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# solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 14

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### solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 14

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 14, Solution 1.The masses are 1350 kg and 5400 kg.A B Cm m m= = =Let , , and be the sought after final velocities, positive to the left.A B Cv v vInitial velocities: ( ) ( ) ( )0 0 00, 8 km/h 2.2222 m/sA B Cv v v= = = =. Truck strikes car . Plastic impact: 0First collision C B e =( )0Let be the common velocity of and after impact.BCv B C:Conservation of momentum for B and C( ) ( ) ( )0 0B C BC B B C Cm m v m v m v+ = +( )( )6750 0 5400 2.2222BCv = + 1.77778 m/sBCv =Car-truck strikes carSecond collision. BC A.Elastic impact. 1e =( ) ( )0 01.77778 m/sA BC A BCv v e v v − = − − = −  (1)Conservation of momentum for A, B, and C.( )( ) ( ) ( )( )0 0A A B C BC A A B C BCm v m m v m v m m v+ + = + +( )( )1350 6750 0 6750 1.77778A BCv v+ = + (2)Solving (1) and (2) simultaneously for and ,A BCv v2.9630 m/s, 1.18519 m/sA BC B Cv v v v= = = =10.67 km/hA =v4.27 km/hB =v4.27 km/hC =v
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 14, Solution 2.Conservation of linear momentum for block, cart, and bullet together.components : ( )0B A B C fm v m m m v= + +( )( )0 0.028 5501.7058 m/s5 0.028 4BfA B Cm vvm m m= = =+ + + +(a) 1.706 m/sfv =Consider block and bullet alone.Principle of impulse and momentum.components ( ) ( ): N t mg t N mg∆ − ∆ =components : ( ) ( )0B k A Bm v N t m m vµ− ∆ = +Just after impact, t∆ is negligible. The velocity then is( )( )000.028 5503.0628 m/s5 0.028BA Bm vvm m= = =+ +Also, just after impact, the velocity of the cart is zero.Accelerations after impact.
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.Block and bullet: :F ma=∑( ) ( )k A B A B ABm m g m m aµ + = +( )( )0.50 9.81AB ka gµ= =24.905 m/s=Cart: :CF ma=∑( ) :k A B C Cm m g m aµ + =( ) ( )( )( )0.50 5.028 9.814k A BCCm m gamµ += =26.1656 m/s=Acceleration of block relative to cart.( ) 2/ 4.905 6.1656 11.0706 m/sAB Ca = − − =Motion of the block relative to the cart.( ) ( )( )( )2 2// /22 2AB CAB C AB Cv va s− =In the final position, / 0AB Cv =( ) ( )( )( )2 2//C3.06280.424 m2 2 11.0706AB CABvsa= − = − = −The block moves 0.424 to the left relative to the cart.(b) This places the block 1.000 0.424 0.576 m− = from the left end of the cart.0.576 m from left end of cart
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 14, Solution 3.( )( )22 20004000 3700The masses are 124.2 slugs, 114.9 slugs, and 1366.5 slugs32.2 32.2 32.2A B Fm m m= = = = = =Let , , and be the sought after velocities in ft/s, positive to the right.A B Fv v vInitial values: ( ) ( ) ( )0 0 00.A B Fv v v= = =Initial momentum of system: ( ) ( ) ( )0 0 00.A A B B F Fm v m v m v+ + =There are no horizontal external forces acting during the time period under consideration. Momentum isconserved.0 A A B B F Fm v m v m v= + +124.2 114.9 1366.5 0A B Fv v v+ + = (1)The relative velocities are given as//7 ft/s3.5 ft/sA F A FB F B Fv v vv v v= − = −= − = −(2)(3)Solving (1), (2), and (3) simultaneously,6.208 ft/s, 2.708 ft/s, 0.7919 ft/s= − = − =A B Fv v v0.792 ft/sF =v
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 14, Solution 4.The masses are , , and .A B FA B FW W Wm m mg g g= = =Let the final velocities be , , and 0.34 ft/s, positive to the right.=A B Fv v vInitial values: ( ) ( ) ( )0 0 00A B Fv v v= = =Initial momentum of system: ( ) ( ) ( )0 0 00A A B B F Fm v m v m v+ + =There are no horizontal external forces acting during the time period under consideration. Momentum isconserved.0 A B FA A B B F F A B FW W Wm v m v m v v v vg g g= + + = + +Solving for ,FW A A B BFFW v W vWv+= − (1)From the given relative velocities,//1.02 7.65 6.63 ft/s1.02 7.5 6.48 ft/sA F A FB F B Fv v vv v v= + = − = −= + = − = −Substituting these values in (1),( )( ) ( )( )4000 6.63 3700 6.4849506 lb1.02FW− + −= − =24.8 tonsFW =
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 14, Solution 5.( ) ( )( )3 33The masses are the engine 80 10 kg , the load 30 10 kg , and the flat car20 10 kg .A BCm mm= × = ×= ×Initial velocities: ( ) ( ) ( )0 0 06.5 km/h 1.80556 m/s, 0.A B Cv v v= = = =No horizontal external forces act on the system during the impact and while the load is sliding relative to the flatcar. Momentum is conserved.Initial momentum: ( ) ( ) ( ) ( )0 00 0A A B C A Am v m m m v+ + = (1)(a) Let v′ be the common velocity of the engine and flat car immediately after impact. Assume that theimpact takes place before the load has time to acquire velocity.Momentum immediately after impact:( ) ( )0A B C A Cm v m m v m m v′ ′ ′+ + = + (2)Equating (1) and (2) and solving for ,v′( ) ( )( )( )30380 10 1.805561.44444 m/s100 10A AA Cm vvm m×′ = = =+ ×5.20 km/h′ =v(b) Let fv be the common velocity of all three masses after the load has slid to a stop relative to the car.Corresponding momentum:( )A f B f C f A B C fm v m v m v m m m v+ + = + + (3)Equating (1) and (3) and solving for ,fv( ) ( )( )( )30380 10 1.805561.11111m/s130 10A AfA B Cm vvm m m×= = =+ + ×4.00 km/hf =v
7. 7. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 6.The masses are m for the bullet and Am and Bm for the blocks.(a) The bullet passes through block A and embeds in block B. Momentum is conserved.( ) ( )0 0Initial momentum: 0 0A Bmv m m mv+ + =Final momentum: B A A B Bmv m v m v+ +0Equating, B A A B Bmv mv m v m v= + +( )( ) ( )( ) 303 3 2.5 543.434 10 kg500 5A A B BBm v m vmv v−++= = = ×− −43.4 gm =(b) The bullet passes through block A. Momentum is conserved.( )0 0Initial momentum: 0Amv m mv+ =1Final momentum: A Amv m v+0 1Equating, A Amv mv m v= +( )( ) ( )( )301 343.434 10 500 3 3292.79 m/s43.434 10A Amv m vvm−−× −−= = =×1 293 m/s=v
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 14, Solution 7.(a) Woman dives first.Conservation of momentum:( )1 1120 300 18016 0v vg g+− − =( )( )1120 163.20 ft/s600v = =Man dives next. Conservation of momentum:( )1 2 2300 180 300 18016v v vg g g+− = − + −( )( )12480 180 169.20 ft/s480vv+= = 2 9.20 ft/s=v(b) Man dives first.Conservation of momentum:( )1 1180 300 12016 v vg g+′ ′− −( )( )1180 164.80 ft/s600v′ = =Woman dives next. Conservation of momentum:( )1 2 2300 120 300 12016v v vg g g+′ ′ ′− = − + −( )( )12420 120 169.37 ft/s420vv′ +′ = = 2 9.37 ft/s′ =v
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 14, Solution 8.(a) Woman dives first.Conservation of momentum:( )1 1120 300 18016 0v vg g+− − + =( )( )1120 163.20 ft/s600v = =Man dives next. Conservation of momentum:( )1 2 2300 180 300 18016v v vg g g+= − + −( )( )12480 180 162.80 ft/s480− += =vv 2 2.80 ft/s=v(b) Man dives first.Conservation of momentum:( )1 1180 300 12016 0v vg g+′ ′− − =( )( )1180 164.80 ft/s600v′ = =Woman dives next. Conservation of momentum:( )1 2 2300 120 300 12016v v vg g g+′ ′ ′− = + −( )( )12420 120 160.229 ft/s420vv′− +′ = = −2 0.229 ft/s′ =v
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 14, Solution 9.The masses are 9 kg.A B Cm m m= = =Position vectors (m): 0.9 , 0.6 0.6 0.9 , 0.3 1.2A B C= = + + = +r k r i j k r i j2In units of kg m /s,⋅ ( ) ( ) ( )O A A A B B B C C Cm m m= × + × + ×H r v r v r v( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )0 0 0.9 0.6 0.6 0.9 0.3 1.2 00 9 0 9 0 0 0 0 98.1 8.1 5.4 10.8 2.78.1 10.8 8.1 2.7 5.4A B CA B B C CA C B C Bv v vv v v v vv v v v v= + += − + − + −= − + + − + −i j k i j k i j ki j k i ji j kBut, OH is given as ( )21.8 kg m /s− ⋅ kEquating the two expressions for OH and resolving into components,: 8.1 10.8 0: 8.1 2.7 0: 5.4 1.8A CB CBv vv vv− + =− =− = −ijk(1)(2)(3)( ) Solving for , , and ,A B Ca v v v 1.333 m/sAv = ( )1.333 m/sA =v j0.333 m/sBv = ( )0.333 m/sB =v i1.000 m/sCv = ( )1.000 m/sC =v kCoordinates of mass center G in m.( )( ) ( )( ) ( )( )9 0.9 9 0.6 0.6 0.9 9 0.3 1.2270.3 0.6 0.6A A B B C CA B Cm m mm m m+ +=+ ++ + + + +== + +r r rrk i j k i ji j kPosition vectors relative to the mass center in m.( ) ( )( ) ( )( ) ( )0.9 0.3 0.6 0.6 0.3 0.6 0.30.6 0.6 0.9 0.3 0.6 0.6 0.3 0.30.3 1.2 0.3 0.6 0.6 0.6 0.6A AB BC C′ = − = − + + = − − +′ = − = + + − + + = +′ = − = + − + + = −r r r k i j k i j kr r r i j k i j k i kr r r i j i j k j k( )( ) ( )( )( ) ( )( )( ) ( )9 1.333 12 kg m/s9 0.333 3 kg m/s9 1 9 kg m/sA AB BC Cmmm= = ⋅= = ⋅= = ⋅v j jv i iv k k
11. 11. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )0.3 0.6 0.3 12 0.3 0.3 3 0.6 0.6 93.6 3.6 0.9 5.4 1.8 0.9 3.6G A A A B B B C C Cb m m m′ ′ ′= × + × + ×= − − + × + + × + − ×= − − + + = + −H r v r v r vi j k j i k i j k ki k j i i j k( ) ( ) ( )2 2 21.800 kg m /s 0.900 kg m /s 3.60 kg m /sG = ⋅ + ⋅ − ⋅H i j k
12. 12. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 10.The masses are 9 kg.A B Cm m m= = =Position vectors (m): 0.9 , 0.6 0.6 0.9 , 0.3 1.2A B C= = + + = +r k r i j k r i jCoordinates of mass center G expressed in m.( )( ) ( )( ) ( )( )9 0.9 9 0.6 0.6 0.9 9 0.3 1.2270.3 0.6 0.6A A B B C CA B Cm m mm m m+ +=+ ++ + + + +== + +r r rrk i j k i ji j kPosition vectors relative to the mass center expressed in m.( ) ( )( ) ( )( ) ( )0.9 0.3 0.6 0.6 0.3 0.6 0.30.6 0.6 0.9 0.3 0.6 0.6 0.3 0.30.3 1.2 0.3 0.6 0.6 0.6 0.6A AB BC C′ = − = − + + = − − +′ = − = + + − + + = +′ = − = + − + + = −r r r k i j k i j kr r r i j k i j k i kr r r i j i j k j kAngular momenta.( ) ( ) ( )( ) ( ) ( )O A A A B B B C C CG A A A B B B C C Cm m mm m m= × + × + ×′ ′ ′= × + × + ×H r v r v r vH r v r v r vSubtracting,( ) ( ) ( ) ( )O G A A A A B B B B C C C Cm m m′ ′ ′− = − × + − × + − ×H H r r v r r v r r v( ) ( ) ( )( )0 A A B B C CA A B B C Cm m mm m m= × + × + ×= × + + = ×r v r v r vr v v v r Lis parallel to .L r 2λ λ= ⋅ = ⋅L r L L r r( )( )22 224550 , 50 N s/m0.9λ λ⋅= = = = ± ⋅⋅L Lr r( )( ) ( )( ) ( )( ) ( )9 9 9 50 0.3 0.6 0.6A A B B C CA B Cm m mv v vv v v rj i k i j kλ+ + =+ + = ± + +
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.( ) Resolve into components and solve for , , and .A B Ca v v v3.333 m/sAv = ( )3.33 m/sA =v j1.6667 m/sBv = ( )1.667 m/sB =v i3.333 m/sCv = ( )3.33 m/sC =v k(b) Angular momentum about O expressed in 2kg m /s.⋅( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )0 0 0.9 0.6 0.6 0.9 0.3 1.2 00 9 0 9 0 0 0 0 98.1 8.1 5.4 10.8 2.78.1 10.8 8.1 2.7 5.49 4.5 9O A A A B B B C C CA B CA B B C CA C B C Bm m mv v vv v v v vv v v v v= × + × + ×= + += − + − + −= − + + − + −= + −H r v r v r vi j k i j k i j ki j k i ji j ki j k( ) ( ) ( )2 2 29.00 kg m /s 4.50 kg m /s 9.00 kg m /sO = ⋅ + ⋅ − ⋅H i j k
14. 14. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 11.Position vectors expressed in ft.3 6 , 6 3 , 3 3A B C= + = + = +r i j r j k r i kMomentum of each particle expressed in lb s.⋅( ) ( )4 142 63 168 252A AWg g g= + = +vi j i j( ) ( )4 142 63 168 252B BWg g g= − + = − +vi j i j( ) ( )28 19 6 252 168C CWg g g= − = − −vj k j k−−−−Angular momentum of the system about O expressed in ft lb s.⋅ ⋅( ) ( ) ( ){ }( )13 6 0 0 6 3 3 0 3168 252 0 168 252 0 0 252 1681252 756 504 1008 756 504 75610 0 0A A B B C CO A B CW W Wg g gggg= × + × + ×  = + +  − − − = − + − − + + + −= + +v v vH r r ri j k i j k i j kk i j k i j ki j kzeroO =H
15. 15. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 12.(a) 4 4 28 36 lbA B CW W W W= + + = + + =( )( ) ( )( ) ( )( ){ }14 3 6 4 6 3 28 3 336A A B B C C A A B B C Cm m m W W mm W+ + + += == + + + + +r r r r r rri j j k i k2.667 1.333 2.667= + +i j k( ) ( ) ( )2.67 ft 1.333 ft 2.67 ft= + +r i j k(b) Linear momentum( )1A A B B C C A A B B C Cm m m m W W Wg= + + = + +v v v v v v v( )( ) ( )( ) ( )( )14 42 63 4 42 63 28 9 6g = + + − + + − − i j i j j k( )1252 16832.2= −j k( ) ( )7.83 lb s 5.22 lb sm = ⋅ − ⋅v j k(c) Position vectors relative to the mass center G (ft).( ) ( )( ) ( )( ) ( )3 6 2.667 1.333 2.6670.333 4.667 2.6676 3 2.667 1.333 2.6672.667 4.667 0.3333 3 2.667 1.333 2.6670.333 1.333 0.333A AB BC C′ = − = + − + += + −′ = − = + − + += − + +′ = − = + − + += − +r r r i j i j ki j kr r r j k i j ki j kr r r i k i j ki j kAngular momentum about the mass center.( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )10.333 4.667 2.667 2.667 4.667 0.333 0.333 1.333 0.3334 42 4 63 0 4 42 4 63 0 0 28 9 28 6A A B B C CG A B CW W Wg g gg     ′ ′ ′= × + × + ×          = − + − + − − − −v v vH r r ri j k i j k i j k( ) ( ) ( ){ }( )1672 448 700 84 56 112 308 56 841896 448 672 27.827 13.913 20.87032.2g= − − + − − + + + −= − − = − −i j k i j k i j ki j k i j k( ) ( ) ( )27.8 ft lb s 13.91 ft lb s 20.9 ft lb sG = ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅H i j k
16. 16. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.From Problem 14.28, O Gm= × +H r v H2.667 1.333 2.6670 7.83 5.2227.826 13.913 20.870 27.827 13.913 20.8700 0 0O =−= − + + + − −= + +i j kHi j k i j ki j kFrom Prob 14.11,( ) ( ) ( )( ) ( ) ( ){ }( ) ( ) ( ){ }( )113 6 0 0 6 3 3 0 3168 252 0 168 252 0 0 252 168252 756 504 1008 756 504 75610 0 0O A A A B B B C C CA A A B B B C C Cm m mW W Wgggg= × + × + ×= × + × + ×  = + +  − − − 1= − + − − + + + −= + +H r v r v r vr v r v r vi j k i j k i j kk i j k i j ki j k
17. 17. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 13.Linear momentum of each particle expressed in kg m/s.⋅3 2 48 66 15 9A AB BC Cmmm= − += += + −v i j kv i jv i j kPosition vectors, (meters): 3 , 3 2.5 , 4 2A B C= + = + = + +r j k r i k r i j k( )2Angular momentum about , kg m /s .O ⋅( ) ( ) ( )( ) ( ) ( )0 3 1 3 0 2.5 4 2 13 2 4 8 6 0 6 15 914 3 9 15 20 18 33 42 4834 65 57O A A A B B B C C Cm m m= × + × + ×= + +− −= + − + − + + + − + += − + +H r v r v r vi j k i j k i j ki j k i j k i j ki j k( ) ( ) ( )2 2 234 kg m /s 65 kg m /s 57 kg m /sO = − ⋅ + ⋅ + ⋅H i j k
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 14, Solution 14.Position vectors, (meters): 3 , 3 2.5 , 4 2A B C= + = + = + +r j k r i k r i j k( ) Mass center:a ( )A B C A A B B C Cm m m m m m+ + = + +r r r r( )( ) ( )( ) ( )( )6 1 3 2 3 2.5 3 4 23 1.5 1.5= + + + + + += + +r j k i k i j kr i j k( ) ( ) ( )3.00 m 1.500 m 1.500 m= + +r i j kLinear momentum of each particle, ( )kg m/s .⋅3 2 48 66 15 9A AB BC Cmmm= − += += + −v i j kv i jv i j k(b) Linear momentum of the system,( )kg m/s.⋅17 19 5A A B B C Cm m m m= + + = + −v v v v i j k( ) ( ) ( )17.00 kg m/s 19.00 kg m/s 5.00 kg m/sm = ⋅ + ⋅ − ⋅v i j kPosition vectors relative to the mass center, (meters).3 1.5 0.51.50.5 0.5A AB BC C′ = − = − + −′ = − = − +′ = − = + −r r r i j kr r r j kr r r i j k(c) Angular momentum about G, ( )2kg m /s .⋅( ) ( ) ( )3 1.5 0.5 0 1.5 1 1 0.5 0.53 2 4 8 6 0 6 15 95 10.5 1.5 6 8 12 3 6 122 24.5 25.5G A A A B B B C C Cm m m′ ′ ′= × + × + ×= − − + − + −− −= + + + − + + + + += + +H r v r v r vi j k i j k i j ki j k i j k i j ki j k( ) ( ) ( )2 2 22.00 kg m /s 24.5 kg m /s 25.5 kg m /sG = ⋅ + ⋅ + ⋅H i j k( ) ( ) ( )2 2 23 1.5 1.517 19 536 kg m /s 40.5 kg m /s 31.5 kg m /sm× =−= − ⋅ + ⋅ + ⋅i j kr vi j k( ) ( ) ( )2 2 234 kg m /s + 65 kg m /s 57 kg m /sG m+ × = − ⋅ ⋅ + ⋅H r v i j k
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.Angular momentum about O.( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2 2 20 3 1 3 0 2.5 4 2 13 2 4 8 6 0 6 15 914 3 9 15 20 18 33 42 4834 kg m /s + 65 kg m /s 57 kg m /sO A A A B B B C C Cm m m= × + × + ×= + +− −= + − + − + + + − + += − ⋅ ⋅ + ⋅H r v r v r vi j k i j k i j ki j k i j k i j ki j kNote thatO G m= + ×H H r v
20. 20. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 15.The mass center moves as if the projectile had not exploded.( ) ( )( ) ( )( )( ) ( )2201 160 2 9.81 22 2120 m 19.62 mt gt   = − = −     = −r v j i ji j( )A B A A B Bm m m m+ = +r r r( )( )( ) ( )1120 120 19.62 8 120 10 2012120 26.033 13.333B A B A ABm m mm = + −  = − − − − = − +r r ri j i j ki j k( ) ( ) ( )120.0 m 26.0 m 13.33 mB = − +r i j k
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 14, Solution 16.There are no external forces. The mass center moves as if the explosion had not occurred.( )( ) ( )0 450 4 1800 mt= = =r v i i( )A B C A A B B C Cm m m m m m+ + = + +r r r r( )( )( ) ( )( )( )( )11500 1800 300 1200 350 60050150 2500 450 9003300 750 900C A B C A A B BCm m m m mm = + + − − = − − −− + + = + +r r r ri i j ki j ki j k( ) ( ) ( )3300 m 750 m 900 mC = + +r i j k
22. 22. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 17.Mass center at time of first collision.( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( ) ( )( )( ) ( )1 1 1 11 1 1 1119600 2800 27.8 3600 38.4 3200 12040 ft 22.508 ftA B C A A B B C CA B C A A B B C Cm m m m m mW W W W W W+ + = + ++ + = + += − + − += −r r r rr r r rr j j ir i jMass center at time of photo.( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( )( )( )( ) ( )2 2 2 22 2 2 2229600 2800 30.3 50.7 3600 30.3 61.23200 59.4 45.640 ft 22.5375 ftA B C A A B B C CA B C A A B B C Cm m m m m mW W W W W W+ + = + ++ + = + += − + + − ++ − −= − +r r r rr r r rr i j i ji jr i jSince no external horizontal forces act, momentum is conserved, and the mass center moves at constantvelocity.( ) ( ) ( ) ( )1 1 1A B C A A B B C Cm m m m m m+ + = + +v v v v (1)2 1 t− =r r v (2)Combining (1) and (2), ( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C Cm m m m m m t + + − = + + r r v v v( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C CW W W W W W t + + − = + + r r v v v( )( ) ( )( ) ( )( )19600 80 45.0455 0 3600 3200 66Bv t − + = + + − i j j iComponents. : 768000 211200t− = −i 3.64 st =: 432440 3600 Bv t=j( )( )( )43244030.0343600 3.6363Bv = = 30.0 ft/sBv =
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 14, Solution 18.Mass center at time of first collision.( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( ) ( )( )( ) ( )1 1 1 11 1 1 1119600 2800 27.8 3600 38.4 3200 12040 ft 22.508 ftA B C A A B B C CA B C A A B B C Cm m m m m mW W W W W W+ + = + ++ + = + += − + − += −r r r rr r r rr j j ir i jMass center at time of photo.( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( )( )( )( ) ( )2 2 2 22 2 2 2229600 2800 30.3 50.7 3600 30.3 61.23200 59.4 45.640 ft 22.5375 ftA B C A A B B C CA B C A A B B C Cm m m m m mW W W W W W+ + = + ++ + = + += − + + − ++ − −= − +r r r rr r r rr i j i ji jr i jSince no external horizontal forces act, momentum is conserved, and the mass center moves at constantvelocity.( ) ( ) ( ) ( )1 1 1A B C A A B B C Cm m m m m m+ + = + +v v v v (1)2 1 t− =r r v (2)Combining (1) and (2), ( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C Cm m m m m m t + + − = + + r r v v v( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C CW W W W W W t + + − = + + r r v v v( )( ) ( )( ) ( )( ) ( )1 19600 80 45.0455 0 3600 3200 3.4B Cv v − + = + + i j j iComponents.( ) ( )1 1: 432440 12240 , 35.33 ft/s,B Bv v= =j24.1mi/hBv =( ) ( )1 1: 768000 10880 , 70.588 ft/s,C Cv v− = − =i48.1mi/hCv =
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 14, Solution 19.Projectile motion 20, 9.81 m/s , 0x y za a g a= = − = − =( ) ( ) ( )00 0165 m/s, 0, 0x y zv v v= = =After the chain breaks the mass center continues the original projectile motion.At 1.5 s,t =( ) ( )( )0 00 165 1.5 247.5 mxx x v t= + = + =( ) ( )( )220 01 115 0 9.81 1.5 3.9638 m2 2yy y v t gt= + − = + − =( )0 00zz z v t= + =Position of first cannon ball at this time is1 1 1240 m, 0, 7 mx y z= = =Definition of mass center: ( )1 2 1 1 2 2m m m m+ = +r r r( )1 22 12 2m m mm m1+= −r r r( ) ( )( ) ( )30 15247.5 3.9638 240 715 15255 m . m 7= + − += + 7 9276 −i j i ki j kPosition of second cannon ball: 2 2 2255 m, 7.93 m, 7 mx y z= = = −
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 14, Solution 20.Place the vertical y axis along the initial vertical path of the rocket. Let the x axis be directed to the right (east).Motion of the mass center: 0, 0, 0x xa v x= = =29.81 m/sya g= − = −0 28 9.81y yv v a t t= + = −2 20 0160 28 4.9052At 5.85 s, 0, 55.939 myy y v t a t t tt x y= + + = + −= = =: A A B BDefinition of mass center m m m= +r r rcomponent: 3 1 20 74.4 2 37.2 mA BB Bx x x xx x= += − + =( )( )component: 3 1 23 55.939 0 2 83.9 mA BB By y y yy y= += + =.Position of part B 37.2 m(east), 83.9 m(up)
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 14, Solution 21.Velocities of pieces C and D after impact and fracture.( ) ( )( ) ( )2.13 m/s, 3tan30 m/s0.72.12.333 m/s, 2.3333tan m/s0.9CC Cx yCDD Dx yDxv vtxv vtθ′ ′= = = = °′ ′= = = = −Assume that during the impact the impulse between spheres A and B is directed along the x axis. Then, the ycomponent of momentum of sphere A is conserved.( )0 A ym v′=Conservation of momentum of system:( ) ( ) ( )0: 0A B A A C C D D xxm v m m v m v m v′ ′ ′+ = + +( ) ( ) ( )4.8 0 3 2.33332 2Am mm mv′+ = + +( )a 2.13 m/sA′ =v( ) ( ) ( ) ( ) ( ): 0 0A B A A C C D Dy yym m m v m v m v′ ′ ′+ = + +( ) ( )0 0 0 3tan30 2.3333tan2 2m mθ+ = + ° −( )b3tan tan30 0.74232.3333θ = ° = 36.6θ = °( ) ( ) ( ) ( )22 2 23 3tan30C C Cx yv v v= + = + o3.46 m/sCv =( ) ( ) ( ) ( )22 2 22.3333 2.3333tan36.6D D Dx yv v v= + = + o2.91m/sDv =
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 14, Solution 22.( )( )( )( )0 0Velocity vectors: cos30 sin30sin7.4 cos7.4sin 49.3 cos49.3cos45 sin 45A AB BC Cvvvv= ° + °= ° + °= ° − °= ° + °v i jv i jv i jv i jConservation of momentum:0A A A B B C Cm m m m= + +v v v vDivide by and substitute data.A B Cm m m= =( ) ( ) ( )( )4 cos30 sin30 sin7.4 cos7.4 sin 49.3 cos49.32.1 cos45 sin 45A Bv v° + ° = ° + ° + ° − °+ ° + °i j i j i ji jResolve into components and rearrange.( ) ( )( ) ( ): sin7.4 sin 49.3 4cos30 2.1cos45: cos7.4 cos49.3 4sin30 2.1sin 45A BA Bv vv v° + ° = − °° − ° = − °ijooSolving simultaneously,(a) 2.01 m/sAv =(b) 2.27 m/sBv =
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 14, Solution 23.( ) ( )0190 mi/h east 278.67 ft/sPlace orgin at point of impact.A = =v i0 0 00, 0, 0x y z= = =After impact the motion is projectile motion.20 012t gt= + −r r v j0012gtt−= +r rv jwhere ( ) ( ) ( )1600 ft 2400 ft 400 ft= − +r i j k0 0=r( )( )( ) ( ) ( )01600 2400 400 132.2 1212 12 12 2133.333 ft/s ft/s 33.333 ft/s = − + +   = − 6.80 +v i j k ji j kImpact: Conservation of momentum.( ) ( ) ( ) 00 0A A B B A Bm m m m+ = +v v v( ) ( )00 0A B AB AB Bm m mm m+= −v v v( ) ( )( ) ( ) ( )23000 10000133.333 6.80 33.333 278.6713000 1300021.537 ft/s 12.031 ft/s 58.975 ft/s= − + −= − +i j k ii j kComponents: ( )21.537 ft/s 14.69 mi/h east=i( )58.974 ft/s 40.2 mi/h south=k( )12.031 ft/s 8.20 mi/h down− =j
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 14, Solution 24.Weight of arrow: 2 oz 0.125 lb.Weight of bird: 6 lb.ABWW= ==Conservation of momentum: Let v be velocity immediately after impact.A B A BA BW W W Wg g g++ =v v v( )( ) ( )( )0.125 180 240 6 306.12529.388 3.6735 4.8980A A B BA B A BW WW W W W+ += + =+ += + +j k iv vvi j k( ) 20 01Vertical motion:2yy y v t gt= + −( ) 2 210 45 3.6735 32.2 or 0.22817 2.7950 02t t t t= + − − − =Solving for , 1.7898 st t =Horizontal motion: ,x zx v t z v t= =( )( )( )( )29.388 1.7898 52.6 ft4.8980 1.7898 8.77 ftxz= == =( ) ( )52.6 ft 8.77 ftP = +r i k
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 14, Solution 25.( )( )( )1 2 1 21 2 1 21 2 1 2Position vectors (mm): 80 80 40 12033 70 10 78.03248 15 50.289A A A AB B B BC C C C= + + == − + − == − =i j ki j kj k1 21 21 2Unit vectors: Along , 0.66667 0.66667 0.33333Along , 0.42290 0.89707 0.12815Along , 0.95448 0.29828ABCA AB BC C= + += − + −= −i j ki j kj kλλλλλλλλλλλλVelocity vectors after the collisions:A A A B B B C C Cv v v= = =v v vλλλλ λλλλ λλλλConservation of momentum:0 0 04 4 4 4A B Cm m m m m m+ + = + +u v v v v vDivide by m and substitute data.( )600 750 800 2400 2400 4 4A A B B C Cv v v− + − + + = + +i j k j j λλλλ λλλλ λλλλResolving into components,: 600 0.66667 1.69160: 5550 0.66667 3.58828 3.81792: 800 0.33333 0.51260 1.19312A BA B CA B Cv vv v vv v v− = −= + +− = − −ijkSolving the three equations simultaneously,919.26 m/s, 716.98 m/s, 619.30 m/sA B Cv v v= = =919 m/sAv =717 m/sBv =619 m/sCv =
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 14, Solution 26.(ft):Position vectors 18D =r k/ // // /7.5 7.5 18 19.518 9 18 9 18 2713.5 13.5 18 22.5A A D A DB B D B DC C D C Drrr= − = − − == + = + − == − = − − =r i r i kr i j r i j kr j r j k( )( )( )///1: Along , 7.5 1819.51Along , 18 9 18271Along , 13.5 1822.5A D AB D BC D CUnit vectors = − −= + −= − −r i kr i j kr j kλλλλλλλλλλλλAssume that elevation changes due to gravity may be neglected. Then, the velocity vectors after theexplosition have the directions of the unit vectors.A A A B B B C C Cv v v= = =v v vλλλλ λλλλ λλλλ0Conservation of momentum: A A B B C Cm m m m= + +v v v v( ) ( ) ( ) ( )18 8 6 460 45 1800 7.5 18 18 9 18 13.5 1819.5 27 22.5A B Cv v vg g g g     − − = − − + + − + − −          i j k i k i j k j kMultiply by g and resolve into components.1080 60 10819.5 27810 54 5227 22.532400 144 108 7219.5 27 22.5A BB CA B Cv vv vv v v   = − +         − = −           − = − − −          Solving, 119.94419.5Av= 2340 ft/sAv =76.63527Bv= 2070 ft/sBv =95.16022.5Cv= 2140 ft/sCv =
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 14, Solution 27.(ft):Position vectors 18D =r k/ // // /7.5 7.5 18 19.518 9 18 9 18 2713.5 13.5 18 22.5A A D A DB B D B DC C D C Drrr= − = − − == + = + − == − = − − =r i r i kr i j r i j kr j r j k( )( )( )///1: Along , 7.5 1819.51Along , 18 9 18271Along , 13.5 1822.5A D AB D BC D CUnit vectors = − −= + −= − −r i kr i j kr j kλλλλλλλλλλλλAssume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the explositionhave the directions of the unit vectors.A A A B B B C C Cv v v= = =v v vλλλλ λλλλ λλλλ/ 19.5where 1950 ft/s0.010A DAArvt= = =/ 271500 ft/s0.018B DBBrvt= = =C/ 22.51875 ft/s0.012DCCrvt= = =( ) ( )so that 750 ft/s 1800 ft/sA = − −v i k( ) ( ) ( )1000 ft/s + 500 ft/s 1000 ft/sB = −v i j k( ) ( )1125 ft 1500 ft/sC = − −v j kConservation of momentum: 0 A A B B C Cm m m m= + +v v v v( ) ( ) ( )0 750 1800 1000 500 1000 1125 1500A B CW W W Wvg g g g − = − − + + − + − −  k i k i j k j k
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.Divide by g and resolve into components.0: 0 750 1000: 0 500 1125: 1800 1000 1500A BB CA B CW WW WWv W W W= − += −− = − − −ijk(1)(2)(3)Since mass is conserved, 6 lbA B CW W W W= + + = (4)Solving equations (1), (2), and (4) simultaneously,(a) 2.88 lb, 2.16 lb, 0.96 lbA B CW W W= = =substituting into (3),( )( ) ( )( ) ( )( )06 1800 2.88 1000 2.16 1500 0.96v− = − − −(b) 0 1464 ft/sv =
34. 34. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 28.( )( )( ) ( )111 1From Eq. (14.7),nO i i iini i iin ni i i i ii iGmmm v r mm=== == ×′ = + × ′= × + ×= × +∑∑∑ ∑H r vr r vr vr v H
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 14, Solution 29.( )( )( )( )( )( )111 111if, and only if, 0i A inA i i iini i A iin ni i A i i ii ini i A Aini i A A AiA A AA A A Ammm mmmmm=== ===′= +′= ×′ ′= × +′ ′= × + ×′ ′= × +′= − × +′= − × +′= − × =∑∑∑ ∑∑∑v v vH r vr v vr v r vr v Hr r v Hr r v HH H r r vThis condition is satisfied if,( ) 0 Point has zero velocity.or ( ) Point coincides with the mass center.or ( ) is parallel to . Velocity is directed along line .AAA A Aa Ab Ac AG==−vr rv r r v
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 14, Solution 30.From equation (1), ( )1nA i i iim=′ ′ ′= ×∑H r v( ) ( )1nA i A i i Aim=′  = − × − ∑H r r v vDifferentiate with respect to time.( ) ( ) ( ) ( )1 1n nA i A i i A i A i i Ai im m= =′    = − × − + − × −   ∑ ∑H r r v v r r v v& & & & &But , , , andi i i i A A A A= = = =r v v a r v v a& && &( ) ( )( ) ( )( ) ( )( )111 1Hence, 0nA i A i i Aini A i i Ain ni A i i i A Ai iA A Ammmm r=== =′  = + − × −  = − × −    = − × − − ×   = − − ×∑∑∑ ∑H r r a ar r F ar r F r r aM r a&( )if, and only if, 0A A A AM m′ = − × =H r r a&This condition is satisfied if( ) 0 The frame is newtonian.or ( ) Point coincides with the mass center.or ( ) is parallel to . Acceleration is directed along line .AAA A Aab Ac AG==−ar ra r r a
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 14, Solution 31.The masses are m for the bullet and Am and Bm for the blocks.The bullet passes through block A and embeds in block B. Momentum is conserved.( ) ( )0 0Initial momentum: 0 0A Bmv m m mv+ + =Final momentum: B A A B Bmv m v m v+ +0Equating, B A A B Bmv mv m v m v= + +( )( ) ( )( ) 303 3 2.5 543.434 10 kg500 5A A B BBm v m vmv v−++= = = ×− −The bullet passes through block A. Momentum is conserved.( )0 0Initial momentum: 0Amv m mv+ =1Final momentum: A Amv m v+0 1Equating, A Amv mv m v= +( )( ) ( )( )301 343.434 10 500 3 3292.79 m/s43.434 10A Amv m vvm−−× −−= = =×(a) Bullet passes through block A. Kinetic energies.( )( )22 30 01 1Before: 43.434 10 500 5429 J2 2T mv −= = × =( )( ) ( )( )2 22 2 31 11 1 1 1After: 43.434 10 292.79 3 3 1875 J2 2 2 2A AT mv m v −= + = × + =0 1Lost: 5429 1875 3554 JT T− = − = energy lost 3550 J=(b) Bullet becomes embedded in block B. Kinetic energies.( )( )22 32 11 1Before: 43.434 10 292.79 1861.7 J2 2T mv −= = × =( ) ( )( )2231 1After: 2.54343 5 31.8 J2 2B BT m m v= + = =2 3Lost: 1862 31.8 1830 JT T− = − = energy lost 1830 J=
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 14, Solution 32.Data and results from Prob. 14.1.Masses: 1350 kg,A Bm m= = 5400 kgcm =Initial velocities: 0 0( ) ( ) 0,A Bv v= = 0( ) 8 km/h = 2.2222 m/scv =Velocities after first collision:1( ) 0,Av = 1 1( ) ( ) 1.77778 m/sB cv v= =Velocities after second collision2.9630 m/s,Av = 1.18519 m/sB cv v= =Initial kinetic energy: 2 2 20 0 0 01 1 1( ) ( ) ( )2 2 2A A B B c BT m v m v m v= + +2 3010 0 (5400)(2.2222) 13.3333 10 J2T = + + = ×Kinetic energy after the first collision:( )22 21 1 111 1 1( ) ( )2 2 2A A B B c cT m v m v m v= + +2 2 31 10 (1350)(1.77778) (5400)(1.77778) 10.6667 10 J2 2= + + = ×Kinetic energy after the second collision:2 2 221 1 12 2 2A A B B c cT m v m v m v= + +2 2 2 31 1 1(1350)(2.9630) (1350)(1.18519) (5400)(1.18519) 10.6668 10 J2 2 2= + + = ×Kinetic energy lost in first collision: 30 1 2.6667 10 JT T− = ×2.67 kJKinetic energies before and after second collision:2 1 10.67 kJT T= =
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 14, Solution 33.( )( )22 20004000 3700The masses are 124.2 slugs, 114.9 slugs, and 1366.5 slugs32.2 32.2 32.2A B Fm m m= = = = = =Let , , and be the sought after velocities in ft/s, positive to the right.A B Fv v vInitial values: ( ) ( ) ( )0 0 00.A B Fv v v= = =Initial momentum of system: ( ) ( ) ( )0 0 00.A A B B F Fm v m v m v+ + =There are no horizontal external forces acting during the time period under consideration. Momentum isconserved.0 A A B B F Fm v m v m v= + +124.2 114.9 1366.5 0A B Fv v v+ + = (1)The relative velocities are given as//7 ft/s3.5 ft/sA F A FB F B Fv v vv v v= − = −= − = −(2)(3)Solving (1), (2), and (3) simultaneously,6.208 ft/s, 2.708 ft/s, 0.7919 ft/sA B Fv v v= − = − =( ) ( ) ( )22 21 0 0 01 1 1Initial kinetic energy: 02 2 2A A B B CT m v m v v= + + =2 2 221 1 1Final kinetic energy:2 2 2A A B B C CT m v m v m v= + +( )( ) ( )( ) ( )( )2 2 221 1 1124.2 6.208 114.9 2.708 1366.5 0.79192 2 23243 ft lbT = + += ⋅Work done by engines: 1 1T U+ 2 2T=1U 2 2 1 3243 ft lbT T= − = ⋅ 1U 2 3240 ft lb= ⋅
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 14, Solution 34.From the solution to Prob. 14.27,Initial velocity of 6-lb shell: 0 1464 ft/sv =Weights of fragments: 2.88 lb,AW = 2.16 lb,BW = 0.96 lbcW =Speeds of fragments: 1950 ft/s,Av = 1500 ft/s,Bv = 1875 ft/scv =Total kinetic energy before the explosion.( )22 30 01 1 61464 199.69 10 ft lb2 2 32.2WT vg = = = × ⋅  Total kinetic energy after the explosion.2 2 211 1 12 2 2A B cA B cW W WT v v vg g g= + +( ) ( ) ( )2 2 21 2.88 1 2.16 1 0.961950 1500 18752 32.2 2 32.2 2 32.2     = + +          3297.92 10 ft lb= × ⋅Increase in kinetic energy. 31 0 98.2 10 ft lbT T− = × ⋅
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 14, Solution 35.Velocity of mass center: ( )A B A A B Bm m m m+ = +v v vA A B BA Bm mm m+=+v vvVelocities relative to the mass center:( )( )B A BA A B BA A AA B A BA A BA A B BB B BA B A Bmm mm m m mmm mm m m m−+′ = − = − =+ +−+′ = − = − =+ +v vv vv v v vv vv vv v v vEnergies:( ) ( )( )( ) ( )( )222212 212 2A B A B A BA A A AA BA B A B A BB B B BA Bm mE mm mm mE mm m− ⋅ −′ ′= ⋅ =+− ⋅ −′ ′= ⋅ =+v v v vv vv v v vv v( ) :a Ratio / /A B B AE E m m=( ) 135 km/h 37.5 m/sAb = =v , 90 km/h 25 m/sB = =v62.5 m/sA B− =v v( )( ) ( )( )( )2 2322400 1350 62.5607.5 10 J2 3750AE = = × 608 kJAE =( ) ( )( )( )( )2 2622400 1350 62.51.08 10 J2 3750BE = = × 1080 kJBE =
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 14, Solution 36.( ): A B A A B BVelocity of mass center m m m m+ = +v v vA A B BA Bm mm m+=+v vvVelocities relative to the mass center:( )( )B A BA A B BA A AA B A BA A BA A B BB B BA B A Bmm mm m m mmm mm m m m−+′ = − = − =+ +−+′ = − = − =+ +v vv vv v v vv vv vv v v vEnergies:( ) ( )( )( ) ( )( )222212 212 2A B A B A BA A A AA BA B A B A BB B B BA Bm mE mm mm mE mm m− ⋅ −′ ′= ⋅ =+− ⋅ −′ ′= ⋅ =+v v v vv vv v v vv v( ) ( )2 20 00 01 1Energies from tests: ,2 2A A B BE m v E m v= =( )( ) ( )( )( )( ) ( )( )22 20 022 20 0Serverities: B A B A BAAA A BA A B A BBBB A BmESE m m vmESE m m v− ⋅ −= =+− ⋅ −= =+v v v vv v v v:Ratio22A BB AS mS m=
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 14, Solution 37.(a) A strikes B and C simultaneously.During the impact, the contact impulses make 30° angles with the velocity 0v( )( )Thus, cos30 sin30cos30 sin30B BC Cvv= ° + °= ° − °v i jv i jBy symmetry, A Av=v i0Conservation of momentum: A B Cm m m m= + +v v v v0component: 0 0 sin30 sin30component: cos30 cos30B C C BA B Cy mv mv v vx mv mv mv mv= + ° − ° == + ° + °( )0 002,cos30 3 3A AB C A B Cv v v vv v v v v v− −+ = = − = =oConservation of energy: 2 2 2 201 1 1 12 2 2 2A B Cmv mv mv mv= + +( )( )( ) ( )( )22 20 022 20 0 0 00 0 0 00 023232 1 5 13 3 3 56 2 355 3A AA A A AA A A AB Cv v v vv v v v v v v vv v v v v v v vv v v v= + −− = − + = −+ = − = − = −= = =00.200A v=v00.693B v=v 30°
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.(b) A strikes B before it strikes C.First impact; A strikes B.During the impact, the contact impulse makes a 30° angle with the velocity 0.v( )Thus, cos30 sin30B Bv= ° + °v i j0Conservation of momentum. A Bm m m= +v v v( ) ( )( ) ( )0 0component: 0 sin30 sin30component: cos30 cos30A B A By yA B A Bx xy m v mv v vx v m v mv v v v′ ′= + ° = − °′ ′= + ° = − °Conservation of energy:( ) ( )( ) ( )( )2 22 202 2 202 2 2 2 2 20 01 1 1 12 2 2 21 1 1cos30 sin302 2 212 cos30 cos 30 sin 302A A Bx yB B BB B B Bmv m v m v mvm v v v vm v v v v v v′ ′= + += − ° + ° += − ° + ° + ° +( )( )20 0 0 00 03 1cos30 , sin 30 ,2 43cos30 sin304B A xA yv v v v v vv v v′= ° = = ° =′ = − ° ° = −Second impact: A strikes C.During the impact, the contact impulse makes a 30oangle with the velocity 0.v( )Thus, cos30 sin30C Cv= ° − °v i jConservation of momentum: A A Cm m m′ = +v v v
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.( ) ( )( ) ( )( ) ( )( ) ( )00component: cos30 ,1cos30 cos304component: sin303sin30 sin304A A Cx xA A C Cx xA A Cy yA A C Cy yx m v m v mvv v v v vy m v m v mvv v v v v′ = + °′= − ° = − °′ = − °′= + ° = − + °Conservation of energy:( ) ( ) ( ) ( )2 2 2 2 2222 2 20 0 0 02 2 20 02 2 2 20 01 1 1 1 12 2 2 2 21 1 3 1 1 3cos30 sin302 16 16 2 4 41 1 1cos30 cos 302 16 23 3sin30 sin 3016 2A A A A Cx y x yC C CC CC C Cm v m v m v m v mvm v v m v v v v vm v v v vv v v v v′ ′+ = + +      + = − ° + − + ° +             = − ° + °+ − ° + ° + 201 30 cos30 sin30 22 2C Cv v v = − ° + ° +   ( )( )0 00 0 00 0 01 3 3cos30 sin304 4 41 3 1cos304 4 83 3 3sin304 4 8CA xA yv v vv v v vv v v v = ° + ° =   = − ° = −= − + ° = −00.250A v=v 60°00.866B v=v 30°00.433C v=v 30°
46. 46. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 38.(a) Velocity of B at maximum elevation. At maximum elevation ball B is at rest relative to cart A. B A=v vUse impulse-momentum principle.components:x ( )0 0A A A B B A B Bm v m v m v m m v+ = + = +0ABA Bm vvm m=+(b) Conservation of energy:( )( )21 0 12 22 2 2 0221, 021 1 12 2 2 2AAA A B B A B BA BBT m v Vm vT m v m v m m vm mV m gh= == + = + =+=( )2 2 1 12 220012 2AB AA BT V T Vm vm gh m vm m+ = ++ =+2 22 0012AAB A Bm vh m vm g m m = − + 202AA Bm vhm m g=+
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 14, Solution 39.Velocity vectors: ( )0 0 cos30 sin30v= ° − °v i j 0 15 ft/sv =A Av= −v j( )( )sin30 cos30cos30 sin30B BC Cvv= ° − °= ° + °v i jv i jConservation of momentum:0 A B Cm m m m= + +v v v vDivide by m and resolve into components.i: 0 cos30 sin30 cos30B Cv v v° = ° + °j: 0 sin30 cos30 sin30A B Cv v v v− ° = − − ° + °Solving for and ,B Cv v( ) ( )0 03 12 2B A C Av v v v v v= − = +Conservation of energy: 2 2 2 201 1 1 12 2 2 2A B Cmv mv mv mv= + +Divide by 12m and substitute for and .B Cv v( ) ( )2 22 20 0 02 20 03 14 42A A AA Av v v v v vv v v v= + − + += + −017.5 ft/s2Av v= = 7.50 ft/sAv =( )315 7.5 6.4952 ft/s2Bv = − = 6.50 ft/sBv =( )115 7.5 11.25 ft/s2Cv = + = 11.25 ft/sCv =
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 14, Solution 40.Velocity vectors: ( )0 0 cos45 sin 45v= ° + °v i j 0 15 ft/sv =A Av=v j( )( )sin60 cos60cos60 sin60B BC Cvv= ° − °= ° + °v i jv i jConservation of momentum:0 A B Cm m m m= + +v v v vDivide by m and resolve into components.i: 0 cos45 sin 60 cos60B Cv v v° = ° + °j: 0 sin 45 cos60 sin 60A B Cv v v v° = − ° + °Solving for and ,B Cv v0 00.25882 0.5 0.96593 0.86603B A C Av v v v v v= − = −Conservation of energy: 2 2 2 201 1 1 12 2 2 2A B Cmv mv mv mv= + +Divide by m and substitute for and .B Cv v( ) ( )2 22 20 0 02 20 00.25882 0.5 0.96593 0.866031.4142 2A A AA Av v v v v vv v v v= + + + −= + +00.70711 10.607 ft/sAv v= = 10.61 ft/sAv =00.61237 9.1856 ft/sBv v= = 9.19 ft/sBv =00.35356 5.303 ft/sCv v= = 5.30 ft/sCv =
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 14, Solution 41.1sin ,3θ =8cos ,3θ = 19.471θ = °Velocity vectors 0 0v= −v j( )cos sinA Av θ θ= −v i j( )/ sin cosB A Bu θ θ= − −v i j/B A B A= +v v vC Cv=v jConservation of momentum: 0 /2A B C A B A Cm m m m m m m= + + = + +v v v v v v vDivide by m and resolve into components.i: 0 2 cos sinA Bv uθ θ= −−j: 0 2 sin cosA B Cv v u vθ θ= + + −Solving for and ,A Bv u ( ) ( )0 010.942816A C B Cv v v u v v= + = +Conservation of energy: 2 2 2 201 1 1 12 2 2 2A B Cmv mv mv mv= + +( )2 2 2 21 1 12 2 2A A B Cmv m v u mv= + + +Divide by 12m and substitute for and .A Bv u( ) ( ) ( )22 222 20 0 012 0.942816C C Cv v v v v v = + + + +  ( )22 20 0 00.94445 0.02857C C Cv v v v v v− = + = 00.0286C v=v[0 00.17143 0.17143A Av v v= =v ]19.471° , 00.1714A v=v 19.5°[0 / 00.96975 0.96975B B Au v v= v ]19.471°/B A B A= +v v v 0.985B =v 80.1°
50. 50. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 42.1 8sin , cos , 19.4713 3θ θ θ= = = °C strikes B.Conservation of momentum:0 0orB C B Cm m m v v v′ ′= + = −v v vConservation of energy:( )22 201 1 12 2 2B Cmv m v mv′= +( )22 20 0 C Cv v v v= − +0Cv =0Bv v′ =Cord becomes taut.Velocity vectors:A Av=v θ/B A Bu=v θConservation of momentum: /B A A B Am m m m′ = + +v v v vDivide by m and resolve into components.+ :θ sin 2B Av vθ′ = 01 1sin2 6A Bv v vθ′= =
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.(a) 06Av=v 19.5°+ :θ cosB Bv uθ′ = 0cos 83B Bvu v θ′= =0119.4716B v = °  v [ 00.94281v+ ]19.471°[ 00.95743B v=v ]80.8° 00.957B v=v 80.5°0C =vInitial kinetic energy: 21 012T mv=Final kinetic energy: 2 2 221 1 12 2 2A B CT mv mv mv= + +( ) ( )222 20 01 1 1.95743 0 0.944442 6 2mv mv  = + + =     (b) Fraction lost: 1 211 0.944440.055551T TT− −= =Fraction of energy lost = 0.0556
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 14, Solution 43.(a) Use part (a) of sample Problem 14.4 with A Bm m m.= =0 012mv v vm m= =+012A Bv v v= =(b) Consider initial position and position when 0θ = again.Conservation of momentum0 A A B Bmv m v m v= +0B Av v v+ = (1)Conservation of energy2 2 201 1 12 2 2A Bmv mv mv= + (2)Substituting (1) into (2),2 2 21 1 1( )2 2 2A B A Bm v v mv mv+ = +0A Bmv v =Either 0Av = with 0Bv v=or 0Bv = with 0Av v=(c) Consider positions when maxθ θ= and min.θ θ=Since / 0B Av = at these statesB Av v=Conservation of momentum0 A Bmv mv mv= +012B Av v v= =
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.Conservation of energy would show that the elevation of B is the same for maxθ θ= and min.θ θ=Both A and B keep moving to the right with A and B stopping intermittently.
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 14, Solution 44.Relative velocity and acceleration./B A B A= + =v v v [vA ] + [vB/A 30° ]/B A B A= + =a a a [aA ] + [aB/A 30° ]Draw free body diagrams and apply Newton’s second law.Block::F ma∑ = 1 cos30 sin30B B AN m g m a− °= − ° (1):F ma∑ = 1 /sin30 cos30B B A B AN m a m a°= °− (2)Wedge::F ma∑ = 1 sin30 A AN m a°= (3)Rearranging (1), (2), and (3) and applying numerical data,1 (6sin30 ) (6)(9.81)cos30AN a+ ° = ° (1)1 /(sin30 ) 6 (6cos30 ) 0A B AN a a° + − ° = (2)1(sin30 ) 10 0AN a° − = (3)Solving (1), (2), and (3) simultaneously,1 44.325N,N = 22.2163 m/s ,Aa = 2/ 6.8243 m/sB Aa =
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.Sliding motion of block relative to wedge.2// /( )2B AB A B Ava s=/ / /2 (2)(6.8243)(1.0) 3.6944 m/sB A B A B Av a s= = =v 3.69440.54136 s6.8243B/AB/Ata= = =Motion of wedge.(2.2163)(0.54136) 1.1998 m/sA Av a t= = =(a) Velocity of B relative to A. / 3.69 m/sB A =v 30°(b) Velocity of A . 1.200 m/sA =v
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 14, Solution 45.There are no external forces. Momentum is conserved.(a) Moments about D : ( )00.9 1.8 0.9A C C A B Bm v m v m m v= + +( )( )( ) ( )00.90.90.5 12 2.5 3.501.8 1.8A BAC BC Cm mmv v vm m+= − = − = 3.50 m/sCv =Moments about C : ( )00.9 0.9 1.8A A B B D Dm v m m v m v= + +( )( )( ) ( )( )0 0.90.90.25 12 0.5 2.5 1.750m/s1.8 1.8A BAD BD Dm mm vv vm m+= − = − = 1.750 m/sDv =(b) Initial kinetic energy:( )221 01 17.5 12 540 N m2 2AT m v= = = ⋅Final kinetic energy:( ) ( ) ( )2 2 222 2 21 1 1( )2 2 21 1 115 2.5 7.5 3.5 15 1.750 115.78 N m2 2 2A B B C C D DT m m v m v m v= + + += + + = ⋅Energy lost: 540 115.78 424.22 N m− = ⋅Fraction of energy lost424.220.786540= =( )1 210.786T TT−=
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 14, Solution 46.There are no external forces. Momentum is conserved.(a) Moments about D : 00.9 1.8 0.9A C C B Bm v m v m v= +( )( ) ( )( )00.9 0.90.5 12 0.5 3.5 4.251.8 1.8A BC BC Cm mv v vm m= − = − = 4.25 m/sCv =Moments about C : 00.9 1.8 0.9A D D B Bm v m v m v= +( )( ) ( )( )00.9 0.90.25 12 0.25 3.5 2.125 m/s1.8 1.8A BD BD Dm mv v vm m= − = − = 2.13 m/sDv =(b) Initial kinetic energy:( )221 01 17.5 12 540 N m2 2AT m v= = = ⋅Final kinetic energy:( ) ( ) ( )2 2 222 2 21 1 12 2 21 1 17.5 3.5 7.5 4.25 15 2.125 147.54 N m2 2 2B B C C D DT m v m v m v= + += + + = ⋅Energy lost: 540 147.54 392.46 N m− = ⋅Fraction of energy lost392.460.727540= =( )1 210.727T TT−=
58. 58. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 47.(a) Linear and angular momentum.0 0A A B Bm m mv= + = +L v v i0mv=L i02( ) ( ) 03 3Gl lmv= × + − ×H j i j i023G lmv= −H kMotion of mass center G. Since there is no external force,0 constantA A B Bm m mv= + = =L v v i03m mv=v i 013v=v iMotion about mass center.( ) ( )G G i i i A A A B B Bm m m′ ′ ′ ′ ′ ′ ′= = Σ × = × + ×H H r v r v r vwhere2 13 3A Bl , l′ ′ ′ ′= = −r j r j2 1,3 3A Bl lθ θ′ ′ ′ ′= = −v i v i& &Thus,2 2 1 1 123 3 3 3 3G l ml l m lθ θ       ′ ′ ′ ′= × + − × ⋅              H j i j i& &223ml θ= − k&But HG is constant.2 002 23 3vml lmvlθ θ− = − =k k &02 23 3Av l vθ′ = =&01 13 3Bv l vθ′ = =&
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.(b) After 180º rotation.0 01 23 3A A v v′= + = −v v v i i013A v= −v i0 01 13 3B B v v′= + = +v v v i i023B v=v i
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 14, Solution 48.Masses: 21253.882 lb s /ft, 2 , 3 .32.2A B A C Am m m m m= = ⋅ = =Conservation of angular momentum about O.240 240 2160 ( ) ( ) ( )A A A x A y A zv v v= + + = + +r i j k v i j k600 1320 3240 500 1100 2200B B= + + = + +r i j k v i j k480 960 1920 400 ( ) ( )C C C y C zv v= − − + = − + +r i j k v i j kSince the three parts pass through O, the angular momentum about O is zero. 0 0=H0 0A A A B B B C C Cm m m= × + × + × =H r v r v r v[ 2 3 ] 0A A A B B C Cm × + × + × =r v r v r vDividing by mA and using determinant form,240 240 2160 1200 2640 6480 1440 2880 5760( ) ( ) ( ) 500 1100 2200 400 ( ) ( )A x A y A z C y C zv v v v v+ + − −−i j k i j k i j k[240( ) 2160( ) ] [2160( ) 240( ) ]A z A y A x A zv v v v= − + −i j6 6[240( ) 240( ) ] 1.320 10 0.6 10A y A xv v+ − − × + ×k i j0 [ 2880( ) 5760( ) ]C z C yv v+ + − −k i6 6[1440( ) 2.304 10 ] [ 1440( ) 1.152 10 ] 0C z C yv v+ − × + − − × =j kDividing by 240 and equating to zero the coefficients of i, j, and k,: ( ) 9( ) 5500 12( ) 24( ) 0A z A y C z C yv v v vi − − − − = (1): 9( ) ( ) 7100 6( ) 0A x A z C zv v vj − − + = (2): ( ) ( ) 6( ) 4800 0A y A x C yv v vk − − − = (3)Conservation of linear momentum.0( )A A B C C C A B Cm m m m m m+ + = + +v v v v0( 2 3 ) 6 ( )A A C C Am m+ + =v v v v
61. 61. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Dividing by mA and substituting given data,( ) + ( ) ( ) (2)(500 1100 2200 ) (3)[ 400 + ( ) ( ) ] (6)(1500)A x A y A z C y C zv v v + v + v+ + + + − =i j k i j k i j k kSeparating into components,: ( ) 1000 1200 0A xvi + − = (4): ( ) 2200 3( ) 0A y C yv vj + + = (5): ( ) 4400 3( ) 9000A z C zv vk + + = (6)From (4), ( ) 200 ft/sA xv =Solving (3) and (5) simultaneously,( ) 200 ft/s ( ) 800 ft/sA y C yv v= = −Solving (2) and (6) simultaneously,( ) 1300 ft/s ( ) 1100 ft/sA z C zv v= =(200 ft/s) (200 ft/s) (1300 ft/s)A = + +v i j k
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 14, Solution 49.Let the system consist of spheres A and B.State 1. Instant cord DC breaks.( ) 013 12 2Am mv = − −   v i j( ) 013 12 2Bm mv = −   v i j( ) ( )1 01 1A Bm m mv= + = −L v v j1012 2vm= = −Lv jMass center lies at point G as shown.( ) ( ) ( )1 1103 32 232G A Bl m l mlmv= × + − ×=H j v j vk2 2 21 0 0 01 12 2T mv mv mv= + =State 2. The cord is taut. Conservation of linear momentum:(a) 012D v= = −v v j 00.500Dv v=Let ( )2andA A B B= + = +v v u v v u2 12 A Bm m m= + + =L v u u LB A B Au u= − =u u( )22G A B Almu lmu lmu= + =H k k k
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.(b) Conservation of angular momentum:( ) ( )2 1G G=H H0322Almu lmv=k k 034A Bu u v= =00.750u v=( ) 2 2 222 20 01 1 122 2 21 1 9 9 132 2 16 16 16A BT m v mu mumv mv= + + = + + =  (c) Fraction of energy lost:131 2 1611 31 16T TT−−= =1 210.1875T TT−=
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 14, Solution 50.The system is spheres A and B and the ring D.Initial velocities: ( )0 cos30 sin30A v= − ° − °v i j( )0 cos30 sin300BDv= − ° − °=v i jvLocate the mass center.( ) ( )004sin30 cos30 sin30 cos3014A Bm m mml mll= += − ° + ° + − ° − °= −r r ri j i jr iVelocity of mass center.( ) ( )0 004cos30 sin30 cos30 sin3014A Bm m mmv mvv= += − ° − ° + ° − °= −v v vi j i jv j(a) Motion of mass center 0 t= +r r v01 14 4l v t= − −r i j(b) / /G A G A B G Bm m= × + ×H r v r v( )( )001cos30 cos30 sin3041cos30 cos30 sin304l l mvl l mv = − + ° × − ° − °   + − − ° × ° − °  i j i ji j i j074G lmv=H k(c) 2 2 201 12 2A BT mv mv mv= + = 20T mv=
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 14, Solution 51.Let m be the mass of one ball.Conservation of linear momentum: 0( ) ( )m mΣ = Σv v0 0 0( ) ( ) ( )A B C A B Cm m m m m m+ + = + +v v v v v vDividing by m and applying numerical data,(0.5 ft/s) [(3.75 ft/s) ( ) ] [( ) ( ) ] (6.5 ft/s) 0 0B y C x C yv v v+ + + + = + +i i j i j iComponents:: 0.5 3.75 ( ) 6.5C xx v+ = + ( ) 2.25 ft/sC xv =: ( ) ( ) 0B y C yy v v+ = (1)Conservation of angular momentum about O:0[ ( )] [ ( )]m mΣ × = Σ ×r v r vwhere rA = 0, rB = 0, (1.5 ft)(cos30 sin30 )C = ° + °r i j( )( )1.5 cos30 + sin 30 [ ( ) ( ) ] 0C x C ym v m vi j i j° ° × + =Since their cross product is zero, the two vectors are parallel.( ) ( ) tan30 2.25tan30 1.2990 ft/sC y C xv v= ° = ° =From (1), ( ) 1.2990 ft/sB yv = −( ) 1.299 ft/sB yv = −(2.25 ft/s) (1.299 ft/s)C +=v i j
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 14, Solution 52.Let m be the mass of one ball.Conservation of linear momentum: 0( ) ( )m mΣ = Σv v0 0 0( ) ( ) ( )A B C A B Cm m m m m m+ + = + +v v v v v vDividing by m and applying numerical data,0 [(6 ft/s) ( ) ] [( ) ( ) ] (8 ft/s) 0 0B y C x C yv v v+ + + + = + +i j i j iComponents:: 6 ( ) 8C xx v+ = ( ) 2 ft/sC xv =: ( ) ( ) 0B y C yy v v+ = (1)Conservation of angular momentum about O:0[ ( )] [ ( )]m mΣ × = Σ ×r v r vwhere rA = 0, rB = 0, (1.5 ft)(cos45 sin 45 )C = ° + °r i j(1.5)(cos45 sin 45 ) [ ( ) ( ) ] 0C x C ym v m v° + ° × + =i j i jSince their cross product is zero, the two vectors are parallel.( ) ( ) tan 45 2tan 45 2 ft/sC y C xv v= ° = ° =From (1), ( ) 2 ft/sB yv = −( ) 2.00 ft/sB yv = −(2.00 ft/s) (2.00 ft/s)C +=v i j
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 14, Solution 53.Conservation of linear momentum: 0A B A BA BW W W Wg g g g + = +  v v vAfter multiplying by g, ( ) ( ) ( )7.2 5.76 1.44 4.8 2.4 2.4A B Bx yv v v+ = + +i j j i ji: ( )41.472 2.4 B xv= ( ) 17.28 ft/sB xv =j: ( )10.348 4.8 2.4A B yv v= − ( ) 4.32 2B Ayv v= −Speeds relative to the mass center: ( ) ( )( )1 13 8 8 ft/s3 3Au lω= = =( ) ( )( )2 23 8 16 ft/s3 3Bu lω= = =Initial kinetic energy: ( ) ( )2 2 2 21 0 01 1 12 2 2A B A BA Bx yW W W WT v v u ug g g g  = + + + +    ( ) ( ) ( )2 22 211 7.2 1 4.8 1 2.45.76 1.44 8 16 18.2517 ft lb2 32.2 2 32.2 2 32.2T     = + + + = ⋅          Final kinetic energy: ( ) ( )2 2221 1 12 2 2A B BA B Bx yW W WT v v vg g g= + +( ) ( )2 2221 4.8 1 2.4 1 2.417.28 4.32 22 32.2 2 32.2 2 32.2A AT v v     = + + −          20.2236 0.6440 11.8234A Av v= − +Conservation of energy: 1 2T T=(a) 20.2236 0.6440 6.4283 0, 6.9919 ft/sA A Av v v− − = = 6.99 ft/sA =v( ) ( )( ) ( ) ( )4.32 2 6.9919 9.6638 ft/s 17.28 ft/s 9.6638 ft/sB Byv = − = − = −v i j19.80 ft/sB =v 29.2°
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.Conservation of angular momentum about O:( ) ( )( ) ( )( ) ( ) ( )( ) ( )( )0 0 0 0123 37.2 4.8 2.47.44 5.76 1.0 8 2.0 16 6.0047 ft lb s32.2 32.2 32.2A B B BO A B G A Bx xW W W l W lH y m m v H y v u ug g g g = − + + = − + + +       = − + + = − ⋅ ⋅          ( ) ( )2( )A BO A A B B A B yyW WH m v a m v b v a v bg g= + = +4.8 2.4(6.9919) ( 9.6638)(24) 1.0423 17.286832.2 32.2a a   = + − = −      ( ) ( )2 11.0423 17.2868 6.0047O OH H a= − = −(b) 10.82 fta = 10.82 fta =
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 14, Solution 54.Conservation of linear momentum: 0A B A BA BW W W Wg g g g + = +  v v v( ) ( )09 6 37.68 10.8 6.7232.2 32.2 32.2   = + −      v j i j(a) ( ) ( )0 3.6 ft/s 2.88 ft/s= +v i j 0 4.61 ft/s=v 38.7°Let Al be the distance from G to A and Bl be the distance from G to B.or 2A B AA B B A ABW W Wl l l l lg g W= = =Let ω be the spin rate.Initial kinetic energy: ( ) ( )2 221 01 1 12 2 2A B A BA BW W W WT v l lg g g gω ω = + + +  ( ) ( ) ( )( )2 22 2121 9 1 6 1 33.6 2.88 22 32.2 2 32.2 2 32.22.9703 0.27950A AAT l llω ωω     = + + +          = +Final kinetic energy: 2 221 12 2A BA BW WT v vg g= +( ) ( )2 2 221 6 1 37.68 10.8 6.72 13.03242 32.2 2 32.2T   = + + =      Conservation of energy: 2 1T T= .( ) ( )213.0324 2.9703 0.27950 6.000 ft/sA Al lω ω= + =Conservation of angular momentum about O:( ) ( ) ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )( )0 0 0 01229 6 30 7.5 3.6 2 232.2 32.2 32.27.5466 0.55901 7.5466 3.3540A B A BO A A B By xA A AA AW W W WH x v y v l l l lg g g gl l ll lω ωω ωω  = + − + +         = − + +           = − + = − +( ) ( ) ( )( ) ( )( )( ) ( )22 16 37.68 5.58 6.72 21.632.2 32.25.5382 ft lb s: 5.5382 7.5466 3.3540 0.600 ftA BO A B yO O A AW WH v a v bg gH H l l   = + = + −      = − ⋅ ⋅= − = − + =(b) 2 1.200 ftB A A Bl l l l l= = = + 1.800 ftl =(c)6.0010.000.600AAllωω = = = 10.00 rad/sω =
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 14, Solution 55.Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.Before impacts: ( ) ( ) ( )00 0 04 , 0A B Cv= = = =v i i v vAfter impacts: ( ) ( )1.92 , ,A B B B C Cx yv v v= − = + =v j v i j v iConservation of linear momentum: 0 A B C= + +v v v vi: ( ) ( )4 0 4B C B Cx xv v v v= + + = −j: ( ) ( )0 1.92 0 1.92B By yv v= − + + =Conservation of energy: 2 2 2 201 1 1 12 2 2 2A B Cv v v v= + +( ) ( ) ( ) ( )22 2 2 21 1 1 1 14 1.92 1.92 42 2 2 2 2C Cv v= + + − +24 3.6864 0C Cv v− + =( ) ( )( )24 4 4 3.68642 0.56 2.56 or 1.442Cv± −= = ± =Conservation of angular momentum about :B′( )( )( ) ( )( )00.75 1.80.75 4 1.8 1.65 1.92 2.7122.712A CCCv a v cvcvcv= − += − − ==If 1.44,Cv = 1.8833 off the table. Rejectc =If 2.56,Cv = 1.059c =Then, ( ) 4 2.56 1.44, 1.44 1.92B Bxv = − = = +v i jSummary.(a) 2.40 m/sB =v 53.1°2.56 m/sC =v(b) 1.059 mc =
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 14, Solution 56.Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.Before impacts: ( ) ( ) ( )00 0 05 , 0A B Cv= = = =v i i v vAfter impacts: ( ) ( ), , 3.2A A B B B Cx yv v v= − = + =v j v i j v iConservation of linear momentum: 0 A B C= + +v v v vi: ( ) ( )5 0 3.2 1.8B Bx xv v= + + =j: ( ) ( )0 0A B B Ay yv v v v= − + + =Conservation of energy: 2 2 2 201 1 1 12 2 2 2A B Cv v v v= + +( ) ( ) ( ) ( ) ( )2 2 2 2 21 1 1 1 15 1.8 3.22 2 2 2 2A Av v= + + +(a) 25.76 2.4A Av v= = 2.40 m/sA =v( ) 2.4B yv = 1.8 2.4B = +v i j 3.00 m/sB =v 53.1°Conservation of angular momentum about :B′( )00.75 1.8 A Cv a v cv= − +01.8 0.75A A Cav v cv v= + −( )( ) ( )( ) ( )( )1.8 2.4 1.22 3.2 0.75 5 4.474= + − =(b)4.474 4.4742.4Aav= = 1.864 ma =
72. 72. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 57.Use a frame of reference that is translating with the mass center G of the system. Let 0v be its velocity.0 0v=v iThe initial velocities in this system are ( ) ( )0 0,A B′ ′v v and ( )0,C′v each having a magnitude of .lω They aredirected 120° apart. Thus,( ) ( ) ( )0 0 00A B C′ ′ ′+ + =v v v(a) Conservation of linear momentum:( ) ( ) ( ) ( ) ( ) ( )0 0 00 0 0A B C A B Cm m m m m m′ ′ ′+ + = − + − + −v v v v v v v v v( ) ( ) ( )0 0 00 A B Cv v v v v v= − + − − + −j i j i i iResolve into components.i: ( )0 01 13 0 4.53 3C Cv v v v− = = = 0 1.500 m/s=vj: 0 2.6 m/sA B B Av v v v− = = =Conservation of angular momentum about G:( ) ( ) ( )20 0 03G A A B B C Cml m rω= = × − + × − + × −H k r v v v v r v v( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( )( )202 2310.260 2.6 0.150 4.5 0.45033 m /s3A B A C C A B CA C A Cl v v va d v av dvlωω= − × + × − + += × + − × = + = + = k r r j r i r r r ii v j j i kConservation of energy: ( )2 2 2 211 332 2T ml mlω ω= =0 00 00 02.6 1.5 3.00 m/s2.6 1.5 3.00 m/s4.5 1.5 3.00 m/sA AB BC C− = − − =− = − − − =− = − − =v v j i v vv v j i v vv v i i v v( ) ( ) ( )2 2 22 0 0 01 1 12 2 2A B CT m m m= − + − + −v v v v v v1 2T T=( ) ( ) ( )2 2 22 23 1 1 13 3 32 2 2 2ml m m mω = + +3 m/slω =
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.(b)2 20.45033 m /s0.1501 m3 m/slllωω= = = 150.1 mml =(c)3 m/s0.1501llωω = = 19.99 rad/sω =
74. 74. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 58.Use a frame of reference that is translating with the mass center G of the system. Let 0v be its velocity.0 0v=v iThe initial velocities in this system are ( ) ( )0 0, ,A B′ ′v v and ( )0,C′v each having a magnitude of .lω They aredirected 120° apart. Thus,( ) ( ) ( )0 0 00A B C′ ′ ′+ + =v v vConservation of linear momentum:( ) ( ) ( ) ( ) ( ) ( )0 0 00 0 0A B C A B Cm m m m m m′ ′ ′+ + = − + − + −v v v v v v v v v( ) ( ) ( )0 0 00 A B Cv v v v v v= − + − − + −j i j i i iResolve into components.i: 03 0Cv v− = ( )( )03 3 0.4 1.2 m/sCv v= = =j: 0A Bv v− = B Av v=Initial kinetic energy: 2 2 21 01 13 32 2T mv ml ω   = +      Final kinetic energy: 2 2 2 2 221 1 1 12 2 2 2A B C A CT mv mv mv mv mv= + + = +Conservation of energy: 2 1T T= Solve for 2Av .( ) ( ) ( ) ( )2 2 2 22 2 203 3 1 3 3 10.4 0.75 1.22 2 2 2 2 2A Cv v l vω= + − = + −(a) 2 20.36375 m /s ,= 0.6031 m/sAv = 0.603 m/sA =v0.603 m/sB =v1.200 m/sC =v
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.Use a frame of reference moving with velocity 0.vConservation of angular momentum about G.( ) ( ) ( )20 0 03G A A B B C Cml m r m mω= = × − + × − + × −H k r v v v v r v v( ) ( ) ( ) ( )( )203 A B A C C A B Cl v v vω = − × + × − + +k r r j r i r r r i( ) ( ) ( ) ( ) ( )3 A C A Cl l a d v av dvω = × + − × = +k i v j j i k( )( )( ) ( )3 0.075 0.75 0.130 1.200Av d= +(b) ( )( )0.1406 0.1083 0.603 0.0753 md = − = 75.3 mmd =
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 14, Solution 59.Mass flow rate. As the fluid moves from section 1 to section 2 in time ∆t, the mass ∆m moved is( )m A lρ∆ = ∆Then, 1( )dm m A lAvdt t tρρ∆ ∆= = =∆ ∆Data: 3 2 6 211000 kg/m , 500 mm 500 10 m , 25 m/sA vρ = = = × =6(1000)(500 10 )(25) 12.5 kg/sdmdt−= × =Principle of impulse and momentum.: 1( ) 0m v P t∆ − ∆ =m dmP v vt dt∆= =∆(12.5)(25)P = 312 NP =
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 14, Solution 60.Consider velocities measured with respect to the plate, which is movingwith velocity V. The velocity of the stream relative to the plate is1u v V= − (1)Mass flow rate. As the fluid moves from section 1 to section 2 in time ∆t,the mass ∆m moved is( )m A lρ∆ = ∆Then( )dm m A lAudt t tρρ∆ ∆= = =∆ ∆(2)Principle of impulse and momentum.( ) ( ) 0m u P t∆ − ∆ =2m dmP u u Aut dtρ∆= = =∆PuAρ=From (1), 1 1PV v u vAρ= − = −Data: 2 6 2400 N, 600 mm 600 10 mP A −= = = ×31 30 m/s, 1000 kg/mv ρ= =640030(1000)(600 10 )V −= −×4.18 m/sV =
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 14, Solution 61.Let F be the force that the wedge exerts on the stream. Assume that the fluid speed is constant. 60 ft/s.v =Volumetric flow rate: 3475 gal/min ft /sQ = = 1.0583Mass flow rate: ( )3 362.4slug/ft 1.0584 ft /s 2.051 slug/s32.2dmQdtρ = = =  Velocity vectors: ( )1, cos30 sin30v v= = ° + °v i v i j( )2 cos45 sin 45v= ° − °v i jImpulse – momentum principle:( ) ( ) 1 22 2m mm t∆ ∆∆ + ∆ = +v F v v( ) ( )( )( )( )( ) ( )1 21 12 21 1cos30 sin30 cos45 sin 452 22.051 60 ft/s 0.21343 0.1035526.26 lb 12.74 lbmtdmvdt∆  = + − ∆   = ° + ° + ° − ° −  = − −= − −F v v vi j i j ii ji jForce that the stream exerts on the wedge:( ) ( )26.26 lb 12.74 lb− = +F i j drag 26.3 lb=lift 12.74 lb=
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 14, Solution 62.For a fixed observer, the upstream velocity is ( )48 ft/s .=v iVolumetric flow rate: 3500 gal/min ft /sQ = = 1.1141Mass flow rate: ( )3 362.4slug/ft 1.1141 ft /s 2.1590 slug/s32.2dmQdtρ = = =  Use a frame of reference that is moving with the wedge to the left at 12 ft/s. In this frame of reference theupstream velocity vector is( ) ( )48 12 60 ft/s .= − − =u i i iFor the moving frame of reference the mass flow rate is ( )602.1590 2.6987 slug/s.48dm u dmdt v dt′= = =Velocity vectors: ( )1, cos30 sin30u u= = ° + °u i u i j( )2 cos45 sin 45u= ° − °u i jLet F be the force that the wedge exerts on the stream.Impulse-momentum principle:( ) ( ) 1 22 2m mm t∆ ∆∆ + ∆ = +u F u u( ) ( )( )( )( )( ) ( )1 21 12 21 1cos30 sin30 cos45 sin 452 22.6987 60 0.21343 0.1035534.6 lb 16.76 lbmtdmudt∆  = + − ∆  ′  = ° + ° + ° − ° −  = − −= − −F u u ui j i j ii ji jForce that the stream exerts on the wedge( ) ( )34.6 lb 16.76 lb− = +F i j drag 34.6 lb=lift 16.76 lb=
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 14, Solution 63.Let F be the force exerted on the chips. Apply the impulse-momentum principle to the chips. Assume thatthe feed velocity is negligible.( ) ( ) Ct m∆ = ∆F v( )cos25 sin 25Cm dmvt dt∆  = = ° + ° ∆  F v i j( )( )1060 cos25 sin 2532.2 = ° + °  i j( ) ( )16.89 lb 7.87 lb= +i j0: 0x x xF D FΣ = − =16.89 lbxD =Force on truck hitch at D:16.89 lbxD− = − 16.89 lbxD− =
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 14, Solution 64.Initial momentum: ( ) 0.Am∆ =vImpulse – momentum principle.( )( ) ( )x yF F t m+ ∆ = ∆i j v( )o ocos35 sin35x ym dmF F vt dt∆  + = = + ∆  i j v i jx component:oEngine thrust cos35xdmF vdt= =Data: 3 388 m /min m /s60Q = = 31000 kg/mρ =( )81000 133.333 kg/s60dmQdtρ = = =  ( )( ) o133.333 50 cos35 5461 NxF = =5.46 kNxF =
82. 82. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 65.Weight: (600)(9.81) 5886 NW mg= = =Principle of impulse and momentum.Moments about F:1 1 1 2( ) (3 ) (2 ) ( ) ( ) ( ) 3( )m v a mv a m v a W t c R t L m v h∆ + ∆ + ∆ + ∆ − ∆ = ∆1 216 3m mR cW av hvL t t∆ ∆ = + + ∆ ∆ Data: 6 m, 4 m, 1.5 m, 0.8 m, 40 kg/sm dmL c a ht dt∆= = = = =∆[ ]1(4)(5886) (6)(1.5)(3)(40) (3)(0.8)(4)(40) 4040 N6R = + − =4040 N=R
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 14, Solution 66.Assume A Bu u u= =Principle of impulse and momentum.Moments about O :( ) ( ) ( )A c BR m u x t R m u∆ + ∆ = ∆F kk r( ) ( )( ) 0c B Ax t R m u u∆ = ∆ − =r F kThe line of action of F passes through point O.Components : ( ) ( ) ( )sin cosA Bm u F t m uα θ∆ + ∆ = ∆sin (1 cos )mF utα θ∆= −∆(1)Components : ( ) ( )0 cos sinF t m uα θ+ ∆ = ∆cos sinmF utα θ∆=∆(2)Dividing (2) by (1),22sin1 cos 2tan tansin 22sin cos2 2θθ θαθ θθ−= = =.2θα =Thus point C lies at the midpoint of arc AB.
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 14, Solution 67.( )( )( )80013.333 L/s 1.000 kg/L 13.333 L/s 13.333 kg/s60dmQ Qdtρ= = = = =( ) ( )( )30 m/s 30 m/s sin 40 cos40B C= = ° + °v j v i jApply the impulse – momentum principle.x components: ( ) ( )( )0 30sin 40xA t m+ ∆ = ∆ °( ) ( )( )30sin 40 13.333 30sin 40xmAt∆= ° = °∆257 NxA =y components: ( )( ) ( ) ( )( )30 30cos40ym A t m∆ + ∆ = ∆ °( ) ( )30cos40 30 13.333 30cos40 30ymAt∆= ° − = ° −∆93.6 N= − 93.6 NyA =moments about :A ( )( )( ) ( )0.060 30 Am M t∆ + ∆( )( )( ) ( )( )( )0.180 30cos40 0.300 30sin 40m m= ∆ ° − ∆ °( ) ( ) ( )1.8 1.6484Am M t m∆ = ∆ − ∆( ) ( )( )3.4484 13.333 3.4484 46.0 N mAmMt∆= − = − = − ⋅∆46.0 N mA = ⋅M274 N=A 20.0°
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 14, Solution 68.Mass flow rate:3 3262.4 lb/ft 40 ft /min1.29193 lb s/ft60 s/min32.2 ft/sdmQdt gγ= = = ⋅75 ft/sA Bv v= =Use impulse - momentum principle.moments about :D ( )( ) ( )( ) ( )15 23 15sin60 cos6012 12 12A Am v m v C t     − ∆ ° + ∆ ° − ∆          ( )312Bm v = ∆   ( )( )( )15 15 23 3sin60 cos60 1.29193 75 0.3742012 12 12 12AmC vt∆   = − ° + ° − = −   ∆   29.006 lbC = − 0, 29.0 lbx yC C= = −x component: ( ) ( ) ( )cos60A x Bm v D t m v∆ ° + ∆ = ∆( ) ( )( )cos60 1.29193 75 75cos60x B AmD v vt∆ = − ° = − ° ∆ 48.4 lbxD =y component: ( ) ( ) ( )sin60 0A ym v C t D t∆ ° + ∆ + ∆ =( )( )sin60 29.006 1.29193 75 sin60y AmD C vt∆= − − ° = + − °∆54.9 lbyD = −
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 14, Solution 69.Volumetric flow rate: 3300 gal/min 0 ft /sQ = = .6684Mass flow rate: ( )3 362.4slug/ft 0.6684 ft /s 1.2954 slug/s32.2dmQdtρ = = =  90 ft/sA Bv v= =Use the impulse – momentum principle.Moments about C: ( ) ( ) ( )cosA P Bm v a W t l m v bθ∆ − ∆ = ∆(a) ( )( )( ) ( )( )( )( )90 4/12 90 1/12cos 1.2954 0.728740 1A Bpm v a v bt W lθ−∆ −= = =∆43.23θ = ° 43.2θ = °x components: ( ) ( ) ( ) cosA x Bm v C t m v θ∆ + ∆ = ∆( ) ( )( )cos 1.2954 90cos 90 31.63 lbx B AmC v vtθ θ∆= − = − = −∆y components: ( ) ( ) ( )0 siny p BC t W t m v θ+ ∆ − ∆ = − ∆( ) ( )( )sin 40 1.2954 90 sin 39.84 lby p BmC W vtθ θ∆= − = − = −∆(b) [31.63 lb=C ] [39.84 lb+ ] 50.9 lb=C 51.6°
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 14, Solution 70.Volumetric flow rate: 3300 gal/min 0 ft /sQ = = .6684Mass flow rate: ( )3 362.4slug/ft 0.6684 ft /s 1.2954 slug/s32.2dmQdtρ = = =  , 45A Av v v θ= = = °Use the impulse-momentum principle.moments about C: ( ) ( ) ( )cosA p Bm v a W t l m v bθ∆ − ∆ = ∆(a)( )( ) ( )( )( )( )cos 40 1 cos4587.338 ft/s4 1/1.295412 12pA BW lv va b m tθ °= = = = − ∆ ∆−  87.3 ft/sv =x components: ( ) ( ) ( ) cosA x Bm v C t m v θ∆ + ∆ = ∆( ) ( )[ ]cos 1.2954 87.338cos45 87.338 33.137 lbx B AmC v vtθ∆= − = ° − = −∆y components: ( ) ( ) ( )0 siny p BC t W t m v θ+ ∆ − ∆ = − ∆( )( )sin 40 1.2954 87.338 sin 45 40.0 lby p BmC W vtθ∆= − = − ° = −∆(b) [33.137 lb=C ] [40 lb+ ] 51.9 lb=C 50.4°
88. 88. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 71.Symbols: mass flow ratedmdt=exhaust relative to the airplaneu =speed of airplanev =drag forceD =Principle of impulse and momentum.( ) ( ) ( )m v D t m u∆ + ∆ = ∆m dm Dt dt u v∆= =∆ −Data: 900 km/h = 250 m/sv =600 m/su =35 kN 35000ND = =35000600 250dmdt=−100 kg/sdmdt=
89. 89. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 14, Solution 72.Let F be the force exerted on the slipstream of one engine. Then, the force exerted on the airplane is −2F asshown.Statics.0BMΣ =( ) ( )( )0.9 4.8 2 0W F− =( )( )( )( )0.9 60002 4.8F =562.5 lb=Calculation of .dmdtmass density volume density area length= × = × ×( ) ( )( )B BB B BA v tm A l A v tgγρ ρ∆∆ = ∆ = ∆ =B Bm dm A vt dt gγ∆= =∆Force exerted on the slipstream: ( )B AdmF v vdt= −Assume that Av , the speed far upstream, is negligible.( ) 2 204B BB BA vF v D vg gγ γ π = − =   ( )( )( )( ) ( )2 2 22 24 562.5 32.247058.9 ft /s6.6 0.075BFgvDπ γ π= = =84.0 ft/sB =v