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

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, 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 =




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 =




Assume 232.2 ft/sg =

Wmg

=

: sWF ma W F ag

Σ = − =

71 or 21 132.2

ss

a FW F W agg

− = = =

− −

7.46 lbW =

27.4635 0.232 lb s /ft32.2

Wmg

= = = ⋅

: sWF ma F W ag

Σ = − =

1

27.46 132.2

saF Wg

= +

= + 7.92 lbsF =

For the balance system B,

0 0: 0w pM bF bFΣ = − =

w pF F=

But, 1w waF Wg

= +

and 1p p

aF Wg

= +

so that and pw p w

WW W m

g= = 20.232 lb s /ftwm = ⋅




Periodic time: 12 h 43200 sτ = =

Radius of Earth: 63960 mi 20.9088 10 ft= = ×R

Radius of orbit: 63960 12580 16540 mi 87.33 10 ftr = + = = ×

Velocity of satellite: ( )( )62 87.33 102

43200rv

ππτ

×= =

312.7019 10 ft/s= ×

It is given that 3750 10 lb s= × ⋅mv

(a) 3

23

750 10 59.046 lb s /ft12.7019 10

mvmv

×= = = ⋅

×

259.0 lb s /ftm = ⋅

(b) ( )( )59.046 32.2 1901 lb= = =W mg

1901 lb=W




+ 40: 10 10 10 20 40

32.2y y yF ma a= + + + − =∑

( )( ) 232.2 108.05 ft/s

40ya = =

ydv dy dv dva vdt dt dy dy

= = =

y yv dv a d=

20 0

12

v vy y yv dv a d v a y= =∫ ∫

( )( )( )2 2 8.05 1.5yv a y= = 4.91 ft/sv =




Data: 0 108 km/h 30 m/s, 75 mfv x= = =

(a) Assume constant acceleration. constant= = =dv dva vdx dt

0

00

fxv v dv a dx=∫ ∫

20

12 fv a x− =

( )( ) ( )

220 30

6 m/s2 2 75f

vax

= − = − = −

0

00

ftv dv a dt=∫ ∫

0 fv a t− =

0 306f

vta

−= − =

− 5.00 sft =

(b) + 0: 0yF N W= − =∑

N W=

:xF ma N maµ= − =∑

ma ma aN W g

µ = − = − = −

( )69.81

µ−

= − 0.612µ =




(a) : sinfF ma F W maα= − + =∑

sin sin= − + = − +f fF FWa gm m m

α α

( )2 27500 N 9.81 m/s sin 4 4.6728 m/s1400 kg

= − + ° = −

24.6728 m/sa = 4°

0 88 km/h 24.444 m/s= =v

From kinematics, dva vdx

=

0

00

fxva dx v dv=∫ ∫

20

12fa x v= −

( )( )( )

220 24.444

2 2 4.6728fvxa

= − = −−

63.9 mfx =

+




(a) Coefficient of static friction.

0: 0yF N WΣ = − = N W=

0 70 mi/h 102.667 ft/sv = =

( )2 2

002 2 t

v v a s s− = −

( )( )( )( )

22 220

0

0 102.66731.001 ft/s

2 2 170tv va

s s−−

= = = −−

For braking without skidding , so that | |s s tN m aµ µ µ= =

:t t s tF ma N maµΣ = − =

31.00132.2

t ts

ma aW g

µ = − = − = 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 k

Na gmµ µ= − = − = −

Since acceleration is constant,

( ) ( )( )( )

22 20

00 102.667

2 2 24.801t

v vs sa

−−− = =

−

0 212 fts s− =




For the thrust phase, : tWF ma F W ma ag

Σ = − = =

( ) 221 32.2 1 289.8 ft/s0.2

tFa gW = − = − =

At 1 s,t =

( )( )289.8 1 289.8 ft/sv at= = =

( )( )221 1 289.8 1 144.9 ft2 2

y 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= + − = + − −

At 289.80, 1 9.00 s, 10.00 s32.2

v t t= − = = =

( ) ( )2 21 1 12 2v v a y y g y y− = − = − −

( )( )( )

22 21

10 289.8

1304.1 ft2 2 32.2

v vy yg

−−− = − = − =

(a) max 1304.1 144.9y h= = + 1449 fth =

(b) As already determined, 10.00 st =




Kinematics: Uniformly accelerated motion. ( )0 00, 0x v= =

( )( )( )

2 20 0 2 2

2 101 2, or 1.25 m/s2 4

xx x v t at at

= + + = = =

0: sin 50 cos 20 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 sin50

ma mgP

µµ

+ ° + °=

° − °

For motion impending, set 0 and 0.30.sa µ µ= = =

( )( ) ( )( )( )40 0 40 9.81 sin 20 0.30cos20593 N

cos50 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 0.25sin50

P+ ° + °

=° − °

612 NP =




Calculation of braking force/mass ( )/bF m from data for level pavement.

0 100 km/hr 27.778 m/sv = =

( )2 2

002 2

v v a x x− = −

( )( )( )( )

22 20

0

2

0 27.7782 2 60

6.43 m/s

v vax x

−−= =

−

= −

br:xF ma F maΣ = − =

2br 6.43 m/sF am

= − =

(a) Going up a 6° incline. ( )6θ = °

br: sinF ma F mg maθΣ = − − =

br sinFa gm

θ= − −

26.43 9.81sin 6 7.455 m/s= − − ° = −

( )( )( )

22 20

00 27.778

2 2 7.455v vx x

a−−

− = =−

0 51.7 mx x− =

(b) Going down a 2% incline. ( )tan 0.02, 1.145θ θ= − = − °

br: sinF ma F mg maθΣ = − − =

br sinFa gm

θ= − −

( ) 26.43 9.81sin 1.145 6.234 m/s= − − − ° = −

( )( )( )

22 20

00 27.778

2 2 6.234v vx x

a−−

− = =−

0 61.9 mx x− =




Let the positive directions of Ax and Bx be down the incline.

Constraint of the cable: 3 constantA Bx x+ =

13 0 or 3A B B Aa a a a+ = = −

For block A: : sin 30A A AF ma m g T m aΣ = ° − = (1)

For block B: : sin 30 3B B B B AF ma m g T m a m aΣ = ° − = = − (2)

Eliminating T and solving for ,Aag

( )3 sin 30 33

− ° = +

BA B A A

mm g m g m a

( ) ( )3 sin 30 30 8 sin 300.33673

3 / 3 30 2.667A BA

A B

m mag m m

− ° − °= = =

+ +

(a) ( ) ( ) 20.33673 9.81 3.30 m/sAa = = 23.30 m/sA =a 30°

( ) 21 3.30 1.101 m/s3

= − = −Ba 21.101 m/sB =a 30°

(b) Using equation (1),

( )( )( )sin 30 10 9.81 sin 30 0.33673AA

aT m gg

= ° − = ° −

16.02 NT =




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 = − ° =

: sin 30x A A A AF ma m g N T m aµΣ = ° − − =

Eliminate .AN

( )sin 30 cos30A A Am g T m aµ° − ° − =

Block B: 0: cos30 0y B BF N m gΣ = − ° =

: sin 30 33B A

B B B Bm aF ma m g N T m aµΣ = ° + − = = −

Eliminate .BN

( )sin 30 cos30 33B A

Bm am g Tµ° + ° − = −

Eliminate T.

( ) ( )3 sin 30 3 cos30 33B

A 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 = and solve for .µ

( )( )

( )( )

3 sin 30 75 20tan 30 0.334.

3 cos30 75 20A B

A B

m mm m

µ− ° −

= = ° =+ ° +

Since 0.25 0.334,sµ = < sliding occurs.

Calculate Aag

for sliding. Use 0.20.kµ µ= =

continued



( ) ( )

( ) ( )( )

3 sin 30 3 cos303 / 3

30 8 sin 30 0.20 30 8 cos300.13525

30 2.667

A B A BA

A B

m m m mag m m

µ− ° − + °=

+

− ° − + °= =

+

(a) ( )( ) 20.13525 9.81 1.327 m/s= =Aa 21.327 m/sA =a 30°

( ) 21 1.327 0.442 m/s3

= − = −

Ba 20.442 m/sB =a 30°

(b) ( )( )( )( ) ( )( )

sin 30 cos30

10 9.81 sin 30 0.20cos30 10 1.327

= ° − ° −

= ° − ° −

A A AT m g m aµ

18.79 N=T




Data: 22

55000 lb 1708.1 lb s / ft32.2 ft/sAm = = ⋅

22

44000 lb 1366.5 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 same acceleration.

:x x b b A x B xF ma F F m a m a= − − = +∑ ∑

27000 7000 4.5534 m/s1708.1 1366.5

b bx

A B

F Fam m

+ += = =

+ +

xdva vdx

=

0

20 0

0 2= =∫ ∫fx

x x fvva dx v dv a x

( )( )( )

220 80.667

751 ft2 2 4.5534f

x

vxa

−= − = − = −

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 F

778 lb tensioncF =




Data: 22

55000 lb 1708.1 lb s / ft32.2 ft/sAm = = ⋅

22

44000 lb 1366.5 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 same acceleration.

:x x b b A x B xF ma F F m a m a= − − = +∑ ∑

27000 2.2767 m/s1708.1 1366.5

bx

A B

Fam m

= = =+ +

xdva vdx

=

0

20 0

0 2= =∫ ∫fx

x x fvva dx v dv a x

( )( )( )

220 80.667

1429 ft2 2 2.2767f

x

vxa

−= − = − =

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 =




Constraint of cable: ( )2 A B A A Bx x x x x+ − = + = constant.

0, or A 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 B

WF ma T N W a

gµ θΣ = − + + =

Eliminate ABN and .Ba

( )sin cos B AB B B

a aT W W W

g 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 A

WF m a T W F F a

gθΣ = − + − − =

( )sin cos sin cosAB B A B

aW W W W

gθ µ θ θ µ θ− + − + −

( )cos AB A A

aW W W

gµ θ− + =

( ) ( ) ( )sin 3 cos AA B A B A B

aW W W W W W

gθ 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



( ) ( )( )0.20 3 0.20 128tan 0.40

64A B

sA B

W W

W Wθ

+= = =

−

21.8 25 .sθ θ= ° < = ° The blocks move.

Calculate Aa

g using 0.15 and 25 .kµ µ θ= = = °

( ) ( )sin 3 cosA B k A BA

A B

W W W Wa

g W W

θ µ θ− − +=

+

( )( )64sin 25 0.15 128 cos 25

0.1004896

° − °= =

( )( ) 20.10048 32.2 3.24 ft/sAa = =

(a) 23.24 ft/sBa = − 23.24 ft/sB =a 25° !

(b) ( )sin cos AB B

aT W W

gθ µ θ= + +

( ) ( )( )16 sin 25 0.15cos 25 16 0.10048= ° + ° +

10.54 lbT = !




Constraint of cable: ( )2 constant.A B A A Bx x x x x+ − = + =

0, or A 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 B

WF ma T N W a

gµ θΣ = − + + =

Eliminate ABN and .Ba

( )sin cos B AB B B

a aT W W W

g 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 A

WF m a T W F F P a

gθ θΣ = − + − − + =

( )sin cos sin cosAB B A B

aW W W W

gθ µ θ θ µ θ− + − + −

( )cos sin cos AB A A

aW W P P W

gµ θ µ θ θ− + + + =

( ) ( ) ( ) ( )sin 3 cos sin cos AA B A B A B

aW W W W P W W

gθ 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



( ) ( )3 cos sin

sin coss A B A B

ss

W W W WP

µ θ θµ θ θ

+ − −=

+

( )( )0.20 128 cos 25 64 sin 25

3.88 lb 10 lb0.20 sin 25 cos 25

° − °= = − <

° + °

Blocks will move with 10 lb.P =

Calculate Aa

g using 0.15, 25 , and 10 lb.k Pµ µ θ= = = ° =

( ) ( ) ( )sin 3 cos sin cosA B k A B kA

A B

W W W W Pa

g W W

θ µ θ µ θ θ− − + + +=

+

( )( ) ( )( )64 sin 25 0.15 128 cos 25 10 0.15sin 25 cos 25

96

° − ° + ° + °=

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 B

aT W W

gθ µ θ= + +

( ) ( )( )16 sin 25 0.15cos 25 16 0.20149= ° + ° +

12.16 lb.T = !




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°




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: , or C 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− − − − =

or 3 60 60 20 0.0769239 60 180 20

B B A C

B A C

a 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 =




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 A

WW T a

g− = or

AA A

WT W a

g= −

Block B: B BB B A

W WP W T a a

g g+ − = = −

A BB A A A

W WP W W a a

g g+ − + = −

Solving for , A BA A

A B

W W Pa a g

W W

− −=+

(1)

( ) ( ) ( )220 0 0

2 with 0A A A A A Av v a y y v − = − =

( )02A A A Av a y y = − (2)

( ) ( )0 0 with 0A A A Av v a t v− = =

A

A

vt

a= (3)

continued



(a) Acceleration of block A.

System (1): 100 lb, 50 lb, 0A BW W P= = =

By formula (1), ( ) ( )1

100 5032.2

100 50Aa−=+

( ) 21

10.73 ft/sA =a !!!!

System (2): 100 lb, 0, 50 lbA BW W P= = =

By formula (1), ( ) ( )2

100 5032.2

100Aa−= ( ) 2

216.10 ft/sA =a !!!!

System (3): 1100 lb, 1050 lb, 0A BW W P= = =

By formula (1), ( ) ( )3

1100 105032.2

1100 1050Aa−=+

( ) 23

0.749 ft/sA =a !!!!

(b) ( )0 at 5 ft. Use formula (2).A A Av y y− =

System (1): ( ) ( )( )( )12 10.73 5Av = ( )1 10.36 ft/sAv = !!!!

System (2): ( ) ( )( )( )22 16.10 5Av = ( )2

12.69 ft/sAv = !!!!

System (3): ( ) ( )( )( )32 0.749 5Av = ( )3

2.74 ft/sAv = !!!!

(c) Time at 10 ft/s. Use formula (3).Av =

System (1): 110

10.73t = 1 0.932 st = !!!!

System (2): 210

16.10t = 2 0.621 st = !!!!

System (3): 310

0.749t = 3 13.35 st = !!!!




(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/sW

am

= = = 227.4 ft/s=a !

For the upper beam, 1: 0.25 xF ma T N maΣ = − =

( )( ) ( )30000.25 0.25 3000 23.37 2927 lb

32.2 = + = + =

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



Case 2: Assume that both beams slip. As before 2 2 .=N W

( ) ( )( ) ( )2 : 0.30 2 2Σ = =F m a W m a

20.30 9.66 ft/sa g= =

The smaller deceleration value governs. 28.05 ft/sa = !!!!




Since both blocks move together, they have a common acceleration. Use blocks A and B together as a free body.

:F maΣ = Σ

( )sin 30 sin 30A B A BP m g m g m m a− ° − ° = +

500sin 30 9.81 sin 3050A B

Pa 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 N fF = ° =

sin 30 : sin 30B B B BF m a N m g m aΣ = ° − = °

( ) ( )sin 30 10 9.81 5.095 sin 30B BN m g a= + ° = + °

123.575 N=

Minimum coefficient of static friction:

min44.124

123.575f

B

FN

µ = = min 0.357µ =




(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 1

1 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 1

22 1

0 4.811.52

2 2 4.6 3.6v vax x

−−= = = −

− − 2

2 11.52 m/s=a

2 12 1

2

0 4.8 0.41667 s11.52

v vt ta− −− = = =

−

(b) Motion of the package.

1. Acceleration phase. Assume no slip. ( ) 21

3.2 m/spa =

0: 0 or yF N W N W mgΣ = − = = =

( )1:x f pF ma F m aΣ = =

The required friction force is .fF

The available friction force is 0.35 0.35sN W mgµ = =

( ) ( )( ) 21

0.35 9.81 3.43 m/sf sp s

F Na gm m

µ µ= < = = =



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. ( ) 22

11.52 m/spa = −

( )2:x f pF ma F m aΣ = − =

( ) 22

11.52 m/sfp

Fa

m= = −

2 23.43 m/s 11.52 m/ss ss

N mg gm mµ µ µ= = = <

Since the available friction force sNµ is less than the required friction force fF for no slip, the package does slip.

( ) 22

11.52 m/s , p f ka F Nµ< =

( ) ( )2 2:x p k pF m a N m aµΣ = − =

( ) ( )( ) 22

0.25 9.81 2.4525 m/skp k

Na gmµ µ= − = − = − = −

( ) ( ) ( ) ( ) ( )( ) 22 12 1 2

4.8 2.4525 0.41667 3.78 m/sp p pv v a t t= + − = + − =

( ) ( ) ( ) ( ) ( ) ( )2 22 1 2 12 1 1 2

12p 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 =




Acceleration 1:a Impending slip. 1 1 10.30 NsF Nµ= =

1 1: sin 65y A y A AF m a N W m aΣ = − = °

1 1 sin 65A AN W m a= + °

( )1 sin 65Am g a= + °

1 1: cos65x A x AF m a F m aΣ = = °

1 sF Nµ= or ( )1 1cos65 0.30 sin 65A Am a m g a° = + °

( )( ) 21

0.30 1.990 9.81 19.53 m/scos65 0.30sin 65

ga = = =° − °

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 sin 65A 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° = − °

( )( ) 22

0.30 0.432 9.81 4.24 m/scos65 0.30sin 65

ga = = =° + °

22 4.24 m/s=a 65°




Let Pa be the acceleration of the plywood, Ta be the acceleration of the truck, 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 with respect to the truck is impending. P Ta a= and 1 1 10.40 NsF Nµ= =

1: cos 20 sin 20y P y P P TF m a N W m aΣ = − ° = − °

( )1 cos 20 sin 20P TN m g a= ° − °

1: sin 20 cos 20x x P P TF ma F W m aΣ = − ° = °

( )1 sin 20 cos 20P TF m g a= ° + °

( ) ( )sin 20 cos 20 0.40 cos 20 sin 20P T P Tm g a m g a° + ° = ° − °

( ) ( )( )0.40cos20 sin 200.03145 9.81 0.309

cos20 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 2

2 12 12.5 m/s ,0.4

P TP T

xat

= = = 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 cos 20 sin 20P P TN W m a− ° = − °

( )2 cos 20 sin 20P TN m g a= ° − °

continued



:x xF maΣ = Σ

2 /sin 20 cos 20P P T P P TF W m a m a− ° = ° −

( )2 /sin 20 cos 20P T P TF m g a a= ° + ° −

For sliding with friction 2 2 20.30 NkF Nµ= =

( ) ( )/sin 20 cos 20 0.30 cos 20 sin 20P T P T P Tm g a a m g a° + ° − = ° + °

( ) /0.30cos 20 sin 20cos 20 0.30sin 20

P TT

g aa

° − ° +=

° + °

( )( ) ( )( )0.05767 9.81 0.9594 12.5 11.43= − + =

211.43 m/sT =a




At maximum speed 0.a = 20 0 0F kv= = 0

20

Fkv

=

When the propellers are reversed, 0F is reversed.

20:xF ma F kv maΣ = − − =

2

0 0 20

vF F mav

− − = ( )2 2002

0

Fa v vmv

− +

( )202 2

0 0

vdv mv vdvdxa F v v

= =+

0

200

2 200 0

xv

mv vdvdxF v v

= −+∫ ∫

( ) ( )0

2 2 202 2 2 20 0 00 0 0

0 0 0

1 ln ln ln 2 ln 22 2 2v

mv mv mvx v v v vF F F

= − + = − − =

2

0 0

00.347 m vx

F=




:F ma P kv maΣ = − =

dv P kvadt m

−= =

( ) ( )0 0 0ln ln ln

vt 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

− −−= = −

/0

0 0

ttt kt mPt P kx v dt e

k k m− = = − − ∫

( ) ( )/ /1 1kt m kt mPt P Pt Pe ek m k m

− −= + − = − −

, which is linear.Pt kvxk m

= −




Let y be the position coordinate of the projectile measured upward from the ground. The velocity and acceleration a taken to positive upward. 2.D kv=

(a) Upward motion. :yF maΣ =

D mg ma− − =

2

22

D kva g gm m

dv kv k mgv g vdy m m k

= − + = − + = − + = − +

2

v dv k dymg mvk

= −+

0

0

02

h

v

v dv k dymg mvk

= −+

∫ ∫

0

021 ln

2 v

mg khvk m

+ = −

20

20

1 1ln ln 12 2

mgkv khk

mg mg mvk

= − + = −

+

( )( )

( )( )( )( )

220 0.0024 904ln 1 ln 1

2 2 0.0024 4 9.81m kvhk mg

= + = +

335.36 m= 335 mh =

continued



(b) Downward motion. :F maΣ =

D mg ma− =

2

22

D kva g gm m

dv kv k mgv g vdy m m k

= − = −

= − = − −

2

v dv k dymg mvk

= −−

0

0

fv

h

v dv k dymg m

= −∫ ∫

2

0

1 ln2

fvmg khvk m

− =

2

1 ln2

fmg v khk

mg mk

− − =

2 2ln 1 fkv kh

mg m

− = −

2

21 f kh/mkve

mg−− =

( )21 kh/mf

mgv ek

−= ± −

( )( ) ( )( )( )2 0.0024 335.36 /44 9.811

0.0024fv e− = −

73.6 m/s= 73.6 m/sfv =




Choose the origin at point C, and let x be positive to the right. Then x is a position coordinate of the slider B, and 0x is its initial value. Let L be the stretched length of the spring. Then, from the right triangle

2 2L x= +

The elongation of the spring is ,e L= − and the magnitude of the force exerted by the spring is

( )2 2sF ke k x= = + −

By geometry, 2 2

cos x

xθ =

+

: cosx x sF ma F maθΣ = − =

( )2 22 2

xk x max

− + − =+

2 2

k xa xm x

= − − +

0

00v

xv dv a dx=∫ ∫

0

00

02 2 2 22 20

1 12 2

v

xx

k x kv x dx x xm mx

= − − = − − + + ∫

2 2 2 2 20 0

1 102 2

kv x xm = − − − + +

( )2 2 2 2 20 02 2kv x x

m= + − +

( )2 2 2 2 20 02k x x

m = + − + +

( )2 20answer: kv x

m= + −




Let yA, yB, and yC be the position coordinates of blocks A, B, and C respectively measured downward from the upper 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 = − =

2

AA

Ta g

m= − (2)

Block B. : B B BF ma m g T m aΣ = − =

BB

Ta g

m= − (3)

Block C. : C C CF ma m g T m aΣ = − =

CC

Ta g

m= − (4)

Substitute (2), (3) and (4) into (1).

2

2 0A B C

T T Tg g g

m m m

− + − + − =

4 1 1

4 0A B C

g Tm m m

− + + =

( )( ) 4 1 14 9.81 0 65.4 N

10 10 10T T

− + + = =

Substitute into (2), ( )( ) 22 65.4

9.81 3.27 m/s 10Aa = − = −

(a) Change in position. ( ) 20

1

2A A Ay y a t− =

( ) ( )( )2

0

13.27 0.5 0.409 m

2A Ay y− = − = −

0.409 my∆ = !!!!

(b) Tension in the cable. 65.4 NT = !




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 friction force 1F is directed as shown. Its magnitude is

1 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: sin 30x B B B B BF m a W F m aΣ = ° − =

( ) ( ) 21

1 1sin 30 49.05sin 30 4.248 4.055 m/s5B B

Ba W F

m= ° − = ° − =

Assume that block A slides downward relative to the fixed plane. The friction force 2F is directed as shown. Its magnitude is

2 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: sin 30x A A A A AF m a W F F m aΣ = ° − + =

( )2 11 sin 30A A

Aa W F F

m= ° − +

( ) 21 98.1sin 30 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 are as 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°




Let the positive direction for position coordinates, velocities, and accelerations 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)

:x xF maΣ =

Block A: 3 33 or 20A A A

A

T TT m a a gm

= = = (2)

Block B: 2 22 or 10B B B

B

T TT m a a gm

= = = (3)

Block C: 4 44 or 20 20C C C

P T P TP T m a a g− −− = = = (4)

Substituting (2), (3), and (4) into (1),

4 2 34 2 3 0 20 10 20

P T T T− − − =

16 4 9 420 10 20 20

PT + + =

( ) ( )( )0.12121 0.12121 50 6.0605 lb T P= = =

(a) From (2), ( )( )( ) 23 6.0605 32.229.3 ft/s

20Aa = = 229.3 ft/sA =a

From (3), ( )( )( ) 22 6.0605 32.239.0 ft/s

10Ba = = 239.0 ft/sB =a

From (4), ( )( ) ( ) 241.5 ft/s

20

50 4 6.0605 32.2Ca

− = = 241.5 m/sC =a

(b) As determined above, 6.06 lbT =




2 220 100.62112 lb s / ft, 0.31056 lb s / ft

32.2 32.2A C Bm m m= = = ⋅ = = ⋅

Let the positive direction for position coordinates, velocities, and accelerations 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 3 0.20k AA

A A

T 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 2 0.20k BB

B B

T 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 0

1 102 2C C C C Cx x v a t a t= + + = +

( ) ( )( )

( )0 2

2 2

2 2 2.430 ft/s

0.4C C

C

x xa

t

− = = = (5)



Substitute (2), (3) and (5) into (1).

( ) ( ) ( ) ( )2 34 30 2 0.20 3 0.200.31056 0.62112

T Tg g − − − −

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 34 30 2 0.2 32.2 3 0.2 32.20.31056 0.62112

T 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/s

0.62112Aa = − =

From (3), ( )( ) ( )( ) 22 5.56080.20 32.2 29.37 ft/s

0.31056Ba = − =

(a) Acceleration vectors. 220.4 ft/sA =a

2 29.4 ft/sB =a

230 ft/sC =a

Since , , and A B Ca a a are to the right, the friction forces , , and A 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




Let the positive direction of x and y be those shown in the sketch, and let the origin lie at the cable anchor.

Constraint of cable: / /constant or 0,A B A A B Ax y a a+ = + = where

the positive directions of / and A B Aa a are respectively the x and the y

directions. Then /B A Aa a= − (1)

First note that (/B A B A Aa= + =a a a ) ( /20 B Aa° + )20°

Block B: ( ) : sin 20 x 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)

( ) /: cos 20y B B B B B AyF m a m g T m aΣ = ° − =

/ cos 20B 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) and subtract 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 = !




Motion of B relative to A. Particle B is constrained to move on a circular path with its center at point A. ( )/B A ta is the component of /B Aa lying

along the circle, say to the left in the diagram and ( )/B A na is directed

toward 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 cos 25x 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 =




The ball moves at constant speed on a circle of radius

sinLρ θ=

Acceleration (toward center of circle). 2vaρ

=

+ :y yF maΣ = cos 0− =T Wθ cos

=WTθ

:y xF maΣ = sinT maθ =

sincosW θθ

2mvρ

= 2

sinmv

L θ=

(a) tan sinθ θ = 2mv

WL=

2vgL

=( )

( )( )

21.50.38226

9.81 0.6=

34.21θ = ° 34.2θ = °

(b) cos cosW mgTθ θ

= = ( )( )2 9.81cos34.21

=°

23.7 N=T




Let ρ be the radius of the horizontal circle. The length of the wire is

1 2sin sinL ρ ρ

θ θ= + . Solving for ,ρ 1 2

1 2

sin sinsin sinL θ θρθ θ

=+

1 20: cos cos 0yF T T mgθ θΣ = + − =

1 2cos cosmgT

θ θ=

+

2

1 2: sin sinx x nmvF ma T T maθ θρ

Σ = + = =

( ) ( )21 2 1 2

1 2 1 2

sin sin sin sincos cos sin sin

mg mvL

θ θ θ θθ θ θ θ

+ +=

+

( )( )2 2 21 2

1 2

sin sin sin 60 sin302 9.81 6.2193 m /scos cos cos60 cos30

v Lg θ θθ θ

° °= = =+ ° + °

2.49 m/sv =




3 1 2 50 25 25θ θ θ= − = ° − ° = °

1 21

2 3 3

sin or sin sin sin

d d θθ θ θ

= =

( )( )( )

2 11 1

3

sin sinsinsin

4 sin 25 sin503.0642 ft

sin 25

d θ θρ θθ

= =

° °= =

°

2 10: cos cos 0y AC BCF T T Wθ θΣ = + − = (1)

2

2 1: sin sinx x AC BCWvF ma T Tg

θ θρ

Σ = + = (2)

Case 1: 0.BCT = 2cos 0ACT Wθ − = or2

cosAC

WTθ

=

2

2 2sin tanACWvT Wg

θ θρ

= =

( )( )22tan 32.2 3.0642 tan 25v gρ θ= = ° 2 246.01 ft /s=

6.78 ft/sv =

Case 2: 0.ACT = 11

cos 0 or cosBC BC

WT W Tθθ

− = =

2

1 1sin tanBCWvT Wg

θ θρ

= =

( )( )2 2 21tan 32.2 3.0642 tan 50 117.59 ft /sv gρ θ= = ° =

10.84 ft/sv = 6.78 ft/s 10.84 ft/sv≤ ≤




(a) 0: sin 0yF T WθΣ = − =

16sin sin 60WTθ

= =°

or 18.48 lbT =

(b) 2

: cosx nvF ma T mθρ

Σ = =

2 cos cossin

T Wvm m

ρ θ ρ θθ

= =

( )( ) 2 23 32.255.77 ft /s

tan tan 60gρθ

= = =°

7.47 ft/sv =




2, 1 tan or 45

2r dyy r

dxθ θ= = = = = °

0: cos 0yF N mgθΣ = − =

cosmgNθ

=

: sinx x nF ma N maθΣ = =

2

tan vmg mr

θ =

2 tanv gr θ=

(a) ( )( )( )2 2 29.81 1 1.0000 9.81 m /sv = = 3.13 m/sv =

(b) ( )( )1 9.81

cos 45 cos 45mgN = =

° ° 13.87 N=N




Geometry

( ) ( ) ( )( )( )

2 2 2

2 2 2

2 cos30

0.3 0.6 2 0.3 0.6 cos30 0.13823 m

0.37179 m

OC OB BC OB OC

OC

= + − °

= + − ° =

=

sin 30 sinOC OB

β=

°

( )0.3 sin 30sin 30sin 0.403450.37179

OB

OCβ

°°= = =

23.79β = °

Acceleration components: 21.3 m/sta =

( )22 221

1.667 m/s0.6

Cn

BC

v vaρ

= = = =

Mass 1 kgm =

( )( ): cos 1 1.3 1.3t tF ma N βΣ = = =

1.3 1.421 Ncos 23.79

= =°

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°




( )( )0.5 9.81 4.905 NW mg= = =

150 mm 0.150 mρ = =

0 : cos 20 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 sin 30

0.150x n DA DEmvF ma T T vρ

Σ = = ° + ° =

2 0.10261 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≤ ≤




( )( )5 9.81 49.05 NW mg= = =

0.9 mρ =

0 : cos 40 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 maρ

Σ = ° + ° = =

( )

( )

2 sin 40 sin15

0.9 sin 40 sin155

0.115702 0.046587

CA CB

CA CB

CA CB

v T Tm

T T

T T

ρ= ° + °

= ° + °

= + (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≤ ≤




(a) Before wire AB is cut. 0a =

0 :xFΣ = cos50 cos70 0AB CDT T− ° + ° = (1)

0 :yFΣ = sin 50 sin 70 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= =

2

0nvaρ

= =

0 :n nF ma= =

cos 20 0CDT W− ° =

cos 20CDT W= ° 0.940CDT W=

+




:y yF maΣ =

2

nmvN mg maρ

− = − = −

2mvN mg

ρ= −

:x xF maΣ =

t tF ma=

At onset of slipping, t sF Nµ=

2s

t smvma mgµρ

= −

( )2 2 20.1500.300 9.81 2.883 m /s0.75

ts

s

av gρµ

= − = − =

1.6979 m/ssv =

Time at slipping. 1.69790.150

ss

t

vta

= = 11.32 sst =




Angle change over arc AB. 8 0.13963 rad180

θ 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 0

2

2 22949

540 0

2 2 2

480 540 or 55892 2

5.475 ft/s

540 or 5.475 29492 2

259300 ft /s 509.2 ft/s

B

t t

t

v Bt

B B

v dv a ds a

a

vv dv a ds

v v

= − =

= −

= − = −

= =

∫ ∫

∫ ∫

Mass of passenger: 2200 6.211 lb s /ft32.2

= = ⋅m

Just before point B. 509.2 ft/s, (4)(5280) = 21120 ftv ρ= =

( )222509.2

12.277 ft/s21120n

vaρ

= = =

( ) ( )( )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 lb

magnitude of change of force = 76.3 lb




( ) ( )23 180 m, 3 180 2 m/s,dss t t v tdt

= − = = −

26 m/s (4)(5280) 21120 ft= = − = =tdvadt

ρ

Length of arc AB. ( ) 821120 2949 ft180AB ABs πρθ °

= = =°

0B AB

A

v stv v dv a ds=∫ ∫

( )( )( )2 2

2 2 22 or 2 540 2 6 29492 2B A

t AB B A t ABv v a s v v a s− = = + = + −

256212,= 506.2 ft/sBv =

For passenger, 2165165 lb, 5.124 lb s /ft32.2

= = = = ⋅WW mg

2: cosn

mvF ma W Nθρ

Σ = − =

2

cos mvN W θρ

= − (1)

: sint tF ma P W maθΣ = − =

sin tP W maθ= + (2)

(a) Just after point A, 0, 180 m/s, 8 t v θ= = = °

From Eq. (1), ( )( )25.124 540165cos8 92.65 lb

21120N = ° − =

From Eq. (2), ( )( )165sin8 5.124 6 7.78 lbP = ° + − = −

2 2 92.6593.0 lb, tan 11.909, 85.27.78

F 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 lb

21120N = − =

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 β

107.3 lb=F 73.4°




(a) 160 km/h 44.44 m/sv = =

Wheels do not touch the road.

2: /y nF ma mg mv ρΣ = − − = −

( )22 44.44201.4

9.81vg

ρ = = =

201 mρ =

(b) 80 km/h 22.22 m/sv = =

70 kgm = for passenger

:y nF maΣ = −

2mvN mgρ

− = −

2vN m gρ

= −

( )222.2270 9.81

201.4

= −

515 NN =




0.2 kg, (0.2)(9.81) 1.962 N= = = =m W mg

: cost t tF ma W maθΣ = =

cos costWa gm

θ θ= =

0 0 0v s

t tv v dv a ds a rdθ θ= =∫ ∫ ∫

2 20 0

1 1 cos sin2 2

v v g rd grθ θ 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 path becomes 0.3 m.ρ =

2max

max:y nmvF ma T W

ρΣ = − =

( )2max maxv T W

mρ

= − (2)

Eliminating 2maxv from equations (1) and (2),

( ) ( )( ) ( )( )( )max20

0.3 10 1.9622 2 9.81 6.0

0.2T W

v grm

ρ − −= − = −

2 20.285 m /s= 0 0.534 m/s=v




Mass of block B. 22

0.5lb 0.015528lb s /ft32.2ft/s

= = ⋅Bm

Acceleration of block B. 2 2

2(9ft/s) 27 ft/s3ft

= = =nvaρ

0ta = 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 0.5

nma PW W

θ ≥ − = −

sin 0.2385θ ≥

13.8 166.2θ° ≤ ≤ °




At position A, the vertical component of apparent weight is shown as .AN

:n A nWF ma N W ag

Σ = − =

( ) 2380 120 32.2 69.77 ft/s120

An

N 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 shown as .CN

:n C nWF ma N W ag

Σ = + =

( ) 280 120 32.2 53.67 ft/s120

Cn

N 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 2 2C A t ACv v a s− =

( )( )2 2 3 3

2

193.2 10 251.2 102 2 11310

2.562 ft/s

C At

AC

v vas− × − ×

= =

= −

At position B, ( )3600 5655 ft2 2ABs π 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= + = × + − = ×



Effective forces at B: 2 3120 222.2 10 230 lb

32.2 3600B

nW vmag ρ

×= = =

( )120 2.562 9.5 lb32.2t t

Wma 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= + = + =

110.5tan230

β = 25.7β = °

255 lb=F 25.7°




The road reaction consists of normal component N and friction component F. The resultant R makes angle sφ with the normal.

Case 1: maxv v=

( )0: cos 0y sF R mgθ φΣ = + − =

( )cos s

mgRθ φ

=+

( ): sinx n s nF ma R maθ φΣ = + =

( )tann sma mg θ φ= +

( )2

max tan sv g

rθ φ= +

( )max tan sv gr θ φ= +

Case 2: minv v=

( )0: cos 0y sF R mgθ φΣ = − − =

( )cos s

mgRθ φ

=−

( ): sinx n s nF ma R maθ φΣ = − =

( )tann sma mg θ φ= −

( )2

min tan sv g

rθ φ= −

( )min tan sv gr θ φ= −

( ) ( )tan tans sgr v grθ φ θ φ− ≤ ≤ +




Weight. W mg=

Acceleration. 2vaρ

=

:∑ =x xF ma sin cosF W = maθ θ+

2

cos sinmvF mgθ θρ

= − (1)

:y yF ma∑ = cos sinN W maθ θ− =

2

sin cosmvN = mgθ θρ

+ (2)

(a) Banking angle. Rated speed 180 km/h 50 m/s.v = = 0F = at rated speed.

2

0 cos sinmv mgθ θρ

= −

2 2(50)tan = 1.2742

(200) (9.81)v

gθ

ρ= =

51.875θ = ° 51.9θ = °

(b) Slipping outward. 320km/h 88.889 m/sv = =

F Nµ= 2

2cos sinsin cos

F v gN v g

θ ρ θµθ ρ θ

−= =

+

2

2(88.889) cos51.875 (200) (9.81)sin 51.875=(88.889) sin 51.875 (200)(9.81)cos51.875

µ °− °°+ °

0.44899= 0.449µ =

continued



(c) Minimum speed. = −F Nµ

2

2cos sinsin cos

v gv g

θ ρ θµθ ρ θ

−− =

+

2 (sin cos )cos sin

gv ρ θ µ θθ µ θ

−=

+

(200)(9.81)(sin 51.875 0.44899cos51.875 )cos51.875 0.44899sin 51.875

° − °=

° + °

2 21029.87 m /s=

32.09m/sv = 115.5 km/h=v




Rated speed: 75 mi/h 110 ft/s, 125 mi/h 183.33 ft/sRv = = =

From Sample Problem 12.6,

( )222 110

tan or 2674 fttan 32.2 tan8

RR

vv g

gρ θ ρ

θ= = = =

°

Let the x-axis be parallel to the floor of the car.

( ) ( ): sin cosx x s nF ma F W maθ φ θ φΣ = + + = +

( )2

cosmv θ φ

ρ= +

(a) 0.φ =

( ) ( )

( )( ) ( )

2

2

cos sin

183.33cos8 sin8

32.2 2674

0.247

sv

F Wg

W

W

θ φ θ φρ

= + − +

= ° − °

= 0.247 sF W= !

(b) For 0,sF =

( ) ( )2

cos sin 0v

gθ φ θ φ

ρ+ − + =

( ) ( )( ) ( )

22 183.33tan 0.39035

32.2 2674

v

gθ φ

ρ+ = = =

21.3θ φ+ = °

21.3 8φ = ° − ° 13.3φ = ° !




Rated speed: 75 mi/h 110 ft/s, 125 mi/h 183.33 ft/sRv = = =

From Sample Problem 12.6,

( )222 110

tan or 2674 fttan 32.2 tan8

RR

vv g

gρ θ ρ

θ= = = =

°

Let the x-axis be parallel to the floor of the car.

( ) ( ): sin cosx x s nF ma F W maθ φ θ φΣ = + + = +

( )2

cosmv θ φ

ρ= +

Solving for ,sF ( ) ( )2

cos sinsv

F Wg

θ φ θ φρ

= + − +

Now ( )

( )( )22 183.33

0.39035 and 0.12 32.2 2674 s

vF W

gρ= = = 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 = ° !!!!




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 2

2( sin )nva rρ ω θ ωρ ρ

= = =

:y yF ma∑ =

cos 0N mgθ − =

cosmgNθ

=

:x xF maΣ =

sinN maθ =

2sin ( sin )cos

mg m rθ θ ωθ

=

Either sin 0θ =

or 2cos gr

θω

=

0 or 180θ = ° °

or 29.81cos 0.3488

(0.5) (7.5)θ = =

0 , 180 , and 69.6θ = ° ° °




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 2

2( sin )nva rρ ω θ ωρ ρ

= = =

FΣ = :ma 2sin ( sin ) cosF mg m rθ θ ω θ− = −

2( cos )sinF m g rω θ θ= −

FΣ = :ma

2cos ( 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) sin 75F = − ° ° 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= =sNµ

Since ,sF Nµ< the collar does not slide.

0.611 NF = 75°

continued



(b) 40 .θ = ° 2(0.25) 9.81 (0.500cos 40 )(7.5) sin 40F = − ° °

1.8858 N= −

2 2(0.25) 9.81cos 40 (0.500sin 40 )(7.5)N = ° + °

4.7839 N=

(0.25)(4.7839) 1.1960 N= =sNµ

Since ,> sF Nµ the collar slides.

Since the collar is sliding, .kF Nµ=

nF∑ = + :ma

cos 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) 9.81cos 40 (0.500sin 40 )(7.5) = °+ °

4.7839 N=

(0.20) (4.7839) 0.957 N= = =kF Nµ

0.957 N=F 40°




Draw the free body diagrams of the block B when the arm is at 150 .θ = °

20, 32.2 ft/stv a g= = =

: sin 30 0t tF ma W NΣ = − ° + =

sin 30N W= °

2 2: cos30n n

v WvF ma W F mgρ ρ

Σ = ° − = =

2cos30 WvF W

gρ= ° −

Form the ratio FN

, and set it equal to sµ for impending slip.

( )22 cos30 4.2 /(1)(32.2)cos30 /sin 30 sin 30s

F v gN

ρµ° −° −

= = =° °

0.636sµ =




Let β be the slope angle of the dish. 1tan6

dy rdr

β = =

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 sins

WNβ µ β

=−

2 2: sin cosn n s

mv WvF ma N Ng

β µ βρ ρ

Σ = + = =

( ) 2sin coscos sin

s

s

W N Wvg

β µ ββ µ β ρ+

=−

( ) ( )2 2 2sin cos sin 45 0.5cos 456 32.2 579.6 ft /scos sin cos 45 0.5sin 45

s

sv g β µ βρ

β µ β+ ° + °

= = =− ° − °

24.1 ft/sv =




Uniformly accelerated motion on a circular path. 55 in. ft2 12Dρ = = =

66 60 1060 10 oz lb

16W

−− ×

= × =

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 −= + = +

lb 10644 6−×=F




Uniformly accelerated motion on a circular path. 8 ftρ =

0 tv v a t= +

( )( )0 0.75 12 9 ft/s= + =

0.75: 0.0233 32.2

tt t t t

W aF ma a F W W Wg g

= = = = =

( )( )( )

22 9: 0.3144

32.2 8n n nWWvF ma F W

gρ= = = =

2 2 0.315 t 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 the vertical direction.

0: 0yF N WΣ = − = N W=

Since sliding is impending, 0.315sFW

µ = = 0.315sµ =




For constant speed, 0ta = 2

with 0.7 m/s, 0.2 mBn B

va v ρρ

= = =

: cosx x nF ma F ma θΣ = =

: sin siny y n nF ma N W ma N mg maθ θΣ = − = − = −

Ratio 2

cos cos cossin sin sin

n

n

n B

F mag gN mg maa v

θ θ θρθ θ θ

= = =− − −

With ( )( )( )2 2

9.81 0.2 cos4.0041, the ratio becomes 4.0041 sin0.7B

g FNv

ρ θθ

= = =−

For no impending slide, cos4.0041 sins

FN

θµ

θ≥ =

−

To find the value of θ for which the ratio is maximum set the derivative with respect to θ equal to zero.

( )2cos 1 4.0041sin 0

4.0041 sin 4.0041 sind

dθ θ

θ θ θ± − = ± = − −

1sin 0.249744.0041

θ = =

cos14.44614.446 , 0.258

4.0041 0.24974FN

θ°

= ° = =−

180 14.446 165.554 , 0.258FN

θ = ° − ° = ° =

(a) Minimum value of sµ for no slip. ( )min 0.258sµ =

(b) Corresponding values of .θ 14.5 and 165.5θ = ° °




For constant speed, 0ta =

2with 0.7 m/s, 0.2 mB

n Bva v ρρ

= = =

: cosx x nF ma F ma θΣ = =

: sin siny y n nF ma N W ma N mg maθ θΣ = − = − = −

Ratio 2

cos cos cossin sin sin

n

n

n B

F mag gN mg maa v

θ θ θρθ θ θ

= = =− − −

Let 2cos so that

sinB

g FuN uv

ρ θθ

= =−

Determine the value of θ at which F/N is maximum.

( )( )( ) ( )

2

2 2cos sin sincos 1 sin 0

sin sin sin

ud ud u u u

θ θ θθ θθ θ θ θ

− − − = = = − − −

The corresponding ratio .FN

2 1

1 2

1 sin tancos1

F u uN u u u

θ θθ

− −

− −

± − ±= = = ± = ±

− −

(a) For impending sliding to the left: tan 0.35sFN

θ µ= = =

( )2

1arctan 0.35 19.29 , sin ,Bvug

θ θρ

−= = ° = =

( )( )2 2 29.81 0.2 sin19.29 0.648 m /sBv = ° =

0.805 m/sBv =



For impending motion to the right: tan 0.35sFN

θ µ= − = =

( )arctan 0.35 160.71θ = − = °

21 2 sin ,vu

gθ

ρ− = = ( )( )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 = °




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 the screen,

2x L= +l

Horizontal motion: There are no horizontal forces acting on the electron so that 0.xa =

Let 1 0t = when the electron passes the left edge of the plate, 1t t= when

it passes the right edge, and 2t t= when it impacts on the screen. For uniform horizontal motion,

0 1 20 0 0

, so that and .2

Lx v t t t

v v v= = = +l l

Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the

electron 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 y

F eVF ma a

m mdΣ = = =

( )0

0y y yeVt

v v a tmd

= + = +

( )2

20 0

10 0

2 2y yeVt

y y v t a tmd

= + + = + +

At ( )2

1 11 11, and

2yeVt eVt

t 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 1

2 1 2 11

2 2

eVt eVt eVtt t t t

md md mdδ = + − = −

0 0 0 0

1

2 2

eV L

mdv v v v

= + −

l l l or

20

eV

mdv

Lδ = l !!!!


Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

© 2007 The McGraw-Hill Companies.


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 the

screen,

2x L= +l

Horizontal motion: There are no horizontal forces acting on the electron so

that 0.x

a =

Let 1

0t = when the electron passes the left edge of the plate, 1

t t= when

it passes the right edge, and 2

t t= when it impacts on the screen. For

uniform horizontal motion,

0 1 2

0 0 0

, so that and .2

Lx v t t t

v v v= = = +l l

Vertical motion: The gravity force acting on the electron is neglected since

we are interested in the deflection produced by the electric force. While the

electron is between the plates ( )10 ,t t≤ ≤ the vertical force on the

electron is / .y

F eV d= After it passes the plates ( )1 2,t t t≤ ≤

it is zero.

For 10 ,t t≤ ≤ :

y

y y y

F eVF ma a

m mdΣ = = =

( )0

0y y y

eVtv v a t

md= + = +

( )2

2

00

10 0

2 2y y

eVty y v t a t

md= + + = + +

At 2

1 2

0 0

, , 2

eVt t y

v mdv= = ll

But 0.075 0.4252

dy d d< − =

So that 2

2

0

0.4252

eVd

mdv<l

2

2 2 2

0 0

11.176

0.425 2

d eV eV

mv mv> =

l

2

0

1.085d eV

mv>

l �






30 , 2 rad/s, 0θ θ θ= ° = =& &&

0.6 m, r W mg= =

Block B: Only force is weight

cos30 , sin 30r

F W F Wθ= ° = − °

(a) ( )2 :F ma m r rθ θ θ θ= = +&& &&

sin302 sin30

F mgr r r g r

m m

θθ θ θ θ°= − = − − = − ° −& && && &&&

( ) ( )( )( )( )

9.81 sin 30 0.6 0sin 301.226 m/s

2 2 2

g rr

θθ

° +° += − = − = −&&

&&

/ rod

1.226 m/sB

=v 60° �

(b) ( )2 :r r

F ma m r rθ= = − &&&

( )( ) ( )

2 2

2 2

cos30cos30

0.6 2 9.81 cos30 10.90 m/s

rF mg

r r r r gm m

θ θ θ°= + = + = + °

= + ° =

& & &&&

2

/ rod10.90 m/s

B=a 60° �




245 , 0.8 m, 10 rad/srθ θ= ° = =

0, 0, rv r v r W mgθ θ= = = = =

(a) ( ): cos 45 2F ma N W m r rθ θ θ θΣ = − ° = +

( )cos 45 2N mg m r rθ θ= ° + +

( )( )(0.5)(9.81)cos 45 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 rm

θ θ= ° + = ° +

( ) 29.81 sin 45 0 6.937 m/s= ° + =

2

/ rod 6.94 m/sB =a 45°




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: 24 0.12422 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 =


(0.12422)( 32)r rF ma= = − 3.98 lbrF = −

(0.12422)(32)F maθ θ= = 3.98 lbFθ =




Use radial and transverse components of acceleration.

6(1 cos 2 ) ft 2 radr t tπ θ π= + =

12 sin 2 ft/s 2 rad/sr tπ π θ π= − =

2 224 cos 2 ft/s 0r tπ π θ= − =

2 2 224 cos 2 (6 6cos 2 )(2 )ra r r t tθ π π π π= − = − − +

2 2 224 48 cos 2 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: 21 0.031056 lb s /ft32.2

Wmg

= = = ⋅

(a) 0.t = 2710.61 ft/sra = −

0aθ =


(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 =


(0.031056)( 236.87)r rF ma= = − 7.36 lbrF = −

(0.031056)(473.74)F maθ θ= = 14.71 lbFθ =




Kinematics: 1.5 ft/s, 0dr r rdt

= = =

0 0 or 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.

or r rma ma a aθ θ− = − =

Using the calculated expressions

3 2 23.60.96 3.6 , 3.75 s0.96

t t t= = =

1.936 st =




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 4

or 171.733 rad /s0.015528 1.5

mr Tmr mr T

mrθ 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 = !




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 2

0.230 12

60 0.230 15mrr

k mθ−

= − = −− −

0.33455 m= 335 mmr =

( ): 2F ma F m r rθ θ θ θ θΣ = = +

2 Fr rmθθ θ= −

( )( )( )9 0 1.304 m/s

2 2 0.230 15F mrr

mθ θ

θ− −

= = =

(b) Radial component of velocity. rv r= 1.304 m/srv =






At point A. 2 2

2(3.8)115.52 m/s

0.125= = =

n

v

a

ρ

125tan 22.62

175 125= − = − °

+θ θ

+ : cos22.6Σ = ° =s t

F ma F ma

2cos22.6 70cos22.6243.077 m/s

1.5

s

t

Fa

m

° °= = =

Acceleration vector.

2 2

n ta a a= +

2 2 2115.52 43.077 123.29 m/s= + =

115.52tan

43.077φ = 69.55φ = °

22.62 46.93φ − ° = °

cos46.93ra a= °

2123.29cos46.93 84.2 m/s= ° = in negative r-direction

284.2 m/s

ra = − �

sin 46.93a aθ = °

123.29sin 46.93= °

290.1 m/saθ = �

Draw the free body diagram of the collar.




Let r and θ be polar coordinates of block A as shown, and let By be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B.

Constraint of cable: constant,Br y+ =

0, 0 or B B Br v r a r a+ = + = = − (1)

For block A, + : cos or sec (2)A Ax A A A A

W WF m a T a T ag g

θ θΣ = = =

For block B, : (3)B By B B B

W WF a W T ag g

Σ = − =

Adding Eq. (1) to Eq. (2) to eliminate T, secA BB A B

W WW a ag g

θ= + (4)

Radial and transverse components of .Aa

Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components.

2 cosr A r Ar r a aθ θ− = = ⋅ = −a e (5)

Noting that initially 0,θ = using Eq. (1) to eliminate ,r and changing signs gives

cosB Aa a θ= (6)

Substituting Eq. (6) into Eq. (4) and solving for ,Aa

( ) ( ) 250 32.217.991 ft/s

sec cos 40sec30 50cos30B

AA B

W 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




Let r and θ be polar coordinates of block A as shown, and let By be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B.

Radial and transverse components of .Av

Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components.

cos30

6cos30 5.19615 ft/sr A r Ar v v= = ⋅ = − °

= − ° = −

v e

sin 30

6sin 30 3 ft/sA Ar v vθ θθ = = ⋅ = − °

= ° =

v e

3 1.25 rad/s2.4

vrθθ = = =

Constraint of cable: constant,Br y+ =

0, 0 or B B Br v r a r a+ = + = = − (1)

For block A, + : cos or secA Ax A A A A

W WF m a T a T ag g

θ θΣ = = = (2)

For block B, :B By B B B

W WF a W T ag g

Σ = − = (3)

Adding Eq. (1) to Eq. (2) to eliminate T, secA BB A B

W WW a ag g

θ= + (4)

Radial and transverse components of .Aa

Use a method similar to that used for the components of velocity.

2 cosr A r Ar r a aθ θ− = = ⋅ = −a e (5)

Using Eq. (1) to eliminate r and changing signs gives

2cosB Aa a rθ θ= − (6)



Substituting Eq. (6) into Eq. (4) and solving for ,Aa

( ) ( ) ( )( )222

50 32.2 2.4 1.2520.086 ft/s

sec cos 40sec30 50cos30B

AA B

W g ra

W W

θ

θ 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




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

00

cos2 cos2r v r v vrr r

θθ θ= = =

Radial component of velocity.

( )0

0 3/ 2sin 2

cos2 cos2r

dr d rv r rd d

θθ θ θθ θ θ θ

= = = =

( )

00 3/ 2

sin 2 cos2cos2

vrr

θ θθ

= 0sin 2cos2rv v θ

θ=

Transverse component of velocity.

0 0

0cos2h r vv

r rθ θ= = 0 cos2v vθ θ=





Using Eq. (12.27), 20 0 0h r h r vθ= = = or 0 0 0 0

2 200

cos2 cos2nr v r v vrr r

θθ θ= = =

Differentiating the expression for r with respect to time,

( ) ( )0 0

0 0 03 / 2 3 / 20

sin 2 sin 2 sin 2cos 2cos 2 cos 2cos 2 cos 2

dr d r vr r r vd d r

θ θ θθ θ θ θθ θ θ θθ θ

= = = = =

Differentiating again,

( )

2 2 2 2 20

0 0 3 / 20

sin 2 2cos 2 sin 2 2cos 2 sin 2cos 2 cos 2cos 2

dr d vr v vd d r

θ θ θ θ θθ θ θθ θ θ θθ

+ + = = = =

(a) 00

0

sin 2 sin 2cos2r

v rv r vr

θ θθ

= = = 0

0cos2v rv r

rθ θ θ= =

( ) ( )2 2 2 20

0sin 2 cos 2r

v rv v vrθ θ θ= + = + 0

0

v rvr

=

2 2 2 22 20 0 0

20 0

2cos 2 sin 2 cos 2cos 2 cos 2r

v r va r rr r

θ θθ θθ θ

+= − = −

2 2 2 2

0 0 02

0 00

cos 2 sin 2cos 2 cos 2

v v v rr rr

θ θθ θ

+= = =

202

0:r r

mv rF mar

= = 202

0r

mv rFr

=

Since the particle moves under a central force, 0aθ =

continued



Magnitude of acceleration: 2

2 2 02

0r

v ra a arθ= + =

Tangential component of acceleration. 2

0 0 02

0 0 0sin 2t

dv d v r v v ra rdt dt r r r

θ

= = = =

Normal component of acceleration. 2 2

22 20 02 2

0 0

cos21 sin 2t tv r v ra a ar r

θθ= − = − =

But 2

0cos 2 rr

θ =

Hence, 2

0n

var

=

(b) But 22 2 2

02 2

0 0 or n

n

v v v r raa r v

ρρ

= = = ⋅ 3

20

rr

ρ =





Using Eq. (12.27), 20 0 0h r h r vθ= = =

or 0 0 0 0 02 2 22

00 coscosr v r v vr rr

θθθ

= = =

Radial component of velocity. ( ) ( )0 0cos sinrdv r r rdt

θ θ θ= = = −

Transverse component of velocity. ( )0 cosv r rθ θ θ θ= =

Speed. 2 2 0 00 2

0 cosrr vv v v r

rθ θθ

= + = = 02cos

vvθ

=




Since the particle moves under a central force, constanth =

Using Eq. (12.27), 20 0 0h r h r vθ= = = 0 0 0 0 0

2 2 2200 coscos

r v r v vr rr

θθθ

= = =

Radial component of velocity. ( ) ( )0cos sinr odv r r rdt

θ θ θ= = = −

Transverse component of velocity. ( )0 cosv r rθ θ θ θ= =

Speed. 2 2 0 0 00 2 2

0 cos cosrr v vv v v r

rθ θθ θ

= + = = =

Tangential component of acceleration.

( )( ) 20 0 0

0 3 3 2 50 0

2 sin 2 sin 2 sincos cos cos cost

dv v v va vdt r r

θ θ θ θθ θ θ θ

− −= = = ⋅ =

Tangential component of force.

20

50

2 sin:cost t t

mvF ma Fr

θθ

= =

(a) 0, 0tFθ = = 0tF =

(b) 05

2 sin 4545 , cos 45t

mvFθ °= ° =

° 2

0

0

8t

mvFr

=




For gravitational force and a circular orbit,

2

2 or rGMm mv GMF v

r rr= = =

Let τ be the periodic time to complete one orbit.

2 or 2GMv r rr

τ π τ π= =

Solving for ,τ 3/ 22 r

GMπτ =

But, 3 3 / 24 , hence 2 3 3

M R GM G Rππ ρ ρ= =

Then, 3/ 23 r

G Rπτρ

=

Using 3r R= as given leads to

3/ 2 33 9G G

π πτρ ρ

= = ( )1/ 29 /Gτ π ρ=




For gravitational force and a circular orbit, 2

2 or rGMm mv GMF v

r rr= = =

Let τ be the period time to complete one orbit.

But 2

2 2 2 22 or 4GMv r v rr

ττ π τ π= = =

Then 1/ 32 2

32 2 or

4 4GM GMr rτ τ

π 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 3

62

398.06 10 86.1624 1042.145 10 m

4r

π

× × = = ×

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 3

62

14.077 10 86.1624 10138.334 10 ft

4r

π

× × = = ×

altitude 6117.425 10 fth r R= − = × 22200 mih =



(b) In SI units:

12

36

398.06 10 3.07 10 m/s42.145 10

GMvr

×= = = ×

× 3.07 km/sv =

In US units:

15

36

14.077 10 10.09 10 ft/s138.334 10

GMvr

×= = = ×

× 310.09 10 ft/sv = ×





2

2 or rGMm mv GMF v

r rr= = =


2 or 2GMv r rr

τ π τ π= =

from which 2 3

24 rGM π

τ=

But 2 3

22 2

4, hence, rGM gR gRπ

τ= = (1)

Solving for ,r 1/ 32 2

24gRr τ

π

=

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 92

2 23 6

4 2.202 1081.5 ft/s

306.8 10 234.4 10g

π ×= =

× × 281.5 ft/sg =




Let M be the mass of the sun and m the mass of Venus.

For the circular orbit of Venus,

2

22 n

GMm 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 2

9 4 44.6792 10 ft /s34.4 10 ft /lb s

GMMG −

×= =

× ⋅

27 2136.0 10 lb s /ftM = × ⋅

(b) At the surface of the sun, 3 9432 10 mi 22.81 10 ftR = × = ×

2GMm mg

R=

( )

21

2 29

4.6792 10

22.81 10

GMgR

×= =

× 2899 ft/sg =





2

2 or rGMm mv GMF v

r rr= = =

But 2

32

2 2, hence, or 4

r r GM GMv rr

π π ττ τ π

= = =

Solving for r, 1/ 32

2 (1)4

GMr τπ

=

( )1/ 31/ 3 222

22 2

44 GMGMv GMr GM

ππτ τ

= = =

1/ 32 (2)GMv π

τ =

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 39

2

126.98 10 617.76 101.071 10 m

4r

π

× × = = ×

61.071 10 kmr = ×

(b) For Callisto: 616.69 days 400.56 h 1.4420 10 sτ = = = ×

By Eq. (2), ( ) 1/ 312

36

2 126.98 108.209 10 m/s

1.4420 10v

π × = = × ×

8.21 km/sv =





2

2 or rGMm mv GMF v

r rr= = =


2 or 2GMv r rr

τ π τ π= =

Solving for r, 1/32

24GMr τ

π

=

For earth, 6 26370 km 6.37 10 m, 9.81 m/sR g= = × =

( )( )22 6 12 3 29.81 6.37 10 398.06 10 m /sGM gR= = × = ×

Data: 120 min 7200 sτ = =

( )( )1/ 3212

62

398.06 10 72008.055 10 m

4r

π

× = = ×

(a) altitude 61.685 10 mh r R= − = × 1685 kmh =

(b) 6

66.37 10cos 0.7908

8.055 10Rr

θ ×= = =

×

37.74θ = °

( )( )75.48 72002 1509.6 s360 360ABt θ τ= = =

°

25.2 minABt =




2 2n nGMm GMma a

r r= =

( )( )9 4 4 21 234.4 10 ft /lb s 5.03 10 lb s /ftGM −= × ⋅ × ⋅

12 3 2173.032 10 ft /s= ×

(a) At the surface of the moon, na g=

61080 mi 5.7024 10 ftr R= = = ×

( )( )

( )9 4 4 21 2

2 26

34.4 10 ft /lb s 5.03 10 lb s /ft

5.7024 10 ft

GMgR

−× ⋅ × ⋅= =

×

25.32 ft/sg =

(b) Orbit of space vehicle.

61080 200 1280 mi 6.7584 10 ftr = + = = ×

2

2nv GMar r

= =

GMvr

=

3/22 2r r

v GMπ πτ = =

( )3/26

312

2 6.7584 108.3923 10 s

173.032 10

πτ

×= = ×

× 2.33 hτ =




2

2 nGMm mvma

rr= = 2 GMv

r=

2 rv πτ

= 2 2

24 r GM

rπτ

=

3

2 2 constant4

r GMτ π

= =

Tethys: 3 6183.3 10 mi 967.8 10 ftr = × = ×

31.888 days 163.12 10 sτ = = ×

( )( )

3615 3 2

2 23

967.8 1034.068 10 ft /s

4 163.12 10

GMπ

×= = ×

×

Rhea: 34.52 days 390.53 10 sτ = = ×

3 224

GMr τπ

=

( )( )215 3 27 334.068 10 390.53 10 5.1958 10 ft= × × = ×

(a) 91.732 10 ftr = × 3328 10 mir = ×

(b) Mass of Saturn.

2

24

4GMM

Gπ

π =

( )2 15

249

4 34.068 1039.1 10

34.4 10

π−

×= = ×

×

24 239.1 10 lb s /ftM = × ⋅




Parabola AB.

66370 km 6.37 10 mR = = ×

66370 960 7330 km 7.33 10 mAr = + = = ×

66370 8300 14670 km 14.67 10 mBr = + = = ×

( )A A B Bmr v mr v θ=

( )( )( )6 3

36

7.33 10 10.4 105.196 10 m/s

14.67 10A A

BB

r vvrθ

× ×= = = ×

×

0, 0A Ax y= =

614.67 10 mB Bx r= = ×

67.33 10 mB Ay r= = ×

For a parabolic trajectory, 2y kx= from which 2B

B

ykx

=

Differentiating with respect to x, 222 B

B

dy y xkxdx x

= =

At point B, ( ) ( )6

6

2 7.33 102tan 1.000014.67 10

B

B

dy ydx x

φ×

= = = =×

45φ = °

( ) 335.196 10 7.349 10 m/s

sin 45 sin 45B

Bv

v θ ×= = = ×

° °

37.35 10 m/sB = ×v 45°




For earth, 63960 mi 20.909 10 ftR = = × ( ) ( )( )22 6 15 3 2earth 32.2 20.909 10 14.077 10 ft /sGM gR= = × = ×

For sun, ( ) ( )( )3 15 21 3 2sun 332.8 10 14.077 10 4.6849 10 ft /sGM = × × = ×

For circular orbit of earth, 6 992.96 10 mi 490.8 10 ftEr = × = ×

( ) 213sun

94.6849 10 97.70 10 ft/s490.8 10E

E

GMv

r×

= = = ××

For transfer orbit AB, 6 9, 141.5 10 mi 747.12 10 ftA E B Mr r r r= = = × = ×

( ) ( )( )397.70 10 1.83 5280 107.36 ft/sA E Av v v= + ∆ = × + =

A A B Bmr v mr v=

( )( )9 3

3

490.8 10 107.36 1070.527 ft/s

747.12 10A A

BB

r vvr

× ×= = =

×

For circular orbit of Mars, 6 9141.5 10 mi = 747.12 10 ftMr = × ×

( ) 213sun

94.6849 10 79.187 10 ft/s747.12 10M

M

GMv

r×

= = = ××

Speed increase at B.

( ) 3 3 379.187 10 70.527 10 8.660 10 ft/sM BBv v v∆ = − = × − × = ×

( ) 1.640 mi/sBv∆ =




Circular orbits: GMvr

=

61400 mi 7.392 10 ftAr = = ×

( ) ( )( )9 213

61

34.4 10 5.03 104.8382 10 ft/s

7.392 10Av−× ×

= = ××

61300 mi 6.864 10 ftBr = = ×

( ) ( )( )9 213

62

34.4 10 5.03 105.0208 10 ft/s

6.864 10Bv−× ×

= = ××

(a) Transfer orbit AB.

( ) ( ) ( ) 3 32 1 4.8382 10 86 4.7522 10 ft/sA A Av v v= + ∆ = × − = ×

( ) ( )2 1A A B Bmr v mr v=

( ) ( ) ( )( )6 332

31

7.392 10 4.7522 105.1178 10 ft/s

6.864 10A A

BB

r vv

r

× ×= = = ×

×

( ) 31 5.12 10 ft/sBv = ×

(b) Speed change at B.

( ) ( ) ( ) 3 32 1 5.0208 10 ft/s 5.1178 10 ft/s 97.0 ft/sB B Bv v v∆ = − = × − × = −

Speed reduction at B. 97.0 ft/sBv∆ =




6 66370 km 6.37 10 m, 6370 610 6980 km 6.98 10 mA DR r r= = × = = + = = ×

6 66370 290 6660 km 6.66 10 m, 6900 km 6.90 10 mB Cr r= + = = × = = ×

( )( )( )262

36circ

9.81 6.37 107.55173 10 m/s

6.98 10AA

gRvr

×= = = ×

×

( )( )( )262

36circ

9.81 6.37 107.73102 10 m/s

6.66 10BB

gRvr

×= = = ×

×

For path BC.

( ) ( ) 3 3circ 7.73102 10 85 7.81602 10 m/sB B Bv v v= + ∆ = × + = ×

( )1B B C Cmr v mr v=

( ) ( )( )6 33

61

6.66 10 7.81602 107.54416 10 m/s

6.90 10B B

CC

r vvr

× ×= = = ×

×

For path CD.

( ) ( ) ( ) 3 32 1 7.54416 10 79 7.62316 10 m/sC C Cv v v= + ∆ = × + = ×

( )2C C D Dmr v mr v=

( ) ( )( )6 3

326

6.90 10 7.62316 107.53579 10 m/s

6.98 10C C

DD

r vv

r

× ×= = = ×

×

( ) ( ) 3 3circ 7.55173 10 7.53579 10 15.94 m/sA DDv v v∆ = − = × − × =

( ) 15.94 m/sDv∆ =




Masses: 22.6 0.08075 lb s /ft32.2A Bm m= = = ⋅

Let y be the position coordinate of B, positive upward with origin at O.

Constraint of the cord: constant or r y y r− = =

(a) Kinematics: ( ) ( ) 2 and B Ay ra y r a r rθ= = = −

Collar B: :y B B B B BF m a T W m y m rΣ = − = = (1)

Collar A: ( ) ( )2:r A A ArF m a T m r rθΣ = − = − (2)

Adding (1) and (2) to eliminate T,

( ) 2B A B AW m m r m rθ− = + +

( )( )( ) ( )22

/ rod0.08075 0.6 10 2.6

0.08075 0.08075A B

AA B

m r Wa rm mθ −−

= = =+ +

2/ rod 13.90 ft/sA =a

From (1), ( ) ( )( )0.08075 13.90 32.2BT m r g= + = +

3.72 lbT =

(b) Conservation of angular momentum of collar A: ( ) ( )0 02 1H H=

1 1 2 2( ) ( )A Am r v m r vθ θ=

2 21 1 1

22 2

( ) (0.6) (10)( ) 4.000.9

r v rvr rθ

θθ

= = = =

( )2 4.00 ft/svθ =




Masses: 22.6 0.08075 lb s /ft32.2A Bm m= = = ⋅

(a) Conservation of angular momentum of collar A: ( ) ( )0 02 1H H=

1 1 2 2( ) ( )A Am r v m r vθ θ= 2 2

1 1 1 12

2 2

( ) (0.6) (12)( ) 3.61.2

r v rvr rθ

θθ

= = = =

( )2 3.60 ft/svθ =

( )22

3.6 3.00 rad/s1.2A

vrθθ = = =

(b) Let y be the position coordinate of B, positive upward with origin at O.

Constraint of the cord: constant or r y y r− = =

Kinematics:

( ) ( ) 2 and B Ay ra y r a r rθ= = = −

Collar B: :y B B B B BF m a T W m y m rΣ = − = = (1)

Collar A: ( ) ( )2:r A A ArF m a T m r rθΣ = − = − (2)

Adding (1) and (2) to eliminate T,

( ) 2B A B AW m m r m rθ− = + +

( )( )( ) ( )222

/ rod0.08075 1.2 3.00 2.6

10.70 ft/s0.08075 0.08075

A BA

A B

m r Wa rm mθ −−

= = = = −+ +

( ) ( )( )0.08075 10.70 32.2BT m r g= + = − + 1.736 lbT =

2/ rod 10.70 ft/sAa = radially inward.




Since friction and mass of the rod are neglected, the resultant force acting on the collar is the spring force, which is a central force. The angular momentum of the collar about the shaft is constant.

20 0 0 0mrv mr v mrθ θ= =

(a) 2 2

0 0 (0.150) (12)0.600

rvrθθ

= = 0.450 m/svθ =

Calculate the spring force.

Elongation: 0.600 0.750 0.150 me = − = −

Force: (5)( 0.150) 0.750 NrF ke= − = − − =


r rF ma=

(b) 0.7500.300

rr

Fam

= = 22.50 m/sra =

0Fθ =

Famθ

θ = 0aθ =

(c) Acceleration of B relative to the rod. 2

2r

va r r rrθθ= − = −

2 2

/(0.450)2.50

0.600B rod rva r arθ= = + = +

2/ 2.84 m/sB roda =




Data from Prob. 12.94.

m = 300 g = 0.300 kg, k = 5 N/m, 0 12 rad/s,θ =

0 0750 mm 0.750 m, 150 mm 0.150 m, 600 mm 0.600 m.= = = = = =l r r

Since friction and mass of the rod are neglected, the resultant force acting on the collar is the spring force, which is a central force. The angular momentum of the collar about the shaft is constant.

2 20 0mrv mr mrθ θ θ= =

20 0

2rrθθ =

(a) Radial component vr of the collar.

Kinematics: 2ra r rθ= −

Newton’s second law: 0( )r rF k r l ma= − − =

Eliminating ar, 20( )kr r r l

mθ− = − −

21 ( )2

dr dr dr dr dr r rdt dr dt dr dr

= = = =

2 20

1 ( ) ( )2

d kr r r ldr m

θ= − −

4 20 0

03 ( )r k r lmr

θ= − −

Separate variables and integrate with respect to r, noting that 00 when .r r r= =

2 2 20 0

0( )2r r kd dr r l dr

r mθ

= − −

0 0

22 2

0 0 030

( )2

rr r

r r

r dr kr r l drmr

θ= − −∫ ∫

continued



0 0

2 4 2 20 0 02

1 1 1 ( )2 22

r r

r r

kr r r lmr

θ = − − −

2 2 2 2 20 0 0 0 02 2

0

1 1 ( ) ( )

= − − − − −

kr r r l r lmr r

θ

4 2 2 22 2

1 1 5(0.150) (12) (0.600 0.750) (0.150 0.750)0.30.150 0.600

= − − − − −

2 23.0375 5.625 8.6625 m /s= + =

2.9432 m/sr = 2.94 m/srv =

(b) Value of .θ

2 0Fa r rmθ

θ θ θ= + = =

2rrθθ = −

where 2 2

0 02 2

(0.150) (12) 0.750 rad/s(0.600)

rrθθ = = =

(2)(2.9432)(0.750)0.600

θ = − 27.36 rad/sθ = −




2

20 0 0

1 2 cos sin cos, ,du d uur r d r rd

θ θ θθ θ

−= = = =

2

2 2 20

2d u Furd mh uθ

+ = = by Eq. ( )12.37

Solving for F, 2 2 2

20 0

2 2mh u mhFr r r

= =

Since m, h, and 0r are constants, F is proportional to 21r

, or inversely proportional to 2.r




2

0 0 0

1 6cos 5 6sin 6cos, ,du d uur r d r d r

θ θ θθ θ

−= = = − =

2

2 2 20

5d u Furd mh uθ

+ = − = by Eq. ( )12.37 .


20 0

5 5mh u mhFr r r

= − = −


, or inversely proportional to 2.r The minus sign

indicates that the force is repulsive, as shown in Fig. P12.97.




From Fig. P12.98, 0sin , cosx r y r rθ θ= = −

But ( )2 22 2 2

00 0 0

1 cossin or cos4 4 4

rx ry r rr r r

θθθ−

= − = =

( )2

20

0

1 cos cos 04

r r rr

θ θ −

+ − =

Solving the quadratic equation for r,

( )

( ) ( )

220

020

002 2

2 1 coscos cos 441 cos

2 1 cos2 cos 11 cos 1 cos

rr rr

rr

θθ θθ

θθ

θ θ

− = − ± − − −

−= − ± =

− −

since 0.r > Simplifying gives

0

0

2 1 1 cos or 1 cos 2

rr ur r

θθ

+= = =+

(1)

2

20 0

sin cos and 2 2

du d ud r rd

θ θθ θ

= − = −

2

2 2 20

12

d u Furd mh uθ

+ = =


20 02 2

mh u mhFr r r

= =


, or inversely proportional to 2.r

By Eq. ( )12.39 ,′ ( )21 1 cosGM ur h

ε θ= + =

Comparing with (1) shows that 1ε = and

20

12

GMrh

= 02h GMr=




0

1 1 bu er r

θ−= =

0

bdu b ed r

θ

θ−= −

2 2

20

bd u b erd

θ

θ−=

2 2

2 2 20

1 bd u b Fu erd mh u

θ

θ−+

+ = =

2 2 2

0

( 1) −+= bb mh u e

rθF

2 2 2 2 2

3( 1) ( 1)b mh u b mh

r r+ +

= =

Since b, m, and h are constants, F is proportional to 31 ,r

or inversely proportional to 3.r




For a parabolic trajectory, the velocity 0v at the perigee is equal to the escape velocity.

00

2esc

GMv vr

= =

Solving for M, 2

0 0

2r vM

G=

Data: 3 60 350 10 km 350 10 mr = × = ×

30 26.9 km/s 26.9 10 m/sv = = ×

12 3 266.73 10 m /kg sG −= × ⋅

6 3 2

12(350 10 )(26.9 10 )

(2)(66.73 10 )M −

× ×=

× 271.898 10 kgM = ×




( ) ( )9 4 4 21 2 15 3 234.4 10 ft /lb 334 10 lb s /ft 11.490 10 ft /ssGM −= × × ⋅ = ×⋅

63761 mi 19.858 10 ftR = = ×

60 3761 175 3936 mi 20.782 10 ftr = + = = ×

(a) Velocity of probe as it approaches A.

( )( )153

0 60

2 11.490 102 33.252 10 ft/s20.782 10

GMvr

×= = = ×

×

or 0 6.30 mi/sv =

15

3cir 6

0

11.490 10 23.513 10 ft/s20.782 10

GMvr

×= = = ×

×

(b) Decrease velocity at A.

3circ 0 9.739 10 ft/s 1.845 mi/sv v v∆ = − = − × = −

or 1.845 mi/sv∆ =




For earth, 66.37 10 mR = × 29.81 m/sg =

( )( )22 6 12 3 29.81 6.37 10 398.06 10 m /sGM gR= = × = ×

For Io, ( )( )12 12 3 20.01496 398.06 10 5.955 10 m /sGM = × = ×

6 30 02.820 10 m, 15 10 m/sr v= × = ×

9 20 0 42.3 10 m /sh r v= = ×

( ) ( )2 20

1 11 cos and 1GM GMr rh h

ε θ ε= + = +

( )

( )( )

292

6 120

42.3 101 106.5

2.820 10 5.955 10h

r GMε

×+ = = =

× ×

105.5ε =




Earth: 66.37 10 mR = × 29.81 m/sg =

2 6 2 12 3 2(9.81)(6.37 10 ) 398.06 10 m /seGM gR= = × = ×

Jupiter: 12 15 3 2(318)(398.06 10 ) 126.583 10 m /sGM = × = ×

Orbit of satellite. 923.6 10 ma = ×

2 20 1

1 1(1 ) (1 )

GM GM

r rh hε ε= + = −

2 2

0 1(1 ) (1 )

h hr r

GM GMε ε= =

+ −

2

0 1 2

22

(1 )

ha r r

GM ε= + =

−

2 2 2 15 9(1 ) (1 0.45 )(126.583 10 )(23.6 10 )h GMaε= − = − × ×

27 4 22.3824 10 m /s= ×

1248.810 10 m/sh = ×

279

0 15

2.3824 1012.980 10 m

(126.583 10 )(1 0.45)r

×= = ×× +

279

1 15

2.3824 1034.220 10 m

(126.583 10 )(1 0.45)r

×= = ×× −

123

max 90

48.810 103.760 10

12.980 10

hv

r

×= = = ××

max 3760 m/sv = !

123

min 91

48.810 101.426 10

34.220 10

hv

r

×= = = ××

min 1426 m/sv = !




Using Eq. ( )12.39 , 2 21 1cos and cos .A BA B

GM GMC Cr rh h

θ θ= + = +

But 180 , so that cos cos .B A A Bθ θ θ θ= + ° = −

Adding, 21

1 1 1 1 2A B o

GMr r r r h

+ = + =




For earth, 63960 mi 20.909 10 ftR = = × ( )( )2 6 15 3 232.2 20.909 10 14.077 10 ft /sGM gR= = × = ×

63960 11870 15830 mi 83.582 10 ftAr = + = = ×

15

3circ 6

14.077 1012.978 10 ft/s

83.582 10A

GMv

r

×= = = ××

3esc circ

22 18.353 10 ft/s

A

GMv v

r= = = ×

(a) Increase in speed at A.

3esc circ 5.375 10 ft/sv v v∆ = − = × 35.38 10 ft/sv∆ = × !!!!

Elliptical orbit with 63960 3960 7920 mi 41.818 10 ft.Br = + = = ×

Using Eq. ( )12.37 , 2 2

1 1cos and cos .A B

A B

GM GMC C

r rh hθ θ= + = +

But 180 , so that cos cosB A A Bθ θ θ θ= + ° = −

Adding, 2

1 1 2A B

A B A B

r r GM

r r r r h

++ = =

( )( )( )( )15 6 6

6

9 2

2 14.077 10 83.582 10 41.818 102

125.4 10

885.848 10 ft /s

A B

A B

GMr rh

r r

× × ×= =

+ ×

= ×

9

36

885.848 1010.599 10 ft/s

83.582 10A

A

hv

r

×= = = ××

(b) Decrease in speed. 3circ 2.379 10 ft/sAv v v∆ = − = × 2380 ft/sv∆ = !!!!

(c) 1 1

cos cos 2A BB A

B A A B

r rC C C

r r r rθ θ−− = = − =

( )( )( )6

9 16 6

41.764 105.974 10 ft

2 2 83.582 10 41.818 10A B

A B

r rC

r r− −− ×= = = ×

× ×

By Eq. (12.40), ( )( )29 92

15

5.974 10 885.848 10

14.077 10

Ch

GMε

−× ×= =

×

0.333ε = !!!!




For earth, 66370 km 6.370 10 mR = = ×

( )( )22 6 12 3 29.81 6.37 10 398.06 10 m /sGM gR= = × = × 66370 10,000 16370 km 16.370 10 m= + = = ×Ar

66370 140,000 146,370 146.37 10 mBr = + = = × For a circular orbit with 0 ,Ar r=

123

circ 60

398.06 10 4.931 10 m/s16.370 10

GMvr

×= = = ×

×

Elliptic orbit.

Using Eq. (12.39), 2 21 1cos and cos .A BA B

GM GMC Cr rh h

θ θ= + = +

But 180 , so that cos cosB A B Aθ θ θ θ= + ° = −

Adding, 21 1 2A B

A B A B

r r GMr r r r h

++ = =

( )( )( )( )12 6 6

6

9 2

2 398.06 10 16.370 10 146.37 102162.74 10

108.27 10 m /s

× × ×= =

+ ×

= ×

A B

A B

GMr rhr r

93

6108.27 10 6.614 10 m/s16.370 10A

A

hvr

×= = = ×

×


3circ 1.683 10 m/sA Av v v∆ = − = × 31.683 10 m/sAv∆ = ×

(b) Speed of observatory at B.

9

6108.27 10146.37 10B

B

hvr

×= =

× 37.40 10 m/sBv = ×




For earth, 66.37 10 mR = ×

( )( )22 6 12 3 29.81 6.37 10 398.06 10 m /sGM gR= = × = ×

For sun: ( )( )3 12 18 3 2332.8 10 398.06 10 132.474 10 m /sGM = × × = ×

For elliptic orbit AB, 9 9325 10 m, 148 10 mA Br r= × = ×

Using Eq. (12.39), 2 21 1cos and cos .A BA B AB

GM GMC Cr rh h

θ θ= + = +


Adding, 21 1 2A B

A B A B AB

r r GMr r r r h

++ = =

( )( )( )( )18 9 9

9

15 2

2 132.474 10 325 10 148 102473 10

5.1907 10 m /s

A BAB

A B

GMr rhr r

× × ×= =

+ ×

= ×

(a) ( )15

391

5.1907 10 15.971 10325 10

ABA

A

hvr

×= = = ×

× ( ) 3

1 15.97 10 m/sAv = ×

For transfer orbit ,AB′ 9 137.6 10 mBr ′ = ×

continued



( )( )( )( )18 9 9

9

15 2

2 132.474 10 325 10 137.6 102462.6 10

5.0609 10 m /s

A BAB

A B

GMr rhr r

′′

′

× × ×= =

+ ×

= ×

( )15

392

5.0609 10 15.572 10 m/s325 10

ABA

A

hvr

′ ×= = = ×

×

(b) Decrease of speed at A.

( ) ( ) ( )1 2 399 m/sA A Av v v∆ = − = 399 m/sAv∆ =

( )15

391

5.0609 10 36.780 10 m/s137.6 10

ABB

B

hvr

′

′

×= = = ×

×

For elliptic orbit ,B A′ ′ 9264.7 10 mAr ′ = ×

( )( )( )( )18 9 9

9

15 2

2 132.474 10 137.6 10 264.7 102402.3 10

4.8977 10 m /s

B AB A

B A

GMr rhr r

′ ′′ ′

′ ′

× × ×= =

+ ×

= ×

( )15

392

4.8977 10 35.594 10 m/s137.6 10

B AB

B

hvr′ ′

′′

×= = = ×

×

Decrease in speed at :B′

( ) ( ) 31 2 1.186 10B Bv v′ ′− = × 31.186 10 m/sBv∆ = ×




For Earth, ( )( )22 6 12 3 2earth 9.81 6.37 10 398.06 10 m /sGM gR= = × = ×

For Venus, 12 3 2earth0.82 326.41 10 m /sGM GM= = ×

For a parabolic trajectory with 615 10 mAr = ×

( )( )( )12

3esc 61

2 326.41 1026.5971 10 m/s

15 10A

A

GMv v

r

×= = = = ×

×

First transfer orbit AB. 6300 10 mBr = ×

At point A, where 180θ = °

2 2

1cos 180= + ° = −

A AB AB

GM GMC C

r h h (1)

At point B, where 0θ = °

2 2

1cos 0

B AB AB

GM GMC C

r h h= + = + (2)

Adding, 2

1 1 2B A

A B A B AB

r r GM

r r r r h

++ = =

Solving for hAB,

( )( )( )( )12 6 6

6

2 326.41 10 15 10 300 102

315 10A B

ABB A

GMr rh

r r

× × ×= =

+ ×

9 296.571 10 m /s= ×

continued



( )9

362

96.571 106.4381 10 m/s

15 10AB

AA

hv

r

×= = = ××

( )9

61

96.571 10321.90 m/s

300 10AB

BB

hv

r

×= = =×

Second transfer orbit BC. 69000 km 9 10 mCr = = ×

At point B, where 0.θ =

2 2

1cos 0

B BC BC

GM GMC C

r h h= + = +

At point C, where 180θ = °

2 2

1cos 180

C BC BC

GM GMC C

r h h= + ° = −

Adding, 2

1 1 2B C

B C B C BC

r r GM

r r r r h

++ = =

( )( )( )( )12 6 6

6

2 326.41 10 300 10 9 102

309 10B C

BCB C

GMr rh

r r

× × ×= =

+ ×

9 275.5265 10 m /s= ×

( )9

62

75.5265 10251.76 m/s

300 10BC

BB

hv

r

×= = =×

( )9

361

75.5265 108.3918 10 m/s

9 10BC

CC

hv

r

×= = = ××

Final circular orbit. 69 10 mCr = ×

( )12

362

326.41 106.0223 10 m/s

9 10C

C

GMv

r

×= = = ××



Speed reductions.

(a) At A. ( ) ( ) 3 31 2

6.5971 10 6.4381 10A Av v− = × − ×

159.0 m/sAv∆ = !!!!

(b) At B. ( ) ( )1 2321.90 251.76B Bv v− = −

70.1 m/sBv∆ = !!!!

(c) At C. ( ) ( ) 3 31 2

8.3918 10 6.0223 10C Cv v− = × − ×

32.37 10 m/sCv∆ = × !!!!




Data from Problem 12.108: earth9000 km, 0.82Cr M M= =

For Earth, ( )( )22 6 12 3 2earth 9.81 6.37 10 398.059 10 m /sGM gR= = × = ×

For Venus, 12 3 2earth0.82 326.409 10 m /sGM GM= = ×

Transfer orbit AB: 66500 m/s, 15 10 mA Av r= = ×

( )( )6 9 215 10 6500 97.5 10 m /sAB A Ah r v= = × = ×

At point A, where 180θ = °

2 21 cos 180A AB AB

GM GMC Cr h h

= + ° = −

At point B, where 0θ = °

2 21 cos0B AB AB

GM GMC Cr h h

= + = +

Adding, 21 1 2

A B AB

GMr r h

+ =

( )( )

( )12

9 12 2 69

2 326.409 101 1 1 2.0057 10 m15 1097.5 10B AAB

GMr rh

− −×2

= − = − = ×××

(a) Radial coordinate rB. 6498.56 10 mBr = × 3499 10 kmBr = ×

( )9

6197.5 10 195.56 m/s

498.56 10AB

BB

hvr

×= = =

×



Second transfer orbit BC. 69 10 mCr = ×

At point B, where 0θ =

2 21 cos0B BC BC

GM GMC Cr h h

= + = +

At point C, where 180θ = °

2 21 cos 180C BC BC

GM GMC Cr h h

= + ° = −

Adding, 21 1 2B C

B C B C BC

r r GMr r r r h

++ = =

( )( )( )( )12 6 6

6

2 326.409 10 498.56 10 9 102507.56 10

B CBC

B C

GMr rhr r

× × ×= =

+ ×

9 275.968 10 m /s= ×

( )9

6275.968 10 152.37 m/s498.56 10

BCB

B

hvr

×= = =

×

( )9

361

75.968 10 8.4409 10 m/s9 10

BCC

C

hvr

×= = = ×

×

Circular orbit with 69 10 m.Cr = ×

( )12

362

326.409 10 6.0223 10 m/s9 10C

C

GMvr

×= = = ×

×

(b) Speed reductions at B and C.

At B ( ) ( )1 2 195.56 – 152.37B Bv v− =

43.2 m/sBv∆ =

At C ( ) ( ) 3 31 2 8.4409 10 6.0223 10C Cv v− = × − ×

32.42 10 m/sCv∆ = ×




For earth, 63960 mi 20.909 10 ftR = = × ( )( )22 6 15 3 232.2 20.909 10 14.077 10 ft /sGM gR= = × = ×

For Mars, ( )( )15 15 3 20.1074 14.077 10 1.51188 10 ft /sGM = × = × 65625 mi 29.7 10 ft,Ar = = ×

6112500 mi 594 10 ftBr = = × For the parabolic approach trajectory at A,

( )( )( )15

361

2 1.51188 102 10.0901 10 ft/s29.7 10A

A

GMvr

×= = = ×

×

First elliptic transfer orbit AB.

Using Eq. (12.39), 2 21 1cos and cos .A BA BAB AB

GM GMC Cr rh h

θ θ= + = +


Adding, 21 1 2A B

A B A B AB

r r GMr r r r h

++ = =

( )( )( )( )15 6 6

6

2 1.51188 10 29.7 10 594 102623.7 10

A BAB

A B

GMr rhr r

× × ×= =

+ ×

9 2292.45 10 m /sABh = ×

( ) ( )6 6 61 1 29.7 10 594 10 311.85 10 ft2 2A Ba r r= + = × + × = ×

( )( )6 6 629.7 10 594 10 132.82 10 ftA Bb r r= = × × = ×

Periodic time for full ellipse: 2 abhπτ =

For half ellipse AB, 12AB

abh

πτ τ= =

( )( )6 6

39

311.85 10 132.82 10444.95 10 s

292.45 10AB

πτ

× ×= = ×

× 123.6 hABτ =




From Keplers third law,

2 3

H H

E E

T aT a

=

Solve for aH.

2/3 2/3

76 years 17.9421 year

HH E E E

E

Ta a r rT

= = =

Also, ( )min max12Ha r r= +

Solve for rmax.

( )( )max min12 2 17.9422H E Er a r r r= − = −

35.3844 Er=

or

( )( )6max 35.3844 92.6 10 mir = ×

9max 3.28 10 mir = ×




For earth, 63960 mi 20.909 10 ftR = = × , 232.2 ft/sg =

( )( )22 6 15 332.2 20.909 10 14.077 10 ft /sGM gR= = × = ×

For circular orbit of satellite, 6

0 3960 310 4270 mi 22.546 10 ft/sr = + = = ×

153

0 60

14.077 10 24.988 10 ft/s22.546 10

GMvr

×= = = ×

×

( )( )630

0 30

2 22.546 102 5.6692 10 s24.988 10

rv

ππτ×

= = = ××

For elliptic orbit of spacecraft it is given that

30

3 8.5038 10 s2

τ τ= = ×

( )1 , 2 A B A Ba r r b r r= + =


GM GMC Cr rh h

θ θ= + = +


Adding, 2

2 21 1 2 2 or A B

A B A B

r r a GM GMbhr r r r ab h

++ = = = =

Periodic time: 3 / 2

2

2 2 2ab ab a ah GMGMb

π π πτ = = =

( )( )215 323 21 3

2 2

14.077 10 8.5038 1025.786 10 ft

4 4GMa τπ π

× ×= = = ×



6 6029.543 10 ft, 22.546 10 ftAa r r= × = = ×

6 62 36.541 10 ft, 28.703 10 ftB A A Br a r b r r= − = × = = ×

( )( )6 69 2

3

2 29.543 10 28.703 102 626.54 10 ft /s8.5038 10

abhππ

τ

× ×= = = ×

×

93

6626.54 10 27.789 10 ft/s22.546 10A

A

hvr

×= = = ×

×


3 30 27.789 10 24.988 10A Av v v∆ = − = × − × 32.80 10 ft/sAv∆ = ×

(b) Periodic time for elliptic orbit.

3As calculated above 8.5038 10 sτ = × 141.7 minτ =




For earth’s orbit about the sun,

3/ 2 3/ 2

0 00 0

2 2 2, or E E E

E

GM R R Rv GMR v GM

π π πττ

= = = = (1)

For the comet Hyakutake,

( ) ( ) 1 02 20 1

1 1 11 , 1 , 1

GM GM r rr rh h

εε εε

+= = + = − =−

( ) 00 1 0 1 0

1 1, 2 1 1

ra r r b r r rεε ε

+= + = = =

− −

( )0 1h GMr ε= +

( )( ) ( )

( ) ( )

( )

( )( )

1/ 220

3/ 20

3/ 2 3/ 20 0 0

3/ 233/ 2

3/ 20

03/ 2

3/ 2 30 03/ 2

2 121 1

2 22 11

1 1

10.230 91.8 10 1 0.999887

E

E

rabh GMr

r rRGM

rR

π επτε ε

π π τπ εε

τε

τ τ

+= =

− +

= =−−

=

−

= = ×−

( )( )30Since 1 yr, 91.8 10 1.000τ τ= = × 391.8 10 yrτ = ×




For earth, 6 23960 mi 20.909 10 ft, 32.2 ft/sR g= = × =

( )( )22 6 15 3 232.2 20.909 10 14.077 10 ft /sGM gR= = × = ×

At point A, 64560 mi 24.077 10 ftAr = = ×

36.48 mi/s 34.214 10 ft/sAv = = ×

from which 9 2823.78 10 ft /sA Ah r v= = ×

For trajectory BAC, ( )21 1 cos with 1GMr h

ε θ ε= + =

At point A, 0 while at and , 90B Cθ θ= = ± °

2

21 1 or 2B C AB C

GM hr r rr r GMh

= = = = =

As the spacecraft travels from B to C, the area swept out is a parabolic area A.

( ) ( )22 6 15 22 8 8 24.077 10 15.458 10 ft3 3 3B C A AA r r r r= + = = × = ×

1 1 1 or 2 2 2

dA h A h dt htdt

= = =∫

( )( )153

9

2 15.458 102 3.753 10 s823.78 10BC

Ath

×= = = ×

× 1.043 hBCτ =




For earth, 2 632.2 ft/s , 3960 mi 20.9088 10 ftg R= = = ×

( )( )22 6 15 3 232.2 20.9088 10 14.077 10 ft /sGM gR= = × = ×

For the orbit, 60 3960 182 4142 mi 21.8698 10 ftr = + = = ×

( )20

11

GM

r hε= + ( )2

1

11

GM

r hε= −

( )6 61 0

1 1.035621.8698 10 23.4844 10 ft

1 0.9644

+= − × = ×−

r rεε

( ) 60 1

122.6770 10 ft

2a r r= + = ×

0 1 22.6627 10 ftb r r 6= = ×

( ) ( )( )( )15 601 1.0356 14.077 10 21.8698 10h GMrε= + = × ×

9 2564.64 10 ft /s= ×

( )( )6 6

9

2 22.6770 10 22.6627 102

564.64 10

ab

h

ππτ× ×

= =×

35.7188 10 s= × 95.3 minτ = !




For the circular orbit, 0 Ar r nR= =

00

GM GMvr nR

= =

The crash trajectory is elliptic.

2

0AGMv v

nRββ= =

2A A Ah r v nRv n GMRβ= = =

2 21GM

h nRβ=

( )2 21 1 cos1 cosGMr h nR

ε θε θβ+= + =

At point A, 180θ = °

2 22

1 1 1 or 1 or 1Ar nR nR

ε β ε ε ββ−= = = − = −

At impact point B, θ π φ= −

1 1Br R=

( )2 2

1 cos1 1 cosR nR nR

ε π φ ε φβ β

+ − −= =

2 22

21 1cos 1 or cos

1n nn β βε φ β φε β

− −= − = =

−

( ) ( )1 2 2cos 1 / 1nφ β β− = − −



Chapter 12, Solution 117

For earth, 63960 mi 20.909 10 ftR = = ×

( )( )22 6 15 3 232.2 20.909 10 14.077 10 ft /sGM gR= = × = ×

For the moon, ( )( )15 12 3 20.01230 14.077 10 173.149 10 ft /sGM = × = ×

For elliptic orbit AB, 6 61110 mi 5.861 10 ft, 2240 mi 11.827 10 ftA Br r= = × = = ×


GM GMC Cr rh h

θ θ= + = +

But 180 , so that cos cosB A B Aθ θ θ θ= + ° = −

Adding, 21 1 2A B

A B A B

r r GMr r r r h

++ = =

( )( )( )( )12 6 69 2

6

2 173.149 10 5.861 10 11.827 102 36.839 10 ft /s17.688 10

A B

A B

GMr rhr r

× × ×= = = ×

+ ×

( )9

361

36.839 10 3.1148 10 ft/s11.827 10

ABB

A

hvr

×= = = ×

×

For crash trajectory BC, ( )21 1 cosGMr h

ε θ= +

At B, ( )21180 , , 1BB BC

GMr rr h

θ ε= ° = = − (1)

At C, ( )2170 , , 1 cos70CC BC

GMr rr h

θ ε= ° = = + ° (2)



Dividing Eq. (2) by Eq. (1),

( )1 cos70 / 1 or

1 / cos70B B C

C B C

r r rr r r

ε εε

+ ° −= =

− + °

2240 /1080 1 0.444552240 /1080 cos70

ε −= =

+ °

From Eq. (1), ( ) ( )( )( )12 6 9 21 173.149 10 0.55545 11.827 10 33.726 10 ft /sBC Bh GM rε= − = × × = ×

( )9

362

33.726 10 2.8516 10 ft/s11.827 10

BCB

B

hvr

×= = = ×

×

( ) ( )2 1 263.2 ft/sB B Bv v v∆ = − = − 263 ft/sBv∆ =




For earth, 63960 mi 20.909 10 ftR = = ×

( )( )22 6 15 3 232.2 20.909 10 14.077 10 ft /sGM gR= = × = ×

For the trajectory, 63960 930 4890 mi 25.819 10 ftCr = + = = × 6

66

25.819 1020.909 10 ft, 1.234820.909 10

CA B

A

rr r Rr

×= = = × = =

×

Range A to B: 63700 mi 19.536 10 ftABs = = × 6

619.536 10 0.9343 rad 53.53420.909 10

ABsR

φ ×= = = = °

×

For an elliptic trajectory, ( )21 1 cosGMr h

ε θ= +

At A, ( )21180 153.233 , 1 cos153.233

2 A

GMr h

φθ ε= ° − = ° = + ° (1)

At C, 180θ = ° , ( )21 1C

GMr h

ε= − (2)

Dividing Eq. (1) by Eq. (2), 1 cos153.233 1.2348

1C

A

rr

εε

+ °= =

−

1.2348 1 0.68661.2348 cos153.233

ε −= =

+ °

From Eq. (2), ( )1 Ch GM rε= −

( )( )( )15 6 9 214.077 10 0.3134 25.819 10 337.500 10 ft /sh = × × = ×

(a) Velocity at C. 9

36

337.5 10 13.07 10 ft/s25.819 10C

C

hvr

×= = = ×

× 2.48 mi/sCv =

(b) Eccentricity of trajectory. 0.687ε =




Radius of Earth 63960 mi 20.9088 10 ftR = = ×

At point A, 6altitude 4160 mi 21.9648 10 ftAr R= + = = ×

For Earth, ( )( )22 6 15 3 232.2 20.9088 10 14.0771 10 ft /sGM gR= = × = ×

For circular orbit at point A,

( )15

361

14.0771 10 25.3159 10 ft/s21.9648 10A

A

GMvr

×= = = ×

×

After speed reduction at A,

( ) ( )( )3 32 0.94 25.3159 10 23.7969 10 ft/sAv = × = ×

For trajectory AB,

( ) ( )( )6 32 21.9648 10 23.7969 10AB A Ah r v= = × ×

9 2522.6946 10 ft /s= ×

At point A, 0.θ =

2 21 cosA AB AB

GM GMC Cr h h

θ= + = +

Solving for C,

( )15

2 6 29

1 1 14.0771 1021.9648 10 522.6946 10A AB

GMCr h

×= − = −

× ×

9 15.997512 10 ft− −= − ×

continued



At point B, 50θ = °

21 cos 50B AB

GM Cr h

= + °

( )( )

15

29

14.0771 10 5.997512 cos 50522.6946 10

×= + − °

×

947.6698 10 ft− −1= ×

620.9777 10 ftBr = ×

Altitude at B. 368.9 10 ftBr R− = × altitude 13.05 mi=





GM GMC Cr rh h

θ θ= + = +


Adding, 21 1 2

A B

GMr r h

+ =

At points A and B the radial direction is normal to the path.

2 2

2nv ha

rρ ρ= =

But 2

2 2n nGMm mhF ma

r r ρ= = =

21 1 1 1

2 A B

GMr rhρ

= = +

0 1

1 1 1 12 r rρ

= +




( )21 1 cosGMr h

ε θ= +

At A, 0θ = ( ) ( )2

21 1 or

1AA

GM hrr GMh

εε

= = + =+

At B, 180θ = ° ( ) ( )2

21 1 or

1BB

GM hrr GMh

εε

= = − =−

1

0

11

B

A

r rr r

εε

+= =−

(a) 1 0

1 0

r rr r

ε −=

+

693 10 miER = ×

60 0.230 21.39 10 miEr R= = ×

(b) ( )61 0

1 1 0.999887 21.39 101 1 0.999887

r rεε

+ += = ×

− − 9

1 379 10 mir = ×




For an ellipse 2 and A B A Ba r r b r r= + =


GM GMC Cr rh h

θ θ= + = +


Adding, 2 21 1 2 2A B

A B A B

r r a GMr r r r b h

++ = = =

GMh ba

=

By Eq. (12.45),

3/ 22 2 2ab ab a ah b GM GMπ π πτ = = =

2 32 4 a

GMπτ =

For orbits 1 and 2 about the same large mass,

2 3 2 32 21 21 2

4 4 and a aGM GMπ πτ τ= =

Forming the ratio, 2 3

1 1

2 2

aa

ττ

=




By Eq. ( )12.39 ,′ ( )21 1 cosGMr h

ε θ= +

At A, 0θ = ° ( ) ( )2

21 1 or

1AA

GM hrr GMh

εε

= = + =+

At B, 180θ = ° ( ) ( )2

21 1 or

1BB

GM hrr GMh

εε

= = − =−

Adding, ( )2 2

21 1 2

1 1 1A B

h hr rGM GMε ε ε

+ = = + = + − −

But for an ellipse, 2A Br r a+ =

( )2

2221ha

GM ε=

− ( )21h GMa ε= −




(a) Front wheel drive with 0.65 .fN W=

: s FWF ma N ag

µΣ = =

( )( )( ) 232.2 0.80 0.6516.74 ft/ss F Wg Na

W Wµ

= = =

For constant acceleration,

( )( )( )2 3 2 2max 2 2 16.74 1320 44.204 10 ft /sv ax= = = ×

max 210 ft/sv = max 143.2 mi/hv =

(b) Rear wheel drive with 0.42 W.RN =

: s RWF ma N ag

µΣ = =

( )( )( ) 232.2 0.80 0.4210.82 ft/ss R Wg Na

W Wµ

= = =

For constant acceleration,

( )( )( )2 3 2 2max 2 2 10.82 1320 28.563 10 ft / sv ax= = = ×

max 169.0 ft/sv = max 115.2 mi/hv =




0 90 km/h 25 m/sv = =

( )3 3: 60 10 16 10 7900 6800xF ma aΣ = − × − × = +

3276 10 5.170 m/s

14700a ×= − = −

For constant acceleration, ( )2 2

002 2

v v a x x− = −

(a) ( )( )( )

22 20

00 25

2 2 5.170v vx x

a−−

− = =−

0 60.4 mx x− =

( )( )3: 16 10 6800 5.170x HF ma FΣ = − − × = −

(b) ( )( ) 3 36800 5.170 16 10 19.16 10 NHF = − × = × 19.16 kNHF =




/ /, where B A B A B A= +a a a a is directed along the inclined contact surface.

Block B: :x xF maΣ = Σ

/sin 25 cos25B B A B B AT W m a m a− ° = ° +

/30 30cos25 50 30sin 25

32.2 32.2A B Aa a ° + = − °

/0.84439 0.93168 37.321A B Aa a+ = (1)

: cos 25 sin 25y B y AB B B AF m a N W m aΣ = Σ − ° = − °

30 sin 25 30cos25 or 0.39374 27.189

32.2 A AB A ABa N a N ° + = ° + =

(2)

Block A: : cos 25 sin 25 x x AB A AF ma T T N m aΣ = − ° + ° =

( ) ( )55 sin 25 50 1 cos2532.2 A ABa N− ° = − °

1.70807 0.42262 4.6846A ABa N− = (3)

Using (2) and (3) to eliminate ABN and solve for ,Aa

(a) Acceleration of block A. 28.63 ft/sA =a

Substituting for Aa into (1) and solving for / ,B Aa

(b) Acceleration of B relative to A. 2/ 32.2 ft/sB A =a 25°




2180 5.59 lb s /ft32.2

Wmg

= = = ⋅

40 ftρ =

2: cosn n n

mvF ma T W maθρ

Σ = − = =

2cosmvT W θ

ρ= +

(a) At top of swing. 30 , 0vθ = ° =

0 180cos30 155.9 T = + ° = 155.9 lbT =

(b) At bottom of swing. 0 , 18.6 ft/svθ = ° =

( )( )25.59 18.6180 228

40T = + = 228 lbT =




The road reaction consists of normal component N and friction component F. The resultant R makes angle sφ with the normal.

tan 0.65 or 33.02s s sφ µ φ= = = °

0 110 km/h 30.56 m/sv = =

(a) 10 .θ = °

( )0 : cos 0y sF R mgθ φΣ = + − =

( )cos s

mgRθ φ

=+

( ): sinx n s nF ma R maθ φΣ = + =

( )tann sma mg θ φ= +

( )2

max tan sv g

rθ φ= +

( ) ( ) ( ) ( )max tan 50 9.81 tan 10 33.02 21.39 m/ssv rg θ φ= + = ° + ° =

Speed reduction: 0 max 30.56 21.39 9.16 m/sv v v∆ = − = − =

33.0 km/hv∆ =

(b) 5 .θ = − °

The equation derived in part (a) applies also for negative value of .θ

( ) ( ) ( ) ( )max tan 50 9.81 tan 5 33.02 16.16 m/ssv rg θ φ= + = − ° + ° =

Speed reduction: 0 max 30.56 16.16 14.40 m/sv v v∆ = − = − =

51.8 km/hv∆ =




40 , 24 in. 2 ft, 0.35srα µ= ° = = =

,sF Nµ= ± where the upper sign applies for downward impending motion and the lower sign for upward impending motion.

0 : sin cos 0yF N F Wα α+Σ = + − =

( )sin cosSN Wα µ α± =

sin cosS

WNα µ α

=±

2: cos sinx n

W vF ma N Fg r

α αΣ = − =

( )2

cos sinsW vNg r

α µ α =∓

( )2 cos sin cos sinsin cos

s s

s

Nv gr gr

Wα µ α α µ α

α µ α= =

±∓ ∓

For impending motion downward,

( )( )2 2 2cos40 0.35sin 4032.2 2.0 38.25 ft /ssin 40 0.35cos40

v ° − °= =

° + °

6.18 ft/sv =

For impending motion upward,

( )( )2 2 2cos40 0.35sin 4032.2 2.0 170.34 ft /ssin 40 0.35cos40

v ° + °= =

° − °

13.05 ft/sv =

Range of speeds for which the collar will not slide:

6.18 ft/s 13.05 ft/sv≤ ≤




( )0.5 0.3sin mr tπ= + ( )22 2 radt tθ π= −

0.3 cos m/sr tπ π= ( )4 1 rad/stθ π= −

20.3 sinr tπ π= − 24 rad/sθ π=

(a) At 0,t = 0.5 mr = 0θ =

0.94248 m/sr = 12.5664 rad/sθ = −

0r = 212.5664 rad/sθ =

( )( )22 20 0.5 12.5664 78.957 m/sra r rθ= − = − − = −

( )( ) ( )( )( )2 0.5 12.5664 2 0.94248 12.5664 17.4040 m/sa r rθ θ θ= + = + − = −

( )( ): 0.4 78.957 31.6r B r rF m a F= = − = − 31.6 NrF = −

( )( ): 0.4 17.4040 6.96BF m a Fθ θ θ= = − = − 6.96 NFθ = −

(b) At 0.8 s,t = 0.67634 mr = 6.0319 radθ = −

0.76248 m/sr = − 2.5133 rad/sθ = −

21.74040 m/sr = − 212.5664 rad/sθ =

( )( )22 21.74040 0.67634 2.5133 6.0126 m/sra r rθ= − = − − − = −

( )( ) ( )( )( ) 22 0.67634 12.5664 2 0.76248 2.5133 12.332 m/sa r rθ θ θ= + = + − − =

( )( ): 0.4 6.0126 2.41r B r rF m a F= = − = − 2.41 NrF = −

( )( ): 0.4 12.332 4.93BF m a Fθ θ θ= = = 4.93 NFθ =




10 rad/s, 0, 9 in. 0.75 ft, 4 oz 0.25 lbb Wθ θ= = = = = =

Kinematics: 2sin,

cos cosb br r θ θθ θ

= =

( )( ) ( )( )( )

( )

22

2 4

2

3

cos cos sin 2cos sinsincos cos

1 sin with 0

cos

bbr

b

θ θ θ θ θθ θ θθ θ

θθ

θ

− − = +

+= =

( )2 22 2 2 2

3 3

2 2

1 sin 2 sincoscos cos

2 tan sec

r

b b ba r r

b

θ θθ θ θ θθθ θ

θ θ θ

+= − = − =

=

2 22

sin2 0 2 2 tan seccosba r r bθ

θθ θ θ θ θ θθ

= + = + =

(a) Radial and transverse components of effective forces. 2 22: tan secr r r

W WbF a Fg g

θ θ θ= =

( )( )( )2 22 0.25 10tan sec

32.2θ θ= 21.165tan sec lbrF θ θ=

2: tan secW WbF a Fg gθ θ θ θ θ θ= =

( )( )( )22 0.25 10tan sec

32.2θ θ= 1.165tan sec lbFθ θ θ=

(b) Forces P and Q exerted on the pin by the arm OA and the wall of the slot DE, respectively.

cos sin cosy rF P F Fθθ θ θΣ = = −

31.165 tan sec lbθ θ=P θ

cosr rF Q FθΣ = =

2 21.165 tan sec lbθ θ=Q





2

2 or rGMm mv GMF v

r rr= = =


2 or 2GMv r rr

τ π τ π= =

Solving for M, 2 3

24 rMGπτ

=

Data: 6384.5 10 mr = ×

627.32 days 655.68 h 2.3604 10 sτ = = = ×

12 3 266.73 10 m /kg sG −= × ⋅

( )( )( )

32 624

212 6

4 384.5 106.04 10

66.73 10 2.3604 10M

π

−

×= = ×

× ×

246.04 10 kgM = ×




For earth: 6 26370 km 6.37 10 m, 9.81 m/se eR g= = × =

For the given orbit: 66370 4500 10870 km 10.870 10 mer = + = = ×

( ) ( )( )22 6 12 3 29.81 6.37 10 398.06 10 m /se eeGM g R= = × = ×

(a) ( ) ( )( )

( )12

2 26

398.06 10 5401819 N

10.870 10eGM m

Fr

×= = =

× 1.819 kNF =

For gravitational force and a circular orbit, 2

2 or rGMm mv GMF v

r rr= = =

Let τ be the periodic time to complete one orbit. 2 3

2 42 or 2 or GM rv r rr GM

πτ π τ π τ= = =

Since earth orbit and moon orbit have the same periodic time, 3 32 2

2 4 4e m

e m

r rGM GMπ πτ = =

(b) ( ) ( )1/ 3

1/3 6 60.01230 10.870 10 2.509 10 mmm e

e

Mr rM

= = × = ×

2510 kmBr = 2 2 and e e e m m mGM g R GM g R= =

2 26 with 1740 km 1.740 10 me e m m

me m

g R g RG RM M

= = = = ×

(c) ( ) ( )22 6

66.37 100.01230 9.81

1.740 10m e

m ee m

M Rg gM R

×= = ×

21.617 m/smg =






Let r and θ be polar coordinates with the origin lying at the shaft.

Constraint of rod: radians; ; .B A B A B A

θ θ π θ θ θ θ θ θ= + = = = =& & & && && &&

(a) Components of acceleration

Sketch the free body diagrams of the balls showing the radial and

transverse components of the forces acting on them. Owing to

frictionless sliding of B along the rod, ( ) 0.B r

F =

Radial component of acceleration of B.

( ) :r B B r

F m a= ( ) 0B r

a = �

Transverse components of acceleration.

( ) 2A A A

a r r raθ θ θ θ= + =&& & &&&

( ) 2B B B

a r rθ θ θ= +&& && (1)

Since the rod is massless, it must be in equilibrium. Draw its free

body diagram, applying Newton’s 3rd Law.

( ) ( ) ( ) ( )00: 0

A A B B A A A B B BM r F r F r m a r m aθ θ θ θΣ = + = + =

( )2 0A A A B B B Br m r r m r rθ θ θ+ + =&& && &&

2 2

2B B

A A B B

r r

m r m r

θθ −=+

&&&&

At 0,t = 0 so that 0.Br θ= =&&&

From Eq. (1), ( ) 0B

a θ = �

(b) Acceleration of B relative to the rod.

At ( ) ( ) 960, 8 ft/s 96 in./s, 9.6 rad/s

10

A

A

A

v

t v

r

θθ θ= = = = = =&

( )20

B B B rr r aθ− = =&&&

( )( )22 28 9.6 737.28 in./s

B Br r θ= = =&&&

261.4 ft/s

Br =&& �

Continued





(c) Speed of A.

Substituting ( )2 for

dmr rF

dtθθ& in each term of the moment equation

gives

( ) ( )2 20

A A B B

d dm r m r

dt dtθ θ+ =& &

Integrating with respect to time,

( ) ( )2 2 2 2

0 0A A B B A A B B

m r m r m r m rθ θ θ θ+ = +& & & &

Applying to the final state with ball B moved to the stop at C,

( )22 2 2

00

A B A BA C f A B

W W W Wr r r r

g g g gθ θ

+ = +

& &

( ) ( )( ) ( )( )( )( ) ( )( )

( )2 2 22

002 2 2 2

1 10 2 89.6 3.5765 rad/s

1 10 2 16

A A B Bf

A A B C

W r W r

W r W r

θ θ+ +

= = =+ +

& &

( ) ( )( )10 3.5765 35.765 in./sA A ffv r θ= = =&

( ) 2.98 ft/sA fv = �




( )( )22 6 12 3 29.81 6.37 10 398.06 10 m /sGM gR= = × = ×

66370 563 6933 km 6.933 10 mAr = + = = × 66370 121 6491 km 6.491 10 mBr = + = = ×

For the circular orbit through point A, 12

3circ 6

398.06 10 7.5773 10 m/s6.933 10A

GMvr

×= = = ×

×

For the descent trajectory, 3 3

circ 7.5773 10 152 7.4253 10 m/sAv v v= + ∆ = × − = ×

( )( )6 3 9 26.933 10 7.4253 10 51.4795 10 m /sA Ah r v= = × × = ×

( )21 1 cosGMr h

ε θ= +

At point A, 180 ,θ = ° Ar r=

( )21 1A

GMr h

ε= −

( )( )( )

292

12 6

51.4795 101 0.96028

398.06 10 6.933 10A

hGM r

ε×

− = = =× ×

0.03972ε =

( )21 1 cos BB

GMr h

ε θ= +

( )( )( )

292

12 6

51.4795 101 cos 1.02567

398.06 10 6.491 10B

B

hGM r

ε θ×

+ = = =× ×

1.02567 1cos 0.6463Bθ ε−

= =

49.7Bθ = ° 180 130.3BAOB θ= ° − = ° 130.3AOB = °

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

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