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solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 12

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solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 12

  1. 1. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 1.220 kg, 3.75 m/sm g= =( )( )20 3.75W mg= = 75 NW =
  2. 2. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 2.At all latitudes, 2.000 kgm =(a) ( )2 20 , 9.7807 1 0.0053 sin 9.7807 m/sgφ φ= ° = + =( )( )2.000 9.7807W mg= = 19.56 NW =(b) ( )2 245 , 9.7807 1 0.0053 sin 45 9.8066 m/sgφ = ° = + ° =( )( )2.000 9.8066W mg= = 19.61 NW =(c) ( )2 260 , 9.7807 1 0.0053 sin 60 9.8196 m/sgφ = ° = + ° =( )( )2.000 9.8196W mg= = 19.64 NW =
  3. 3. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 3.Assume 232.2 ft/sg =Wmg=: sWF ma W F agΣ = − =71 or21 132.2ssa FW F Wagg − = = =   − −7.46 lbW =27.46350.232 lb s /ft32.2Wmg= = = ⋅: sWF ma F W agΣ = − =127.46 132.2saF Wg = +   = +  7.92 lbsF =For the balance system B,0 0: 0w pM bF bFΣ = − =w pF F=But, 1w waF Wg = +  and 1p paF Wg = +  so that andpw p wWW W mg= = 20.232 lb s /ftwm = ⋅
  4. 4. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 4.Periodic time: 12 h 43200 sτ = =Radius of Earth: 63960 mi 20.9088 10 ft= = ×RRadius of orbit: 63960 12580 16540 mi 87.33 10 ftr = + = = ×Velocity of satellite:( )( )62 87.33 10243200rvππτ×= =312.7019 10 ft/s= ×It is given that 3750 10 lb s= × ⋅mv(a)323750 1059.046 lb s /ft12.7019 10mvmv×= = = ⋅×259.0 lb s /ftm = ⋅(b) ( )( )59.046 32.2 1901 lb= = =W mg1901 lb=W
  5. 5. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 5.+40: 10 10 10 20 4032.2y y yF ma a= + + + − =∑( )( ) 232.2 108.05 ft/s40ya = =ydv dy dv dva vdt dt dy dy= = =y yvdv a d=20 012v vy y yvdv a d v a y= =∫ ∫( )( )( )2 2 8.05 1.5yv a y= = 4.91 ft/sv =
  6. 6. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 6.Data: 0 108 km/h 30 m/s, 75 mfv x= = =(a) Assume constant acceleration. constant= = =dv dva vdx dt000fxvvdv adx=∫ ∫2012fv a x− =( )( )( )220 306 m/s2 2 75fvax= − = − = −000ftvdv adt=∫ ∫0 fv at− =0 306fvta−= − =−5.00 sft =(b) + 0: 0yF N W= − =∑N W=:xF ma N maµ= − =∑ma ma aN W gµ = − = − = −( )69.81µ−= − 0.612µ =
  7. 7. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 7.(a) : sinfF ma F W maα= − + =∑sinsin= − + = − +f fF FWa gm m mαα( )2 27500 N9.81 m/s sin 4 4.6728 m/s1400 kg= − + ° = −24.6728 m/sa = 4°0 88 km/h 24.444 m/s= =vFrom kinematics,dva vdx=000fxvadx vdv=∫ ∫2012fa x v= −( )( )( )220 24.4442 2 4.6728fvxa= − = −−63.9 mfx =+
  8. 8. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 8.(a) Coefficient of static friction.0: 0yF N WΣ = − = N W=0 70 mi/h 102.667 ft/sv = =( )2 2002 2tv va s s− = −( )( )( )( )22 22000 102.66731.001 ft/s2 2 170tv vas s−−= = = −−For braking without skidding , so that | |s s tN m aµ µ µ= =:t t s tF ma N maµΣ = − =31.00132.2t tsma aW gµ = − = − = 0.963sµ =(b) Stopping distance with skidding.Use ( )( )0.80 0.963 0.770kµ µ= = =:t k tF ma N maµΣ = = −224.801 ft/skt kNa gmµµ= − = − = −Since acceleration is constant,( )( )( )( )22 2000 102.6672 2 24.801tv vs sa−−− = =−0 212 fts s− =
  9. 9. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 9.For the thrust phase, : tWF ma F W ma agΣ = − = =( ) 221 32.2 1 289.8 ft/s0.2tFa gW   = − = − =      At 1 s,t =( )( )289.8 1 289.8 ft/sv at= = =( )( )221 1289.8 1 144.9 ft2 2y at= = =For the free flight phase, 1 s.t > 32.2 ft/sa g= − = −( ) ( )( )1 1 289.8 32.2 1v v a t t= + − = + − −At289.80, 1 9.00 s, 10.00 s32.2v t t= − = = =( ) ( )2 21 1 12 2v v a y y g y y− = − = − −( )( )( )22 2110 289.81304.1 ft2 2 32.2v vy yg−−− = − = − =(a) max 1304.1 144.9y h= = + 1449 fth =(b) As already determined, 10.00 st =
  10. 10. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 10.Kinematics: Uniformly accelerated motion. ( )0 00, 0x v= =( )( )( )2 20 0 2 22 101 2, or 1.25 m/s2 4xx x v t at at= + + = = =0: sin50 cos20 0yF N P mgΣ = − ° − ° =sin50 cos20N P mg= ° + °: cos50 sin 20xF ma P mg N maµΣ = ° − ° − =or ( )cos50 sin 20 sin50 cos20P mg P mg maµ° − ° − ° + ° =( )sin 20 cos20cos50 sin50ma mgPµµ+ ° + °=° − °For motion impending, set 0 and 0.30.sa µ µ= = =( )( ) ( )( )( )40 0 40 9.81 sin 20 0.30cos20593 Ncos50 0.30sin50P+ ° + °= =° − °For motion with 21.25 m/s , use 0.25.ka µ µ= = =( )( ) ( )( )( )40 1.25 40 9.81 sin 20 0.25cos20cos50 0.25sin50P+ ° + °=° − °612 NP =
  11. 11. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 11.Calculation of braking force/mass ( )/bF m from data for level pavement.0 100 km/hr 27.778 m/sv = =( )2 2002 2v va x x− = −( )( )( )( )22 20020 27.7782 2 606.43 m/sv vax x−−= =−= −br:xF ma F maΣ = − =2br6.43 m/sFam= − =(a) Going up a 6° incline. ( )6θ = °br: sinF ma F mg maθΣ = − − =brsinFa gmθ= − −26.43 9.81sin 6 7.455 m/s= − − ° = −( )( )( )22 2000 27.7782 2 7.455v vx xa−−− = =−0 51.7 mx x− =(b) Going down a 2% incline. ( )tan 0.02, 1.145θ θ= − = − °br: sinF ma F mg maθΣ = − − =brsinFa gmθ= − −( ) 26.43 9.81sin 1.145 6.234 m/s= − − − ° = −( )( )( )22 2000 27.7782 2 6.234v vx xa−−− = =−0 61.9 mx x− =
  12. 12. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 12.Let the positive directions of Ax and Bx be down the incline.Constraint of the cable: 3 constantA Bx x+ =13 0 or3A B B Aa a a a+ = = −For block A: : sin30A A AF ma m g T m aΣ = ° − = (1)For block B: : sin30 3B B B B AF ma m g T m a m aΣ = ° − = = − (2)Eliminating T and solving for ,Aag( )3 sin30 33 − ° = +  BA B A Amm g m g m a( ) ( )3 sin30 30 8 sin300.336733 /3 30 2.667A BAA Bm mag m m− ° − °= = =+ +(a) ( )( ) 20.33673 9.81 3.30 m/sAa = = 23.30 m/sA =a 30°( ) 213.30 1.101 m/s3= − = −Ba 21.101 m/sB =a 30°(b) Using equation (1),( )( )( )sin30 10 9.81 sin30 0.33673AAaT m gg = ° − = ° −  16.02 NT =
  13. 13. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 13.Let the positive directions of Ax and Bx be down the incline.Constraint of the cable: 3 constantA Bx x+ =3 0A Ba a+ =13B Aa a= −Block A: 0: cos30 0y A AF N m gΣ = − ° =: sin30x A A A AF ma m g N T m aµΣ = ° − − =Eliminate .AN( )sin30 cos30A A Am g T m aµ° − ° − =Block B: 0: cos30 0y B BF N m gΣ = − ° =: sin30 33B AB B B Bm aF ma m g N T m aµΣ = ° + − = = −Eliminate .BN( )sin30 cos30 33B ABm am g Tµ° + ° − = −Eliminate T.( ) ( )3 sin30 3 cos30 33BA B A B A Amm g m g m g m g m aµ − ° − + ° = +  Check the value of sµ required for static equilibrium. Set 0Aa = andsolve for .µ( )( )( )( )3 sin30 75 20tan30 0.334.3 cos30 75 20A BA Bm mm mµ− ° −= = ° =+ ° +Since 0.25 0.334,sµ = < sliding occurs.Calculate Aagfor sliding. Use 0.20.kµ µ= =continued
  14. 14. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( )( ) ( )( )3 sin30 3 cos303 /330 8 sin30 0.20 30 8 cos300.1352530 2.667A B A BAA Bm m m mag m mµ− ° − + °=+− ° − + °= =+(a) ( )( ) 20.13525 9.81 1.327 m/s= =Aa 21.327 m/sA =a 30°( ) 211.327 0.442 m/s3 = − = −  Ba 20.442 m/sB =a 30°(b) ( )( )( )( ) ( )( )sin30 cos3010 9.81 sin30 0.20cos30 10 1.327= ° − ° −= ° − ° −A A AT m g m aµ18.79 N=T
  15. 15. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 14.Data: 2255000 lb1708.1 lb s /ft32.2 ft/sAm = = ⋅2244000 lb1366.5 lb s /ft32.2 ft/sBm = = ⋅0 55 mi/h 80.667 ft/sv = − = −(a) Use both cars together as a free body. Consider horizontal force components only. Both cars have sameacceleration.:x x b b A x B xF ma F F m a m a= − − = +∑ ∑27000 70004.5534 m/s1708.1 1366.5b bxA BF Fam m+ += = =+ +xdva vdx=020 002= =∫ ∫fxx x fvva dx vdv a x( )( )( )220 80.667751 ft2 2 4.5534fxvxa−= − = − = −715 ft to the left(b) Use car A as free body. Fc = coupling force.:x x c b A xF ma F F m a= − =∑ ∑( )( )1708.1 4.5534 7000 778 lb= − = + =c A x bF m a F778 lb tensioncF =
  16. 16. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 15.Data: 2255000 lb1708.1 lb s /ft32.2 ft/sAm = = ⋅2244000 lb1366.5 lb s /ft32.2 ft/sBm = = ⋅0 55 mi/h 80.667 ft/sv = − = −(a) Use both cars together as a free body. Consider horizontal force components only. Both cars have sameacceleration.:x x b b A x B xF ma F F m a m a= − − = +∑ ∑270002.2767 m/s1708.1 1366.5bxA BFam m= = =+ +xdva vdx=020 002= =∫ ∫fxx x fvva dx vdv a x( )( )( )220 80.6671429 ft2 2 2.2767fxvxa−= − = − =1429 ft to the left(b) Use car B as a free body. Fc = coupling force.:x x c B xF ma F m a= − =∑ ∑( )( )1366.5 2.2767 3110 lbcF− = =3110 lb. compressioncF =
  17. 17. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 16.Constraint of cable: ( )2 A B A A Bx x x x x+ − = + = constant.0, orA B B Aa a a a+ = = −Assume that block A moves down and block B moves up.Block B: 0: cos 0y AB BF N W θΣ = − =: sin Bx AB B BWF ma T N W agµ θΣ = − + + =Eliminate ABN and .Ba( )sin cos B AB B Ba aT W W Wg gθ µ θ− + + = = −Block A: 0: cos 0y A AB AF N N W θΣ = − − =( )cos cosA AB A B AN N W W Wθ θ= + = +: sin Ax A A A AB A AWF m a T W F F agθΣ = − + − − =( )sin cos sin cosAB B A BaW W W Wgθ µ θ θ µ θ− + − + −( )cos AB A AaW W Wgµ θ− + =( ) ( ) ( )sin 3 cos AA B A B A BaW W W W W Wgθ µ θ− − + = +Check the condition of impending motion.0.20, 0,s A B sa aµ µ θ θ= = = = =( ) ( )sin 0.20 3 cos 0A B s A B sW W W Wθ θ− − + =continued
  18. 18. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( )( )0.20 3 0.20 128tan 0.4064A BsA BW WW Wθ+= = =−21.8 25 .sθ θ= ° < = ° The blocks move.Calculate Aagusing 0.15 and 25 .kµ µ θ= = = °( ) ( )sin 3 cosA B k A BAA BW W W Wag W Wθ µ θ− − +=+( )( )64sin 25 0.15 128 cos250.1004896° − °= =( )( ) 20.10048 32.2 3.24 ft/sAa = =(a) 23.24 ft/sBa = − 23.24 ft/sB =a 25° !(b) ( )sin cos AB BaT W Wgθ µ θ= + +( ) ( )( )16 sin 25 0.15cos25 16 0.10048= ° + ° +10.54 lbT = !
  19. 19. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 17.Constraint of cable: ( )2 constant.A B A A Bx x x x x+ − = + =0, orA B B Aa a a a+ = = −Assume that block A moves down and block B moves up.Block B: 0: cos 0y AB BF N W θΣ = − =x : sin Bx AB B BWF ma T N W agµ θΣ = − + + =Eliminate ABN and .Ba( )sin cos B AB B Ba aT W W Wg gθ µ θ− + + = = −Block A: 0: cos sin 0y A AB AF N N W Pθ θΣ = − − + =cos sinA AB AN N W Pθ θ= + −( )cos sinB AW W Pθ θ= + −: sin cos Ax A A A AB A AWF m a T W F F P agθ θΣ = − + − − + =( )sin cos sin cosAB B A BaW W W Wgθ µ θ θ µ θ− + − + −( )cos sin cos AB A AaW W P P Wgµ θ µ θ θ− + + + =( ) ( ) ( ) ( )sin 3 cos sin cos AA B A B A BaW W W W P W Wgθ µ θ µ θ θ− − + + + = +Check the condition of impending motion.0.20, 0, 25s A Ba aµ µ θ= = = = = °( ) ( ) ( )sin 3 cos sin cos 0A B s A B s sW W W W Pθ µ θ µ θ θ− − + + + =continued
  20. 20. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.( ) ( )3 cos sinsin coss A B A BssW W W WPµ θ θµ θ θ+ − −=+( )( )0.20 128 cos25 64 sin 253.88 lb 10 lb0.20 sin 25 cos25° − °= = − <° + °Blocks will move with 10 lb.P =Calculate Aagusing 0.15, 25 , and 10 lb.k Pµ µ θ= = = ° =( ) ( ) ( )sin 3 cos sin cosA B k A B kAA BW W W W Pag W Wθ µ θ µ θ θ− − + + +=+( )( ) ( )( )64 sin 25 0.15 128 cos25 10 0.15sin 25 cos2596° − ° + ° + °=0.20149=( )( ) 20.20149 32.2 6.49 ft/sAa = =(a) 26.49 ft/s ,Ba = − 26.49 ft/sB =a 25° !(b) ( )sin cos AB BaT W Wgθ µ θ= + +( ) ( )( )16 sin 25 0.15cos25 16 0.20149= ° + ° +12.16 lb.T = !
  21. 21. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 18.Assume B A>a a so that the boxes separate. Boxes are slipping.kµ µ=0: cos15 0yF N mgΣ = − ° =cos15N mg= °: sin15x kF ma N mg maµΣ = − ° =cos15 sin15kmg mg maµ ° − ° =( )cos15 sin15 ,ka g µ= ° − ° independent of m.For box A, 0.30kµ =( )9.81 0.30cos15 sin15Aa = ° − ° or 20.304 m/sA =a 15°For box B, 0.32kµ =( )9.81 0.32cos15 sin15Ba = ° − ° or 20.493 m/sB =a 15°
  22. 22. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 19.Let y be positive downward position for all blocks.Constraint of cable attached to mass A: 3 constantA By y+ =3 0A Ba a+ = or 3A Ba a= −Constraint of cable attached to mass C: constantC By y+ =0C Ba a+ = or C Ba a= −For each block :F maΣ =Block A: , or 3A A A A A A A A A A BW T m a T W m a W m a− = = − = −Block C: , orC C C C C C C C C C BW T m a T W m a W m a− = = − = −Block B: 3B A C B BW T T m a− − =( ) ( )3 3B A A B C C B B BW W m a W m a m a− − − − =or3 60 60 200.0769239 60 180 20B B A CB A Ca W W Wg W W W− − − −= = = −+ + + +(a) Accelerations. ( )( ) 20.076923 32.2 2.477 ft/s= − = −Ba 22.48 ft/sB =a( )( ) 23 2.477 7.431 ft/sAa = − − = 27.43 ft/sA =a( ) 22.477 2.477 ft/sCa = − − = 22.48 ft/sC =a(b) Tensions. ( )2020 7.4332.2AT = − 15.38 lbAT =( )2020 2.47732.2CT = − 18.46 lbCT =
  23. 23. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 20.Let y be positive downward for both blocks.Constraint of cable: constantA By y+ =0A Ba a+ = or B Aa a= −For blocks A and B, :F maΣ =Block A: AA AWW T ag− = or AA AWT W ag= −Block B: B BB B AW WP W T a ag g+ − = = −A BB A A AW WP W W a ag g+ − + = −Solving for , A BA AA BW W Pa a gW W− −=+(1)( ) ( ) ( )220 0 02 with 0A A A A A Av v a y y v − = − = ( )02A A A Av a y y = −  (2)( ) ( )0 0with 0A A A Av v a t v− = =AAvta= (3)continued
  24. 24. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(a) Acceleration of block A.System (1): 100 lb, 50 lb, 0A BW W P= = =By formula (1), ( ) ( )1100 5032.2100 50Aa−=+( ) 2110.73 ft/sA =a !!!!System (2): 100 lb, 0, 50 lbA BW W P= = =By formula (1), ( ) ( )2100 5032.2100Aa−= ( ) 2216.10 ft/sA =a !!!!System (3): 1100 lb, 1050 lb, 0A BW W P= = =By formula (1), ( ) ( )31100 105032.21100 1050Aa−=+( ) 230.749 ft/sA =a !!!!(b) ( )0at 5 ft. Use formula (2).A A Av y y− =System (1): ( ) ( )( )( )12 10.73 5Av = ( )110.36 ft/sAv = !!!!System (2): ( ) ( )( )( )22 16.10 5Av = ( )212.69 ft/sAv = !!!!System (3): ( ) ( )( )( )32 0.749 5Av = ( )32.74 ft/sAv = !!!!(c) Time at 10 ft/s. Use formula (3).Av =System (1): 11010.73t = 1 0.932 st = !!!!System (2): 21016.10t = 2 0.621 st = !!!!System (3): 3100.749t = 3 13.35 st = !!!!
  25. 25. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 21.(a) Maximum acceleration. The cable secures the upper beam; only the lower beam can move.For the upper beam, 10: 0yF N WΣ = − =1N W mg= =For the lower beam, 2 10: 0yF N N WΣ = − − = or 2 2N W=( )1 2: 0.25 0.30 0.25 0.60xF ma N N W maΣ = + = + =( )( ) 20.85 0.85 32.2 23.37 ft/sWam= = = 227.4 ft/s=a !For the upper beam, 1: 0.25xF ma T N maΣ = − =( )( ) ( )30000.25 0.25 3000 23.37 2927 lb32.2 = + = + =  T W ma 2930 lbT = !(b) Maximum deceleration of trailer.Case 1: Assume that only the top beam slips. As in Part (a) 1 .N mg=: 0.25F ma W maΣ = =20.25 8.05 ft/sa g= =continued
  26. 26. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Case 2: Assume that both beams slip. As before 2 2 .=N W( ) ( )( ) ( )2 : 0.30 2 2Σ = =F m a W m a20.30 9.66 ft/sa g= =The smaller deceleration value governs. 28.05 ft/sa = !!!!
  27. 27. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 22.Since both blocks move together, they have a common acceleration. Useblocks A and B together as a free body.:F maΣ = Σ( )sin30 sin30A B A BP m g m g m m a− ° − ° = +500sin30 9.81 sin3050A BPa gm m= − ° = − °+25.095 m/s=Use block B as a free body.cos30 : cos30B f BF m a F m aΣ = ° = °( )( )10 5.095 cos30 44.124 NfF = ° =sin30 : sin30B B B BF m a N m g m aΣ = ° − = °( ) ( )sin30 10 9.81 5.095 sin30B BN m g a= + ° = + °123.575 N=Minimum coefficient of static friction:min44.124123.575fBFNµ = = min 0.357µ =
  28. 28. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 23.(a) Kinematics of the belt. 0ov =1. Acceleration phase with 21 3.2 m/sa =( )( )1 1 1 0 3.2 1.5 4.8 m/sov v a t= + = + =( )( )221 1 1 11 10 0 3.2 1.5 3.6 m2 2o ox x v t a t= + + = + + =2. Deceleration phase. 2 0v = since the belt stops.( )2 22 1 2 2 12v v a x x− = −( )( )( )22 22 122 10 4.811.522 2 4.6 3.6v vax x−−= = = −− −22 11.52 m/s=a2 12 120 4.80.41667 s11.52v vt ta− −− = = =−(b) Motion of the package.1. Acceleration phase. Assume no slip. ( ) 213.2 m/spa =0: 0 oryF N W N W mgΣ = − = = =( )1:x f pF ma F m aΣ = =The required friction force is .fFThe available friction force is 0.35 0.35sN W mgµ = =( ) ( )( ) 210.35 9.81 3.43 m/sf sp sF Na gm mµµ= < = = =
  29. 29. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Since 2 23.2 m/s 3.43 m/s ,< the package does not slip.( ) ( )11 14.8 m/s and 3.6 m.p pv v x= = =2. Deceleration phase. Assume no slip. ( ) 2211.52 m/spa = −( )2:x f pF ma F m aΣ = − =( ) 2211.52 m/sfpFam= = −2 23.43 m/s 11.52 m/ss ssN mggm mµ µµ= = = <Since the available friction force sNµ is less than the requiredfriction force fF for no slip, the package does slip.( ) 2211.52 m/s ,p f ka F Nµ< =( ) ( )2 2:x p k pF m a N m aµΣ = − =( ) ( )( ) 220.25 9.81 2.4525 m/skp kNa gmµµ= − = − = − = −( ) ( ) ( ) ( ) ( )( ) 22 12 1 24.8 2.4525 0.41667 3.78 m/sp p pv v a t t= + − = + − =( ) ( ) ( ) ( ) ( ) ( )2 22 1 2 12 1 1 212p p p px x v t t a t t= + − + −( )( ) ( )( )213.6 4.8 0.41667 2.4525 0.41667 5.387 m2= + + − =Position of package relative to the belt( ) 225.387 4.6 0.787px x− = − = /belt 0.787 mpx =
  30. 30. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 24.Acceleration 1:a Impending slip. 1 1 10.30 NsF Nµ= =1 1: sin65y A y A AF m a N W m aΣ = − = °1 1 sin 65A AN W m a= + °( )1 sin65Am g a= + °1 1: cos65x A x AF m a F m aΣ = = °1 sF Nµ= or ( )1 1cos65 0.30 sin65A Am a m g a° = + °( )( ) 210.301.990 9.81 19.53 m/scos65 0.30sin65ga = = =° − °21 19.53 m/s=a 65°Deceleration 2 :a Impending slip. 2 2 20.30 NSF Nµ= =1 2: sin 65y y A AF ma N W m aΣ = − = − °1 2 sin65A AN W m a= − °2 2: cos65x x AF ma F m aΣ = = °2 2SF Nµ= or ( )2 2cos65 0.30 cos65A Am a m g a° = − °( )( ) 220.300.432 9.81 4.24 m/scos65 0.30sin 65ga = = =° + °22 4.24 m/s=a 65°
  31. 31. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 25.Let Pa be the acceleration of the plywood, Ta be the acceleration of thetruck, and /P Ta be the acceleration of the plywood relative to the truck.(a) Find the value of Ta so that the relative motion of the plywood withrespect to the truck is impending. P Ta a= and1 1 10.40 NsF Nµ= =1: cos20 sin 20y P y P P TF m a N W m aΣ = − ° = − °( )1 cos20 sin 20P TN m g a= ° − °1: sin 20 cos20x x P P TF ma F W m aΣ = − ° = °( )1 sin 20 cos20P TF m g a= ° + °( ) ( )sin 20 cos20 0.40 cos20 sin 20P T P Tm g a m g a° + ° = ° − °( )( )( )0.40cos20 sin 200.03145 9.81 0.309cos20 0.40sin 20Ta g° − °= = =° + °20.309 m/sT =a s(b) ( ) ( ) 2 2/ / / / /1 10 02 2P T P T P T P T P Tox x v t a t a t= + + = + +( )( )( )2// 2 22 1212.5 m/s ,0.4P TP Txat= = = 2/ 12.5 m/sP T =a 20°( )/P T P T Ta= + = → +a a a ( 212.5 m/s )20°:y P yF m a=2 cos20 sin 20P P TN W m a− ° = − °( )2 cos20 sin 20P TN m g a= ° − °continued
  32. 32. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.:x xF maΣ = Σ2 /sin 20 cos20P P T P P TF W m a m a− ° = ° −( )2 /sin 20 cos20P T P TF m g a a= ° + ° −For sliding with friction 2 2 20.30 NkF Nµ= =( ) ( )/sin 20 cos20 0.30 cos20 sin 20P T P T P Tm g a a m g a° + ° − = ° + °( ) /0.30cos20 sin 20cos20 0.30sin 20P TTg aa° − ° +=° + °( )( ) ( )( )0.05767 9.81 0.9594 12.5 11.43= − + =211.43 m/sT =a
  33. 33. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 26.At maximum speed 0.a = 20 0 0F kv= = 020Fkv=When the propellers are reversed, 0F is reversed.20:xF ma F kv maΣ = − − =20 0 20vF F mav− − = ( )2 20020Fa v vmv− +( )202 20 0vdv mv vdvdxa F v v= =+02002 200 0xvmv vdvdxF v v= −+∫ ∫( ) ( )02 2 202 2 2 20 0 00 0 00 0 01ln ln ln 2 ln 22 2 2vmv mv mvx v v v vF F F = − + = − − = 20 000.347m vxF=
  34. 34. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 27.:F ma P kv maΣ = − =dv P kvadt m−= =( ) ( )0 0 0ln ln lnvt v m dv m mdt P kv P kv PP kv k k = = − − = − − − −∫ ∫ln or lnm P kv P kv kttk P m m− −= − = −( )/ /or 1kt m kt mP kv Pe v em k− −−= = −/00 0ttt kt mPt P kx v dt ek k m− = = − −  ∫( ) ( )/ /1 1kt m kt mPt P Pt Pe ek m k m− −= + − = − −, which is linear.Pt kvxk m= −
  35. 35. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 28.Let y be the position coordinate of the projectile measured upwardfrom the ground. The velocity and acceleration a taken to positiveupward. 2.D kv=(a) Upward motion. :yF maΣ =D mg ma− − =222D kva g gm mdv kv k mgv g vdy m m k  = − + = − +          = − + = − +       2vdv kdymg mvk= −+0002hvvdv kdymg mvk= −+∫ ∫0021ln2 vmg khvk m + = −  20201 1ln ln 12 2mgkv khkmg mg mvk = − + = −   +( )( )( )( )( )( )220 0.0024 904ln 1 ln 12 2 0.0024 4 9.81m kvhk mg   = + = +       335.36 m= 335 mh =continued
  36. 36. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) Downward motion. :F maΣ =D mg ma− =222D kva g gm mdv kv k mgv g vdy m m k= − = − = − = − −  2vdv kdymg mvk= −−00fvhvdv kdymg m= −∫ ∫201ln2fvmg khvk m − =  21ln2fmgvkhkmg mk − − =    22ln 1fkv khmg m  − = −  221f kh/mkvemg−− =( )21 kh/mfmgv ek−= ± −( )( ) ( )( )( )2 0.0024 335.36 /44 9.8110.0024fv e− = −  73.6 m/s= 73.6 m/sfv =
  37. 37. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 29.Choose the origin at point C, and let x be positive to the right. Then x is aposition coordinate of the slider B, and 0x is its initial value. Let L be thestretched length of the spring. Then, from the right triangle2 2L x= +The elongation of the spring is ,e L= − and the magnitude of the forceexerted by the spring is( )2 2sF ke k x= = + −By geometry,2 2cosxxθ =+: cosx x sF ma F maθΣ = − =( )2 22 2xk x max− + − =+2 2k xa xm x = − −  + 000vxv dv a dx=∫ ∫00002 2 2 22 201 12 2vxxk x kv x dx x xm mx   = − − = − − +      + ∫2 2 2 2 20 01 102 2kv x xm = − − − + +  ( )2 2 2 2 20 02 2kv x xm= + − +( )2 2 2 2 20 02kx xm = + − + +  ( )2 20answer:kv xm= + −
  38. 38. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 30.Let yA, yB, and yC be the position coordinates of blocks A, B, and C respectively measured downward from theupper support. Then the corresponding velocities and accelerations are positive downward.Constraint of cable: 2 constantA B A B Cy y y y y− + + + =Differentiating twice, 2 0A B A B Ca a a a a− + + + =2 0A B Ca a a+ + = (1)Draw free body diagrams of each of the blocks.Block A. : 2A A AF ma m g T m aΣ = − =2AATa gm= − (2)Block B. : B B BF ma m g T m aΣ = − =BBTa gm= − (3)Block C. : C C CF ma m g T m aΣ = − =CCTa gm= − (4)Substitute (2), (3) and (4) into (1).22 0A B CT T Tg g gm m m    − + − + − =         4 1 14 0A B Cg Tm m m − + + =  ( )( )4 1 14 9.81 0 65.4 N10 10 10T T − + + = =  Substitute into (2),( )( ) 22 65.49.81 3.27 m/s10Aa = − = −(a) Change in position. ( ) 2012A A Ay y a t− =( ) ( )( )2013.27 0.5 0.409 m2A Ay y− = − = −0.409 my∆ = !!!!(b) Tension in the cable. 65.4 NT = !
  39. 39. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 31.98.1 N= =A AW m g 49.05 NB BW m g= =Assume that block B slides downward relative to block A. Then the frictionforce 1F is directed as shown. Its magnitude is1 1 10.10 N .kF Nµ= =1 10: cos30 0, cos30 49.05cos30 42.48 N.y B BF N W N WΣ = − ° = = ° = ° =( )( )1 0.10 42.48 4.248 N.F = =1: sin30x B B B B BF m a W F m aΣ = ° − =( ) ( ) 211 1sin30 49.05sin30 4.248 4.055 m/s5B BBa W Fm= ° − = ° − =Assume that block A slides downward relative to the fixed plane. Thefriction force 2F is directed as shown. Its magnitude is2 2 20.20 N .kF Nµ= =2 1 20: cos30 0, 42.48 98.1cos30 127.44 N.y AF N N W N= − − ° = = + ° =( )( )2 0.20 127.44 25.49 NF = =2 1: sin30x A A A A AF m a W F F m aΣ = ° − + =( )2 11sin30A AAa W F Fm= ° − +( ) 2198.1sin30 25.49 4.248 2.781 m/s10= ° − + =2/ 4.055 2.781 1.274 m/sB A B Aa a a= − = − =Since both /B Aa and Aa are positive, the directions of relative motion areas assumed above.(a) Acceleration of block A. 22.78 m/sA =a 30°(b) Velocity of B relative to A at 0.5 s.t =( )( )/ / 1.274 0.5= =B A B Av a t / 0.637 m/sB A =v 30°
  40. 40. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 32.Let the positive direction for position coordinates, velocities, and accelerations be to the right. Let the originlie at the fixed anchor.Constraint of cable: ( ) ( ) ( )3 constantC A C B Bx x x x x− + − + − =4 2 3 0C B Aa a a− − = (1):x xF maΣ =Block A:3 33 or20A A AAT TT m a a gm= = = (2)Block B:2 22 or10B B BBT TT m a a gm= = = (3)Block C:4 44 or20 20C C CP T P TP T m a a g− −− = = = (4)Substituting (2), (3), and (4) into (1),4 2 34 2 3 020 10 20P T T T−     − − =          16 4 9 420 10 20 20PT + + =  ( ) ( )( )0.12121 0.12121 50 6.0605 lbT P= = =(a) From (2),( )( )( ) 23 6.0605 32.229.3 ft/s20Aa = = 229.3 ft/sA =aFrom (3),( )( )( ) 22 6.0605 32.239.0 ft/s10Ba = = 239.0 ft/sB =aFrom (4),( )( ) ( ) 241.5 ft/s2050 4 6.0605 32.2Ca − = = 241.5 m/sC =a(b) As determined above, 6.06 lbT =
  41. 41. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 33.2 220 100.62112 lb s / ft, 0.31056 lb s / ft32.2 32.2A C Bm m m= = = ⋅ = = ⋅Let the positive direction for position coordinates, velocities, andaccelerations be to the right. Let the origin lie at the fixed anchor.Constraint of cable: ( ) ( ) ( )3 constantC A C B Bx x x x x− + − + − =4 2 3 0C B Aa a a− − = (1)Block A: 0: 0y A AF N WΣ = − =,A A A k A AN W F N Wκµ µ= = =: 3x A x A A AF m a T F m aΣ = − =3 30.20k AAA AT W Ta gm mµ−= = − (2)Block B: ,B B B k BN W F Wµ= =: 2x B B B B BF m a T F m aΣ = − =2 20.20k BBB BT W Ta gm mµ−= = − (3)Block C: ,C C C k CN W F Wµ= =: 4x C A C C CF m a P T F m aΣ = − − = (4)Kinematics: ( ) ( ) 2 20 01 102 2C C C C Cx x v a t a t= + + = +( ) ( )( )( )0 22 22 2 2.430 ft/s0.4C CCx xat − = = = (5)
  42. 42. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Substitute (2), (3) and (5) into (1).( )( ) ( ) ( )2 34 30 2 0.20 3 0.200.31056 0.62112T Tg g   − − − −      ( )( ) ( ) ( )( ) ( ) ( )( )2 34 30 2 0.2 32.2 3 0.2 32.20.31056 0.62112T T   = − − − −      120 27.37 32.2 0T= − + = 5.5608 lbT =From (4), 4 C C CP T F m a= + +( )( ) ( )( ) ( )( )4 5.5608 0.20 20 0.62112 30= + +44.877 lb=From (2),( )( )( )( ) 23 5.56080.20 32.2 20.42 ft/s0.62112Aa = − =From (3),( )( )( )( ) 22 5.56080.20 32.2 29.37 ft/s0.31056Ba = − =(a) Acceleration vectors. 220.4 ft/sA =a229.4 ft/sB =a230 ft/sC =aSince , , andA B Ca a a are to the right, the friction forces, , andA B CF F F are to the left as assumed.(b) Tension in the cable. 5.56 lbT =(c) Force P. 44.9 lb=P
  43. 43. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 34.Let the positive direction of x and y be those shown in the sketch, and letthe origin lie at the cable anchor.Constraint of cable: / /constant or 0,A B A A B Ax y a a+ = + = wherethe positive directions of /andA B Aa a are respectively the x and the ydirections. Then /B A Aa a= − (1)First note that (/B A B A Aa= + =a a a ) ( /20 B Aa° + )20°Block B: ( ) : sin 20x B B B AB B AxF m a m g N m aΣ = ° − =sin 20B A AB Bm a N m g+ = °15 50.328A ABa N+ = (2)( ) /: cos20y B B B B B AyF m a m g T m aΣ = ° − =/ cos20B B A Bm a T m g+ = °/15 138.276B Aa T+ = (3)Block A: AAABAAAx amTNgmamF =++°=Σ 20sin:sin 20A A AB Am a N T m g− + = °25 83.880A ABa N T− + = (4)Eliminate /B Aa using Eq. (1), then add Eq. (4) to Eq. (2) andsubtract Eq. (3).2 255 4.068 or 0.0740 m/s , 0.0740 m/sA A Aa a= − = − =a 20° !!!!From Eq. (1), 2/ 0.0740 m/sB Aa =From Eq. (3), 137.2 NT = 137.2 NT = !
  44. 44. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 35.Motion of B relative to A. Particle B is constrained to move on a circularpath with its center at point A. ( )/B A ta is the component of /B Aa lyingalong the circle, say to the left in the diagram and ( )/B A na is directedtoward point A. Initially, / 0,B A =a since the system starts from rest.(a) (/B A B A Aa= + =a a a ) ( /25 B Aa° + )Crate B: /: 0 cos25x x B B A B AF ma m a m aΣ = Σ = − °/ cos25 0.4cos25 0.363B A Aa a= ° = ° =2/ 0.363 m/sB A =a( ) : sin 25y B B AB B B AyF m a T m g m a= − = °( )sin 25AB B AT m g a= + °( )250 9.81 0.4sin 25 2495 N= + ° =(b) Trolley A: :A AF m aΣ =( )sin 25CD AB A A AT T m g m a− + ° =( )sin 25CD AB A A AT T m g m a= + ° +( )( ) ( )( )2495 20 9.81 sin 25 20 0.4 = + ° + 1145 NCDT =
  45. 45. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 36.The ball moves at constant speed on a circle of radiussinLρ θ=Acceleration (toward center of circle).2vaρ=+ :y yF maΣ = cos 0− =T Wθcos=WTθ:y xF maΣ = sinT maθ =sincosWθθ2mvρ=2sinmvL θ=(a) tan sinθ θ =2mvWL=2vgL=( )( )( )21.50.382269.81 0.6=34.21θ = ° 34.2θ = °(b)cos cosW mgTθ θ= =( )( )2 9.81cos34.21=°23.7 N=T
  46. 46. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 37.Let ρ be the radius of the horizontal circle. The length of the wire is1 2sin sinLρ ρθ θ= + . Solving for ,ρ 1 21 2sin sinsin sinL θ θρθ θ=+1 20: cos cos 0yF T T mgθ θΣ = + − =1 2cos cosmgTθ θ=+21 2: sin sinx x nmvF ma T T maθ θρΣ = + = =( ) ( )21 2 1 21 2 1 2sin sin sin sincos cos sin sinmg mvLθ θ θ θθ θ θ θ+ +=+( )( )2 2 21 21 2sin sin sin60 sin302 9.81 6.2193 m /scos cos cos60 cos30v Lgθ θθ θ° °= = =+ ° + °2.49 m/sv =
  47. 47. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 38.3 1 2 50 25 25θ θ θ= − = ° − ° = °1 212 3 3sinorsin sin sind d θθ θ θ= =( )( )( )2 11 13sin sinsinsin4 sin 25 sin503.0642 ftsin 25d θ θρ θθ= =° °= =°2 10: cos cos 0y AC BCF T T Wθ θΣ = + − = (1)22 1: sin sinx x AC BCWvF ma T Tgθ θρΣ = + = (2)Case 1: 0.BCT = 2cos 0ACT Wθ − = or2cosACWTθ=22 2sin tanACWvT Wgθ θρ= =( )( )22tan 32.2 3.0642 tan 25v gρ θ= = ° 2 246.01 ft /s=6.78 ft/sv =Case 2: 0.ACT = 11cos 0 orcosBC BCWT W Tθθ− = =21 1sin tanBCWvT Wgθ θρ= =( )( )2 2 21tan 32.2 3.0642 tan50 117.59 ft /sv gρ θ= = ° =10.84 ft/sv = 6.78 ft/s 10.84 ft/sv≤ ≤
  48. 48. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 39.(a) 0: sin 0yF T WθΣ = − =16sin sin60WTθ= =°or 18.48 lbT =(b)2: cosx nvF ma T mθρΣ = =2 cos cossinT Wvm mρ θ ρ θθ= =( )( ) 2 23 32.255.77 ft /stan tan60gρθ= = =°7.47 ft/sv =
  49. 49. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 40.2, 1 tan or 452r dyy rdxθ θ= = = = = °0: cos 0yF N mgθΣ = − =cosmgNθ=: sinx x nF ma N maθΣ = =2tanvmg mrθ =2tanv gr θ=(a) ( )( )( )2 2 29.81 1 1.0000 9.81 m /sv = = 3.13 m/sv =(b)( )( )1 9.81cos45 cos45mgN = =° °13.87 N=N
  50. 50. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 41.Geometry( ) ( ) ( )( )( )2 2 22 2 22 cos300.3 0.6 2 0.3 0.6 cos30 0.13823 m0.37179 mOC OB BC OB OCOC= + − °= + − ° ==sin30 sinOC OBβ=°( )0.3 sin30sin30sin 0.403450.37179OBOCβ°°= = =23.79β = °Acceleration components: 21.3 m/sta =( )22 2211.667 m/s0.6CnBCv vaρ= = = =Mass 1 kgm =( )( ): cos 1 1.3 1.3t tF ma N βΣ = = =1.31.421 Ncos23.79= =°N( )( ): sin 1 1.667 1.667n nF ma T N βΣ = − = =(a) 1.667 1.421 sin 23.79T = + ° 2.24 N=T(b) Force exerted by rod on collar is 1.421 N ( )30 53.8β° + = °Force exerted by collar on rod: 1.421 N 53.8°
  51. 51. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 42.( )( )0.5 9.81 4.905 NW mg= = =150 mm 0.150 mρ = =0: cos20 cos30 0y DA DEF T T WΣ = ° − ° − =0.93969 0.86603 4.905DA DET T− =0.92160 5.2198DA DET T= + (1a)1.08506 5.6638DE DAT T= − (1b)220.5: sin 20 sin300.150x n DA DEmvF ma T T vρΣ = = ° + ° =20.10261 0.15DA DEv T T= + (2)Try 75 N. By Eq. (1 ), 75.72 N 75 N (unacceptable)DA DET b T= = >Try 75 N. By Eq. (1 ), 74.34 N 75 N (acceptable)DE DAT a T= = <By Eq. (2), ( )( ) ( )( )2 2 20.10261 74.34 0.15 75 18.877 m /sv = + =4.34 mv =Try ( ) ( )0. By Eq. 1 , 5.6638 unacceptableDA DET b T= = −Try ( ) ( )0. By Eq. 1 , 5.2198 acceptableDE DAT a T= =By Eq. (2), ( )( ) ( )( )2 2 20.10261 5.2198 0.15 0 0.5356 m /sv = + =0.732 m/sv =For 0 , , , 75 N,BA BC DA DET T T T≤ ≤ 0.732 m/s 4.34 m/sv≤ ≤
  52. 52. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 43.( )( )5 9.81 49.05 NW mg= = =0.9 mρ =0: cos40 cos15 0y CA CBF T T WΣ = ° − ° − =0.76604 0.96593 49.05CA CBT T W− = =0.79307 50.780CB CAT T= − (1a)1.26093 64.030CA CBT T= + (1b)2: sin 40 sin15x x CA CB nmvF ma T T maρΣ = ° + ° = =( )( )2sin40 sin150.9sin40 sin1550.115702 0.046587CA CBCA CBCA CBv T TmT TT Tρ= ° + °= ° + °= + (2)Try 116 N. By (1 ), 210.3 N (unacceptable)CB CAT b T= =Try 116 N. By (1 ), 41.216 N (acceptable)CA CBT a T= =By (2), ( )( ) ( )( )2 2 20.115702 116 0.046587 41.216 15.34 m /sv = + =3.92 m/sv =Try ( ) ( )0. By 1 , 50.78 N unacceptableCA CBT a T= = −Try ( ) ( )0. By 1 , 64.03 N acceptableCB CAT b T= =By (2), ( )( )2 2 20.115702 64.030 0 7.408 m /sv = + =2.72 m/sv =For 0 , 116 N,CA CBT T≤ ≤ 2.72 m/s 3.92 m/sv≤ ≤
  53. 53. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 44.(a) Before wire AB is cut. 0a =0 :xFΣ = cos50 cos70 0AB CDT T− ° + ° = (1)0 :yFΣ = sin50 sin70 0° + ° − =AB CDT T W (2)Solving (1) and (2) simultaneously,0.395=ABT W 0.742=CDT W(b) Immediately after wire AB is cut. 0, 0ABT v= =20nvaρ= =0 :n nF ma= =cos20 0CDT W− ° =cos20CDT W= °0.940CDT W=+
  54. 54. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 45.:y yF maΣ =2nmvN mg maρ− = − = −2mvN mgρ= −:x xF maΣ =t tF ma=At onset of slipping, t sF Nµ=2st smvma mgµρ = −   ( )2 2 20.1500.300 9.81 2.883 m /s0.75tssav gρµ   = − = − =     1.6979 m/ssv =Time at slipping.1.69790.150sstvta= = 11.32 sst =
  55. 55. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 46.Angle change over arc AB.80.13963 rad180θ π°= =°Length of arc: ( ) ( )4 (5280) 0.13963 2949 ftABs ρθ= = =(0.5)(5280) 2640 ft, 2949 2640 5589 ftBC ACs s= = = + =( )( )( )2 2480 5589540 022 22949540 02 2 2480 540or 55892 25.475 ft/s540or 5.475 29492 2259300 ft /s 509.2 ft/sBt ttv BtB Bv dv a ds aavv dv a dsv v= − == −= − = −= =∫ ∫∫ ∫Mass of passenger: 22006.211 lb s /ft32.2= = ⋅mJust before point B. 509.2 ft/s, (4)(5280) = 21120 ftv ρ= =( )222509.212.277 ft/s21120nvaρ= = =( ) ( )( )1 11: 200 6.211 12.277 123.75 lby nF N W m a NΣ = − = − = − =( ) ( )( )1: 0.6221 5.474 34.01 lbx t t tF F ma FΣ = = = − = −Just after point B. 509.2 ft/s, , 0nv aρ= = ∞ =20 : 0yF N WΣ = − = 2 200 lbN W= =( )( ): 6.211 5.475 34.01 lbx t t tF ma F maΣ = = = − = −tF does not change.N increases by 76.25 lbmagnitude of change of force = 76.3 lb
  56. 56. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 47.( ) ( )23 180 m, 3 180 2 m/s,dss t t v tdt= − = = −26 m/s (4)(5280) 21120 ft= = − = =tdvadtρLength of arc AB. ( )821120 2949 ft180AB ABsπρθ°= = =°0B ABAv stvv dv a ds=∫ ∫( )( )( )2 22 2 22 or 2 540 2 6 29492 2B At AB B A t ABv va s v v a s− = = + = + −256212,= 506.2 ft/sBv =For passenger, 2165165 lb, 5.124 lb s /ft32.2= = = = ⋅WW mg2: cosnmvF ma W NθρΣ = − =2cosmvN W θρ= − (1): sint tF ma P W maθΣ = − =sin tP W maθ= + (2)(a) Just after point A, 0, 180 m/s, 8t v θ= = = °From Eq. (1),( )( )25.124 540165cos8 92.65 lb21120N = ° − =From Eq. (2), ( )( )165sin8 5.124 6 7.78 lbP = ° + − = −2 2 92.6593.0 lb, tan 11.909, 85.27.78F N P β β= + = = = = °8 77.2β − ° = ° 93.0 lb=F 77.2°(b) At point B. 0, 506.2 ft/svθ = =From Eq. (1),( )( )25.124 506.2165 102.83 lb21120N = − =From Eq. (2), ( )( )5.124 6 30.74 lbP = − = −( ) ( )2 2 102.83102.83 30.74 107.3 lb, tan 3.345, 73.430.74= + = = = = °F β β107.3 lb=F 73.4°
  57. 57. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 48.(a) 160 km/h 44.44 m/sv = =Wheels do not touch the road.2: /y nF ma mg mv ρΣ = − − = −( )22 44.44201.49.81vgρ = = =201 mρ =(b) 80 km/h 22.22 m/sv = =70 kgm = for passenger:y nF maΣ = −2mvN mgρ− = −2vN m gρ = −   ( )222.2270 9.81201.4 = −   515 NN =
  58. 58. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 49.0.2 kg, (0.2)(9.81) 1.962 N= = = =m W mg: cost t tF ma W maθΣ = =cos costWa gmθ θ= =0 0 0v st tvv dv a ds a rdθθ= =∫ ∫ ∫2 20 01 1cos sin2 2v v g rd grθθ θ θ− = =∫2 20 2 sin , where 0.6 m for 0 90v v gr rθ θ= + = ≤ ≤ °max when the cord touches the peg or 90 .v v θ= = °2 2max 0 2v v gr= + (1)When the cord touches the peg, the radius of curvature of the pathbecomes 0.3 m.ρ =2maxmax:y nmvF ma T WρΣ = − =( )2max maxv T Wmρ= − (2)Eliminating 2maxv from equations (1) and (2),( ) ( )( )( )( )( )max200.3 10 1.9622 2 9.81 6.00.2T Wv grmρ − −= − = −2 20.285 m /s= 0 0.534 m/s=v
  59. 59. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 50.Mass of block B. 220.5lb0.015528lb s /ft32.2ft/s= = ⋅BmAcceleration of block B.2 22(9ft/s)27ft/s3ft= = =nvaρ0ta = since v = constant.:F ma∑ =sin nQ P W θ ma− − = −sin nQ = P + W θ ma−But, 0.Q ≥ sin 0nP W maθ+ − >(0.015528)(27) 0.3sin0.5 0.5nma PW Wθ ≥ − = −sin 0.2385θ ≥13.8 166.2θ° ≤ ≤ °
  60. 60. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 51.At position A, the vertical component of apparent weight is shown as .AN:n A nWF ma N W agΣ = − =( ) 2380 12032.2 69.77 ft/s120AnN Wa gW− −= = =( )( )2 3 2 23600 69.77 251.2 10 ft /sA nv aρ= = = ×At position C, the vertical component of apparent weight is shownas .CN:n C nWF ma N W agΣ = + =( ) 280 12032.2 53.67 ft/s120CnN Wa gW+ += = =( )( )2 3 2 23600 53.67 193.2 10 ft /sC nv aρ= = = ×Length of arc ABC:( )3600 11310 ftACs πρ π= = =Calculate ,ta using 2 22C A t ACv v a s− =( )( )2 2 3 32193.2 10 251.2 102 2 113102.562 ft/sC AtACv vas− × − ×= == −At position B, ( )3600 5655 ft2 2ABsπ πρ= = =( )( )( )2 2 3 3 2 22 251.2 10 2 2.562 5655 222.2 10 ft /sB A t ABv v a s= + = × + − = ×
  61. 61. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Effective forces at B:2 3120 222.2 10230 lb32.2 3600BnW vmag ρ×= = =( )1202.562 9.5 lb32.2t tWma ag= = − = −:tF maΣ =or 120 9.5 110.5 lbt tP W ma P W ma− = = + = − =: 230 lbn B nF ma N maΣ = = =Force exerted by seat:2 2 2 2230 110.5 255 lbBF N P= + = + =110.5tan230β = 25.7β = °255 lb=F 25.7°
  62. 62. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 52.The road reaction consists of normal component N and frictioncomponent F. The resultant R makes angle sφ with the normal.Case 1: maxv v=( )0: cos 0y sF R mgθ φΣ = + − =( )cos smgRθ φ=+( ): sinx n s nF ma R maθ φΣ = + =( )tann sma mg θ φ= +( )2maxtan svgrθ φ= +( )max tan sv gr θ φ= +Case 2: minv v=( )0: cos 0y sF R mgθ φΣ = − − =( )cos smgRθ φ=−( ): sinx n s nF ma R maθ φΣ = − =( )tann sma mg θ φ= −( )2mintan svgrθ φ= −( )min tan sv gr θ φ= −( ) ( )tan tans sgr v grθ φ θ φ− ≤ ≤ +
  63. 63. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 53.Weight. W mg=Acceleration.2vaρ=:∑ =x xF ma sin cosF W = maθ θ+2cos sinmvF mgθ θρ= − (1):y yF ma∑ = cos sinN W maθ θ− =2sin cosmvN = mgθ θρ+ (2)(a) Banking angle. Rated speed 180 km/h 50 m/s.v = = 0F = at rated speed.20 cos sinmvmgθ θρ= −2 2(50)tan = 1.2742(200)(9.81)vgθρ= =51.875θ = ° 51.9θ = °(b) Slipping outward. 320km/h 88.889 m/sv = =F Nµ=22cos sinsin cosF v gN v gθ ρ θµθ ρ θ−= =+22(88.889) cos51.875 (200)(9.81)sin51.875=(88.889) sin51.875 (200)(9.81)cos51.875µ°− °°+ °0.44899= 0.449µ =continued
  64. 64. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(c) Minimum speed. = −F Nµ22cos sinsin cosv gv gθ ρ θµθ ρ θ−− =+2 (sin cos )cos singvρ θ µ θθ µ θ−=+(200)(9.81)(sin51.875 0.44899cos51.875 )cos51.875 0.44899sin51.875° − °=° + °2 21029.87m /s=32.09m/sv = 115.5 km/h=v
  65. 65. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 54.Rated speed: 75 mi/h 110 ft/s, 125 mi/h 183.33 ft/sRv = = =From Sample Problem 12.6,( )222 110tan or 2674 fttan 32.2 tan8RRvv ggρ θ ρθ= = = =°Let the x-axis be parallel to the floor of the car.( ) ( ): sin cosx x s nF ma F W maθ φ θ φΣ = + + = +( )2cosmvθ φρ= +(a) 0.φ =( ) ( )( )( )( )22cos sin183.33cos8 sin832.2 26740.247svF WgWWθ φ θ φρ = + − +    = ° − °  = 0.247sF W= !(b) For 0,sF =( ) ( )2cos sin 0vgθ φ θ φρ+ − + =( )( )( )( )22183.33tan 0.3903532.2 2674vgθ φρ+ = = =21.3θ φ+ = °21.3 8φ = ° − ° 13.3φ = ° !
  66. 66. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 55.Rated speed: 75 mi/h 110 ft/s, 125 mi/h 183.33 ft/sRv = = =From Sample Problem 12.6,( )222 110tan or 2674 fttan 32.2 tan8RRvv ggρ θ ρθ= = = =°Let the x-axis be parallel to the floor of the car.( ) ( ): sin cosx x s nF ma F W maθ φ θ φΣ = + + = +( )2cosmvθ φρ= +Solving for ,sF ( ) ( )2cos sinsvF Wgθ φ θ φρ = + − +  Now( )( )( )22183.330.39035 and 0.1232.2 2674svF Wgρ= = = so that( ) ( )0.12 0.39035 cos sinW W θ φ θ φ = + − + Let ( ) ( ) 2sin . Then, cos 1 .u uθ φ θ φ= + + = −2 20.12 0.39035 1 or 0.39035 1 0.12u u u u= − − − = +Squaring both sides, ( )2 20.15237 1 0.0144 0.24u u u− = + +or 21.15237 0.24 0.13797 0u u+ + =The positive root of the quadratic equation is 0.2572.u =Then, 1sin 14.90uθ φ −+ = = °14.90 14.90 8φ θ= ° − = ° − ° 6.90φ = ° !!!!
  67. 67. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 56.If the collar is not sliding, it moves at constant speed on a circle of radius sin .rρ θ= v ρω=Normal acceleration.2 2 22( sin )nva rρ ωθ ωρ ρ= = =:y yF ma∑ =cos 0N mgθ − =cosmgNθ=:x xF maΣ =sinN maθ =2sin( sin )cosmgm rθθ ωθ=Either sin 0θ =or 2cosgrθω=0 or 180θ = ° °or 29.81cos 0.3488(0.5)(7.5)θ = =0 , 180 , and 69.6θ = ° ° °
  68. 68. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 57.If the collar is not sliding, it moves at constant speed on a circle of radius sin .rρ θ= v ρω=From Prob. 12.56 500mm 0.500m, 7.5rad/s,r ω= = =250g 0.250 kg.m = =Normal acceleration:2 2 22( sin )nva rρ ωθ ωρ ρ= = =FΣ = :ma2sin ( sin ) cosF mg m rθ θ ω θ− = −2( cos )sinF m g rω θ θ= −FΣ = :ma2cos ( sin ) sinN mg m rθ θ ω θ− =2 2( cos ) sin )N m g rθ ω θ= +(a) 75 .θ = ° 2(0.25) 9.81 (0.500 cos75 )(7.5) sin75F  = − ° ° 0.61112 N=2 2(0.25) 9.81cos75 (0.500sin 75 )(7.5)N  = °+ ° 7.1950 N=(0.25)(7.1950) 1.7987 N= =sNµSince ,sF Nµ< the collar does not slide.0.611 NF = 75°continued
  69. 69. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) 40 .θ = ° 2(0.25) 9.81 (0.500cos40 )(7.5) sin 40F  = − ° ° 1.8858 N= −2 2(0.25) 9.81cos40 (0.500sin 40 )(7.5)N  = ° + ° 4.7839 N=(0.25)(4.7839) 1.1960 N= =sNµSince ,> sF Nµ the collar slides.Since the collar is sliding, .kF Nµ=nF∑ = + :macos sinnN mg maθ θ− =2cos ( sin ) sinN mg m rθ θ ω θ= +2 2cos ( sin )m g rθ θ ω = + 2 2(0.25) 9.81cos40 (0.500sin 40 )(7.5) = °+ ° 4.7839 N=(0.20)(4.7839) 0.957 N= = =kF Nµ0.957 N=F 40°
  70. 70. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 58.Draw the free body diagrams of the block B when the arm is at 150 .θ = °20, 32.2 ft/stv a g= = =: sin30 0t tF ma W NΣ = − ° + =sin30N W= °2 2: cos30n nv WvF ma W F mgρ ρΣ = ° − = =2cos30WvF Wgρ= ° −Form the ratioFN, and set it equal to sµ for impending slip.( )22cos30 4.2 /(1)(32.2)cos30 /sin30 sin30sF v gNρµ° −° −= = =° °0.636sµ =
  71. 71. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 59.Let β be the slope angle of the dish.1tan6dyrdrβ = =At 6ft, tan 1 or 45r β β= = = °Draw free body sketches of the sphere.0: cos sin 0y sF N N Wβ µ βΣ = − − =cos sinsWNβ µ β=−2 2: sin cosn n smv WvF ma N Ngβ µ βρ ρΣ = + = =( ) 2sin coscos sinssW N Wvgβ µ ββ µ β ρ+=−( )( )2 2 2sin cos sin 45 0.5cos456 32.2 579.6 ft /scos sin cos45 0.5sin 45ssv gβ µ βρβ µ β+ ° + °= = =− ° − °24.1 ft/sv =
  72. 72. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 60.Uniformly accelerated motion on a circular path.55 in. ft2 12Dρ = = =66 60 1060 10 oz lb16W−− ×= × =232.2 ft/sg =9 2116.46 10 lb s /ftWmg−= = ×(a) For uniformly accelerated motion,( )( )0 0 12 4tv v a t= + = +48 ft/sv =(b) ( )( )9 6: 116.46 10 12 1.3975 10 lb.t t tF ma F − −Σ = = × = ×( )( )292 116.46 10 48:5/12n n n nmvF ma F maρ−×Σ = = = =6644.0 10 lb−= ×Magnitude of force:( ) ( )2 22 2 610 644.0 1.3975t nF F F −= + = +lb10644 6−×=F
  73. 73. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 61.Uniformly accelerated motion on a circular path. 8 ftρ =0 tv v a t= +( )( )0 0.75 12 9 ft/s= + =0.75: 0.023332.2tt t t tW aF ma a F W W Wg g= = = = =( )( )( )229: 0.314432.2 8n n nWWvF ma F Wgρ= = = =2 20.315t nF F F W= + =This is the friction force available to cause the trunk to slide.The normal force N is calculated from equilibrium of forces in thevertical direction.0: 0yF N WΣ = − = N W=Since sliding is impending, 0.315sFWµ = = 0.315sµ =
  74. 74. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 62.For constant speed, 0ta =2with 0.7 m/s, 0.2 mBn Bva v ρρ= = =: cosx x nF ma F ma θΣ = =: sin siny y n nF ma N W ma N mg maθ θΣ = − = − = −Ratio2cos cos cossin sin sinnnn BF mag gN mg maa vθ θ θρθ θ θ= = =− − −With( )( )( )2 29.81 0.2 cos4.0041, the ratio becomes4.0041 sin0.7Bg FNvρ θθ= = =−For no impending slide,cos4.0041 sinsFNθµθ≥ =−To find the value of θ for which the ratio is maximum set thederivative with respect to θ equal to zero.( )2cos 1 4.0041sin04.0041 sin 4.0041 sinddθ θθ θ θ± − = ± = −  −1sin 0.249744.0041θ = =cos14.44614.446 , 0.2584.0041 0.24974FNθ°= ° = =−180 14.446 165.554 , 0.258FNθ = ° − ° = ° =(a) Minimum value of sµ for no slip. ( )min0.258sµ =(b) Corresponding values of .θ 14.5 and 165.5θ = ° °
  75. 75. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 63.For constant speed, 0ta =2with 0.7 m/s, 0.2 mBn Bva v ρρ= = =: cosx x nF ma F ma θΣ = =: sin siny y n nF ma N W ma N mg maθ θΣ = − = − = −Ratio2cos cos cossin sin sinnnn BF mag gN mg maa vθ θ θρθ θ θ= = =− − −Let 2cosso thatsinBg FuN uvρ θθ= =−Determine the value of θ at which F/N is maximum.( )( )( ) ( )22 2cos sin sincos 1 sin0sin sin sinud ud u u uθ θ θθ θθ θ θ θ− − − = = = −  − −The corresponding ratio .FN2 11 21 sintancos1F u uN u u uθθθ− −− −± − ±= = = ± = ±− −(a) For impending sliding to the left: tan 0.35sFNθ µ= = =( )21arctan 0.35 19.29 , sin ,Bvugθ θρ−= = ° = =( )( )2 2 29.81 0.2 sin19.29 0.648 m /sBv = ° =0.805 m/sBv =
  76. 76. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.For impending motion to the right: tan 0.35sFNθ µ= − = =( )arctan 0.35 160.71θ = − = °21 2sin ,vugθρ−= = ( )( )2 2 29.81 0.2 sin160.71 0.648 m /sBv = ° =0.805 m/s=(b) For impending sliding to the left, 19.3θ = °For impending sliding to the right, 160.7θ = °
  77. 77. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 64.Consider the motion of one electron. For the horizontal motion, let 0x =at the left edge of the plate and x = l at the right edge of the plate. At thescreen,2x L= +lHorizontal motion: There are no horizontal forces acting on the electron sothat 0.xa =Let 1 0t = when the electron passes the left edge of the plate, 1t t= whenit passes the right edge, and 2t t= when it impacts on the screen. Foruniform horizontal motion,0 1 20 0 0, so that and .2Lx v t t tv v v= = = +l lVertical motion: The gravity force acting on the electron is neglected sincewe are interested in the deflection produced by the electric force. While theelectron is between plates ( )10 ,t t≤ ≤ the vertical force on the electron is/ .yF eV d= After it passes the plates ( )1 2 ,t t t≤ ≤ it is zero.For 10 ,t t≤ ≤ :yy y yF eVF ma am mdΣ = = =( )00y y yeVtv v a tmd= + = +( )220 010 02 2y yeVty y v t a tmd= + + = + +At ( )21 11 11, and2yeVt eVtt t v ymd md= = =For 1 2, 0yt t t a≤ ≤ =( ) ( )1 11yy y v t t= + −At 2t t= ( ) ( )2 1 2 11yy y v t tδ= = + −( )21 1 12 1 2 112 2eVt eVt eVtt t t tmd md mdδ = + − = −  0 0 0 012 2eV Lmdv v v v = + −  l l lor 20eVmdvLδ =l!!!!
  78. 78. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 65.Consider the motion of one electron. For the horizontal motion, let 0x =at the left edge of the plate and x = l at the right edge of the plate. At thescreen,2x L= +lHorizontal motion: There are no horizontal forces acting on the electron sothat 0.xa =Let 1 0t = when the electron passes the left edge of the plate, 1t t= whenit passes the right edge, and 2t t= when it impacts on the screen. Foruniform horizontal motion,0 1 20 0 0, so that and .2Lx v t t tv v v= = = +l lVertical motion: The gravity force acting on the electron is neglected sincewe are interested in the deflection produced by the electric force. While theelectron is between the plates ( )10 ,t t≤ ≤ the vertical force on theelectron is / .yF eV d= After it passes the plates ( )1 2 ,t t t≤ ≤it is zero.For 10 ,t t≤ ≤ :yy y yF eVF ma am mdΣ = = =( )00y y yeVtv v a tmd= + = +( )220 010 02 2y yeVty y v t a tmd= + + = + +At21 20 0, ,2eVt t yv mdv= =llBut 0.075 0.4252dy d d< − =So that2200.4252eVdmdv<l22 2 20 011.1760.425 2d eV eVmv mv> =l 201.085d eVmv>l
  79. 79. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 66.30 , 2 rad/s, 0θ θ θ= ° = =& &&0.6 m,r W mg= =Block B: Only force is weightcos30 , sin30rF W F Wθ= ° = − °(a) ( )2 :F ma m r rθ θ θ θ= = +&& &&sin302 sin30F mgr r r g rm mθθ θ θ θ°= − = − − = − ° −& && && &&&( ) ( )( )( )( )9.81 sin30 0.6 0sin301.226 m/s2 2 2g rrθθ° +° += − = − = −&&&&/ rod 1.226 m/sB =v 60°(b) ( )2:r rF ma m r rθ= = − &&&( )( ) ( )2 22 2cos30cos300.6 2 9.81 cos30 10.90 m/srF mgr r r r gm mθ θ θ°= + = + = + °= + ° =& & &&&2/ rod 10.90 m/sB =a 60°
  80. 80. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 67.245 , 0.8 m, 10 rad/srθ θ= ° = =0, 0,rv r v r W mgθ θ= = = = =(a) ( ): cos45 2F ma N W m r rθ θ θ θΣ = − ° = +( )cos45 2N mg m r rθ θ= ° + +( )( )(0.5)(9.81)cos45 0.5 0.8 10 0 = ° + + 7.468= 7.47 NN = 45°(b) ( )2: sin 45r rF ma mg m r rθΣ = ° = −2 2sin 45 sin 45mgr r g rmθ θ= ° + = ° +( ) 29.81 sin 45 0 6.937 m/s= ° + =2/ rod 6.94 m/sB =a 45°
  81. 81. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 68.Use radial and transverse components of acceleration.2 3 23 ft 2 radr t t tθ= − =26 3 ft/s 4 rad/sr t t tθ= − =2 26 6 ft/s 4 rad/sr t θ= − =2 2 3 26 6 (3 )(16 )ra r r t t t tθ= − = − − −5 4 216 48 6 6 ft/st t t= − − +2 3 22 (3 )(4) (2)(6 3 )(4 )a r r t t t t tθ θ θ= + = − + −2 3 260 28 ft/st t= −Mass: 240.12422 lb s /ft32.2= = = ⋅Wmg(a) 0.t = 26 ft/secra =0aθ =Apply Newton’s second law.(0.12422)(6)r rF ma= = 0.745 lbrF =(0.12422)(0)F maθ θ= = 0Fθ =(b) 1t = s. 232 ft/sra = −232 ft/saθ =Apply Newton’s second law.(0.12422)( 32)r rF ma= = − 3.98 lbrF = −(0.12422)(32)F maθ θ= = 3.98 lbFθ =
  82. 82. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 69.Use radial and transverse components of acceleration.6(1 cos2 ) ft 2 radr t tπ θ π= + =12 sin 2 ft/s 2 rad/sr tπ π θ π= − =2 224 cos2 ft/s 0r tπ π θ= − =2 2 224 cos2 (6 6cos2 )(2 )ra r r t tθ π π π π= − = − − +2 2 224 48 cos2 ft/stπ π π= − −2 0 (2)( 12 sin 2 )(2 )a r r tθ θ θ π π π= + = + −248 sin 2 tπ π= −Mass: 210.031056 lb s /ft32.2Wmg= = = ⋅(a) 0.t = 2710.61 ft/sra = −0aθ =Apply Newton’s second law.(0.031056)(710.61)r rF ma= = 22.1 lbrF = −0F maθ θ= = 0Fθ =(b) 0.75 s.t = 2236.87 ft/sra = −2473.74 ft/saθ =Apply Newton’s second law.(0.031056)( 236.87)r rF ma= = − 7.36 lbrF = −(0.031056)(473.74)F maθ θ= = 14.71 lbFθ =
  83. 83. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 70.Kinematics: 1.5 ft/s, 0drr rdt= = =0 0or 1.5 ftr tdr r dt r t= =∫ ∫20.8 rad/s, 0.8 rad/stθ θ= =( )( )22 3 20 1.5 0.8 0.96 ft/sra r r t t tθ= − = − = −( )( ) ( )( )( ) 22 1.5 0.8 2 1.5 0.8 3.6 ft/sa r r t t tθ θ θ= + = + =Kinetics: Sketch the free body diagrams for the collar.:r r rF ma T maΣ = − =:F ma Q maθ θ θΣ = =Set T Q= to obtain the required time.orr rma ma a aθ θ− = − =Using the calculated expressions3 2 23.60.96 3.6 , 3.75 s0.96t t t= = =1.936 st =
  84. 84. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 71.210 rad/s, 10 rad/stθ θ= =& &&20.5/32.2 0.015528 lb s /ftm = = "Before cable breaks: and 0.rF T r= − =&&( )2:r rF ma T m r rθ= − = − &&&( )( )2 2 2 20 4or 171.733 rad /s0.015528 1.5mr Tmr mr Tmrθ θ+ += + = = =&&& &&&13.105 rad/sθ =&Immediately after the cable breaks: 0, 0rF r= =&(a) Acceleration of B relative to the rod.( ) ( )( )22 2 20 or 1.5 13.105 257.6 ft/sm r r r rθ θ− = = = =& &&& &&2/ rod 258 ft/sB =a radially outward !(b) Transverse component of the force.( ): 2F ma F m r rθ θ θ θ θ= = +&& &&( ) ( )( ) ( )( )( )0.015528 1.5 10 2 0 12 0.233 + =  0.233 lbFθ = !
  85. 85. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 72.215 rad/s, 230 g 0.230 kg, 0, 9 N, 12 m/sm F rθθ θ= = = = = = −Due to the spring, , 60 N/mrF kr k= − =( )2:r r rF F ma kr m r rθΣ = = − = −( )2k m r mrθ− = −(a) Radial coordinate.( )( )( )( )2 20.230 1260 0.230 15mrrk mθ−= − = −− −0.33455 m= 335 mmr =( ): 2F ma F m r rθ θ θ θ θΣ = = +2Fr rmθθ θ= −( )( )( )9 01.304 m/s2 2 0.230 15F mrrmθ θθ− −= = =(b) Radial component of velocity. rv r= 1.304 m/srv =
  86. 86. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 73.At point A.2 22(3.8)115.52 m/s0.125= = =nvaρ125tan 22.62175 125= − = − °+θ θ+ : cos22.6Σ = ° =s tF ma F ma2cos22.6 70cos22.6243.077 m/s1.5stFam° °= = =Acceleration vector.2 2n ta a a= +2 2 2115.52 43.077 123.29 m/s= + =115.52tan43.077φ = 69.55φ = °22.62 46.93φ − ° = °cos46.93ra a= °2123.29cos46.93 84.2 m/s= ° = in negative r-direction284.2 m/sra = −sin 46.93a aθ = °123.29sin 46.93= °290.1 m/saθ =Draw the free body diagram of the collar.
  87. 87. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 74.Let r and θ be polar coordinates of block A as shown, and let By be theposition coordinate (positive downward, origin at the pulley) for therectilinear motion of block B.Constraint of cable: constant,Br y+ =0, 0 orB B Br v r a r a+ = + = = − (1)For block A, + : cos or sec (2)A Ax A A A AW WF m a T a T ag gθ θΣ = = =For block B, : (3)B By B B BW WF a W T ag gΣ = − =Adding Eq. (1) to Eq. (2) to eliminate T, secA BB A BW WW a ag gθ= + (4)Radial and transverse components of .AaUse either the scalar product of vectors or the triangle constructionshown, being careful to note the positive directions of the components.2cosr A r Ar r a aθ θ− = = ⋅ = −a e (5)Noting that initially 0,θ = using Eq. (1) to eliminate ,r and changingsigns givescosB Aa a θ= (6)Substituting Eq. (6) into Eq. (4) and solving for ,Aa( )( ) 250 32.217.991 ft/ssec cos 40sec30 50cos30BAA BW gaW Wθ θ= = =+ ° + °From Eq. (6), 217.991cos30 15.581 ft/sBa = ° =(a) From Eq. (2), ( )( )40/32.2 17.991 sec30 25.81T = ° = 25.8 lbT =(b) Acceleration of block A. 217.99 ft/sA =a(c) Acceleration of block B. 215.58 ft/sB =a
  88. 88. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 75.Let r and θ be polar coordinates of block A as shown, and let By be theposition coordinate (positive downward, origin at the pulley) for therectilinear motion of block B.Radial and transverse components of .AvUse either the scalar product of vectors or the triangle constructionshown, being careful to note the positive directions of the components.cos306cos30 5.19615 ft/sr A r Ar v v= = ⋅ = − °= − ° = −v esin306sin30 3 ft/sA Ar v vθ θθ = = ⋅ = − °= ° =v e31.25 rad/s2.4vrθθ = = =Constraint of cable: constant,Br y+ =0, 0 orB B Br v r a r a+ = + = = − (1)For block A, + : cos or secA Ax A A A AW WF m a T a T ag gθ θΣ = = = (2)For block B, :B By B B BW WF a W T ag gΣ = − = (3)Adding Eq. (1) to Eq. (2) to eliminate T, secA BB A BW WW a ag gθ= + (4)Radial and transverse components of .AaUse a method similar to that used for the components of velocity.2cosr A r Ar r a aθ θ− = = ⋅ = −a e (5)Using Eq. (1) to eliminate r and changing signs gives2cosB Aa a rθ θ= − (6)
  89. 89. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Substituting Eq. (6) into Eq. (4) and solving for ,Aa( ) ( ) ( )( )22250 32.2 2.4 1.2520.086 ft/ssec cos 40sec30 50cos30BAA BW g raW Wθθ θ ++   = = =+ ° + °From Eq. (6), ( )( )2 220.086cos30 2.4 1.25 13.645 ft/sBa = ° − =(a) From Eq. (2), ( )( )40/32.2 20.086 sec30 28.8T = ° = 28.8 lbT =(b) Acceleration of block A. 220.1 ft/sA =a(c) Acceleration of block B. 213.65 ft/sB =a
  90. 90. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 76.Since the particle moves under a central force, constant.h =Using Eq. (12.27), 20 0 0h r h r vθ= = =or 0 0 0 0 02 200cos2cos2r v r v vrr rθθ θ= = =Radial component of velocity.( )00 3/ 2sin 2cos2 cos2rdr d rv r rd dθθ θ θθ θ θ θ = = = =  ( )00 3/ 2sin 2cos2cos2vrrθθθ= 0sin 2cos2rv vθθ=Transverse component of velocity.0 00cos2h r vvr rθ θ= = 0 cos2v vθ θ=
  91. 91. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 77.Since the particle moves under a central force, constant.h =Using Eq. (12.27), 20 0 0h r h r vθ= = = or 0 0 0 02 200cos2cos2nr v r v vrr rθθ θ= = =Differentiating the expression for r with respect to time,( ) ( )0 00 0 03/ 2 3/ 20sin 2 sin 2 sin 2cos2cos2 cos2cos2 cos2dr d r vr r r vd d rθ θ θθ θ θ θθ θ θ θθ θ = = = = =  Differentiating again,( )2 2 2 2 200 0 3/ 20sin 2 2cos 2 sin 2 2cos 2 sin 2cos2 cos2cos2dr d vr v vd d rθ θ θ θ θθ θ θθ θ θ θθ+ + = = = =  (a) 000sin2sin 2cos2rv rv r vrθθθ= = = 00cos2v rv rrθ θ θ= =( ) ( )2 2 2 200sin 2 cos 2rv rv v vrθ θ θ= + = + 00v rvr=2 2 2 22 20 0 020 02cos 2 sin 2cos 2cos2 cos2rv r va r rr rθ θθ θθ θ+= − = −2 2 2 20 0 020 00cos 2 sin 2cos2 cos2v v v rr rrθ θθ θ+= = =2020:r rmv rF mar= =2020rmv rFr=Since the particle moves under a central force, 0aθ =continued
  92. 92. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Magnitude of acceleration:22 2 020rv ra a arθ= + =Tangential component of acceleration.20 0 020 0 0sin 2tdv d v r v v ra rdt dt r r rθ = = = =  Normal component of acceleration.2 222 20 02 20 0cos21 sin 2t tv r v ra a ar rθθ= − = − =But20cos 2rrθ =   Hence,20nvar=(b) But22 2 202 20 0ornnv v v r raa r vρρ= = = ⋅320rrρ =
  93. 93. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 78.Since the particle moves under a central force, constant.h =Using Eq. (12.27), 20 0 0h r h r vθ= = =or 0 0 0 0 02 2 2200 coscosr v r v vr rrθθθ= = =Radial component of velocity. ( ) ( )0 0cos sinrdv r r rdtθ θ θ= = = −Transverse component of velocity. ( )0 cosv r rθ θ θ θ= =Speed. 2 2 0 00 20 cosrr vv v v rrθ θθ= + = = 02cosvvθ=
  94. 94. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 79.Since the particle moves under a central force, constanth =Using Eq. (12.27), 20 0 0h r h r vθ= = = 0 0 0 0 02 2 2200 coscosr v r v vr rrθθθ= = =Radial component of velocity. ( ) ( )0cos sinr odv r r rdtθ θ θ= = = −Transverse component of velocity. ( )0 cosv r rθ θ θ θ= =Speed. 2 2 0 0 00 2 20 cos cosrr v vv v v rrθ θθ θ= + = = =Tangential component of acceleration.( )( ) 20 0 00 3 3 2 50 02 sin 2 sin 2 sincos cos cos costdv v v va vdt r rθ θ θ θθ θ θ θ− −= = = ⋅ =Tangential component of force.20502 sin:cost t tmvF ma Frθθ= =(a) 0, 0tFθ = = 0tF =(b) 052 sin 4545 ,cos 45tmvFθ°= ° =°2008tmvFr=
  95. 95. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 80.For gravitational force and a circular orbit,22orrGMm mv GMF vr rr= = =Let τ be the periodic time to complete one orbit.2 or 2GMv r rrτ π τ π= =Solving for ,τ3/ 22 rGMπτ =But, 3 3/ 24, hence 23 3M R GM G Rππ ρ ρ= =Then,3/ 23 rG Rπτρ =   Using 3r R= as given leads to3/ 2 33 9G Gπ πτρ ρ= = ( )1/ 29 /Gτ π ρ=
  96. 96. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 81.For gravitational force and a circular orbit,22orrGMm mv GMF vr rr= = =Let τ be the period time to complete one orbit.But22 2 2 22 or 4GMv r v rrττ π τ π= = =Then1/32 232 2or4 4GM GMr rτ τπ π = =    Data: 323.934 h 86.1624 10 sτ = = ×(a) In SI units: 2 69.81 m/s , 6.37 10 mg R= = ×( )( )22 6 12 3 29.81 6.37 10 398.06 10 m /sGM gR= = × = ×( )( )1/3212 362398.06 10 86.1624 1042.145 10 m4rπ × × = = ×   altitude 635.775 10h r R= − = × 35800 kmh =In US units: 2 632.2 ft/s , 3960 mi 20.909 10 ftg R= = = ×( )( )22 6 15 3 232.2 20.909 10 14.077 10 ft /sGM gR= = × = ×( )( )1/3215 36214.077 10 86.1624 10138.334 10 ft4rπ × × = = ×   altitude 6117.425 10 fth r R= − = × 22200 mih =
  97. 97. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.(b) In SI units:1236398.06 103.07 10 m/s42.145 10GMvr×= = = ××3.07 km/sv =In US units:153614.077 1010.09 10 ft/s138.334 10GMvr×= = = ××310.09 10 ft/sv = ×
  98. 98. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 82.For gravitational force and a circular orbit,22orrGMm mv GMF vr rr= = =Let τ be the periodic time to complete one orbit.2 or 2GMv r rrτ π τ π= =from which2 324 rGMπτ=But2 322 24, hence,rGM gR gRπτ= = (1)Solving for ,r1/ 32 224gRrτπ =    Data: 9417,000 mi 2.202 10 ftr = = ×644,400 mi 234.4 10 ftR = = ×33.551 days 85.224 h 306.8 10 sτ = = = ×Using (1),( )( ) ( )32 922 23 64 2.202 1081.5 ft/s306.8 10 234.4 10gπ ×= =× ×281.5 ft/sg =
  99. 99. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 83.Let M be the mass of the sun and m the mass of Venus.For the circular orbit of Venus,222 nGMm mvma GM rvrr= = =where r is radius of the orbit.Data: 6 967.2 10 mi 354.8 10 ftr = × = ×3 378.3 10 mi/h 114.84 10 ft/sv = × = ×( )( )29 3 21 3 2354.8 10 114.84 10 4.6792 10 ft /sGM = × × = ×(a) Mass of sun.21 3 29 4 44.6792 10 ft /s34.4 10 ft /lb sGMMG −×= =× ⋅27 2136.0 10 lb s /ftM = × ⋅(b) At the surface of the sun, 3 9432 10 mi 22.81 10 ftR = × = ×2GMmmgR=( )212 294.6792 1022.81 10GMgR×= =×2899 ft/sg =
  100. 100. COSMOS: Complete Online Solutions Manual Organization SystemVector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell© 2007 The McGraw-Hill Companies.Chapter 12, Solution 84.For gravitational force and a circular orbit,22orrGMm mv GMF vr rr= = =But2322 2, hence, or4r r GM GMv rrπ π ττ τ π= = =Solving for r,1/ 322(1)4GMrτπ =    ( )1/31/3 22222 244 GMGMv GMr GMππτ τ   = = =       1/ 32(2)GMvπτ =   For earth: 6 26.37 10 m, 9.81 m/sR g= × =( )( )22 6 12 3 29.81 6.37 10 398.06 10 m /sGM gR= = × = ×For Jupiter: ( )( )12 15319 398.06 10 126.98 10GM = × = ×(a) For Ganymede: 37.15 days 171.6 h 617.76 10 sτ = = = ×By Eq. (1),( )( )1/ 3215 392126.98 10 617.76 101.071 10 m4rπ × × = = ×   61.071 10 kmr = ×(b) For Callisto: 616.69 days 400.56 h 1.4420 10 sτ = = = ×By Eq. (2),( )1/ 312362 126.98 108.209 10 m/s1.4420 10vπ × = = × × 8.21 km/sv =

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