Dynamics Answers 2009-11-2

Engineering Mechanics: Dynamics 1e
Gray, Costanzo, Plesha
Answers to Selected Even-Numbered Problems

We are in the process of adding content (e.g., FBDs) to this document. The answers reported here, along with those
in the solutions manual, are being accuracy checked by several professors. This document, as well as the solutions
manual, will be updated and completed by September 2009.
Please note that answers are not provided for the following type of even-numbered problems:
Concept Problems.
Computer Problems.
Design Problems.
For Concept and Computer problems, please consult the solutions manual.
Chapter 1
1.2
.rC =D /` D 3:88 ft
1.4
rEB=A
rEB=A
rE
D .4:00 {O 1:00 |O/ ft

D .3:31 uOp 2:46 uO q / ft
D rE
B=A xy system
B=A pq system
D 4:12 ft
1.6
The angle is dimensionless (or nondimensional) Observe that since angles can be expressed in a variety of units,
such as radian or degree, this problem illustrates the idea that some nondimensional quantities still require proper
use of units.
1.8
Ixx D Iyy D I D M L2
Units of Ixx , Iyy , and I in the SI system: kgm2 ;
Units of Ixx , Iyy , and I in the U.S. Customary system: slugft2 D lbs2 ft.
1.10
Concept problem.
Chapter 2
2.2
2.4
2.6
2.8
vE.15 s/ D .72:1 {O C 6:24 |O/ ft=s

aE .15 s/ D . 15:8 {O 1:39 |O/ ft=s2
.15 s/ D 4:95
.15 s/ D 180
Concept problem.

y.x/ D 2:00 C 3:00x C 2:00x 2 m
Concept problem.
2.10
Er .t1 ; t2 / D .0:899 {O C 41:3 |O/ ft

vEavg .t1 ; t2 / D .0:450 {O C 20:7 |O/ ft=s
2.12
v.t/ D r.t/
P D .4:08e 13:6t / {O C . 0:720 1:28e 13:6t / |O m=s
a.t/ D v.t/
P D . 55:5e 13:6t / {O C .17:4e 13:6t / |O m=s2
The impact angle is the slope of the stones trajectory at the time that the stone enters the water. Then, recalling
that the velocity is always tangent to the trajectory, we have D 26:1

y.x/ D 5:25 C 3:25x C 1:00x 2 m
xmax D 3:00 m and xmin D 1:00 m
2.14
Last modified: November 2, 2009

2.16
vE D .29:3 ft=s/1 C cos.25:5 rad=s/t {O .29:3 ft=s/1 C sin.25:5 rad=s/t |O

p
v D .29:3 ft=s/ 2 C 2 cos.25:5 rad=s/t

aE D 748 ft=s2 sin.25:5 rad=s/t {O
748 ft=s2 cos.25:5 rad=s/t |O
2.18
Computer problem.
2.20
Computer problem.
2.22
.E
vP =O /1 D .1 C cos t/ {O1 C .4 2t/ |O1 m=s
.E
vP =O /2 D .1 C cos t/ cos 2.t 2/ sin {O2
.E
aP =O /1 D . sin t {O1
2.t
2 |O1 / m=s2
.E
aP =O /2 D . cos sin t 2 sin / {O2 C .sin sin t
jE
vP =O .t D 2 s/j1 D 0:584 m=s
jE
vP =O .t D 2 s/j2 D 0:584 m=s
2.24

2/ cos C .1 C cos t/ sin |O2 m=s
2 cos / |O2 m=s2
vEP =B D . 6:14 {OB C 23:7 |OB / ft=s

vP =A D 24:5 ft=s
vP =B D 24:5 ft=s
aEP =A D .1:78 {OA C 5:96 |OA / ft=s2
aP =A D 6:22 ft=s2
aP =B D 6:22 ft=s2
2.26
vEA D . 20:0 {O 12:8 |O/ ft=s

p
p
2 {O C 3 2 |O ft=s2 D . 1:41 {O C 4:24 |O/ ft=s2
aEA D
2.28
Computer problem.
2.30
Computer problem.
2.32
xR D
8a3 v02
4a2 C y 2
yR D
2.34
2.36
2
4a2 v02 y
4a2 C y 2
2
Computer problem.
p

v0
vE D p
3a {O C b |O
3a2 C b 2
5
s D 7:85 s
2
2.38
tbraking D
2.40
tjajmax D 3:93 s and tjajmax D 11:8 s

samax D 12:8 m and samax D 129 m
2.42
v0 D 10:1 m=s
2.44
2.46
2.48
2.50
where jajmax D 3:60 m=s2

a0
sin 2!t C cos !v.t/ D 7:00 sin t C 1:50 cos.0:500t/
!
x.t/ D 7:001:00 1:50 cos t C 3:00 sin.0:500t/ m
r
mg
vterm D
D 5:00 m=s
Cd

Cd t
mg
v.t/ D
1 e m
Cd
mg
vterm D
Cd
x.t/
P
D

1:50 m=s
vf D 3:56 m=s

2.52
Computer problem.
v D 10:1 ft=s
2.54
s D 186 ft
t D 3:90 s
r
2g
P
./ D P02 C
cos
L
2.56
2.58
2.60
2.62
2.66
k 2
x
m
x0 /
x02
txmax D 0:243 s
"
r0
(a) t D p
sin
2G .mA C mB /
(b) t D 106 h
t D 106 h
v
tstop D 0 D 2:88 s
gk
.vT /20 aT
2.68
k D
2.70
t D 8:63 s
yP D 77:7 m=s
2.72
v D v0 C ac .t
s D s0 C v0 .t
2.74
cos 0
.P0 /min D 4:93 rad=s

s

kL0
xP D v02 C 2 g C
.x
m
3=2
2.64
g .vT /20
2daT
r
r0
r
r0

1
r
r0

2
D 0:301
t0 /
t0 / C 21 ac .t
ap
hvp2
s D
C
2 r 3
r
s rDr D 1:44 rad=s2
and
t 0 /2
s rDr D 3:63 rad=s2

2
2.76
D tan
2.78
vEgun D .14:7 ft=s/ {O
2.80
y D h C tan .x
2.82
The golfers chip shot is successful.
2.84
v0 D 52:8 ft=s
2.86
The two firing ranges are

30:4 56:8 and 69:4 74:5
The sizes of these ranges are 1 D 26:4 and 2 D 5:09
The angle subtended by the target as seen from an observer at point O is D 21:3
Unlike Example 2.11, the difference between the angle subtended by the target and 1 or 2 is significant. In
addition, we see that the value of 1 is much closer to than 2 .
2.88
dAB D 82:0 m
2.90
v0 D 62:5 ft=s
2.92
Computer problem.
d D 4:70 m
w/
g
2v02 cos2
.x
w/2
and D 20:6
vEinitial D v0 cos {O C v0 sin |O D .6:02 {O C 7:67 |O/ m=s

where v0 is the initial speed of the tiger and is the angle formed with the horizontal by the initial velocity of
the tiger.

r
2.94
v0 D
gR
cos
p
2
sin cos.
2.96
max D

D 45
4
2.98
v02
D 447 ft
g
2v sin max
tD 0
D 5:27 s
g
h

y D 59:9103 ln 1 2:8410
/
RD

i
x C 17:7x m
2.100
Computer problem.
tI D 6:19 s
xI D 268 m
2.102
y D v0 sin
2.104
(a) aE bE D 1 6 C 2 3 C 3 0 D 0

(b) aE aE bE D 84 {O 42 |O C 0 kO
x cos y sin
v0 cos. C /
1
2 g cos
x cos y sin
v0 cos. C /
2
(c)
2.106

(a) !E 1 D 105 kO rad=s; !E 2 D 105 {O rad=s and !E 3 D 105 |O rad=s

100
(b) !E ` D p
{O C |O C kO rad=s ) 60:5 {O C |O C kO rad=s
3 3
2.108
vEA D 5 kO 2 {O D 10 |O m=s;

vEC D 5 kO 2 {O C 2 kO D 10 |O m=s;
v0 O
kD
R

vEB D 5 kO 2 {O C 1 kO D 10 |O m=s;

vE D 5 kO 2 {O C 3 kO D 10 |O m=s
D
.65:7 rad=s/ kO
2.110
!E D
2.112
jE
v j D 65;600 ft=s
Referring to the below,
let denote the angle formed by the velocity vector and the x axis such that tan D vy =vx . Then,
D 70:0
P D
2.114
rP D v0 cos. C /
2.116
vEB D .4:00 uO r C 0:800 uO / ft=s

aE D . 0:320 uO r C 3:20 uO / ft=s
2.118
rP D 0
rR D
v02
d
and
P D
D 9:00 ft=s2
and
v0
D
d
and
2.120
aE D . 112 {O C 5 |O/ m=s2
2.122
2! rP C !r
P D0
2.124
aE D . 262 uO C
v0
sin. C /
r
1:50 rad=s
R D 0 rad=s2
200 uO B / ft=s

2.126
Concept problem.
2.128
Concept problem.
2.130
Concept problem.
2.132
Concept problem.
.0:292/s uO n g ft=s2
2.134
aE .s/ D f 19:0 uO t C 69:9
2.136
2.138
Concept problem.

aE D .33:710 3 cos / m=s2
2.140
v0 D 19:2 ft=s
2.142
D
2.144
min D
2.146
vP D
2.148
s D 304 ft
v02
D 282 ft
g
v02
D 565 ft
2g
v02
D 4:74 s
tf D
2g vf C v0
13:8 ft=s2
2.150
Concept problem.
2.152
Concept problem.
2.154
rP D 466 mph
P D 41:6 rad=h
rR D 12;100 mi=h2
R D 5550 rad=h2
2.156
P
r.0/
P
D 438 ft=s and .0/
D 34710 6 rad=s
2
R
r.0/
R
D 14:9 ft=s
and .0/
D 10:610 6 rad=s2
2.158
rP D 0:265 ft=s
q
P 2 D 5490 ft=s
vE D rP 2 C .L/
v
!2
u

2
2
u
2
P
aE D t vs
L
C 2vs P D 1:00106 ft=s2
2L
2.160
2.162
vE D . 1:3 uO r C 0:22 r uO / m=s

. 0:0484 r uO r 0:572 uO / m=s2
2.164
PA D
PH D
vA L
D
.d=2/2 C L2
s
2 C v2
1 vA
B
L
1
RA D 2
r
R D
2.166
2.168
2
vA
"
2
vA
D
2

2 L
vB
2d
2
vB
2dL

D
1:1610
rad=s
4:29 rad=s
#
2 dL
vA
D
r2
1:6810
rad=s2
0:0755 rad=s2
rR D 5:78 ft=s2 and R D 1:42 rad=s2

h

i
aEP D
r 3r P R 3rP P 2 uO r C r P 3 C 3rR P C 3rP R uO

2.170
.h C sin /2 C .d C cos /2

v0 .h cos d sin /
rP D q
.d C cos /2 C .h C sin /2
rD
rR D
R D
2.172
v0 . C d cos C h sin /
:
d 2 C h2 C 2 C 2d cos C 2h sin

v02 . C d cos C h sin / d 2 C d cos C h2 C h sin
3=2
d 2 C 2d cos C h2 C 2 C 2h sin

v02 d 2 C h2 2 .d sin h cos /
2
d 2 C 2d cos C h2 C 2 C 2h sin
and
P D
P D 0
R D p
2
R!AB
h2
R2
15:8 rad=s2
2.174
vE D .37:710 3 uO r C 0:534 uO / m=s

aE D . 1680 uO r C 237 uO / m=s2
2.176
vEfollower D .1:76 m=s/ |O

aEfollower D .137g/ |O
2.178
v D 0:119 ft=s
jE
aj D 0:0864 ft=s2
2.180
Concept problem.
2.182
Concept problem.
rEB=A
vEA
jErB=A j
2.184
ROS D vEB
2.186
vEC =A D .6:86 {O C 30:3 |O/ m=s

aEC =A D .1:38 {O C 1:00 |O/ m=s2
2.188
Computer problem.
For vB=W D 10 s, the boat reaches A in 6:86 s
When vB=W D 7 s, the boat does not reach A
2.190
2.192
vrain D 16:4 ft=s

aEB D L1 R cos L1 P 2 sin C L2 R cos L2 P 2 sin {O

C L1 P 2 cos C L1 R sin C L2 P 2 cos C L2 R sin |O
2.194
D C sin
2.196
D 14:9
2.198
vpmax D vs C
2.200
vEB D
3:09 ft=s
aEB D
0:319 ft=s2
`
2
.vA `
vB d / cos
vP =A d
D 63:6 :
g
D 331 ft=s D 225 mph
2h
2.202
vEB D . 2:00 m=s/ |O
2.204
aEB D . 11:1 m=s2 / {O

s
d
tD
D 0:164 s
2a0
@

2.206
2.208
2.210
2.212
2.214
26d
D 0:937 ft=s2
3t 2
p
v0 w 2 C h 2
vEA D
{O D . 3:08 ft=s/ {O
p
2
2
wC
w pC h

h

i
p
v02 w
h2 C w 2
a0 h2 C w w C h2 C w 2
{O D . 0:893 ft=s2 / {O
aEA D
2

p
w C h2 C w 2
a0 D
Concept problem.

0:292 cos
|O
vEC D . 6:61 ft=s/ sin 1 C p
2
0:195
0:0851
sin

0:292 cos
aEC D 1:28104 ft=s2 cos 1 C p
0:195 0:0851 sin2

2
0:292 sin
0:0248 cos sin
|O
C sin
p
.0:195 0:0851 sin2 /3=2
0:195 0:0851 sin2
q
r D .d C cos /2 C .h C sin /2
P D
v0 . C d cos C h sin /
2.216
Concept problem.
2.218
Concept problem.
2.220
and
v0 .h cos d sin /
rP D p
Referring to the figure below and keeping in mind that the coordinate system is defined in such a way that the
triad .uO R ; uO ; uO / is right-handed
we have
vEC D . 6:5 uO r C 5:52 uO C 5:3 uO / ft=s
aEC D . 0:662 uO r
2.222
1:56 uO / ft=s2
K
vE D P tan uO R C
uO C P uO
tan

K2
aE D R tan
uO R C R uO
3
tan3
2.224
P R sin C 2rP P sin C 2r P P cos /
r .r
r P sin .rR r P 2 r P 2 sin2 /
r.r
P R C 2rP P
r P sin .r R C 2rP P r P 2 sin cos / D 0

r.r
P R sin C 2rP P sin C 2r P P cos / D 0
r P 2 sin cos /
P rR
r .
r P 2
r P 2 sin2 / D 0
2.226 rP D 0
P D 0
v0
D 0:0177 rad=s
r sin
v2
rR D 0 D 11:3 ft=s2
r
P D

v02 cos
D 78:110
r 2 sin
R D 0
R D
2.228
2.230
2.232
2.234
2.238
rad=s2
Computer problem.
xland D 3:35 m;
yland D 1:42 m;
1:26 |O2 / m=s2
Computer problem.
h D 29:410 6 m

mg
s.t/ D 2 Cd t C m e
Cd
Cd
m
2.244
D 240 ft
2.246
f D 629 m
tf D 5:24 s
rP D180 D 0
P D180 D 2:94 rad=s
rR D180 D 40:8 m=s2 and
P D30 D 0:0539 rad=s
R D30 D 0:243 rad=s2
2.250

1
r

k
cos
t
m
2.242
2.248
land D 0:
vEP =B .10:2 {O2 C 10:3 |O2 / m=s

gm
x D L0 C
1
k
p
v0 gR
2.240
and
Concept problem.
aEP =B D .7:82 {O2

2.236
R D180 D 0
2.252
jE
aj D 2L! 2 0 D 1:24 ft=s2
2.254
a t D 3:33 ft=s2
2.256
jE
aA j D 2:58 m=s2
s
2 2

d
d! 2
aE D
C
C a02 D 173;000 ft=s2
2
2
q

1 P4
4 2
aE
D
L
! 2 P 2 1 D 2:53 m=s2
2 C ! 1

2.258
2.260
min
Chapter 3
3.2
Concept problem.
3.4
ay D
3.6
xR C
3.8
ay D
3.10
dD
g .2P W /
W
EA
xD0
mL
Cd 2
v
m
v02
2k g
g
L D 50:8 ft

3.12
3.14
3.16
3.18
3.20
3.22

Cd t
mg
D 0:701 m=s
1 e m
Cd
9
8
>
2
=
<
2
v0 .` h/
m
T D
D 251 lb
gC h
i
3=2 >
2
;
:
2 xA2 C .` h/2
yP D
.Fs /max D 2mg

Concept problem.
v v0
D 7:59 s
k g
s
W
xstop D v0
D 0:737 ft
gk
tD
3.24
k D 50;800 lb=ft
3.26
D 0:167 m
max
Fsp
D kmax D 586 N
max
3.28
3.30
Computer problem.
max D 0:568 ft; when t D 0:177 s
s
2
4 2mvi
xstop D
D 0:160 m
3.32
k D 5:93 lb=ft
3.34
k D 9:03 kN=m
v "
u

u k x2
0
v.0/ D 2t
C L0 L
m 2
3.36
3.38
Concept problem.
3.40
Concept problem.
3.42
Concept problem.
3.44
3.46
D
x02
C L2
#
W v2
D 1:40104 ft
g .N W /
v2
D 35:51 m=s2 D 3:62 g

mv 2
D 43;100 N
FL D mg C

an D
3.48
k
xR C x 1
m
k
yR C y 1
m
ru
p
D 0;
x2 C y2
ru
x2 C y2
!
D0
3.50
rR
r P 2 C
k
.r ru / g cos D 0;
m
r R C 2rP P C g sin D 0
3.52
3.54
3.56

m
m2 g
tan 0 e Cd x=m 1
e Cd x=m
2
2
2
Cd
2v0 cos 0 Cd
!
v02
1 2
s D cos
C
D 47:7
3
3gR
yD
vmax D
s
g
2
s C tan
D 55:36 m=s D 199 km=h
1 s tan
tan s
D 35:33 m=s D 127 km=h
1 C s tan
127 km=h v 199 km=h
p
vmin D
g
3.58
(a) aE D .319103 uO r C 603 uO / m=s2

(b) P D mar C mg sin D 5:90106 N
(c) R D ma C mg cos D 11:3103 N
3.60
! D 9:94 rad=s
3.62
3.64
3.66
Concept problem.
RR C 2RP P D 0
R C cot2 / RP 2 D g cot
R.1

2
B D sin 1
D 41:8
3
3.68
me
D 0;
r2
r R C 2rP P D 0
r P 2 C G
rR
3.70
3.72
3.74
kD
.dmax =2
L0 /

2
s D cos 1
D 132
3
.dmax
L0 /2
D 4:74 lb=ft
P2 2
rL
R 2
P 2 C R L R C p gR
R

D0
.L2 R2 /
.L2 R2 /2
L2 R 2
RR C 2RP P D 0
3.76
3.78
3v0 m
2
Computer problem.
rP D 0:471 m=s
3.80
Computer problem.
3.82
Concept problem.
3.84
Concept problem.
3.86
Concept problem.
F
C g D 16:1 ft=s2
2mA
3.88
aAy D
3.90
rRA D G
rRB=A
mB
; and rRB D
r2

mA C mB
D G
r2
mA
r2
3.92
During the 1 s time interval: T D 49:3 lb; after 1 s T D 47:3 lb
3.94
ax D
k1 g D
4:41 m=s2
10

3.96
s D tan D 0:424
3.98
aEA D 1 g {O D .8:05 {O/ ft=s2
aEB D .145 {O/ ft=s2

s
1

k m C mp
3.100
vmax D mg
3.102
vimpact D 3:69 m=s
3.104
gd
3.106
tcontact D 78:5103 s D 21:8 h
2mB g
j.yB /max j D
k
j.yB /max j D 2 ft
.vB /max D vB .yB /max D 8:58 ft=s
3.108
3.110
4R2
d 2 cos
g.4R2
d 2 / sin D 4R3 R
Defining a Cartesian coordinate system with the y axis opposite tothe direction of gravity and with origin at the
2
.mA sin C mB /g D 1:84 ft
initial position of B, we have j.yB /max j D
k
.vB /max D vB j.y D 0:9202/ D 4:79 ft=s
B
3.112
3.114
vEA .tf / D
.3:16 {O C 3:13 |O/ m=s
and vEB .tf / D .3:26 {O/ m=s
Computer problem.
9mG g 39P
|O D . 9:46 ft=s/ |O
mB C mC C 25mD C 9mG
3g .mC C 10mD / mG C .4mB 9mC 30mD C 36mG / P
D 1190 lb
T4 D
mB C mC C 25mD C 9mG
3.116
aEG D
3.118
Computer problem.
Cd D 0:0888 lbs=ft
s

3.120
vc D
3.122
ab D g .1 cos / D 1:94 ft=s2

an D g sin D 11 ft=s2
v2
D 101103 ft D 19:2 mi
D
g sin
3.124
Defining a Cartesian coordinate system with origin at O, y axis pointing opposite to gravity and x axis pointing

x
x
mg
m2 g
to the right, we have y D
C v0 sin C 2 ln 1
v0 cos
mv0 cos

3.126
Verification problem
3.128
WA D
3.130
m1 L1 R D m2 g cos sin
2G .mA C mB /
2
dA C dB
1
r0

D 27610
ft=s
WB .g aB /
D 161 lb
2g C 4aB
L2 R D
g sin
L1 R cos
h
m2 sin L1 R sin
L1 P 2 cos
L2 P 2
m1 g sin
L1 P 2 sin
11
Chapter 4
4.2
Concept problem.
4.4
9:02104 ftlb
4.6
U1-2 D
2
1
2 m.v2
4.8
U1-2 D
1
2m
4.10
(a) 283 m
(b) 283 m
(c) 14:8 km
v12 / D 180 kJ

v12 D 4:30105 J

v22
4.12 d D 15:8 m
4.14 4:9810
11
4.16 43:5106 ft lb
23:3106 ft lb
50:8103 lb
4.18
Concept problem.
4.20
Computer problem.
5
lb=ft3
4.22
1:2910
4.24
k D 2:74105 N=m
p
v0 D 2gL .1 cos / D 6:40 ft=s
4.26
4.28 k D
mv 2
D 44:8 lb=ft
2
4.30
.U1-2 /Engine D 4:64105 ftlb
4.32
Favg D mg C
4.34
m 2
.v
v22 / D 1:47 kN
2h 1

D sin 1 23 D 41:8
VB D 14 x 4
d D 0:420 ft
s
2

4.38 vE A D vEP
4.36

2GmE

4.40
V D
4.42
1:92 ftlb
4.44
D cos
4.46
v2 D 24:1 ft=s
4.48
k D 276 lb=ft
r
m
.vy /max D g
k
4.50
4.52
4.54
L0 D
vA2 D
P0 s0 A ln
2
3
s
s0
1
RP
1
RA
D 2:34103 m=s
D 132
2
5R mvB
D 1:06 m
3
4kR
v

u
1
u 2gd m
B
t
3 mA
mA C 9mB
D 1:09 m=s
and vB2
12
v

u
1
u 2gd m
B
t
3 mA
D3
D 3:28 m=s
mA C 9mB
4.56
2 C k 2
mA vB2
mB D
2g
2
vB2
D 0:685 kg
4.58
4.60
q
2gR dg 1 d 2 = .2R/2

q
d
NA D mg 3
1 d 2 = .2R/2
R
s
2hg mA .1 0:35k / mB
D 3:69 m=s
vA D
mA C mB
v2 D
s
4.62
4.64
vB D
8g
mA ÒA mB ÒB
.sin 35 C sin 75 / D 26:6 ft=s
mA C 4mB
1
.m sin C mB /g
k A
.vB /max D vB .yB D 2/ D 9:42 ft=s
dB D 0 ft
yB D
4.66
W D 2mg
4.68
2
Part (a): T D 2m.vO
C R2 ! 2 / D 580 J
2
2
1
Part (b): T D 2 .4m/vO
C 12 m.4R2 ! 2 / D 2m.vO
C R2 ! 2 / D 580 J
4.70
U D 5760 kJ
4.72
Concept problem.
4.74
P1 D P2 D 7:48 kW
4.76
v D 54:7 m=s D 197 km=h

CR D 0:0171 L=s D 61:7 L=h:
4.78
Energy Burned D 274 C
4.80
Pi D
4.82
D 1:07109 kg=.m2 s2 /

0:99 mg 2
D 0:0361 J
U1-2 D 12 m
Cd
4.84
4.86
kD
mgv .k cos C sin /

D 1:51103 ftlb=s D 2:75 hp

mv 2
2
u2
2
C l2
3
2 mgL
2
2
u1
l1
D 9030 N=m
3EIcs 2
2L3
s
2g.mA sin C mB k mA cos /
.v2 /max D
mA C mB
The distance between B and the ground is 0 ft
4.88 V D
4.90
4.92
Eb D
k 2
D 9:67 ft=s
mgvt sin
D 5:395103 kJ D 1290 C

Chapter 5
5.2
xe D 28:510
xe D 4:1510
xe D 1:4210
24
m
m
6
m
12
13
5.4
5.6
5.8
5.10
5.12
5.14
t2
E
F dt D mv2 D 2:64 lbs
t1
mv2
E
D 2400 lb
Favg D
t2 t1
Concept problem.
mvx2
D 10:6 s
FT
d D 1280 ft
t2 D
Concept problem.
(a) FEavg D 8080 N
FE
avg
(b) s
D 0:515
mg
(a) Impulse D .3:17 {O C 1:20 |O/ lbs

(b) FEb avg D .3170 {O C 1200 |O/ lb
(c) Within the accuracy of our calculation (4 significant figures), remains equal to 31 .
5.16
Concept problem.
5.18
Concept problem.
5.20
n D 9:170106 bees
Slowdown D 48:5 ft=s D 33:0 mph
5.22
.xfp /2 D
5.24
vP 4 D 1:86 ft=s
5.26
vEG D . 2580 {O
5.28
dD
5.30
.E
vP /final D .1:52 ft=s/ {O
5.32
(a) Amplitude D
Lfp mp
mp C mfp
mA Lfp
mA C mC
(b) .vB /max
3030 |O/ m=s;

D 4:95 m
mA r
D 2:1410
mA C mB
mA r!
D
D 0:0135 m=s
mA C mB
5.34
vEA2 D .5500 {O C 9520 |O/ m=s
5.36
.vA /max D 25:2 ft=s
5.38
and vEB2 D .5500 {O C 9530 |O/ m=s
and .vB / D 0 for
60
mB ! 1
Computer problem.
5.40
C
C
vEA
D vEB
D . 55:9 ft=s/ {O
5.42
vB D
mA C mB p
2gL.1
mB
5.44
vB D
mA C mB p
2k gd D 199 m=s
mB
5.46
0:817 e 0:882
5.48
dD
2 .v /2
mB
B
2k g.mA C mB /2
cos m / D 210 m=s
D 2:89 m
14
q
2 gk 2 2hk C .m C m /g
.mA C mB /mA
A
B
5.50
Wmax D mA g C
5.52
C
C
D .8:33 {O C 78:0 |O/ ft=s
D vEB
vEA
.mA C mB /k
D 819 lb
rE D .14:5 {O C 136 |O/ ft

5.54
C
D 77:0 |O ft=s
vEA
rEA D 132 |O ft
C
D .73:3 {O C 85:8 |O/ ft=s
and vEB
and rEB D .184 {O C 215 |O/ ft
5.56
Given the assumption that the masses are not identical, it is not possible to have a moving ball A
hit a stationary ball B so that A stops right after the impact.
1
5.58
D tan
5.60
The answer is an explanation.
5.62
5.64
.cot /
Concept problem.
C
vEA
D . 23:6 {O C 18:5 |O/ m=s
C
and vEB
D . 0:717 {O
6:20 |O/ m=s
5.66
s
dD
2
.1 C e/h1 cos sin
gh1
r
gh1
.e
2
C
5.68
5.70
5.72
5.74
5.76
5.78
5.80
5.82
d D 3:50 ft

C
vN
D 2N
1/ C .1 C e/ cos 2
q
2gh2 C 12 gh1 .e

1/ C .1 C e/ cos 22 D 6:24 ft
p
1 2gh

D 593103 kO slugft2 =s

hEO A D 79:9103 kO slugft2 =s
hEO
hEQ D
1:44106 kO slugft2 =s
Concept problem.
!D
.L sin 0 C d /2
!
.L sin C d /2 0
P
O it is indeed true that M
E O D mP gvP .0/t cos k,
E O D hEP O .
Since hEO D mP gvP .0/t cos and M
s
p
2L3
hEO D W
.cos cos 33 / kO D .3:99 lbfts/ cos 0:839 kO
@ 57
g
5.84
s
v1 D sin 2
2gL.cos 2
sin2 1
s
v2 D sin 1
5.86
MA D 2m!02 r
q
r2
2gL.cos 2
sin2 1
cos 1 /
sin2 2
cos 1 /
sin2 2
D 6:85 ft=s;
D 2:32 ft=s
r02
15
5.88
Computer problem.
time to reach end of arm D 0:150 s
5.90
!satellite D 0:0612 rad=s
5.92
The ratio of the orbital sectors is equal to 1.

p
b D rP rA
s
gre2 p
v D
. 2 1/ D 3:23103 m=s
rc
5.94
5.96
5.98
.1:631010 m3=2 =s/

vesc D
p
r

vesc Earth D 4:21104 m=s D 151;000 km=h
4 2 a3
D 3:971014 m3 =s2
2
5.100
Gme D
5.102
rg D 1:39108 ft D 26;200 mi; hg D 1:18108 ft D 22;300 mi;

s

2
1
vP D gre2
D 7:71103 m=s D 27;800 km=h
rP
a
s

2
1
vA D gre2
D 7:70103 m=s D 27;700 km=h
rA a
r
e D A 1 D 7:4510 4
as
5.104
D 2
5.106
5.108
5.110
5.112
a3
gre2
vc D 10;100 ft=s D 6870 mph
D 5470 s D 91:2 min
vA D 4710 ft=s D 3210 mph

vA D 6830 ft=s D 4660 mph
s
ae3
D 10;600 s D 2:95 h
t D
gre2
Concept problem.
(a) vbo D 5530 ft=s D 3770 mph
(b) vLM D 54:0 ft=s D 37:0 mph
(c) t D 3420 s D 0:950 h
(d) D 180 D 5:30
s
2
rP vP
2GmB
(a) v1 D
rP
2
(b) rP vP
> 2GmB
5.114
Concept problem.
5.116
Concept problem.
2
1
4 pA dA
RD
5.120
vB D 29:2 ft=s

mf
vmax D vo ln 1 C
mb
5.122
5.124
kD

4 Q2 1
g
dA2
5.118
2
d 2 vw
1
4gmax
1
dB2

D 2120 lb
cos.=2/ D 370 lb=ft
16

5.126
v D vo ln
M
M m
P ot

gt
5.128
N D
2
1
4 PA d
4Q2
D 53:4 N;
d 2
4Q2
D 0:354 N;
d 2
4Q2 `
D 0:0707 Nm
M D
d 2

F D gL C v02 D 1:10 lb
V D
5.130
5.132
Computer problem.
5.134
m
P o D 0:0150 slug=s
vo D 4530 ft=s

mf
vmax D vo ln 1 C
mb
5.136
max power D 0
v0 D
5.138
5.140
1
2 vw
p

m q
FEavg D
2ghf C 2ghi |O D .654 N/ |O
t
q
p
2ghf C 2ghi
C
E
j
F
j
jpE
pE j
avg
D
D
D 111
jt. mg |O/j
mg
gt
C
vEA
D . 13:6 m=s/ {O
TC
T
T
5.142
gtbo D 1320 ft=s
and
C
vEB
D . 21:3 m=s/ {O
100% D 45:3%
C
vEA
D .4:61 {O C 75:3 |O/ ft=s
C
vEA
rEA D dA
C
vA
rEB D dB
C
vEB
C
vB
C
and vEB
D .37:4 {O C 98:8 |O/ft=s
D .7:71 {O C 126 |O/ ft

D .87:6 {O C 231 |O/ ft
5.144
hA D rA re D 7:02106 ft D 1330 mi
D 6860 s D 1:91 h
5.146
ve D 8770 m=s D 31;600 km=h

vj D 5640 m=s D 20;300 km=h
5.148
D 8:64107 s D 1000 days

8
q
< 1 K 1 K 2 4v 2 D 0:219 m=s;
0
2q
vw D 2
: 1 K C 1 K 2 4v 2 D 18:3 m=s;
2
5.150
vEA D .0:813 m=s/ {O
and
vEB D .3:25 m=s/ {O
Chapter 6
6.2
!EB D
6.4
!B D
1800 kO rpm
and !E C D 1160 kO rpm
RA
! D 600 rad=s
RB A
17

6.6
Concept problem.
6.8
Concept problem.
6.10
Concept problem.
6.12
.!OA /max D 2:48 rad=s
6.14
j!E ram j D 0
6.16
vEG D . 0:150 uO r 3:38 uO / m=s

aEG D . 20:4 uO r 3:66 uO / m=s2
6.18
!E s D
10:0 kO rad=s
Es D
1:33 kO rad=s2
6.20
vEC D . 0:179 {O C 0:492 |O/ m=s

aEC D .2:58 {O C 0:938 |O/ m=s2
6.22
!s D
6.24
!EB D 8:00 kO rad=s;
6.26
!E C D 40:0 kO rad=s
2 rad 1 h
D 72:710
24 h 3600 s
rad=s
EB D 2:00 kO rad=s2 ;
!ED D 1:60 kO rad=s;
ED D 0:400 kO rad=s2
EC D 5:20 kO rad=s2
6.28
Letting !E OB and EOB denote the angular velocity and acceleration of the turbine,
!E OB D 1:60 kO rad=s and EOB D 0:461 kO rad=s2
aEA D . 93:0 {O 110 |O/ m=s2
6.30
Chain ring/sprocket combination: C1/S3

R
52:6 mm
!W D !S D C !C D
68 rpm D 126 rpm
RS
28:3 mm
6.32
vEB D .9:88 {O C 3:66 |O/ m=s
6.34
!EP D 9:60 kO rad=s

vEO D 2:00 {O ft=s
6.36
Concept problem.
6.38
!EP D 1:60 kO rad=s

vEC D 5:33 {O m=s
6.40
1 D 67:7
2 D 231
6.42
!EA D
6.44
!E C D 60:0 kO rad=s
6.46
vEB D 10:6 {O ft=s

vEC D .5:31 {O 4:00 |O/ ft=s
6.48
!EP D 8:80 kO rad=s

rEIC =Q D 0:307 |O rad=s
6.50
vEC D 82:2 |O ft=s
6.52
!E OC D 2:86 kO rad=s
vEP D .0:0286 uO r 0:119 uO / m=s
7:00 kO rad=s
18

6.54
vEC D 16:3 {O ft=s

the cable is unwinding at 12:2 ft=s
6.56
!E CD D
6.58
!EAB D 5:09 kO rad=s
6.60
!E spool D 3:33 kO rad=s

vEO D 5:00 {O rad=s
6.62
!E gate D
6.64
!EAB D
42:4 kO rad=s
0:0467 kO rad=s
7:21 kO rad=s
D 8:58 kO rad=s
!EBC
6.66
!E R D 35:0 kO rpm
vED D 17:2 |O in.=s
6.68
!EAB D 50:6 kO rad=s

vEB D 1:92 |O m=s
#
"
P
R.H
R cos / sin
P
|O
vEB D R cos p
L2 .H R cos /2
6.70
6.72
Computer problem.
5:38 |O/ ft=s2
6.74
aEB D .17:0 {O
6.76
aEC D . 16:8 {O C 87:7 |O/ ft=s2
6.78
aEQ D
6.80
EAB D 48:6 kO rad=s2
6.82
Letting Q denote the point on the ball in contact with the bowl, at the instant shown
aEA D . 1020 uO r C 24:0 uO / ft=s2 and aEQ D 3480 uO r ft=s2
6.84
EAB D 27:1 kO rad=s2

aEB D 267 {O ft=s2
6.86
!EAB D
6.88
EW D 7:97 kO rad=s2
6.90
EBC D 1640 kO rad=s2
6.92
EBC D
6.94
Es D
6.96
Egate D 4:50 kO rad=s2
6.98
EAB D
6.100
Letting Q denote the point of contact between the wheel W and the cylinder S , aEQ D 82:0 uO r ft=s2
6.102
For C accelerating downward and to the right: EAB D
9:16 uO r m=s2
0:458 kO rad=s
and
EBC D 0E
and EAB D
0:258 kO rad=s2
HRP 2 .R2 H 2 / sin O

k
.H 2 C R2 2HR cos /2
0:400 kO rad=s2
92:1 kO rad=s2
and
and
aEO D 2:00 {O ft=s2
EBC D
191 kO rad=s2
For C accelerating downward and to the left: EAB D

6.104
For D 10 and D 56:11 aED D

For D 10 and D 303:9 aED D
551 kO rad=s2 and EBC D 320 kO rad=s2

729 kO rad=s2 and EBC D 320 kO rad=s2
.18;300 {O C 34;900 |O/ ft=s2

.18;300 {O C 32;900 |O/ ft=s2
19
6.106
aEC
6.108

R sin
D RAB cos 1 C p
L2 R2 cos2
"
2
R.L
R2 / cos2
R sin2
2
C R!AB
p
.L2 R2 cos2 /3=2
L2 R2 cos2
sin
|O
Computer problem.

D sR s!02 {O C s0 C 2Ps !0 |O
6.110
aEP
6.112
(a) !E CD D
.R= h/!AB
D 3:14 kO rad=s
1 C R= h
(b) ECD D 0E and
and vEbar D
.R= h/d!AB O
I D 0:377 IO m=s
1 C R= h
aEbar D dCD IO D 0E
!AB . C sin / O
k
1 C 2 C 2 sin
!AB . C sin / O
dI
vEbar D
1 C 2 C 2 sin

2 cos
1 2 !AB
(b) ECD D
2 kO
1 C 2 C 2 sin
2
1 2 !AB
cos
aEbar D d
2 kO
1 C 2 C 2 sin
6.114
(a) !E CD D
6.116
(a) vEbar D
6.118
vEP D . 23:6 {OA C 26:1 |OA / ft=s

aEP D . 52:9 {OA 40:3 |OA / ft=s2
6.120
aEC D . 30:0 {O C 51:8 |O/ ft=s2
6.122
vEC D .7:50 {O C 7:51 |O/ ft=s

aEC D .105 {O 161 |O/ ft=s2
6.124
#)
! d IO
1 C AB
(b) vEbar D
! d IO
1 AB
(c) vbar D 2:90 m=s
vbar D 5:39 m=s
(d) vbar D 5:59 m=s
vbar D 50:3 m=s
Concept problem.
6.126
dPAB D 1:12 ft=s

dRAB D 2:01 ft=s2
6.128
!E bar D
6.130
!EA D
6.132
aEE D . 1:50 {O
6.134
dPE C D 0:214 ft=s
6.136
vED D .vC C !OA ` sin / {O C .d!OA `!C cos / |O `!C sin kO

2
2
2
aED D . d!OA
C 2!OA `!C cos C ÒA sin / {O C .dOA C .!C
C !OA
/` sin C 2!OA vC / |O
2
O
!C cos k:
6.138
vEC D .21:5 {O C 7:51 |O/ ft=s

aEC D .107 {O 166 |O/ ft=s2
14:3 kO rad=s
481 kO rad=s
0:320 |O/ ft=s2
and dRE C D
0:0192 ft=s2
20
Chapter 7
and dstop D v02 =.2gk / D 16:8 ft
7.2
tstop D v0 =.k g/ D 1:86 s
7.4
F D 179 N
7.6
D 18:9
7.8
tmin D v=.aGx / D v=.gs / D 1:12 s
7.10
The FBD for the problem is
and aGx D 4:71 m=s2
F D 577 lb; Nf D 784 lb; Nr D 783 lb; and NB D 2230 lb

HE D Hx {O C Hy |O D .653 {O C 166 |O/ lb and HE D 674 lb
.s /min D 0:736
7.12
m1 D
m1 D
1
4 m;
1
2 m;
m2 D
m2 D
3
4 m;
1
2 m;
xP D
2
3L
.IG /extra D
2
1
6 mL
7.14
Rx D
.0:112 N=s2 /t 2 ;
Ry D 0:0101 N;
M D 3:1110
7.16
Nm
Or D
7.18
7.20
16:8 lb and O D 5:00 lb

p g
p
jbar jmax D 3
for d D 61 3
3 L or d D
L
3C
p
3 L
OE D Ox {O C Oy |O D . 1320 {O C 48:5 |O/ N

7.22
1
6
and ET-bar D . 6:92 rad=s2 / kO
Concept problem.
21

7.24
.vG /final D
tfinal
2gL.sin pk cos C sin / D 18:4 ft=s

2gL.sin k cos C sin /
D 1:08 s
D .vG /final =aGx D
g.sin k cos /
7.26
tr D 2:41 s
and d D 50:7 ft
7.28
aGx D k g
7.30
and b D
TCD D 101 N:
and
grk
2
kG
aEG D aGx {O C aGy |O D . 6:45 m=s2 / |O
7.32
The solution of this problem under the stated assumptions leads to conclusion that the crate starts rotating counterclockwise. Since this fact runs contrary to the fact that the force system applied to the crate can only induce
a clockwise rotation of the crate, then we conclude that the solution obtained under the assumptions given in the
problem is not physically admssible.
7.34
separation D
7.36

D 60:0
3
Under the assumption that the contact between the sphere and the semicylinder is frictionless, the solution in terms
of contact force N , second time derivative of the angle , and angular acceleration of the sphere is as follows:
g sin
N D mg cos m.R C /P 2 ; R D
; and s D 0:
RC
Comparing these equations to Eqs. (5) and (6) in Example 3.6 on p. 216 for a particle, we can see that the sphere in
question behaves like a particle of mass m sliding over a semicylinder of radius R C and therefore if we shrink
the sphere to a particle by letting ! 0, we will recover exactly the same solution obtained for a particle.
7.38
(a) F D
IG vf
r 2 tf
(b) .tf /new D

7.40
AB D
D 12:1 lb
mc r 2 tf
2IG C r 2 mc
D 6:85 s
3mAB g cos
L.2mAB C 9mB
sin2 /
aEB D
3mAB g sin cos

2mAB C 9mB
22
sin2
{O;
w D
3mAB g sin cos

R.2mAB C 9mB sin2 /

7.42
3.2mA C mAB /g cos
AB D
L.6mA cos2 C 2mAB C 9mB sin2 /

3.2mA C mAB /g sin cos
{O;
aEB D
6mA cos2 C 2mAB C 9mB sin2
3.2mA C mAB /g sin cos
w D
R.6mA cos2 C 2mAB C 9mB sin2 /
7.44
IG D mR2
7.46
ac D 1:35 m=s2 ;
7.48
xR D
7.50
(a)
T D 3130 N;
.3g cos C 2LP 2 / sin
and
R D
and
1 C 3 sin2
s D
1:68 rad=s
3.2g C LP 2 cos / sin

L.1 C 3 sin2 /

1
2hac
Wc 1 C
D 736 lb;
4
g`

2hac
1
D 549 lb;
Nf D Wc 1
4
g`
W a
Fr D c c D 503 lb
2g
Nr D
(b)
Ax D
ac Ww
g
Ay D Wr
Fr D
Nr D
485 lb
689 lb
d
2a I
MA D Fr C c B D 522 ftlb
2
d
7.52
Computer problem.
!
k2
m 1C G
xR C kx D mg sin C kL0
r2
7.54
AB D
7.56 aAx D
g
2L
cos
1
3
C cos2 .cos = sin /2 C sin cos .cos = sin /
60
g
181
b D
and
5
O;
7 g sin u
7.58
aEG D
7.60
R C
7.62
(a) aEG D .1
D 3:74 rad=s2
45g
181R
b D
5g sin
;
7
and FEdue to bowl D
mg cos uO r C 27 mg sin uO
5g
sin D 0
7.R /

(b) aEG D
k /
P
.1
m
PR2
{O;
2/
m.R2 C kG

2k / k g {O;
and
2
k R 2 C k G
P
2 P C mg
R2 C kG
k

k R P
d D 2
.1 k / g
kG m
and .s /min D
7.64
R.2mA R2 C IO /!P s C 2mA `.`P C R!s /2 D 0;
``R C `P2 C `R!P s R2 !s2 D 0
7.66
7.68
and By D 54 mg

P
(b) D 2 tan 1
; 0 D 0;
mg
(a) Ay D
3
4 mg
TCD D 2060 lb;
and
aEE D
and
1 D
.5:65 ft=s2 / |O
23

7.70
25:9 |O/ ft=s2
aEE D .13:2 {O
Ax D
3g.mb C mr /.d C h/mb C 2d mr sin cos C g.h

2.d C h/.2mb C 3mr /
d /mb mr tan
D 437 lb;
gmb h C d.mb C mr /mb C 3.mb C 2mr / cos2 C 3ghmb mr sin2

D 256 lb;
3g.mb C mr /.d C h/mb C 2hmr sin cos g.h d /mb mr tan
Cx D
D 629 lb;
Ay D
Cy D
7.72
7.74
gmb d C h.mb C mr /mb C 3.mb C 2mr / cos2 C 3gd mb mr sin2

D 353 lb
aEE D .7:00 {O C 7:06 |O/ ft=s2

Ax D 229 lb; Ay D 1330 lb;
Cx D 336 lb;
and
Cy D 1330 lb
Pmax D 1600 lb
1
vc2
gR
7.76
D tan
7.78
AC D R D 1:46 rad=s2
7.80
Computer problem.
tclosure D 1:29 s
.vE /at closure D 1:87 m=s
and CE D
R D
1:46 rad=s2
7.82
gR sin g cos R cos.
R D
2 C 2R sin.
R2 C 2 C kG
F D mg sin C maGx

D mg sin m RC sin.
N D mg cos C maGy

D mg cos C m cos.
/P 2
/
/gR sin g cos R cos. /P 2

2 C2R sin. /
R2 C2 CkG
/gR sin g cos R cos. /P 2

2 C2R sin. /
R2 C2 CkG

/ ;
C P 2 cos.
P 2 sin.

/
7.84
.mAB C mC /xRA C 12 .mAB C 2mC /L cos R C 21 .2d C h/mC cos R
1
P 2 1 .2d C h/m sin P 2 D 0;
C
2 .mAB C 2mC /L sin
2
2
1
1
1
1
Lm cos xR C m
C m cos C . m
C m / sin2 L2 R
12
1
2 .2d
7.86
7.88
7.90
AB
C
2 C
4 AB
1
C 4 .2d C h/LmC .2 sin sin C cos cos /R
C 41 .mAB C 2mC /L2 sin cos P 2
C 41 .2d C h/LmC .2 sin cos sin cos /P 2
C 21 .mAB C 2mC /gL sin
C h/mC cos xRA C 21 .2d C h/LmC cos. /R
1
C 12
.h2 C w 2 / C 41 .2d C h/2 mC R
1
sin cos /P 2
2 .2d C h/LmC .sin cos
C 21 g.2d C h/mC sin
D 0;
D 0:
(a) F D 43 mg.3 cos 2/ sin and N D 14 mg.1 3 cos /2

(b) Given any (finite) value of s , there always is a value of for which the rod will slip.
!
!
R2
md
R2
Pmax D mc gs 1 C
C 2 ; aC x D g 1 C 2 ; and aGx D s g
mc
kG
kG
P D 7:59 N
and .s /min D 0:0462
24
Chapter 8
8.2
TA D TB D 518 J
8.4
TR1 D 7760 ftlb
8.6
T D 324;000 ftlb
8.8
T D
8.10
T D 28:9 ftlb
8.12
and TR2 D 15;000 ftlb
1
2
I R2 C H 2 .IA C mBC R2 /!AB
D 25:3 J
2H 2 D
Computer problem.
8.14
Lf D 34:1 ft
8.16
Concept problem.
s
.2Md=R/ kd 2
!2 D
2 C R2 /
m.kG
2M
ds D
kR
8.18
8.20
8.22
M D 2:06106 ftlb
p
vperson D vG D 2gH.1 C cos /
v
D 10:8 m=s
person D0
8.24
vO D 10:9 ft=s
8.26
nmin D 26
(a) vperson D 26:5 ft=s
8.28
Ufracture D
8.30
Ufriction D
8.32
kD
8.34
D 1:03 ft
8.36
WP D
8.38
vmax D 6:39 ft=s
8.40
vC D 4:96 m=s
8.42
vS D 7:53 m=s
8.44
vB D 2:21 m=s
In B is moving to the left.
8.46
M D 32:9 Nm
Pmax D 132 W
8.48
.s /max D 0:371 and
8.50
Concept problem.
8.52
hEA
1
2 gLA .mA
C 2mS / C 2mS rS .cos f
cos i / D
318 J
364 ftlb
2 C g.L C d
W vmax
2H /
D 11:3 lb=ft
2g .H d /.H C 20 d /
2
3gH C 6vA2
AB
(a) hEA
2
vA2
3gH
(b) hED
BC
W D 800 lb
and !ED D .44:1 rad=s/ kO
slide D 35:1
1
m R2 !AB kO D .1:72 kgm2 =s/ kO
3 AB
D m ! R2 kO D .7:20 kgm2 =s/ kO
CD
BC AB
1
3 mCD HR!AB
kO D .7:75 kgm2 =s/ kO
25

8.54
8.56
8.58
8.60
hEC
hEO
W
W
D .91:910
D .66910
kgm2 =s/ kO
@
kgm =s/ kO
@
pE D WAB vA D 3:23 lbs

AB
2g cos
W
LvA O
AB
k D .2:42 ftlbs/ kO
hEG D
12g cos
W vfinal
{O D .878 lb/ {O
FEavg D
gt
E avg D 2IG vfinal kO D . 10:7 lbft/ kO
M
dt
Concept problem.
v0
2k g
8.62
tr D
8.64
!A end of slip D 61:3 rad=s
8.66
(a) Collar modeled as a particle: vimpact D 0:990 ft=s,

(b) Collar modeled as a rigid body: vimpact D 0:943 ft=s,
"
#
g .1 C w /g
t !0 kO
!.t/
E D
.1 C g w /r
8.68
8.70
vG t D2 s D 21:7 ft=s
.s /min D 0:0588
8.72
Concept problem.
and !B end of slip D
8.74
The system rotates clockwise.

m a e
D B 0 tf2 ; (clockwise)
2IO
8.76
dD
4r 2 d
.2r b/2
8.78
Concept problem.
8.80
Concept problem.
2
3`
8.82
dD
8.84
d D 2:09 ft
!EB D . 26:6 rad=s/ kO
8.86
8.88
8.90
40:3 rad=s
and
vbC D 226 ft=s
!E bC D .10:7 rad=s/ kO
C
vEG
D . 1:31 m=s/ {O and !EAC D .3:43 rad=s/ kO
s
4g` sin
v0 D
D 7:48 ft=s
3 cos2
8.92
C
!EAB
D . 0:851 rad=s/ kO
8.94
!EAC D . 0:000 715 rad=s/ kO
8.96
T D 0:0257 ftlb
8.98
d D 1:98 m
C
and !EBD
D .2:55 rad=s/ kO
C
and !EB
D . 0:000 024 1 rad=s/ kO
26

8.100
Concept problem.
8.102
Concept problem.
8.104
C
D 2v0 {O D .12 ft=s/ {O
vEP
8.106
!EAC D .1:03 rad=s/ kO

max D 14:8
Chapter 9
9.2
9.4
mgL 2
4 2
s
7L
D 2
6g
IO D
r
9.6
`D
9.8
!n D
9.10
9.12
9.14
9.16
IG
m
s
Gr 4
LR4 t
r
2d m
D
h
k
s
1
.h3 C d 3 / C md 2
D 2 3
kh2
Wpayload D 90;700 lb
r
g
!
d
f D n D
D 0:566 Hz
2
4 m
9.18
(a) myR C ky D 0
(b) mR C k D 0
9.20
mn D 35:6 g
9.22
xR G C
9.24
9.26
9.28
17k
x D0
6m G
8k
x D0
3m G r
2
3m
D 2
D
D 2:22 s
!n
8k
s
2g
!n D
3.R r/
s
3.R r/
D 2
2g
xR G C
2
D
D 2
!n
s
.3
L
2g
2/R
9.30
D 2
9.32
x.t / D A sin !n t C B cos !n t C
3g
F0 =k
cos !0 t
.!0 =!n /2
27
9.34
mu D 0:01 kg
mu D 0:1 kg
mu D 1 kg
q
2:1910
q
jym j D 1:1010
q
jym j D 9:8010
jym j D
6 cos.4:37t/ C 1:8910 5 m;
5 cos.4:37t/ C 1:1110 5 m;
4 cos.4:37t/ C 1:0110 3 m
amp D 0:001 76 rad=s

s
3EI
9.38 !n D
D 312 rad=s
.mu C me /d 3 C 43 mw h2 d
!
f D n D 49:7 Hz
2
jp j.mu C me /d
.!p =!n /2
D 0:820
MF D
D
mu R
1 .!p =!n /2
9.36
9.40
9.42
9.44
Concept problem.
x.t / D 0:000 501 sin 10t C 0:100 cos 10t

L0
myR C 2k 1
y D F0 sin !0 t
L

sin 200t m

!0
sin
!
t
C
sin
!
t
;
n
0
1 .!0 =!n /2 !n
s

L0
2k
where keq D 2k 1
and !n D
1
L
m

2
1
1
3 mB C 2 mA xR C 2kx D 2 F0 sin !0 t
F0
xamp D
2
4
4k
m
3 B C mA !0
y.t/ D
9.46
F0 =keq
2:5110
9.48
Concept problem.
9.50
Concept problem.
9.52
no peak in MF for
9.54
yR C 2!n yP C !n2 y D
mu "!r2
sin !r t
m
9.56
jym j D 0:002 19 m
9.58

IG R C cro2 P C k1 ro2 C k2 ri2 D 0
h
.t / D 0:0504e 3:80t 0:000386e
L0
L
1=2
496t
rad
9.60 x.t / D e 0:5t .A sin 3:12t C B cos 3:12t/

q

p
L2 C .L C x/2
2L p
9.62 mxR C c xP C 2k
LCx
L2
C .L C x/2
D F0 sin !0 t
mxR C c xP C kx D F0 sin !0 t
DD p
F0 =k
1
.!0 =!n /2 2 C .2!0 =!n /2
r
;
!n D
where
9.64
k > 1:35109 N=m
9.66
(a) mRs C c sP C ks D mA!02 sin !0 t

"
#

mA!02 k m!02
(b) y.t/ D
C A sin !0 t
2
k m!02 C c 2 !02
v
u
u
k 2 C c 2 !02
(c) DT D t
2
k m!02 C c 2 !02
k
m
and
c
D p
2 km
and c D 0
"
k
28
#
mcA!03
cos !0 t
2
m!02 C c 2 !02
k1 k2
xD0
k1 C k2
9.68
mxR C
9.70
k D
9.72
.mm C mp /yRm C keq ym D mu "!r2 sin !r t
5:91%
9.74
mu D 0:01 kg W
F t D 14:8mu sin.125:7t C 0:842/ C 17:7mu cos.125:7t C 0:842/ N;
mu D 0:1 kg W
F t D 148mu sin.125:7t C 0:842/ C 177mu cos.125:7t C 0:842/ N;
mu D 1 kg W
9.76 myR C 2k 1
9.78
F t D 1480mu sin.125:7t C 0:842/ C 1770mu cos.125:7t C 0:842/ N

!
L0
p
y 2 C L2

L0
yD0
L

myR C 2k 1
yD0
Chapter 10
10.2
.d C ` cos /!arm kO
i
2 h

!arm
2
!arm
sin .d C ` cos / |O
d 2 C 2d ` cos C `2 C R2 cos2 kO
R
vEE D R!arm cos {O C .d C ` cos /!arm |O

aEE D
2
!arm
cos .d C ` cos / {O
10.4
vEP D 2.d C ` cos /!arm kO

2
2
aEE D !arm
)
( cos .d C ` cos / {O !arm sin .d C ` cos / |O
2 h
2 i
!arm
2
2
2
.d C ` cos /arm C
d C 2d ` cos C ` C R cos kO
R
10.6

aEA D 231 |O C 108 kO ft=s2

EAB D 18:3 {O C 15:1 |O C 34:9 kO rad=s2
10.8
vEP D
aEP
10.10
R!d cos {O .h!b C R!d sin / |O C R!b cos kO

h

i
D R !d2 sin !P d cos {O R !b2 C !d2 cos C !P d sin |O

C R!P b cos h!b2 2R!b !d sin kO
vEA D
2L!0 cos2 kO
aEA D
2L!02 cos2 {O
!s |O
L!02 cot |O
2L0 cos2 kO
P kO
R kO
10.12
!EAB D
10.14
P {O
EAB D !
s

L
LR
L
L P2
sin C d C cos !s2 C P 2 cos {O C
sin
aEG D
2
2
2
2

vEA D 1:432 |O 0:6677 kO m=s
10.16

aEA D 103 |O C 47:8 kO m=s2
10.18

!EAB D 0:0161 {O C 2:69 |O C 0:917 kO rad=s

E
D 1:73 {O 4:59 |O C 15:8 kO rad=s2

R cos |O
P sin kO
L!
s
AB
10.20
!E bit D !d {O C !a |O C !b kO
10.22
Ebit D .!P d !a !b / {O C .!P a C !d !b / |O C .!P b !d !a / kO

h

i
aEB D a t C h!P a .` C r t / !a2 C !b2 {O C 2.v t C h!a / !b C .` C r t / !P b |O
.2v C h! / ! C .` C r / !P kO
t
29

Dynamics Answers 2009-11-2

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Engineering Mechanics: Dynamics 1e

Gray, Costanzo, Plesha

Answers to Selected Even-Numbered Problems

D .4:00 {O 1:00 |O/ ft

vE.15 s/ D .72:1 {O C 6:24 |O/ ft=s

Er .t1 ; t2 / D .0:899 {O C 41:3 |O/ ft

Last modified: November 2, 2009

Engineering Mechanics: Dynamics 1e

vE D .29:3 ft=s/1 C cos.25:5 rad=s/t {O .29:3 ft=s/1 C sin.25:5 rad=s/t |O

Gray, Costanzo, Plesha

2 cos / |O2 m=s2

vEP =B D . 6:14 {OB C 23:7 |OB / ft=s

vEA D . 20:0 {O 12:8 |O/ ft=s

tjajmax D 3:93 s and tjajmax D 11:8 s

where jajmax D 3:60 m=s2

Last modified: November 2, 2009

Engineering Mechanics: Dynamics 1e

.P0 /min D 4:93 rad=s

Gray, Costanzo, Plesha

s rDr D 1:44 rad=s2

s rDr D 3:63 rad=s2

vEgun D .14:7 ft=s/ {O

The golfers chip shot is successful.

The two firing ranges are

vEinitial D v0 cos {O C v0 sin |O D .6:02 {O C 7:67 |O/ m=s

Last modified: November 2, 2009

Engineering Mechanics: Dynamics 1e

Gray, Costanzo, Plesha

vEB D .4:00 uO r C 0:800 uO  / ft=s

aE D . 112 {O C 5 |O/ m=s2

Last modified: November 2, 2009

Engineering Mechanics: Dynamics 1e

Gray, Costanzo, Plesha

aE .s/ D f 19:0 uO t C 69:9

vE D . 1:3 uO r C 0:22 r uO  / m=s

rR D 5:78 ft=s2 and R D 1:42 rad=s2

Last modified: November 2, 2009

Engineering Mechanics: Dynamics 1e

.h C  sin /2 C .d C  cos /2

Gray, Costanzo, Plesha

vE D .37:710 3 uO r C 0:534 uO  / m=s

vEfollower D .1:76 m=s/ |O

vEC =A D .6:86 {O C 30:3 |O/ m=s

vrain D 16:4 ft=s

vEB D . 2:00 m=s/ |O

aEB D . 11:1 m=s2 / {O

Last modified: November 2, 2009

Engineering Mechanics: Dynamics 1e

Gray, Costanzo, Plesha

r P sin .r R C 2rP P r P 2 sin  cos / D 0

Last modified: November 2, 2009

Engineering Mechanics: Dynamics 1e

1:26 |O2 / m=s2

P  D180 D 2:94 rad=s

rR  D180 D 40:8 m=s2 and

P D30 D 0:0539 rad=s

2 cos / |O2 m=s2

.P0 /min D 4:93 rad=s

vEB D .4:00 uO r C 0:800 uO / ft=s

vE D . 1:3 uO r C 0:22 r uO / m=s

rR D 5:78 ft=s2 and R D 1:42 rad=s2

.h C sin /2 C .d C cos /2

vE D .37:710 3 uO r C 0:534 uO / m=s

r P sin .r R C 2rP P r P 2 sin cos / D 0

P D180 D 2:94 rad=s

rR D180 D 40:8 m=s2 and

P D30 D 0:0539 rad=s

R D30 D 0:243 rad=s2

(a) aE D .319103 uO r C 603 uO / m=s2

aEA D 1 g {O D .8:05 {O/ ft=s2

ab D g .1 cos / D 1:94 ft=s2

mgv .k cos C sin /

cos m / D 210 m=s

cos.=2/ D 370 lb=ft

D 8:64107 s D 1000 days