Solution Manual for Engineering Mechanics: Dynamics, 2nd Edition

Name: Solution Manual for Engineering Mechanics: Dynamics, 2nd Edition
Brand: Academic Documents Platform
Price: 9.99 USD
Availability: InStock
Author: Madison Taylor

Strengthen your problem-solving skills with Solution Manual for Engineering Mechanics: Dynamics, 2nd Edition, your essential study tool.

Madison Taylor

Contributor

4.4

about 2 months ago

Preview (13 of 41)

Engineering Mechanics: Dynamics 2e 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 3:883 ft
1.4 ErB=A
ˇ
ˇxy system D .4:000 O{ 1:000 O| / ft
ErB=A
ˇ
ˇpq system D .3:313 Oup 2:455 Ouq / ft
ˇ
ˇErB=A
ˇ
ˇxy system D ˇ
ˇErB=A
ˇ
ˇpq system D 4:123 ft
1.6 Ev D .79:68 O{ C 7:190 O| / ft=s
1.8 D 3:229 rad
D 0:087 31 rad
1.10 vr D 504:6 ft=s and v D 353:3 ft=s
1.12 D 101:1ı
1.14 xP 2 D 0:7679 ft and yP 2 D 5:330 ft
1.16 EvA D .20:02 O{ C 12:81 O| / ft=s and EaA D .1:414 O{ C 4:243 O| / ft=s2
1.18 r D 26:36 mm
1.20 ŒIxx ç D ŒIyy ç D ŒI´´ç D ML2
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.
1.22 Concept problem.
1.24 the units of E are kg=ms2.
Chapter 2
2.2 Concept Problem
2.4 Concept Problem
1 Last modified: August 16, 2012

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.6 Er1 D 8:747 Our m and Er2 D 13:73 Our m
Evavg

1 D 5:467 Our m=s and Evavg

2 D 8:579 Our m=s
2.8 Ev D .154:0 O{ C 266:7 O| / ft=s
2.10 Ev D .145:4 O{ C 69:81 O| / ft=s
2.12 Ev1 D .0:4623 O{ C 0:08155 O| / m=s
Ev2 D .0:002550 O{ C 0:0004499 O| / m=s
2.14 1 D 12:21ı and 2 D 17:30ı
1 D 102:2ı and 2 D 72:70ı
2.16 .x/ D cos1
3 C 4x
p10 C 24x C 16x2

2.18 Eaavg D .0:04588 O{ C 3:886 O| / ft=s2
Eaavg Ea.5 s/ D .0:001154 O{ 0:003175 O| / ft=s2
2.20 Er D E0
Evavg D E0
d D 1:394 ft
vavg D 0:3485 ft=s
2.22 Computer Problem
2.24 vmax D 2v0 D 58:67 ft=s and vmin D 0 ft=s
yvmin D 0 ft and yvmax D 2R D 2:300 ft
Eavmin D v2
0
R O| D .748:2 ft=s2/ O| and Eavmax D v2
0
R O| D .748:2 ft=s2/ O|
2.26 Rx D 8v2
0 a3
y2 C 4a22
Ry D 4v2
0 a2y
y2 C 4a22
2.28 Ev D .203:2 O{ C 117:3 O| / ft=s
Ea D .27:53 O{ C 47:69 O| / ft=s2
2.30 vmax D 80:82 ft=s
2.32 Ev D .51:32 O{ 1:046 O| / ft=s
Ea D .0:3873 O{ C 19:00 O| / ft=s2
2 Last modified: August 16, 2012

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.34 v D !
p
d 2 x2
s
1 C h2
x
d 2 x3
d 4
2
v D 0:5236 ft=s; v D 0:4554 ft=s; and v D 0
2.36 EvA D .72:00 O{ C 96:00 O| / ft=s
EaA D .460:8 O{ 345:6 O| / ft=s2
2.38 Computer Problem
2.40 tbraking D 7:854 s
2.42 jajmax D 3:600 m=s2
tjamaxj

1 D 3:927 s and tjamaxj

2 D 11:78 s
sjamaxj

1 D 12:84 m and sjamaxj

2 D 128:5 m
2.44 Largest distance traveled in 1 s is d D 0:2667 m, corresponding to a D ˇ1
pt
2.46 v.0/ D 0:080 00 ft=s
2.48 v0 D 10:10 m=s
2.50 ac D 22:36 g
2.52 tstop D 2:880 s
2.54 D 98:10 s1
2.56 Px D ˚7:000 sinŒ.1:000 rad=s/tç C 10:50 cosŒ.0:5000 rad=s/tç 10:50 m=s
x D ˚7:000 7:000 cosŒ.1:000 rad=s/tç C 21:00 sinŒ.0:5000 rad=s/tç .10:50 s1/t m=s2
2.58 tstop D 0:2233 s
2.60 vterm D 4:998 m=s
2.62 jvjmax D 1:128 m=s
sjvjmax D 0:1250 m and sjvjmax D 0:1250 m
2.64 v.t/ D mg
Cd

1 eCd t=m
vterm D mg
Cd
2.66 vf D 3:563 m=s
2.68 Computer Problem
3 Last modified: August 16, 2012

Loading page 4...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.70 swet D 26:31 m
.swet sdry/
sdry
.100%/ D 65:21%
2.72 P./ D ˙
r
P2
0 C 2 g
L
cos cos 0

2.74 Pmin D 4:930 rad=s
2.76 Px D ˙
s
v2
0 C 2

g C kL0
m

.x x0/ k
m
x2 x2
0

2.78 txmax D 0:2433 s
2.80 (a) Pr D p2G .mA C mB /
r r0 r
rr0
(b) (i) Pr D 5:980 105 ft=s and (ii) Pr ! 1
2.82 Concept Problem
2.84 d D 186:4 ft
2.86 ˛s D ap
r C hv2
p
2 r3
˛s
ˇ
ˇrDr1 D 1:442 rad=s2 and ˛s
ˇ
ˇrDr2 D 3:631 rad=s2
2.88 v D v0 C ac .t t0/
s D s0 C v0.t t0/ C 1
2 ac .t t0/2
2.90 vmin D 113:5 ft=s
2.92 ti D 3:497 s
EvB D .33:33 O{ 34:31 O| / m=s
2.94 R D 5:047 m
2.96 tflight D 6:263 s
Erland D .281:8 O{ 4:500 O| / m
2.98 d D 24:09 m
2.100 (a) Rmax D
4 rad D 45ı
(b) Rmax D 233:5% of the actual maximum range
(c) tRmax D 119:5 s
2.102 1 D 1
2 sin1.gR=v2
0 / and 2 D 90ı 1
2 sin1.gR=v2
0 /
1 D 24:86ı and 2 D 65:14ı
4 Last modified: August 16, 2012

Loading page 5...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.104 .v0/min D 50:75 ft=s
2.106 Evinitial D .5:920 O{ C 7:786 O| / m=s:
dy
dx
ˇ
ˇ
ˇ
ˇxDxB
D 0:2244
2.108 hmax D v2
0 sin2 ˇ
2g cos
2.110 Computer Problem
2.112 tfRmax
D 5:211 s
Rmax D 437:1 ft and Rmax D 45:66ı
2.114 tmax D 3:000 s
Hmax D 49:21 m
percent increase in height with no air resistance D 2:363%
2.116 Computer Problem
2.118 (a) Ea Eb D 1 6 C 2 3 C 3 0 D 0
(b) Ea

Ea Eb

D

84 O{ 42 O| C 0 Ok

(c) they are the same
2.120 (a) E!1 D 104:7 Ok rad=s; E!2 D 104:7 O{ rad=s and E!3 D 104:7 O| rad=s
(b) E!` D 100
3p3

O{ C O| C Ok

rad=s D 60:46

O{ C O| C Ok

rad=s
2.122 Derivation, so no answer needed.
2.124 EvA D 10:00 O| m=s; EvB D 10:00 O| m=s; EvC D 10:00 O| m=s; and EvD D 10:00 O| m=s
2.126 E!wheel D 65:71 Ok rad=s
2.128 jEaP j D 19;300 ft=s2
Orientation of EaP from x axis D 110:0ı .ccw/
2.130 v0 D 263:7 ft=s and r D 6875 ft
2.132 Pr D 1:204 m=s and P D 0:8504 rad=s
2.134 Ev D .0:037 70 Our C 0:5341 Ou / m=s
Ea D 1678 Our C 236:9 Ou
m=s2
2.136 Concept Problem
2.138 Prˇ
ˇD90ı D 8:825 ft=s and Pˇ
ˇD90ı D 0:7746 rad=s
5 Last modified: August 16, 2012

Loading page 6...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.140 EaP D .252:9 OuC 399:8 OuB / ft=s2
2.142 Concept Problem
2.144 Concept Problem
2.146 D 49:69 ft
2.148 Pv D 3:213 m=s2
2.150 Südkurve D 50:56 m
Nordkurve D 92:54 m
2.152 D 327:3 ft
2.154 d D 831:0 ft
2.156 Ea D E0
2.158 Ea D .39:40 O{ C 42:88 O| / m=s2
2.160 Concept Problem
2.162 ˇ
ˇEaˇ
ˇ D .33:69103 cos / m=s2
2.164 Ea D .121:2 O{ C 230:0 O| / ft=s2
2.166 Ea D 18:96 ft=s2 Out C 69:91 ft=s20:2917 s2s Oun
2.168 Pv D 5:343 m=s2 and D 17:50 m
2.170 D 282:2 ft
2.172 min D 564:5 ft
tf t0 D 4:742 s
2.174 jEaj D 1

v0 2v0
s
s
4v2
0
2 C

v0 2v0
s
2
2.176 d D 304:4 ft and tf D 3:106 s
2.178 Concept Problem
2.180 Concept Problem
2.182 Pr D 684:1 ft=s and P D 0:011 57 rad=s
Rr D 4:944 ft=s2 and R D 0:000 428 1 rad=s2
2.184 v D 1:816 m=s
ˇ
ˇEaˇ
ˇ D 4:573 m=s2
6 Last modified: August 16, 2012

Loading page 7...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.186 Rr D 1:599 m=s2
2.188 0 D 0
r0 D 0:014 86 m
2.190 Ev D .1:3 m=s/ Our C 0:22 s1r Ou
Ea D 0:048 40 s2r Our 0:5720 m=s2 Ou
2.192 Ev D .0:1500 Our C 1:732 Ou / ft=s and Ea D .0:3638 Our C 0:1125 Ou / ft=s2
2.194 Er D 30:20 Our ft
Ev D .3 Our C 7:550 Ou / ft=s
Ea D .1:888 Our C 1:500 Ou / ft=s2
2.196 D 5:602 m
! D 2630 rad=s D 25;120 rpm
2.198 v D 5490 ft=s
ˇ
ˇEaˇ
ˇ D 1:005106 ft=s2
2.200 Ev D v0
p2 C .r0 C /2 Our C v0.r0 C /
p2 C .r0 C /2 Ou
Ea D v2
0 .r0 C /3
r2
0 C 2r0 C .1 C 2/22 Our C v2
0
˚.r0 C /2 C 22 C .r0 C /2
Œ2 C .r0 C /22 Ou
2.202 PA D 0:001 157 rad=s and PH D 4:287 rad=s
RA D 1:684104 rad=s2 and RH D 0:075 47 rad=s2
2.204 PEa D

«r 3r P R 3Pr P2
Our C
h
r
« P3
C 3Rr P C 3Pr R
i
Ou
2.206 Rr D 5:781 ft=s2
R D 1:421 rad=s2
2.208 r D
q
h2 C .d C /2
Pr D v0h
ph2 C .d C /2 and P D v0.d C /
h2 C .d C /2
Rr D v2
0

. C d /d 2 C h2 C d
d 2 C h2 C 2 C 2d3=2 and R D v2
0

hd 2 C h2 2
d 2 C h2 C 2 C 2d2
2.210 Ev D 1:758 O| m=s and Ea D 136:9g O|
2.212 v D 0:1185 ft=s and ˇ
ˇEaˇ
ˇ D 0:086 42 ft=s2
7 Last modified: August 16, 2012

Loading page 8...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.214 ar D K2a
b2

1
r2

2.216 Computer Problem
2.218 Concept Problem
2.220 vH=F D
h
5:630105 0:8091 cos
x
2
i
m=s
aH=F D 5:237 sin
x
2

m=s2
2.222 vA=B D 144:7 m=s
2.224 EvC =A D .6:857 O{ C 30:27 O| / m=s:
EaC =A D .1:376 O{ 1:000 O| / m=s2:
2.226 .vP /avg D 17:30 ft=s
2.228 Computer Problem
2.230 vpmax D 330:7 ft=s D 225:4 mph
2.232 EvB D 2:928 O{ ft=s
2.234 vrain D 16:37 ft=s
2.236 EaB D L1 R cos L1 P2 sin C L2 R cos L2 P2 sin O{
C L1 R sin C L1 P2 cos C L2 R sin C L2 P2 cos O|
2.238 D 63:57ı
2.240 D 14:90ı
2.242 EvB D 3:088 O| ft=s:
EaB D 0:3192 O| ft=s2:
2.244 EvB D 2:000 O| m=s:
2.246 vAy D 4:596 ft=s:
2.248 EaB D 11:10 O{ m=s2: @
td D 0:1644 s
2.250 a0 D 0:9374 ft=s2
2.252 P` D dv0
ph2 C d 2
2.254 Component of EvC =B along BC D 0
8 Last modified: August 16, 2012

Loading page 9...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.256 r D
q
.d C cos /2 C .h C sin /2
P D v0. C d cos C h sin /
.d C cos /2 C .h C sin /2 and Pr D v0.h cos d sin /
p.d C cos /2 C .h C sin /2
2.258 t D 0:1556 s
2.260 Rye D 9:81 m=s2
RyC
ˇ
ˇtD0:4391 s D 44:51 m=s2
2.262 Concept Problem
2.264 Concept Problem
2.266 t D 83:82 s
2.268 EvC D .6:500 OuR C 5:520 Ou C 5:300 Ou´/ ft=s
EaC D .0:6624 OuR 1:560 Ou / ft=s2
2.270 r P.r R sin C 2Pr P sin C 2r P P cos / r P sin .r R C 2Pr P r P2 sin cos / D 0;
r P sin .Rr r P2 r P2 sin2 / Pr.r R sin C 2Pr P sin C 2r P P cos / D 0;
Pr.r R C 2Pr P r P2 sin cos / r P.Rr r P2 r P2 sin2 / D 0
2.272 Ev D .0:04869 OuR C 7:961 Ou 0:2999 Ou´/ m=s
Ea D .79:79 OuR C 0:9738 Ou 0:8120 Ou´/ m=s2
2.274 ˇ
ˇEamax
ˇ
ˇ D 12:36 ft=s2
2.276 Ev D df
d´ P´ Our C K
f .´/ Ou C P´ Ou´
Ea D

R´ C P´2 d 2f
d´2 K2
f 3.´/

OuR C R´ Ou´
2.278 ar D 0:1598 m=s2; a D 0:1884 m=s2; and a D 0:2232 m=s2
2.280 EvP =B D .10:17 O{B C 10:33 O|B / m=s and vP =B D vP =A D 14:50 m=s
EaP =B D .7:822 O{B 1:258 O|B / m=s2 and ˇ
ˇEaP =B
ˇ
ˇ D ˇ
ˇEaP =A
ˇ
ˇ D 7:923 m=s2
2.282 Concept Problem
2.284 Computer Problem
2.286 s D mg
C 2
d

Cd t C m

e Cd
m t 1

2.288 x D

L0 C gm
k

1 cos
r k
m t

2.290 v0 pgR
9 Last modified: August 16, 2012

Loading page 10...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
2.292 D 240:5 ft
2.294 The car will lose contact with the ground.
ac D 1:606 m=s2
2.296 Prˇ
ˇD180ı D 0
Pˇ
ˇD180ı D 2:941 rad=s
Rrˇ
ˇD180ı D 40:85 m=s2 and Rˇ
ˇD180ı D 0
2.298 P D 0:053 91 rad=s
R D 0:2380 rad=s2
2.300 vmax D 6:020 ft=s
2.302 hmax D 0:8898 ft
2.304 jEaAj D 18:65 m=s2
2.306 ˇ
ˇEaˇ
ˇ D 172;700 ft=s2
2.308 ˇ
ˇEaˇ
ˇmin D 2:534 m=s2
Chapter 3
3.2 Concept Problem
3.4 EaA D 2:147 O| ft=s2:
3.6 amax D 17:75 ft=s2
3.8 aBmax D 0:1650 m=s2 D 0:5417 ft=s2
3.10 Rx C EA
mL x D 0
3.12 aBmax D 5:048 m=s2 D 16:57 ft=s2
3.14 d D 6:630 ft
3.16 d D 417:5 ft
If an entire train weighing 30106 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.
3.18 Ea D
Cd
m v2 g

O| :
3.20 .s /min D 0:3436
3.22 vi D 5:297 ft=s
10 Last modified: August 16, 2012

Loading page 11...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
3.24 Computer Problem
3.26 vi D 32:68106 m=s
3.28 vmax D g
r m
k
3.30 Spring compression D 0:7367 ft
3.32 k D 5:084104 lb=ft
3.34 ımax D 0:1675 m
.Fs /max D 586:1 N
3.36 Computer Problem
3.38 xstop D 0:1597 m
3.40 k D 5:930 lb=ft
3.42 k D 9034 kg=s2
3.44 v.0/ D
s
2 k
m

x2
0 C 2L0

L
q
x2
0 C L2

3.46 Concept Problem
3.48 Concept Problem
3.50 Concept Problem
3.52 D 1:401104 ft
3.54 No value of !c can be found that would cause the person to slide up the wall
3.56 .FOA/before release D mg
cos ; .FOA/after release D mg cos ; .% change in FOA/D30ı D 25:00%
3.58 jEaj D 3:620 g
jFLj D 43;060 N
3.60 Rx C k
m x 1 ru
px2 C y2
!
D 0;
Ry C k
m y 1 ru
px2 C y2
!
D 0
3.62 Rr r P2 C k
m .r ru/ g cos D 0 and r R C 2Pr P C g sin D 0
3.64 ts D 0:7466 s
3.66 y D m
Cd
tan 0

eCd x=m 1

m2g
2C 2
d v2
0 cos2 0

eCd x=m 1
2
11 Last modified: August 16, 2012

Loading page 12...

Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha
3.68
a D 0 D a´;
aR D g.s C tan /
1 s tan D 120:1 ft=s2;
F D 0;
F D s mg
cos s sin D .83:16 ft=s2/ m;
N D mg
cos s sin D .92:40 ft=s2/ m;
vmax D pg
s
s C tan
1 s tan D 363:4 ft=s D 247:8 mph
3.70 F D 5971 lb
3.72 .!c /max D 2:733 rad=s
3.74 (a) Ea D .318:8103 Our C 603:0 Ou / m=s2
(b) P D 5:898106 N and R D 11:33103 N
3.76 ! D 9:939 rad=s
3.78 B D 41:81ı
3.80 Concept Problem
3.82 Rr r P2 C G me
r2 D 0 and r R C 2Pr P D 0
3.84 ımin D
r 5mgr
k
vB D p2gr
3.86 Computer Problem
3.88 hmax D 0:3618 ft
3.90 RxL2 y2 C Ryxy C x Px2L2 y2 C Py2L2 x2 C 2xy Px Py
L2 x2 y2 C gx
q
L2 x2 y2 D 0;
RyL2 x2 C Rxxy C y Px2L2 y2 C Py2L2 x2 C 2xy Px Py
L2 x2 y2 C gy
q
L2 x2 y2 D 0
3.92 R P2 sin cos C .g=L/ sin D 0 and R sin C 2 P P cos D 0
3.94 Rr r P2 D 0 and r R C 2Pr P D 0
3.96 M´ D 2m!2
0 r
q
r2 r2
0
jvr j D !0
q
d 2 C 2dr0
v D !0
q
2d 2 C 4dr0 C r2
0
3.98 Concept Problem
12 Last modified: August 16, 2012

Loading page 13...

12 more pages available. Scroll down to load them.

Preview Mode

100%

Study Now!

XY-Copilot AI

Unlimited Access

Secure Payment

Instant Access

24/7 Support

AI Assistant

Document Details

Subject

Mechanical Engineering

Solution Manual for Engineering Mechanics: Dynamics, 2nd Edition

Study Now!

Document Details

Related Documents

Biofluid Mechanics: An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation, 3rd Edition (2021)

Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition

Solution Manual for Introduction to Finite Elements in Engineering, 4th Edition

Solution Manual for Engineering Mechanics Dynamics, 13th Edition

Solution Manual For Introduction To Engineering Experimentation, 3rd Edition

Solution Manual for Mechanics of Materials SI, 9th Edition

Solution Manual for Engineering Mechanics Statics, 13th Edition

Solution Manual For Modern Compressible Flow: With Historical Perspective, 3rd Edition