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solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 13

1. 1. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 1.Given: Weight of satellite, 1000 lbW =Speed of satellite, 14,000 mi/hv =Find: Kinetic energy, T( )( )h14,000 mi/h 5280 ft/mi 20,533 ft/s3600 sv = =  ( )( )221000 lbMass of satellite 31.0559 lbs /ft32.2 ft/s= =( )( )22 91 131.0559 20,533 6.5466 10 lb ft2 2T mv= = = × ⋅96.55 10 lb ftT = × ⋅Note: Acceleration of gravity has no effect on the mass of the satellite.96.55 10 lb ftT = × ⋅ !
2. 2. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 2.CircumferenceTimev   =( )( )( )( )( ) ( )( )2 6370 km 35,800 km 1000 m/km3075.2 m/s23 hr 3600 s/hr 56 min 60 s/hrvπ += =+3075.2 m/sv =( )( )221 1Kinetic energy, 500 kg 3075.2 m/s2 2T mv= =2.36 GJT = !
3. 3. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 3.Given: Mass of stone, 2 kgm =Velocity of stone, 24 m/sv =Acceleration of gravity on the moon, 21.62 m/smg =Find:(a) Kinetic energy, THeight h, from which the stone was dropped(b) T and h on the Moon(a) On the Earth ( )( )221 12 kg 24 m/s 576 N m2 2T mv= = = ⋅ 576 JT = !( )( )22 kg 9.81 m/s 19.62 NW mg= = =1 1 2 2 1 1 2 20 576 JT U T T U Wh T− −= = = = =( )( )22576 N m29.36 m19.62 NTWh T hW⋅= = = =29.4 mh = !(b) On the MoonMass is unchanged. 2 kgm =Thus T is unchanged. 576 JT = !Weight on the moon is, ( )( )22 kg 1.62 m/sm mW mg= =N24.3=mW( )576 N m177.8 m3.24 NmmThW⋅= = =177.8 mmh = !
4. 4. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 4.21 lb 11.620316 oz 32.2 ft/sm  =      (a) 2 21 1 1.62(160 ft/s)2 2 16(32.2)T mv = =   40.2 ft-lbT = !At maximum height,(160 ft/s) cos25xv v= = °(b) 21(160cos25 )2T m= °33.1 ft-lbT = !
5. 5. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 5.Use work and energy with position 1 at A and position 2 at C.At 10yFΣ =11cos30 0cos30N mgN mg⇒ − °== °0yFΣ =22= 0N mgN mg⇒ −=Work and energy1 1 2 2T V T→+ = (1)Where2 21 11 1(4ft/s) 82 2T mv m m+ = =1 2 1 2 (20) ( sin30 )k kV N d N mg dµ µ→ = − − + °2 22 21 1(8) 322 2T mv m m= = =Into (1)8 cos30 (20) sin30 32k km mgd mg mgd mµ µ− ° − + °=Solve for32 8 20 32 8 (0.25)(32.2)(20)20.3 ftcos30 sin30 ( 0.25) (32.2)(0.866 32.2(0.5))kkgdg gµµ− + − += = =− °+ ° − +!At 2
6. 6. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 6.(a) Use work and energy from A to B. 1 1 2 2T V T→+ =2 21 11 1 50(40) 1242.24 ft lb2 2 32.2T mv = = = ⋅  2 0T = (Stops at top)1 2 sin 20U Nx mg xµ→ = − − °N is neededy 0FΣ = cos20 (50 lb)cos20 46.985 lbN W⇒ = °= °=So1 2 0.15(46.985) 50sin 2024.149U x xx→ = − − °= −Substitute1242.24 24.149 0x− =51.44 ftx = 51.4 ftx = !(b) Package returns to A – use work and energy from B to A2 2 3 3T U T→+ =Where 2 0T = (At B)2 3 sin 20 kU W x Nxµ→ = ° −(50) sin 20 (51.44) 0.15(46.985)(51.44)= ° −517.13 ft lb= ⋅
7. 7. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.2 2 23 3 3 31 1 500.77642 2 32.2T mv v v = = =  Substitute230 517.13 0.7764v+ =3 25.81 ft/sv = 3 25.8 ft/sv = 20°!(c) Energy dissipated is equal to change of kinetic energy2 21 3 1 21 12 2T T mv mv− = −2 21 50(40 25.81 )2 32.2 = −  725 ft lb= ⋅Energy dissipated 725 ft lb= ⋅ !
8. 8. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 7.Given: Automobile Weight W = mg = (2000 kg) (9.81)19,620 NW =Initial Velocity A, 0m/sAv =Incline Angle, 6α = °Vehicle brakes at impending slip for 20 m from B to C0Cv =Find; speed of automobile at point B, vBCoefficient of static friction, µ(a) (19620 N) (150 m)sin6A B A BU Wh→ →= − °3307.63 10 N m= × ⋅2102A B B AU T T mv→ = − = −3 21307.63 10 N m (2000kg) 02Bv× ⋅ = −17.54 m/sBv = !(b) 0A C A C B C C AU Wh Fd T T→ → →= − = − =20 mB Cd → = NF µ=Where coefficient of static frictionµ =(19620N)(sin6 )(170m) (20m)A CU F→ = ° −(19620N) cos6F µ= °(19620N)(sin6 )(170m) (19620N)(cos6 ) (20m) 0µ° − ° =170tan6 0.89320µ = °= 0.893µ = !
9. 9. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 8.Given: Automobile weight, (2000)(9.81) 19620 NW = =Initial velocity at A, 0 m/sAv =Incline Angle, 6α = °Vehicle costs 150 m from A to BVehicle skids 20 m from B to CDynamic friction coefficient, 0.75µ =Find: Work done on automobile by air resistance and rolling resistance between points A and C.(20 m) 0A C R A C C AU U Wh F T T→ →= + − = − =N 0.75 (19620 N) cos6F µ= = °UR = Resistance work0.75 (19620 N)cos6 (20 m) (19620 N)sin6 (170 m)= ° − °356.0 10 N mRU = − × ⋅356.0 10 Jor − × !
10. 10. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 9.90cos20 sin50 0NF N P= − °− °=∑90cos20 sin50N P= °+ °1 2 [ cos50 90sin 20 0.35 N](3 ft)U P→ = °− °−22 21 90 lb(2 ft/s)2 32.2 ft/sT =   2( cos50 ) 3 (90sin 20 ) (3) 0.35 (90cos20 sin50 )3P P T° − ° − °+ ° =21 90(3 cos50 0.35(3) sin50 ) 90sin 20 (3) 0.35 (90cos20 )(3) (2)2 32.2P °− ° = ° + ° +   186.736166.1 lb1.12402P = = !
11. 11. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 10.(a) First stage: 1 2 2(5.5 3)(50) 125 lb ft =U T→ = − = ⋅22 21 32 32.2T v =   2 51.8 ft/sv = !(b) At the top: 2 3 3( 50) 0 125U h→ = − − = −275,3h∴ = 91.7 fth = !(c) At the return: 23 4 4 41 33(91.6667)2 32.2U T v→ = + = =   4 76.8 ft/sv = !
12. 12. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 11.WA = 7(9.81) = 68.67 NGiven: Block A is released from rest and moves up incline 0.6 m.Friction and other masses are neglectedFind: Velocity of the block after 0.6 m, vFrom the Law of Cosines2 2 2(1.2) (0.6) 2(1.2)(0.6)cos 15d = + − °2 20.4091 md =0.63958 md =C CU W= (Distance pulley C lowered)1140 N (1.2 0.63958) m 39.229 N m2 = − = ⋅  68.67 N (sin15 )(0.6 m) 10.6639 N mAU = − ° = − ⋅2 1 C AU T T U U= − = −2102A C Am v U U− = −21(7kg) (39.229 10.6639) N m2v = − ⋅28.1615v =2.857 m/sv = 2.86m/sv = 15°!
13. 13. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 12.WA = 7(9.81) = 68.67 NGiven: Block A is released at the position shown at a velocity of1.5 m/s up.After moving 0.6 m the velocity is 3 m/s.Find: work done by friction force on the block, Vf JFrom the Law of Cosines2 2 2(1.2) (0.6) 2(1.2)(0.6)(cos 15 )d = + − °2 20.4091md =0.63958md =UC1140N (1.2 0.63958) m 39.229 N m2 = − = ⋅  68.67 N(sin15 )(0.6 m) 10.664 N mAU = − ° = − ⋅2 22 1 2 11[ ]2C A friction AU U U T T m v v+ − = − = −2 2 2139.229 10.664 (7 kg)[(3) (1.5) ] m2frictionU− − = −39.229 10.664 23.625frictionU− = − + +4.94 J= −4 94 JfrictionU .= − !
14. 14. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 13.Given: At A, 0v v=For AB, 0.40kµ =At B, 2 m/sv =Find: 0v( )22 201 1 12 m/s2 2 2A B BT mv T mv m= = =2 mBT =( )( )sin15 N 6 mA B kU W µ− = ° −0 cos15 0F N WΣ = − ° =cos15N W= °( )( )sin15 0.40cos15 6 mA BU W− = ° − °( )0.76531 0.76531A BU W mg− = − = −A A B BT U T−+ =2010.76531 2 m2mv mg− =( ) ( )( )( )2 20 2 2 0.76531 9.81 m/sv = +20 19.0154v =0 4.36 m/sv = !
15. 15. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 14.Given: At A, 0v v=At B, 0v =For AB, 0.40kµ =Find: 0v20102A BT mv T= =( )( )sin15 6 mA B kU W Nµ− = ° −0 cos15 0F N WΣ = − ° =cos15N W= °( )( )sin15 0.40cos15 6 mA BU W− = ° − °( )0.76531 0.76531A BU W mg− = − = −A A B BT U T−+ =2010.76531 02mv mg− =( )( )( )2 20 2 0.76531 9.81 m/sv =20 15.015v =0 3.87 m/sv = !Down to the left.
16. 16. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 15.Car C 3 2 30 0 (35 10 kg) (9.81 m/s ) 343.35 10 Ny C C CF N M g NΣ − ⇒ − = ⇒ = × = ×So, 3 3(0.35)(343.35 10 ) 120.173 10 NCF = × = ×Car B 3 2 30 0 (45 10 kg)(9.81m/s ) 441.45 10 Ny B B BF N M g NΣ = ⇒ − = ⇒ = × = ×So, 3 30.35(441.45 10 ) 154.508 10 NBF = × = ×Also,1 (54 km/h)(1h/3600s)(1000 m/km) 15 m/sv = =(a) Work and energy for the train1 1 2 2T U T→+ =3 3 3 2 3 31(35 10 45 10 35 10 )(15) (120.173 10 154.508 10 ) 02x× + × + × − × + × =47.10 mx =47.1 mx = !
17. 17. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) Force in each couplingCar ACar C1 1 2 2T U T→+ =( )( ) ( )23135 10 15 47.10 02ABF× − =383.599 10 NABF = ×83.6 kNABF = !Tension1 1 2xT U T→+ =( )( ) ( )( )23 3135 10 15 120.173 10 47.10 02BCF× + − × =Solve for FBC336.6 10 NBCF = ×36.6 kNBCF = !Tension
18. 18. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 16.0 0y A AF N M gΣ = ⇒ − =( )( )3 335 10 9.81 343.35 10 NAN = × = ×so,( )( )3 30.35 343.35 10 120.173 10 NAF = × = ×( )1 54 km/h 15 m/sv = =(a) Work - energy for the entire train1 1 2 2T U T→+ =( ) ( ) ( ) ( )23 3 3 3135 10 45 10 35 10 15 120.173 10 02x × + × + × − × = 107.66 mx =107.7 mx = !(b) Force in each couplingCar A1 1 2 2T U T→+ =( )( ) ( )( )23 3135 10 15 120.173 10 107.66 02ABF× − + × =383.60 10 NABF = − ×83.6 kNABF = !Compression1 1 2 2T U T−+ =( )( ) ( )23135 10 15 107.66 02BCF× + =336.57 10 NBCF = − ×36.6 kNBCF = !CompressionCar ACar C
19. 19. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 17.Car B:Car A:Given: Car B towing car A uphill at A constant speed of 30 ft/sCar B skids to a stop. µk = 0.9Car A strikes rear of car B.Find: Speed of car A before collision, vALet d = Distance traveled by car B after braking.1 2 2 1U T T− = − ( )21sin52Bm v mg F d− = − °−( )21302sin5 0.9 cos5mdmg mg= −°+ °( ) ( )( )450 45032.2 sin5 0.9cos5 32.2 0.9837d = =°+ °14.206 ft traveled byd B=For car A, travel to contact2 21 1 11 12 2C A AU T T mv mv→ = − = −( )( ) ( )221 1sin5 15 302 2Amg d mv m− ° + = −( )( )21450 32.2sin5 14.206 152Av − = − ° +21368.0362Av =27.13Av = 27.1 ft/sAv = !
20. 20. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 18.F = 0.8 NAGiven: Car B tows car A at 30 ft/s uphill.Car A brakes for 4 wheels skid µk = 0.8Car B continues in same gear and throttle setting.Find: (a) distance, d, traveled to stop(b) tension in cable(a) 1 Traction force (from equilibrium)F =( ) ( )1 3000 sin5 2500 sin5F = ° + °5500sin5= °For system A + B( )1 2 1 3000 sin5 2500sin5U F F d→  = − °− ° − ( )222 11 1 55000 302 2 32.2A BT T m v+ − = − = −   Since ( )1 3000 sin5 2500 sin5 0F − ° − ° =( )0.8 3000cos5 76863 ft lbFd d− = − ° = − ⋅32.1 ftd = !(b) cable tension, T( )( )1 2 2 10.8 sin5 32.149A AU T N W T T→ = − − ° = −( )( )( ) ( )( )( )23000 300.8 3000 cos5 3000 sin5 32.1492 32.2T − ° − ° =( )2652.3 1304T − = −1348 lb=1348 lbT = !NA = 3000 cos 5°NB = 2500 cos 5°
21. 21. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 19.Given: Blocks A, B released from rest and friction and masses of pulleysneglected.Find: (a) Velocity of block A, vA, after moving down dA = 1.5 ft.(b) The tension in the cable.(a) constraint 3 0A Bv v+ =13B Av v=Also,13B Ad d=( ) ( )1 2 sin30 sin30A A B BU W d W d→ = ° − °( )( ) ( )1.520 sin30 1.5 16 sin303 = ° − °   11 ft lb= ⋅2 21 210,2 2A A B BT T m v m v1= = +22 21 20 1 160.338162 32.2 2 32.2 3AA Avv v     = + =          21 2 2 1 ; 11 0.33816 AU T T v→ = − =5.703Av = 5.70 ft/sAv = 30° !(b) For A alone( ) ( ) ( )21 21sin302A A A A AU W d T d m v→ = ° − =( )( ) ( ) ( )21 2020 0.5 1.5 1.5 5.703 10.1022 32.2T − = =  3.265 ft lbT = ⋅3.27 ft lbT = ⋅ !
22. 22. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 21.Given: System at rest when 500 N force is applied to collar A. Nofriction. Ignore pulleys mass.Find: (a) Velocity, Av of A just before it hits C.(b) Av If counter weight B is replaced by a 98.1 Ndownward force.Kinematics2B AX X=2B Av v=(a) Blocks A and B2 21 21 102 2B B A AT T m v m v= = +( )( ) ( )( )2 221 110 kg 2 20 kg2 2A AT v v= +( )( )22 30 kg AT v=( ) ( )( ) ( )( )1 2 500 A A A B BU X W X W X− = + −( )( ) ( )( )21 2 500 N 0.6 m 20 kg 9.81 m/s 0.6 mU − = + ×( )( )210 kg 9.81 m/s 1.2 m− ×1 2 300 117.72 117.72 300 JU − = + − =
23. 23. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) 21 1 2 2 0 300 J 30 kg AT U T v−+ = + =210Av =3.16 m/sAv = !(b) Since the 10 kg mass at B is replaced by a 98.1 N force, kineticenergy at 2 is,( )2 22 11 120 kg 02 2A A AT m v v T= = =The work done is the same as in part (a)1 2 300 JU − =( ) 21 1 2 2 0 300 J 10 kg AT U T v−+ = + =230Av =5.48 m/sAv = !
24. 24. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 21.Given: System at rest when 500 N force is applied to collar A. Nofriction. Ignore pulleys mass.Find: (a) Velocity, Av of A just before it hits C.(b) Av If counter weight B is replaced by a 98.1 Ndownward force.Kinematics2B AX X=2B Av v=(a) Blocks A and B2 21 21 102 2B B A AT T m v m v= = +( )( ) ( )( )2 221 110 kg 2 20 kg2 2A AT v v= +( )( )22 30 kg AT v=( ) ( )( ) ( )( )1 2 500 A A A B BU X W X W X− = + −( )( ) ( )( )21 2 500 N 0.6 m 20 kg 9.81 m/s 0.6 mU − = + ×( )( )210 kg 9.81 m/s 1.2 m− ×1 2 300 117.72 117.72 300 JU − = + − =
25. 25. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.PROBLEM 13.21 CONTINUED( ) 21 1 2 2 0 300 J 30 kg AT U T v−+ = + =210Av =3.16 m/sAv = !(b) Since the 10 kg mass at B is replaced by a 98.1 N force, kineticenergy at 2 is,( )2 22 11 120 kg 02 2A A AT m v v T= = =The work done is the same as in part (a)1 2 300 JU − =( ) 21 1 2 2 0 300 J 10 kg AT U T v−+ = + =230Av =5.48 m/sAv = !
26. 26. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 22.Given: 10 kg; 4 kg; 0.5 mA Bm m h= = =System released from rest.Block A hits the ground without rebound.Block B reaches a height of 1.18 m.Find: (a) Av just before block A hits the ground.(b) Energy, ,PE dissipated by the pulley friction.(a) Bv at 2 Av= at 2 just before impact.from 2 to 3; Block B( )2 2 23 21 10 4 22 2B B B BT T m v v v= = = =Tension in the cord is zero, thus( )( )( )22 3 4 kg 9.81 m/s 0.18 m 7.0632 JU − = − = −2 22 2 3 3; 2 7.0632; 3.5316B BT U T v v−+ = = =2 23.5316 1.8793B A B Av v v v= = = = 1.879 m/sAv = !(b) From 1 to 2 Blocks A and B,( ) 21 2 2102A BT T m m v= = +Just before impact 2 1.793 m/sB Av v v= = =( )( )22110 4 1.8793 24.722 J2T = + =( ) ( )1 2 0.5 0.5 ;A B PU W W E− = − −Energy dissipated by pulleyPE =( )( ) ( )21 2 9.81 m/s 10 4 kg 0.5 m 29.43P PU E E− = − − = − 1 1 2 2; 0 29.43 24.722PT U T E−+ = + − =4.708PE = 4.71 JPE = !
27. 27. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 23.Given: 8 kg; 10 kg; 6 kgA B Cm m m= = =System released from rest.Collar C removed after blocks move 1.8 m.Find: Av , just before it strikes the ground.Position 1 to position 21 10 0v T= =At 2, before C is removed from the system( ) 22 212A B CT m m m v= + +( ) 2 22 2 2124 kg 122T v v= =( ) ( )1 2 1.8 mA C BU m m m g− = + −( ) ( )1 2 8 6 10 g 1.8 m 70.632 JU − = + − =21 1 2 2 2; 0 70.632 12T U T v−+ = + =22 5.886v =Position 2 to position 3( ) ( )22 21 185.886 52.9742 2A BT m m v′ = + = =( ) 2 23 3 3192A BT m m v v= + =( ) ( ) ( )( )( )22 3 2 0.6 2 kg 9.81 m/s 1.4 mA BU m m g′− = − − = −2 3 27.468 JU ′− = −22 2 3 3 352.974 27.468 9T U T v′−′ + = = − =23 32.834 1.68345v v= = 1.683 m/sAv = !
28. 28. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 23.Given: 8 kg; 10 kg; 6 kgA B Cm m m= = =System released from rest.Collar C removed after blocks move 1.8 m.Find: Av , just before it strikes the ground.Position 1 to position 21 10 0v T= =At 2, before C is removed from the system( ) 22 212A B CT m m m v= + +( ) 2 22 2 2124 kg 122T v v= =( ) ( )1 2 1.8 mA C BU m m m g− = + −( ) ( )1 2 8 6 10 g 1.8 m 70.632 JU − = + − =21 1 2 2 2; 0 70.632 12T U T v−+ = + =22 5.886v =Position 2 to position 3( ) ( )22 21 185.886 52.9742 2A BT m m v′ = + = =( ) 2 23 3 3192A BT m m v v= + =( ) ( ) ( )( )( )22 3 2 0.6 2 kg 9.81 m/s 1.4 mA BU m m g′− = − − = −2 3 27.468 JU ′− = −22 2 3 3 352.974 27.468 9T U T v′−′ + = = − =23 32.834 1.68345v v= = 1.683 m/sAv = !
29. 29. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 24.Given: Conveyor is disengaged, packages held by friction and system is released from rest. Neglect mass ofbelt and rollers. Package 1 leaves the belt as package 4 comes onto the belt.Find: (a) Velocity of package 2 as it leaves the belt at A.(b) Velocity of package 3 as it leaves the belt at A.(a) Package 1 falls off the belt, and 2, 3, 4 move down.2.40.8 m3=22 2132T mv =   ( ) 22 233 kg2T v=22 24.5T v=( )( )( ) ( )( ) ( )21 2 3 0.8 3 3 kg 9.81 m/s 0.8U W− = = ×1 2 70.632 JU − =21 1 2 2 20 70.632 4.5T U T v−+ = + =22 15.696v =2 3.9618v = 2 3.96 m/sv = !
30. 30. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 25.Work and energy 1 1 2 2T U T→+ = (1)Where 1 20; 0T T= =WorkOuter spring ( )221 2 11 1 N3000 1.5 m2 2 mV k x− = − = −   33.75 J= −Inner spring ( )221 2 21 1 N10,000 0.06 m2 2 mU k x− = − = −   18 J= −Gravity ( )1 2 0.15U mg h− = +( )( )( )8 9.81 0.15 78.48 11.722h h= + = +Total work 1 2 33.75 18 78.48 11.772U h− = − − + +39.978 78.48 h= − +Substituting into (1)0 39.978 78.48 0h− + =0.5094 mh =509 mmh = !
31. 31. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 26.Work and energyAssume 0.09 mx >1 1 2 2T U T→+ = (1)Where 1 20; 0T T= =WorkOuter spring ( )12 2 21 2 11 13000 15002 2U k x x x→ = − = = −Inner spring ( ) ( )22 21 2 210.09 5000 0.092U k x x→ = − − = − −Gravity ( )1 2 0.6U mg x→ = +( )( )( )8 9.81 0.6 78.48 47.09x x= + = +Total work( )221 2 1500 5000 0.09 78.48 47.09U x x x→ = − − − + +Substitute into (1)26500 978.48 6.588 0x x− + + =Solve0.1570 mx = or –0.00646Reject negative solution 157.0 mmx = !
32. 32. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 27.(a)Given: A 0.7 lb block rests on a 0.5 lb block which is not attached to aspring of constant 9 lb/ft; upper block is suddenly removed.Find: (a) maxv of 0.5 lb block(b) maximum height reached by the 0.5 lb blockAt the initial position (1), the force in the spring equals the weight of bothblocks, i.e., 1.2 lb.Thus at a distance x, the force in the spring is,1.2sF kx= −1.2 9sF x= −Max velocity of the 0.5 lb block occurs while the spring is still in contactwith the block.2 2 21 21 1 0.5 0.2502 2T T mv v vg g = = = =  ( ) 21 2 091.2 9 0.5 0.72xU x dx x x x− = − − = −∫2 21 1 2 29 0.250.72T U T x x vg−+ = = − =2 294 0.72v g x x = −  max when 0 0.7 9 0.077778 ftdvV x xdx= = − ⇒ =( ) ( )22max94 0.7 0.077778 0.0777782v g = −  2max 3.5063v =max 1.87249v =max 1.872 ft/sv = !
33. 33. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b)0 Initial compressionx =01.2 lb0.133333 ft9 lb/ftx = =1.2 9sF x= −1 30, 0T T= =01 3 00.5xsU F dx h− = −∫( )01 3 01.2 9 0.5xU x dx h− = − −∫20 091.2 0.52x x h= − −( ) ( )291.2 0.133333 0.133333 0.52h= − −0.08 0.5h= −( )1 1 3 3: 0 0.08 0.5 0T U T h−+ = + − =0.16 fth =1.920 in.h = !
34. 34. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 28.(a) Same as 13.25 solution for Part (a) max 1.872 ft/sv = !(b) With 0.5 lb block attached to the spring, refer to figure in (b) of Problem 13.27.( )1 3 1 3 00 0 1.2 9 0.5hT T U x dx h−= = = − −∫Since the spring remains attached to the 0.5 lb block,the integration must be carried out for the total distance, h.21 1 3 390 0.7 02T U T h h−+ = + − =( )20.7 lb 0.155556 ft9 lb/fth = =  1.86667 in.h =1.867 in.h = !
35. 35. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 29.Position 1, initial conditionPosition 2, spring deflected 5 inchesPosition 3, initial contact of spring with collar( )21 218 5 1 5 18 560 + 7.5 sin3012 2 12 12(Friction) (Spring) (Gravity)U F→+ +     = − − °          1 2 1 20, 0T T U −= = ∴ =( ) ( ) ( ) ( )223 1 5 230 7.5 0.866 60 7.5 0.512 2 12 12µ     = − − +          (a) 0.1590µ = !(b) Max speed occurs just before contact with the spring( ) ( ) ( ) 21 3 3 max18 18 1 7.57.5 0.866 7.5 0.512 12 2 32.2U T vµ→     = − + = =          max 5.92 ft/sv = !
36. 36. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 30.(a) W = Weight of the block ( )10 9.81 98.1 N= =12B Ax x=( ) ( ) ( )2 21 21 12 2A A A B BU W x k x k x− = − −(Gravity) (Spring A) (Spring B)( )( ) ( )( )21 2198.1 N 0.05 m 2000 N/m 0.05 m2U − = −( ) ( )212000 N/m 0.025 m2−( ) ( )2 21 21 110 kg2 2U m v v− = =( ) 214.905 2.5 0.625 102v− − =0.597 m/sv = !(b) Let Distance moved down by the 10 kg blockx =( ) ( ) ( )22 21 21 1 12 2 2 2A BxU W x k x k m v− = − − =  ( ) ( ) ( )210 22 8BAd km v W k x xdx = = − −  
37. 37. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( ) ( )20000 98.1 2000 2 98.1 2000 2508x x x= − − = − +( )0.0436 m 43.6 mmx =For ( ) 210.0436, 4.2772 1.9010 0.4752 102x U v= = − − =max 0.6166 m/sv =max 0.617 m/sv = !
38. 38. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 31.(a) W = Weight of the block = (10)(9.81) = 98.1 N12B Ax x=Block moves down at release after spring A is stretched 25 mm2 21 21 1( ) ( ) ( )2 2A A A B BU W x k x k x− = + −(Gravity) (Spring A) (Spring B)21 21(98.1 N)(0.025 m) (2000 N/m)(0.025 m)2U − = +21(2000 N/m)(0.0125 m)2−2 21 1( ) (10 kg)2 2m v v= =21 212.4525 0.625 0.15625 (10)2U v− = + − =0.764 m/sv = 0.764 m/sv = !(b) Let x = Distance moved down by the 10 kg block(for x > 25 mm)21 21 1(0.025) ( 0.025)2 2A AU Wx k k x− = + − −221 1( )2 2 2Bxk M v − =  22 21 21 1 198.1 (2000)(0.025) (2000)( 0.025) (2000)2 2 2 2xU x x− = + − − −   21(10)2v=
39. 39. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.210 (10) 98.1 2000( 0.025) 20002 4d xv xdx   = = − − −      98.1 2000 50 500x x= − + −148.10.05924 m ( 59.24 mm)2500x = = =For 0.05924 mx =21 2 98.1(0.05924) 0.625 1000(0.03424) 0.87734U − = + − −215.8114 0.625 1.1724 0.87734 (10)2v+ − − =max 0.937 m/sv =max 0.937 m/sv = !
40. 40. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 32.(a)Assume auto stops in 1.5 4 m.d< <( )( )221 1 11 127.778 m/s 1000 kg 27.778 m/s2 2v T mv= = =385809 J 385.81 kJ= =2 20 0T v= =( )( ) ( )( )1 2 80 kN 1.5 m 120 kN 1.5 120 120 180 120 60U d d d− = + − = + − = −1 1 2 2 385.81 120 60 3.715 mT U T d d−+ = = − =3.72 md = !Assumption that 4 md < is O.K.(b) Maximum deceleration occurs when F is largest.For 3.3401 m, 120 kN, thus Dd F F ma= = =( ) ( )( )120,000 N 1000 kg Da=2120 m/sDa =2120.0 m/sDa = !
41. 41. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 33.Pressures vary inversely as the volumeLLP Aa PaPP Ax x= =( ) ( )2 2RRP Aa PaPP A a x a x= =− −Initially at 1 02av x= =1 0T =At 2, 221,2x a T mv= =( )2 21 21 12a aa aL Ru P P Adx PaA dxx a x− = − = − − ∫ ∫( )1 22ln ln 2aau paA x a x−  = + − 1 23ln ln ln ln2 2a au paA a a−    = + − −        221 23 4ln ln ln4 3au paA a paA−   = − =     21 1 2 24 10 ln3 2T U T paA mv− + = + =  242 ln30.5754paApaAvm m   = = 0.759paAvm= !
42. 42. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 34.( ) ( )22 2/1E EhhRGM m GM m RF mgh R= = =+ +At earths’ surface ( )0h = 02EGM mmgR=( )202 21EGME RhhRGMg gR= =+Thus( )021hhRgg =+6370 kmR =At altitude h, “true” weight h TF mg W= =Assume weight 0 0W mg=0 0 00 0 0Error T h hW W mg mg g gEW mg g− − −= = = =( )( )( )020102 20111 1hRghh hR Rggg Eg+−   = = = − + +  (a)( )21637011 km: 100 100 11h P E  = = = − +  0.0314%P = !(b)( )21000637011000 km: 100 100 11h P E  = = = − +  25.3%P = !
43. 43. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 35.Newtons law of gravitation2 21 0 21 12 2T mv T mv= =( )21 2 2m nmR h m mn nRmg Ru F dr Fr+− = − =∫21 2 2m nmR hm m Rdru mg Rr+− = − ∫21 21 1m mm m nu mg RR R h− = − + 1 1 2 2T U T−+ =2 201 12 2mm mm nRmv mg R mvR h + − = + ( )( )2 202 2022mnv vmm gmv v RhgR− −  =   −  (1)Uniform gravitational field2 21 0 212T mv T mv= =( ) ( )1 2m nmR hu m m u m uRu F dr mg R h R mgh+− = − = − + − = −∫2 21 1 2 2 01 12 2m uT u T mv mg h mv−+ = − =( )2 202umv vhg−= (2)Divide (1) by (2)( )( )2 20211m mnv vug Rhh −=−!
44. 44. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 36.6(3960 mi)(5280 ft/mi) 20.9088 10 ftR = = ×1 2 2 12 1GMm GMmU T Tr r− = − = −Since 1r is very large,110 thus 0GMmTr≈ ≈2 222 21,2 2GMm v Rmv gr r= =( )( )22 6222 22 32.2 ft/s 20.9088 10 ft2gRvr r×= =(a) For 62 20.9088 10 ft ((620 mi)(5280 ft/mi))r = × +624.1824 10 ft= ×34,121 ft/sv = 6.46 mi/sv = !(b) For 62 20.9088 10 ft ((30 mi)(5280 ft/mi))r = × +621.0672 10 ft= ×36,557 ft/sv = 6.92 mi/sv = !(c) For 62 20.9088 10 ft/sr = ×36,695 ft/sv = 6.95 mi/sv = !
45. 45. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 37.0 620 mi 3960 mi 4580 mi 24,182,400 ftr = + = =3960 mi 5200 mi 9160 mi = 48,364,800 ftBr = + =Parabola: 2y Kx=At B: ( )220 24,182,400 ft 48,364,800 ftBr Kr K= =9 110.3381 10 ftK − −= ×At A: ( ) ( )20sin 45 , cos45A A A A Ax r y Kx r r= ° = = − °( ) ( )220 cos45 sin 45A Ar r Kr− ° = °20 0A AKx x r+ − =0 (6.5)(5280) 34,320 ft/sv = =( )011 1 42Ax KrK= − + +6 620.0334 10 ft, 14.1657 10 ft.A Ax r= × = ×(a) 2 20 001 12 2A AAGMm GMmU mv mvr r→ = − = −( )26 2 2001 132.2 20.9088 10 , 2AAGM v v GMr r = × = + −  ( )226 61 134320 214.1657 10 24.1824 10Av GM = + − × × 44734 ft/sAv = 8.47 mi/sAv = !(b) 2 2001 12BBv v GMr r = + −  ( )226 61 134320 248.3648 10 28.1824 10Bv GM = + − × × 24408 ft/sBv = 4.62 mi/sBv = !
46. 46. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 38.(a) 21.62 m/sg = Use work and energy 1 1 2 2T U T→+ = (1)where 2 21 11 1(600) 180,0002 2T mv m m= = =2 0T = (maximum elevation)1 2U mgh→ = −(1.62) 1.62m h mh= − = −Substituting into (1) 180,000 1.62 0m mh− =3111.11 10 mh = ×111.1 kmh = !(b) 12so 180,000 (same as above)GMmF T mR= =22GMmW mg GM gRR= = ⇒ =At some elevation r 2GMmFr= −so,1 2 2R hRGMmU drr+→ = −∫2221 1 1rRGMm gR mr r R  = = −     6 2621 1(1.62)(1.740 10 )r 1.740 10m = × − × 
47. 47. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Substituting into (1)12621 1180,000 4.9047 10 m 01.74 10mr + × − = × Solve for r262 1.8587 10 mr = ×1858.7 km=2so, 1858.7 1740h r R= − = −118.7 km=118.7 kmh = !
48. 48. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 39.Assume the blocks move as one: 2 1 1 2T T U −− =2 21 2friction1 1( )2 2A Bm m v kx U −+ = −2 21 1(3 kg) (180 N/m)(0.1 m) (3)(9.81)(0.1)(0.1 m)2 2v = −20.4038 0.63545 m/sv v= =Check assumption at release, s18 N > (3)(9.81) 4.41 Nµ =∴ Slips at the horizontal surfaceAt release18 NsF = x xF maΣ =18 2.94 3a− = 25.02 m/sa =For A alone:x xF maΣ =18 (1.5)(5.02)fF− =10.47 NfF =s18 7.53 10.47 N < (1.5)(9.81) (0.95)(14.175) 13.98 NfF µ= − = = =∴ A and B move as one, thus (a) 0.635 m/sv = !(b) vmax is max at a = 0,0 180 2.943, 0.01635 ms fF F x x= = = − =2 2 2max 0 01 1( ) ( ) ( )2 2A B fm m v k x x F x x+ = − − −2 2 2max1 1(3) (180)[(0.1) (0.01635) ] 2.943(0.1 0.01635)2 2v = − − −max 0.648 m/sv = !0
49. 49. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 40.From problem 13.39 assuming the blocks move together,25.02 m/s at release.a =2(1.5 kg)(9.81 m/s ) 14.715 NA BW W= = =s18 7.53 10.47 N < (1.5)(9.81) 0.35(14.715) 5.15 NfF µ= − = = =∴ Block A slides on Block BA alone:(a) 22 1 1 2 1 2 friction1 1, ( )2 2AT T U m v kx U− −− = = −2 21 1(1.5)( (180 N/m)(0.1 m ) 14.175(0.3)(0.1 m)2 2v = −20.6114v = 0.782 m/sv = !(b) max at acceleration 0,v v= =0s f k AF F kx Wµ− = = −180 (0.30)(14.715) 4.4145, 0.0245 mx x= = =2 22 1 0 01( ) ( )2A kT T k x x W x xµ− = − − −2max1(1.5) 0.84598 0.33329 0.512682v = − =max 0.827 m/sv = !00
50. 50. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 41.21 012T mv=2212T mv=1 2U mgl− = −2 21 1 2 2 01 12 2T U T mv mgl mv−+ = − =2 20 2v v gl= +Newtons’ law at 2(a) For minimum v, tension in the cord must be zero.Thus 2v gl=2 20 2 3v v gl gl= + =0 3v gl= !(b) Force in the rod can support the weight so that v can be zero.Thus 20 0 2v gl= +0 2v gl= !
51. 51. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 42.( )221 01 116 128 m2 2T mv m= = =2212T mv=( )1 2 6sinU mg θ− =21 1 2 21: 128 6 sin2T U T m mg mvθ−+ = + =2256 12 sing vθ+ = (a)Newton’s law2: 2 sin6nmvF ma mg mg θΣ = − =Using (a) ( )12 6sin 256 12 sing gθ θ− = +18 sin 12 256g gθ = −( )( )12 32.2 256sin 0.2249818 32.2θ−= = 13.00θ = ° !
52. 52. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 43.Use work - energy : position 1 is bottom and position 2 is the top1 1 2 2T U T→+ = (1)where,21 012T mv=22 212T mv=1 2 (0.5)U mgh mg→ = − = −Substituting into (1)2 20 21 1(0.5)2 2mv mg mv− =so2 20 2v v g= + (2)(a) Slender rod 2 00v v g= ⇒ =0 3.13 m/sv = !(b) Cord, so the critical condition is tension = 0 at the top2222nmvF ma mg v gρρΣ = ⇒ = ⇒ =Substituting into (2)20 9.81(0.25 1)v g gρ= + = +12.2625=0 3.502 m/sv =0 3.50 m/sv = !
53. 53. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 44.Use work - energy : position 1 is at A, position 2 is at B.1 1 2 2T U T→+ = (1)Where 21 1 2 210; sin ;2BT U mg l T mvθ→= = =Substitute210 sin2Bmg l mvθ+ =22 sinBv g l θ= (2)For T = 2 W use Newtons 2nd law.22 sin Bn nmvF ma W WlθΣ = ⇒ − = (3)Substitute (2) into (3)sin2 sin 2lmg mg mglθθ− =2 3sinθ=2or sin 41.813θ θ= ⇒ = °41.8θ = °!
54. 54. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 45.( )2 2 21 10 0 250 kg 1252 2A A B B B Bv T T mv v v= = = = =( ) ( )21250 kg 9.81 m/sW = ×( )( )27 1 cos40A BU W− = − °( )( )( )2250 kg 9.81m/s 27 m 0.234A BU − = ×15495 JA BU − =20 15495 125A A B A BT U T v−+ = + =( )( )2 15495 J125 kgBv =2 2 2124.0 m /sBv =Newtons Law at B22 2 2cos40 ; 124.0 m /sBBmvN W vR−− ° = =( )( )( )( )2 22250 kg 124.0 m /s250 kg 9.81m/s cos4027 mN = × ° −1879 1148 731 NN = − = 731 NN = !
55. 55. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 46.Normal force at BSee solution to Problem 13.45, 731.0 NBN =Newtons LawFrom B to C (car moves in a straight line)cos40 0BN W′ − ° =( )2250 kg 9.81m/s cos40 1878.7 NBN′ = × ° =At C and D (car in the curve at C)At C2cos CCW vN Wg Rθ− =( )22250 kg 9.81m/s cos CCvNgRθ = × +   At D2DDW vN Wg R− = +continued
56. 56. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )2250 9.81 1 DDvNgR = × +   Since and cos 1,D C D Cv v N Nθ> < >Work and energy from A to D2 210, 0 1252A A D D Dv T T mv v= = = =( ) ( )( )( )227 18 250 kg 9.81m/s 45 m 110362.5A DU W− = + = =20 110362.5 125A A D D DT U T v−+ = + =2882.90Dv =( )( )2882.90250 1 250 9.81 1 5518.1 N72 72 9.81DDvN gg  = + = + =        min max731 N; 5520 NB DN N N N= = = = !
57. 57. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 47.Kinematics: 21constant, ,2a v at x at= = =( )2 21110 5.4 , 7.5446 ft/s2x a a= = =( )22150 lbconstant 7.5446 ft/s 35.1456 lb32.2 ft/sF ma = = = =  7.5446v t=( )2150Power 7.544632.2Fv mav t = = =   ( )( ) ( )2 25.40150/32.2 7.5446 5.41Average power 05.4 5.4 2Fv dt  = = −  ∫Average power 715.93 ft lb/s= ⋅Average power 1.302 hp= !
58. 58. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 48.3tan100θ =1.718θ = °( ) 27 60 kg 9.81m/sB WW W W= + = + ×657.3 NW =( )( )sinWP W v W vθ= ⋅ =( )( )( )657.3 sin1.718 2WP = °39.41 WWP =39.4 WWP = !( ) 29 90 kg 9.81m/sB mW W W= + = + ×971.2 NW =Brake must dissipate the power generated by the bike and the man goingdown the slope at 6 m/s.( )( )sinBP W v W vθ= ⋅ =( )( )( )971.2 sin1.718 6 174.701BP = ° =174.7 WBP = !
59. 59. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 49.(a) ( ) ( )( ) ( )( ) ( )2800 650p A A L A AAP F v W W v v= = + = +6.5 fts/t 0.40625 ft/s16 sAv = = =( ) ( )( )3450 lb 0.40625 ft/s 1401.56 lb ft/sp AP = = ⋅( )1hp 550 ft lb/s, 2.548 hpp AP= ⋅ =( ) 2.55 hpp AP = !(b) ( )( ) 2.550.82p AE APPη= =( ) 3.11hpE AP = !
60. 60. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 50.(a) Material is lifted to a height b at a rate, ( )( ) ( )2kg/h m/s N/hm g mg =  Thus,( ) ( )( )N/hN m/s3600 s/h 3600mg b mU mgbt   ∆     = = ⋅ ∆  1000 N m/s 1kw⋅ =Thus, including motor efficiency, η( )( )( ) ( )N m/skw1000 N m/s3600kwmgbPη⋅=⋅   ( ) 6kw 0.278 10mgbPη−= × !(b)( )( ) ( )tons/h 2000 lb/ton ft3600 s/hW bUt   ∆    =∆ft lb/s; 1hp 550 ft lb/s1.8Wb= ⋅ = ⋅With ,η ( )1hp 1ft lb/s1.8 550 ft lb/sWbhpη    = ⋅      ⋅    31.010 10 Wbhpη−×= !
61. 61. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 51.For constant power, P:dvP Fv mav m vdt= = =Separate variables12 21 00 02 2t v m dv m v vdt tP v P = ⇒ = −   ∫ ∫ (1)Distance2dv dx dvP mv mvdx dt dx= =Separate variables103 32 1 003 3x vvm m v vdx v dv xP P = ⇒ = −   ∫ ∫ (2)with numbers(a) 0 136 km/h 10 m/s; 54 km/h 15 m/s, so,v v= = = =( )( ) ( )32 2315 10 kg15 m/s 10 m/s2 50 ×10 Wt×  = −  18.75 st⇒ = !( )( )33 3315 1015 10 237.5 m3 50 10x×  = − = ×238 mx⇒ = !(b) 0 154 km/h 15 m/s; 72 km/h 20 m/sv v= = = =( )( )32 2315 1020 15 26.25 s2 50 10t×  = − = ×26.2 st⇒ = !( )( )33 3315 1020 15 462.5 m3 50 10x×  = − = ×462 mx⇒ = !
62. 62. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 52.(a) constantdvP F v m vdt = ⋅ = =  4.3 52.0 0m v dv P dt∴ =∫ ∫( ) ( )2 24.3 m/s 20 m/s60 kg 5 , 86.94 W2P P −  = =  86.9 WP = !(b) constantdvP F v mv vdx = ⋅ = =  4.3 22.0 0xm v dv P dx∴ =∫ ∫( ) ( )3 34.3 2.060 86.943x −  =  16.45 mx = !
63. 63. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 53.Motion is determined as a function of time as,( )12000 ln cosh 0.03x t=Velocity ( )( )112000 sinh 0.03 0.03cosh 0.03dxv tdt t = =   360 sinh 0.03cosh 0.03tvt=Power dissipated ( )2 30.01 0.01P Dv v v v= = =( )33 0.03 0.033 30.03 0.03sinh 0.030.01 360 466.56 10cosh 0.03t tt tt e ePt e e−−   −= = ×   +   (a) 10 s,t = 11534 ft lb/sP = ⋅ 21.0 hpP = !(b) 15 s,t = 35037 ft lb/sP = ⋅ 63.7 hpP = !
64. 64. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 54.Motion is defined by the following function:0.000511 x dva e vdx−= =0.00050.00050 0 011110.0005v x xx uvdv e dx e du−− −= =∫ ∫ ∫( )20.000522000 12xve−= −( )2 0.000544000 1 xv e−= −( )10.0005 2209.76 1 xv e−= −3Power dissipated 0.01P Dv v= =30.0005 292295 1 xP e− = − (a) 600 ftx = , 12178 ft lb/sP = ⋅ 22.1 hpP = !(b) 1200 ft,x = 27971 ft lb/sP = ⋅ 50.9 hpP = !
65. 65. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 55.System is in equilibrium in deflected 0x position.Case (a) Force in both springs is the same P=0 1 2x x x= +0ePxk=1 21 2P Px xk k= =Thus1 2eP P Pk k k= +1 21 1 1ek k k= +1 21 2ek kkk k=+!Case (b) Deflection in both springs is the same 0x=1 0 2 0P k x k x= +( )1 2 0P k k x= +0eP k x=Equating the two expressions for( )1 2 0 0eP k k x k x= + =1 2ek k k= + !
66. 66. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 56.Use conservation of energyLet position 1 be where A is compressed 0.1 m; position 2 when B is compressed a maximum distanceSo1 1 2 2T V T V+ = + (1)Where 1 0;T = ( )( )221 11 11600 N/m 0.1 m 8 J2 2AV k x= = =2 0;T = ( )2 2 22 2 2 21 12800 N/m 14002 2BV k x x x= = =Substituting into (1)22 20 8 0 1400 0.07559 mx x+ = + ⇒ =This answer is independent of massDistance traveled 0.5 m 0.05 m 0.07559 m 0.526 m= − + =The maximum velocity will occur when the mass is between the two springswhere 1 1 2 2T V T V+ = +1 0;T = ( )1 8 J same as beforeV =22 max1;2T mv= 2 0V =Substituting into (1)2max10 8 0;2mv+ = + 2max16vm=For 1 kgm = 2max 16v = (a) max 4 m/sv = !For 2.5 kgm = 2max166.42.5v = = (b) max 2.53 m/sv = !
67. 67. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 57.2 21 6 9 10.817 in.= + =l( ) ( )2 20 6 8 10 in. 0.8333 ft= + = =lStretch 10.817 10 0.817 in.= − =1 0.06805 ftS =( ) ( )2 22 7 6 9.215 in.= + =lStretch 9.2195 10 0.7805 in.= − = −2 0.06504 ftS =1 20, 0T V= =2 22 2 21 1 42 2 32.2T mv v = =   ( )( )2 21 1 2133,600 lb/ft2V S S= +( )( )1 16,800 0.008861 148.86 ft lbV = = ⋅1 1 2 2T V T V+ = +22 2396.7v = 2 49.0 ft/sv = !0 0
68. 68. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 58.2800 lb/in. 33,600 lb/ftk = =( ) ( )221 1 11 1 10 33,6002 2 12T V k l = = ∆ =   116.667 ft lb= ⋅2 21 6 9 10.817 in. 0.9014 ft= + = =l1 Stretch 10.817 10 0.817 in. 0.06808 ftS = = − = =2 22 6 7 9.2195 in.= + =l2 Stretch 9.2195 9 0.2195 in. 0.018295 ftS = = − = =2 2 22 2 2 21 1 40.06212 2 32.2T mv v v = = =  ( )( )2 22 1 2133,6002V S S= +( ) ( )2 216800 0.06808 0.018295 = +  83.489 ft lb= ⋅1 1 2 2T V T V+ = +22116.667 0.06211 83.489v= +2 23.1 ft/sv = !0
69. 69. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 59.Use conservation of energy 1 1 2 2T V T V+ = + (1)(a) Position 1 is at A and position 2 is at B1 0;T = 21 112V kx= where 1 0x = −l l12 2 2 2500 400 350 729.726 mm = + + = lSo 0 429.726 mm− =l l( )( )2111500 N/m 0.429726 m 13.8498 J2V = =At B12 2 20350 400 531.507 mm 231.507 mm = + = ⇒ − = l l l( )2 2 22 2 2 21 10.75 0.3752 2T mv v v= = =( )( )221150 0.231507 4.01966 J2V = =Substituting into (1) 220 13.8498 0.375 4.01966v+ = +2 5.12 m/sBv v= = !(b) At E ( )1 10; 13.8498 same as beforeT V= =At E12 2 20350 500 610.328 mm 310.328 mm = + = ⇒ − = l l l2 2310.3752E ET mv v= =( )( )221 1150 0.310328 7.223 J2 2Ev kx= = =Substituting into (1)20 13.8498 0.375 7.223 4.20 m/sE Ev v+ = + ⇒ = !
70. 70. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 60.Conservation of energy 1 1 2 2T V T V+ = +At A12 2 2 20500 400 350 729.726 mm 729.726 450 279.726 mm = + + = ⇒ − = − = l l lSo ( )( )221 10; 150 0.279726 5.8685 J2 2A AT V kx= = = =At B12 2 20350 400 531.507 81.507 mm = + = ⇒ − = l l l( )( )22 2 21 1 10.375 ; 150 0.081507 0.49825 J2 2 2B B B BT mv v V kx= = = = =Substituting into (1) 20 5.8685 0.375 0.49825Bv+ = +3.78 m/sBv = !At E ( ) ( )12 2 20350 500 610.328 mm 160.328 mm = + = ⇒ − =  l l lSo ( )( )22 2 21 1 10.375 ; 150 0.160328 1.9279 J2 2 2E E E ET mv v V kx= = = = =Substituting into (1) 20 5.8685 0.375 1.9279EV+ = +3.24 m/sEv = !The fact the cord becomes slack doesn’t matter.
71. 71. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 61.(a) Maximum height when 2 0v =1 2 0T T∴ = =g eV V V= +Position 1 ( )10gV =16 lb6 in. 0.4 6 6.4 in.15 lb/in.x = + = + =( ) ( )( )22111 115 lb/in. 6.4 in.2 2eV kx= =307.2 lb in. 25.6 lb ft= ⋅ = ⋅Position 2 ( ) ( )266 0.512gV mg h h = + = +  ( )20eV =( ) ( ) ( ) ( )1 1 2 2 1 21 2: g e g eT V T V V V V V+ = + + = +( )25.6 6 0.5 h= +3.767 fth = 45.2 in.h = !(b) Maximum velocity occurs when acceleration is 0, equilibriumposition2 2 23 3 3 31 1 60.0931672 2 32.2T mv v v = = =  ( ) ( ) ( ) ( ) ( )2 23 13316 6 6 36 7.5 6.4 62g eV V V k x= + = + − = + −37.2 lb in. 3.1lb ft= ⋅ = ⋅21 1 3 3 3: 25.6 0.093167 3.1T V T V v+ = + = +max 15.54 ft/sv = !
72. 72. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 62.(a) Collar is in equilibrium.( )15 lb/in. 6 lbF δΣ = −( )( )6 lb0.4 in.15 lb/in.δ = =max 0.4 in.δ = !(b) Maximum compression occurs when velocity at 2 is zero.1 10 0T V= =22 2 max max102T V W kδ δ= = − +max12W kδ=( )( )max2 6 lb0.8 in.15 lb/in.δ = =max 0.8 in.δ = !
73. 73. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 63.30 rad6πθ = ° =0.3 mR =(a) Maximum heightAbove B is reached when the velocity at E is zero0CT =0ET =e gV V V= +Point C( )0.3 m rad6BCLπ ∆ =   m20BCLπ∆ =( ) ( ) ( )221 140 N/m m 0.4935 J2 2 20C BCeV k Lπ = ∆ = =  ( ) ( ) ( )( )( )21 cos 0.2 kg 9.81m/s 0.3 m 1 cos30C gV WR θ= − = × − °( ) 0.07886 JC gV =( ) 0 (spring is unattached)E eV =( ) ( )( ) ( )0.2 9.81 1.962 JE gV WH H H= = × =C C E ET V T V+ = +0 0.4935 0.07886 0 0 1.962H+ + = + +0.292 mH = !continued
74. 74. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) The maximum velocity is at B where the potential energy iszero, maxBv v=0 0.4935 0.07886 0.5724 JC CT V= = + =( )2 2max1 10.2 kg2 2B BT mv v= =2max0.1BT v=0BV =( ) 2max0 0.5724 0.1C C B BT V T V v+ = + + =2 2 2max 5.72 m /sv =max 2.39 m/sv = !
75. 75. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 64.0.3 mR =( ) ( )20.2 kg 9.81m/sW = ×1.962 N=(a) Smallest angleθ occurs when the velocity at D is close to zero0 0C Dv v= =0 0C DT T= =e gV V V= +Point C( )0.3 m 0.3 mBCL θ θ∆ = =( ) ( )212C BCeV k L= ∆( ) 21.8C eV θ=( ) ( )1 cosC gV WR θ= −( ) ( )( )( )1.962 N 0.3 m 1 cosC gV θ= −( ) ( ) ( )21.8 0.5886 1 cosC C Ce gV V V θ θ= + = + −Point D( ) 0 (spring is unattached)D eV =( ) ( ) ( )( )( )2 2 1.962 N 0.3 m 1.1772 JD gV W R= = =( )2; 0 1.8 0.5586 1 cos 1.1772 JC C D DT V T V θ θ+ = + + + − =( ) ( )21.8 0.5886 cos 0.5886θ θ− =By trial 0.7522 radθ =43.1θ = ° !continued
76. 76. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) Velocity at APoint D( )0 0 1.1772 J see Part ( )D D DV T V a= = =Point A( )2 21 10.2 kg2 2A A AT mv v= =20.1A AT v=( ) ( ) ( )( )1.962 N 0.3 m 0.5886 JA A gV V W R= = = =A A D DT V T V+ = +20.1 0.5886 0 1.1772Av + = +2 2 25.886 m /sAv =2.43 m/sAv = !
77. 77. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 65.Conservation of energyPosition (1) is at the top of the incline; position (2) is when the spring has maximum deformation1500 lb/ftk =Where 1 1 2 2T V T V+ = +At (1) ( )221 11 1 2008 198.76 ft lb2 2 32.2T mv + = = ⋅  ( )21 1 1 1 11datum at point 22g eV V V mgz k x= + = =( ) ( )( )21200 25 sin 20 1500 0.52x= − ° +Deformation of the springx =1 1710.1 68.404 187.5V x= + +At (2) ( )( )222 2 2 2 21 10; 1500 0.52 2g eT V V V k x x= = + = = +Substituting into (1) ( )2198.78 1710.1 68.404 187.5 750 0.5x x+ + + = +Solve 2750 681.596 1908.9 0x x+ − =2.11 or +1.2044 ftx = −1.204 ftx = !14.45 in.= !0
78. 78. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 66.Spring length 2 214 28= +31.305 in.=Stretch31.305 in. 14 in.12 in./ft−=1.44208 ft=( )( )20 010 0 48 lb/ft 1.44208 ft 49.910 lb ft2T V+ = + + = ⋅(a) At A: ( )2221 10 lb 1 14 2 1449.910 48 lb/ft2 2 1232.2 ft/sAv   −= +        16.89 ft/sAv = !(b) At B: ( )221 10 1 14 1449.910 48 102 32.2 2 12 12Bv     = + −          13.64 ft/sBv = !
79. 79. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 67.( ) ( )1 1 10, 0 Constraint: 2e g B AT V V y x= = = ↓ = →2 221 12 2A A B BT M v M v= +( ) ( )22 21 14 kg 1.5 kg 1.252 2 2BB Bvv v = + =  (a) ( )( )2210.15 m, 0.075 m, 300 N/m 0.075 m 0.84375 N m2B A ey x V= = = = ⋅( )( )( )2 1.5 9.81 0.15 m 2.2073 JgV = − = −21 1 2 2 ; 0 1.25 0.84375 2.20725BT V T V v+ = + = + −1.044 m/sBv = !(b) Maximum velocity when acceleration 0=; 2 0; Cord tensile forcex x AF ma k x T T= − = =∑( )2 14.7152; 0.0981 m; 2 0.1962 m300A A B ATx x x xk= = = = =( )( )2 14.715 N 0.1962 m 2.8871gV = − = −( )( )221300 N/m 0.0981 m 1.44352eV = =21 1 2 2 ; 0 1.25 1.44354BT V T V v+ = + = − 1.075 m/sBv = !(c) ( )22 210; 0 300 14.7152 2BByT V y = = = −  0.392 m 392 mmBy = = !
80. 80. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 68.(a) Calculate spring lengths after deflection.Original spring length 0.75 m,= collar moved 100 mm 0.1 m=( ) 2 21 1 210, 4 22T V T v v= = = =( )( )( )22 4 kg 9.81 m/s 0.1 m 3.924 JgV = − = −( ) ( ) ( )2 221300 0.8322 0.75 0.6727 0.752eV  = − + − 1.9098 J=21 1 2 2 : 0 2 3.924 1.9098T V T V v+ = + = − +1.0035 m/sv = 1.004 m/sv = !(b) Calculate spring lengths after deflection, collar moved190 mm 0.19 m=( )( )( )22 4 kg 9.81 m/s 0.19 m 7.4556 JgV = − = −( ) ( ) ( )2 221300 0.90918 0.75 0.60877 0.752eV  = − + − 6.7927 J=21 1 2 2 : 0 2 7.4556 6.7927T V T V v+ = + = − +0.576 m/sv = 0.576 m/sv = !
81. 81. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 69.(a) 0, 0C Cv T= =212B BT mv=( ) 210.2 kg2B BT v=20.1B BT v= ( ) ( )C C Ce gV V V= +arc BCBC L Rθ= ∆ =( )( )( )0.3 m 30180BCLπ∆ = °°0.15708 mBCL∆ =( ) ( ) ( )( )2 21 140 N/m 0.15708 m 0.49348 J2 2C BCeV k L= ∆ = =( ) ( ) ( )( )( )( )21 cos 0.2 kg 9.81 m/s 0.3 m 1 cos30C gV WR θ= − = − °( ) 0.078857 JC gV =( ) ( ) 0.49348 J 0.078857 J 0.57234 JC C Ce gV V V= + = + =( ) ( ) 0 0 0B B Be gV V V= + = + =2; 0 0.57234 0.1C C B B BT V T V v+ = + + =2 2 25.7234 m /sBv = 2.39 m/sBv = !(b)2BRmvF F WRΣ = − =( )( )( )2 25.7234 m /s1.962 N 0.2 kg0.3 mRF = +1.962 N 3.8156 N 5.7776 NRF = + = 5.78 NRF = !
82. 82. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 70.(a) Speed at C( ) ( ) ( )2 2 2300 150 75 343.69318 mmABL = + + =320 N/mk =At B 0 0B Bv T= =( ) ( )B B Be gV V V= +343.69318 mm 200 mmABL∆ = −143.69318 mm 0.14369318 mABL∆ = =( ) ( ) ( )( )2 21 1320 N/m 0.1436932 m2 2B ABeV k L= ∆ =( ) 3.303637 JB eV =( ) ( )( )( )20.5 kg 9.81m/s 0.15 m 0.73575 JB gV Wr= = =( ) ( ) 3.303637 J 0.73575 J 4.03939 JB B Be gV V V= + = + =At C ( )( )2 21 10.5 kg2 2C C CT mv v= =20.25C CT v=( ) ( )212C ACeV k L= ∆309.23 mm 200 mm 109.23 mm 0.10923 mACL∆ = − = =( ) ( )( )21320 N/m 0.10923 m 1.90909 J2C eV = =
83. 83. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.B B C CT V T V+ = +20 4.0394 0.25 1.90909Cv+ = +2 2 24.0394 1.909098.5212 m /s0.25Cv−= = 2.92 m/sCv = !(b) Force of rod on collar AC0zF = (no friction)x yF F= +F i j1 75tan 14.04300θ −= = °( )( )cos sinACk L θ θ= ∆ +eF i k( )( )( )320 0.10923 cos14.04 sin14.04= ° + °eF i k33.909 8.4797 (N)= +eF i k( ) ( )233.909 4.905 8.4797x ymvF F mgrΣ = + + − + = +F i j k j k( )( )2 28.5212 m /s33.909 N 0 4.905 N 0.50.15 mx yF F+ = = +33.909 NxF = −33.309 NyF =33.9 N 33.3 N= − +F i j !
84. 84. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 71.Datum at point C.222.5 lb 1b s0.07764ft32.2 ft/sm  ⋅= =  ( ) ( ) 21 2.50, 0, 0,2 32.2C C C A Ag eT V V T v = = = =   ( ) ( )( )2.5 lb 7/12 ft 1.4583A gV = − = −(a) ( ) ( )( )2120 lb/ft 0.63465 ft 0.3333 ft 0.908122A CV = − =From conservation of energy:( ) 210 0.07764 1.4583 0.908122Av= − +3.7646 ft/sAV = 3.76 ft/sAv = !At point A,( )( )20 lb/ft 0.63465 ft 0.3333 ft 6.0263 lbS CAF k L= ∆ = − =( )( )( )222 2.5 lb/32.2 ft/s 3.76551 ft/s1.8872 lb7/12 ftAmvr= =
85. 85. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )27; 6.0263 1.88727.61577AAmvF Nr= − =∑3.65 lbAN = !(b) Datum at C.( ) ( )( )2.5 lb 14/12 ft 2.9167 ft lbB gV = − = − ⋅( ) ( ) ( )( )2 21 120 lb/ft 0.5 ft 2.5 ft lb2 2B CBeV k L= ∆ = = ⋅From Conservation of energy:21 2.50 2.9167 2.52 32.2Bv + − +  3.2762 ft/sBv = 3.28 ft/sBv = !( ) ( )( )20 0.5 10 lbS CBF k L= ∆ = =( )22.51.4286 lb32.2 7/12Bv =  10 2.5 1.4286y BF N= − + − =∑6.07 lbBN = !
86. 86. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 72.(a) For maximum velocity,7.61577 4 3.61577 in. 0.30131 ftL∆ = − = =0ta s= =&& ( )sin 3/7.61577θ =2.5 lbW =( )0 0.30131 3/7.61577 2.5 0yF k= = − =∑21.063 lb/ftk = 21.1 lb/ftk = !(b) Put datum at C( ) ( ) 0,C C Cg eT V V= = = 21 2.52 32.2AAT v =   ( ) ( )2.5 7/12 1.4583A gV = − = −( ) ( )( )2121.063 0.30131 0.95612A eV = =Conservation of energy: 21 2.50 1.4583 0.95612 32.2Av = − +  3.597Av = 3.60 ft/sAv = !
87. 87. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 73.Loop 1(a) The smallest velocity at B will occur when the force exerted by thetube on the package is zero.20 BmvF mgrΣ = + =( )2 21.5 ft 32.2 ft/sBv rg= =248.30Bv =At A2012AT m v=A0.50 8 oz 0.5 lb 0.0155332.2V = = ⇒ = =  At B ( )21 148.30 24.15 m2 2B BT mv m= = =( ) ( )7.5 1.5 9 9 0.5 4.5 lb ftBV mg mg= + = = = ⋅( ) ( )201: 0.01553 24.15 0.01553 4.52A A B BT V T V v+ = + = +20 0627.82 25.056v v= = 0 25.1 ft/sv = "At C( )2 210.007765 7.5 7.5 0.5 3.752C C C CT mv v v mg= = = = =2 20: 0.007765 0.007765 3.75A A C C CT V T V v v+ = + = +( )2 20.007765 25.056 3.75 0.007765 Cv− =2144.87Cv =continued
88. 88. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Loop 2(b)( )144.87: 0.015531.5nF ma NΣ = =1.49989N ={Package in tube} 1.500 lbCN = "(a) At B, tube supports the package so,0Bv ≈( )0, 0 7.5 1.5B B Bv T V mg= = = +4.5 lb ft= ⋅A A B BT V T V+ = +( ) 210.01553 4.5 24.0732A Av v= ⇒ =24.1 ft/sAv = "(b) At C 20.007765 , 7.5 3.75C C CT v V mg= = =( )2 2: 0.007765 24.073 0.007765 3.75A A C C CT V T V v+ = + = +296.573Cv =96.5730.01553 0.999851.5CN = =  {Package on tube} 1.000 lbCN = "
89. 89. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 74.(a) Loop 1From 13.75, at B2 2 248.3 ft /s 6.9498 ft/sB Bv gr v= = ⇒ =( )( )21 10.01553 48.3 0.375052 2B BT mv= = =( ) ( )( )7.5 1.5 0.5 9 4.5 lb ftBV mg= + = = ⋅( )2 2 21 10.01553 0.0077652 2C C C CT mv v v= = =( )7.5 0.5 3.75 lb ftCV = = ⋅2: 0.37505 4.5 0.007765 3.75B B C C CT V T V v+ = + + = +2144.887 12.039 ft/sC Cv v= ⇒ =12.04 ft/s 10 ft/s> ⇒ Loop (1) does not work !(b) Loop 2 at A 2 20 010.0077652AT mv v= =0AV =At C assume 10 ft/sCv =( )2210.007765 10 0.77652C CT mv= = =( )7.5 0.5 3.75Cv = =20: 0.007765 0.7765 3.75A A C CT V T V v+ = + = +0 24.144v = 0 24.1 ft/sv = !
90. 90. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 75.Use conservation of energy from the point of release (A) and the top of the circle.1 1 2 2T V T V+ = + (1) (datum at lowest point)where1 10;T V mg= = lAt 2 ( )( )22 21; 22T mv V mgz mg a= = = −lSubstituting into (1) ( )210 22mg mv mg a+ = + −l l (2)We need another equation – use Newton’s 2ndlaw at the top. ( )0Tension, 0 at topT =n nF ma m= ⇒∑mg =2vρ( )2v g g aρ= = −lSubstituting into (2)( ) ( )122mg mg a mg a= − + −l l l2 4 45 3a aa= − + −=l l ll35a = l !
91. 91. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 76.( ) ( )( )( )2 210, 70 , 70 kg 9.81 m/s 40 m 1 0.70712A A B B BT V T v V= = = = − −( ) 2170 ,2C CT v= 2C BV V= −Conservation of energy: 0, 15.161 m/sB B BT V v+ = =0, 21.441 m/sC C CT V v+ = =(a)(b)(c)( )270 /40 + 70402.26 686.7BN v gN== +1 1089 NBN = !( )270 /40 70402.26 686.7BN v gN= − += − +2 = 284 NBN !( ) 270 kg 0.7071 70 /40CN v− = −1 804.5 485.6 0CN = − + <Skier airborne? Yes!,,,
92. 92. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 77.For a conservative force, Equation (13.22) must be satisfiedx y zV V VF F Fx y z∂ ∂ ∂= − = − = −∂ ∂ ∂We now write2 2yxFF V Vy x y x y x∂∂ ∂ ∂= − = −∂ ∂ ∂ ∂ ∂ ∂Since2 2:yxFV V Fx y y x y x∂∂ ∂ ∂= =∂ ∂ ∂ ∂ ∂ ∂!We obtain in a similar wayy z z xF F F Fz y x z∂ ∂ ∂ ∂= =∂ ∂ ∂ ∂!
93. 93. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 78.(a) x yyz zxF Fxyz xyz= =( ) ( )110 0yyxxFFy y x x∂∂ ∂∂= = = =∂ ∂ ∂ ∂ThusyxFFy x∂∂=∂ ∂The other two equations derived in Problem 13.80 are checked in a similar way.(b) Recall that , ,x y zv v vF F Fx y z∂ ∂ ∂= − = − = −∂ ∂ ∂( )1ln ,xvF V x f y zx x∂= = − = − +∂(1)( )1ln ,yvF V y g z xy y∂= = − = − +∂(2)( )1ln ,zvF V z h x yz z∂= = − = − +∂(3)Equating (1) and (2)( ) ( )ln , ln ,x f y z y g z x− + = − +Thus ( ) ( ), lnf y z y k z= − + (4)( ) ( ), lng z x x k z= − + (5)Equating (2) and (3)( ) ( )ln , ln ,z h x y y g z x− + = − +( ) ( ), lng z x z l x= − +From (5)( ) ( ), lng z x x k z= − +Thus( ) lnk z z= −( ) lnl x x= −continued
94. 94. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.From (4)( ), ln lnf y z y z= − −Substitute for ( ),f y z in (1)ln ln lnV x y z= − − −lnV xyz= − "
95. 95. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 79.(a)( ) ( )3 12 2 2 2 2 22 2x yx yF Fx y z x y z= =+ + + +( )( )( )( )( )3 32 25 52 2 2 2 2 22 22 2yxx y y yFFy xx y z x y z− −∂∂= =∂ ∂+ + + +Thus yxFFy x∂∂=∂ ∂The other two equations derived in Problem 13.79 are checked in a similar fashion(b) Recalling that , ,x y zV V VF F Fx y z∂ ∂ ∂= − = − = −∂ ∂ ∂( )32 2 2 2xV xF V dxxx y z∂= − = −∂+ +∫( ) ( )12 2 2 2 ,V x y z f y z−= + + +Similarly integratingVy∂∂andVz∂∂shows that the unknown function ( ),f x y is a constant.( )12 2 2 21Vx y z=+ +!
96. 96. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 80.(a)20 2aABaU kxdx k= =∫,x yF F F= ∴ is normal to BC, 0BCU =( )20 2aCAaU a u du−= − − =∫( )21 ,2ABCAaU k= − not conservative !(b) From Problem 13.77, 1yxFFy x∂∂= =∂ ∂Conservative, 0ABCAU = !
97. 97. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 81.(a) ( )2 21 131 2 1 2x xx xU Fdx k x k x dx→ = − = − +∫ ∫( ) ( )2 2 4 41 22 1 2 12 4k kx x x x= − − −1 2 1 2e eU V V→ = −2 41 21 12 4eV k x k x= + !(b) Conservation of energy: 21 210,2T T mv= =2 41 1 0 2 0 21 1, 02 4e eV k x k x V= + =2 2 41 0 2 01 1 12 2 4mv k x k x= +2 41 20 02k kv x xm m   = +      !
98. 98. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 82.(a) ( )2 21 131 2 1 2x xx xU Fdx k x k x dx→ = − = − +∫ ∫( ) ( )2 2 4 41 22 1 2 12 4k kx x x x= − − + −1 2 1 2:e eU V V→ = −2 41 21 12 4eV k x k x= − !(b) Conservation of energy: 21 210,2T T mv= =2 41 1 0 2 0 21 1, 02 4e eV k x k x V= − =2 2 41 0 2 01 1 12 2 4mv k x k x= −2 41 20 02k kv x xm m   = +      !1022Requireskxk <   
99. 99. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 83.Circular orbit velocity63960 mi = 20.9088 10 ftR = ×6930 mi = 4.9104 10 ftAor = ×222,Cv GMGM gRr r= =( )( )( )26226 632.2 ft/s 20.9088 10 ft20.9088 10 ft + 4.9104 10 ftCGM gRvr r×= = =× ×2 6 2 2545.22 10 ft /sCv = ×23350 ft/sCv =Velocity reduced to 60% of 14010 ft/sCv =Conservation of energy:A A B BT V T V+ = +2 21 12 2A BA BGMm GMmmv mvr r− = −( )( )( )( )( )2 26 6226 632.2 20.9088 10 32.2 20.9088 101140102 225.819 10 20.9088 10Bv× ×− = −× ×21269 ft/sBv = 4.03 mi/sBv = !
100. 100. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 84.Distance 6OA 3960 mi + 376 mi = 4336 mi = 22.894 10 ft= ×Distance 6EOO 10,840 mi 4336 mi = 6504 mi = 34.341 10 ft= − ×( ) ( )2 228670 6504Cr = +610,838.4 mi = 57.2268 10 ftCr = ×( )221= 15681.62 CgR mT V mr+ −( ) 6 2 21constant = 123.032 10 ft /sT Vm+ = − ×63960 mi = 20.9088 10 ft; 2.97 mi/s = 15681.6 ft/sCR v= × =(a) At point A,2261=2 22.894 10 ftAT V gRvm+−×( )( )( )( )22 66 2632.2 ft/s 20.9088 10 ft1123.032 102 22.894 10 ftAv×− × = −×31364 ft/sAv = 5.94 mi/sAv = !(b) At point B, ( ) ( )2 10,840 mi 4336 miBr = −617344 mi = 91.5763 10 ft= ×( )( )( )( )266 2632.2 20.9088 101123.032 102 91.5763 10Bv×− × = −×7834.3 ft/sBv = 1.484 mi/sBv = !
101. 101. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 85.-At A, 32.5 /h 9028 m/sAv Mm= =( )2 619028 m/s 40.752 102AT m m= = ×2AA AGMm gR mVr r−= − =610.67 M 10.67 10 mAr m= = ×66370 km 6.37 10 mR = = ×( )( )( )22 6669.81m/s 6.37 10 mm 37.306 10 m10.67 10 mAV×= − = − ××At B221;2B B BB BGMm gR mT mv Vr r−= = − =619.07 M 19.07 10 mBr m= = ×( )( )( )22 6669.81m/s 6.37 10 m20.874 10 m19.07 10 mBmV×= − = − ××6 6 2 61; 40.752 10 m 37.306 10 m 20.874 10 m2A A B B BT V T V mv+ = + × − × = − ×2 6 6 62 40.752 10 37.306 10 20.874 10Bv  = × − × + × 2 6 2 248.64 10 m /sBv = ×36.9742 10 m/s 25.107 Mm/hBv = × = 25.1Mm/hBv = !4.3 6.37A Ar h R Mm Mm= + = +10 67Ar Mm=72.7 6.37B Br h R Mm Mm= + = +19.07Br Mm=
102. 102. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 86.Note: moon earth0.0123GM GM=By Equation 12.30 2moon 0.0123 EGM gR=At ∞ distance from moon: 2 2, assume 0r v= ∞ =lem2 2 2 0 0 0 0mGM mE T V= + = − = − =∞(a) On surface of moon: 61740 km 1.74 10 mMR = = ×61 10, 0 6370 km 6.37 10 mEv T R= = = = ×2lem lem1 1 1 10.01230m EM mGM m gR mV E T VR R−= = + = −( )( )( )( )22 6lem1 60.0123 9.81m/s 6.37 10 m1.740 10 mmE×= −×Where lem mass of the lemm =( )6 2 21 lem2.814 10 m /sE m= − ×( )6 2 22 1 lem0 2.814 10 m /sE E E m∆ = − = + ×Energy per kilogram:lem2810 kj/kgEm∆= !(b) 1 80 kmmr R= +( ) 61 1740 km 80 km 1820 km 1.82 10 mr = + = = ×Newton’s second law:2lem lem 1lem lem 211: mGM m m vF m arr= =2 2 lem1 1 lem 11 11 12 2m mGM m GMv T m vr r= = =
103. 103. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.lem lem lem1 1 1 11 1 112m m mGM m GM m GM mV E T Vr r r−= = + = −( ) 2lemlem11 10.01231 12 2Em gR mGM mEr r= − = −( )( )( )2 3lem1 60.0123 9.81m/s 6.37 10 m12 1.82 10 mmE×= −×( )6 2 21 lem1.345 10 m /sE m= − ×( )6 2 22 1 lem0 1.345 10 m /sE E E m∆ = − = + ×Energy per kilogram:lem1345 kJ/kgEm∆= !
104. 104. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 87.Total energy per unit weight2 20 01 1;2 2A A p p A pA pGMm GMmE T V T V E mv mvr r= + = + = − = −Unit weightW mg=2 202 2A pA pW W GM W W GME v vg g r g g r= − = −2202 2pAA pvE v GM GMW g gr g gr= − = − (1)2221 112pAA pAvv GMg g r rv    − = −     2221 2p p AAA pAv r rv GMr rv   − − =      ( )givenpAp Arvv r=2221 2p AAAA ppr rrv GMr rr   − − =      2 2222p A p AAA ppr r r rv GMr rr   − −  =      2 12pAA p Arv GMr r r =   + (2)
105. 105. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Substitute Av from (2) into (1)0 1 1 1 122p pA p A A A p A Ar rE GM GMGMW g r r r gr g r r r r         = − = −      + +         ( )1p p ApA p A A p Ar r rrGM GMr g r r r g r r   − + = − =  + +    ( )A BGMg r r−=+22 0 EEA pE RGM gRW r r= ⇒ = −+( )( )260 3960 mi 5280 ft/mi1.65598 10 ft lb/lb50,000 mi 5280 ft/miEW×= − = − × ⋅×On earth: , 0, 0,E E E E E EEWGME T V V T VgR= + = = = −2620.9088 10 ft lb/lbE EEE EE GM gRRW gR gR= − = − = − = − × ⋅For propulsion: 0p EE E EW W W= −( )6 61.65598 10 20.9088 10= − × − − ×619.25 10 ft lb/lbpEW= × ⋅ "
106. 106. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 88.Geosynchronous orbit61 3960 200 4160 mi 21.965 10 ftr = + = = ×62 3960 22,000 25,960 mi 137.07 10 ftr = + = = ×Total energy 212GMmE T V mvr= + = −mass of earthM =mass of satellitem =Newton’s second law222;nGMm mv GMF ma vr rr= = ⇒ =212 2GM GMmT mv m Vr r= = = −1 12 2GMm GMm GMmE T Vr r r= + = − = −( )2 22 1 1where2 2E EEgR m R WGM gR E W mgr r= = − = − =( )( )26 186000 20.9088 10 ft1 1.3115 10lb ft2Er r× ×= − = − ⋅Geosynchronous orbit at 62 137.07 10 ftr = ×18961.3115 109.5681 10 lb ft137.07 10GsE− ×= = − × ⋅×(a) At 200 mi, 61 21.965 10 ftr = ×1810200 61.3115 105.9709 1021.965 10E×= − = − ××10300 200 5.0141 10GsE E E∆ = − = ×9300 50.1 10 ft lbE∆ = × ⋅ !
107. 107. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) Launch from earthAt launch pad2EE EE EGMm gR mE WRR R−= − = = −( ) 116000 3960 5280 1.25453 10EE = − × = − ×9 99.5681 10 125.453 10E Gs EE E E∆ = − = − × + ×9115.9 10 ft lbEE∆ = × ⋅ !
108. 108. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 89.We knowGeosynchronous orbit: 2 35780 6370 42,150 kmr = + =Orbit of shuttle: 1 6370 km + 296 km = 6666 kmr =Radius of Earth: 26370km also =R GM = gR=For any circular orbit222nGMm mv GMF ma vr rr= ⇒ = ⇒ =Energy221 1;2 21 1 12 2 2GMm GMmT mv vr rGMm GMm GMm gR mE T Vr r r r= = = − = + = − = − = −    For Geosynchronous orbit( )( ) ( )22 692 69.81 m/s 6.370 10 m 3600 kg116.999 10 J2 42.150 10 mE×= − = − ××For orbit of shuttle( )( ) ( )2661 69.81 6.370 10 36001107.487 10 J2 6.666 10E×= − = − ××On the launching pad vo = 0( )( )( )6 93600 9.81 6.370 10 224.963 10 JoGMmE mgRR= − = − = − × = − ×(a) From shuttle to orbit( )9 92 1 16.999 10 J 107.487 10 JE E E∆ = − = − × − − ×990.5 10 JE∆ = × "(b) From surface to orbit( )9 92 0 16.999 10 J 224.963 10 JE E E∆ = − = − × − − ×9208 10 JE∆ = × "
109. 109. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 90.(a) Potential energy2constantGMm gR mVr r= − = − +(cf. Equation 13.17)Choosing the constant so that 0 for :V r R= =1RV mgRr = −  !(b) Kinetic energyNewton’s second law22:nGMm vF ma mrr= =22 GM gRvr r= =212T mv=212mgRTr= !Energy(c) Total energy2112gR RE T V m mgr r = + = + −  12RE mgRr = −  !
110. 110. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 91.In a circular orbit,22Nmv GMmFr r= =mass of Venus, mass of Sunm M= =2,GMvr∴ = 212 2GMmT mvr= =,2GMm GMmV T Vr r= − + = −(a)23 621988(78.3 10 ) (67.2 10 )(5280)6034.4 10v rMG −  × ×    = =×27 2136.0 10 lb s /ftM = × ⋅ !(b) 212 2GMmT V mvr+ = − = −22731 136.029 10 8878.3 102 60407 10T V 3   ×  + = − ×      ×   332.20 10 lb ftT V+ = − × ⋅ !