Dynamics 1e Answers

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

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 .rB=A /` D 3:88 ft
1.4 rEB=A D .4:00 {O 1:00 |O/ ft

rE D .3:31 uOp 2:46 uO q / ft
ˇ B=A ˇ ˇ ˇ
ˇrE D ˇrE D 4:12 ft
B=A xy system B=A pq system
ˇ ˇ
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 ŒIxx D ŒIyy D ŒI´´ D ŒM ŒL2
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 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 4:95ı
.15 s/ D 180ı
y.x/ D 2:00 C 3:00x C 2:00x 2 m

2.6
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 P D Œ.4:08e 13:6t / {O C . 0:720 1:28e 13:6t / |O m=s

v.t / D r.t/
P D Œ. 55:5e 13:6t / {O C .17:4e 13:6t / |O m=s2
a.t / D v.t/
The impact angle is the slope of the stone’s trajectory at the time that the stone enters the water. Then, recalling
that the velocity is always tangent to the trajectory, we have D 26:1ı
y.x/ D 5:25 C 3:25x C 1:00x 2 m

2.14
xmax D 3:00 m and xmin D 1:00 m
2.16 vE D .29:3 ft=s/Œ1 C cosŒ.25:5 rad=s/t {O .29:3 ft=s/ sinŒ.25:5 rad=s/t |O

p
v D .29:3 ft=s/ 2 C 2 cosŒ.25:5 rad=s/t
aE D 748 ft=s2 sinŒ.25:5 rad=s/t {O 748 ft=s2 cosŒ.25:5 rad=s/t |O

2.18 Computer problem.
1 Last modified: January 20, 2011

2.22 vP =O /1 D Œ.1 C cos t/ {O1 C .4 2t/ |O1 m=s

.E
˚
vP =O /2 D Œ.1 C cos t/ cos 2.t 2/ sin {O2
.E Œ2.t 2/ cos C .1 C cos t/ sin |O2 m=s
aP =O /1 D . sin t {O1
.E 2 |O1 / m=s2
aP =O /2 D Œ. cos sin t 2 sin / {O2 C .sin sin t
.E 2 cos / |O2 m=s2
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 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
aEA D 2 {O C 3 2 |O ft=s2 D . 1:41 {O C 4:24 |O/ ft=s2
8a3 v02
2.32 xR D 2
4a2 C y 2
4a2 v02 y
yR D 2
4a2 C y 2

v0 p
2.36 vE D p 3a {O C b |O
3a2 C b 2
5
2.38 tbraking D s D 7:85 s
2
2.40 tjajmax D 3:93 s and tjajmax D 11:8 s where jajmax D 3:60 m=s2
samax D 12:8 m and samax D 129 m
2.42 v0 D 10:1 m=s

a0
2.44 P /D
x.t sin 2!t C ˇ cos !v.t/ D 7:00 sin t C 1:50 cos.0:500t/ 1:50 m=s
!
x.t / D 7:00Œ1:00 1:50 cos t C 3:00 sin.0:500t/ m
mg
r
2.46 vterm D D 5:00 m=s
Cd

mg Cd t
2.48 v.t / D 1 e m
Cd
mg
vterm D
Cd
2.50 vf D 3:56 m=s

v D 10:1 ft=s
2.54 s D 186 ft
t D 3:90 s

r
P / D ˙ P 2 C 2g cos

2.56 . 0 cos 0
L
2.58 .P0 /min D ˙4:93 rad=s
s
2 kL0 k 2
x02

2.60 xP D ˙ v0 C 2 g C .x x0 / x
m m
2.62 txmax D 0:243 s
3=2
" s #
r0

r r r
r
1
2.64 (a) t D p sin 1
2G .mA C mB / r0 r0 r0 2
(b) (i) t D 380;000 s ; (ii) t D 383;000 s
v0
2.66 tstop D D 2:88 s
gk
.vT /20 aT
2.68 k D D 0:301
g .vT /20

2daT
2.70 t D 8:63 s
yP D 77:7 m=s
2.72 v D v0 C ac .t t0 /
s D s0 C v0 .t t0 / C 21 ac .t t 0 /2
ap hvp2
2.74 ˛s D C
ˇ r 2 r 3
˛s rDr D 1:44 rad=s2 ˛s ˇrDr D 3:63 rad=s2
ˇ
ˇ and
1 2

1 w h
2.76 D tan
d
2.78 vEgun D .14:7 ft=s/ {O

g
2.80 y D h C tan ˇ.x w/ .x w/2
2v02 cos2 ˇ
2.82 The golfer’s chip shot is successful.
2.84 v0 D 52:8 ft=s
2.86 The two firing ranges are 30:4ı 56:8ı and 69:4ı 74:5ı :
The sizes of these intervals are, respectively, 1 D 26:4ı and 2 D 5:09ı
The angle subtended by the target as seen from an observer at point O is ˇ D 21:3ı
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 ˇ than 2 .
2.88 dAB D 82:0 m
2.90 v0 D 62:5 ft=s and ˇ D 20:6ı

d D 4:70 m
vEinitial D v0 cos ˇ {O C v0 sin ˇ |O D .6:02 {O C 7:67 |O/ m=s
where v0 is the initial speed of the tiger and ˇ is the angle formed with the horizontal by the initial velocity of the
tiger.
r
gR cos
2.94 v0 D p
2 sin ˇ cos.ˇ /


2.96 max D D 45ı
4
v02
RD D 447 ft
g
2v sin max
tD 0 D 5:27 s
g
h i
2.98 y D 59:9103 ln 1 2:8410 4
x C 17:7x m

tI D 6:19 s
xI D 268 m
2
x cos ˛ y sin ˛ 1 x cos ˛ y sin ˛
2.102 y D v0 sin ˇ 2 g cos ˛
v0 cos.˛ C ˇ/ v0 cos.˛ C ˇ/
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
(c)
(a) !E 1 D 105 kO rad=s; !E 2 D 105 {O rad=s and !E 3 D 105 |O rad=s

2.106
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; vEB D 5 kO 2 {O C 1 kO D 10 |O m=s;

vEC D 5 kO 2 {O C 2 kO D 10 |O m=s; vE D 5 kO 2 {O C 3 kO D 10 |O m=s
D
v0 O
2.110 !E D kD .65:7 rad=s/ kO
R
2.112 jE
v j D 65;600 ft=s
Referring to the diagram below,
let denote the angle formed by the velocity vector and the x axis such that tan D vy =vx . Then,
D 70:0ı
v0
2.114 rP D v0 cos. C / and P D sin. C /
r
2.116 vEB D .4:00 uO r C 0:800 uO / ft=s
aEB D . 0:320 uO r C 3:20 uO / ft=s2
v0
2.118 rP D 0 and P D D 1:50 rad=s
d
v02
rR D D 9:00 ft=s2 and R D 0 rad=s2
d
aEA ˇrD0 D . 112 {O C 5:00 |O/ m=s2 :
ˇ
2.120
2.122 2! rP C !r
P D0
2.124 aE D . 262 uO C 200 uO B / ft=s

2.134 aE .s/ D f 19:0 uO t C Œ69:9 .0:292/s uO n g ft=s2

ˇaE ˇ D .33:710 3 cos / m=s2
ˇ ˇ
2.138
2.140 v0 D 19:2 ft=s
v02
2.142 D D 282 ft
g
2.144 Letting min and tf t0 denote the tightest radius of curvature and the time needed to perform the turn, respectively,
then min D 565 ft and tf t0 D 4:74 s:
2.146 vP D 13:8 ft=s2
2.148 s D 304 ft
2.154 rP D 684 ft=s; P D 0:0116 rad=s; rR D 4:94 ft=s2 ; and R D 0:000 428 rad=s2
2.156 P
r.0/ P
D 438 ft=s and .0/ D 34710 6 rad=s
2
R
r.0/ D 14:9 ft=s R
and .0/ D 10:610 6 rad=s2
2.158 rP D 0:265 ft=s
2.160 v D 5490 ft=s

ˇaE ˇ D 1:00106 ft=s2
ˇ ˇ
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
vA L 3
2.164 PA D D 1:1610 rad=s
.d=2/2 C L2
s
2 C v2
1 vA B
PH D D 4:29 rad=s
L 2
2 2 L 2 dL
" #
1 vA vB vA 4
RA D 2 D 1:6810 rad=s2
r 2d r2
2 2

vA vB
R D D 0:0755 rad=s2
2dL
2.166 rR D 5:78 ft=s2 and R D 1:42 rad=s2
h i
2.168 aEP D «r 3r P R 3rP P 2 uO r C r « P 3 C 3rR P C 3rP R uO
q
2.170 rD .h C sin /2 C .d C cos /2
v0 .h cos d sin / v0 . C d cos C h sin /
rP D q and P D :
d2
C h2 C 2 C 2d cos C 2h sin
.d C cos /2 C .h C sin /2
v02 . C d cos C h sin / d 2 C d cos C h2 C h sin

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

R D 2
d 2 C 2d cos C h2 C 2 C 2h sin

2.172 P D 0
2
R!AB
R D p D 15:8 rad=s2
h2 R2
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
v D 0:119 ft=s and ˇaE ˇ D 0:0864 ft=s2

ˇ ˇ
2.178
rEB=A
2.184 ROS D vEB vEA
jErB=A j
2.186 vEC =A D .6:86 {O C 30:3 |O/ m=s
aEC =A D .1:38 {O 1:00 |O/ m=s2

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 vrain D 16:4 ft=s

2.192 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

1 .vA ` vB d / cos
2.194 D C sin D 63:6ı :
vP =A d
2.196 D 14:9ı
r
` g
2.198 vpmax D vs C D 331 ft=s D 225 mph
2 2h
2.200 vEB D . 3:09 ft=s/ |O
aEB D 0:319 ft=s2 |O

2.202 vEB D . 2:00 m=s/ |O
2.204 aEB D . 11:1 m=s2 / {O @

td D 0:164 s
26d
2.206 a0 D D 0:937 ft=s2
3t 2
p
v0 w 2 C h 2
2.208 vEA D p {O D . 3:08 ft=s/ {O
wC 2 2
w pC h h p i
v02 w h2 C w 2 a0 h2 C w w C h2 C w 2
aEA D p 2 {O D . 0:893 ft=s2 / {O
2
wC h Cw 2


0:292 cos
2.212 vEC D . 61:1 ft=s/ sin 1 C p |O
0:195 0:0851 sin2
0:292 cos
aEC D 1:28104 ft=s2 cos 1 C p

0:195 0:0851 sin2
0:0248 cos2 sin

0:292 sin
C sin |O
.0:195 0:0851 sin2 /3=2
p
0:195 0:0851 sin2
q
2.214 r D .d C cos /2 C .h C sin /2
v0 . C d cos C h sin / v0 .h cos d sin /
P D and rP D p
.d C cos /2 C .h C sin /2 .d C cos /2 C .h C sin /2
2.220 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 1:56 uO / ft=s2
K
2.222 vE D Ṕ tan ˇ uO R C uO C Ṕ uO ´
´ tan ˇ
K2

aE D Ŕ tan ˇ uO R C Ŕ uO ´
´ tan3 ˇ
3
2.224
P R sin C 2rP P sin C 2r P P cos /
r .r r P sin .r R C 2rP P r P 2 sin cos / D 0
r P sin .rR r P 2 r P 2 sin2 / P R sin C 2rP P sin C 2r P P cos / D 0
r.r
P R C 2rP P
r.r 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
P D D 0:0177 rad=s
r sin
v2
rR D 0 D 11:3 ft=s2
r
v02 cos
R D 2 D 78:110 6 rad=s2
r sin
R D 0
2.230 xland D 3:35 m; yland D 1:42 m; and ´land D 0:
2.234 vEP =B .10:2 {O2 C 10:3 |O2 / m=s

aEP =B D .7:82 {O2 1:26 |O2 / m=s2


h D 29:410 6 m

mg Cd
t
2.238 s.t / D 2 Cd t C m e m 1
Cd
r
gm k
2.240 x D L0 C 1 cos t
k m
p
2.242 v0 gR
2.244 D 240 ft
2.246 f D 629 m
tf D 5:24 s
ˇ
2.248 rP ˇ D180ı D 0
P ˇ D180ı D 2:94 rad=s
ˇ
rR ˇ D180ı D 40:8 m=s2 and R ˇ D180ı D 0

ˇ ˇ
P ˇD30ı D 0:0539 rad=s

ˇ
2.250
R ˇD30ı D 0:243 rad=s2
ˇ
2.252 aj D 2L! 2 ˇ0 D 1:24 ft=s2

jE
2.254 a t D 3:33 ft=s2
2.256 aA j D 2:58 m=s2

jE
s
2 2
d! 2

d˛
C a02 D 173;000 ft=s2
ˇ ˇ
2.258 ˇaE ˇ D C
2 2
q
1 P4 4 2 ! 2 P 2 1 D 2:53 m=s2
ˇ ˇ
ˇaE ˇ D
2.260 min
L 2 C ! 1
Chapter 3
3.4 aEA D 2:15 |O ft=s2

EA
3.6 xR C xD0
mL

Cd 2
3.8 aE D v g |O
m
3.10 d D 50:8 ft
3.12 v D 0:701 m=s
3.14 Letting Fc denote the tension in the cord, Fc D 251 lb
3.16 The reading of the scale corresponds to the value of the spring force supporting the platform. Denoting this force by
Fs , the maximum reading of the scale is .Fs /max D 2mg:
3.20 Letting d and ts denote the stopping distance and time, respectively,
d D 418 ft and ts D 7:59 s:
If an entire train weighing 30 106 lb was skidding to a stop, instead of a locomotive, we would have the same
stopping distance because the weight does not appear in our solution.
3.22 Spring compression D 0:737 ft

3.24 k D 5:08104 lb=ft
3.26 Denoting by ımax the maximum displacement of the vase relative to the packaging and viewing this quantity as being
positive, ımax D 0:167 m.
Denoting the maximum elastic force by .Fs /max and the overall maximum force by Fmax D .Fs /max mg, .Fs /max D
586 N and Fmax D 557 N.

ımax D 0:569 ft and tstop D 0:177 s
3.30 The origin of the x axis is placed where the railcar first comes into contact with the bumper. Therefore, the com-
pression of the bumper required to stop the railcar coincides with xstop , the stopping position of the railcar, and
xstop D 0:160 m:
3.32 k D 5:93 lb=ft
3.34 k D 9:03103 kg=s2

v "
u k x2
u q #
0
3.36 Letting v.0/ denote the speed of the mass for x D 0, v.0/ D 2t C L0 L x02 C L2
m 2
3.44 D 1:40104 ft
3.46 Letting aE and FL denote the acceleration of the airplane and the lift force acting on the airplane, respectively, we
have jE
aj D 3:62 g and jFL j D 43;100 N
3.48
!
k ru
xR C x 1 p D 0;
m x2 C y2
!
k ru
yR C y 1 p D0
m x2 C y2
3.50
k
rR r P 2 C .r ru / g cos D 0;
m
r R C 2rP P C g sin D 0
m m2 g 2
3.52 yD tan 0 e Cd x=m 1 e Cd x=m 1
Cd 2v02 cos2 0 Cd2
3.54 s D 47:7ı
3.56 35:3 m=s v 55:4 m=s
3.58 (a) aE D .319103 uO r C 603 uO / m=s2 @

(b) Denoting by P the pressure force acting on the back of the projectile, taken positive in the direction of uO r (see
component system in the answer of Part (a)), P D 5:90106 N
(c) Denoting by R the normal force due to the projectile, taken positive in the direction of uO (see component system
in the answer of Part (a)), R D 11:3103 N
3.60 ! D 9:94 rad=s
3.64 RR C 2RP P D 0

R C cot2 / RP 2 D
R.1 g cot


1 2
3.66 B D sin D 41:8ı
3
me
3.68 rR r P 2 C G D 0 and r R C 2rP P D 0
r2
3.70 k D 94:8 lb=ft
3.72 s D 132ı
L2 RRP 2 p
3.74 L2 RR RP 2 .L2 R2 / C 2 2
C gR L2 R2 D 0
L R
RR C 2RP P D 0
3.78 jvr j D 0:471 m=s and v D 0:500 m=s

F
3.88 aAy D C g D 16:1 ft=s2
2mA

mA C mB
3.90 rRB=A D G
r2
(
49:3 lb for 0 t 1 s,
3.92 T D
47:3 lb for t > 1 s.
3.94 ax D k1 g D 4:41 m=s2
3.96 s D tan D 0:424
3.98 aEA D 8:05 {O ft=s2
aEB D 145 {O ft=s2

mg
3.100 q
k.m C mp /
3.102 vimpact D 3:69 m=s

p
3.104 gd 4R2 d 2 cos g.4R2 d 2 / sin D 4R3 R
3.106 tcontact D 78:5103 s D 21:8 h

ˇ ˇ
ˇ 2mB g ˇ
3.108 j.yB /max j D ˇ
ˇ ˇ
k ˇ
j.yB /max j D 2 ft
.vB /max D vB Œ.yB /max D 8:58 ft=s
3.110 Defining a Cartesian coordinate system with ˇ the y axis opposite tot
ˇ he direction of gravity and with origin at the
ˇ 2 ˇ
initial position of B, we have j.yB /max j D ˇ
ˇ .mA sin ˛ C mB /g ˇˇ D 1:84 ft
k
.vB /max D vB j.y D 0:9202/ D 4:79 ft=s
B
3.112 vEA D .3:16 {O C 3:13 |O/ m=s and vEB D .3:26 {O/ m=s

9mG g 39P
3.116 aEG D |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
T4 D D 1190 lb
mB C mC C 25mD C 9mG

Cd D 0:0888 lbs=ft
s
2 1 6
3.120 vc D 2G .mA C mB / D 27610 ft=s
dA C dB r0
3.122 ab D g .1 cos / D 1:94 ft=s2

an D g sin D 11 ft=s2
v2
D D 101103 ft D 19:2 mi
g sin
3.124 Defining a Cartesian coordinate system with origin at O, y axis pointing opposite to gravity and x axis pointing to
m2 g

x mg x
the right, we have y D C v0 sin ˇ C 2 ln 1
v0 cos ˇ mv0 cos ˇ
3.126 Verification problem
WB .g aB /
3.128 WA D D 161 lb
2g C 4aB
h i
3.130 m1 L1 R D m2 g cos sin m2 sin L1 R sin L1 P 2 cos L2 P 2 m1 g sin
L2 R D g sin L1 R cos L1 P 2 sin
Chapter 4
4.4 9:02104 ftlb

2
4.6 U1-2 D 1
2 m.v2 v12 / D 180 kJ
4.8 U1-2 D 430 kJ
4.10 (a) dmin D 283 m

(b) dmin D 283 m
(c) dmin D 14;800 m
4.12 d D 15:8 m
11
4.14 4:9810 J
4.16 U1-2 D 4:35107 ftlb

.U1-2 /C D 2:33107 ftlb
FC D 50;800 lb

5
4.22 ˇ D 1:2910 lb=ft3
4.24 k D 2:74105 N=m
4.26 v0 D 6:40 ft=s
4.28 k D 44:8 lb=ft
4.30 .U1-2 /Engine D 4:64105 ftlb

m 2
4.32 Favg D mg C .v v22 / D 1:47 kN
2h 1

4.34 D sin 1 23 D 41:8ı
4.36 VB D 14 ˇx 4
d D 0:420 ft
4.38 vA D 2:34103 m=s

s
4.40 V D P0 s0 A ln
s0
4.42 1:92 ftlb

1
4.44 D cos 3
2
D 132ı
4.46 v2 D 24:1 ft=s
4.48 k D 276 lb=ft

r
m
4.50 .vy /max D g
k
2
mvB
5R
4.52 L0 D D 1:06 m
3 4kR
4.54 vA2 D 1:09 m=s and vB2 D 3:28 m=s
2 C kı 2
mA vB2
4.56 mB D 2
D 0:685 kg
2gı vB2
r q
4.58 v2 D 2gR dg 1 d 2 = .2R/2

d
q
NA D mg 3 1 d 2 = .2R/2
R
4.60 Letting vA2 denote the speed of A at impact, vA2 D 3:69 m=s
s
m ` mB `OB
4.62 vB D 8g A OA .sin 35ı C sin 75ı / D 26:6 ft=s
mA C 4mB
1
4.64 yB D .m sin ˛ C mB /g
k A
.vB /max D vB .yB D 2/ D 9:42 ft=s
dB D 0 ft
4.66 W D 2mg
2
4.68 Part (a): T D 2m.vO C R2 ! 2 / D 580 J
2
1
Part (b): T D 2 .4m/vO C 12 m.4R2 ! 2 / D 2m.vO
2
C R2 ! 2 / D 580 J
4.70 U D 5760 kJ
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

mgv .k cos C sin /
4.80 Pi D D 1:51103 ftlb=s D 2:75 hp

4.82 ˇ D 1:07109 kg=.m2 s2 /

2
1 0:99 mg
4.84 U1-2 D 2m D 0:0361 J
Cd
3
mv 2 2 mgL
4.86 kD 2 C ı2 2 2
D 9030 N=m
ıu2 l2
ıu1 ıl1
3EIcs 2
4.88 V D ı
2L3
s
2gı.mA sin ˛ C mB k mA cos ˛/ kı 2
4.90 .v2 /max D D 9:67 ft=s
mA C mB
The distance between B and the ground is 0 ft
mgvt sin
4.92 Eb D D 5:395103 kJ D 1290 C

Chapter 5
5.2 xe D 28:510 24 m
xe D 4:1510 12 m
xe D 1:4210 6 m
ˇZ ˇ
ˇ t2 ˇ
5.4 FE dt ˇ D mv2 D 2:64 lbs
ˇ ˇ
ˇ
ˇ t1 ˇ
ˇ
ˇE ˇ
ˇ mv2
ˇFavg ˇ D D 2400 lb
t2 t1

mvx2
5.8 t2 D D 10:6 s
FT
d D 1280 ft
(a) ˇFEavg ˇ D 8080 N

ˇ ˇ
5.12
ˇFE ˇ
ˇ ˇ
avg
(b) s D 0:515
mg
5.14 (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 (four significant figures), ˛ remains equal to 31ı .
5.20 n D 9:170106 bees

Slowdown D 48:5 ft=s D 33:0 mph
Lfp mp
5.22 .xfp /2 D
mp C mfp
5.24 vP 4 D 1:86 ft=s
5.26 vEG D . 2580 {O 3030 |O/ m=s;

mA Lfp
5.28 dD D 4:95 m
mA C mC
5.30 vP /final D .1:52 ft=s/ {O

.E

mA r 6
5.32 (a) Amplitude D D 2:1410 m
mA C mB
mA r!
(b) .vB /max D D 0:0135 m=s
mA C mB
5.34 vEA2 D .5500 {O C 9520 |O/ m=s and vEB2 D .5500 {O C 9530 |O/ m=s @ 60ı
5.36 .vA /max D 25:2 ft=s and .vB / D 0 for mB ! 1
C C
5.40 vEA D vEB D . 55:9 ft=s/ {O
mA C mB p
5.42 vB D 2gL.1 cos m / D 210 m=s
mB
mA C mB p
5.44 vB D 2k gd D 199 m=s
mB
5.46 0:817 e 0:882
2 .v /2
mB B
5.48 dD D 2:89 m
2k g.mA C mB /2
q
.mA C mB /mA 2 gk 2 Œ2hk C .m C m /g
A B
5.50 Wmax D mA g C D 819 lb
.mA C mB /k
C C
5.52 vEA D vEB D .8:33 {O C 78:0 |O/ ft=s
rE D .14:5 {O C 136 |O/ ft
C C
5.54 vEA D 77:0 |O ft=s and vEB D .73:3 {O C 85:8 |O/ ft=s
rEA D 132 |O ft 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 ˇ D tan .cot ˛/
5.60 The answer is an explanation.
C C
5.64 vEA D . 23:6 {O C 18:5 |O/ m=s and vEB D . 0:717 {O 6:20 |O/ m=s
5.66
s r
2 gh1
dD .1 C e/h1 cos ˛ sin ˛ Œ.e 1/ C .1 C e/ cos 2˛
gh1 2
q
C 2gh2 C 21 gh1 Œ.e 1/ C .1 C e/ cos 2˛2 D 6:24 ft
5.68 d D 3:50 ft
p
5.70 C
vN D 2N 1 2gh
hEO D 593103 kO slugft2 =s

5.72 A
hEO A D 79:9103 kO slugft2 =s


5.74 hEQ D 1:44106 kO slugft2 =s
.L sin 0 C d /2
5.78 !D !
.L sin C d /2 0
P E O D hEP O .
5.80 Since hEO D mP gvP .0/t cos and M O it is indeed true that M
E O D mP gvP .0/t cos k,
s
2L3 p
5.82 E
hO D ˙W .cos cos 33ı / kO D ˙.3:99 lbfts/ cos 0:839 kO @ 57ı
g
5.84
s
2gL.cos 2 cos 1 /
v1 D sin 2 D 6:85 ft=s;
sin2 1 sin2 2
s
2gL.cos 2 cos 1 /
v2 D sin 1 D 2:32 ft=s
sin2 1 sin2 2
q
5.86 MA D 2m!02 r r2 r02

time to reach end of arm D 0:150 s
ˇ ˇ
5.90 ˇ!E ˇ D 0:0612 rad=s
s
5.92 The ratio of the orbital sectors is equal to 1.

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

5.98 vesc D p
r
vesc Earth D 4:21104 m=s D 151;000 km=h

4 2 a3
5.100 Gme D D 3:971014 m3 =s2
2
5.102 rg D 1:39108 ft D 26;200 mi; hg D 1:18108 ft D 22;300 mi; vc D 10;100 ft=s D 6870 mph
s
2 1
5.104 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
a3
D 2 D 5470 s D 91:2 min
gre2
5.106 vA D 4710 ft=s D 3210 mph

vA D 6830 ft=s D 4660 mph
s
ae3
t D D 10;600 s D 2:95 h
gre2

5.110 (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 180ı ˇ D 5:30ı
s
2
rP vP 2GmB
5.112 (a) v1 D
rP
2
(b) rP vP > 2GmB
4 Q2 1

1 2 1
5.118 RD 4 pA dA C D 2120 lb
g dA2 dB2
5.120 vB D 29:2 ft=s

mf

5.122 vmax D vo ln 1 C
mb
d 2 vw2
5.124 kD Œ1 cos.=2/ D 370 lb=ft
4gımax

M
5.126 v D vo ln gt
M m P ot
1 4Q2
2 4Q2 4Q2 `
5.128 N D 4 PA d C
D 53:4 N; V D D 0:354 N; M D D 0:0707 Nm
d 2 d 2 d 2

5.130 F D gL C v02 D 1:10 lb
5.134 mP o D 0:0150 slug=s

vo D 4530 ft=s
mf

vmax D vo ln 1 C gtbo D 1320 ft=s
mb
5.136 max power D 0ı

1
v0 D 2 vw
m q
FEavg D
p
5.138 2ghf C 2ghi |O D .654 N/ |O
t q p
jpEC pE j jFEavg j 2ghf C 2ghi
D D D 111
jt . mg |O/j mg gt
C C
5.140 vEA D . 13:6 m=s/ {O and vEB D . 21:3 m=s/ {O
T TC
100% D 45:3%
T
C C
5.142 vEA D .4:61 {O C 75:3 |O/ ft=s and vEB D .37:4 {O C 98:8 |O/ft=s
C
vEA
rEA D dA C
D .7:71 {O C 126 |O/ ft
vA
C
vEB
rEB D dB C
D .87:6 {O C 231 |O/ ft
vB
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
D 8:64107 s D 1000 days
8 q
< 1 K 1 K 2 4v 2 D 0:219 m=s;
2 2q 0
5.148 vw D
: 1 K C 1 K 2 4v 2 D 18:3 m=s;
2 2 0
5.150 vEA D .0:813 m=s/ {O and vEB D .3:25 m=s/ {O
Chapter 6
6.2 !EB D 1800 kO rpm and !E C D 1160 kO rpm
RA
6.4 !B D ! D 600 rad=s
RB A
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
2 rad 1 h 6
6.22 !s D D 72:710 rad=s
24 h 3600 s
6.24 !EB D 8:00 kO rad=s; ˛EB D 2:00 kO rad=s2 ; !ED D 1:60 kO rad=s; ˛ED D 0:400 kO rad=s2
6.26 !E C D 40:0 kO rad=s
˛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
6.38 !EP D 1:60 kO rad=s

vEC D 5:33 {O m=s

6.40 1 D 67:7ı
2 D 231ı
6.42 !EA D 7:00 kO rad=s
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
6.54 vEC D 16:3 {O ft=s

the cable is unwinding at 12:2 ft=s
6.56 !E CD D 42:4 kO rad=s
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 0:0467 kO rad=s

!EBC D 8:58 kO rad=s
6.66 !E R D 35:0 kO rpm

vED D 17:2 |O in.=s

vEB D 1:92 |O m=s
" #
P
R.H R cos / sin
6.70 P
vEB D R cos p |O
L2 .H R cos /2
6.74 aEB D .17:0 {O 5:38 |O/ ft=s2
6.76 aEC D . 16:8 {O C 87:7 |O/ ft=s2
6.78 aEQ D 9:16 uO r m=s2
6.80 ˛EAB D 48:6 kO rad=s2 and ˛EBC D 0E
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 0:458 kO rad=s and ˛EAB D 0:258 kO rad=s2
6.88 ˛EW D 7:97 kO rad=s2
6.90 ˛EBC D 1640 kO rad=s2

HRP 2 .R2 H 2 / sin O

6.92 ˛EBC D k
.H 2 C R2 2HR cos /2
6.94 ˛Es D 0:400 kO rad=s2 and aEO D 2:00 {O ft=s2
6.96 ˛Egate D 4:50 kO rad=s2
6.98 ˛EAB D 92:1 kO rad=s2 and ˛EBC D 191 kO rad=s2
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 551 kO rad=s2 and ˛EBC D 320 kO rad=s2
For C accelerating downward and to the left: ˛EAB D 729 kO rad=s2 and ˛EBC D 320 kO rad=s2
6.104 For D 10ı and D 56:11ı aED D .18;300 {O C 34;900 |O/ ft=s2
For D 10ı and D 303:9ı aED D .18;300 {O C 32;900 |O/ ft=s2
( " #)
R.L2 R2 / cos2 R sin2

R sin 2
6.106 aEC D R˛AB cos 1 C p C R!AB p sin |O
L2 R2 cos2 .L2 R2 cos2 /3=2 L2 R2 cos2
D sR s!02 {O C s˛0 C 2Ps !0 |O

6.110 aEP
.R= h/!AB O .R= h/d!AB O

6.112 (a) !E CD D k D 3:14 kO rad=s and vEbar D I D 0:377 IO m=s
1 C R= h 1 C R= h
(b) ˛ECD D 0E and aEbar D d˛CD IO D 0E
ı!AB .ı C sin / O ıd!AB .ı C sin / O

6.114 (a) !E CD D k and vEbar D I
1 C ı 2 C2ı sin 1 C ı 2 C 2ı sin
ı 1 ı 2 !AB 2 cos ıd 1 ı 2 !AB 2 cos

(b) ˛ECD D 2 kO and aEbar D 2 IO
1 C ı 2 C 2ı sin 1 C ı 2 C 2ı sin
6.116 (a) position 1: vEbar D 0:348 IO m=s and position 2: vEbar D 0:646 IO m=s
(b) position 1: vEbar D 0:670 IO m=s and position 2: vEbar D 6:03 IO m=s
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
6.126 dPAB D 1:12 ft=s

dRAB D 2:01 ft=s2
6.128 !E bar D 14:3 kO rad=s
6.130 !EA D 481 kO rad=s
6.132 aEE D . 1:50 {O C 0:0336 |O/ ft=s2
6.134 dPE C D 0:214 ft=sand dRE C D 0:0192 ft=s2
D vC C `!OA sin {O C d!OA `!C cos |O `!C sin kO

6.136 vED
h i
2 2 2 2
aED D 2`!C !OA cos C `˛OA sin d!OA {O C d˛OA C 2vC !OA C ` !C C !OA sin |O `!C cos kO
vEC D .21:5 {O C 7:51 |O/ ft=s D 21:5 IO C 7:51 JO ft=s

6.138
aEC D .107 {O C 166 |O/ ft=s2 D 107 IO C 166 JO ft=s2


Chapter 7
7.2 tstop D v0 =.k g/ D 1:86 s and dstop D v02 =.2gk / D 16:8 ft
7.4 F D 179 N and aGx D 4:71 m=s2
7.6 D 18:9ı
7.8 tmin D v=.aGx / D v=.gs / D 1:12 s
7.10 The FBD for the problem is
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 ˇHE ˇ D 674 lb
ˇ ˇ
.s /min D 0:736

1 3 2
7.12 m1 D 4 m; m2 D 4 m; xP D 3L
1 1 1 2
m1 D 2 m; m2 D 2 m; .IG /extra D 6 mL
7.14
Rx D .0:112 N=s2 /t 2 ;
Ry D 0:0101 N;
5
M´ D 3:1110 Nm
Or D 16:8 lb and O D 5:00 lb

p g p p
7.18 j˛bar jmax D 3 for d D 16 3 3 L or d D 1
6 3C 3 L
L
OE D Ox {O C Oy |O D . 1320 {O C 48:5 |O/ N and ˛ET-bar D . 6:92 rad=s2 / kO

p
7.24 .vG /final D 2gL.sin pk cos C sin / D 18:4 ft=s
2gL.sin k cos C sin /
tfinal D .vG /final =aGx D D 1:08 s
g.sin k cos /
7.26 tr D 2:41 s and d D 50:7 ft
grk
7.28 aGx D k g and ˛b D 2
kG
TCD D 101 N: and 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 counter-
clockwise. 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 D 60:0ı
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 3.8 on p. 218 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.
IG vf
7.38 (a) F D D 12:1 lb
r 2 tf
mc r 2 tf
(b) .tf /new D D 6:85 s
2IG C r 2 mc
3mAB g cos 3mAB g sin cos 3mAB g sin cos
7.40 ˛AB D ; aEB D {O; ˛w D
L.2mAB C 9mB sin2 / 2mAB C 9mB sin2 R.2mAB C 9mB sin2 /

3.2mA C mAB /g cos

7.42 ˛AB D
L.6mA cos2 C 2mAB C 9mB sin2 /
3.2mA C mAB /g sin cos
aEB D {O;
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 ; T D 3130 N; and ˛s D 1:68 rad=s
.3g cos C 2LP 2 / sin 3.2g C LP 2 cos / sin

7.48 xR D and R D
1 C 3 sin2 L.1 C 3 sin2 /
7.50 (a)

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

1 2hac
Nf D Wc 1 D 549 lb;
4 g`
W a
Fr D c c D 503 lb
2g
(b)
ac Ww
Ax D Fr D 485 lb
g
Ay D Wr Nr D 689 lb
d 2a I
MA D Fr C c B D 522 ftlb
2 d

k2
!
m 1C G xR C kx D mg sin C kL0
r2
g cos
7.54 ˛AB D 1
D 3:74 rad=s2
2L 3 C cos2 .cos = sin /2 C sin cos .cos = sin /
60 45g
7.56 aAx D g and ˛b D
181 181R
5 5g sin
7.58 aEG D O;
7 g sin u ˛b D ; and FEdue to bowl D mg cos uO r C 72 mg sin uO
7
5g
7.60 R C sin D 0
7.R /
2
k R2 C kG
PR2 P
7.62 (a) aEG D .1 k / 2/
{O; and .s /min D 2 P C mg
m.R2 C kG R2 C kG k

P k R P
(b) aEG D .1 2k / k g {O; and ˛d D 2 .1 k / g
m kG m
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
3
7.66 (a) Ay D 4 mg and By D 54 mg

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

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

Ax D D 437 lb;
2.d C h/.2mb C 3mr /
gŒmb h C d.mb C mr /Œmb C 3.mb C 2mr / cos2 C 3ghmb mr sin2
Ay D D 256 lb;
3g.mb C mr /Œ.d C h/mb C 2hmr sin cos g.h d /mb mr tan
Cx D D 629 lb;
gŒmb d C h.mb C mr /Œmb C 3.mb C 2mr / cos2 C 3gd mb mr sin2
Cy D D 353 lb
7.72 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
7.74 Pmax D 1600 lb
!
1 vc2
7.76 D tan
gR
7.78 ˛AC D R D 1:46 rad=s2 and ˛CE D R D 1:46 rad=s2

tclosure D 1:29 s
.vE /at closure D 1:87 m=s
7.82
gR sin g cos R cos. /P 2
R D 2 C 2R sin.
;
R2 C 2 C kG /
F D mg sin C maGx

/ŒgR sin g cos R cos. /P 2
D mg sin m ŒRC sin. 2 C2R sin. / C P 2 cos. / ;
R2 C2 CkG
N D mg cos C maGy

/ŒgR sin g cos R cos. /P 2
D mg cos C m cos. 2 C2R sin. / P 2 sin. /
R2 C2 CkG
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;
2 .mAB C 2mC /L sin 2 C
2
1
2 C
1
Lm cos xR C Œ m
A
1
C m cos C . m
12 AB
1
2 C C m / sin2 L2 R
4 AB C
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 D 0;
1
2 .2d C h/mC cos xRA C 21 .2d C h/LmC cos. /R
1
CŒ 12 .h2 C w 2 / C 41 .2d C h/2 mC R
1
2 .2d C h/LmC .sin cos sin cos /P 2
C 21 g.2d C h/mC sin D 0:
7.86 (a) F D 34 mg.3 cos 2/ sin and N D 41 mg.1 3 cos /2

(b) Given any (finite) value of s , there always is a value of for which the rod will slip.
! !
md R2 R2
7.88 Pmax D mc gs 1 C C 2 ; aC x D g 1 C 2 ; and aGx D s g
mc kG kG
7.90 P D 7:59 N and .s /min D 0:0462

Chapter 8
8.2 TA D TB D 518 J
8.4 TR1 D 7760 ftlb and TR2 D 15;000 ftlb
8.6 T D 324;000 ftlb

1
8.8 T D ŒI R2 C H 2 .IA C mBC R2 /!AB
2
D 25:3 J
2H 2 D
8.10 T D 28:9 ftlb
8.14 Lf D 34:1 ft

s
.2Md=R/ kd 2
8.18 !2 D 2 C R2 /
m.kG
2M
ds D
kR
8.20 M D 2:06106 ftlb
p
8.22 vperson D vG D 2gH.1 C cos /
ˇ
v ı D 10:8 m=s
ˇ
person D0
8.24 vO D 10:9 ft=s
8.26 nmin D 26
(a) vperson D 26:5 ft=s
1
8.28 Ufracture D 2 gŒLA .mA C 2mS / C 2mS rS .cos f cos i / D 318 J
8.30 Ufriction D 364 ftlb

2 C g.L C d
W vmax 2H /
8.32 kD D 11:3 lb=ft
2g .H d /.H C 2ı0 d /
8.34 ı D 1:03 ft
3gH 2
vA2
8.36 WP D 2
W D 800 lb
3gH C 6vA2
8.38 vmax D 6:39 ft=s
8.40 vC D 4:96 m=s and !ED D .44:1 rad=s/ kO
8.42 vS D 7:53 m=s
8.44 vB D 2:21 m=s

In Á 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 slide D 35:1ı

1
hEA m R2 !AB kO D .1:72 kgm2 =s/ kO

8.52 D
AB 3 AB
(a) hEA D m ! R2 kO D .7:20 kgm2 =s/ kO

BC BC AB
1
(b) hED kO D .7:75 kgm2 =s/ kO

CD D 3 mCD HR!AB

6
hEC kgm2 =s/ kO

8.54 W D .91:910 @
6
hEO kgm =s/ kO
2

W D .66910 @
ˇpE ˇ D WAB vA D 3:23 lbs

ˇ ˇ
8.56 AB
2g cos
WAB LvA O
hEG D k D .2:42 ftlbs/ kO
12g cos
W vfinal
8.58 FEavg D {O D .878 lb/ {O
gt
ME avg D 2IG vfinal kO D . 10:7 lbft/ kO
dt

v0
8.62 tr D
2k g
ˇ ˇ
8.64 !A ˇend of slip D 61:3 rad=s and !B ˇend of slip D 40:3 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
8.68 E /D
!.t t !0 kO
.1 C g w /r
ˇ
8.70 vG ˇ tD2 s D 21:7 ft=s
.s /min D 0:0588
8.74 The system rotates clockwise.

m a e
D B 0 tf2 ; (clockwise)
2IO
4r 2 d
8.76 dD
.2r b/2

2
8.82 dD 3`
8.84 d D 2:09 ft
!EB D . 26:6 rad=s/ kO and vbC D 226 ft=s
8.86 !E bC D .10:7 rad=s/ kO
8.88 C
vEG D . 1:31 m=s/ {O and !EAC D .3:43 rad=s/ kO
s
4g` sin ˇ
8.90 v0 D D 7:48 ft=s
3 cos2 ˇ
8.92 C
!EAB D . 0:851 rad=s/ kO and C
!EBD D .2:55 rad=s/ kO
8.94 !EAC D . 0:000 715 rad=s/ kO and C

!EB D . 0:000 024 1 rad=s/ kO
8.96 T D 0:0257 ftlb
8.98 d D 1:98 m

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

max D 14:8ı
Chapter 9
mgL 2
9.2 IO D
4 2
s
7L
9.4 D 2
6g
r
IG
9.6 `D
m
s
Gr 4
9.8 !n D
LR4 t
r
2d m
9.10 D
h k
s
1
.h3 C d 3 / C md 2
9.12 D 2 3
kh2
9.14 Wpayload D 90;700 lb
r
! d g
9.16 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
17k
9.22 xR G C x D0
6m G
8k
9.24 xR G C x D0
3m G r
2 3m
D D 2 D 2:22 s
!n 8k
s
2g
9.26 !n D
3.R r/
s
3.R r/
D 2
2g
s
2 L
9.28 D D 2
!n 2g
s
.3 2/R
9.30 D 2
3g
F0 =k
9.32 x.t / D A sin !n t C B cos !n t C cos !0 t
1 .!0 =!n /2

9.34
q
mu D 0:01 kg ) jym j D 2:1910 6 cos.4:37t/ C 1:8910 5 m;
q
mu D 0:1 kg ) jym j D 1:1010 5 cos.4:37t/ C 1:1110 5 m;
q
mu D 1 kg ) jym j D 9:8010 4 cos.4:37t/ C 1:0110 3 m
9.36 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
MF D D D 0:820
mu R 1 .!p =!n /2

5

9.42 x.t / D 0:000 501 sin 10t C 0:100 cos 10t 2:5110 sin 200t m

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

!0
y.t/ D sin ! n t C sin ! 0 t ;
1 .!0 =!n /2 !n
s
L0 2k L0
where keq D 2k 1 and !n D 1
L m L
2 1 1

9.46 3 mB C 2 mA xR C 2kx D 2 F0 sin !0 t
F0
xamp D 4
2
4k 3 B C mA !0
m

p
9.52 no peak in MF for 1=2
mu "!r2
9.54 yR C 2!n yP C !n2 y D sin !r t
m
9.56 jym j D 0:002 19 m
IG R C cro2 P C k1 ro2 C k2 ri2 D 0

9.58
h i
.t / D 0:0504e 3:80t 0:000386e 496t
rad
9.60 x.t / D e 0:5t .A sin 3:12t C B cos 3:12t/

p
q
LCx
9.62 mxR C c xP C 2k L2 C .L C x/2 2L p D F0 sin !0 t
L2 C .L C x/2
mxR C c xP C kx D F0 sin !0 t
r
F0 =k k c
DD p ; where !n D and D p
Œ1 .!0 =!n /2 2 C .2!0 =!n /2 m 2 km
9.64 k > 1:35109 N=m and c D 0

9.66 (a) mRs C c sP C ks D mA!02 sin !0 t
" # " #
mA!02 k m!02 mcA!03

(b) y.t/ D 2 C A sin !0 t 2 cos !0 t
k m!02 C c 2 !02 k m!02 C c 2 !02
v
k 2 C c 2 !02
u
u
(c) DT D t 2
k m!02 C c 2 !02

k1 k2
9.68 mxR C xD0
k1 C k2
9.70 k D 5:91%
9.72 .mm C mp /yRm C keq ym D mu "!r2 sin !r t
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 F t D 1480mu sin.125:7t C 0:842/ C 1770mu cos.125:7t C 0:842/ N
!
L
9.76 myR C 2k 1 p 0 yD0
y 2 C L2

L0
9.78 myR C 2k 1 yD0
L
Chapter 10
10.2 vEE D R!arm cos {O C .d C ` cos /!arm |O .d C ` cos /!arm kO
2 h
!arm i
2 2
d 2 C 2d ` cos C `2 C R2 cos2 kO

aEE D !arm cos .d C ` cos / {O !arm sin .d C ` cos / |O
R
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
!arm 2 i
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 R!d cos ˇ {O .h!b C R!d sin ˇ/ |O C R!b cos ˇ kO

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

C R!P b cos ˇ h!b2 2R!b !d sin ˇ kO
10.10 vEA D 2L!0 cos2 ˇ kO

aEA D 2L!02 cos2 ˇ {O L!02 cot ˇ |O 2L˛0 cos2 ˇ kO
10.12 !EAB D !s |O ˇP kO
P {O
˛EAB D ˇ! ˇR kO
s
LR L L L P2
aEG D ˇ sin ˇ C d C cos ˇ !s2 C ˇP 2 cos ˇ {O C ˇ sin ˇ ˇR cos ˇ |O P sin ˇ kO
Lˇ!s
2 2 2 2
vEA D 1:432 |O 0:6677 kO m=s

10.14
aEA D 103 |O C 47:8 kO m=s2

10.16
!EAB D 0:0161 {O C 2:69 |O C 0:917 kO rad=s

10.18
D 1:73 {O 4:59 |O C 15:8 kO rad=s2

˛E
AB
10.20 !E bit D !d {O C !a |O C !b kO
˛Ebit D .!P d !a !b / {O C .!P a C !d !b / |O C .!P b !d !a / kO
h i
10.22 aEB D a t C h!P a .` C r t / !a2 C !b2 {O C Œ2.v t C h!a / !b C .` C r t / !P b |O
Œ.2v C h! / ! C .` C r / !P kO
t a a t a

Dynamics 1e Answers

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Engineering Mechanics: Dynamics 1e Gray, Costanzo, Plesha

Answers to Selected Even-Numbered Problems

1.4 rEB=A D .4:00 {O 1:00 |O/ ft

1.8 ŒIxx  D ŒIyy  D ŒI´´  D ŒM ŒL2

Units of Ixx , Iyy , and I´´ in the SI system: kgm2 ;

1.10 Concept problem.

2.4 Concept problem.

y.x/ D 2:00 C 3:00x C 2:00x 2 m

2.8 Concept problem.

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

2.12 P D Œ.4:08e 13:6t / {O C . 0:720 1:28e 13:6t / |O m=s

y.x/ D 5:25 C 3:25x C 1:00x 2 m

2.16 vE D .29:3 ft=s/Œ1 C cosŒ.25:5 rad=s/t {O .29:3 ft=s/ sinŒ.25:5 rad=s/t |O

2.18 Computer problem.

1 Last modified: January 20, 2011

2.20 Computer problem.

2.22 vP =O /1 D Œ.1 C cos t/ {O1 C .4 2t/ |O1  m=s

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

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

2.28 Computer problem.

2.30 Computer problem.

2.34 Computer problem.

2.42 v0 D 10:1 m=s

2.52 Computer problem.

2 Last modified: January 20, 2011

2.62 txmax D 0:243 s

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

2.82 The golfer’s chip shot is successful.

2.84 v0 D 52:8 ft=s

2.88 dAB D 82:0 m

2.90 v0 D 62:5 ft=s and ˇ D 20:6ı

2.92 Computer problem.

3 Last modified: January 20, 2011

2.100 Computer problem.

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

2.124 aE D . 262 uO C 200 uO B / ft=s

2.126 Concept problem.

2.128 Concept problem.

4 Last modified: January 20, 2011

2.130 Concept problem.

2.132 Concept problem.

2.134 aE .s/ D f 19:0 uO t C Œ69:9 .0:292/s uO n g ft=s2

2.136 Concept problem.

2.140 v0 D 19:2 ft=s

2.146 vP D 13:8 ft=s2

2.150 Concept problem.

2.152 Concept problem.

2.158 rP D 0:265 ft=s

2.160 v D 5490 ft=s

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

5 Last modified: January 20, 2011

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

v D 0:119 ft=s and ˇaE ˇ D 0:0864 ft=s2

2.180 Concept problem.

2.182 Concept problem.

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

aEC =A D .1:38 {O 1:00 |O/ m=s2

2.188 Computer problem.

2.190 vrain D 16:4 ft=s

1.8 ŒIxx D ŒIyy D ŒI´´ D ŒM ŒL2

2.12 P D Œ.4:08e 13:6t / {O C . 0:720 1:28e 13:6t / |O m=s

2.16 vE D .29:3 ft=s/Œ1 C cosŒ.25:5 rad=s/t {O .29:3 ft=s/ sinŒ.25:5 rad=s/t |O

2.22 vP =O /1 D Œ.1 C cos t/ {O1 C .4 2t/ |O1 m=s

2.134 aE .s/ D f 19:0 uO t C Œ69:9 .0:292/s uO n g ft=s2

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

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

rR ˇ D180ı D 40:8 m=s2 and R ˇ D180ı D 0

P ˇD30ı D 0:0539 rad=s

3.56 35:3 m=s v 55:4 m=s

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

3.64 RR C 2RP P D 0

3.94 ax D k1 g D 4:41 m=s2

3.96 s D tan D 0:424

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