               !!!! !!0.310559 16.667 0, 7.326 rad/snx x ω+ = =!!(a) 0.858 sτ = !(b) sin , cosn n nx A t x A tω ω ω= =!( )0.5 7.326 3.66 in./snAω = = !
48. 48. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 41.Kinematics2cos sinNon-linear terms: sin2Ax Aa l lmlF N maθ θ θ θθ θ= ++Σ = − & &&&&2 2sin cos sin sin2 3 2A Akl ml lM Nl maθθ θ θ θΣ = − + = −&&or, letting cos 1,θ = sin cos , sin 0, since 1θ θ θ θ θ= = <<2 22 30,3 2 2nml kl kmθ θ ω+ = =&&For 1.5 ft,l = 26 lb, 23 lb/in. 276 lb/ft32.2 ft/sm k= = =(a)1 37.50 Hz2 2kfmπ= = !(b) 2 2 maxmax maxn nxLθ ω θ ω= − = −&&( )( )2max0.83 2763 1298.747 rad/s622 1.532.2k xm L −   = − = =   2max 98.7 rad/sθ =&& !
49. 49. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 42.(a)( )eff:A AM MΣ = Σ( ) ( ) ( )1.2 0.825 2 0.75 0.825 1.2R S R t tG Cmg m g F I m a m aα+ − = + + (1)Where ( )0.75S stF k θ δ= +At equilibrium, 0θ =And( )0: 1.2 0.825 2 0.75 0A R stM mg m g kδΣ = + − = (2)Substitute (2) into (1)( ) ( ) ( ) ( )0.825 1.2 2 0.75 0.75R t tG CI m a m a kα θ+ + = −( ) ( ), 0.825 , 1.2t tG Ca aα θ α α= = =&&( )( )2221.2 kg 1.65 m0.2725 kg m12 12mlI = = = −( ) ( ) ( ) ( ) ( )2 2 20.2725 0.825 1.2 1.2 2 0.75 450 0m θ θ + + + =  &&[ ]1.08925 1.44 506.25 0m θ θ+ + =&&506.2501.08925 1.44mθ θ+ =+&&
50. 50. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Now ( )( )( )2 22222506.25 22 109.661.08925 1.44 0.6n nnfmπ πω πτ = = = = = +  ( )506.25 109.66 1.08925 1.44m= +2.4495 kgm =2.45 kgm =(b) ( ) ( )max max0.061.2 , 0.03636 rad1.65 1.65θ θ= = = =& BC myv( )( )1220.03636 rad 109.66 s 0.38075 rad/sm m nθ θ ω −= = =&( ) ( )max1.2 0.38075 rad/s 0.45691 m/sCv = =( )max457 mm/sCv =
51. 51. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 43.2mx kx P= −&& Mx P F= −&&212Mr Frθ =&&Eliminate , andP F θ32 02Mm x kx + + =  &&( )2 2 5000 N/m2769.233 4 kg 9 kg2nkMmω = = =++27.735 rad/snω =(a)20.22654 snnπτω= =0.227 snτ = !(b) ( )max 12 mm 333 mm/snv ω= = !
52. 52. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 44.From Problem 19.43( )2 2 3500 N/m2538.46; 23.2048 rad/s3 4 kg 9 kg2n nkMmω ω= = = =++( )( ) ( )( )2 26 kg 9.81 m/s 4 kg 9.81 m/s 98.1 NN W Mg mg= = + = + =2 21 12 2xFr Mr Mrrθ = =   &&&&( )1226 kgmax , ; Amplitude98.1 Nω= = =&& &&m m nFx x A AN( )( )216 kg20.5 23.2048 rad/s98.1 Ns Aµ = =or 0.03036 mA =30.4 mmA = !
53. 53. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 45.222422 30.021440 m162r rrbrrπππ  −     = =−( )216 0.22546 kg2m tr tπρ ρ = − =  ( )24 4 24 8 44 26 4 9 2 3ot r r r rI r t rρ π πρπ π  = − − + −     ( )4 136 90.0093423 4tr tπρ ρ = − =  (a) 20,o notmgbI mgbIρθ θ ω+ = = =&& ( )( )( )0.22546 9.81 0.021440tρ ( )0.00934225.07602, 2.2530 rad/sn nω ω= =22.79 snπτω= = !(b)222 2 29.81 m/s, 1.9326 m5.07602 rad /snng gllωω= = = =1.933 ml = !
54. 54. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 46.Kinematics: / ,G O G Oa a a GO d= + =0.3 m, 4 2 /3 0.180063 mr d r π= = =( )2PM mgd I m r dθ θ θ Σ = − = + −  && &&Where2 22322 9mr mrIπ= −2nmω =gdm 2322r m−229rmπ+ ( )265.515r d= −  20.776264 snπτω= = 0.776 sτ = !
55. 55. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 47.212AI mr=(a)2402 3mr rmgθ θπ + =  &&or803grθ θπ+ =&&83ngrωπ=2 26.8283nnrggrπ πτωπ= = = !(b)22 2 43rb rπ = +   2 22 2 21 4 42 3 3B Gr rI I mb mr m mr mmπ   = + = − + +      232mr=2302mr mgbθ θ+ =&&or 223gbrθ =&&223ngbrω =22 27.38162 193nnrggrπ πτωπ= = =+!
56. 56. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 48.( )( )222 1.61 1.62 0.63 3.6 3.6 12Abm mI b b      = + +           21125mb=( )2 0.33.6 6b b byb= =130.66 30bAG b b= − =22 211 130.6425 30 36BbI mb m b m b  = − + +     22325mb=24.046bGB =(a)211 13 20, 6.3325 30nnmb b bmggπθ θ τω+ = = =&& !(b)223 24.04 20, 6.6725 6nnmb bmg bgπθ θ τω + = = =   && !
57. 57. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 49.( )2 2112I m b c= +12ta r cα θ= = &&( )eff1 1: sin2 2A Α tM M mg c I ma cθ α   Σ = Σ − = +      ( )22 21 1sin2 12 2mmgc b c m cθ θ θ − = + +   && &&2 2sin 023 12g cc bθ θ  + =  +  &&22 2 2624 13 12cg c g bbc b cbω   = =  + +     (1)
58. 58. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(a) Since2,πτω= minimum τ occurs when 2ω is maximum. We considercb   as the variable.( )22224 1 864 1c c cd b b bgc bd cbbω        + −               = =      +      2 24 1 8 0c cb b   + − =      214cb =  0.5cb= !(b) For a simple pendulum of length c we have 2 8.cω = We equate the value of 2ω to that of Eq. (1).264 1cg gbb ccb    =  +     2 2 214 1;2     = + =          c c cbb b b0.707cb= !
59. 59. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 50.Consider general pendulum of centroidal radius of gyration k.( ) ( ) 20 eff: sinM Μ mgr mr r mkθ θ θ0Σ = Σ = +&& &&2 2sin 0grr kθ θ + = + &&For small oscillations, sin ,θ θ≈ we have2 222 2 2 220 ; ; 2gr gr r kgrr k r kπθ θ ω τ πω  ++ = = = = + + &&Connecting rod suspended at A: 1.03 sτ =160 mm 0.16 mar r= = =Thus( )( )( )2 20.161.03 29.81 0.16kπ+=( ) ( )( )22 2 1.030.16 9.81 0.16 0.042182kπ + = =  20.04218 0.0256 0.01658 ; 0.12876 mk k= − = = 128.8 mmk = !
60. 60. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 51.221 1 10m 0.334722 m2 2 12I mr = = =  α θ= &&(a) The disk is free to rotate and is in curvilinear translation.Thus 0Iα =Then ( )effB BM MΣ = Σsin , sintmgl lmaθ θ θ− = ≈20ml mglθ θ+ =&&And22 2261232.2 ft/s14.861 sftnglω −= = =13.8551 snω −=21.6299 snnπτω= =1.630 snτ =
61. 61. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) When the disk is riveted at A, it rotates at an angular acceleration α( )effB BM MΣ = Σ21sin , , sin2tmgl I lma I mrθ α θ θ− = + = ≈2 2102mr ml mglθ θ + + =  &&Then( )22 22 2222632.2 ft/s ft1213.838 s10ft2 2612ft2 12ω −   = = =    +           +       nglrl13.7199 snω −=21.6891 snnπτω= =1.689 snτ =
62. 62. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 52.( )0 0 eff: sin tM M W r I ma rθ αΣ = Σ − = +2 2sinmgr mk mrθ θ θ− = +&& &&2 2sin 0grr kθ θ+ =+&& (1)For a simple pendulum of length OA l=0glθ θ+ =&& (2)Comparing Equations (1) and (2)2 2r klr+=2kGA l rr= − = (Q.E.D) !
63. 63. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 53.See Solution to Problem 19.52 for derivation of2 2sin 0grr kθ θ+ =+&&For small oscillations sinθ θ≈ and2 2 22 22nnr k krgF rgπ πτ πω+= = = +For smallest nτ we must have2krr+ a minimum2221 0kd rr kdr r +    = − =2 2r k=r k= (Q.E.D) !
64. 64. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 54.Same derivation as in Problem 19.52 with r replaced by .RThus 20gRR kθ θ+ =+&&Length of the equivalent simple pendulum is2 2 2R k kL RR R+= = +( )22= − + =kL l r lkrThus the length of the equivalent simple pendulum is the same as inProblem 19.52. It follows that the period is the same and that the newcenter of oscillation is at O (Q.E.D)
65. 65. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 55.2 22 254 4l ld l= + =Ignore static terms: 22ClM mg klθ θΣ = − −22 2 2 5212 4 4mlmgl kl ml mlθ θθ θ θ − − = + +    && &&&&232253mgklmω+=1 6 92 5 10nk gfm lπ= + !
66. 66. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 56.( )eff:4 4 2D DkL r mrL mLrM M mgrθθ θ θΣ = Σ − − = +&& &&304 2mLr kLrmgrθ θ + + =  &&2 4 23 3ng kL mω = +1 2 4Hz2 3 3nk gfm Lπ= + !
67. 67. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 57.2221 16 lb 1 12.5 in.12 12 12 in./ft32.2 ft/sI ml   = = ⋅     20.04493 lb ft s= ⋅ ⋅α θ= !!20.1666712ta α θ = =  !!sinθ θ≈( )effC CM MΣ = Σ( ) ( )2 20.6875 0.16667 0.16667k mg I mθ θ θ θ− + = +!! !!( )( )160.04493 0.02778 0.47266 0.16667 16 032.2kθ θ   + + − =      !![ ]0.05693 0.47266 2.6667 0kθ θ+ − =!! (1)(a) For 50 lb/ft,k = 368.28 0θ θ+ =!!368.283.007 Hz2 2nnfωπ π= = =3.01Hznf = !
68. 68. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) For ,nτ → ∞ 0nω → oscillations will not occurFrom Equation (1)2 0.47266 2.666700.05873nkω−= =5.642 lb/ftk =5.64 lb/ftk = !
69. 69. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 58.2 333 2 3a ab = =   4 423 396 96Oa aI t mbρ = + +   23 5,2 2 12Ot am a I maρ  = =   With220 lb s 16, ft, 14 lb/ft32.2 ft 12m a k⋅ = = =  So with sinθ θ≈ for small ,θ ( )effO ΟΜ ΜΣ = Σ yields2252 012 3ma mgakaθ θ + + =  &&22122 05 3mgakamaθ θ + + =  &&2 2212 12 24255 3 5 3nmga g kkaa mmaω θ     = + = +         2 212 32.2 24 14141.655 s16 2055 312 32.2nω −    = + =          11.90 rad/snω =1.894 Hz2nfωπ= = !
70. 70. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 59.Ignore static terms( ) ( ) ( ) ( )( ) ( )( )2 2 2 21 150 0.6 50 0.9 1.2 0.3 1.2 0.6 1.8 0.93 3BM gθ θ θ θ Σ = − − + = +  &&54.96840.63 54.9684 0, 9.3408 rad/s0.63nθ θ ω+ = = =&&015cos cos900t tθ θ ω ω = =   At 2 200.7 s, cos 0.7 1.407 rad/sn nt θ ω θ ω= = − = −&&21.407 rad/sα =
71. 71. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 60.Determine location of the centroid G.Let mass per unit lengthρ =Then total mass ( ) ( )2 2m r r rρ π ρ π= + = +About C 220 2rmgc r g r gπ ρ ρπ = + =  2ryπ=( ) 22 2r c rρ π ρ+ =( )22rcπ=+( )0 0 eff tM M a c cα θ α θΣ = Σ = = =&& &&sin sinnmgc I mcaθ α θ θ− = + ≈( )200 0I mc mgc I mgcθ θ θ θ+ + = + =&& &&
72. 72. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.But 20I mc I+ =( ) ( )( )220 0 0 s-circ linesemi circ line212rI I I m r m= + = +( )s-circ line 22mm r m rrρπ ρ ρπ= = =+( )2 2202 23 2 3r r mrI r rρ π ππ ⋅  = ⋅ + = +   +   ( ) ( )22 202 3 2mr rmgπ θ θπ π + + = + + &&( )( )( )( )222 23 32 9.81 m/s20.32 mngrωπ π= =+ +2 216.0999 s 4.0125 rad/sn nω ω−= =( )sinm n Bt y rθ θ ω φ θ= + =( ) ( ) ( )sin sinB m n B nmy r t y tθ ω φ ω φ= + = +At 0t = 0.03 m 0B By y= =&( ) ( ) ( )0 0 cos 02B B mt y yπφ φ= = = + =&( ) ( )0.03 sin 0 0.03 m2B B Bm my y yπ = = + =  10.03sin 4.0125 s2B n ny tπω ω − = + =  ( )( ) ( )0.03 cos 0.03 sin2B n n n ny t tπω ω ω ω = + = −  &( )( )2 20.03 sin 0.03 cos2B n n n ny t tπω ω ω ω = − + =  &&At 10 s:t = ( ) ( ) ( )( )2 20.03 4.0125 cos 4.0125 10 0.36437 m/sB Bta y= = = −&&( ) ( ) ( )( )0.03 4.0125 sin 4.0125 10 0.07902 m/sB Bv y= = =&( ) ( )( )112 22 2222 2 20.079020.36437 0.365 m/s0.32BB B tva ar          = + = − + =             
73. 73. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 61.( )221 10.2502 2 32mI mr m α θ= = = = &&0.650ta lα θ= = &&(a) The disk is free to rotate and is in curvilinear translation.Thus 0I α =( )effB BM MΣ = Σsin tmgl lmaθ− =2 20 ngml mgllθ θ ω+ = =&&From 19.17, the solution to this Equation is( )sinm ntθ θ ω φ= +At 0,t = 2 rad, 0180 90π πθ θ= ⋅ = =&( )cosm n ntθ θ ω ω φ= +&0 0 cos2m ntπθ ω φ φ= = =sin 090 2mπ πθ = +  rad90mπθ =
74. 74. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Thus sin90 2ntπ πθ ω = +  ( ) maxmaxA m nv l lθ θ ω= =&With 26 in. 2.1667 ft, rad,90m ngllπθ ω= = = =( ) ( )( )2max32.2 ft/s2.1667 ft rad90 2.1667 ftAvπ =   0.29156 ft/s=3.4987 in./s=( )max3.50 in./sAv =(b) For disk riveted at A ( I α included)( )effB BM MΣ = Σ sin tmgl I lmaθ α− = +2 2102mr ml mglθ θ + + =  &2222nrgllω =+sin90 2ntπ πθ ω = +  (See (a))( ) maxmaxA m nv l lθ θ ω= =&( )( )( )22 232.2 ft/s 2.1667 ft2.1667 ft90 1 10 in. 26 in.2 12 in./ft 12 in./ftradπ =       +      0.28135 ft/s=3.376 in./s=( )max3.38 in./sAv =
75. 75. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 62.( )18 0.38 in. = ft, 2 ,12 32.2r m rπ   = =      ( )20.332.2m r =   21 22 0.045767 ftrmbm m= =+2 2 21 210.58813 lb s ft3= + = ⋅ ⋅AI m r m r( )1 2 0AI m m gbθ θ+ + =&&( )1 22nAm m gbIω+=1.9105 rad/s,nω = 0 cos ntθ θ ω=22 rad,180πθ° = ° = 0 sinn ntω θ ω ω= −20 cosn ntα θ ω ω= −At 5 s,t = 0.00849 rad/sω = !0.1264 rad/sα = !
76. 76. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 63.( )effM MΣ = ΣK Iθ θ− = &&0KIθ θ+ =&&2,nKIω = 2IKτ π= (1)For 50-mm-diameter sphere, 0.025 mr =2 3 2 52 2 4 85 5 3 15SI mr r r rπ ρ π ρ = = =  ( ) ( )5 3 6 280.025 7850 kg/m 128.45 10 kg m15π −= = × ⋅Solve Eq. (1) For :I22;4KI τπ=2R RS SIIττ =   (2)Data: 4.1 s, 6.2 s, 0.45 kgR S Rmτ τ= = =Eq. (2):24.1 s0.43736.2 sRSII = =  ( )6 20.4373 0.4373 128.45 10 kg m−= = × ⋅R SI I6 256.17 10 kg m−= × ⋅I6 22 56.17 10 kg m0.45 kgRRRIkm−× ⋅= =2 6 2124.82 10 mk −= ×311.173 10 mk −= × 11.17 mmRk = !
77. 77. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 64.( )effG GM MΣ = ΣK Iθ θ− = &&0KIθ θ+ =&&2nKIω = (1)2IKτ π= (2)Data: 1.95 N m/radK = ⋅3 kgm =( )( )22 3 21 13 kg 0.5 m 62.5 10 kg m12 12I ml −= = = × ⋅(a) Eq. (2)3 262.5 10 kg m21.95 N m/radτ π−× ⋅=⋅1.125 sτ = !(b) Max velocityEq. (1) 23 21.95 N m/rad31.262.5 10 kg mω −⋅= = =× ⋅nKI5.586 rad/snω =Simple harmonic motionm m nω θ ω= 180 radmθ π= °=( )( )rad 5.586 rad/s 17.548 rad/smω π= =( ) ( ) ( )( )0.25 m 17.548 rad/s 4.387 m/sA mmv AG ω= = =( ) 4.39 m/sA mv = !
78. 78. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 65.Equivalent torsional spring constant( )2 1 1 1, ,eT K T K T Kθ θ θ θ= = − =( )2 1 2 1K K Kθ θ= +211 2KK Kθ θ=+1eT K Kθ θ= =211 2eKK KK Kθ θ=+1 21 2eK KKK K=+Newton’s Law( )effc cM MΣ = ΣeK Jθ θ− = &&212J mr=2102emr Kθ θ+ =&&( )( )224 402 2 32.2 1223 1.5223 1.5nn eKmrπ πτ πω      = = =  + 5.2197 s=5.22 snτ = !
79. 79. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 66.2 214ABI mr mb= +[ ]2 0ABI k mgbθ θ+ + =&&22224nk mgbrm bω+= +   2120 lb 8, ft1232.2 ft/sm r   = =     6.4ft,12b =   150 lb ft/radk = ⋅(a)( ) ( )222120 6.42 150 32.232.2 12246.938120 6.41232.2 4 12nω   +       = =         +          15.7139 rad/snω =2 20.3998 s15.7139nπ πτω= = =0.400 sτ = !(b) 0 0 max 0sin ; cos ;n n n nt tθ θ ω θ θ ω ω θ θ ω= = =& &So max max 0 nv b bθ θ ω= =&( )max6.4 2ft 15.7139 rad/s 0.2925 ft/s12 180vπ  = =    max 3.51 in./sv = !
80. 80. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 67.( )effG GM MΣ = Σ 0K J J Kθ θ θ θ− = + =&& &&Empty platform centroidal of platformPJ J=( )( )222 227 N m/rad 2.2 s2 24 4nn PnPKJKJπ π ττω π π⋅= = = =23.31 N m sPJ = ⋅ ⋅Platform with object A centroidal ofAJ J A=( )( )222 24nn P AnP AKJ JKJ Jτπ πτω π′′ = = + =′+( )( )22227 N m/rad 3.8 s3.31 N m s4AJπ⋅= − ⋅ ⋅29.88 3.31 6.57 N m sAJ = − = ⋅ ⋅26.57 N m sAJ = ⋅ ⋅ !
81. 81. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 68.Geometry3blθφ =3 3mg bFlθ=Then3G GbM I Fθ θ Σ = = −   &&or2 2012 3mb mgbLθ θ+ =&&(a) 2 4 2,nng ll gπω τ πω= = = !
82. 82. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b)3mx F= −&&mgxl= −2nglω =2lgτ π= !
83. 83. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 69.( )2 2 212T m b c θ= + &2 212V kc θ=( )222 2nkcm b cω =+720 N/m, 1.5 kgk m= =17.5271 rad/snω =00.015 m0.033333 rad0.45 mθ = =0max 0.351 m/sD nv cω θ= =max 0.351 m/sDv = !
84. 84. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 70.From Problem 19.6917.5271 rad/snω =( )00.25 m/s0.5555 rad/s0.45 mθ = =&0max 0.01902 mDncxθω= =&max 19.02 mmDx = !
85. 85. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 71.22 2 21 12 2 2BlT I ml mθ θ  = = +     & &212 2lV kθ =   222858nklmlω =5km=1Hz2 5nkfmπ= !
86. 86. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 72.Equilibrium:0002220 lb1ft9k yTT k yy=+ =∴ =Vibration: 22 21 20,2T vg =   ( ) ( )22 11 11 1144 1442 2 2xV x = +   1 2:V T= 2 211 202 32.2n xω   ( ) ( ) 211 11442 8x = +  2289.8, 17.0235 rad/sn nω ω= =(a)20.36909 s,nπτω= = 0.369 sτ = !(b)1ft 1.8915 ft/s9nmv ω = =  1.892 ft/smv = !
87. 87. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 73.Datum at 1Position 121 0 1102mT J Vθ= =&Position 22 20T V mgh= =( ) 21 cos 2sin2mmh r rθθ= − =22mrθ≈222mV mgrθ=Conservation of energy221 1 2 2 010 02 2mmT V T V J mgrθθ+ = + + = +&2 2 20m n m n m mJ mgrθ ω θ ω θ θ= =&( )2202 22044n nnJmgrJ mgrππω τω= = =( )( )22 20 24n mgrJ J mr J mrτπ= + + =( )( )2224n mgrJ mrτπ= −( ) ( )( )( )( )( )2 2221.26 s 38 kg 9.81 m/s 0.175 m38 kg 0.175 m4π= −2 2 22.62345 N m s 1.16375 N m s 1.4597 N m sJ = ⋅ ⋅ − ⋅ ⋅ = ⋅ ⋅21.460 kg mJ = ⋅ !
88. 88. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 74.Find nω as a function of c.Datum at 2Position 1 1 10T V mgh= =( )1 1 cos mV mgc θ= −221 cos 2sin2 2m mmθ θθ− = ≈212mV mgcθ=Position 22212C mT I θ= &2 2 2112CI I mc ml mc= + = +22 22 2102 12mlT m c Vθ = + =   &2 2 221 1 2 2 0 02 12 2m mlT V T V mgc m cθ θ + = + + = + +   &m n mθ ω θ=&22 212nlgc m c ω = +   22212ngclcω = +   
89. 89. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Maximum c, when22 22222120 012nlg c c gddc lcω + −   = = = +   22012lc− =12lc = !
90. 90. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 75.Consider a general pendulum of centroidal radius of gyration .kDatum at 1Position 121 012mT J θ= !1 0V =Position 22 20T V mgh= =( ) 21 cos 2sin2mmh r Fθθ= − =2 222 2m mV mg Fθ θ= ≈1 1 2 2T V T V+ = +2 201 10 02 2m mJ mgrθ θ+ = +!m n mθ ω θ=!2 2 20 n m mJ mgrω θ θ=2 0022n nnmgr JJ mgrπω τ πω= = =2 2 20J J mr mk mr= + = +
91. 91. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(a)2 22nk rgrτ π+=For a rod suspended at A:2 20.895 2 ,an aak rr rgrτ π+= = = (1)For a rod suspended at B:2 20.805 2 ,bn bbk rr rgrτ π+= = = (2)10.5 in. 0.875 fta br r+ = = (3)From (1) and (2)( )22 220.8954aagrk rπ+ = ( )1′( )22 220.8054bbgrk rπ+ = ( )2′Taking the difference ( ) ( )1 2′ ′− :[ ]2 220.80125 0.6480254a b a bgr r r rπ− = −( )( ) [ ]20.80125 0.6480254a b a b a bgr r r r r rπ− + = −( )( )[ ]232.20.80125 0.6480254 0.875a b a br r r rπ− = −( ) 0.746685 0.60406a b a br r r r− = −0.25332 0.39594 0.63979a b b ar r r r= ⇒ =0.63979 0.875 0.533605 fta a ar r r+ = ⇒ =6.4033 in.= 6.40 in.ar = !(b)22 20.8952a ak gr rπ = −  ( )( )( )( )22220.895 s32.2 ft/s 0.533605 ft 0.533605 ft2π= −2 2 20.34863 ft 0.28473 ft 0.0639 ft= − =0.25278 ft 3.0334 in.k = =3.03 in.k = !
92. 92. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 76.22PmrI >2 224 41 0.715 0.715 ft3 3rd r rπ π   = + − = =      2224 22 3Omr rI m mdπ   = − +     0.65117 m=221,2 2OT I V mgdθθ= =&2 2 35.35640.651172ndmgmω = =5.9461 rad/snω =21.057 snπτω= = !
93. 93. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 77.Since vibration takes place about the position of equilibrium, we shall neglect the effect of weight and thestatic deflection.Position of Max DisplacementSpring Elongations2B mlx θ= C mx lθ=Position 1: 1 0T =( )222 211 1 12 2 2 2B B C C B m C mlV k x k x k k lθ θ = + = +  2 211 12 4B C mV k k l θ = +  Position 2: 2 0V =22 2 2 221 1 1 1 12 2 2 12 2 2m m m mlT I mv ml mω ω ω   = + = +      2 2216mT ml ω=Conservation of Energy2 2 2 21 1 2 21 1 1: 02 4 6B C m mT V T V k k l mlθ ω + = + + + =  For simple harmonic motion, ( )22 2 21 1 1;2 4 6m n m B C m n mk k l mlω ω θ θ ω θ = + =  2 3 14n B Ck kmω = +  (1)
94. 94. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Note: Result is independent of length of the rod.Data6 lb,W = 3 lb/in. 36 lb/ft,k = = 5 lb/in. = 60 lb/ft.Ck =223 36 lb/ft60 lb/ft 1110.946 lb32.2 ft/snω = + =      33.33 rad/snω =33.332 2nfωπ π= = 5.30 Hzf = !
95. 95. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 78.2 212 2 4l lV k mgθ θ = +  2 2 2 2 221 12 4 2 2 4L mr LT mrθ θ   = +         & &2 2316mL θ=&Then2228 4316nkL mgLmLω+=2 23k gm L = +  1 2 4Hz2 3 3nk gfm Lπ= + !
96. 96. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 79.(a) Position 1( )21 1102st mT V k rδ θ= = +Position 2( )22 22 21 1 12 2 2m m stT J mv V mgh kθ δ= + = +!1 1 2 2T V T V+ = +( ) ( )2 22 21 1 1 102 2 2 2st m m m stk r J mv mgh kδ θ θ δ+ + = + + +!2stkδ 2 2 2 2 22 2st m m m m stk r kv J mv mgh kδ θ θ θ δ+ + = + + +! (1)When the disk is in equilibrium0 sinC stM mg r k rβ δΣ = = −Also sin mh r βθ=Thus 0stmgh k rδ− = (2)Substitute (2) into (1)2 2 2 2m m mkr J mvθ θ= +!m n m m m n mv r rθ ω θ θ ω θ= = =! !( )2 2 2 2 2m m nkr J mrθ θ ω= +
97. 97. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.222nkrJ mrω =+212J mr=222 221 32nkr kmmr mrω = =+( )2 20.71983 s800 N/m23 7 kgnnπ πτω= = =0.720 snτ = !(b)m mv rθ= !m m nθ θ ω=!0.01 mm m n mv r rθ ω θ= =( )20.01 m 0.0873 m/s0.720 smvπ = =  !
98. 98. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 80.sinθ θ≈221 cos 2sin2 2m mmθ θθ− = ≈Position12211 122 2 2m mlT J mθ θ   = +      & &1 0V =Position 2( )22 220 1 cos2 2 2m ml lT V W kθ θ = = − − +   2 2222 2 4mmWl klVθθ= − +Conservation of Energy1 1 2 2T V T V+ = +( )2 2 22 2 21 12 0 02 2 4 2 2 4mm m ml wl klJ mθθ θ θ+ + = − +& &2112m n mWJ lgθ ω θ= =&
99. 99. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.22 2 2 26 4 2 2n m mW W Wl kllg gω θ θ   −+ = +       2 62 25512nW klg kWlWlggω −+  − = = +       26 9.81 m/s 120 N/m5 0.160 m 0.6 kg −= +   2166.43 s−=12.901 rad/snω =12.0532 s2nnfωπ−= =2.05 Hznf = !
100. 100. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 81.( )2212 4lV k l mgθθ = +    22 21 12 12 2lT ml m θ  = +     &2222 46nkl mglmlω+=2 3 32nk gm lω = +With 1500 N/m, 0.6 m, 10 kgk l m= = =2 2 2474.525 rad /snω =20.288 snπτω= = !
101. 101. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 82.Let m be the mass of rod and Cm be mass of each collarThen2 2 2 22 4 2Ckl mgl lV m gθ θ θ = + +   ( )2 221 12 3 2ClT m m lθ θ = +   & &22222 4 26 2CnCmgl m glklm lmlω+ +=+2 4 26 2CCmg m gkl lmm+ +=+( )( )( )( )( )( )2 225 kg 9.81 m/s 2.5 kg 9.81 m/s1500 N/m2 4 0.6 m 2 0.6 m5 kg 2.5 k6 2nω  + +  = +  2397.62 s−=19.4838 rad/snω =20.322 snπτω= = !
102. 102. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 83.2 21 20, 22 2 2lV V mgl mgθ θ = = +    2mglθ=22 2 22 11 10, 22 2 3mlT T ml θ θ = = +   & &2 256ml θ=&22 2 2 2 25 6: ,6 5n n nml gmgllθ ω θ ω ω= = =&( )( )2 21.586 s9.81 m/s60.75 m5nπ πτω= = = !
103. 103. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 84.6.87 NA AW m g= =4.91 NC CW m g= =9.81 NAC ACW m g= =Position 1 ( ) ( ) ( )2 2 2 211 1 1 10.1 0.16 0.032 2 2 2A m C m AC m AC mT m m m Iθ θ θ θ= + + +& & & &( )210.2612AC ACI m=So ( ) ( ) ( ) ( )( )2 2 2 2 211 10.7 0.1 0.5 0.16 1 0.03 1 0.262 12mT θ = + + +  &( )2 210.02633 kg m2mθ= ⋅ &1 0V =Position 2( )( ) ( )( ) ( )( )2 20, 0.1 1 cos 0.16 1 cos 0.03 1 cosA m C m AC mT V W W Wθ θ θ= = − − + − + −
104. 104. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.With221 cos 2sin2 2m mmθ θθ− = "( )( ) ( )( ) ( )( )2 22 6.87 0.1 4.91 0.16 9.81 0.03 0.39292 2m mVθ θ = − + + = ( ) ( )221 1 2 21 1: 0.02633 0 0 0.39292 2 2mmT V T Vθθ+ = + + = +&m n mθ ω θ=&So 2 20.392914.922 s0.02633nω −= =2 21.6266 s14.922nnπ πτω= = =1.627 snτ = !
105. 105. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 85.Position 1( ) ( ) ( )( )222 2 21 01 1 1 1 222 2 2 2 2 4A C m mm mm lT m v m v I θ θ = + + +    & &( ) ( )2112 2G A C mm mmI l v v lθ= = = &2 22 2 21724 8 6m m mml mlT ml mlθ θ θ = + + =   & & &152 cos cos cos2 2lV mgl mg mglβ β β= − − = −Position 22 0T =( ) ( )2 cos cos2 2m mm lV mgl gβ θ β θ= − − − −( ) ( )cos cos2 2m mm lmgl β θ β θ− + − +[ ]25cos cos sin sin cos cos sin sin4m m m mV mgl β θ β θ β θ β θ= − + + −
106. 106. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.25cos cos2mV mgl β θ= −( )2cos 1 small angles2mmθθ ≈ −225cos 12 2mV mglθβ = − −  1 1 2 2T V T V+ = +22 27 5 5cos 0 cos 16 2 2 2mmml mgl mglθθ θ β − = − −   &m n mθ ω θ=&2 2 27 5cos6 4n m ml gω θ βθ= ⋅2 15cos14nglω β=22 15 32.2 ft/scos4025 in.1412 in./ftnω   = °   212.686 s−=13.5617 snω −=10.56686 s2nnfωπ−= =0.567 Hznf = !
107. 107. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 86.D = DiskR = Rod2ABl r=For small oscillations: ( ) 211 cos2m mh r rθ θ= − =Position 1: 1 0T =2112R R mV W h m grθ= =Position 2: 2 0V =( )22 221 1 12 2 2D m D D R mmT I m v Iω ω= + +( )22 2 2 2 21 1 1 1 122 2 2 2 12D m D m R mm r m r m rω ω ω   = + +      2 21 3 12 2 3D R mm m r ω = +  Conservation of Energy.1 1 2 2 :T V T V+ = +2 2 21 1 3 102 2 2 3R m D R mm g m m rθ ω + = +  But for simple harmonic motion: m n mω ω θ=( )22 21 1 3 12 2 2 3R m D R n mm gr m m rθ ω θ = +  
108. 108. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.23 12 3RnD Rm grm mω =   +or 23 12 3RnD RW grW Wω =   +(1)Data:43 lb, 5 lb, ft12R DW W r= = =( ) ( )2 3 32.234.0943 1 45 32 3 12nω  = =  +   5.839 rad/snω =2 25.839nπ πτω= = 1.076 sτ = !
109. 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 19, Solution 87.Position 1: 1 0,T = 2112mV kx=Position 2: 2 0,V = 2 2 221 1 122 2 2AB m m Disk mT m v I m vω = + +  22 2 221 12 2mAB m Disk Disk mvT m v m r m vr  = + +    ( ) 22132AB Disk mT m m v= +Conservation of Energy1 1 2 2:T V T V+ = + ( )2 21 10 32 2m AB Disk mkx m m v+ = +But for simple harmonic motion, :m n mv xω=( )( )221 132 2m AB Disk n mkx m m xω= +23nAB Diskkm mω =+Note: Result is independent of rData: 5 kN/m, 9 kg, 6 kgAB Diskk m m= = =( )2 5000 N/m185.1859 kg 3 6 kgnω = =+13.608 rad/snω =13.6082 2nfωπ π= = 2.17 Hzf = !
110. 110. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 19, Solution 88.( )141 cos3rV mgh mg θπ = = −  21 cos2mθθ− ≈21 23mV mgrθπ=2 2 2 221 1 1 12 2 2 2A A B BT I I mrω ω ω = + +   Where2 212 8 4 256A Bm r mrI I  = = =    and 4A Bω ω ω= =2 2 2 222516 4 16mr mr mrTωω ∴ = + =   1 22,mV T=2mgrθ 53mπ=2 2 2n mr ω θ162 32,15ngrωπ=1 322 15ngfrπ π =   
111. 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 19, Solution 89.Kinematics:Data0.15 , 0.05m m m AB m mv r v bω ω ω ω= = = = 10kgABm =4kgDm =Conservation of Energy1 1 2 2T V T V+ = + (1)Where( )110Position 1, Max. DisplacementcosAB mTV m g r b θ= = − − ( )0.10 cos 9.81cosAB m mm g θ θ= − = −2 2 221 1 122 2 2D m D m AB ABT m v I m vω = + +  ( )( ) ( )( ) ( )( )2 22 21 1 1 12 4 0.15 4 0.15 10 0.052 2 2 2m m mω ω ω  = + +    20.1475 mω=( ) ( )2 0.10 9.81AB ABV m g r b m g= − − = − = −Into (1)20 9.81cos 0.1475 9.81m mθ ω− = −( )20.1475 9.81 1 cosm mω θ= −But 211 cos2m mθ θ− ≈2 20.1475 4.905m mω θ= (2)
112. 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.m n mω ω θ=Into (2)( )2 2 20.1475 4.905n m mω θ θ=So2 4.90533.2540.1475nω = =5.7666 rad/snω =0.9178 Hz2nfωπ= = 0.918 Hznf =
113. 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 19, Solution 90.22 21 12 72 2 6xT mx my m = + =  && &Equilibrium of BC:302 2 2Dmg l lM ku Σ = = ⋅ −    Where 0, natural length2 3mgu l l uk= = = +( ) ( )2 2012 22 2kkV k l x l x u = − − = + ,gV mgy= − where222 32 2l ll x y  = − + +       And ( )( )2 22 23 3 43 0,2l l x lxy ly x lx y− + − −+ + − = =So 21 1 8H.O.T.23 3 3y x xl   = + − +      Then22 42 2 constant3 3 3x mgxV kx kux mgl= + + − +2423 376nmgklmω+=1 12 8Hz2 7 7 3nk gfm lπ= + !
114. 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 19, Solution 91.Position 1( )22 2 2 211 1 12 22 2 2D m D m r mT I m r c m rθ θ θ   = + − +      & & &For one disk ( )2 2 22 2 221 4 160.319872 3 2 9D D D D D D DAr r rI I m c m r m m m rπ π  = − = − = − =     ( )22 2 241 0.33133D D Dm r c m r m rπ − = − =  ( ) 2 2 21 0.3199 0.3313 0.5D r mT m r m r θ = + + &( ) 2 20.6512 0.5D r mm m r θ= + &But ,D rD rW Wm mg g= =So ( ) ( )22 211 60.65117 6 0.5 4 0.04586632.2 12m mT θ θ  = + =   & &
115. 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.Position 2 ( )22 220, 2 1 cos , , 1 cos 12 3 2m mD m mrT V m gc cθ θθ θπ = = − = − ≅  "( )22 2 224 4 4 62 6 1.273243 2 3 3 12mD D m m mrV m g W rθθ θ θπ π π   = = = =      &2 21 1 2 2 : 0.045866 0 0 1.27324m mT V T V θ θ+ = + + = +&2 1.27324: 27.7470.045866m n m nθ ω θ ω= = =&25.2677, 1.192 sn nnπω τω= = =1.192 snτ = !
116. 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 19, Solution 92.With 2 24 lb 6 lb 40, , 20 lb/ft, ft1232.2 ft/s 32.2 ft/sm M k l = = = =   2 2 212 2 2 2 2l lV k mg Mglθ θ θ    = + +             And for small :θ( )222 2 21 1 1 1 12 3 2 2 2lT ml M l Mrrθθ θ    = + +         && &So222 28 4 2 23.0593536 4nkl l lmg Mgl lm Mω+ += =+Then21.308 snπτω= = !
117. 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 19, Solution 93.2 221 12 12T ml θ =   &Ignoring static terms:2 211 12 2 2 2a aV k kθ θ   = +      224kaθ=22 2 2 22 11:24 4n m maT V ml kω θ θ = =    2226nkamlω =62na kfl mπ=
118. 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 19, Solution 94.2 2m m m ma aAA BB llθ α α θ′ ′= = = =Position 1( )1 10 1 coscT V mgy mgl α= = = −For small angles2 2221 cos 2sin2 2 8m mm malα αα θ− = ≈ =221 28maV mgllθ =    Position 2 2 2 22 21 1 102 2 12m mT I ma Vθ θ = = =  & &m n mθ ω θ=&1 1 2 2T V T V+ = +22 2 2210248n mamgl malω θ + +   2 3nglω =1 32ngflπ= !
119. 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 19, Solution 95.( ) 21 cos2m mn r rθ θ1= − = ( )m mv r r ω= −Position 1: 21 110,2mT V Wh mgrθ= = =Position 2: 2 0V =( )22 2 2 2 221 1 1 12 2 2 2m m m mT I mv mk m r rω ω ω= + = + −Conservation of energy1 1 2 2T V T V+ = +( )22 2 21 102 2m mmgr m k r rθ ω + = + −  But for simple harmonic motion, m n mω ω θ=( ) ( )222 21 12 2m n mmgr m k r rθ ω θ = + −  ( )222nrgk r rω =+ −(1)For half section of pipe2 21rr r r rπ π = − = −  Parallel-Axis Theorem: 20I I mr= +22 202;rI mr mr I mπ = = +   