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.
2
20 kg, 3.75 m/sm g= =
( )( )20 3.75W mg= = 75 NW =

At all latitudes, 2.000 kgm =
(a) ( )2 2
0 , 9.7807 1 0.0053 sin 9.7807 m/sgφ φ= ° = + =
( )( )2.000 9.7807W mg= = 19.56 NW =
(b) ( )2 2
45 , 9.7807 1 0.0053 sin 45 9.8066 m/sgφ = ° = + ° =
( )( )2.000 9.8066W mg= = 19.61 NW =
(c) ( )2 2
60 , 9.7807 1 0.0053 sin 60 9.8196 m/sgφ = ° = + ° =
( )( )2.000 9.8196W mg= = 19.64 NW =

Assume 2
32.2 ft/sg =
W
m
g
=
: s
W
F ma W F a
g
Σ = − =
7
1 or
2
1 1
32.2
s
s
a F
W F W
ag
g
 
− = = = 
  − −
7.46 lbW =
27.4635
0.232 lb s /ft
32.2
W
m
g
= = = ⋅
: s
W
F ma F W a
g
Σ = − =
1
2
7.46 1
32.2
s
a
F W
g
 
= + 
 
 
= + 
 
7.92 lbsF =
For the balance system B,
0 0: 0w pM bF bFΣ = − =
w pF F=
But, 1w w
a
F W
g
 
= + 
 
and 1p p
a
F W
g
 
= + 
 
so that and
p
w p w
W
W W m
g
= = 2
0.232 lb s /ftwm = ⋅

Periodic time: 12 h 43200 sτ = =
Radius of Earth: 6
3960 mi 20.9088 10 ft= = ×R
Radius of orbit: 6
3960 12580 16540 mi 87.33 10 ftr = + = = ×
Velocity of satellite:
( )( )6
2 87.33 102
43200
r
v
ππ
τ
×
= =
3
12.7019 10 ft/s= ×
It is given that 3
750 10 lb s= × ⋅mv
(a)
3
2
3
750 10
59.046 lb s /ft
12.7019 10
mv
m
v
×
= = = ⋅
×
2
59.0 lb s /ftm = ⋅
(b) ( )( )59.046 32.2 1901 lb= = =W mg
1901 lb=W

+
40
: 10 10 10 20 40
32.2
y y yF ma a= + + + − =∑
( )( ) 232.2 10
8.05 ft/s
40
ya = =
y
dv dy dv dv
a v
dt dt dy dy
= = =
y yvdv a d=
2
0 0
1
2
v v
y y yvdv 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 dv
a v
dx dt
0
0
0
fx
v
vdv adx=∫ ∫
2
0
1
2
fv a x− =
( )
( )( )
2
20 30
6 m/s
2 2 75f
v
a
x
= − = − = −
0
0
0
ft
v
dv adt=∫ ∫
0 fv at− =
0 30
6
f
v
t
a
−
= − =
−
5.00 sft =
(b) + 0: 0yF N W= − =∑
N W=
:xF ma N maµ= − =∑
ma ma a
N W g
µ = − = − = −
( )6
9.81
µ
−
= − 0.612µ =

(a) : sinfF ma F W maα= − + =∑
sin
sin= − + = − +
f fF FW
a g
m m m
α
α
( )2 27500 N
9.81 m/s sin 4 4.6728 m/s
1400 kg
= − + ° = −
2
4.6728 m/sa = 4°
0 88 km/h 24.444 m/s= =v
From kinematics,
dv
a v
dx
=
0
0
0
fx
v
adx vdv=∫ ∫
2
0
1
2
fa x v= −
( )
( )( )
22
0 24.444
2 2 4.6728
f
v
x
a
= − = −
−
63.9 mfx =
+

(a) Coefficient of static friction.
0: 0yF N WΣ = − = N W=
0 70 mi/h 102.667 ft/sv = =
( )
2 2
0
0
2 2
t
v v
a s s− = −
( )
( )
( )( )
22 2
20
0
0 102.667
31.001 ft/s
2 2 170
t
v v
a
s s
−−
= = = −
−
For braking without skidding , so that | |s s tN m aµ µ µ= =
:t t s tF ma N maµΣ = − =
31.001
32.2
t t
s
ma a
W g
µ = − = − = 0.963sµ =
(b) Stopping distance with skidding.
Use ( )( )0.80 0.963 0.770kµ µ= = =
:t k tF ma N maµΣ = = −
2
24.801 ft/sk
t k
N
a g
m
µ
µ= − = − = −
Since acceleration is constant,
( )
( )
( )( )
22 2
0
0
0 102.667
2 2 24.801t
v v
s s
a
−−
− = =
−
0 212 fts s− =

For the thrust phase, : t
W
F ma F W ma a
g
Σ = − = =
( ) 22
1 32.2 1 289.8 ft/s
0.2
tF
a g
W
   
= − = − =   
   
At 1 s,t =
( )( )289.8 1 289.8 ft/sv at= = =
( )( )221 1
289.8 1 144.9 ft
2 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
289.8
0, 1 9.00 s, 10.00 s
32.2
v t t= − = = =
( ) ( )2 2
1 1 12 2v v a y y g y y− = − = − −
( )
( )( )
22 2
1
1
0 289.8
1304.1 ft
2 2 32.2
v v
y y
g
−−
− = − = − =
(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 2
0 0 2 2
2 101 2
, or 1.25 m/s
2 4
x
x x v t at a
t
= + + = = =
0: sin50 cos20 0yF N P mgΣ = − ° − ° =
sin50 cos20N P mg= ° + °
: cos50 sin 20xF ma P mg N maµΣ = ° − ° − =
or ( )cos50 sin 20 sin50 cos20P mg P mg maµ° − ° − ° + ° =
( )sin 20 cos20
cos50 sin50
ma mg
P
µ
µ
+ ° + °
=
° − °
For motion impending, set 0 and 0.30.sa µ µ= = =
( )( ) ( )( )( )40 0 40 9.81 sin 20 0.30cos20
593 N
cos50 0.30sin50
P
+ ° + °
= =
° − °
For motion with 2
1.25 m/s , use 0.25.ka µ µ= = =
( )( ) ( )( )( )40 1.25 40 9.81 sin 20 0.25cos20
cos50 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
0
0
2 2
v v
a x x− = −
( )
( )
( )( )
22 2
0
0
2
0 27.778
2 2 60
6.43 m/s
v v
a
x x
−−
= =
−
= −
br:xF ma F maΣ = − =
2br
6.43 m/s
F
a
m
= − =
(a) Going up a 6° incline. ( )6θ = °
br: sinF ma F mg maθΣ = − − =
br
sin
F
a g
m
θ= − −
2
6.43 9.81sin 6 7.455 m/s= − − ° = −
( )
( )( )
22 2
0
0
0 27.778
2 2 7.455
v v
x 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
sin
F
a g
m
θ= − −
( ) 2
6.43 9.81sin 1.145 6.234 m/s= − − − ° = −
( )
( )( )
22 2
0
0
0 27.778
2 2 6.234
v v
x 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+ =
1
3 0 or
3
A B B Aa a a a+ = = −
For block A: : sin30A A AF ma m g T m aΣ = ° − = (1)
For block B: : sin30 3B B B B AF ma m g T m a m aΣ = ° − = = − (2)
Eliminating T and solving for ,Aa
g
( )3 sin30 3
3
 
− ° = + 
 
B
A B A A
m
m g m g m a
( ) ( )3 sin30 30 8 sin30
0.33673
3 /3 30 2.667
A BA
A B
m ma
g m m
− ° − °
= = =
+ +
(a) ( )( ) 2
0.33673 9.81 3.30 m/sAa = = 2
3.30 m/sA =a 30°
( ) 21
3.30 1.101 m/s
3
= − = −Ba 2
1.101 m/sB =a 30°
(b) Using equation (1),
( )( )( )sin30 10 9.81 sin30 0.33673A
A
a
T m g
g
 
= ° − = ° − 
 
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+ =
1
3
B Aa a= −
Block A: 0: cos30 0y A AF N m gΣ = − ° =
: sin30x A A A AF ma m g N T m aµΣ = ° − − =
Eliminate .AN
( )sin30 cos30A A Am g T m aµ° − ° − =
Block B: 0: cos30 0y B BF N m gΣ = − ° =
: sin30 3
3
B A
B B B B
m a
F ma m g N T m aµΣ = ° + − = = −
Eliminate .BN
( )sin30 cos30 3
3
B A
B
m a
m g Tµ° + ° − = −
Eliminate T.
( ) ( )3 sin30 3 cos30 3
3
B
A B A B A A
m
m 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 sin30 75 20
tan30 0.334.
3 cos30 75 20
A B
A B
m m
m m
µ
− ° −
= = ° =
+ ° +
Since 0.25 0.334,sµ = < sliding occurs.
Calculate Aa
g
for sliding. Use 0.20.kµ µ= =
continued

( ) ( )
( ) ( )( )
3 sin30 3 cos30
3 /3
30 8 sin30 0.20 30 8 cos30
0.13525
30 2.667
A B A BA
A B
m m m ma
g m m
µ− ° − + °
=
+
− ° − + °
= =
+
(a) ( )( ) 2
0.13525 9.81 1.327 m/s= =Aa 2
1.327 m/sA =a 30°
( ) 21
1.327 0.442 m/s
3
 
= − = − 
 
Ba 2
0.442 m/sB =a 30°
(b) ( )
( )( )( ) ( )( )
sin30 cos30
10 9.81 sin30 0.20cos30 10 1.327
= ° − ° −
= ° − ° −
A A AT m g m aµ
18.79 N=T

Data: 2
2
55000 lb
1708.1 lb s /ft
32.2 ft/s
Am = = ⋅
2
2
44000 lb
1366.5 lb s /ft
32.2 ft/s
Bm = = ⋅
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/s
1708.1 1366.5
b b
x
A B
F F
a
m m
+ +
= = =
+ +
x
dv
a v
dx
=
0
2
0 0
0
2
= =∫ ∫
fx
x x fv
v
a dx vdv a x
( )
( )( )
22
0 80.667
751 ft
2 2 4.5534
f
x
v
x
a
−
= − = − = −
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: 2
2
55000 lb
1708.1 lb s /ft
32.2 ft/s
Am = = ⋅
2
2
44000 lb
1366.5 lb s /ft
32.2 ft/s
Bm = = ⋅
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/s
1708.1 1366.5
b
x
A B
F
a
m m
= = =
+ +
x
dv
a v
dx
=
0
2
0 0
0
2
= =∫ ∫
fx
x x fv
v
a dx vdv a x
( )
( )( )
22
0 80.667
1429 ft
2 2 2.2767
f
x
v
x
a
−
= − = − =
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, orA B B Aa a a a+ = = −
Assume that block A moves down and block B moves up.
Block B: 0: cos 0y AB BF N W θΣ = − =
: sin B
x AB B B
W
F ma T N W a
g
µ θΣ = − + + =
Eliminate ABN and .Ba
( )sin cos B A
B B B
a a
T W W W
g g
θ 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 A
x A A A AB A A
W
F m a T W F F a
g
θΣ = − + − − =
( )sin cos sin cosA
B B A B
a
W W W W
g
θ µ θ θ µ θ− + − + −
( )cos A
B A A
a
W W W
g
µ θ− + =
( ) ( ) ( )sin 3 cos A
A B A B A B
a
W W W W W W
g
θ 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 0.20 128
tan 0.40
64
A B
s
A B
W W
W W
θ
+
= = =
−
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 cos25
0.10048
96
° − °
= =
( )( ) 2
0.10048 32.2 3.24 ft/sAa = =
(a) 2
3.24 ft/sBa = − 2
3.24 ft/sB =a 25° !
(b) ( )sin cos A
B B
a
T W W
g
θ µ θ= + +
( ) ( )( )16 sin 25 0.15cos25 16 0.10048= ° + ° +
10.54 lbT = !

Constraint of cable: ( )2 constant.A B A A Bx x x x x+ − = + =
0, orA B B Aa a a a+ = = −
Assume that block A moves down and block B moves up.
Block B: 0: cos 0y AB BF N W θΣ = − =
x : sin B
x AB B B
W
F ma T N W a
g
µ θΣ = − + + =
Eliminate ABN and .Ba
( )sin cos B A
B B B
a a
T W W W
g g
θ 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 A
x A A A AB A A
W
F m a T W F F P a
g
θ θΣ = − + − − + =
( )sin cos sin cosA
B B A B
a
W W W W
g
θ µ θ θ µ θ− + − + −
( )cos sin cos A
B A A
a
W W P P W
g
µ θ µ θ θ− + + + =
( ) ( ) ( ) ( )sin 3 cos sin cos A
A B A B A B
a
W W W W P W W
g
θ 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 cos
s A B A B
s
s
W W W W
P
µ θ θ
µ θ θ
+ − −
=
+
( )( )0.20 128 cos25 64 sin 25
3.88 lb 10 lb
0.20 sin 25 cos25
° − °
= = − <
° + °
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 cos25 10 0.15sin 25 cos25
96
° − ° + ° + °
=
0.20149=
( )( ) 2
0.20149 32.2 6.49 ft/sAa = =
(a) 2
6.49 ft/s ,Ba = − 2
6.49 ft/sB =a 25° !
(b) ( )sin cos A
B B
a
T W W
g
θ µ θ= + +
( ) ( )( )16 sin 25 0.15cos25 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 2
0.304 m/sA =a 15°
For box B, 0.32kµ =
( )9.81 0.32cos15 sin15Ba = ° − ° or 2
0.493 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: , orC C C C C C C C C C BW T m a T W m a W m a− = = − = −
Block B: 3B A C B BW T T m a− − =
( ) ( )3 3B A A B C C B B BW W m a W m a m a− − − − =
or
3 60 60 20
0.076923
9 60 180 20
B B A C
B A C
a W W W
g W W W
− − − −
= = = −
+ + + +
(a) Accelerations. ( )( ) 2
0.076923 32.2 2.477 ft/s= − = −Ba 2
2.48 ft/sB =a
( )( ) 2
3 2.477 7.431 ft/sAa = − − = 2
7.43 ft/sA =a
( ) 2
2.477 2.477 ft/sCa = − − = 2
2.48 ft/sC =a
(b) Tensions. ( )
20
20 7.43
32.2
AT = − 15.38 lbAT =
( )
20
20 2.477
32.2
CT = − 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: A
A A
W
W T a
g
− = or A
A A
W
T W a
g
= −
Block B: B B
B B A
W W
P W T a a
g g
+ − = = −
A B
B A A A
W W
P W W a a
g g
+ − + = −
Solving for , A B
A A
A B
W W P
a a g
W W
− −
=
+
(1)
( ) ( ) ( )22
0 0 0
2 with 0A A A A A Av v a y y v − = − = 
( )0
2A A A Av a y y = −  (2)
( ) ( )0 0
with 0A A A Av v a t v− = =
A
A
v
t
a
= (3)
continued

(a) Acceleration of block A.
System (1): 100 lb, 50 lb, 0A BW W P= = =
By formula (1), ( ) ( )1
100 50
32.2
100 50
Aa
−
=
+
( ) 2
1
10.73 ft/sA =a !!!!
System (2): 100 lb, 0, 50 lbA BW W P= = =
By formula (1), ( ) ( )2
100 50
32.2
100
Aa
−
= ( ) 2
2
16.10 ft/sA =a !!!!
System (3): 1100 lb, 1050 lb, 0A BW W P= = =
By formula (1), ( ) ( )3
1100 1050
32.2
1100 1050
Aa
−
=
+
( ) 2
3
0.749 ft/sA =a !!!!
(b) ( )0
at 5 ft. Use formula (2).A A Av y y− =
System (1): ( ) ( )( )( )1
2 10.73 5Av = ( )1
10.36 ft/sAv = !!!!
System (2): ( ) ( )( )( )2
2 16.10 5Av = ( )2
12.69 ft/sAv = !!!!
System (3): ( ) ( )( )( )3
2 0.749 5Av = ( )3
2.74 ft/sAv = !!!!
(c) Time at 10 ft/s. Use formula (3).Av =
System (1): 1
10
10.73
t = 1 0.932 st = !!!!
System (2): 2
10
16.10
t = 2 0.621 st = !!!!
System (3): 3
10
0.749
t = 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 = + = + =
( )( ) 2
0.85 0.85 32.2 23.37 ft/s
W
a
m
= = = 2
27.4 ft/s=a !
For the upper beam, 1: 0.25xF ma T N maΣ = − =
( )( ) ( )
3000
0.25 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 = =
2
0.25 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
2
0.30 9.66 ft/sa g= =
The smaller deceleration value governs. 2
8.05 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
( )sin30 sin30A B A BP m g m g m m a− ° − ° = +
500
sin30 9.81 sin30
50A B
P
a g
m m
= − ° = − °
+
2
5.095 m/s=
Use block B as a free body.
cos30 : cos30B f BF m a F m aΣ = ° = °
( )( )10 5.095 cos30 44.124 NfF = ° =
sin30 : sin30B B B BF m a N m g m aΣ = ° − = °
( ) ( )sin30 10 9.81 5.095 sin30B BN m g a= + ° = + °
123.575 N=
Minimum coefficient of static friction:
min
44.124
123.575
f
B
F
N
µ = = min 0.357µ =

(a) Kinematics of the belt. 0ov =
1. Acceleration phase with 2
1 3.2 m/sa =
( )( )1 1 1 0 3.2 1.5 4.8 m/sov v a t= + = + =
( )( )22
1 1 1 1
1 1
0 0 3.2 1.5 3.6 m
2 2
o ox x v t a t= + + = + + =
2. Deceleration phase. 2 0v = since the belt stops.
( )2 2
2 1 2 2 12v v a x x− = −
( )
( )
( )
22 2
2 1
2
2 1
0 4.8
11.52
2 2 4.6 3.6
v v
a
x x
−−
= = = −
− −
2
2 11.52 m/s=a
2 1
2 1
2
0 4.8
0.41667 s
11.52
v v
t t
a
− −
− = = =
−
(b) Motion of the package.
1. Acceleration phase. Assume no slip. ( ) 2
1
3.2 m/spa =
0: 0 oryF 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 = =
( ) ( )( ) 2
1
0.35 9.81 3.43 m/sf s
p s
F N
a g
m m
µ
µ= < = = =

Since 2 2
3.2 m/s 3.43 m/s ,< the package does not slip.
( ) ( )11 1
4.8 m/s and 3.6 m.p pv v x= = =
2. Deceleration phase. Assume no slip. ( ) 2
2
11.52 m/spa = −
( )2
:x f pF ma F m aΣ = − =
( ) 2
2
11.52 m/sf
p
F
a
m
= = −
2 2
3.43 m/s 11.52 m/ss s
s
N mg
g
m m
µ µ
µ= = = <
Since the available friction force sNµ is less than the required
friction force fF for no slip, the package does slip.
( ) 2
2
11.52 m/s ,p f ka F Nµ< =
( ) ( )2 2
:x p k pF m a N m aµΣ = − =
( ) ( )( ) 2
2
0.25 9.81 2.4525 m/sk
p k
N
a g
m
µ
µ= − = − = − = −
( ) ( ) ( ) ( ) ( )( ) 2
2 12 1 2
4.8 2.4525 0.41667 3.78 m/sp p pv v a t t= + − = + − =
( ) ( ) ( ) ( ) ( ) ( )2 2
2 1 2 12 1 1 2
1
2
p p p px x v t t a t t= + − + −
( )( ) ( )( )21
3.6 4.8 0.41667 2.4525 0.41667 5.387 m
2
= + + − =
Position of package relative to the belt
( ) 22
5.387 4.6 0.787px x− = − = /belt 0.787 mpx =

Acceleration 1:a Impending slip. 1 1 10.30 NsF Nµ= =
1 1: sin65y A y A AF m a N W m aΣ = − = °
1 1 sin 65A AN W m a= + °
( )1 sin65Am g a= + °
1 1: cos65x A x AF m a F m aΣ = = °
1 sF Nµ= or ( )1 1cos65 0.30 sin65A Am a m g a° = + °
( )( ) 2
1
0.30
1.990 9.81 19.53 m/s
cos65 0.30sin65
g
a = = =
° − °
2
1 19.53 m/s=a 65°
Deceleration 2 :a Impending slip. 2 2 20.30 NSF Nµ= =
1 2: sin 65y y A AF ma N W m aΣ = − = − °
1 2 sin65A AN W m a= − °
2 2: cos65x x AF ma F m aΣ = = °
2 2SF Nµ= or ( )2 2cos65 0.30 cos65A Am a m g a° = − °
( )( ) 2
2
0.30
0.432 9.81 4.24 m/s
cos65 0.30sin 65
g
a = = =
° + °
2
2 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: cos20 sin 20y P y P P TF m a N W m aΣ = − ° = − °
( )1 cos20 sin 20P TN m g a= ° − °
1: sin 20 cos20x x P P TF ma F W m aΣ = − ° = °
( )1 sin 20 cos20P TF m g a= ° + °
( ) ( )sin 20 cos20 0.40 cos20 sin 20P T P Tm g a m g a° + ° = ° − °
( )
( )( )
0.40cos20 sin 20
0.03145 9.81 0.309
cos20 0.40sin 20
Ta g
° − °
= = =
° + °
2
0.309 m/sT =a s
(b) ( ) ( ) 2 2
/ / / / /
1 1
0 0
2 2
P T P T P T P T P To
x x v t a t a t= + + = + +
( )( )
( )
2/
/ 2 2
2 12
12.5 m/s ,
0.4
P T
P T
x
a
t
= = = 2
/ 12.5 m/sP T =a 20°
( )/P T P T Ta= + = → +a a a ( 2
12.5 m/s )20°
:y P yF m a=
2 cos20 sin 20P P TN W m a− ° = − °
( )2 cos20 sin 20P TN m g a= ° − °
continued

:x xF maΣ = Σ
2 /sin 20 cos20P P T P P TF W m a m a− ° = ° −
( )2 /sin 20 cos20P T P TF m g a a= ° + ° −
For sliding with friction 2 2 20.30 NkF Nµ= =
( ) ( )/sin 20 cos20 0.30 cos20 sin 20P T P T P Tm g a a m g a° + ° − = ° + °
( ) /0.30cos20 sin 20
cos20 0.30sin 20
P T
T
g a
a
° − ° +
=
° + °
( )( ) ( )( )0.05767 9.81 0.9594 12.5 11.43= − + =
2
11.43 m/sT =a

At maximum speed 0.a = 2
0 0 0F kv= = 0
2
0
F
k
v
=
When the propellers are reversed, 0F is reversed.
2
0:xF ma F kv maΣ = − − =
2
0 0 2
0
v
F F ma
v
− − = ( )2 20
02
0
F
a v v
mv
− +
( )
2
0
2 2
0 0
vdv mv vdv
dx
a F v v
= =
+
0
2
00
2 20
0 0
x
v
mv vdv
dx
F v v
= −
+
∫ ∫
( ) ( )0
2 2 20
2 2 2 20 0 0
0 0 0
0 0 0
1
ln ln ln 2 ln 2
2 2 2v
mv mv mv
x v v v v
F F F
 = − + = − − =
 
2
0 0
0
0.347
m v
x
F
=

:F ma P kv maΣ = − =
dv P kv
a
dt m
−
= =
( ) ( )0 0 0
ln ln ln
vt v m dv m m
dt P kv P kv P
P kv k k
 = = − − = − − − −
∫ ∫
ln or ln
m P kv P kv kt
t
k P m m
− −
= − = −
( )/ /
or 1kt m kt mP kv P
e v e
m k
− −−
= = −
/
0
0 0
tt
t kt mPt P k
x v dt e
k k m
− 
= = − −  ∫
( ) ( )/ /
1 1kt m kt mPt P Pt P
e e
k m k m
− −
= + − = − −
, which is linear.
Pt kv
x
k 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
2
2
D kv
a g g
m m
dv kv k mg
v g v
dy m m k
  
= − + = − +       
   
= − + = − +       
2
vdv k
dy
mg mv
k
= −
+
0
0
02
h
v
vdv k
dy
mg mv
k
= −
+
∫ ∫
0
0
21
ln
2 v
mg kh
v
k m
 
+ = − 
 
2
0
2
0
1 1
ln ln 1
2 2
mg
kv khk
mg mg mv
k
 
= − + = −  
 +
( )( )
( )( )
( )( )
22
0 0.0024 904
ln 1 ln 1
2 2 0.0024 4 9.81
m kv
h
k mg
  
 = + = +       
335.36 m= 335 mh =
continued

(b) Downward motion. :F maΣ =
D mg ma− =
2
2
2
D kv
a g g
m m
dv kv k mg
v g v
dy m m k
= − = −
 
= − = − − 
 
2
vdv k
dy
mg mv
k
= −
−
0
0
fv
h
vdv k
dy
mg m
= −∫ ∫
2
0
1
ln
2
fv
mg kh
v
k m
 
− = 
 
2
1
ln
2
f
mg
v
khk
mg m
k
 
− 
− = 
  
 
2
2
ln 1
fkv kh
mg m
 
 − = −
 
 
2
2
1
f kh/mkv
e
mg
−
− =
( )2
1 kh/m
f
mg
v e
k
−
= ± −
( )( ) ( )( )( )2 0.0024 335.36 /44 9.81
1
0.0024
fv 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 2
L x= +
The elongation of the spring is ,e L= − and the magnitude of the force
exerted by the spring is
( )2 2
sF ke k x= = + −
By geometry,
2 2
cos
x
x
θ =
+
: cosx x sF ma F maθΣ = − =
( )2 2
2 2
x
k x ma
x
− + − =
+
2 2
k x
a x
m x
 
= − −  
+ 
0
0
0
v
x
v dv a dx=∫ ∫
0
0
0
02 2 2 2
2 2
0
1 1
2 2
v
x
x
k x k
v x dx x x
m mx
   
= − − = − − +      + 
∫
2 2 2 2 2
0 0
1 1
0
2 2
k
v x x
m
 
= − − − + + 
 
( )2 2 2 2 2
0 02 2
k
v x x
m
= + − +
( )2 2 2 2 2
0 02
k
x x
m
 = + − + +
  
( )2 2
0answer:
k
v 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
A
A
T
a g
m
= − (2)
Block B. : B B BF ma m g T m aΣ = − =
B
B
T
a g
m
= − (3)
Block C. : C C CF ma m g T m aΣ = − =
C
C
T
a g
m
= − (4)
Substitute (2), (3) and (4) into (1).
2
2 0
A B C
T T T
g g g
m m m
    
− + − + − =    
     
4 1 1
4 0
A B C
g T
m m m
 
− + + = 
 
( )( )
4 1 1
4 9.81 0 65.4 N
10 10 10
T T
 
− + + = = 
 
Substitute into (2),
( )( ) 22 65.4
9.81 3.27 m/s
10
Aa = − = −
(a) Change in position. ( ) 2
0
1
2
A A Ay y a t− =
( ) ( )( )2
0
1
3.27 0.5 0.409 m
2
A 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: sin30x B B B B BF m a W F m aΣ = ° − =
( ) ( ) 2
1
1 1
sin30 49.05sin30 4.248 4.055 m/s
5
B B
B
a 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: sin30x A A A A AF m a W F F m aΣ = ° − + =
( )2 1
1
sin30A A
A
a W F F
m
= ° − +
( ) 21
98.1sin30 25.49 4.248 2.781 m/s
10
= ° − + =
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. 2
2.78 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 3
3 or
20
A A A
A
T T
T m a a g
m
= = = (2)
Block B:
2 2
2 or
10
B B B
B
T T
T m a a g
m
= = = (3)
Block C:
4 4
4 or
20 20
C C C
P T P T
P T m a a g
− −
− = = = (4)
Substituting (2), (3), and (4) into (1),
4 2 3
4 2 3 0
20 10 20
P T T T−     
− − =     
     
16 4 9 4
20 10 20 20
P
T
 
+ + = 
 
( ) ( )( )0.12121 0.12121 50 6.0605 lbT P= = =
(a) From (2),
( )( )( ) 23 6.0605 32.2
29.3 ft/s
20
Aa = = 2
29.3 ft/sA =a
From (3),
( )( )( ) 22 6.0605 32.2
39.0 ft/s
10
Ba = = 2
39.0 ft/sB =a
From (4),
( )( ) ( ) 2
41.5 ft/s
20
50 4 6.0605 32.2
Ca
 − = = 2
41.5 m/sC =a
(b) As determined above, 6.06 lbT =

2 220 10
0.62112 lb s / ft, 0.31056 lb s / ft
32.2 32.2
A 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 A
A
A A
T W T
a g
m m
µ−
= = − (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 B
B
B B
T W T
a g
m m
µ−
= = − (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 2
0 0
1 1
0
2 2
C C C C Cx x v a t a t= + + = +
( ) ( )( )
( )
0 2
2 2
2 2 2.4
30 ft/s
0.4
C C
C
x x
a
t
 − = = = (5)

Substitute (2), (3) and (5) into (1).
( )( ) ( ) ( )
2 3
4 30 2 0.20 3 0.20
0.31056 0.62112
T T
g g
   
− − − −      
( )( ) ( ) ( )( ) ( ) ( )( )
2 3
4 30 2 0.2 32.2 3 0.2 32.2
0.31056 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 5.5608
0.20 32.2 20.42 ft/s
0.62112
Aa = − =
From (3),
( )( )
( )( ) 22 5.5608
0.20 32.2 29.37 ft/s
0.31056
Ba = − =
(a) Acceleration vectors. 2
20.4 ft/sA =a
2
29.4 ft/sB =a
2
30 ft/sC =a
Since , , andA B Ca a a are to the right, the friction forces
, , andA B CF F F are to the left as assumed.
(b) Tension in the cable. 5.56 lbT =
(c) Force P. 44.9 lb=P

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 /andA 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 20x B B B AB B Ax
F m a m g N m aΣ = ° − =
sin 20B A AB Bm a N m g+ = °
15 50.328A ABa N+ = (2)
( ) /: cos20y B B B B B Ay
F m a m g T m aΣ = ° − =
/ cos20B B A Bm a T m g+ = °
/15 138.276B Aa T+ = (3)
Block A: AAABAAAx amTNgmamF =++°=Σ 20sin:
sin 20A A AB Am a N T m g− + = °
25 83.880A ABa N T− + = (4)
Eliminate /B Aa using Eq. (1), then add Eq. (4) to Eq. (2) and
subtract Eq. (3).
2 2
55 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 t
a is the component of /B Aa lying
along the circle, say to the left in the diagram and ( )/B A n
a 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 cos25x x B B A B AF ma m a m aΣ = Σ = − °
/ cos25 0.4cos25 0.363B A Aa a= ° = ° =
2
/ 0.363 m/sB A =a
( ) : sin 25y B B AB B B Ay
F 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).
2
v
a
ρ
=
+ :y yF maΣ = cos 0− =T Wθ
cos
=
W
T
θ
:y xF maΣ = sinT maθ =
sin
cos
W
θ
θ
2
mv
ρ
=
2
sin
mv
L θ
=
(a) tan sinθ θ =
2
mv
WL
=
2
v
gL
=
( )
( )( )
2
1.5
0.38226
9.81 0.6
=
34.21θ = ° 34.2θ = °
(b)
cos cos
W mg
T
θ θ
= =
( )( )2 9.81
cos34.21
=
°
23.7 N=T

Let ρ be the radius of the horizontal circle. The length of the wire is
1 2sin sin
L
ρ ρ
θ θ
= + . Solving for ,ρ 1 2
1 2
sin sin
sin sin
L θ θ
ρ
θ θ
=
+
1 20: cos cos 0yF T T mgθ θΣ = + − =
1 2cos cos
mg
T
θ θ
=
+
2
1 2: sin sinx x n
mv
F ma T T maθ θ
ρ
Σ = + = =
( ) ( )2
1 2 1 2
1 2 1 2
sin sin sin sin
cos cos sin sin
mg mv
L
θ θ θ θ
θ θ θ θ
+ +
=
+
( )( )2 2 21 2
1 2
sin sin sin60 sin30
2 9.81 6.2193 m /s
cos cos cos60 cos30
v Lg
θ θ
θ θ
° °
= = =
+ ° + °
2.49 m/sv =

3 1 2 50 25 25θ θ θ= − = ° − ° = °
1 2
1
2 3 3
sin
or
sin sin sin
d d θ
θ θ θ
= =
( )( )( )
2 1
1 1
3
sin sin
sin
sin
4 sin 25 sin50
3.0642 ft
sin 25
d θ θ
ρ θ
θ
= =
° °
= =
°
2 10: cos cos 0y AC BCF T T Wθ θΣ = + − = (1)
2
2 1: sin sinx x AC BC
Wv
F ma T T
g
θ θ
ρ
Σ = + = (2)
Case 1: 0.BCT = 2cos 0ACT Wθ − = or
2cos
AC
W
T
θ
=
2
2 2sin tanAC
Wv
T W
g
θ θ
ρ
= =
( )( )2
2tan 32.2 3.0642 tan 25v gρ θ= = ° 2 2
46.01 ft /s=
6.78 ft/sv =
Case 2: 0.ACT = 1
1
cos 0 or
cos
BC BC
W
T W Tθ
θ
− = =
2
1 1sin tanBC
Wv
T W
g
θ θ
ρ
= =
( )( )2 2 2
1tan 32.2 3.0642 tan50 117.59 ft /sv gρ θ= = ° =
10.84 ft/sv = 6.78 ft/s 10.84 ft/sv≤ ≤

(a) 0: sin 0yF T WθΣ = − =
16
sin sin60
W
T
θ
= =
°
or 18.48 lbT =
(b)
2
: cosx n
v
F ma T mθ
ρ
Σ = =
2 cos cos
sin
T W
v
m m
ρ θ ρ θ
θ
= =
( )( ) 2 23 32.2
55.77 ft /s
tan tan60
gρ
θ
= = =
°
7.47 ft/sv =

2
, 1 tan or 45
2
r dy
y r
dx
θ θ= = = = = °
0: cos 0yF N mgθΣ = − =
cos
mg
N
θ
=
: sinx x nF ma N maθΣ = =
2
tan
v
mg m
r
θ =
2
tanv gr θ=
(a) ( )( )( )2 2 2
9.81 1 1.0000 9.81 m /sv = = 3.13 m/sv =
(b)
( )( )1 9.81
cos45 cos45
mg
N = =
° °
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
= + − °
= + − ° =
=
sin30 sin
OC OB
β
=
°
( )0.3 sin30sin30
sin 0.40345
0.37179
OB
OC
β
°°
= = =
23.79β = °
Acceleration components: 2
1.3 m/sta =
( )22 2
21
1.667 m/s
0.6
C
n
BC
v v
a
ρ
= = = =
Mass 1 kgm =
( )( ): cos 1 1.3 1.3t tF ma N βΣ = = =
1.3
1.421 N
cos23.79
= =
°
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: cos20 cos30 0y DA DEF T T WΣ = ° − ° − =
0.93969 0.86603 4.905DA DET T− =
0.92160 5.2198DA DET T= + (1a)
1.08506 5.6638DE DAT T= − (1b)
2
20.5
: sin 20 sin30
0.150
x n DA DE
mv
F ma T T v
ρ
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 2
0.10261 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 2
0.10261 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: cos40 cos15 0y CA CBF T T WΣ = ° − ° − =
0.76604 0.96593 49.05CA CBT T W− = =
0.79307 50.780CB CAT T= − (1a)
1.26093 64.030CA CBT T= + (1b)
2
: sin 40 sin15x x CA CB n
mv
F ma T T ma
ρ
Σ = ° + ° = =
( )
( )
2
sin40 sin15
0.9
sin40 sin15
5
0.115702 0.046587
CA CB
CA CB
CA CB
v T T
m
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 2
0.115702 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 2
0.115702 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 = sin50 sin70 0° + ° − =AB CDT T W (2)
Solving (1) and (2) simultaneously,
0.395=ABT W 0.742=CDT W
(b) Immediately after wire AB is cut. 0, 0ABT v= =
2
0n
v
a
ρ
= =
0 :n nF ma= =
cos20 0CDT W− ° =
cos20CDT W= °
0.940CDT W=
+

:y yF maΣ =
2
n
mv
N mg ma
ρ
− = − = −
2
mv
N mg
ρ
= −
:x xF maΣ =
t tF ma=
At onset of slipping, t sF Nµ=
2
s
t s
mv
ma mgµ
ρ
 
= −  
 
( )2 2 20.150
0.300 9.81 2.883 m /s
0.75
t
s
s
a
v gρ
µ
   
= − = − =     
1.6979 m/ssv =
Time at slipping.
1.6979
0.150
s
s
t
v
t
a
= = 11.32 sst =

Angle change over arc AB.
8
0.13963 rad
180
θ π
°
= =
°
Length of arc: ( ) ( )4 (5280) 0.13963 2949 ftABs ρθ= = =
(0.5)(5280) 2640 ft, 2949 2640 5589 ftBC ACs s= = = + =
( )
( )( )
2 2
480 5589
540 0
2
2 2
2949
540 0
2 2 2
480 540
or 5589
2 2
5.475 ft/s
540
or 5.475 2949
2 2
259300 ft /s 509.2 ft/s
B
t t
t
v B
t
B B
v dv a ds a
a
v
v dv a ds
v v
= − =
= −
= − = −
= =
∫ ∫
∫ ∫
Mass of passenger: 2200
6.211 lb s /ft
32.2
= = ⋅m
Just before point B. 509.2 ft/s, (4)(5280) = 21120 ftv ρ= =
( )22
2509.2
12.277 ft/s
21120
n
v
a
ρ
= = =
( ) ( )( )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

( ) ( )2
3 180 m, 3 180 2 m/s,
ds
s t t v t
dt
= − = = −
2
6 m/s (4)(5280) 21120 ft= = − = =t
dv
a
dt
ρ
Length of arc AB. ( )
8
21120 2949 ft
180
AB ABs
π
ρθ
°
= = =
°
0
B AB
A
v s
tv
v dv a ds=∫ ∫
( )( )( )
2 2
2 2 2
2 or 2 540 2 6 2949
2 2
B A
t AB B A t AB
v v
a s v v a s− = = + = + −
256212,= 506.2 ft/sBv =
For passenger, 2165
165 lb, 5.124 lb s /ft
32.2
= = = = ⋅
W
W m
g
2
: cosn
mv
F ma W Nθ
ρ
Σ = − =
2
cos
mv
N W θ
ρ
= − (1)
: sint tF ma P W maθΣ = − =
sin tP W maθ= + (2)
(a) Just after point A, 0, 180 m/s, 8t v θ= = = °
From Eq. (1),
( )( )2
5.124 540
165cos8 92.65 lb
21120
N = ° − =
From Eq. (2), ( )( )165sin8 5.124 6 7.78 lbP = ° + − = −
2 2 92.65
93.0 lb, tan 11.909, 85.2
7.78
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),
( )( )2
5.124 506.2
165 102.83 lb
21120
N = − =
From Eq. (2), ( )( )5.124 6 30.74 lbP = − = −
( ) ( )
2 2 102.83
102.83 30.74 107.3 lb, tan 3.345, 73.4
30.74
= + = = = = °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 44.44
201.4
9.81
v
g
ρ = = =
201 mρ =
(b) 80 km/h 22.22 m/sv = =
70 kgm = for passenger
:y nF maΣ = −
2
mv
N mg
ρ
− = −
2
v
N m g
ρ
 
= −  
 
( )
2
22.22
70 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 cost
W
a g
m
θ θ= =
0 0 0
v s
t tv
v dv a ds a rd
θ
θ= =∫ ∫ ∫
2 2
0 0
1 1
cos sin
2 2
v v g rd gr
θ
θ θ θ− = =∫
2 2
0 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 2
max 0 2v v gr= + (1)
When the cord touches the peg, the radius of curvature of the path
becomes 0.3 m.ρ =
2
max
max:y n
mv
F ma T W
ρ
Σ = − =
( )2
max maxv T W
m
ρ
= − (2)
Eliminating 2
maxv from equations (1) and (2),
( ) ( )( )
( )( )( )max2
0
0.3 10 1.962
2 2 9.81 6.0
0.2
T W
v gr
m
ρ − −
= − = −
2 2
0.285 m /s= 0 0.534 m/s=v

Mass of block B. 2
2
0.5lb
0.015528lb s /ft
32.2ft/s
= = ⋅Bm
Acceleration of block B.
2 2
2(9ft/s)
27ft/s
3ft
= = =n
v
a
ρ
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) 0.3
sin
0.5 0.5
nma P
W W
θ ≥ − = −
sin 0.2385θ ≥
13.8 166.2θ° ≤ ≤ °

At position A, the vertical component of apparent weight is shown as .AN
:n A n
W
F ma N W a
g
Σ = − =
( ) 2380 120
32.2 69.77 ft/s
120
A
n
N W
a g
W
− −
= = =
( )( )2 3 2 2
3600 69.77 251.2 10 ft /sA nv aρ= = = ×
At position C, the vertical component of apparent weight is shown
as .CN
:n C n
W
F ma N W a
g
Σ = + =
( ) 280 120
32.2 53.67 ft/s
120
C
n
N W
a g
W
+ +
= = =
( )( )2 3 2 2
3600 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 10
2 2 11310
2.562 ft/s
C A
t
AC
v v
a
s
− × − ×
= =
= −
At position B, ( )3600 5655 ft
2 2
ABs
π π
ρ= = =
( )( )( )2 2 3 3 2 2
2 251.2 10 2 2.562 5655 222.2 10 ft /sB A t ABv v a s= + = × + − = ×

Effective forces at B:
2 3
120 222.2 10
230 lb
32.2 3600
B
n
W v
ma
g ρ
×
= = =
( )
120
2.562 9.5 lb
32.2
t t
W
ma a
g
= = − = −
: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 2
230 110.5 255 lbBF N P= + = + =
110.5
tan
230
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
mg
R
θ φ
=
+
( ): sinx n s nF ma R maθ φΣ = + =
( )tann sma mg θ φ= +
( )
2
max
tan s
v
g
r
θ φ= +
( )max tan sv gr θ φ= +
Case 2: minv v=
( )0: cos 0y sF R mgθ φΣ = − − =
( )cos s
mg
R
θ φ
=
−
( ): sinx n s nF ma R maθ φΣ = − =
( )tann sma mg θ φ= −
( )
2
min
tan s
v
g
r
θ φ= −
( )min tan sv gr θ φ= −
( ) ( )tan tans sgr v grθ φ θ φ− ≤ ≤ +

Weight. W mg=
Acceleration.
2
v
a
ρ
=
:∑ =x xF ma sin cosF W = maθ θ+
2
cos sin
mv
F mgθ θ
ρ
= − (1)
:y yF ma∑ = cos sinN W maθ θ− =
2
sin cos
mv
N = mgθ θ
ρ
+ (2)
(a) Banking angle. Rated speed 180 km/h 50 m/s.v = = 0F = at rated speed.
2
0 cos sin
mv
mgθ θ
ρ
= −
2 2
(50)
tan = 1.2742
(200)(9.81)
v
g
θ
ρ
= =
51.875θ = ° 51.9θ = °
(b) Slipping outward. 320km/h 88.889 m/sv = =
F Nµ=
2
2
cos sin
sin cos
F v g
N v g
θ ρ θ
µ
θ ρ θ
−
= =
+
2
2
(88.889) cos51.875 (200)(9.81)sin51.875
=
(88.889) sin51.875 (200)(9.81)cos51.875
µ
°− °
°+ °
0.44899= 0.449µ =
continued

(c) Minimum speed. = −F Nµ
2
2
cos sin
sin cos
v g
v g
θ ρ θ
µ
θ ρ θ
−
− =
+
2 (sin cos )
cos sin
g
v
ρ θ µ θ
θ µ θ
−
=
+
(200)(9.81)(sin51.875 0.44899cos51.875 )
cos51.875 0.44899sin51.875
° − °
=
° + °
2 2
1029.87m /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,
( )22
2 110
tan or 2674 ft
tan 32.2 tan8
R
R
v
v g
g
ρ 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
cos
mv
θ φ
ρ
= +
(a) 0.φ =
( ) ( )
( )
( )( )
2
2
cos sin
183.33
cos8 sin8
32.2 2674
0.247
s
v
F W
g
W
W
θ φ θ φ
ρ
 
= + − + 
 
 
 = ° − °
  
= 0.247sF W= !
(b) For 0,sF =
( ) ( )
2
cos sin 0
v
g
θ φ θ φ
ρ
+ − + =
( )
( )
( )( )
22
183.33
tan 0.39035
32.2 2674
v
g
θ 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,
( )22
2 110
tan or 2674 ft
tan 32.2 tan8
R
R
v
v g
g
ρ 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
cos
mv
θ φ
ρ
= +
Solving for ,sF ( ) ( )
2
cos sins
v
F W
g
θ φ θ φ
ρ
 
= + − + 
 
Now
( )
( )( )
22
183.33
0.39035 and 0.12
32.2 2674
s
v
F W
gρ
= = = so that
( ) ( )0.12 0.39035 cos sinW W θ φ θ φ = + − + 
Let ( ) ( ) 2
sin . Then, cos 1 .u uθ φ θ φ= + + = −
2 2
0.12 0.39035 1 or 0.39035 1 0.12u u u u= − − − = +
Squaring both sides, ( )2 2
0.15237 1 0.0144 0.24u u u− = + +
or 2
1.15237 0.24 0.13797 0u u+ + =
The positive root of the quadratic equation is 0.2572.u =
Then, 1
sin 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 )n
v
a r
ρ ω
θ ω
ρ ρ
= = =
:y yF ma∑ =
cos 0N mgθ − =
cos
mg
N
θ
=
:x xF maΣ =
sinN maθ =
2sin
( sin )
cos
mg
m r
θ
θ ω
θ
=
Either sin 0θ =
or 2
cos
g
r
θ
ω
=
0 or 180θ = ° °
or 2
9.81
cos 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 )n
v
a r
ρ ω
θ ω
ρ ρ
= = =
FΣ = :ma
2
sin ( sin ) cosF mg m rθ θ ω θ− = −
2
( cos )sinF m g rω θ θ= −
FΣ = :ma
2
cos ( sin ) sinN mg m rθ θ ω θ− =
2 2
( cos ) sin )N m g rθ ω θ= +
(a) 75 .θ = ° 2
(0.25) 9.81 (0.500 cos75 )(7.5) sin75F  = − ° ° 
0.61112 N=
2 2
(0.25) 9.81cos75 (0.500sin 75 )(7.5)N  = °+ ° 
7.1950 N=
(0.25)(7.1950) 1.7987 N= =sNµ
Since ,sF Nµ< the collar does not slide.
0.611 NF = 75°
continued

(b) 40 .θ = ° 2
(0.25) 9.81 (0.500cos40 )(7.5) sin 40F  = − ° ° 
1.8858 N= −
2 2
(0.25) 9.81cos40 (0.500sin 40 )(7.5)N  = ° + ° 
4.7839 N=
(0.25)(4.7839) 1.1960 N= =sNµ
Since ,> sF Nµ the collar slides.
Since the collar is sliding, .kF Nµ=
nF∑ = + :ma
cos sinnN mg maθ θ− =
2
cos ( sin ) sinN mg m rθ θ ω θ= +
2 2
cos ( sin )m g rθ θ ω = + 
2 2
(0.25) 9.81cos40 (0.500sin 40 )(7.5) = °+ ° 
4.7839 N=
(0.20)(4.7839) 0.957 N= = =kF Nµ
0.957 N=F 40°

Draw the free body diagrams of the block B when the arm is at 150 .θ = °
2
0, 32.2 ft/stv a g= = =
: sin30 0t tF ma W NΣ = − ° + =
sin30N W= °
2 2
: cos30n n
v Wv
F ma W F m
gρ ρ
Σ = ° − = =
2
cos30
Wv
F W
gρ
= ° −
Form the ratio
F
N
, and set it equal to sµ for impending slip.
( )22
cos30 4.2 /(1)(32.2)cos30 /
sin30 sin30
s
F v g
N
ρ
µ
° −° −
= = =
° °
0.636sµ =

Let β be the slope angle of the dish.
1
tan
6
dy
r
dr
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
W
N
β µ β
=
−
2 2
: sin cosn n s
mv Wv
F ma N N
g
β µ β
ρ ρ
Σ = + = =
( ) 2sin cos
cos sin
s
s
W N Wv
g
β µ β
β µ β ρ
+
=
−
( )( )2 2 2sin cos sin 45 0.5cos45
6 32.2 579.6 ft /s
cos sin cos45 0.5sin 45
s
s
v g
β µ β
ρ
β µ β
+ ° + °
= = =
− ° − °
24.1 ft/sv =

Uniformly accelerated motion on a circular path.
5
5 in. ft
2 12
D
ρ = = =
6
6 60 10
60 10 oz lb
16
W
−
− ×
= × =
2
32.2 ft/sg =
9 2
116.46 10 lb s /ft
W
m
g
−
= = ×
(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
:
5/12
n n n n
mv
F ma F ma
ρ
−
×
Σ = = = =
6
644.0 10 lb−
= ×
Magnitude of force:
( ) ( )2 22 2 6
10 644.0 1.3975t nF F F −
= + = +
lb10644 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
t
t t t t
W a
F ma a F W W W
g g
= = = = =
( )
( )( )
22
9
: 0.3144
32.2 8
n n n
WWv
F ma F W
gρ
= = = =
2 2
0.315t nF F F W= + =
This is the friction force available to cause the trunk to slide.
The normal force N is calculated from equilibrium of forces in the
vertical direction.
0: 0yF N WΣ = − = N W=
Since sliding is impending, 0.315s
F
W
µ = = 0.315sµ =

For constant speed, 0ta =
2
with 0.7 m/s, 0.2 mB
n B
v
a v ρ
ρ
= = =
: cosx x nF ma F ma θΣ = =
: sin siny y n nF ma N W ma N mg maθ θΣ = − = − = −
Ratio
2
cos cos cos
sin sin sin
n
n
n B
F ma
g gN mg ma
a v
θ θ θ
ρθ θ θ
= = =
− − −
With
( )( )
( )2 2
9.81 0.2 cos
4.0041, the ratio becomes
4.0041 sin0.7B
g F
Nv
ρ θ
θ
= = =
−
For no impending slide,
cos
4.0041 sin
s
F
N
θ
µ
θ
≥ =
−
To find the value of θ for which the ratio is maximum set the
derivative with respect to θ equal to zero.
( )2
cos 1 4.0041sin
0
4.0041 sin 4.0041 sin
d
d
θ θ
θ θ θ
± − 
= ± = 
−  −
1
sin 0.24974
4.0041
θ = =
cos14.446
14.446 , 0.258
4.0041 0.24974
F
N
θ
°
= ° = =
−
180 14.446 165.554 , 0.258
F
N
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 =
2
with 0.7 m/s, 0.2 mB
n B
v
a v ρ
ρ
= = =
: cosx x nF ma F ma θΣ = =
: sin siny y n nF ma N W ma N mg maθ θΣ = − = − = −
Ratio
2
cos cos cos
sin sin sin
n
n
n B
F ma
g gN mg ma
a v
θ θ θ
ρθ θ θ
= = =
− − −
Let 2
cos
so that
sinB
g F
u
N uv
ρ θ
θ
= =
−
Determine the value of θ at which F/N is maximum.
( )( )
( ) ( )
2
2 2
cos sin sincos 1 sin
0
sin sin sin
ud u
d u u u
θ θ θθ θ
θ θ θ θ
− − − 
= = = 
−  − −
The corresponding ratio .
F
N
2 1
1 2
1 sin
tan
cos1
F u u
N u u u
θ
θ
θ
− −
− −
± − ±
= = = ± = ±
− −
(a) For impending sliding to the left: tan 0.35s
F
N
θ µ= = =
( )
2
1
arctan 0.35 19.29 , sin ,Bv
u
g
θ θ
ρ
−
= = ° = =
( )( )2 2 2
9.81 0.2 sin19.29 0.648 m /sBv = ° =
0.805 m/sBv =

For impending motion to the right: tan 0.35s
F
N
θ µ= − = =
( )arctan 0.35 160.71θ = − = °
2
1 2
sin ,
v
u
g
θ
ρ
−
= = ( )( )2 2 2
9.81 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,
2
x 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 2
0 0 0
, so that and .
2
L
x 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≤ ≤ :
y
y y y
F eV
F ma a
m md
Σ = = =
( )0
0y y y
eVt
v v a t
md
= + = +
( )
2
2
0 0
1
0 0
2 2
y y
eVt
y y v t a t
md
= + + = + +
At ( )
2
1 1
1 11
, and
2
y
eVt eVt
t t v y
md 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= = + −
( )
2
1 1 1
2 1 2 1
1
2 2
eVt eVt eVt
t t t t
md md md
δ
 
= + − = − 
 
0 0 0 0
1
2 2
eV L
mdv v v v
 
= + − 
 
l l l
or 2
0
eV
mdv
L
δ =
l
!!!!

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,
2
x 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 2
0 0 0
, so that and .
2
L
x 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 / .yF eV d= After it passes the plates ( )1 2 ,t t t≤ ≤
it is zero.
For 10 ,t t≤ ≤ :
y
y y y
F eV
F ma a
m md
Σ = = =
( )0
0y y y
eVt
v v a t
md
= + = +
( )
2
2
0 0
1
0 0
2 2
y y
eVt
y y v t a t
md
= + + = + +
At
2
1 2
0 0
, ,
2
eV
t t y
v mdv
= =
ll
But 0.075 0.425
2
d
y d d< − =
So that
2
2
0
0.425
2
eV
d
mdv
<
l
2
2 2 2
0 0
1
1.176
0.425 2
d eV eV
mv mv
> =
l 2
0
1.085
d eV
mv
>
l

30 , 2 rad/s, 0θ θ θ= ° = =& &&
0.6 m,r W mg= =
Block B: Only force is weight
cos30 , sin30rF W F Wθ= ° = − °
(a) ( )2 :F ma m r rθ θ θ θ= = +&& &&
sin30
2 sin30
F mg
r r r g r
m m
θθ θ θ θ
°
= − = − − = − ° −& && && &&&
( ) ( )( )
( )( )
9.81 sin30 0.6 0sin30
1.226 m/s
2 2 2
g r
r
θ
θ
° +° +
= − = − = −
&&
&
&
/ rod 1.226 m/sB =v 60°
(b) ( )2
:r rF ma m r rθ= = − &&&
( )( ) ( )
2 2
2 2
cos30
cos30
0.6 2 9.81 cos30 10.90 m/s
rF mg
r r r r g
m m
θ θ θ
°
= + = + = + °
= + ° =
& & &&&
2
/ rod 10.90 m/sB =a 60°

2
45 , 0.8 m, 10 rad/srθ θ= ° = =
0, 0,rv r v r W mgθ θ= = = = =
(a) ( ): cos45 2F ma N W m r rθ θ θ θΣ = − ° = +
( )cos45 2N mg m r rθ θ= ° + +
( )( )(0.5)(9.81)cos45 0.5 0.8 10 0 = ° + + 
7.468= 7.47 NN = 45°
(b) ( )2
: sin 45r rF ma mg m r rθΣ = ° = −
2 2
sin 45 sin 45
mg
r r g r
m
θ θ= ° + = ° +
( ) 2
9.81 sin 45 0 6.937 m/s= ° + =
2
/ rod 6.94 m/sB =a 45°

Use radial and transverse components of acceleration.
2 3 2
3 ft 2 radr t t tθ= − =
2
6 3 ft/s 4 rad/sr t t tθ= − =
2 2
6 6 ft/s 4 rad/sr t θ= − =
2 2 3 2
6 6 (3 )(16 )ra r r t t t tθ= − = − − −
5 4 2
16 48 6 6 ft/st t t= − − +
2 3 2
2 (3 )(4) (2)(6 3 )(4 )a r r t t t t tθ θ θ= + = − + −
2 3 2
60 28 ft/st t= −
Mass: 24
0.12422 lb s /ft
32.2
= = = ⋅
W
m
g
(a) 0.t = 2
6 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. 2
32 ft/sra = −
2
32 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 cos2 ) ft 2 radr t tπ θ π= + =
12 sin 2 ft/s 2 rad/sr tπ π θ π= − =
2 2
24 cos2 ft/s 0r tπ π θ= − =
2 2 2
24 cos2 (6 6cos2 )(2 )ra r r t tθ π π π π= − = − − +
2 2 2
24 48 cos2 ft/stπ π π= − −
2 0 (2)( 12 sin 2 )(2 )a r r tθ θ θ π π π= + = + −
2
48 sin 2 tπ π= −
Mass: 21
0.031056 lb s /ft
32.2
W
m
g
= = = ⋅
(a) 0.t = 2
710.61 ft/sra = −
0aθ =
(0.031056)(710.61)r rF ma= = 22.1 lbrF = −
0F maθ θ= = 0Fθ =
(b) 0.75 s.t = 2
236.87 ft/sra = −
2
473.74 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, 0
dr
r r
dt
= = =
0 0
or 1.5 ft
r t
dr r dt r t= =∫ ∫
2
0.8 rad/s, 0.8 rad/stθ θ= =
( )( )22 3 2
0 1.5 0.8 0.96 ft/sra r r t t tθ= − = − = −
( )( ) ( )( )( ) 2
2 1.5 0.8 2 1.5 0.8 3.6 ft/sa r r t t tθ θ θ= + = + =
Kinetics: Sketch the free body diagrams for the collar.
:r r rF ma T maΣ = − =
:F ma Q maθ θ θΣ = =
Set T Q= to obtain the required time.
orr rma ma a aθ θ− = − =
Using the calculated expressions
3 2 23.6
0.96 3.6 , 3.75 s
0.96
t t t= = =
1.936 st =

2
10 rad/s, 10 rad/stθ θ= =& &&
2
0.5/32.2 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 /s
0.015528 1.5
mr T
mr mr T
mr
θ θ
+ +
= + = = =
&&& &&&
13.105 rad/sθ =&
Immediately after the cable breaks: 0, 0rF r= =&
(a) Acceleration of B relative to the rod.
( ) ( )( )22 2 2
0 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 = !

2
15 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 = = − = −
( )2
k m r mrθ− = −
(a) Radial coordinate.
( )( )
( )( )2 2
0.230 12
60 0.230 15
mr
r
k mθ
−
= − = −
− −
0.33455 m= 335 mmr =
( ): 2F ma F m r rθ θ θ θ θΣ = = +
2
F
r r
m
θ
θ θ= −
( )( )( )
9 0
1.304 m/s
2 2 0.230 15
F mr
r
m
θ 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
125
tan 22.62
175 125
= − = − °
+
θ θ
+ : cos22.6Σ = ° =s tF ma F ma
2cos22.6 70cos22.62
43.077 m/s
1.5
s
t
F
a
m
° °
= = =
Acceleration vector.
2 2
n ta a a= +
2 2 2
115.52 43.077 123.29 m/s= + =
115.52
tan
43.077
φ = 69.55φ = °
22.62 46.93φ − ° = °
cos46.93ra a= °
2
123.29cos46.93 84.2 m/s= ° = in negative r-direction
2
84.2 m/sra = −
sin 46.93a aθ = °
123.29sin 46.93= °
2
90.1 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 orB B Br v r a r a+ = + = = − (1)
For block A, + : cos or sec (2)A A
x A A A A
W W
F m a T a T a
g g
θ θΣ = = =
For block B, : (3)B B
y B B B
W W
F a W T a
g g
Σ = − =
Adding Eq. (1) to Eq. (2) to eliminate T, secA B
B A B
W W
W a a
g 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 32.2
17.991 ft/s
sec cos 40sec30 50cos30
B
A
A B
W g
a
W Wθ θ
= = =
+ ° + °
From Eq. (6), 2
17.991cos30 15.581 ft/sBa = ° =
(a) From Eq. (2), ( )( )40/32.2 17.991 sec30 25.81T = ° = 25.8 lbT =
(b) Acceleration of block A. 2
17.99 ft/sA =a
(c) Acceleration of block B. 2
15.58 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/s
r A r Ar v v= = ⋅ = − °
= − ° = −
v e
sin30
6sin30 3 ft/s
A Ar v vθ θθ = = ⋅ = − °
= ° =
v e
3
1.25 rad/s
2.4
v
r
θ
θ = = =
Constraint of cable: constant,Br y+ =
0, 0 orB B Br v r a r a+ = + = = − (1)
For block A, + : cos or secA A
x A A A A
W W
F m a T a T a
g g
θ θΣ = = = (2)
For block B, :B B
y B B B
W W
F a W T a
g g
Σ = − = (3)
Adding Eq. (1) to Eq. (2) to eliminate T, secA B
B A B
W W
W a a
g 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
2
cosB Aa a rθ θ= − (6)

Substituting Eq. (6) into Eq. (4) and solving for ,Aa
( ) ( ) ( )( )22
2
50 32.2 2.4 1.25
20.086 ft/s
sec cos 40sec30 50cos30
B
A
A B
W g r
a
W W
θ
θ θ
 ++   = = =
+ ° + °
From Eq. (6), ( )( )2 2
20.086cos30 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. 2
20.1 ft/sA =a
(c) Acceleration of block B. 2
13.65 ft/sB =a

Since the particle moves under a central force, constant.h =
Using Eq. (12.27), 2
0 0 0h r h r vθ= = =
or 0 0 0 0 0
2 2
00
cos2
cos2
r v r v v
rr r
θ
θ θ= = =
Radial component of velocity.
( )
0
0 3/ 2
sin 2
cos2 cos2
r
dr d r
v r r
d d
θ
θ θ θ
θ θ θ θ
 
= = = = 
 
( )
0
0 3/ 2
sin 2
cos2
cos2
v
r
r
θ
θ
θ
= 0
sin 2
cos2
rv v
θ
θ
=
Transverse component of velocity.
0 0
0
cos2
h r v
v
r r
θ θ= = 0 cos2v vθ θ=

Using Eq. (12.27), 2
0 0 0h r h r vθ= = = or 0 0 0 0
2 2
00
cos2
cos2nr v r v v
rr r
θ
θ θ= = =
Differentiating the expression for r with respect to time,
( ) ( )
0 0
0 0 03/ 2 3/ 2
0
sin 2 sin 2 sin 2
cos2
cos2 cos2cos2 cos2
dr d r v
r r r v
d d r
θ θ θ
θ θ θ θ
θ θ θ θθ θ
 
= = = = = 
 
Differentiating again,
( )
2 2 2 2 2
0
0 0 3/ 2
0
sin 2 2cos 2 sin 2 2cos 2 sin 2
cos2 cos2cos2
dr d v
r v v
d d r
θ θ θ θ θ
θ θ θ
θ θ θ θθ
+ + 
= = = = 
 
(a) 0
0
0
sin2
sin 2
cos2
r
v r
v r v
r
θ
θ
θ
= = = 0
0
cos2
v r
v r
r
θ θ θ= =
( ) ( )2 2 2 20
0
sin 2 cos 2r
v r
v v v
r
θ θ θ= + = + 0
0
v r
v
r
=
2 2 2 2
2 20 0 0
2
0 0
2cos 2 sin 2
cos 2
cos2 cos2
r
v r v
a r r
r r
θ θ
θ θ
θ θ
+
= − = −
2 2 2 2
0 0 0
2
0 00
cos 2 sin 2
cos2 cos2
v v v r
r rr
θ θ
θ θ
+
= = =
2
0
2
0
:r r
mv r
F ma
r
= =
2
0
2
0
r
mv r
F
r
=
Since the particle moves under a central force, 0aθ =
continued

Magnitude of acceleration:
2
2 2 0
2
0
r
v r
a a a
r
θ= + =
Tangential component of acceleration.
2
0 0 0
2
0 0 0
sin 2t
dv d v r v v r
a r
dt dt r r r
θ
 
= = = = 
 
Normal component of acceleration.
2 2
22 20 0
2 2
0 0
cos2
1 sin 2t t
v r v r
a a a
r r
θ
θ= − = − =
But
2
0
cos 2
r
r
θ
 
=  
 
Hence,
2
0
n
v
a
r
=
(b) But
22 2 2
0
2 2
0 0
orn
n
v v v r r
a
a r v
ρ
ρ
= = = ⋅
3
2
0
r
r
ρ =

Using Eq. (12.27), 2
0 0 0h r h r vθ= = =
or 0 0 0 0 0
2 2 22
00 coscos
r v r v v
r rr
θ
θθ
= = =
Radial component of velocity. ( ) ( )0 0cos sinr
d
v r r r
dt
θ θ θ= = = −
Transverse component of velocity. ( )0 cosv r rθ θ θ θ= =
Speed. 2 2 0 0
0 2
0 cos
r
r v
v v v r
r
θ θ
θ
= + = = 0
2
cos
v
v
θ
=

Since the particle moves under a central force, constanth =
Using Eq. (12.27), 2
0 0 0h r h r vθ= = = 0 0 0 0 0
2 2 22
00 coscos
r v r v v
r rr
θ
θθ
= = =
Radial component of velocity. ( ) ( )0cos sinr o
d
v r r r
dt
θ θ θ= = = −
Transverse component of velocity. ( )0 cosv r rθ θ θ θ= =
Speed. 2 2 0 0 0
0 2 2
0 cos cos
r
r v v
v v v r
r
θ θ
θ θ
= + = = =
Tangential component of acceleration.
( )( ) 2
0 0 0
0 3 3 2 5
0 0
2 sin 2 sin 2 sin
cos cos cos cos
t
dv v v v
a v
dt r r
θ θ θ θ
θ θ θ θ
− −
= = = ⋅ =
Tangential component of force.
2
0
5
0
2 sin
:
cos
t t t
mv
F ma F
r
θ
θ
= =
(a) 0, 0tFθ = = 0tF =
(b) 0
5
2 sin 45
45 ,
cos 45
t
mv
Fθ
°
= ° =
°
2
0
0
8
t
mv
F
r
=

For gravitational force and a circular orbit,
2
2
orr
GMm mv GM
F v
r rr
= = =
Let τ be the periodic time to complete one orbit.
2 or 2
GM
v r r
r
τ π τ π= =
Solving for ,τ
3/ 2
2 r
GM
π
τ =
But, 3 3/ 24
, hence 2
3 3
M R GM G R
π
π ρ ρ= =
Then,
3/ 2
3 r
G R
π
τ
ρ
 
=  
 
Using 3r R= as given leads to
3/ 2 3
3 9
G G
π π
τ
ρ ρ
= = ( )1/ 2
9 /Gτ π ρ=

2
2
orr
GMm mv GM
F v
r rr
= = =
Let τ be the period time to complete one orbit.
But
2
2 2 2 2
2 or 4
GM
v r v r
r
τ
τ π τ π= = =
Then
1/3
2 2
3
2 2
or
4 4
GM GM
r r
τ τ
π π
 
= =   
 
Data: 3
23.934 h 86.1624 10 sτ = = ×
(a) In SI units: 2 6
9.81 m/s , 6.37 10 mg R= = ×
( )( )
2
2 6 12 3 2
9.81 6.37 10 398.06 10 m /sGM gR= = × = ×
( )( )
1/3
2
12 3
6
2
398.06 10 86.1624 10
42.145 10 m
4
r
π
 
× × = = × 
  
altitude 6
35.775 10h r R= − = × 35800 kmh =
In US units: 2 6
32.2 ft/s , 3960 mi 20.909 10 ftg R= = = ×
( )( )
2
2 6 15 3 2
32.2 20.909 10 14.077 10 ft /sGM gR= = × = ×
( )( )
1/3
2
15 3
6
2
14.077 10 86.1624 10
138.334 10 ft
4
r
π
 
× × = = × 
  
altitude 6
117.425 10 fth r R= − = × 22200 mih =

(b) In SI units:
12
3
6
398.06 10
3.07 10 m/s
42.145 10
GM
v
r
×
= = = ×
×
3.07 km/sv =
In US units:
15
3
6
14.077 10
10.09 10 ft/s
138.334 10
GM
v
r
×
= = = ×
×
3
10.09 10 ft/sv = ×

2
2
orr
GMm mv GM
F v
r rr
= = =
2 or 2
GM
v r r
r
τ π τ π= =
from which
2 3
2
4 r
GM
π
τ
=
But
2 3
2
2 2
4
, hence,
r
GM gR g
R
π
τ
= = (1)
Solving for ,r
1/ 3
2 2
2
4
gR
r
τ
π
 
=   
 
Data: 9
417,000 mi 2.202 10 ftr = = ×
6
44,400 mi 234.4 10 ftR = = ×
3
3.551 days 85.224 h 306.8 10 sτ = = = ×
Using (1),
( )
( ) ( )
3
2 9
2
2 2
3 6
4 2.202 10
81.5 ft/s
306.8 10 234.4 10
g
π ×
= =
× ×
2
81.5 ft/sg =

Let M be the mass of the sun and m the mass of Venus.
For the circular orbit of Venus,
2
2
2 n
GMm mv
ma GM rv
rr
= = =
where r is radius of the orbit.
Data: 6 9
67.2 10 mi 354.8 10 ftr = × = ×
3 3
78.3 10 mi/h 114.84 10 ft/sv = × = ×
( )( )
2
9 3 21 3 2
354.8 10 114.84 10 4.6792 10 ft /sGM = × × = ×
(a) Mass of sun.
21 3 2
9 4 4
4.6792 10 ft /s
34.4 10 ft /lb s
GM
M
G −
×
= =
× ⋅
27 2
136.0 10 lb s /ftM = × ⋅
(b) At the surface of the sun, 3 9
432 10 mi 22.81 10 ftR = × = ×
2
GMm
mg
R
=
( )
21
2 2
9
4.6792 10
22.81 10
GM
g
R
×
= =
×
2
899 ft/sg =

2
2
orr
GMm mv GM
F v
r rr
= = =
But
2
3
2
2 2
, hence, or
4
r r GM GM
v r
r
π π τ
τ τ π
= = =
Solving for r,
1/ 3
2
2
(1)
4
GM
r
τ
π
 
=   
 
( )
1/31/3 222
2
2 2
44 GMGM
v GM
r GM
ππ
τ τ
  
 = = =       
1/ 3
2
(2)
GM
v
π
τ
 
=  
 
For earth: 6 2
6.37 10 m, 9.81 m/sR g= × =
( )( )
2
2 6 12 3 2
9.81 6.37 10 398.06 10 m /sGM gR= = × = ×
For Jupiter: ( )( )12 15
319 398.06 10 126.98 10GM = × = ×
(a) For Ganymede: 3
7.15 days 171.6 h 617.76 10 sτ = = = ×
By Eq. (1),
( )( )
1/ 3
2
15 3
9
2
126.98 10 617.76 10
1.071 10 m
4
r
π
 
× × = = × 
  
6
1.071 10 kmr = ×
(b) For Callisto: 6
16.69 days 400.56 h 1.4420 10 sτ = = = ×
By Eq. (2),
( )
1/ 3
12
3
6
2 126.98 10
8.209 10 m/s
1.4420 10
v
π ×
 = = ×
 ×
 
8.21 km/sv =

2
2
orr
GMm mv GM
F v
r rr
= = =
2 or 2
GM
v r r
r
τ π τ π= =
Solving for r,
1/3
2
2
4
GM
r
τ
π
 
=   
 
For earth, 6 2
6370 km 6.37 10 m, 9.81 m/sR g= = × =
( )( )
2
2 6 12 3 2
9.81 6.37 10 398.06 10 m /sGM gR= = × = ×
Data: 120 min 7200 sτ = =
( )( )
1/ 3
212
6
2
398.06 10 7200
8.055 10 m
4
r
π
 ×
 = = ×
 
 
(a) altitude 6
1.685 10 mh r R= − = × 1685 kmh =
(b)
6
6
6.37 10
cos 0.7908
8.055 10
R
r
θ
×
= = =
×
37.74θ = °
( )( )75.48 72002
1509.6 s
360 360
ABt
θ
τ= = =
°
25.2 minABt =

2 2n n
GMm GM
ma a
r r
= =
( )( )9 4 4 21 2
34.4 10 ft /lb s 5.03 10 lb s /ftGM −
= × ⋅ × ⋅
12 3 2
173.032 10 ft /s= ×
(a) At the surface of the moon, na g=
6
1080 mi 5.7024 10 ftr R= = = ×
( )( )
( )
9 4 4 21 2
2 2
6
34.4 10 ft /lb s 5.03 10 lb s /ft
5.7024 10 ft
GM
g
R
−
× ⋅ × ⋅
= =
×
2
5.32 ft/sg =
(b) Orbit of space vehicle.
6
1080 200 1280 mi 6.7584 10 ftr = + = = ×
2
2n
v GM
a
r r
= =
GM
v
r
=
3/2
2 2r r
v GM
π π
τ = =
( )
3/2
6
3
12
2 6.7584 10
8.3923 10 s
173.032 10
π
τ
×
= = ×
×
2.33 hτ =

2
2 n
GMm mv
ma
rr
= = 2 GM
v
r
=
2 r
v
π
τ
=
2 2
2
4 r GM
r
π
τ
=
3
2 2
constant
4
r GM
τ π
= =
Tethys: 3 6
183.3 10 mi 967.8 10 ftr = × = ×
3
1.888 days 163.12 10 sτ = = ×
( )
( )
3
6
15 3 2
2 2
3
967.8 10
34.068 10 ft /s
4 163.12 10
GM
π
×
= = ×
×
Rhea: 3
4.52 days 390.53 10 sτ = = ×
3 2
2
4
GM
r τ
π
=
( )( )
2
15 3 27 3
34.068 10 390.53 10 5.1958 10 ft= × × = ×
(a) 9
1.732 10 ftr = × 3
328 10 mir = ×
(b) Mass of Saturn.
2
2
4
4
GM
M
G
π
π
 
=  
 
( )2 15
24
9
4 34.068 10
39.1 10
34.4 10
π
−
×
= = ×
×
24 2
39.1 10 lb s /ftM = × ⋅

Parabola AB.
6
6370 km 6.37 10 mR = = ×
6
6370 960 7330 km 7.33 10 mAr = + = = ×
6
6370 8300 14670 km 14.67 10 mBr = + = = ×
( )A A B Bmr v mr v θ
=
( )
( )( )6 3
3
6
7.33 10 10.4 10
5.196 10 m/s
14.67 10
A A
B
B
r v
v
rθ
× ×
= = = ×
×
0, 0A Ax y= =
6
14.67 10 mB Bx r= = ×
6
7.33 10 mB Ay r= = ×
For a parabolic trajectory, 2
y kx= from which 2
B
B
y
k
x
=
Differentiating with respect to x, 2
2
2 B
B
dy y x
kx
dx x
= =
At point B,
( )( )6
6
2 7.33 102
tan 1.0000
14.67 10
B
B
dy y
dx x
φ
×
= = = =
×
45φ = °
( ) 3
35.196 10
7.349 10 m/s
sin 45 sin 45
B
B
v
v θ ×
= = = ×
° °
3
7.35 10 m/sB = ×v 45°

For earth, 6
3960 mi 20.909 10 ftR = = × ( ) ( )( )
2
2 6 15 3 2
earth
32.2 20.909 10 14.077 10 ft /sGM gR= = × = ×
For sun, ( ) ( )( )3 15 21 3 2
sun
332.8 10 14.077 10 4.6849 10 ft /sGM = × × = ×
For circular orbit of earth, 6 9
92.96 10 mi 490.8 10 ftEr = × = ×
( ) 21
3sun
9
4.6849 10
97.70 10 ft/s
490.8 10
E
E
GM
v
r
×
= = = ×
×
For transfer orbit AB, 6 9
, 141.5 10 mi 747.12 10 ftA E B Mr r r r= = = × = ×
( ) ( )( )3
97.70 10 1.83 5280 107.36 ft/sA E A
v v v= + ∆ = × + =
A A B Bmr v mr v=
( )( )9 3
3
490.8 10 107.36 10
70.527 ft/s
747.12 10
A A
B
B
r v
v
r
× ×
= = =
×
For circular orbit of Mars, 6 9
141.5 10 mi = 747.12 10 ftMr = × ×
( ) 21
3sun
9
4.6849 10
79.187 10 ft/s
747.12 10
M
M
GM
v
r
×
= = = ×
×
Speed increase at B.
( ) 3 3 3
79.187 10 70.527 10 8.660 10 ft/sM BB
v v v∆ = − = × − × = ×
( ) 1.640 mi/sB
v∆ =

Circular orbits:
GM
v
r
=
6
1400 mi 7.392 10 ftAr = = ×
( )
( )( )9 21
3
61
34.4 10 5.03 10
4.8382 10 ft/s
7.392 10
Av
−
× ×
= = ×
×
6
1300 mi 6.864 10 ftBr = = ×
( )
( )( )9 21
3
62
34.4 10 5.03 10
5.0208 10 ft/s
6.864 10
Bv
−
× ×
= = ×
×
(a) Transfer orbit AB.
( ) ( ) ( ) 3 3
2 1
4.8382 10 86 4.7522 10 ft/sA A A
v v v= + ∆ = × − = ×
( ) ( )2 1A A B Bmr v mr v=
( )
( ) ( )( )6 3
32
31
7.392 10 4.7522 10
5.1178 10 ft/s
6.864 10
A A
B
B
r v
v
r
× ×
= = = ×
×
( ) 3
1
5.12 10 ft/sBv = ×
(b) Speed change at B.
( ) ( ) ( ) 3 3
2 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∆ =

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

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

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