solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 13
Upcoming SlideShare
Loading in...5
×
 

Like this? Share it with your network

Share

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

on

  • 17,349 views

 

Statistics

Views

Total Views
17,349
Views on SlideShare
17,349
Embed Views
0

Actions

Likes
7
Downloads
563
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

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

  • 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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+ = − × ⋅ !
  • 111. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 92.2 2setting :1g g yRWR WR WRV r R y Vr R y= − = + = − = −+ +( ) ( )( )1 21 1 21 11 1 2gy y yV WR WRR R R−  − − −   = − + = − + + +    ⋅     !We add the constant WR, which is equivalent to changing the datum from to :r r R= ∞ =2gy yV WRR R  = − +     !(a) First order approximation:gyV WR WyR = =  ![ ]Equation 13.16(b) Second order approximation:2gy yV WRR R  = −     2gWyV WyR= − !
  • 112. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 93.Use Conservation of energy and Conservation of angular momentum.Conservation of angular momentum.( )( )21 11 1 2 220.2 60.5= ⇒ = =rvrmv r mv vrθθ θ θ22.4 m/s=vθ !Conservation of energy1 1 2 2T V T V+ = + (1)At 1 ( )( )221 11 14 kg 6 m/s 72 J2 2T mv= = =( )( )221 11 11500 N/m 0.2 m 0.4 m 72 J2 2V kx= = − =At 2 ( )( ) ( )22 2 22 2 2 21 1 1 14 2.4 42 2 2 2r rT mv mv v= + = +θ2211.52 2 rv= +( )( )222 21 11500 N/m 0.5 m 0.4 m 7.5 J2 2V kx= = − =Substituting into (1) 2272 30 11.52 2 7.5rv+ = + +222 82.98rv =2 6.44 m/srv = !
  • 113. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 94.Conservation of angular momentum.1r m 1 maxv r mθ = 2v θso( )( )1 12max max max0.2 6 1.2rvvr r r= = =θθEnergy1 1 2 2T V T V+ = + (1)where,At 1 ( )( )221 11 14 kg 6 m/s 72 J2 2T mv= = =( )( )221 11 11500 0.2 0.4 30 J2 2V kx= = − =At 2 ( )22 22 2 2 2max max1 1 1 1.2 28842 2 2rT mv mvr r = + = =  θ( )( )222 2 max1 11500 0.42 2V kx r= = −Substituting into (1) ( )2max2max2.8872 30 750 0.4rr+ = + −Solve by trial for maxr ⇒ max 0.760 mr = !Solve for 2v θ2max1.2 1.20.760vr= = ⇒ 2 1.580 m/sv = !0
  • 114. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 95.Initial stateF maΣ =220 0, AAmr θkx mr θ xk= =!!( )( )( )204/32.2 3 ft 5 rad/s1.331 ft7 lb/ft= =xUnstretched length = 4.5 ft + 1.331 ft = 5.831 ftConservation of angular momentum(A) ( ) ( )22= 3 5 45= =!A Ah r θ (1)(B) ( ) ( ) ( )2 22= 9 = 7.5 5 , = 3.4722= ! ! !B Bh r θ θ θ (2)(a) Substitute into (1) gives 3.6 ftAr = !(b) = 3.47 rad/sθ! !Conservation of energy ( )( )20 0 01, 7 lb/ft 1.331 ft 6.200 ft lb2T V T V V+ = + = = ⋅( ) ( )2 201 4 1 45 3 5 7.5 101.320 ft lb2 32.2 2 32.2T      = + = ⋅         ( ) ( )217 9 3.6 5.831 0.6501 ft lb2V  = − − = ⋅ 0 0= + −T T V V101.32 6.200 0.6501= + −(c) T = 106.9 ft ⋅ lb!
  • 115. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 96.Initial stateF maΣ =20 Akx mr θ= !( )( )( )2204/32.2 3 51.331 ft7Amr θxk= = =!Unstretched length = 4.5 + 1.331 = 5.831 ftConservation of energy( ) ( )2 201 4 1 45 3 5 7.52 32.2 2 32.2T      = +         101.320 ft lb= ⋅( )( )2017 1.331 6.200 ft lb2V = = ⋅( ) 211100 7 101.32 6.2002T V x+ = + = +1 1.4658 ftx = ±For compression:( ) 5.831 1.4658B Ar r− − = −Conservation of angular momentum(A) ( ) ( )22= 3 5 45A Ah r θ= =!(B) ( ) ( )22= 7.5 5 = 281.25B Bh r θ= !continued
  • 116. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )22 22.5 , 2.5B A B Ar r r r= =5.831 2.5 5.831 1.5 5.831B A A A Ar r r r r− − = − − = −(a) 1.5 5.831 1.4658Ar∴ − = − 2.91 ftAr = !(b) 2.5 7.2753 ftB Ar r= = 7.28 ftBr = !(c) 2= 45, = 5.31 rad/sAr θ θ! ! = 5.31 rad/sθ! !
  • 117. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 97.6370 kmR =0 500 km 6370 kmr = +0 6870 kmr =66.87 10 m= ×0 36,900 km/hv =6336.9 10 m3.6 10 s×=×310.25 10 m/s= ×Conservation of angular momentum0 0 1 0 min 1 max, ,Ar mv rmv r r r r= = =( )63001 16.870 1010.25 10ArV vr r′   ×= = ×       9170.418 10AVr′×= (1)Conservation of energyPoint A30 10.25 10 m/sv = ×( )22 301 110.25 102 2AT mv m= = ×( )( )( )652.53 10 JAT m= ×( )( )22 2 609.81m/s 6.37 10 mAGMmV GM gRr= − = = ×12 3 2398 10 m /sGM = ×60 6.87 10 mr = ×continued
  • 118. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )( )( )12 3 266398 10 m /s57.93 10 m J6.87 10 mAmV×= − = − ××Point A′212A AT mv′ ′=( )121 1398 10 mJAGMmVr r′×= − = −A A A AT V T V′ ′+ = +652.53 10 m× 657.93 10 m− ×12m=122 398 10Amv ′×−1rSubstituting for Av ′ from (1)( )( )( )29 12621170.418 10 398 105.402 102 rr× ×− × = −( )21 1262112.4793 10 398 105.402 10rr× ×− × = −( ) ( )6 2 12 211 15.402 10 398 10 2.4793 10 0r r× − × + × =6 61 66.7 10 m, 6.87 10 mr = × ×max 66,700 kmr = !
  • 119. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 13, Solution 98.The cord will not go slack if 2v is perpendicular to the undeformed cordlength, 0,L at 2Conservation of angular momentum1 2 2 1 00.80.8 0.6 1.3330.6v v v v v= = =Conservation of energyPoint 1 2 21 0 1 0 010.352v v T mv v= = =( ) ( )( )2 21 01 1150 N/m 0.8 m 0.6 m2 2V k L L= − = −1 3JV =Point 2 2 22 2 210.352T mv v= =2 21 1 2 2 20 0 : 0.35 3 0.35 0BL V T V T V v v∆ = = + = + + = +From conservation of angular momentum 2 1.3158 Bv v=( )2200.35 1.3158 1 3v  − =  ( )( )( )2 2 203J11.72 m /s0.35 kg 0.7313v = =0 3.42 m/sv = "continued
  • 120. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.The ball travels in a straight line after the cord goes slack.Conservation of angular momentum( )( )0.8 1.71 dv=1.368dv=Conservation of energy1 1.71m/sv =Point 1( )( )221 11 10.7 kg 1.71m/s 1.0234J2 2T mv= = =( ) ( )( )2 21 01 1150 N/m 0.8 m 0.6 m 3J2 2V k L L= − = − =Point 3 2 23 310.352T mv v= =3 0V =21 1 3 3 : 1.0234 3 0.35 0T V T V v+ = + + = +3.39 m/sv =From conservation of momentum1.368 1.368404 mm3.39dv= = =404 mmd = "
  • 121. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 99.(a) Conservation of angular momentum: About O00.8 0.27v v=02.963v v=Conservation of energyPoint 1 2 21 0 1 0 00.352v v T mv v1= = =( ) ( )( )2 21 1 01 1150 N/m 0.8 m 0.6 m2 2V k L L= − = −1 3 JV =Point 2 2 22 210.352v v T mv v= = =( )2 0 cord is slackV =2 21 1 2 2 0: 0.35 3 0.35 0T V T V v v+ = + + = +From conservation of angular momentum, 03.125v v=( )2200.35 3.125 1 3v  − =  ( )( )( )203J0.35 kg 8.7656v =2 2 20 0.9779 m /sv =0 0.989 m/sv = "(b) Maximum velocity occurs when the ball is at its minimum distancefrom O, (when 0.27 m)d =( )( )03.125 3.125 0.9889 3.09 m/smv v= = =3.09 m/smv = "
  • 122. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 100.Conservation of angular momentumAA A B B B ABrmr v mr v v vr= ⇒ = (1)Conservation of energy2 21 12 2A BA BGMm GMmmv mvr r− = − (2)Put (1) into (2) and solve for Av( )2 2 BAA B AGMrvr r r=+(3)Given data 31740 8 1748 km = 1.748 10 mA Ar R h= + = + = ×At B 31740 140 1880 km = 1.880 10 mB Br R h= + = + = ×moon earth0.0123M M=( ) ( ) 2moon earth0.0123 0.0123GM GM gR= =( )( )26 12 3 20.0123 9.81 6.370 10 4.8961 10 m /s= × = ×(a) Speed at A( )( )( )12 626 6 62 4.8961 10 1.88 102.90291.748 10 1.880 10 1.748 10Av× ×= =× × + ×1703.8 m/sAv = 1704 m/sAv = !
  • 123. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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) At B( )17481703.8 1584.2 m/s1880AB ABrv vr= = =The command module is in a circular orbitAt 61.880 10 mBr = ×112 2circ 64.8961 101613.8 m/s1.880 10BGMvr ×= = = × ( )Relative velocity 1613.8 1584.2 29.6 m/scirc Bv v= − = − =Relative velocity 29.6 m/s=
  • 124. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 101.From 13.100 AB ABrv vr= (1) and 2 2( )BAA A BGMrvr r r=+(2)Given data 264 km;Ah = 6370 km + 264 km 6634 kmAr = =66.634 10 m= ×35780 km;Bh = 6370 km 35780 km 42,150 kmBr = + =642.150 10 m= ×( )( )22 6 129.81 6.37 10 398.06 10GM gR= = × = ×Substitute into (2)( )( )( )( )12 626 6 62 398.06 10 42.150 106.634 10 42.150 10 6.634 10Av× ×=× × + ×6 2 2103.69 10 m /s= ×10,183 m/sAv =Substitute into (1) ( )663410,183 1602.7 m/s42,150Bv = =  At A we have a circular orbit126398.06 107746.2 m/s6.634 10circAGMvr×= = =×Relative velocity ( )10,183 7746.2A circv v− = −Relative velocity 2.44 km/s= "At B112 26398.06 103073.1 m/s42.15 10circBGMvr ×= = = × Increase in velocity 3073.1 1602.7 1470 m/scirc Bv v= − = − =Increase in 1.470 km/sv = "
  • 125. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 102.M = Mass of the sun; 2;EGM gR= R = 3960 mi 620.9088 10 ft= ×( ) ( )( )22 2 6 21 3 2= 332.8 10 32.2 ft/s 20.9088 10 ft 4.6849 10 ft /sGM × = ×Circular orbits Earth 697,677 ft/s(93 10 )(5280)EGMv = =×Mars 679187 ft/s(141.5 10 )(5280)MGMv = =×Conservation of angular momentumElliptical orbit ( ) ( )93 141.5A Bv v=Conservation of energy( )( ) ( )( )2 26 61 12 293 10 5280 141.5 10 5280A BGM GMv v− = −× ×141.51.521593A B Bv v v = =  ( )( )( ) ( )( )21 212 2 26 61 4.6849 10 1 4.6849 101.52152 293 10 5280 141.5 10 5280B Bv v× ×− = −× ×2 90.6575 3.270 10Bv = ×70524 ft/s;Bv = 107,303 ft/sAv =(a) Increase at A, 107303 97677 9626 ft/s = 1.823 mi/sA Ev v− = − = !(b) Increase at B, 79187 70524 8663 ft/s = 1.641 mi/sB Mv v− = − = !
  • 126. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 103.R = planet radius;A A B Br R h r R h= + = +Conservation of angular momentum( ) ( )A A B Bv R h v R h+ = +( ) ( )5 1200 mi 1.2 16300 mi+ = +R R5 6000 1.2 19560; 3.8 13560+ = + =R R R(a) 3568.4 miR = = 3570 miR !Conservation of energy mass of spacecraft=sm( )( )( )( )( )( )( )2 25 5280 1.2 52802 3568.4 1200 5280 2 3568.4 16300 5280GM GM− = −+ +6 66 6348.48 10 20.072 1025.177 10 104.905 10× − = × −× ×GM GM15 3 210.879 10 ft /s= ×GMUsing 9 4 434.4 10 ft /lb sG −= × ⋅(b) 21316.26 10 slugs= ×M 21316 10 slugs= ×M !(planet is Venus)( ) ( )2 21 1;2 2+ = + − = −+ +s sA A B B s A s BA BGMm m GMT V T V m v m vR h R h
  • 127. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 104.Elliptical orbit between A and BConservation of angular momentumA A B Bmr v mr v=7.1706.690BA B BArv v vr= =66370 km 320 km 6690 km, 6.690 10 mA Ar r= + = = ×1.0718A Bv v= (1)66370 km 800 km 7170 km, 7.170 10 mB Br r= + = = ×( ) 66370 km 6.37 10 mR = = ×Conservation of energy( )( )22 2 6 12 3 29.81 m/s 6.37 10 m 398.060 10 m /sGM gR= = × = ×Point A( )( )1226398.060 1012 6.690 10A A AAmGMmT mv Vr×= = − = −×659.501 10 mAV = ×Point B( )( )1226398.060 1012 7.170 10B B BBmGMmT mv Vr×= = − = −×655.5 10 mBV = ×A A B BT V T V+ = +continued
  • 128. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 6 2 61 159.501 10 m 55.5 10 m2 2A Bmv mv− × = − ×2 2 68.002 10A Bv v− = ×From (1) ( )22 61.0718 1.0718 1 8.002 10A B Bv v v  = − = ×  2 6 2 253.79 10 m /s , 7334 m/sB Bv v= × =( )( )1.0718 7334 m/s 7861 m/sAv = =Circular orbit at A and B(Equation 12.44)( )126398.060 107714 m/s6.690 10A CAGMvr×= = =×( )126398.060 107451 m/s7.170 10B CBGMvr×= = =×(a) Increases in speed at A and B( ) 7861 7714 147 m/sA A A Cv v v∆ = − = − = !( ) 7451 7334 117 m/sB B BCv v v∆ = − = − = !(b) Total energy per unit mass( ) ( ) ( ) ( )2 2 2 21/2A A B BC CE m v v v v = − + −  ( ) ( ) ( ) ( )2 2 2 21/ 7861 7714 7451 73342E m  = − + −  6/ 2.01 10 J/kgE m = × !
  • 129. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 105.(a) 6185 10 mAr = ×6295 10 mBr = ×Speed of spacecraft in the elliptical orbit after its speedAv′ =has been decreasedElliptical orbit between A and B conservation of energyPoint A 2 sat1,2A A AAGM mT mv Vr= = −sat mass of saturnM =Determine satGM from the speed of Tethys in its circular orbit.(Equation 12.44) 2satcirc sat circBGMv GM r vr= =( )( )26 3 15 3 2sat 295 10 m 11.3 10 m/s 37.67 10 m /sGM = × × = ×( )( )15 3 29637.67 10 m /s0.2036 10 m185 10 mAmV×= − = − ××Point B( )( )15 3 22 sat637.67 10 m /s12 295 10 mB B BBmGM mT mv Vr×= = − = −×90.1277 10BV = − ×;A A B BT V T V+ = +2 9 2 91 10.2036 10 0.1277 102 2A Bmv m mv m− × = − ×2 2 90.1518 10A Bv v′ − = ×continued
  • 130. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 momentum66185 100.6271295 10AA A B B B A A ABrr mv r mv v v v vr×= = = =′ ′ ′ ′×( )22 91 0.6271 0.1518 10 , 15817 m/sA Av v ′ ′− = × =  21000 15817 5183 m/s 5.18 km/sA A Av v v′∆ = − = − = = !(b) ( )( )0.6271 15817 9919 m/s,AB ABrv vr′= = =9.92 km/sBv = !
  • 131. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 106.Conservation of energy2 201 12 2AB AGMm GMmmv mvr r− = −So 2 2021 BAB AGM rv vr r = − −  (1)Given66370 km = 6.37 10 mR = ×( )( )22 6 129.81 6.37 10 398 10GM gR= = × = ×66370 360 6730 km 6.73 10 mAr = + = = ×66370 60 6430 km 6.43 10 mBr = + = = ×Substitute into (1)( )12 62 20 6 62 398 10 6.43 1016.43 10 6.73 10Av v×  ×= − −  × × 2 2 60 5.518 10Av v= − × (2)We need another equation → conservation of angular momentumsinB A Ar mv r mvφ =6006sin 6.43 10sin506.73 10BAAr vv vrφ  ×= = °  × 00.7319Av v=Substitute into (2)( )2 2 60 00.7319 5.518 10v v= − ×2 600.46433 5.518 10v = ×0 3477 m/sv =0 3450 m/sv = "
  • 132. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 107.21Let constant2AAT V GME vm r+= = = −km m 130,000 1000hr km 3600 s/hrAv   =       8333.33 m/sAv =( )( )22 2 69.81 m/s 6.37 10 mGM gR= = ×12 3 2398.059 10 m /sGM = ×6 6 64.3 10 m 4.3 10 m + 6.37 10 mAr R= × + = × ×610.67 10 mAr = ×( )12261 398.059 108333.32 10.67 10E×∴ = −×3 2 22584.19 10 m /sE = − ×/ constant = sin60A Ah m v r= °( )( )( )6/ 8333.33 m/s 10.67 10 m 3/2h m = ×9 2/ 77.0041 10 m /sh m = ×At min or max altitude, / ,h m r v=/h mvr ∴ =   Eliminate22/ 1 1 /:2 2h m GM h m GMv E vr R r r   = = − = −      Multiply by r2:22 12hEr GMrm = −  
  • 133. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.And rearrange ( )22 102hEr GM rm + − =  Quadratic formula for minimum r:( ) ( )2 2min2 /2GM GM E h mrE− + +=( ) ( )( )( )2 212 24 3 18min 3398.06 10 398.06 10 2 2584.2 10 77.004 102 2584.2 10r− × + × + − × ×=− ×( )12 12min 3398.06 10 357.50 102 2584.2 10r− × + ×=− ×6min 7.848 10 mr = ×Minimum altitude 67.848 10 R= × −6 67.848 10 m 6.37 10 m= × − ×61.478 10 m= ×Minimum altitude 1478 km= !
  • 134. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 108.At A: ( )( ) ( )( )6.5 5280 3960 mi + 567 mi 5280 ft/miAh vr    = =    9 2820.336 10 ft /sAh = ×( )( ) 63960 mi 5280 ft/mi 20.9088 10 ftR = = ×( ) 21 12A AGMT V vm r+ = −( )( )( )( )( )262 32.2 20.9088 1016.5 5280 02 3960 567 5280 ×  = − ≅   + Parabolic orbitAt B: ( ) 21 102B B BBGMT V vm r+ = − =( )( )26232.2 20.9088 1012 3960 5190 5280Bv × = + 2 6582.76 10 ; 24140 ft/sB Bv v= × =(a) 4.57 mi/sBv = !9sin 820.336 10B B B Bh v rφ= = ×( )( )9820.336 10sin24140 3960 5190 5280Bφ×= + 0.7034=(b) 44.7Bφ = ° !
  • 135. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 109.6, 3960 mi 377 mi = 4337 mi = 22.8994 10 ftA A A Ah v r r= = + ×( )( )2.97 mi/s 5280 ft/mi 15681.6 ft/sCv = =( )( ) 63960 5280 20.9088 10 ftR = = ×615681.6 22.8994 10C C Ah v b b v  = = = ×  (1)( )2612 22.8994 10AA Av GMT Vm+ = −×2632.2 20.9088 10GM  = × 262 22.8994 10Av GM= −×(2)Two equations in two unknowns: , .Av bSolve ;31361 ft/sAv =645.8 10 ftb = ×(a) 5.94 mi/sAv = "(b) 8670 mib = "
  • 136. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 110.61080 87 1167 mi 6.1618 10 ftAr = + = = ×61080 mi 5.7024 10 ftCr R= = = ×2moon 0.0123 0.0123E EGM GM gR= =( )( )20.0123 32.2 3960 5280= ×14 3 21.7315 10 ft /s= ×At 87 mi: mooncirc 5301.0 ft/sAGMvr= =(a) An elliptic trajectory between A and C, where the lem is just tangentto the surface of the moon, will give the smallest reduction of speedat A which will cause impact.2 6128.101 10 m2mA A AAGM mT mv Vr= = − = − ×2 6130.364 10 m2mC C CCGM mT mv Vr= = − = − ×2 61: 28.101 10 m2A A C C AT V T V mv+ = + − ×2 6130.364 10 m2Cmv= − ×2 2 64.526 10A Cv v= − × (1)
  • 137. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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: A A C Cr mv r mv=6.16181.08065.7024AC A A ACrv v v vr= = =( )22 61.0806 4.526 10 5195.1 ft/sA A Av v v= − × ⇒ =( )circ5343.9 5195.1 148.8A A Av v v∆ = − = − =148.8 ft/sAv∆ = !(b) Conservation of energy (A and B)Since B Cr r= conservation of energy is the same as betweenA and CConservation of angular momentum:sin ; 45A A B Br mv r mv φ φ= = °6.16181.5281sin 45 5.7024 0.70711A A AB ABr v Vv vr  = = =  °   From (1)( )22 61.5281 4.526 10 1841.4 ft/sA A Av v v= − × ⇒ =( )circ5343.9 1841.4 3487.3A A Av v v∆ = − = − =3503 ft/sAv∆ = !
  • 138. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 111.For circular orbit of radius 0r20200nGMm vF ma mrr= =200GMvr=But 0v forms an angle α with the intended circular pathFor elliptic orbitConservation of angular momentum0 0 cos A Ar mv r mvα =00cosAArv vrα =   (1)Conservation of energy2 2001 12 2AAGMm GMmmv mvr r= = =2 2 00021AAGM rv vr r − = −  Substitute for Av from (1)22 20 00021 cos 1A Ar GM rvr r rα     − = −        But 200GMvr= thus220 01 cos 2 1A Ar rr rα   − = −      22 0 0cos 2 1 0A Ar rr rα   − + =      
  • 139. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Solving for 0Arr202 22 4 4cos 1 sin2cos 1 sinArrα αα α+ ± − ±= =−( )( )( )0 01 sin 1 sin1 sin1 sinAr r rα ααα+ −= =±∓also valid for point A′Thus( )max 01 sinr rα= + ( )min 01 sinr rα= − !
  • 140. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 112.6370 km 362 km 6732 kmAr = + =66.732 10 mAr = ×6370 km 64.4 km 6434.4 kmBr = + =66.4344 10 mBr = ×66370 km 6.37 10 mR = = ×( )( )22 2 69.81 m/s 6.37 10 mGM gR= = ×9 3 2398.06 10 m /sGM = ×Conservation of energy92 661 398.06 1059.130 10 m2 6.732 10A A AAGMm mT mv Vr×= = − = − = − ××92 661 398.06 1061.86 10 m2 6.4344 10B B BBGMm mT mv Vr×= = − = − = − ××2 6 2 61 1: 59.130 10 61.86 102 2A A B B A BT V T V mv m mv m+ = + − × = − ×2 2 65.4609 10A Bv v= − × (1)
  • 141. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 momentumsinA A B B Br mv r mv φ=( )( )( )6732 11.208sin 6434.4 sin60A AB A B AB Br vv v v vr φ = = = ° (2)Substitute Bv from (2) in (1)( ) ( )2 22 6 2 61.208 5.4609 10 ; 1.208 1 5.4609 10A A Av v v  = − × − = ×  2 6 2 211.8905 10 m /sAv = ×3.448 km/sAv =(a) 3.45 km/sAv = !From (2) ( )1.208 1.208 3.45 km/s 4.1655 km/sB Av v= = =(b) 4.17 km/sBv = !
  • 142. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 113.6370 km 362 kmAr = +6732 km=6370 km 64.4 kmBr = +6434.4 km=( )22 2 69.81 m/s 6.37 10 mGM gR  = = × 15 3 20.39806 10 m /sGM = ×Velocity in circular orbit at 362 km altitude2AGMmFr=( )2circ2AnAm var=Newton’s second law( )2circ2AnAAm vGMmF marr= =( )1536circ0.39806 107.69 10 m/s6.732 10AAGMvr×= = = ××Energy expenditureFrom Problem 13.112, 33.448 10 m/sAv = ×
  • 143. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Energy, ( )2 2112 circ1 12 2A AE m v mv∆ = −( ) ( )2 23 31121 17.690 10 3.448 102 2E m m∆ = × − ×6112 23.624 10 m JE∆ = ×( )6113 11223.624 10 m0.50 J2E E×∆ = ∆ =Thus, additional kinetic energy at A is,( )( )6211323.624 10 m12 2Am v E×∆ = ∆ = (1)Conservation of energy between A and B( ) ( )2 2circ12A A A AAGMmT m v v Vr = + ∆ = −  212B B BBGMmT mv Vr= = −A A B BT V T V+ = +( )6 152361 23.624 10 m 0.3980 10 m7.690 102 2 6.732 10m× ×× + −×15261 0.39806 10 m2 6.434 10Bmv×= −×2 6 6 6 659.136 10 23.624 10 118.26 10 123.74 10Bv = × + × − × + ×2 688.240 10Bv = ×9.39 km/sBv = !Conservation of angular momentum between A and B( )circsinA A B B Br m v r mv φ=( )( )( )( )( )( )3circ37.690 106732sin 0.85656434.4 9.394 10AABB Bvrr vφ× = = = × 58.9Bφ = ° !
  • 144. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 114.Conservation of angular momentumA A P Pr mv r mv=PA PArv vr= (1)Conservation of energy2 21 12 2P AP AGMm GMmmv mvr r− = − (2)Substituting for Av from (1) into (2)22 22 2PP PP A AGM r GMv vr r r − = −  22 1 11 2PPA P Arv GMr r r     − = −        2 2222A P A PPA PAr r r rv GMr rr− −=With ( )( )2 2A P A P A Pr r r r r r− = − +2 2 APA P PGM rvr r r =  +  (3) !Exchanging subscripts P and A( )2 2QEDPAA P AGM rvr r r =  +  !
  • 145. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 115.See solution to Problem 13.113 (above) for derivation of Equation (3)( )2 2 APA P PGM rvr r r=+Total energy at point P is212P P PPGMmE T V mvr= + = −( )01 22AA P PGMm r GMmr r r r = − +  ( )1AP A P PrGMmr r r r = − +  ( )( )A A PP A Pr r rGMmr r r− −=+A PGMmEr r= −+"Note: Recall that gravitational potential of a satellite is defined as beingzero at an infinite distance from the earth.
  • 146. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 116.(a) For a circular orbit of radius r22:nGMm vF ma mrr= =2 GMvr=21 12 2GMm GMmE T V mvr r= + = − = − (1)Thus E∆ required to pass from circular orbit of radius 1r to circularorbit of radius 2r is1 21 21 12 2GMm GMmE E Er r∆ = − = − +( )2 11 22GMm r rErr−∆ = (2) (Q.E.D.)(b) For an elliptic orbit we recall Equation (3) derived inProblem 13.113 ( )1with Pv v=( )2 211 2 12Gm rvr r r=+At point A: Initially spacecraft is in a circular orbit of radius 1r2circ1GMvr=2circ circ11 12 2GMT mv mr= =
  • 147. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.After the spacecraft engines are fired and it is placed on asemi-elliptic path AB, we recall( )2 211 2 12GM rvr r r= ⋅+And( )2 21 11 1 21 1 22 2GMrT mv mr r r= =+At point A, the increase in energy is( )21 circ1 1 2 11 2 12 2AGMr GME T T m mr r r r∆ = − = −+( )( )( )( )2 1 2 2 11 1 2 1 1 222 2AGMm r r r GMm r rEr r r r r r− − −∆ = =+ +( )2 121 2 1 22AGMm r rrEr r rr −∆ =  +  Recall Equation (2):( )( )21 2Q.E.DArE Er r∆ = ∆+A similar derivation at point B yields,( )( )11 2Q.E.DBrE Er r∆ = ∆+
  • 148. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 117.If the point of intersection 0P of the circular and elliptic orbits is at anend of the minor axis, then 0v is parallel to the major axis. This will bethe case only if 090 ,α θ+ ° = that is if 0cos sinθ α= − we musttherefore prove that0cos sinθ α= − (1)We recall from Equation (12.39):21cosGMCr hθ= + (2)When 0,θ = ( )min min 0and 1 sinr r r r α= = −( ) 2011 sinGMCr hα= +−(3)For 180 ,θ = ° ( )max 0 1 sinr r r α= = +( ) 2011 sinGMCr hα= −+(4)Adding (3) and (4) and dividing by 2:2 20 01 1 1 12 1 sin 1 sin cosGMrh rα α α = + = − + Subtracting (4) from (3) and dividing by 2:20 01 1 1 1 2sin2 1 sin 1 sin 2 1 sinCr r  = − =   − + −   αα α α20sincosCrαα=Substitute for 2GMhand C into Equation (2)( )201 11 sin coscosr rα θα= + (5)
  • 149. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Letting 0 0andr r θ θ= = in Equation (5), we have20cos 1 sin cosα α θ= +2 20cos 1 sincos sinsin sinα αθ αα α−= = − = −This proves the validity of Equation (1) and thus 0P is an end of theminor axis of the elliptic orbit.
  • 150. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 118.(a) Ar R=Conservation of angular momentum0 0sin B BRmv r mvφ =( )1Br R R Rα α= + = +( ) ( )0 0 0 0sin sin1 1BRv vvRφ φα α= =+ +(1)Conservation of Energy( )2 201 12 2 1A A B B BGMm GMmT V T V mv mvR Rα+ = + − = −+2 202 1 211 1BGMm GMmv vR Rαα α   − = − =   + +   Substitute for Bv from (1)( )22 00 2sin 2111GMmvRφ ααα    − =    + + From Equation (12.43): 2esc2GMvR=( )22 200 esc2sin111v vφ ααα    − =    + + ( )220 esc20sin111vvφ ααα = −  ++  (2)( )2esc00sin 1 1 Q.E.D.1vvαφ αα = + −  +  
  • 151. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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) Allowable values of 0v ( )for which maximum altitude Rα=200 sin 1φ≤ ≤For 0sin 0,φ = from (2)2esc00 11vvαα = −  + 0 esc1v vαα=+For 0sin 1,φ = from (2)( )2esc201111vvααα = −  ++  ( )2 2esc01 1 1 2 1 211 1 1vvα α ααα α α α α  + + − + = + − = =   + + +  0 esc12v vαα+=+esc 0 esc11 2v v vα αα α+≤ ≤+ +!
  • 152. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 119.60 60m dt m+ =∫v F v( ) ( )66030 20sin 2 24cos232.2t t dt + + = ∫ i j v[ ]66030 10cos2 12sin 232.2t t+ − + =i j v[ ] [ ] 610 cos12 cos0 12 sin12 sin 0 0.09317− − + − =i j v61.5615 6.4389 0.09317− =i j v[ ]6 16.759 69.109 ft/s= −v i j6 71.1 ft/sv = !76.4° !
  • 153. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 120.0dt m m= −∫F v v( )( ) ( )204 8 2 4 2 8 2tt dt t t t− − = = − −∫ i j v i j( )20.5 2 0.5 m/st t t= − −v i j( ) ( ) ( )22 2 22speed 0.5 2 0.5v t t t= = − +4 3 2 20.25 2 4 0.25t t t t= − + +2 20.25 2 4.25t t t = − + ( ) [ ]2 2 23 2speed 2 0.25 2 4.25 0.5 26 8.5 0dt t t t tdtt t t = − + + − = − + =Roots:( )6 36 4 8.50, 0,2t v t± −= = =( )6 22.2929 s, 3.7071 s outside interval2t±= =At 2.2929 s, 1.9571 1.1464t v= = − −i j2.27 m/s, maxv = "0
  • 154. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 121.1 1 212 km/h 3.33 m/s 10 sv t −= = =2 18 km/h 5.00 m/sv = =1 1 2 2impulsemv mv−+ =( ) ( ) ( )333 m/s 10 s 5.00 m/sNm F m+ =( )( )440 kg 5.00 m/s 3.3333 m/s73.33 N10 sNF−= =73.3 NNF = !Note: NF is the net force provided by the sails. The force on the sails is actually greater and includes the forceneeded to overcome the water resistance of the hull.
  • 155. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 122.(a) 0θ = °k kFt Wt mgtµ µ= =A k Bmv mgt mvµ− =( )9 0.30 9.81 0t− =3.06 st = !(b) 20θ = °cos20 cos20Nt Wt mgt= ° = °kFt Ntµ=cos20 sin 20 0A kmv mg t mgtµ− ° − ° =( )9 0.65 9.81 cos20 9.81sin 20 0t − ° + ° = ( )9 9.81 0.9528 0t− =0.963 st = !
  • 156. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 123.20,000 lbW =220,000621.118 lb s ft32.2m = = ⋅Momentum in the x direction( )0 1: sin15x mv F mg t mv− + ° =( ) ( ) ( )( )621.118 108 sin15 6 621.118 36F mg− + ° =sin15 7453.4F mg+ ° =(a) 7453.4 20,000 sin15 2277 lbF = − ° =2280 lbF = !(b) ( )0 sin15 0mv F mg t− + ° = t = total time( )621.118 108 7453.4 0;t− = t = 9.00 sAdditional time = 9 – 6 = 3 s !
  • 157. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 124.20,000 lbW =220,000621.118 lb s ft32.2m = = ⋅Momentum, ( )0 1: sin15x mv F mg t mv− + ° =( ) ( )( ) 00621.118 sin15 5.5 621.1182vv F mg − + ° =   ( )( )0310.559 sin15 5.5v F mg= + ° (1)Conservation of energy:( )2 20 11 1sin152 2mv F mg x mv− + ° =( )( )22 00310.559310.559 sin15 5404vv F mg− + ° =( ) ( )( )203 310.559sin15 5404vF mg= + °Using (1) eliminate ( )sin15 :F mg+ °( )2003 310.559 5.5310.5594 180vv =   (a)( )( )( )04 540130.909 ft s 89.3 mi h3 5.5v = = = !(b) ( )( )( )0 310.559 130.909310.559+ sin15 7391.85.5 5.5vF mg ° = = =( )7391.8 20000 sin15 2215 lbF = − ° =F = 2220 lb !
  • 158. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 125.100 km h 27.777 m s2v = =80 km h 22.222 m s1v = =(a) ( ) ( ) ( )82 1010008 100 803600F dt F m v v m = = − = −   ∫0.69444F m= on the levelon the up grade( ) ( )10100: sin6 27.777x F mg dt m v− ° = −∫( ) ( ) ( ) ( )100.69444 10 9.81 sin6 10 27.777m m m v− ° = −(a) 10 1027.777 3.3098, 24.468 m sv v− = − =10 88.1 km hv = !(b) ( ) ( )01000sin 6 60 1003600tF mg dt m − ° = −   ∫0.69444 m m− ( )9.81 sin6 t m ° =  ( )11.111−t = 33.6 s !F = 0.69444 m
  • 159. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 126.Impulse diagonally (assume sliding)[ ]( ) ( ): cos20 196.2 sin 20 0.3 N 6 20 15x P ° − ° − =[ ]( ): N 196.2 cos20 sin 20 6 0y P− ° − ° =[ ]15cos20 196.2 sin 20 0.3 196.2 cos20 + sin 20 20 506P P ° − ° − ° ° = =  ( )cos20 0.3 sin 20 50 67.104 55.310P ° − ° = + +205.97 NP = 206 NP = !Check sµStatic value ( )0.4 N = 0.4 196.2 cos20 sin 20P° + °cos20 196.2 sin 20 0.4 N = 0P ° − ° −( )cos20 0.4 sin 20 196.2 sin 20 78.48 cos20 140.85P ° − ° = ° + ° =static = 175.4 N 206 NP <W = (20)(9.81) = 196.2 N
  • 160. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 127.Use impulse momentum 1 55mi h 80.667 ft sv = =x-Directionm 1 sv µ m− 0gt =( )( )12s80.667 ft s0.4 32.2 ft svtµ g= =6.26 st = !Since this is the shortest time the load can be brought to rest and the load does not slide it is also the shortesttime the rig can be brought to rest.
  • 161. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 128.( )42 10ttF dt m v v=== −∫( ) ( ) ( )( )405 40 40 32.2 2 3P dt− = − −∫( ) ( )( )20 40 4 40 32.2 5P − =(a) ( )= 6.2112 + 160 20P8.31 lbP = !(b) ( ) ( ) ( )( )05 40 40 32.2 0 3tP dt− = − −∫( )( )5 t 40 40 32.2 3P t =−2.4 st = !
  • 162. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 129.(a) Combined90km h 25 m sv = =( )( ) ( )( )1 26500 9.81 63765 N; 3600 9.81 35316 NW W= = = =1 1 2 2 1; 0.75N W N W F N= = =+⊗← Impulse = 0 – mv0( )10.75 10,100 kg 25 m sN t− = −( )( )( )10,100 255.2798 s0.75 63765t = = 5.28 st = !(b) Second trailer alone+⊗← Impulse 2= Ct m v− =( ) ( )5.2798 3600 kg 25 m sC− = −= 17046 NC17.05 kNC = !Compression
  • 163. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 130.(a) Entire train 1 72 km/h 20 m/sv = =318 13 31 Mg 31 10 kgA Bm m+ = + = = ×( ) ( )( )31 20 19000 N 19000 N 31 10 kg 20 m/st −= − + + ×( )( )31 231 10 kg 20 m/s16.3158 s38000 Nt −×= =1 2 16.32 st − = !(b) Car A 31 218 Mg 18 10 kg; 16.32 sAm t −= = × =( ) [ ] ( )( )0 19000 N 16.32 s 18000 kg 20 m/sCF = + + 3058.8 NCF =3.06 kN TCF = !
  • 164. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 131.(a) Entire train 1 72 km/h 20 m/sv = =18 13 31 Mg 31000 kgA Bm m+ = + = =( ) ( )( )1 20 19000 N 31000 kg 20 m/st −= − +1 2 32.63 st − =1 2 32.6 st − = !(b) Car A1 2 1 1 20 ; 32.63 sC AF t m v t− −= − + =( )( )( )18000 kg 20 m/s11033 N32.63 sCF = =11.03 kN TCF = !
  • 165. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 132.Impulse diagramsConstraint: Av 3 Bv=A: x ( ) ( )2020 0.5 sin30 0.532.2AT v° − =B: x ( ) ( )( )16 163 0.5 16 0.5 sin3032.2 32.2 3ABvT v− ° = =Substituting for T(0.5) from the equation for A into the equation for BFrom A:( )0.5 5 0.62112 AT v= −A0.496915 1.8634 43Avv− − =2.029 11Av =vA = 5.4214(a) 5.42 ft/sAv = 30° !( ) ( )0.5 5 0.62112 5.4214T = −3.2653 lbT =(b) 3.27 lbT = !
  • 166. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 133.( ) ( ) ( )3 C A C B Bl x x x x d x≅ − + − + −Constraint: 4 2 3 0C B Av v v− − = (1)For , andA B Cv v v +At 0,t = ( ) ( )4 15ft/s 2 3 9ft/s 0, 16.5 ft/sB Bv v− − = =(2) ( )203 4 932.2ATt t v− = − (3) ( )204 4 1532.2CTt t v− − = − (4) ( )102 2 16.532.2BTt t v− = −Given 0.5 s 4t = ⇒ Equations in T, vA, vB, vC3 2 4 0A B Cv v v− − + = (1)1.5 0.6211 AT v− 3.5901= − (2)2T− 0.6211 7.3168Cv− = − (3)0.31056 BT v− 4.1242= − (4)From the solution of the above equations.(a) 6.07 ft/sAv = "13.7 ft/sBv = "11.4 ft/sCv = "(b) 0.1212 lbT = "
  • 167. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 134.KinematicsLength of cable is constant.2 A BL X X= +2 0A BdLv vdt= + =2B Av v= −( )20.6 m/sAv =Collar A15 kgAm =( ) ( )( ) ( )1 2 1 21 22A A A Am v T t W t m v− −+ − =( ) ( )( )1 20 2 15 9.81 15 0.6T t − + − × = ( ) 1 273.575 4.5T t −− = (1)Collar B10 kgBm =( ) ( )2 22 1.2 m/sB Av v= =( ) ( ) ( ) ( )1 2 1 21 2B A B B Bm v T t W t m v− −− + =( ) ( ) ( )1 20 10 9.81 10 1.2T t − + × − =  (2)Add Equation (1) and Equation (2) (eliminating T)( ) 1 298.1 73.575 4.5 12t −− = +1 216.50.673 s24.52t − = = 0.673 st = !
  • 168. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 135.Lets find out if they slide – assume they don’t slide and find the required angle for impending motion0; cos 0; cosy A A i A A iF N m g N m gθ θ= ⇒ − = ⇒ =∑0; sin 0x s A A iF N m gµ θ= ⇒ − =∑s mµ A g cos i mθ = A g sin iθtan 0.3i sθ µ= =so 16.7iθ = ° so they slideAssume they slide at the same velocity (remain in contact)impulse – momentum-x dir( ) ( ) ( )0 sinA B kA A kB B A Bm g m g t N N t m m vθ µ µ+ + − + = +(a) Solve for v( ) [ ]sin cos cosA B k A A k B BA Bm g m g t m g m g tvm mθ µ θ µ θ+ − +=+( )( )( ) ( )( ) ( ) ( )6 9 9.81 3 sin 20 0.25 6 0.15 9 9.81 cos20 36 9 + ° − + ° =+4.811 m/sv = 4.81 m/sA Bv v= = !
  • 169. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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) Just look at ABx 0 + sinB AB k B B Bm gt F t N t m vθ µ− − =So( )sin cosB k B B BABm gt m g t m vFtθ µ θ− −=( )( )( ) ( )( ) ( ) ( )( )9 9.81 3 sin 20 0.15 9 9.81 cos20 3 4.81 93°− ° −== 3.319 N3.32 NABF = !Since this is positive our assumption that the blocks stay in contact is correct
  • 170. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 136.The block does not move until 22 kg 9.81m/s 19.62 NP = × =From 0 to 2 st t= = 20P t=Thus, the block starts to move when19.620.981s20t = =(a) For 0 2 st< <20P t=1 2 10.981s 2 s, 0t t v= = =( )211 2 1 2ttmv Pdt W t t mv+ − − =∫( )( )220.9810 20 2 9.81 2 0.981 2t dt v+ − × − =∫( ) ( ) ( )( )2 221 20 N2 s 0.981s 19.62 N 2 s 0.981s2 kg 2 sv  = − − −    2 5.1918 m/sv =2 5.19 m/sv = !(b) From 2 s to 3 st t= =2 5.19 m/s,v = from (a)40 N 2 s 3 sP t= ≤ ≤2 32 s 3 st t= =( )322 3 2 3ttmv Pdt W t t mv+ − − =∫( ) ( )( )3322 5.1918 40 19.62 3 2 2dt v+ − − =∫( )( )( )( )315.1918 m/s 20.38 N 1s2 kgv  = +  3 5.1918 10.19 15.3818 m/sv = + =3 15.38 m/sv = !
  • 171. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 137.(a) Determine time at which collar starts to move.20 , 0 2 sP t t= < <Collar moves when2 19.622 kg 9.81m/s , or 0.981s20 20PP t= × = = =1 20.981 0.981t tmv Pdt Wdt mv+ − =∫ ∫For ( )2 s 20 Nt P t< =2 s 3 s 40 Nt P< < =3 s 0t P> =For 23 s 2 kg 9.81m/st W< = ×The maximum velocity occurs when the total impulse is maximum.area maximum impulseABCD =( )( ) ( )( )120.38 N 1.019 s 20.38 N 1s2= +area 30.76 N sABCD = ⋅( ) max0 30.76 N s 2 kg v+ ⋅ =max 15.38 m/sv = !(b) Velocity is zero when total impulse is zero at .t t+ ∆For ( )0.981s 3 s, impulse 19.62 N st t< < = − ∆ ⋅Thus, total impulse 0 30.76 19.62 t= = − ∆1.57 st∆ =Time to zero velocity– 3 s 1.57 s 4.57 st = + = !
  • 172. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 138.(a)20 4P t= −1 20sin30tmv W t Pdt mv− + =∫( ) ( )5201212sin30 5 20 432.2t dt v− ° + − =∫522030 20 2 0.37267t t v − + − = 2 53.7 ft/sv = !(b) After 5 s,t = 50 53.7 ft/s,P v t′= = is time after 5 s5 12sin30 0mv t′− =( )1253.7 6 3.34 s; 532.2t t t t′ ′ ′= ⇒ = = +8.34 st = !0
  • 173. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 139.201 2 1 0 3, where ,1.6 10pp C C t C p C −= − = =×16 oz1 lb0.70 oz0.04375mg g    = =( )2 20.4 0.12566 m4Aπ= =( )( )( )31.6 101 20 20.04375 2100 ft/s0.12566 2.8532632.2 ft/sC C t dt−×− = =∫31.6 1021 20122.7062C t C t−× − =  ( )( )233 00 31.6 10 s1.6 10 s 22.7062 1.6 10 spp−−−×× − =×( )301.6 10 s22.7062p−×=20 28.383 lb/inp =0 28.4 ksip = !
  • 174. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 140.( ) 2212 oz 1 lb/16 oz 0.003882 lb s /ft32.2 ft/sm = = ⋅  Conservation of energy (before impact)( )2 2 21 1 11 12 2Aym v mgh m v v+ = +( ) ( ) ( ) ( )2 2 21 154 32.2 4.5 542 2Aym m m v+ = +17.0235 ft/sAyv = (Just before impact)Conservation of energy (after impact)( ) ( )22 22 2 21 12 2Aym v v m v mg h+ = +′( ) ( ) ( )( )2 221 130 30 32.2 32 2Aym v m m + = +′ 13.8996 ft/sAyv =′ (Just after impact)( ) ( ) ( ): 0.003882 54 0.004 0.003882 30 , 23.292 lbH Hx F F− = =( ) ( ) ( ): 0.003882 17.0235 0.004 0.003882 13.8996 , 30.011 lbv vy F F− + = =impulsive forceIF =
  • 175. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.38.0 lbIF = 52.2° !
  • 176. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 141.1 230 ft/s 36 ft/s 0.18 sv v t= = ∆ =( )1 2mv P W t mv+ − ∆ =Vertical components( )( ) ( )0 0.18 36sin50VWP Wg+ − = °4.758VP W W− =5.76VP W= !
  • 177. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 142.Use impulse–momentum for bulletKnowns: 0.028 kg,m = 1 650 m/s,v = 2 500 m/s,v =-dirx1 2cos20 cos10xmv F t mv°− ∆ = °So, 1 2cos20 cos10xF t mv mv∆ = °− °( ) ( )0.028 650 cos20 0.028 500 cos10 3.3151 N s= °− °= ⋅-diry1 2sin 20 sin10ymv F t mv− °+ ∆ = °So, 2 1sin10 sin 20yF t mv mv∆ = °+ °( ) ( )0.028 500 sin10 0.028 650 sin 20 8.6558 N s= °+ °= ⋅We need ∆t. The average velocity is 600 m/s6aveave0.05 m; 83.33 10 s600 m/sxx v t tv−∆∆ = ∆ ∆ = = = ×So663.315139.78 kN83.33 10111.2 kN8.6558F 103.87 kN83.33 10xyFF−−= = × == =× !
  • 178. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 143.Conservation of angular momentum( ) ( ) ( ) ( ) ( )220.15 18 0.15 0.5 0.51.62 rad/sm m θθ==&&Conservation of energy:( ) ( )( )22 21 10.15 18 0.5 1.622 2m m R   = +    &2.5756 m/sR =&Motion relative to the rod:( ) ( )1.5 2.5756 0F t− ∆ =3.86 N sF t∆ = ⋅ !
  • 179. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 144.2 38 m/sv =01 20tmv Fdt mv+ =∫( )( )30.5 10300 sin 0.045 kg 38 m/s0.5 10mtF dtπ−×−+ =×∫5.37 kNmF = !
  • 180. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 145.(a) Force on the belt is opposite to the direction shown.1 72 km/h 20 m/s, 100 kgv m= = =1 2 avemv Fdt mv Fdt F t− = = ∆∫ ∫( )( ) ( )ave100 kg 20 m/s 0.110 s 0 0.110 sF t− = ∆ =( )( )( )ave100 2018182 N0.110F = =ave 18.18 kNF = !(b) Impulse area under diagramF t= −( )10.110 s2mF=From (a) aveImpulse F t= ∆( )( )18182 N 0.110 s=( ) ( )10.110 18182 0.1102mF =36.4 kNmF = !
  • 181. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 146.( ) ( ) 21100 2000 6211.18 lb s /ft32.2Bm = = ⋅  ( ) ( ) 21120 2000 7453.42 lb s /ft32.2Tm = = ⋅  ( )0 6211.18B B BF t m v v+ ∆ = =( )6T T Tm F t m v− ∆ =( ) ( ) ( )7453.42 6 7453.42 TF t v− ∆ =constraint: / 5.4 ft/sT B T Bv v v= − =Solving; ( ) ( ) [ ]7453.42 6 6211.18 5.4 7453.42T Tv v= − +( )13664.6 78260.9Tv =5.7273 ft/sTv =(a) 5.73 ft/sTv = !5.7273 ft/s 5.4 ft/s 0.3273 ft/sBv = − =( ) ( )6211.18 0.3273 2032 lb sB BF t m v∆ = = = ⋅(b) 2030 lb sF t∆ = ⋅ !
  • 182. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 147.21 1 1lb 0.0625 lb; 0.001941 lb s /ft16 16 32.2B BW m   = = = = ⋅      2block 8 lb; 0.248447 lb s /ftBW m= = ⋅Initial impact (Bullet + Block)( )0 block: cos30 0B Bx m v m m v°+ = + ′ (1)0: sin30 0By m v F t− °+ ∆ = (2)After impact( ) ( )( ) ( )block block: 32.2 1.2 sin15 0B Bx m m v m m′+ − + °=sin15 10.001 ft/sv gt− °=′From (1)( )( ) ( )block08.0625 10.0010.0625cos30cos30BBW WvgvWg +′  = =  °°  (a) 0 1489.7 ft/sv = 0 1490 ft/sv = !
  • 183. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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) Bullet alone( ) ( ) ( ) ( )0 0.001941 1489.7 2.8915; 0.001941 10.001 0.01941B Bm v m v′= = = =x : 0 cos15 cos15B x Bm v F t m v° + ∆ = °′y : 0 sin15 sin15B y Bm v F t m v− ° + ∆ = °′Solve: 2.7742xF t∆ = −0.7534yF t∆ = 2.87 lb sF t∆ = ⋅ !
  • 184. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 148./30:3.59 m/sA A B B A°= + = +v v v v Bv[ ] ( )0 2 3.59cos30 10 0A Ax B Bx B Bm v m v v v+ = °− + − =0.518 m/sBv = Just before impactAfter impact, 0A Bv v= =(a) A AF t m v∴ ∆ = − ="(b) ( ) ( )2 21 1Loss , just before impact 2 102 2A BT v v= = +( )226.30385 0.51817 11.28 J2T = + =  "
  • 185. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 149.( )75 kg, 50 kg, 200 kg BoatA B Cm m m= = =(a) Swimmers dive simultaneously( ) 20 C C A Bm v m m v= + + (1)Relative velocity of swimmers with respect to the boat is 3 m/s2 23 m/s 3C Cv v v v− = ⇒ = +Substitute into (1)( ) ( )0 3C C A B Cm v m m v= + + +Solve( ) ( )( )3 3 75 5075 50 200A BCA B Cm mvm m m− + − += =+ + + +1.154 m/sCv = !(b) A dives first and then Bx-dir( ) 2 20 C B C Am m v m v= + − (2)continuedmC vC
  • 186. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Relative velocity 2 223 3C C Cv v v v− = ⇒ = +Substitute into (2)( ) ( )2 20 3C B C A Cm m v m v= + − +Solve for 2Cv23 ACA B Cmvm m m−=+ +(3)Now look at C and B.-dir.x ( ) 2 3 3C B C C C Bm m v m v m v+ = + (4)Relative velocity2 33 33 3C Cv v v v− = ⇒ = +Substitute into (4)( ) ( )2 3 23C B C C C B Cm m v m v m v+ = + +so 3 23C B BC CC B C Bm m mv vm m m m+= −+ +(5)Substituting (3) into (5)33 3A BCA B C C Bm mvm m m m m−= −+ + +(6)with numbers375 503 1.292375 50 200 200 50Cv = − + = − + + + 31.292 m/sCv = !(c) Swimmer B dives first – solution is the same as for (b) except switch mA and mB33 3B ACA B C C Am mvm m m m m−= −+ + +50 753 1.28075 50 200 200 75 = − + = − + + + 31.280 m/sCv = !
  • 187. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 150.ball3 10.00582316 32.2m   = =      plate14 10.02717416 32.2m   = =      ( )2 4.8 17.582 ft/syv g= = ( )2 1.8 10.7666 ft/syv g= =′(a) Conservation of momentumball ball plate plate0y ym v m v m v+ = − +′ ′( ) ( ) ( ) ( ) ( ) plate0.005823 17.582 0 0.005823 10.7666 0.027174 v+ = − + ′plate 6.0747 ft/sv =′ plate 6.07 ft/sv =′ !(b) Energy lossInitial energy ( ) ( ) ( ) ( ) ( )2110.005823 6 0.005823 4.8 1.00482T V g+ = + =Final energy ( ) ( ) ( ) ( ) ( ) ( ) ( )2 221 10.005823 6 0.005823 1.8 0.027174 6.07472 2T V g+ = + +0.9437=Energy lost ( )1.0048 0.9437 ft lb= − ⋅ = 0.0611 ft lb⋅ !
  • 188. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 151.Before impact( )( )( )21 10, 30 kg 9.81m/s 2 m 588.6 NT V mgh= = = =22 21, 02T mv V= =( ) 21 1 2 21: 588.6 30 6.2642 m/s2T V T V v v+ = + = ⇒ =(a) Rigid columns0mv F t− + ∆ =( )30 6.2642 F t= ∆187.93 N sF t∆ = ⋅ on the block187.9 N sF t∆ = ⋅ !All of the kinetic energy of the block is absorbed by the chain.( )( )2130 6.2642 588.6J2T = =589JE = !
  • 189. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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) Elastic columnsMomentum of system of block and beam is conserved( ) ( )306.2642 1.2528 m/s150mmv M m v v vm M′ ′= + = − = =+Referring to figure in Part (a) mv F t mv′− + ∆ = −( ) ( )30 6.2642 1.2528 150.34F t m v v′∆ = − = − =150.3 N sF t∆ = ⋅ !( ) ( ) ( )2 2 22 21 1 30 1206.2642 1.2528 1.25282 2 2 2E mv mv  ′= − = − −  565.06 94.170 470.89= − =471JE = !
  • 190. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 152.Before impact 2 21 1(0.75) (6) 13.5J2 2H HT m v= = =(a) For ,Am = ∞ 2 0T = So,Energy absorbed = 13.5 J!Impulse 2= ( ) (0.75)(6) 4.5N sH Hm v v− = = ⋅ !(b) 4kgAm =y-dir : 2( )H H A Hm v m m v= +2(0.75)(6)0.9474 m/s(4 0.75)H HA Hm vvm m= = =+ +So 2 22 21 1( ) (4 0.75)(0.9474) 2.1316 J2 2A HT m m v= + = + =Energy absorbed 1 2 13.5 2.1316T T= − = −11.37 JE∆ = !System = hammer0
  • 191. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.y-dir : 2H H Hm v F t m v− ∆ =So 2( )H HF t m v v∆ = −0.75(6 0.9474)= −3.79= 3.79 N sF t∆ = ⋅ !
  • 192. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 153.96 mi/h 140.8 ft/sv = =5/ 0.3125/16m g g= =AVE812 0.02667 s25dtv= = =( ) ( )AVE0.3125140.8 0.02667 0Fg+ − =AVE 51.2 lbF = !
  • 193. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 154.For the sphere at A′ immediately before and after the cord becomes taut0 Amv F t mv ′+ ∆ =0 sin 0 0.8 lb smv F t F tθ − ∆ = ∆ = ⋅4mg=( ) 04sin65.38 0.8vg° =0 7.08 ft/sv = !
  • 194. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 155.(a)Conservation of total momentum2A Bmv mv mv′− =( )12A Bv v v′ = − !(b) Energy loss( )L A B A BE T T T T′ ′= + − +( ) ( )2 2 2 21 12 2L A BE m v v m v v′ ′= + − +From (a)( )12A Bv v v′ = −( ) ( )22 21 1 12 2 2L A B A BE m v v m v v = + − −  ( ) ( )2 2 2 21 122 4L A B A A B BE m v v m v v v v= + − − +( )22 21 124 4L A A B B A BE m v v v v m v v = + + = +  !
  • 195. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 156.Before impact After impactA B A Bm v m v m v m v′ ′+ = + (1)( )B A A Bv v e v v′ ′− = − (2)From (1) and (2) solve for ,A Bv v′ ′( ) 0.5 ( )2A B A BAv v v vv− − −′ =( ) 0.5( )2A B A BBv v v vv+ + −′ =( ) ( 3 )/ 4A A Ba v v v′ = + !(3 )/ 4B A Bv v v′ = + !(b) Loss of energy 2 2 2 2( ) ( )2 2A B A Bm mv v v v′ ′= + − +Loss of energy = 2 2 2 2 2 21( 6 9 9 6 )2 16A B A A B B A A B Bmv v v v v v v v v v + − + + + + +  Loss of energy 23( )16A Bmv v= − !
  • 196. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 157.System = A + B(a) x-dirA A B B A A B Bm v m v m v m v′ ′− = + (1)Unknowns ,B Av v′ ′Coefficient of restitution( )Br Ar Ar Brv v e v v′ ′− = −For our problem( )B A A Bv v e v v′ ′− = + (2)With numbers 0.6; 0.9; 4 m/s; 2 m/sA B A Bm m v v= = = =Solve 2 equations and 2 unknowns2.3 m/s; 2.2 m/sA Bv v′ ′= − = 2.3 m/sAv =′ !2.2 m/sBv =′ !(b) Energy lost2 2 2 211 1 1 1(0.6)(4) (0.9)(2) 6.6 J2 2 2 2A A B BT m v m v= + = + =( ) ( )2 2 2 221 1 1 1(0.6)(2.3) (0.9)(2.2) 3.765 J2 2 2 2A A B BT m v m v= + = + =′ ′1 2 2.835JE T T∆ = − =2.84 JE∆ = !
  • 197. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 158.System = A + Bx-dirA A B B A A B Bm v m v m v m v′ ′− = + (1)Unknowns , Ae v′Coefficient of restitution( )B A A Bv v e v v′ ′− = + (2)Where, 4 m/s; 2 m/s; 2.5 m/sA B Bv v v′= = =0.6 kg; 0.9 kgA Bm m= =Solve 2 equations and 2 unknowns2.75 m/s; 0.875Av e= − =′0.875e = !
  • 198. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 159.From conservation of momentumA A B B A A B Bm v m v m v m v′ ′+ = +1.2 2.4 1.20A Agv g gv− = − +′g’s cancel (1)From restitution0.8 , 0.8 14.418AA AAvv vv′′= = ++(2)(a) Velocity of A before impact from equations (1) and (2)1.2 43.2 1.2(0.8 14.4) 0.96 17.28A A Av v v− = − + = − −2.16 25.92Av =12 ft/sAv = !(b) Velocity of A after impact0.8(12) 14.4Av = +′24 ft/sAv′ = !
  • 199. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 160.From conservation of momentumA A B B A B B Bm v m v m v m v′ ′+ = + g’s cancel1.2 2.4 2.4(24 ft/s) 0B Bv vg g g     ′− = +          (1)From restitution0.2 , 4.8 0.224BB BBvv vv′′= = ++(2)(a) Velocity of B before impact from equations (1) and (2)28.8 2.4 2.4(4.8 0.2 ) 11.52 0.48B B Bv v v− = + = +2.88 17.28Bv = 6 ft/sBv = !(b) Velocity of B after impact4.8 0.2(6)Bv′ = + 6 ft/sBv′ = !
  • 200. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 161.(a) Total momentum conservedA A B B A A Bm v m v m v m v′ ′+ = +( ) ( )0 0 066 kg 01B BBv m v m v v vm ′ ′+ − = + ⇒ =  − (1)Relative velocities( ) 02A B B Av v e v v v v e′ ′ ′− = − ⇒ = (2)From equations (1) and (2)( )0 0 0 06 62 2 0.51 1B Bv e v v vm m   = ⇒ =   − −   3 kgBm = !(b) Using 0 0621Bv e vm =  − Gives,6 62 12 1BBe mm e+ = ⇒ =+0, 6 kgBe m= =1, 2 kgBe m= =2 kg 6 kgBm≤ ≤ !
  • 201. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 162.(a) First collision (between A and B)The total momentum is conservedA B A Bmv mv mv mv′ ′+ = +0 A Bv v v′ ′= + (1)Relative velocities( ) ( )A B B Av v e v v′ ′− = −0 B Av e v v′ ′= − (2)Solving equations (1) and (2) simultaneously( )0 12Av ev−′ = !( )0 12Bv ev+′ = !(b) Second collision (Between B and C)The total momentum is conserved.B C B Cmv mv mv mv′ ′′ ′+ = +Using the result from (a) for Bv′( )0 102B Cv ev v+′′ ′+ = + (3)Relative velocities( )0B C Bv e v v′ ′ ′′− = −Substituting again for Bv′ from (a)( )( )012C Bev e v v+′ ′′= − (4)continued
  • 202. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Solving equations (3) and (4) simultaneously( )( )( )001112 2 2Cv e ev v e +′ = + +  ( )20 14Cv ev+′ = !( )20 14Bv ev−′′ = !(c) For n spheresn Balls1 collisionn th−We note from the answer to part (b), with 3n =( )20314n Cv ev v v+′ ′ ′= = =or( )( )( )3 103 3 112v ev−−+′ =Thus for n balls( )( )( )10112nn nv ev−−+′ = !(d) For 8, 0.90n e= =From the answer to part (c) with 8n =( )( )( )( )( )8 1 70 078 11 0.9 1.92 2Bv vv−−+′ = =8 00.698v v′ = !
  • 203. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 163.(a) Packages A and BTotal momentum conservedA A B B A A B Bm v m v m v m v′ ′+ = +( )16 16 86 16 8 96A B A Bv v v vg g g′ ′ ′ ′= + ⇒ + =2 12A Bv v′ ′+ = (1)Relative velocities( ) ( )0.3 6 1.8A B B A B Av v e v v v v′ ′ ′ ′− = − ⇒ − = = (2)Solving Equations (1) and (2) simultaneously3.4 ft/sAv′ =5.2 ft/sBv′ =Packages B and CB B C C B B C Cm v m v m v m v′ ′′ ′′+ = +( )8 8 125.2 B Cv vg g g′′ ′= + 4 6 20.8B Cv v′′ ′+ = (3)Relative velocities( ) ( )0.3 5.2 1.56B C C B C Bv v e v v v v′ ′ ′′ ′ ′′− = − ⇒ − = = (4)continued
  • 204. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Solving (3) and (4) simultaneously2.70 ft/sCv′ = !1.144 ft/sBv′′ =(b) Packages A and B (second time)A A B B A A B Bm v m v m v m v′ ′′ ′′ ′′′+ = +( ) ( )16 8 16 83.4 1.144 ; 2 7.944A B A Bv v v vg g g g′′ ′′ ′′ ′′′+ = + + = (5)( )A B B Av v e v v′ ′′ ′′′ ′− = −( )( )3.4 1.144 0.3 0.6768 ; 0.6768B A A Bv v v v′′′− = = − − + = (6)Solving Equations (5) and (6) simultaneously2.42 ft/sAv′′ = !
  • 205. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 164.Impact2.5 cos40 1.915 m/sAnv = − °= −2.5 sin 40 1.607 m/sAtv = − °= −2 m/sBnv =0Btv =Impulse-momentumUnknowns , , ,An Bn At Btv v v v′ ′ ′ ′System A + B n-dirA An B Bn A An B Bnm v m v m v m v′ ′+ = + (1)Coefficient of restitution( )Bn An An Bnv v e v v′ ′− = − (2)Solve (1) and (2) for An Bnv v′ ′+0.7493 m/s; 2.1870 m/sAn Bnv v′ ′= = −System A t-dir1.607 m/sA At A At Atm v m v v= ⇒ = −′ ′continued
  • 206. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.System B t-dir0 m/sB Bt B Bt Btm v m v v′ ′= ⇒ =Resolve into componentsBall A2 2(0.7493) (1.607) 1.773 m/sAv′ = + =1 0.7493tan 25.01.607β −  = = °  40 25 15.0θ = ° − °= °So 1.773 m/sAv′ = !Ball B2.19m/sBv′ = !
  • 207. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 165.Ball A t-dir 0 0sin sinAt Atmv mv v vθ θ′ ′= ⇒ =Ball B t-dir0 0B Bt Btm v v′ ′= ⇒ =Ball A + B n-dir0 cos 0 An Bnmv m v m vθ ′ ′+ = + (1)Coefficient of restitution( )Bn An An Bnv v e v v− = −′ ′0( cos 0)Bn Anv v e v θ′ ′− = − (2)Solve (1) and (2)0 01 1cos ; cos2 2An Bne ev v v vθ θ− +   ′ ′= =      With numbers0.8; 45e θ= = °0 0sin 45 0.707Atv v v= °=′0 01 0.8cos45 0.07072Anv v v− ′ = ° =  0Btv′ =0 01 0.8cos45 0.63642Bnv v v+ ′ = °=  continued
  • 208. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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)12 2 20 0(0.707 ) (0.0707 )Av" v v = + 00.711v=1 0.0707tan 5.71060.707β −  = = °  So 45 5.7106 39.3θ = − = °(B)00.711Av v′ = "00.636Bv v′ = "
  • 209. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 166.Angle of impulse force from geometry1 6cos 22.626.5θ −= = °Total momentum conservedBall A:x : ( )cos 0A A A Am v F t m vθ ′− ∆ + = (1)Ball B:B BF t m v′∆ =Restitution ( )0Av v=continued
  • 210. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Approach Separation( ) ( )0 0cos cos ; cosB A B Av v e v v v evθ θ θ′ ′ ′ ′− = = +Using equations (1) and (2) x : cosA A B B A Am v m v m vθ′ ′= +( )( ) ( ) ( )617.5/ 6 ft/s 1.6/ 17.5/ v6.5B Ag g v g ′ ′= +  g’s cancelSubstituting for6 6 6; 105 1.6 4.8 17.56.5 6.5 6.5B A Av v v      ′ ′ ′= + +            5.22 ft/sAv′ = !9.25 ft/sBv′ = 22.6°!
  • 211. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 167.Angle of impulse force from geometry1 6cos 22.626.5θ −= = °Total momentum conservedBall A:x : ( )cos 0A A A Am v F t m vθ ′− ∆ + = (1)Ball B:x : cos B BF t m vθ ′∆ = (2)Restitution ( )0Av v−continued
  • 212. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Approach Separation( )cos cos cos ; 4.8B A A B Av v e v v vθ θ θ′ ′ ′ ′− = − = (3)Using Equations (1) and (2) x : A A B B A Am v m v m v′ ′= +( )( ) ( ) ( )17.5/ 6 1.6/ 17.5/B Ag g v g v′ ′= + g’s cancelSubstituting for Bv′ from (3) ( )105 1.6 4.8 17.5A Av v′ ′= + +5.10 ft/sAv′ = !9.90 ft/sBv′ = !
  • 213. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 168.Angle of impulse force from geometry of A and B1 6cos 22.626.5θ −  = = °  Total momentum conservedBall A:Ball B:RestitutionApproach Separation( )6cos 22.66.51666.5A BBn AnAn Bnv vv ve ev vθ ′ ′ ′− + ° +  ′ ′−  = ⇒ = = −  (1)A: ( )sin sin 22.6A A A Am v m vθ θ′ ′= + °( )2.56 sin 22.66.5Av θ ′ ′= + °  (2)continued
  • 214. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 B+ cosA A A A B Bm v m v m vθ′ ′ ′= +( )( ) ( ) ( )58/ 6 58/ cos 5.3/A Bg g v g vθ′ ′ ′= + g’s cancelEquations (1), (2), and (3) in , andA Bv v θ′ ′ ′5.027 ft/s; 10.838 ft/s; 0.08218 rad 4.71A Bv v θ′ ′ ′= = = = °5.03 ft/sAv′ = 4.71° !10.84 ft/sBv′ = !
  • 215. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 169.(a)Before Aftert-DirectionMomentum of A is conserved.( )sinA A tmv m vθ ′=( ) sinA Atv v θ′ =Momentum of B is conserved.( )cosB B tmv m vθ ′=( ) cosB Btv v θ′ =n-DirectionTotal momentum is conserved.( ) ( )cos sinA B A Bn nmv mv m v m vθ θ ′ ′− = +( ) ( ) cos sinA B A Bn nv v v vθ θ′ ′+ = − (1)Relative velocities (coefficient of restitution)1e = ( ) ( ) ( )( )1 cos sinB A A Bn nv v v vθ θ′ ′− = + (2)Adding Equation (1) and (2)( ) cosB Anv v θ′ =(1) (2)− ( ) sinA Bnv v θ′ = −continuedA Bm m m= =
  • 216. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Thus, after impacttan tanA AB Bv vv vα β= Thus and A Bv vα β ′ ′= ⊥ !(b) Using the results from (a)( ) ( )2 2 2 2 2 2sin sinA A A A Bt nv v v v vθ θ′ ′ ′= + = +( ) ( )2 2sin30 30 40 25 ft/sAv′ = ° + = !( ) ( )2 2 2 2 2 2cos cosB B B B At nv v v v vθ θ′ ′ ′= + = +( ) ( )2 2cos30 40 30 43.3 ft/sBv′ = ° + = !1 1 30tan tan 36.940ABvvα β − −= = = = °( )180 90 90γ α β = − + − = ° !
  • 217. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 170.(a) Since Bv′ is in the x-direction and (assuming no friction), thecommon tangent between A and B at impact must be parallel to they-axisThus250tan150 Dθ =−1 250tan 70.20150 60θ −= = °−70.2θ = ° !(b) Conservation of momentum in x(n) direction( ) ( )cosA B A Bn nmv m v m v mvθ ′ ′+ = +( ) ( ) ( )1 cos 70.20 0 A Bnv v′ ′+ = +( ) ( )0.3387 A Bnv v′ ′= + (1)Relative velocities in the n direction( )( ) ( )0.9 cosA B B An ne v v e v vθ ′ ′= − = −( )( ) ( )0.3387 0 0.9 B A nv v′ ′− = − (2)(1) + (2)( )2 0.3387 1.9 0.322 m/sB Bv v′ ′= = !
  • 218. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 171.Momentum: cos 0A Ax Bv v vθ ′+ = +Restitution: 0.9 cosB Ax Av v v θ′ ′− =( )cos 0.9 cosA Ax Ax Av v v vθ θ′ ′− − =( )10.1 1 cos 70.2 0.016936 m/s2Axv′  = ° = ( )1 sin 70.2 0.94089, 0.32178Ay Bv v′ ′= ° = =(a) 0.941 m/sAv′ = !(b) Fraction of Initial Energy loss = F. L.12F. L. =( )2 112m − ( )2 12Bm v′ − ( )212Am v′( )21mF. L. 1 0.1035 0.8856 0.01090= − − = !
  • 219. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 172.Rebound at AProjectile motion between A and BConservation of momentum – t-direction( ) ( )0 sin30 5sin30A At tmv m v v′ ′° = ⇒ = °( ) 2.5 m/sA tv′ =Relative velocities in the n-direction( ) ( ) ( ) ( )( )0 cos30 0 0 5cos30 0.8A An nv e v v′ ′− ° − = − ⇒ =( ) 3.464 m/sA nv′ =After rebound ( ) ( ) ( )0cos30 sin30x A At nv v v′ ′= ° + °2.5cos30 3.464sin30= ° + °( )03.897 m/sxv =( ) ( ) ( )0sin30 cos30y A At nv v v′ ′= − ° + °2.5sin30 3.464cos30= − ° + °( )01.75 m/syv =x-direction: ( ) ( )0 03.897 , 3.897 m/sx x xx v t t v v= = = =y-direction: ( ) 2 2011.75 4.9052yy v t gt t t= − = −( )01.75 9.81y yv v gt t= − = −At B: 0 1.75 9.81 0.17839 sy A B A Bv t t− −= = − ⇒ =( ) ( ) ( )2204.905 1.75 0.17839 4.905 0.17839y A B A By h v t t− −= = − = −0.15609 mh =156.1 mmh = !
  • 220. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 173.22 209.817.5 m, 0.6 m, m/s 0.6 m 0.34975 s2 2AA A A Agtx v t t t= = = = ⇒ =0 21.444 m/sv =(a) First bounce:0 1.5 m, 0.06995 sB Bv t t= =( )229.81 m/s0.12 m 3.431 m2A B Be t t= −( )( ) ( )( )20.12 3.431 0.06995 4.905 0.06995Ae= −0.24 0.024Ae= −0.600Ae = !
  • 221. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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) Second bounce:BeforeAfter( )0.6 3.431 9.81By Bv t= −1.3724 m/s=( )0 21.444 m/sB Be v e=2 20.12 4.905 0 0.12 1.3724 4.905C By C C C Cy v t t t t= + − = = + −0.3497 sCt =( )( )06.75 21.444 0.3497C B C Bx e v t e= = =0.900Be = !
  • 222. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 174.Momentum in t direction is conservedsin30 tmv mv′° =( )( )25 sin30 tv′° =12.5 ft/stv′ =Coefficient of restitution in n-direction( )cos30 nv e v′° =( )( )( )25 cos30 0.9 19.49 ft/sn nv v′ ′° = =Write v′ in terms of x and y components( ) ( ) ( ) ( ) ( )0cos30 sin30 19.49 cos30 12.5 sin30x n tv v v′ ′ ′= ° − ° = ° − °10.63 ft/s=( ) ( ) ( ) ( ) ( )0sin30 cos30 19.49 sin30 12.5 cos30y n tv v v′ ′ ′= ° + ° = ° + °20.57 ft/s=
  • 223. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Projectile motion( ) ( ) ( )22 20 013 ft 20.57 ft/s 32.2 ft/s2 2yty y v t gt t′= + − = + −At B, 20 3 20.57 16.1 ; 1.4098 sB B By t t t= = + − =( ) ( )0 00 10.63 1.4098 ; 14.986 ftB x B Bx x v t x′= + = + =( ) ( )3cos60 14.986 ft 3 ft cot 60 13.254 ftBd x= − ° = − ° =13.25 ftd = !
  • 224. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 175.Find x and y components of vx yv v v= −i jAfter the first impact x component is multiplied by e and the y component is unchangedx yv ev v= − −′ i jAfter rebound at C the y component is multiplied by e and the x component is unchanged( )x y x yv ev ev e v v= − + = − −′′ i j i jso v ev′′ = − And the final velocity is parallel to the original velocity !
  • 225. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 176.Velocities just after impactTotal momentum in the horizontal direction is conserved( ) ( )1.5 2.5 1.5 2.5; 0 6A A B B A A B B A Bm v m v m v m v v vg g g g′ ′ ′ ′+ = + + = +15 1.5 2.5A Bv v′ ′= + (1)Relative velocities( ) ( )( ): 0 6 0.8 4.8A B B A B A B Av v e v v v v v v′ ′ ′ ′ ′ ′− = − − = − ⇒ − = − (2)Solving (1) and (2) simultaneously6.75 ft/s 1.95 ft/sA Bv v′ ′= =(a)Conservation of energy21 1102A AT m v V= =( )( )21 21.5 lb 6.75 ft/s11.061242 32.2 ft/sT = =continued
  • 226. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 0T =2 1.5AV m gh h= =1 1 2 2 : 1.06124 1.5T V T V h+ = + =0.70749 ft 8.9899 in.h = =8.49 in.h = !(b) Work and energy2 0T =( )221 1 2.51.95 0.147612 2B BT m vg ′= = =  ( )1 2 0.6 2.5 1.5f k BU F x W x x xµ− = − = − = − = −1 1 2 2 : 0.14761 1.5 0T U T x−+ = − =0.0984 ft 1.1808 in.x = =1.181 in.x = !
  • 227. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 177.(a) Impact between A and BTotal momentum conserved80A Bm mg= =30Cmg=15A A B B A A B B A Bm v m v m v m v v v′ ′ ′ ′+ = + ⇒ + = (1)Relative velocities( ) ( )( )15 0 0.8 : 12A B AB B A B A B Av v e v v v v v v′ ′ ′ ′ ′ ′− = − ⇒ − = − − = (2)Solving (1) & (2) 13.5 ft/sBv′ =Impact between B and C (after A hits B)Total momentum conserved( ) ( )80 30 80 30: 13.5 0B B C C B B C C B Cm v m v m v m v v vg g g g′ ′ ′′ ′′ ′′ ′′+ = + + = +1080 80 30B Cv v′′ ′′= + (3)continued
  • 228. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Relative velocities( ) ( )( ): 13.5 0 0.3B C BC C B C Bv v e v v v v′ ′ ′′ ′′ ′′ ′′− = − − = −4.05 C Bv v′′ ′′= − (4)Solving (3) and (4) 8.7136 ft/s 12.7636 ft/sB Cv v′′ ′′= =8.71 ft/sBv′′ = !(b) ( ) ( )L B C B CT T T T T′ ′ ′′ ′′∆ = + − +( ) ( )2 221 1 80 lb13.5 ft/s 226.39 lb ft2 2 32.2 ft/sB B BT m v ′ ′= = = ⋅  0CT′ =( ) ( )2 221 1 80 lb8.7136 ft/s 94.319 lb ft2 2 32.2 ft/sB B BT m v ′′ ′′= = = ⋅  ( ) ( )2 221 1 30 lb12.764 ft/s 75.894 lb ft2 2 32.2 ft/sC C CT m v ′′ ′′= = = ⋅  ( ) ( )226.39 0 94.319 75.894 56.177LT∆ = + − + =56.2 lb ftLT∆ = ⋅ !
  • 229. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 178.(a)Before AfterA Bm m m= =5 km/h 1.3889 m/s=Conservation of total momentumA A B B A A B Bm v m v m v m v′ ′+ = +1.3889 1.3889B A B A B Bv v v v v v′ ′ ′ ′− = − − + = − (1)Work and energy – car A (after impact)2112A AT m v′=2 0T =( )1 2 4fU F− =( )4k Am gµ= −21 1 2 21; 4 02A A k AT U T m v m gµ− ′+ = − =continued
  • 230. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 2 22 4 m 0.3 9.81 m/s 23.544 m /sAv′ = =4.852 m/sAv′ =Car B – (after impact)21 21, 02B BT m v T′= =( )1 2 1k BU m gµ− = −( )21 1 2 21: 1 02B B k BT U T m v m gµ− ′+ = − =( )( )( )2 2 2 22 0.3 1 m 9.81 m/s 5.886 m /s ; 2.426 m/sB Bv v′ ′= = =From (1) 1.3889 4.852 2.426 1.38B A Bv v v′ ′= + + = + +31.2 km/hBv = !(b) Relative velocities( )A B B Av v e v v′ ′− − = −( )1.3889 8.667 2.426 4.852e− − = −( )10.0559 2.426e− = −0.241e = !
  • 231. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 179.(a) Work and energyVelocity of A just before impact with B( )2 21 0 221 12 2A A AT m v T m v= =( )1 2 0.3 mk AU m gµ− = −f k k AF N m gµ µ= =( )( ) ( )( )( ) ( )( )2 2 21 1 2 221 1: 0.4 kg 3 m/s 0.3 0.4 kg 9.81 m/s 0.3 m 0.4 kg2 2AT U T v−+ = − =( ) ( )2227.2342 2.6896 m/sA Av v= =Velocity of A after impact with B ( )2Av′( )22 32102A AT m v T′= =( )2 3 0.075 mk AU m gµ− = −continued
  • 232. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 3 3T U T−+ =( )( ) ( )( )( )2 2210.4 kg 0.3 0.4 kg 9.81 m/s 0.075 m 02Av′ − =( )20.6644 m/sAv′ =Conservation of momentum as A hits B( )22.6896 m/sAv =( )20.6644 m/sAv′ =( ) ( ) ( ) ( )2 2A A B B A A B B A Bm v m v m v m v m m′ ′+ = + =2.6896 0 0.6644 2.0252 m/sB Bv v′ ′+ = + =Relative velocities (A and B)( ) ( )2 2A B AB B Av v e v v  ′ ′− = − ( )2.6896 0 2.0252 0.6644ABe− = − 0.506ABe = !Work and energyVelocity of B just before impact with C( ) ( )2 22 21 0.42.0252 0.82032 2B BT m v′= = =( ) ( )2 24 4 41 0.42 2B B BT m v v′ ′= =( )2 4 0.30 0.35316k BU m gµ− = − = −( )21 2 4 4 4: 0.8203 0.35316 0.2 BT U T v− ′+ = − =( )41.5283 m/sBv′ =
  • 233. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 momentum as B hits C1.2 kg, 0.4 kgC Bm m= =( ) ( )4 4B B C C B B C Cm v m v m v m v′ ′′ ′+ = +( ) ( )40.4 1.5283 0 0.4 1.2B Cv v′′ ′+ = +Velocity of B after B hits C, ( )40Bv′′ =With( )40; 0.61132 1.2 0.5094 m/sB C Cv v v′′ ′ ′= = ⇒ =Relative velocities (B and C)( ) ( ) ( )4 4; 1.5283 0 0.5094 0B C BC C B BCv v e v v e ′ ′ ′′− = − − = − 0.333BCe = !(b) Work and energy – Block C5 0T =( )( )2411.2 0.5094 0.155692T = =( )( )4 5 0.3 1.2 9.81 3.5316kU mgx x xµ− = − = − = −4 4 5 5 : 0.15569 3.5361 0 0.044 mT U T x x−+ = − = ⇒ =44.0 mmx = !
  • 234. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 180.Conservation of energy before impact( )( ) ( )2 2010 8 kg 9.81 m/s 0.15 m 8 kg2T V v+ = + =0 1.7155 m/sv =0 11 220Cylinder C: 8Platform A: 0 54 unknownsCounterweight B: 0 5Restitution: 0.8CAAA Cv F dt vF dt F dt vF dt vv v v′− =′+ − = ′+ = ′ ′− = ∫∫ ∫∫Simultaneous solution 4 Equations and 4 unknowns( ) 0 0, 1.372 m/sAa v v′ ′= = !( ) 2 6.86 N sb F dt = ⋅∫ !
  • 235. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 181.(a) A and B after the first impact2 212 22 2 21 21 1(5 kg)(1.372 m/s) (5 kg)(1.372 m/s)2 21 1(5) (5)2 20TT v vU Td Td→= + = += + − =2 Av v∴ =1.372 m/sAv = !1 1.372d v t t= =C after the first impact20 9.81 , 0.2798 s2td t = + ∴ =   At which time 9.81Cv t= 2.74 m/sCv = !(b) Second impact: (As before)C: 18(2.7448) 8 CF dt v ′′− =∫A: 1 25(1.3724) 5 AF dt F dt v ′′+ − =∫ ∫B: 25(1.3724) 5 AF dt v ′′+ =∫Restitution: 0.8[2.7448 1.3724]A Cv v′′ ′′− = −With 4 unknowns and four equations, solve for2.47 m/sAv ′′ = !1.372 m/sCv ′′ = !
  • 236. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 182.After impactVelocity of A just before impact, 0v( )( )20 2 2 32.2 ft/s 3.6 ft sin30v gh= = °( )( )( )2 32.2 3.6 0.5 10.7666 ft/s= =Conservation of momentumA B B B A Am v m v m v= −00.6 0.6(1.8 ) B Av g v vg g   = −      (1)g’s cancelRestitution( ) ( )0 00 0.9A Bv v e v v+ = + = (2)Substituting for vB from (2) in (1)0 00.6 1.8(0.9 ) 0.6 ; 2.4 1.02A A A Bv v v v v v= − − =(a) A moves up the distance d where,2 2 21 1sin30 ; (4.5758 ft/s) (32.2 ft/s ) (0.5)2 2A A Am v m gd d= ° =0.65025 ft 7.80 mAd = = !(b) Static deflection 0,x= B moves downConservation of energy (1) to (2)Position (1) – spring deflected, 0x0 sin30Bkx m g= °
  • 237. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 1 21; , 02B BT V T V T m v T+ = + = =21 01sin302e g B BV V V kx m gd= + = + °( )0 2 22 0 00122Bx de g B BV V V kxdx k d d x x+= + = = + +′ ′ ∫( )2 2 2 20 0 01 1 1sin30 2 0 02 2 2B B B B Bkx mgd m v k d d x x+ ° + = + + + +2 2 2 21.8; 34 (5.1141)32.2B B B Bkd m v d ∴ = =   0.20737 ftBd =2.49 in.Bd = !
  • 238. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 183.After impactVelocity of A just before impact, 0v( )( )20 2 2 32.2 ft/s 3.6 ft sin30v gh= = °0 2(32.2)(3.6)(0.5) 10.7666 ft/sv = =Conservation of momentum0 00.6 1.8;A B B A A Bm v m v m v v vg g   = − =      (1)g’s cancelRestitution( ) ( )0 00 ;A B Bv v e v v ev+ = + = (2)From (1) ( )00.6 0.610.7666 ft/s 3.5889 ft/s1.8 1.8Bv v   = = =      From (2) ( )01/ ,3Be v v e= =(a) 0.333e = !(b) Energy loss( ) 21Energy 3.6 sin302A B Bm g m v∆ = ° −( )( )( )0.6 lb 3.6 ft 0.5= ( )21 1.83.5889 ft/s2 32.2 −   1.08 0.36 0.72 ft lb= − = ⋅Loss 0.720 ft lb= ⋅ !(c) Static deflection 0,x= B moves downConservation of energy 1 to 2Position 1-spring deflected, 0x0 sin30Bkx m g= °
  • 239. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 1 21; , 02B BT V T V T m v T+ = + = =21 01sin302e g B BV V V kx m gd= + = + °( )0 2 22 0 00122Bx de g B BV V V kxdx k d d x x+= + = = + +′ ′ ∫( )2 2 2 20 0 01 1 1sin30 2 0 02 2 2B B B B Bkx mgd m v k d d x x+ ° + = + + + +2 2 2 21.8; 34 (3.5889)32.2B B B Bkd m v d ∴ = =   0.1455 ftBd =1.746 in.Bd = !
  • 240. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 184.Ball A falls1 1 2 2T V T V+ = + (Put datum at 2)2122A Amgh mv v gh= ⇒ =(2)(9.81)(0.2) 1.9809 m/s= =Impact1sin 302rrθ −= = °Impulse–MomentumUnknowns , ,B At Anv v v′ ′ ′x-dir0 0 sin30 cos30B B A An A Atm v m v m v′ ′ ′+ = + ° + ° (1)We need more equations0 0
  • 241. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Coefficient of restitution( )Bn An An Bnv v e v v′ ′− = −For our problemsin30 ( cos30 0)B An Av v e v′ ′° − = ° − (2)System = At-dir ( sin30 )A A A Atm v m v′− ° = (3)Solve 3 equations and 3 unknowns (maple) using A Bm m m= =1.3724 m/sBv′ =1.029 m/sAnv′ = −0.9905 m/sAtv′ = −Now lets look at B after impact.1 1 2 2T V T V+ = +21( )2B Bm v mgh′ =So2 2( ) (1.3724)2 (2)(9.81)BBvhg′= =0.0960 m=96.0 mmBh = !0 0
  • 242. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 185.Momentum: 1 1 20 (0.7071)n nmv mv Mv′ ′+ = +1 2(2 kg)(5 m/s)(0.7071) 2 kg 9 kg (0.7071)nv v= +′ ′Restitution: 2 1 1(0.7071) 0.6 0.6(5)(0.7071)n nv v v− = =′ ′Solve for 2 116 17m/s, (0.7071)11 11nv v   = = −′ ′      1 1.092801 m/snv′ = −"(b) Conservation of energy – cylinder + spring:2 2 20 2 21 1 1( )2 2 2kx M v kx′+ =22 2220,000 1 16 20,000(0.05) (9) 34.522 2 11 2x + = =  2 0.05875 m,x = 2N20,000 (0.0587 m) = 1175 NmF kx= = "
  • 243. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 186.Angle of impulse force from geometry of A and B1 6cos 22.626.5θ −  = = °  Total momentum conservedBall A:Ball B:(1)Restitution( ) ( )0cos cos sin6 ft/scosB A Ax yAAv v ve v vvθ θ θθ′ ′ ′− += = =( ) ( ) ( ) ( ) ( )256tan2B A A B A Ax y x yAv v v v v vevθ′ ′ ′ ′ ′ ′− + − += =continued
  • 244. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( ) ( ) ( ): sin sin cosA A A A A Ax yA m v m v m vθ θ θ′ ′= +2.5 2.5tan ( ) tan ( ) ; 6 ( ) ( )6 6A A x A y A x A yv v v v vθ θ   ′ ′ ′ ′= + = +      ( ) ( )15 2.5 6A Ax yv v′ ′= + (2)( )50 50 4.6: ; (6 ft/s) ( )A A A A B B A x BxA B m v m v m v v vg g g     ′ ′ ′ ′+ = + = +          (3)g’s cancelFrom equation (1) 22(32.2 ft/s )(0.75 ft) 6.9498 ft/sBv′ = =From equation (3) (50)(6) 50( ) 4.6(6.9498)A xv′= +( ) 5.3606 ft/sA xv′ =From equation (2) 15 2.5(5.3606) 6( )A yv′= +( ) 0.2664 ft/sA yv′ =2.56.9498 5.3606 0.266460.28346e − +   = =0.283e = !
  • 245. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 187.Angle of impulse force from geometry of A and B1 6cos 22.626.5θ −  = = °  Momentum considerationBall A:Ball B:2B B Bm v m gh′ = (1)RestitutionApproach Separation( ) ( )0cos cos sin6 ft/scosB A Ax yAAv v ve v vvθ θ θθ′ ′ ′− += = =( ) ( ) ( ) ( ) ( )2.56tan6B A A B A Ax y x yAv v v v v vevθ′ ′ ′ ′ ′ ′− + − += =continued
  • 246. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( ) ( ) ( ): sin sin cosA A A A A Ax yA m v m v m vθ θ θ′ ′= +2 2tan ( ) tan ( ) ; 6 ( ) ( )6 6A A x A y A x A yv v v v vθ θ   ′ ′ ′ ′= + = +      ( ) ( )12 2.5 6A Ax yv v′ ′= + (2)( )20 20 2: ; (6) ( )A A A A B B A x BxA B m v m v m v v vg g g+ = + = +′ ′ ′ ′6 ( )10BA xvv′′= + (3)From the equation for e2.50; ( ) ( ) 06B A x A ye v v v ′ ′ ′= − + =  (4)2.51; ( ) ( ) 66B A x A ye v v v ′ ′ ′= − + =  (5)Simultaneous solution of equations (2), (3) and (4) for e = 0 and equations (2), (3) and (5) for e = 1 yields0 : ( ) 5.463 ft/s, ( ) 0.224 ft/s, 5.370 ft/sA x A y Be v v v′ ′ ′= = = =1 : ( ) 4.926 ft/s, ( ) 0.4475 ft/s, 10.740 ft/sA x A y Be v v v′ ′ ′= = = =2( )0.4478 ft, 1.791 ft2(32.2)Bvh′= =5.37 in. 21.5 in.h≤ ≤ !
  • 247. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 188.Ball A aloneMomentum in t-direction conserved( ) ( )A A A At tm v m v′=( ) ( )0A At tv v′= =Thus ( )A Anv v′ ′= 60°Total momentum in the x-direction is conserved.( ) ( )sin60 sin60A A B B A A B Bxm v m v m v m v′ ′° + = − +( )0 1.5 m/s 0A B xv v v= = =( )( ) ( )( )( ) ( )0.17 1.5 sin60 0 0.17 sin60 0.34A Bv v′ ′° + = − ° +0.2208 0.1472 0.34A Bv v′ ′= − + (1)Relative velocity in the n-direction( ) cos30 ;A B B Anv v e v v  ′ ′− − = − ° − ( )( )1.5 0 1 0.866 B Av v′ ′− − = − − (2)Solving Equations (1) and (2) simultaneously0.9446 m/s, 0.6820 m/sB Av v′ ′= =Conservation of energy ball B( )2112B BT m v′=( )21 213.0232 02BWT Tg= =continued
  • 248. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 20 BV V W h= =( )21 1 2 21; 0.9446 0 ;2BBWT V T V W hg+ = + = +( )( )( )20.94460.0455 m2 9.81h = =45.5 mmh = !
  • 249. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 189.(a) Momentum of the sphere A alone is conserved in the t-direction.( ) ( ) ( ) 0A A A A At t tm v m v v′= =( ) ( )0A A At nv v v′ ′ ′= = 50°Total momentum is conserved in the x-direction.( )cos50 cos50A A B B A A B Bm v m v m v m v′ ′° + = − ° +0 4 m/sB Av v= =( ) ( )2 4 cos50 0 2 cos50 6A Bv v′ ′° + = − ° +5.1423 1.2855 6A Bv v′ ′= − + (1)Relative velocities in the n-direction( ) ( )cos50 ; 0, 4 m/sA B B A B Av v e v v v v′ ′− = ° + = =( )4 0.5 0.6428 ; 2 0.6428B A B Av v v v′ ′ ′ ′= + = + (2)Solving Equation (1) and Equation (2) simultaneously1.2736 m/s; 1.1299 m/sA Bv v′ ′= =1.274 m/sAv′ = 50° !1.130 m/sBv′ = !(b) ( ) ( )2 22lost1 12 2A A A A B BT m v m v m v ′ ′= − +  ( )( ) ( )( )2 212 kg 4 m/s 2 kg 1.274 m/s2= −( )( )26 kg 1.130 m/s 10.546 J− =lost 10.55 JT = !
  • 250. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 190.First 18 m: Since all the cars’ weight is on the rear wheels which skid, theforce on the car is( )k kF N Wµ µ= =( )( )181 hr58 km/h 1000 m/km3600 sv =   16.1 m/s=( ) ( )221 2 181 10 16.1 m/s 129.62 2W WT T mvg g = = = =  ( )( ) ( )( )1 2 18 m 18 mkU F Wµ− = =1 1 2 2T U T−+ =( )0 18 129.6kWWgµ + =   ( )( )129.60.7339518 9.81kµ = =For 400 m: Force moving the car is for the first 18 m,( )( ) ( )1 0.73395kF W Wµ= =For the remaining 382 m, with 75% of weight on rear drive wheels andimpending sliding,( )( ) ( ) ( )( )2 0.75 0.80 0.73395 0.80 0.91744s s kF Wµ µ µ= = = =( )( )( )2 0.91744 0.75 0.68808F W= =( )21 2 400102WT T vg = =   
  • 251. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 2 1 218 m 382 mU F F− = +( )( )( ) ( )( )( )0.73395 18 m 0.68808 328 mW W= +13.21 262.8 276.01W W W= + =( )21 1 2 2 40010 276.012WT U T W vg− + = + =   ( ) ( )( )( )4002 22 276.01 2 9.81 m/s 276.01v g= =2400 4005415.3 73.6 m/sv v= =400 265 km/hv = !
  • 252. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 191.(a) Spring constants 3 N/mm 3000 N/m=2 N/mm 2000 N/m=Max deflection at 2 when velocity of 0C =1 1 2 20, 0, 0, 0v T v T= = = =1 2 e gU U U− = +( ) ( ) ( )0.051 2 0 01 20.15mye e C mU F dx F dx W y− = − + +∫ ∫( )( )( )( )221 23000 N/m 2000 N/m0.05 m2 2mU y− = −( )( )( )23 kg 9.81 m/s 0.15 my+ +( ) ( )23.750 1000 4.4145 29.43m my y= − + +( ) ( )21 1 2 2 : 0 1000 29.43 8.1645 0m mT U T y y−+ = − + + =0.10626 mmy = 106.3 mmmy = !(b) Maximum velocity occurs as the lower spring is compressed adistance y′( ) ( )2 2 21 21 10; 3 kg 1.52 2CT T m v v v= = = =( )( ) ( ) ( )2 21 1 2 2; 0 1000 29.43 8.1645 1.5T U T y y v−+ = − + + =′ ′Substitute 0.014715 my′ =( )20 2000 29.43 0; 0.014715 mdvy ydy′ ′= − + = =′20.21653 0.43306 8.1645 1.5v− + + =25.5873 m/sv = 2.36 m/sv = !
  • 253. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 192.(a)Block leaves surface at C when the normal force 0N =cos nmg maθ =2cos Cvghθ = (1)2cosCv gh gyθ= =Work-energy principle( ) ( )212B C B C CT mv U W h y mg h y−= = − = −B B C CT U T−+ =Use Equation (1) ( ) 214.52Cm mg h y mv+ − =( )14.52Cg h y gy+ − = (2)34.52Cgh gy+ =( )4.532Cghyg+=   ( )( )( )( )24.5 9.81 139.812y+=0.97248 my = (3)continued
  • 254. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.0.97248cos cos 0.972481mCCyy hhθ θ= = = =1cos 0.97248 13.473θ −= = ° 13.47θ = ° !(b)From Equations (1) and (3)( )9.81 0.97248 3.0887 m/sCv gy= = =At C; ( ) cos 3.0887cos13.47C Cxv v θ= = °3.0037 m/s=( ) sin 3.0887sin13.47C Cyv v θ= − = °0.71947 m/s= −( ) ( )2 21 10.97248 0.71947 9.812 2C C yy y v t gt t t= + − = − −At E: 20: 4.905 0.7194 0.97248 0Ey t t= + − =0.37793 st =At E: ( ) ( ) ( )cos 1 sin13.47 3.0037 0.37793C xx h v tθ= + = ° +0.23294 1.3519 1.3681m= + =1.368 mx = !
  • 255. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 193.Find unstretched length of the spring1 0.3tan 71.5650.1θ θ−  = = °  ( ) ( )2 20.3 0.1 0.3162 mBDL = + == length at equilibriumEquilibrium: ( )0.1 sin 0.6 10 g 0A sM F θΣ = − =63.25 gsF =( )( ): 63.25 g 8000 N/m 0.07756 ms BD BD BDF k L L L= ∆ = ∆ ⇒ ∆ =Unstretched length 0 0.3162 0.07756BD BDL L L= − ∆ = −0.23864 m=Spring elongation, BDL′∆ when 90φ = °( ) 00.3 m 0.1m 0.4 0.23864BDL L′∆ = + − = −0.16136 m=At 1 1 190 0, 0V Tφ = ° = =( ) ( )1 1 1e gV V V= +( ) ( ) ( )2 211 80000.161362 2BDeV k L′= ∆ =104.15 N m= ⋅continued
  • 256. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/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 10g 0.6 58.86 N mgV = − = − ⋅1 104.15 58.86 45.29 N mV = − = ⋅At 2 ( ) ( ) ( )2 221 80000 N/m 0.07756 m2 2BDeV k Lφ = = ∆ =   24.06 N m= ⋅2 2 22 2 2 21 10 kg52 2T mv v v = = =  21 1 2 2 2: 0 45.29 5 24.06T V T V v+ = + + = +22 4.246v = 2 2.06 m/sv = "
  • 257. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 194.63960 1500 5460 mi 28.829 10 ftAr = + = = ×63960 6000 9960 mi 52.589 10 ftBr = + = = ×Conservation of momentum A A B Br mv r mv=28.8290.5481952.589AB A A ABrv v v vr = = =  (1)Conservation of energy2 21 1, , ,2 2A A A B B BA BGMm GMmT mv V T mv Vr r= = − = = −( )( )22 232.2 ft/s 3960 mi 5280 ft/miGM gR= = ×14 3 2140.77 10 ft /s= ×1466140.77 10488.29 10 m28.829 10AmV×= − = − ××1466140.77 10267.68 10 m52.589 10BmV− ×= = − ××2 61: 488.29 10 m2A B B B AT T T V mv+ = + − ×2 61267.68 10 m2Bmv= − ×2 6 21 1220.61 102 2A Bv v− × =Using (1) ( )22 61 1220.61 10 0.548192 2A Av v− × =2 60.34974 220.61 10Av = ×25115.39Av =325.1 10 ft/sAv = × !
  • 258. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 195.s sFt Nt mgtµ µ= =1 260 mi/h 88 ft/s 20 mi/h 29.333 ft/sv v= = = =1 2 0.65s smv mgt mvµ µ− = =( )88 0.65 32.2 29.333 2.803t t− = ⇒ =2.80 st = !
  • 259. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 196.0.22t∆ =( )1 2mv P W t mv+ − ∆ =Horizontal components( ) ( )84 9.14cos35 0.22 0HP° − =22858.69 kg m/sHP = ⋅2.86 kNHP = !
  • 260. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 197.(a) Total momentum of the two cars is conserved.( ), : cos30 cos10A A A Amv x m v m m vΣ ° = + ° (1)( ), : sin30 sin10A A B B A Bmv y m v m v m m vΣ ° − = + ° (2)Dividing (1) into (2)sin30 sin10cos30 cos30 cos10B BA Am vm v° °− =° ° °( )( )tan30 tan10 cos30ABA Bmvv m° − ° °=3600 28000.3473B AA BA Bv mm mv m g g= = =36001.28572800ABmm= =( )( )0.3473 1.2857 0.4465B Av v= =Car A was going faster !(b) Since B was the slower car, 30 mi/hBv =2.2396A Bv v=67.2 mi/hAv = !
  • 261. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 198.A B Cm m m m= = =Collision between B and CThe total momentum is conserved.4.5B C B C B Cmv mv mv mv v v′ ′ ′ ′+ = + ⇒ + = (1)Relative velocities( ) ( )0.8 4.5 3.6C B B C B Cv v e v v v v′ ′ ′ ′− = − = − ⇒ − = (2)Solving (1) and (2) simultaneously4.05 ft/sBv′ = !0.450 ft/sCv′ = !Since ,B Cv v′ ′> Car B collides with Car ACollision between A and B4.05A Bv v′ ′′+ = (3)Relative velocities( ) ( )0.5 4.05 ; 2.025A B B A B A A Bv v e v v v v v v′ ′′ ′ ′′ ′ ′ ′′− = − ⇒ − = − − = (4)Solving (3) and (4) simultaneously1.013 ft/sBv′′ = !3.04 ft/sAv′ = !C B Av v v′ ′′ ′< < ⇒ No more collisions
  • 262. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 199.BeforeAfter( ) ( )18 ft/s; 18cos40 13.79 ft/s; 18sin 40 11.57 ft/sA A An tv v v= = ° = = − ° = −( ) ( )12 ft/s; 0B B Bn tv v v= = − =t-directionTotal momentum conserved( ) ( ) ( ) ( )A A B B A A B Bt t t tm v m v m v m v′ ′+ = +( )( )( )( )( )( )1.5 lb 1.5 lb 2.5 lb11.57 ft/s 0 A At tv vg g g′ ′− + = +( ) ( )17.36 1.5 2.5A Bt tv v′ ′− = + (1)Ball A alone momentum conserved( ) ( ) ( ) 11.57 ft/sA A A A At t tm v m v v′ ′= ⇒ = − (2)Replace( )A tv′ in (2) in equation (1)( ) ( ) ( )17.36 1.5 11.57 2.5 ; 0B Bt tv v′ ′− = − + =
  • 263. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.n-Direction Relative velocities( ) ( ) ( ) ( )A B B An n nv v e v v  ′ ′− = − ( ) ( ) ( )13.79 12 0.8 B An nv v′ ′ − − = − ( ) ( ) 20.632B An nv v′ ′− = (3)Total momentum conserved( ) ( ) ( ) ( )A A B B A A B Bn n n nm v m v m v m v′ ′+ = +( )( )( )( )( )( )( )( )1.5 lb 2.5 lb 1.5 lb 2.5 lb13.79 ft/s 12 ft/s A Bn nv vg g g g′ ′+ − = +( ) ( )1.5 2.5 9.315A Bn nv v′ ′+ = − (4)Solve (3) and (4): ( )4 21.633B nv′ =( ) 5.408 ft/sB nv′ =( ) 15.224 ft/sA nv′ = −A( ) ( )2 215.224 11.57 19.12 ft/s, 37.23Av θ= + = = °19.12 ft/sAv = 72.2° !B5.41ft/sBv = 40° !
  • 264. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 200.(a) Rebound at AProjectile motion between A and BConservation of momentum – t direction( ) ( )0 cos60 6cosA At tmv m v v θ′ ′° = ⇒ =( ) 3 m/sA tv′ =Coefficient of restitution in the n direction( )( ) ( ) ( )( ) ( )0 0 : 6sin 60 0.6A A An n nv e v v′ ′− − = − ° =( ) 3.12 m/sA nv′ =After rebound ( ) ( )03 m/sx A tv v′= − = −( ) ( )03.12 m/sy A nv v′= =( )03 , 3 m/sx xx v t t v= = − = −( ) 2 2013.12 4.905 ;2yy v t gt t t= − = −( )03.12 9.81y yv v gt t= − = −At B, 0: 3.12 9.81 0 0.318 sy A B A Bv t t− −= − = ⇒ =2: 3.12 4.905 0.496 mB A B A By h h t t− −= = − =3B A Bx d t −= − = −0.496 mh = !0.953 md = !(b) ( )03 m/sB xv v= = −3.00 m/sBv = !
  • 265. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 201.(a) Momentum of sphere A alone is conserved in the t-direction.0 cos sinA A Am v m vθ θ′=0 tanAv v θ′= (1)Total momentum is conserved in the x-direction.( ) ( )0 0, 0B B A B B A A B Ax xm v m v m v m v v v′ ′ ′+ = + = =01.5 4.50 0Bv vg g′+ = +03Bvv′ = (2)Relative velocities in the n-direction( )0 sin 0 sin cosB Av e v vθ θ θ′ ′− − = − −( )( )0 0.6 cotB Av v v θ′ ′= + (3)Substituting Bv′ from (2) into (3)0 00.6 0.333 cotAv v v θ′= +00.267 cotAv v θ′= (4)Divide (4) into (1)21 tantan0.267 cotθθθ= =tan 1.935θ = 62.7θ = ° !(b) From (1) ( )0 tan 1.935A Av v vθ′ ′= =000.5168 ,3A Bvv v v′ ′= = (2)continued
  • 266. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( )( )22 2lost1 12 2A A A A B BT m v m v m v′= − +( ) ( )22 2 0lost 0 01 1.5 1 1.5 4.50.51682 2 2 3vT v vg g    = − +         [ ]2 20 00.31.5 0.40 0.502v vg g= − − =2lost 00.00932 ft lbT v= ⋅ !(For 0v in ft/s).