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

    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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.
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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+ + =+ + = ± + +
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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= = = −
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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)
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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= =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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= ⋅
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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− = × ⋅
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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°
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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°
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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=+
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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°
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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′= =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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= =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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.
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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−=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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−=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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′ = =&
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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−=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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°
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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− =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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.
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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°
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 = −
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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°
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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°
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 73.Let F be the force exerted on the slipstream of one engine.( )B AdmF v vdt= −Calculation of .dmdtmass density volume density area length= × = × ×( ) ( )( )B BB B BA v tm A l A v tgγρ ρ∆∆ = ∆ = ∆ =2or4B BBm A v dmD vt g dt gγ γ∆ π = =  ∆  Assume that ,Av the velocity far upstream, is negligible.( ) ( ) ( )2 22 0.0750 6.6 60 286.87 lb4 32.2 4B BF D v vgγ π π    = − = =        The force exerted by two slipstreams on the airplane is 2 .F− 2 573.74 lbF− =Statics.0:BMΣ =( )0.9 4.8 2 9.3 0W F A− − − =( )( ) ( )( )10.9 6000 4.8 573.749.3A  = − 284.5 lb= 285 lb=A0: 2 0x xF F B= − − =2 573.74 lbxB F= − =( )0: 284.5 6000 0y y yF A B W BΣ = + − = + − =5715.5 lbyB =[573.74 lb=B ] [5715.5 lb+ ] 5740 lb=B 84.3°
    • 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 14, Solution 74.Use a frame of reference moving with the plane. Apply the impulse-momentum principle. Let F be the force thatthe plane exerts on the air.x components: ( ) ( ) ( )A Bm u F t m u∆ + ∆ = ∆( ) ( )B A B Am dmF u u u ut dt∆= − = −∆(1)moments about B: ( ) ( ) 0A Be m u M t− ∆ + ∆ =B AdmM e udt= (2)Let d be the distance that the line of action is below B.BFd M= B AB AM eudF u u= =−(3)Data: 90 kg/s,dmdt= 600 m/s,Bu = 4 me =(a) 480 km/h 133.333 m/sAu = =From (1), ( )( ) 390 600 133.333 42 10 NF = − = × 42.0 kNF =From (2),( )( )( )4 133.333600 133.333d =−1.143 md =(b) 960 km/h 266.67 m/sAu = =From (1), ( )( ) 390 600 266.67 30 10 NF = − = × 30.0 kNF =From (2),( )( )( )4 266.67600 266.67d =−3.20 md =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 75.The thrust on the fluid is ( )B AdmF v vdt= −Calculation ofdmdt. mass density volume density area length= × = × ×( ) ( )B B Bm A l A v tρ ρ∆ = ∆ = ∆B Bm dmA vt dtρ∆= =∆where BA is the area of the slipstream well below the helicopter and Bv is the corresponding velocity in theslipstream. Well above the blade, 0.Av ≈Hence, 2BF Avρ=( ) ( ) ( )2 233 21.21 kg/m 9 m 24 m/s444.338 10 kg m/sπ =   = × ⋅44.3 kNF =The force on the helicopter is 44.3 kN .Weight of helicopter: 15kNH =WWeight of payload: P PW=WStatics: 0y H PF F W WΣ = − − =44.3 15 29.3 kN.P HW F W= − = − = 29.3 kNW =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 76.Letdmdt= mass flow rate, u = discharge velocity relative to the airliner, speed of airliner,v = andthrustF = of the engines.0F D− = ( ) 0dmu v Ddt− − =Configuration before control surface malfunction:1720= 22.36 slug/s, 1860 ft/s, 560 mi/h 821.33 ft/s32.2dmu vdt= = = =( )( ) 1 122.36 1860 821.33 0 23225 lbD D− − = =Drag force factor:( )2 2 211 1 1 1 2 21232250.03443 lb s /ft821.33DD k v kv= = = = ⋅After control surface malfunction: 2 22 11.2 0.04131 lb s /ftk k= = ⋅When the new cruising speed is attained,( ) 22 2 2 0dmu v k vdt− − =( )( ) 22 222.36 1860 0.04131 0v v− − =Solving for v2, 2 768.6 ft/sv = 2 524 mi/hv =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 77.Apply the impulse - momentum principle to the moving air. Use a frame of reference that is moving with theairplane. Let F be the force on the air.270 km/h 75 m/s600 m/svu= ==( ) ( )( )2 sin 202mm v F t u∆− ∆ + ∆ = °( ) ( )( )( ) 3sin 20 sin 20120 75 600sin 20 33.6 10 Nm dmF v u v ut dtF∆= + ° = + °∆= + ° = ×Force on airplane is .−F 33.6 kN=F
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 78.Symbols: n = number engines operatingdmdt= mass flow rate for one engineu = discharge velocity relative to jetliner = 800 m/sv = speed of jetlinerF = thrust force ( )dmn u vdt= −D = drag force 2kv=Force balance.0F D− = 2( )dmn u v kvdt− =21( )v dmn u v k dt=−All 3 engines operating: 3, 900km/h = 250 m/sn v= =2(250) 137.879 m/s3(800 250)dmk dt= =−(a) One engine lost: n = 2237.8792(800 )vv=−275.758 60606 0v v+ − =211.2m/sv = 760 km/hv =(b) 2 engines lost: n = 1237.879800vv=−237.879 30303 0v v+ − =156.16 m/sv = 562 km/hv =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 79.Let u be the velocity of the stream relative to the velocity of the blade. ( )u v V= −Mass flow rate: AmAvtρ∆=∆Principle of impulse and momentum.( ) ( ) ( ) costm u F t m u θ∆ − ∆ = ∆( )1 cos ( )(1 cos )t A AmF u Av v Vtθ ρ θ∆= − = − −∆where Ft is the tangential force on the fluid.The force Ft on the fluid is directed to the left as shown. By Newton’s law of action and reaction, thetangential force on the blade is Ft to the right.Output power: out ( ) (1 cos )t A AP FV Av v V Vρ θ= = − −(a) V for maximum power output.out( 2 ) (1 cos ) 0AdPA v VdVρ θ= − − =12Av V=(b) Maximum power.out max1 1( ) (1 cos )2 2A A A AP Av v v vρ θ  = − −    3out max1( ) (1 cos )4AP Avρ θ= −Input power = rate of supply of kinetic energy of the stream2 2 3in1 1 1 1( )2 2 2A A AmP m v v Avt tρ∆ = ∆ = = ∆ ∆ 
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(c) Efficiency. outinppη =3( ) (1 cos )12A AAAv v V VAvρ θηρ− −=2 1 (1 cos )A AV Vv vη θ = − −  
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 80.Data:240slugs/s,32.2dmdt= = 7.4534 2200 ft/s,u = 570 mi/h 836 ft/sv = =( ) ( )( )7.4534 2200 836 10166 lbdmF u vdt= − = − =(a) Power used to propel airplane:( )( ) 61 10166 836 8.499 10 ft lb/sP Fv= = = × ⋅propulsion power 15450 hp=Power of kinetic energy of exhaust:( ) ( )( )2212P t m u v∆ = ∆ −( ) ( )( )2 2 621 17.4534 2200 836 6.934 10 ft lb/s2 2dmP u vdt= − = − = × ⋅(b) Total power: 61 2 15.433 10 ft lb/sP P P= + = × ⋅total power 28060 hp=(c) Mechanical efficiency:6168.499 100.55115.433 10PP×= =×mechanical efficiency 0.551=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 81.Kinetic energy of fluid in slipstream passing in time .t∆2 21 1mass speed density volume speed2 2T∆ = × = × ×21density area length speed2= × × ×( ) ( )2 21 12 2A l v Av t vρ ρ= ∆ = ∆312TAvtρ∆=∆Input power 312dTAvdtρ= =Data: 31.2 kg/mρ =( )22 26.5 33.183 m4 4A dπ π= = =30 km/h 8.333 m/sv = =(a) ( )( )( )3 311.2 33.183 8.333 11.521 10 N m/s2dTdt= = × ⋅11.52 kJ/sInput power 11.521 kWdTdt= =(b) Output power ( )( )0.4 11.521 4.61 kW= =output power 4.61 kW=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 82.Kinetic energy of fluid in slipstream passing in time .t∆2 21 1mass speed density volume speed2 2T∆ = × = × ×21density area length speed2= × × ×( ) ( )2 21 12 2A l v Av t vρ ρ= ∆ = ∆312TAvtρ∆=∆(1)Data: output power 3.5 kW 3500 W= =3500input power 10000 W0.35= =input power 10,000 W, 36 km/h 10 m/sdTvdt= = = =Using (1),( )( )( )( )23 32 10000216.667 m1.2 10dTAdtvρ= = =(a)( )( )2 4 16.66744Ad A dππ π= = =4.61 md =(b) From above, 10.00 kJ/sdTdt=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 83.Mass flow rate:mass density volumedensity area length= ×= × ×( ) ( )m bd l bdv tρ ρ∆ = ∆ = ∆dm mbdvdt tρ∆= =∆1 dmQ bdvdtρ= =Continuity of flow: 1 2Q Q Q= =1 21 2,Q Qv vbd bd=Resultant pressure forces:1 1 2 2p d p dγ γ= =21 1 1 11 12 2F p bd bdγ= =22 2 2 21 12 2F p bd bdγ= =Apply impulse-momentum principle to water betweensections 1 and 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.( ) ( ) ( ) ( )1 1 2 2m v F t F t m v∆ + ∆ − ∆ = ∆( )1 2 2 1mv v F Ft∆− = −∆( )2 22 11 212Q QQ b d dbd bdρ γ ⋅ − = −  ( )( )( )22 11 2 2 11 212Q d db d d d dbd dργ−= + −Noting that ,gγ ρ=( )1 2 1 212Q b gd d d d= +
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 84.Mass flow rate:mass density volumedensity area length= ×= × ×( ) ( )m bd l bdv tρ ρ∆ = ∆ = ∆dm mbdvdt tρ∆= =∆1 dmQ bdvdtρ= =Continuity of flow: 1 2Q Q Q= =1 21 2,Q Qv vbd bd=Resultant pressure forces:1 1 2 2p d p dγ γ= =21 1 1 11 12 2F p bd bdγ= =22 2 2 21 12 2F p bd bdγ= =Apply impulse-momentum principle to water betweensections 1 and 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.( ) ( ) ( ) ( )1 1 2 2m v F t F t m v∆ + ∆ − ∆ = ∆( )1 2 2 1mv v F Ft∆− = −∆( )2 22 11 212Q QQ b d dbd bdρ γ ⋅ − = −  ( )( )( )22 11 2 2 11 212Q d db d d d dbd dργ−= + −Noting that ,gγ ρ=( )1 2 1 212Q b gd d d d= +Data: 1 29.81 m/s, 3 m, 1.25 m, 1.5 mg b d d= = = =( )( )( )( ) 313 9.81 1.25 1.5 1.25 1.5 15.09 m /s2Q = + = 315.09 m /sQ =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 85.From hydrostatics, the pressure at section 1 is 1 .p rh ghρ= =The pressure at section 2 is 2 0.p =Calculate the mass flow rate using section 2.mass density volume density area length= × = × ×( ) ( )2 2A Am l v tρ ρ∆ = ∆ = ∆2Adm mvdt tρ∆= =∆Apply the impulse-momentum principle to fluid between sections 1 and 2.( ) ( ) ( )1 1 1m v p A t m v∆ + ∆ = ∆1 1 1dm dmv p A vdt dt+ =( ) ( )1 1 1 2 1dmp A v v A v v vdtρ= − = −But 1v is negligible, 1 ,p ghρ= and 2v gh=( )1 2 2ghA A ghρ ρ= or 1 22A A=2 224 4D dπ π =    2Dd =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 86.The flow through each arm is 1.25 gal/min.33 3gal 1 ft 1 min5 11.1408 10 ft /smin 7.48 gal 60 sQ −    = = ×         ( )33 32362.4 lb/ftQ 11.1408 10 ft /s32.2 ft/s21.590 10 lb s/ftdmQdt gγρ− = = = ×   = × ⋅Consider the moment about O exerted on the fluid stream of one arm.Apply the impulse-momentum principle. Compute moments about O.First, consider the geometry of triangle OAB. Using first the law ofcosines,( ) ( )( )( )2 2 2 o6 4 2 6 4 cos120OA = + −76 in. 0.72648 ftOA = =Law of sines.osin sin1204 76β=o23.413 ,β = o o60 36.587α β= − =Moments about O:( )( )( ) ( ) ( )( ) ( )( )( )0 sinO O sm v M t OA m v OA m OAα ω∆ + ∆ = ∆ − ∆( ) ( )( ) ( )( ) ( )( )223sin21.590 10 0.72648 60 sin36.587 0.726480.56093 0.011395 lb ftO smM OA v OAtα ωωω−∆  = −  ∆ = × −  = − ⋅Moment that the stream exerts on the arm is .OM−Friction couple for one arm:( )10.275 0.06875 lb ft4FM = = ⋅Balance of moments on one arm:0F O F OM M M M− = =0.06875 0.56093 0.011395ω= −43.19 rad/s 412 rpmω = =412 rpmω =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 87.Consider the conservation of the horizontal component of momentum of the railroad car of mass 0m and thesand mass .qt( ) 0 00 0 00m vm v m qt v vm qt= + =+(1)0 00dx m vvdt m qt= =+Integrating, using 0 0x = and x L= when ,Lt t=( )0 0 0 00 00 000 0 00ln lnlnL Lt tLLm v m vL vdt dt m qt mm qt qm v m qtq m = = = + − ++=∫ ∫00 0 0ln Lm qt qLm m v+= 0 0/00qL m vLm qtem+=(a) Final mass of railroad car and sand 0 0/0 0qL m vLm qt m e+ =(b) Using (1), 0 0/0 0 0 00 0qL m vLLm v m vv em qt m−= =+0 0/0qL m vLv v e−=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 88.Apply the impulse-momentum principle.Moments about C: ( )( ) ( ) ( ) ( )( )0.9 3 1.8 1.65A Bm v D t W t m v− ∆ + ∆ − ∆ = − ∆( )( )( )( ) ( )( ) ( )( )1.8 10.9 1.653 31.8 4000 1100 0.9 4.5 1.65 4.5 2287.5 N3 3A BmD W v vt∆= + −∆ = + − = 2.29 kN=Dx components: ( ) ( ) ( )A x Bm v C t m v∆ + ∆ = ∆( ) ( )( )100 4.5 4.5 0x B AmC v vt ∆= − = − = ∆ y components: ( ) ( ) ( )0 0yC t D t W t+ ∆ + ∆ − ∆ =4000 2287.5 1712.5 NyC W D= − = − =1.712 kN=C
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 89.Let ρ be the mass per unit length of chain. Apply the impulse - momentum to the entire chain. Assume that thereaction from the floor it equal to the weight of chain still in contact with the floor.Calculate the floor reaction.( )R g l yρ= − 1yR mgl = −  Apply the impulse-momentum principle.( ) ( ) ( ) ( )yv P t R t gl t y y vρ ρ ρ+ ∆ + ∆ − ∆ = + ∆( ) ( ) ( )P t y v gl t R tρ ρ∆ = ∆ + ∆ − ∆(a) ( )yP v gl l y gtρ ρ ρ∆= + − −∆Lety dyvt dt∆= =∆2P v gyρ ρ= + ( )2mP v gyl= +(b) From above, 1ymgl = −  R
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 90.(a) Let ρ be the mass per unit length of chain. The force P supports the weight of chain still off the floor.P gyρ=mgyPl=(b) Apply the impulse-momentum principle to the entire chain.( ) ( ) ( ) ( )yv P t R t gl t g y y vρ ρ ρ− + ∆ + ∆ − ∆ = − + ∆( ) ( ) ( ) ( )R t gl t P t g y vρ ρ∆ = ∆ − ∆ − ∆yR gl gy vtρ ρ ρ∆= − −∆Let 0.t∆ → Then,y dyvt dt∆= = −∆( ) 2R g l y vρ ρ= − + ( ) 2mg l y vl = − + R
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 91.Let ρ be the mass per unit length of chain. Consider the impulse-momentum applied to the link being brought torest at point C.Calculation of .m∆( ) ( )m l v tρ ρ∆ = ∆ = ∆Impulse-momentum principle:( ) 0m v C t− ∆ + ∆ =( ) 0v t v C tρ− ∆ + ∆ =2C vρ=Impulse – momentum applied to the moving portion of the chain. Consider only the changes in momentum andforces contributing to moments about O in the diagram.Moments about O:( ) [ ] ( )m v gh C t m vρ∆ + − ∆ = ∆C ghρ=Equating the two expressions for C,2v ghρ ρ=2vhg=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 92.Let ρ be the mass per unit length of chain. Assume that the weight of any chain above the hole is supported bythe floor. It and the corresponding upward reaction of the floor are not shown in the diagrams.Case 1. Apply the impulse-momentum principle to the entire chain.( )( )( ) ( ) ( )( )( )( )yv gy t y y v vyv y v y v y vy vy vgy v yt t tρ ρ ρρ ρ ρ ρρ ρ ρ ρ+ ∆ = + ∆ + ∆= + ∆ + ∆ + ∆ ∆∆ ∆∆ ∆= + +∆ ∆ ∆Let 0.t∆ → ( )dy dv dgy v y yvdt dt dtρ ρ ρ ρ= + =Multiply both sides by .yv ( )2 dgy v yv yvdtρ ρ=Letdyvdt= on left hand side. ( )2 dy dgy yv yvdt dtρ ρ=Integrate with respect to time. ( ) ( )2g y dy yv d yvρ ρ=∫ ∫( )231 13 2gy yvρ ρ= or 2 23v gy= (1)Differentiate with respect to time.2 223 3dv dyv g gvdt dt= =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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)13dva gdt= = 0.333g=a(b) Set in (1)y l= 2 23v gl= 0.817 gl=vCase 2. Apply conservation of energy using the floor as the level for from which the potential energy ismeasured.1 10, 0T V= =22 21,2 2yT mv V gyρ= = −1 1 2 2T V T V+ = +2 21 102 2mv gyρ= −2 22 gy gyvm lρ= = (1)Differentiating with respect to y,22dv gyvdy l=(a) Acceleration.dv gya vdy l= =gyl=a(b) Setting in (1),y l= 2v gl= gl=vNote: The impulse-momentum principle may be used to obtain the force that the edge of the hole exertson the chain.
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 93.750 lb/s,dWdt=1 75023.292 lb s/ft32.2dm dWdt g dt= = = ⋅Thrust of one engine: ( )( ) 312500 23.292 291.15 10 lbdmP udt= = = ×For 3 engines, 33 873.4 10 lbP = ×Thrust 873 kips=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 94.Thrust of one engine:331200 10400 10 lb.3P×= = ×But,dmP udt=3400 1032 lb s/ft12500dm Pdt u×= = = ⋅( )( )32.2 32 1030 lb/sdW dmgdt dt= = = 1030 lb/sdWdt=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 95.Thrust, .dmP udt=F P mg maΣ = − =P u dma g gm m dt= − = −Data: 15 kg/sdmdt=As rocket is fired: 1500 kgm =0159.81 0.01 9.811500ua u= − = − (1)As all the fuel is consumed: 1 1500 1200 300 kg.m = − =1159.81 0.05 9.81300ua u= − = − (2)From the given data, 21 0 220 m/sa a− = (3)Using (1) and (2) for 1a and 0a and substituting into (3),0.04 220u = 5500 m/su =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 96.Thrust:dmP u uqdt= =Since u anddmdtare constant, P is also constant.:F maΣ =P mg ma− =( )P m a g= +( )min maxm a g= +( )( )1500 25 9.81= +352.215 10 N= ×(a) Fuel consumption rate.352.215 10450Pqu×= =116.0 kg/sq =(b) Mass of fuel consumed.( )( )fuel 116.0 15.6m qt= =fuel 1810 kgm =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 97.Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time ∆t.: ( )( ) ( )( )mv m m v v m v v v= − ∆ + ∆ + ∆ + ∆ −( ) ( ) ( )( ) ( ) ( )( ) ( )mv m v m v m v m v m v m v= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆( ) ( ) 0m v u m∆ − ∆ =( )dmm tdt∆ = − ∆v dv u dmt dt m dt∆= = −∆1 1 10 00v t mv mu dm dmdv dt um dt m= − = −∫ ∫ ∫1 01 00 1ln lnm mv v u um m− = − =0 1 01expm v vm u− =   Data: 1 0 2430 m/s,v v− = 04200 m/s 5000 kgu m= =15000 2430exp 1.78354200m= =1 2800 kgm =fuel 0 1 5000 2800m m m= − = − fuel 2200 kgm =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 98.Data from Problem 14.97: 0 5000 kg, 4200 m/sm u= =1 0 fuel 5000 1500 3500 kgm m m= − = − =Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time ∆t.: ( )( ) ( )( )mv m m v v m v v v= − ∆ + ∆ + ∆ + ∆ −( ) ( ) ( )( ) ( ) ( )( ) ( )mv m v m v m v m v m v m v= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆( ) ( ) 0m v u m∆ − ∆ =( )dmm tdt∆ = − ∆v dv u dmt dt m dt∆= = −∆1 1 10 00v t mv mu dm dmdv dt um dt m= − = −∫ ∫ ∫1 01 00 1ln lnm mv v u um m− = − =1 050004200 ln3500v v v∆ = − = 1498 m/sv∆ =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 99.Apply conservation of momentum to the rocket plus the fuel.( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )mv m m v v m v v vmv m v m v m v m v m v m v= − ∆ + ∆ + ∆ + ∆ −= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆( ) ( ) 0m v u m∆ − ∆ =( )dmm tdt∆ = − ∆v dv u dmt dt m dt∆= = −∆1 1 10 00v t mv mu dm dmdv dt um dt m= − = −∫ ∫ ∫1 0 01 00 1 1ln ln lnm m Wv v u u um m W− = − = =Data: 1 0 360 ft/sv v− =( )0 111,600 lb, 11,600 1000 10,600 lbW W= = − =11,600360 ln10,600u= 3990 ft/su =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 100.Apply conservation of momentum to the rocket plus the fuel.( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )mv m m v v m v v vmv m v m v m v m v m v m v= − ∆ + ∆ + ∆ + ∆ −= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆( ) ( ) 0m v u m∆ − ∆ =( )dmm tdt∆ = − ∆v dv u dmt dt m dt∆= = −∆1 1 10 00v t mv mu dm dmdv dt um dt m= − = −∫ ∫ ∫1 01 00 1ln lnm mv v u um m− = − =0 1 0 01 1expm v v Wm u W− = =  Data: 1 0 450 ft/s,v v− = 5400 ft/su =0 1 fuel 1 1200 lbW W W W= + = +111200 450exp 1.086905400WW+= =112000.08690W= 1 13810 lbW =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 101.See sample Problem 14.8 for derivation of0 00 0ln lnm m qtv u gt u gtm qt m−= − = − −−(1)Note that g is assumed to be constant.Setdyvdt= in (1) and integrate with respect to time.00 0 002000ln1ln2h t ttmh dy vdt u gt dtm qtm qtu dt gtm = = = − − −= − −∫ ∫ ∫∫Let 00m qtzm−=0qdz dtm= − or 0mdt dzq= −( )0 02 20 020 0 0 0 00 0 0 020 00 00 0 00 01 1ln ln2 21ln 1 ln 1211 ln 1 12ln 1 1 ln 1zzz zm u m uh z dz gt z z z gtq qm u m qt m qt m mgtq m m m mm u qt m qtgtq m mm u m qt m qtutq m m = − = + −     − −= − − − −            −= − − + −          − −= − + − −     ∫220 00121ln2gtm u m qtut ut gtq m−  −= + − −  20 001ln2m mh u t t gtq m qt  = − − −  −  (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.Data: 2007300226.7 lb s /ft32.2Wmg= = = ⋅2608.0745 lb s/ft32.2wqg= = = ⋅&2fuelfuel4000124.22 lb s /ft32.2Wmg= = = ⋅fuel 124.2215.385 s8.0745mtq= = =20 226.7 124.22 102.48 lb s /ftm qt− = − = ⋅1500 ft/su =( )( )2226.7 226.7 11500 15.385 15.385 ln 32.2 15.3858.0745 102.48 2h  = − − −    4150 fth =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 102.Thrust forcedmP u uqdt= =Mass of rocket plus unspent fuel 0m m qt= −Acceleration:0P uqam m qt= =−Integrating with respect to time to obtain the velocity,0 00 00t t qv v adt v u dtm qt= + = +−∫ ∫( ) 00 0 0 00ln ln lnm qtv u m qt m v um− = − − − = −  (1)Integrating again to obtain the displacement,00 0 00lnt m qts s v t u dtm−= + − ∫Let 0 00 0orm qt q mz dz dt dt dzm m q−= = − = −( )]0 00 00 000 0 0 00 00 0 0 00 00 000 0 00 00 0ln lnln 1 ln 11 ln 1 1ln 1 1 ln 1zzz zom u m us s v t zdz z z zq qmm u m qt m qt ms v tq m m m mm u qt m qts v tq m mm u m qt m qts v t utq m m= + + = −    − −= + + − − −            −= + + − − +          − −= + + − + − −     ∫0 00 00lnm u m qts v t ut utq m  −= + + + −  0 00 00lnm ms s v t u t tq m qt  = + + − −  −  (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.Data: 0 7500 ft/s 7500 360 7860 ft/sv v= = + =( )( )fuel fuel 100060 s 0.5176 slug/s32.2 60m Wt qt gt= = = = =0011600360.25 slugs32.2Wmg= = =011600 1000329.19 slugs32.2 32.2m qt− = − =0 0s =From (1),329.197860 7500 ln360.25u= −360.25360 ln329.19u= 3993 ft/su =From (2), ( )( )360.25 360.250 7500 60 3993 60 60 ln0.5176 329.19s  = + + − −    6460.6 10 ft= × 87.2 mis =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 103.See sample Problem 14.8 for derivation of0 00 0ln lnm m qtv u gt u gtm qt m−= − = − −−(1)Note that g is assumed to be constant.Setdyvdt= in (1) and integrate with respect to time.00 0 002000ln1ln2h t ttmh dy vdt u gt dtm qtm qtu dt gtm = = = − − −= − −∫ ∫ ∫∫Let 0 00 0orm qt q mz dz dt dt dzm m q−= = − = −( )]0 02 20 020 0 0 0 00 0 0 020 00 00 0 00 01 1ln ln2 21ln 1 ln 1211 ln 1 121ln 1 1 ln 12zzz zm u m uh zdz gt z z z gtq qm u m qt m qt m mgtq m m m mm u qt m qtgtq m mm u m qt m qtutq m m= − = − −    − −= − − − −            −= − − + −          − −= − + − − −      ∫220 001ln2gtm u m qtut ut gtq m  −= + − −  20 001ln2m mh u t t gtq m qt  = − − −  −  (2)Data: 20 960 kg, 10 kg/s, 3600 m/s, 9.81 m/sm q u g= = = =fuelfuel800800 kg, 80 s10mm tq= = = =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 3960 960 13600 80 80 ln 9.81 80 153.4 10 m10 960 800 2h  = − − − = ×  −  153.4 kmh =(b) From equation (1), ( )( )9603600ln 9.81 80960 800v = −−5670 m/s=v
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 104.Data from Problem 14.97: 0 5000 kg, 4200 m/sm u= =0 00, 0v s= =Thrust forcedmP u uqdt= =Mass of satellite plus unspent fuel 0m m qt= −Acceleration:0P uqam m qt= =−Integrating with respect to time to obtain the velocity,0 00 00t t qv v adt v u dtm qt= + = +−∫ ∫ (1)( ) 00 0 0 00ln ln lnm qtv u m qt m v um− = − − − = − Integrating again to obtain the displacement,00 0 00lnt m qts s v t u dtm−= + − ∫Let 0 00 0orm qt q mz dz dt dt dzm m q−= = − = −( )]0 00 00 000 0 0 00 00 0 0 00 00 000 0 00 00 0ln lnln 1 ln 11 ln 1 1ln 1 1 ln 1zzz zom u m us s v t zdz z z zq qmm u m qt m qt ms v tq m m m mm u qt m qts v tq m mm u m qt m qts v t utq m m= + + = −    − −= + + − − −            −= + + − − +          − −= + + − + − −     ∫0 00 00lnm u m qts v t ut utq m  −= + + + −  0 00 00lnm ms s v t u t tq m qt  = + + − −  −  (2)Using the data,( )( )5000 50000 0 4200 80 80 ln18.75 5000 18.75 80s  = + + − −  −   356.367 10 m= × 56.4 kms =
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 105.Let F be the thrust force, anddmdtbe the mass flow rate.Absolute velocity of exhaust: ev u v= −Thrust force: ( )dmF u vdt= −Power of thrust force: ( )1dmP Fv u v vdt= = −Power associated with exhaust: ( ) ( ) ( )( )2221 12 2eP t m v m u v∆ = ∆ = ∆ −( )2212dmP u vdt= −Total power supplied by engine: 1 2P P P= +( ) ( ) ( )2 2 21 12 2dm dmP u v v u v u vdt dt = − + − = −  Mechanical efficiency: 1useful powertotal powerPPη = =( )2 22 u v vu vη−=−( )2vu vη =+1η = when .u v= The exhaust, having zero velocity, carries no power away.
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 106.Let F be the thrust force anddmdtbe the mass flow rate.Absolute velocity of exhaust: ev u v= −Thrust force:dmF udt=Power of thrust force: 1dmP Fv uvdt= =Power associated with exhaust: ( ) ( ) ( )( )2221 12 2eP t m v m u v∆ = ∆ = ∆ −( )2212dmP u vdt= −Total power supplied by engine: 1 2P P P= +( ) ( )2 2 21 12 2dm dmP uv u v u vdt dt = + − = +  Mechanical efficiency: 1useful powertotal powerPPη = =( )2 22uvu vη =+1η = when .u v= The exhaust, having zero velocity, carries no power away.
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 107.The weights are 30 lb, 40 lb, and 50 lb.A B CW W W= = =Initial velocities: ( )09 ft/sAv = , ( )06 ft/sBv = , ( )0and 0.Cv =There are no horizontal external forces acting during the impacts, and the baggage carrier is free to coastbetween the impacts.(a) Suitcase A is thrown first.Let 1v be the common velocity of suitcase A and the carrier after the first impact and 2v be the commonvelocity of the two suitcases and the carrier after the second impact.Initial momenta: ( ) ( )0 0, , and 0.A BA BW Wv vg gSuitcase A impacts carrier. Conservation of momentum:( )( ) ( )( )01 1030 90 3.375 ft/s80A AA A CAA CW vW W Wv v vg g W W++ = = = =+Suitcase B impacts on suitcase A and carrier. Conservation of momentum:( ) 1 20B A C A B CBW W W W W Wv v vg g g+ + ++ =( ) ( ) ( )( ) ( )( )10240 6 80 3.3754.25 ft/s120B B A CA B CW v W W vvW W W+ + += = =+ +2 4.25 ft/s=v(b) Suitcase B is thrown first.Let 3v be the common velocity of suitcase B and the carrier after the first impact and 4v be the commonvelocity of all after the second impact.Suitcase B impacts the carrier. Conservation of momentum:( )( ) ( )( )03 3040 60 2.6667 ft/s90B BB B CBB CW vW W Wv v vg g W W++ = = = =+Suitcase A impacts on suitcase B and carrier. Conservation of momentum:( ) 3 40A B C A B CAW W W W W Wv v vg g g+ + ++ =( ) ( ) ( )( ) ( )( )30430 9 90 2.66674.25 ft/s120A A B CA B CW v W W vvW W W+ + += = =+ +4 4.25 ft/s=v
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 108.The weights are 30 lb, ?, and 50 lb.A B CW W W= = =Initial velocities: ( ) ( )0 07.2 ft/sA Bv v= = ( )0, 0Cv =Final velocity: 3.6 ft/sfv =(a) Conservation of momentum:( ) ( )0 00A B A B CA B fW W W W Wv v vg g g+ ++ + = (1)( )( ) ( ) ( )( )30 7.2 7.2 30 50 3.6B BW W+ = + +20.0 lbBW =(b) Equation (1) shows that the final velocity is independent of the order in which the suitcases are thrown.3.60 ft/sf =v
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 109.Linear momentum of each particle expressed in kg m/s.⋅12 6 68 68 16 8A AB BC Cmmm= + += += − + +v i j kv i jv i j kPosition vectors, (meters): 3 , 1.2 2.4 3 , 3.6A B C= = + + =r j r i j k r i( )2Angular momentum about , kg m /s .O ⋅( ) ( ) ( )( ) ( ) ( )0 3 0 1.2 2.4 3 3.6 0 012 6 6 8 6 0 8 16 818 36 18 24 12 28.8 57.60 4.8 9.6O 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 k i j k j ki j k( ) ( )2 24.80 kg m /s 9.60 kg m /sO = − ⋅ + ⋅H j k
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 110.Position vectors, (meters): 3 , 1.2 2.4 3 , 3.6A B C= = + + =r j r i j k r i( ) Mass center:a ( )A B C A A B B C Cm m m m m m+ + = + +r r r r( )( ) ( )( ) ( )( )9 3 3 2 1.2 2.4 3 4 3.61.86667 1.53333 0.66667= + + + += + +r j i j k ir i j k( ) ( ) ( )1.867 m 1.533 m 0.667 m= + +r i j kLinear momentum of each particle, ( )2kg m /s .⋅12 6 68 68 16 8A AB BC Cmmm= + += += − + +v i j kv i jv i j k(b) Linear momentum of the system,( )kg m/s.⋅12 28 14A A B B C Cm m m m= + + = + +v v v v i j k( ) ( ) ( )12.00 kg m/s 28.0 kg m/s 14.00 kg m/sm = ⋅ + ⋅ + ⋅v i j kPosition vectors relative to the mass center, (meters).1.86667 1.46667 0.666670.66667 0.86667 2.333331.73333 1.53333 0.66667A AB BC C′ = − = − + −′ = − = − + +′ = − = − −r r r i j kr r r i j kr r r i j k
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(c) Angular momentum about G, ( )2kg m /s .⋅( ) ( )( )1.86667 1.46667 0.66667 0.66667 0.86667 2.3333312 6 6 8 6 01.73333 1.53333 0.666678 16 812.8 3.2 28.8 14 18.6667 10.93331.6 8.5333 15.46672.8 13.3333G A A A B B B C C Cm m m′ ′ ′= × + × + ×= − − + −+ − −−= + − + − + −+ − − += − + −H r v r v r vi j k i j ki j ki j k i j ki j ki j 24.2667k( ) ( ) ( )2 2 22.80 kg m /s 13.33 kg m /s 24.3 kg m /sG = − ⋅ + ⋅ − ⋅H i j k( ) ( ) ( )2 2 21.86667 1.53333 0.6666712 28 142.8 kg m /s 18.1333 kg m /s 33.8667 kg m /sm× == ⋅ − ⋅ + ⋅i j kr vi j k( ) ( )2 24.8 kg m /s 9.6 kg m /sG m+ × = − ⋅ + ⋅H r v j kAngular momentum about O.( ) ( ) ( )( ) ( ) ( )( ) ( )2 20 3 0 1.2 2.4 3 3.6 0 012 6 6 8 6 0 8 16 818 36 18 24 12 28.8 57.64.8 kg m /s 9.6 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 k i j k j kj kNote thatO G m= + ×H H r v
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 111.Choose x axis pointing east, y axis north, and z axis vertical.Velocities before collision:( )05.5 mi 5280 ft/miHelicoptor: 121ft/s4 min 60 s/minHv = ⋅ =( ) 010 mi 5280 ft/miAirplane: 220 ft/s4 min 60 s/minA xv  = ⋅ = ( )07.5 mi 5280 ft/mi165 ft/s4 min 60 s/minA yv  = − ⋅ = − ( ) ( ) ( ){ }( ) ( )( ) ( )0 0 0 0Mass center:6000 3000121 220 1659000 9000154 ft/s 55 ft/sH AH A Ax yA H A Hm mv v vm m m m  = + +   + + = + −  = −v i i ji i ji jNo external forces act during impact. Assume that only gravity acts after the impact. Motion of mass centerafter impact:( )2 20 01154 55 3600 16.12t z gt t t t = + − = − + −  r v k i j kTime of fall. 2 360014.953 s16.1t t= =( )( ) ( )( ) ( ) ( )154 14.953 55 14.953 2302.8 ft 822.42 ft= − = −r i j i j( ) ( ) ( ) ( ) ( )1 1 2 2H A H H H H A Am m m m m+ = + +r r r r( ) ( ) ( ) ( ) ( )( )( )( )( ) ( )( )1 1 2 2119000 2302.8 822.4230002000 1500 300 4000 1800 1500A H A H H H HAm m m mm = + − − = −− − − − r r r ri ji j i j( ) ( )3510 ft 267 ft= −i j( )Coordinates of point : 3510 ft, 267 ftA −
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 112.Choose x axis pointing east, y axis north, and z axis vertical.Velocities before collision:( )05.5 mi 5280 ft/miHelicoptor: 121ft/s4 min 60 s/minHv = ⋅ =( ) 010 mi 5280 ft/miAirplane: 220 ft/s4 min 60 s/minA xv  = ⋅ = ( )07.5 mi 5280 ft/mi165 ft/s4 min 60 s/minA yv  = − ⋅ = − ( ) ( ) ( ){ }( ) ( )( ) ( )0 0 0 0Mass center:6000 3000121 220 1659000 9000154 ft/s 55 ft/sH AH A Ax yA H A Hm mv v vm m m m  = + +   + + = + −  = −v i i ji i ji jNo external forces act during impact. Assume that only gravity acts after the impact. Motion of mass centerafter impact:( )2 20 01154 55 3600 16.12t z gt t t t = + − = − + −  r v k i j kTime of fall. 2 360014.953 s16.1t t= =( )( ) ( )( ) ( ) ( )154 14.953 55 14.953 2302.8 ft 822.42 ft= − = −r i j i j( ) ( ) ( ) ( ) ( )1 1 2 2H A H H H H A Am m m m m+ = + +r r r r( )( )( ) ( ) ( )( )( )( )( ) ( )( )2 1 12119000 2302.8 822.4240003000 3600 240 2000 1200 600H H A A A H HHr m m m mm = + − − = −− + − − r r ri ji j i j( ) ( )1881 ft 1730 ft= −i j ( )2Coordinates of point : 1881ft, 1730 ftH −
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 113.Use a frame of reference moving with the mass center.Conservation of momentum:0 A A B BBA BAm v m vmv vm′ ′= − +′ ′=Conservation of energy:( ) ( ) ( )( )( )( )22 2 221 1 1 12 2 2 222BA A B B A B B BAB A BBAABB A BmV m v m v m v m vmm m mvmm Vvm m m ′ ′ ′ ′= + = +  +′=′ =+Data: 2 25 30.15528 lb s /ft, 0.09317 lb s /ft32.2 32.2A Bm m= = ⋅ = = ⋅90 ft lbV = ⋅( )( )( )( )( )2 0.15528 9034.75 34.75 ft/s0.09317 0.24845B Bv v′ ′= = = 30°( )0.0931734.75 20.85 20.85 ft/s0.15528A Av v′ ′= = = 30°Velocities of A and B:[24 ft/sA =v ] [20.85 ft/s+ ]30° 12.00 ft/sA =v 60.3°[24 ft/sB =v ] [34.75 ft/s+ ]30° 56.8 ft/sB =v 17.8°
    • 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 114.Locate the mass center.Let l be the distance between A and B.( )B A B Am l m m l= +2757BAA BBml l lm ml l= =+=(a) Linear momentum.( )( )0 2.5 3.5 8.75AL m v= = = 8.75 kg m/s= ⋅LAngular momentum about G:( )( )( )0 02 20.210 2.5 3.5 0.5257 7G A A AH l m v lm v= = = =20.525 kg m /sG = ⋅H(b) There are no resultant external forces acting on the system;therefore, L and GH are conserved.:L A A B Bm v m v L+ = 2.5 1.0 8.75A Bv v+ = (1):GH B B B A A A Gl m v l m v H− =( )( ) ( )( )5 20.210 1.0 0.210 2.5 0.5257 7B Av v− =0.15 0.15 0.525B Av v− = (2)Solving (1) and (2) simultaneously, 1.5 m/s, 5 m/sA Bv v= =1.500 m/sA =v5.00 m/sB =v
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 115.For steady flow 1 2Q Q Q+ = (1)Assume that the fluid speed is constant.Velocity vectors: ( ) 1 2sin cos , ,v v vθ θ= − = − =v i j v i v iLet Pj be the force that plate C exerts on the fluid.Impulse-momentum principle:( ) ( ) ( ) ( )1 21 2m P t m m∆ + ∆ = ∆ + ∆v j v v( ) ( )1 21 2m m mPt t t∆ ∆ ∆= + −∆ ∆ ∆j v v v( ) ( ) ( )1 2sin cosdm dm dmv v v vdt dt dtθ θ     = − + − −          i i i j( )1 2 sin cosQ v Q v Qvρ ρ ρ θ θ= − + − −i i i jResolve into components.i: 1 2 2 10 sin sinQ v Q v Qv Q Q Qρ ρ ρ θ θ= − + − = − (2)j: cosP Qvρ θ= (3)Data: 2 462.41.93789 lb s /ft , 90 ft/s32.2vgγρ = = = ⋅ =3 3 3 31 226 gal/min 57.932 10 ft /s, 130 gal/min 289.66 10 ft /sQ Q− −= = × = = ×
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 (1), 3 3347.59 10 ft /sQ −= ×(a) From (2), 2 1sin 0.66667 41.810Q QQθ θ−= = = °41.8θ = °From (3), ( )( )( )31.93789 347.59 10 90 cos41.810 45.2 lb.P −= × ° =(b) Force that stream exerts on plate C:45.2 lbP− =j
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 116.Calculation of or .m dmt dt∆∆mass density volume density area length= × = × ×( ) ( )mm A l Av t Avtρ ρ ρ∆∆ = ∆ = ∆ =∆3 22 2 262.4 lb/ft 1.5 in60 ft/s 1.21118 lb s/ft32.2 ft/s 144 in /ftdmAvdtρ= = ⋅ ⋅ = ⋅Apply the principle of impulse-momentum.moments about D: ( ) ( ) ( ) ( )12 16 10 412 12 12 12A Bm v C t W t m v − ∆ − ∆ + ∆ = − ∆  ( )16 10 4 12B AmC W v vt∆= + −∆( )( ) ( ) ( )( ) ( )( )10 10 1.21118 4 60 12 60 481.37 = + − = − 30.085 lbC = − 30.1 lb=Cx components: ( ) ( ) ( )A x Bm v D t m v∆ + ∆ = ∆( ) ( ) ( )( )1.21118 60 60 0x B A B Am dmD v v v vt dt∆= − = − = − =∆0xD =y components: ( ) ( ) ( )0 0yC t D t W t+ ∆ + ∆ − ∆ =( )10 30.085 40.085 lbyD W C= − = − − = 40.1 lb=D
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 117.Calculation ofdmdtat a section in the airstream:mass density volume density area length= × = × ×( )m A l Av tρ ρ∆ = ∆ = ∆m dmAvt dtρ∆= =∆(a) ( )Thrust B Admdt= −v v where Bv is the velocity just downstreamof propeller and Av is the velocity far upstream. Assume Av isnegligible.( ) 2 2Thrust4Av v D vπρ ρ = =   ( ) ( )2 2 23600 1.21 2 3.8014v vπ = =  30.774m/sv = 30.8 m/sv =(b) ( ) ( )2212 30.7744 4dmQ Av D vdtπ πρ = = = =  396.7 m /sQ =(c) Kinetic energy of mass :m∆( ) ( ) ( )2 2 21 1 12 2 2T m v A l v Av t vρ ρ∆ = ∆ = ∆ = ∆3 2 31 12 2 4T dTAv D vt dtπρ ρ∆  = = =  ∆  ( ) ( ) ( )2 3 311.21 2 30.774 55.4 10 N m/s2 4π = = × ⋅  55.4 kWdTdt=
    • COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 118.From Eq. (14.44) of the text book, the thrust is( )( ) 3 2 310 kg/s 3600 m/s 36 10 kg m/s 36 10 NdmP udt= = = × ⋅ = ×F maΣ =P mg ma− =Pa gm= − (1)(a) At the start of firing, 20 960 kg, 9.81 m/sm m g= = =From (1),3236 109.81 27.69 m/s960a×= − = 227.7 m/s=a(b) As the last particle of fuel is consumed,960 800 160 kg,m = − = ( )29.81 m/s assumedg =From (1),3236 109.81 215.19 m/s160a×= − = 2215 m/s=a