Dynamics Answers 2009-11-2

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Engineering Mechanics: Dynamics 1e Gray, Costanzo, Plesha

Answers to Selected Even-Numbered ProblemsWe 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 ErB=A D .4:00 O{ 1:00 O| / ft

ErB=A D .3:31 Oup 2:46 Ouq/ ftˇErB=Aˇxy system D ˇErB=Aˇpq system D 4:12 ft

1.6 The angle Â 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 ŒI xx � D ŒI yy � D ŒI ´� D ŒM�ŒL�2

Units of I xx , I yy , and I ´ in the SI system: kgm2;

Units of I xx , I yy , and I ´ in the U.S. Customary system: slug ft2 D lbs2 ft.

1.10 Concept problem.

Chapter 2

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

Ea.15 s/ D .15:8 O{ 1:39 O| / ft=s2

Â.15 s/ D 4:95ı

.15 s/ D 180ı


2.6 y.x/ D 2:00 C 3:00x C 2:00x2

m


2.10 Er.t1; t2/ D .0:899 O{ C 41:3 O| / ftEvavg.t1; t2/ D .0:450 O{ C 20:7 O| / ft=s

2.12 v.t/ D Pr.t/ D Œ.4:08e13:6t/ O{ C .0:720 1:28e13:6t/ O| � m=s

a.t/ D Pv.t/ D Œ.55:5e13:6t/ O{ C .17:4e13:6t/ O| � m=s2

The impact angle Â 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 Â D 26:1ı

2.14 y.x/ D 5:25 C 3:25x C 1:00x2

m

xmax D 3:00 m and xmin D 1:00 m

1 Last modified: November 2, 2009




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

p 2 C 2 cosŒ.25:5 rad=s/t�

Ea 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.


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

.EvP=O /2 D ˚Œ.1 C cos t / cos Â 2.t 2/ sin Â� O{2 Œ2.t 2/ cos Â C .1 C cos t / sin Â� O|2«

m=s

.EaP=O /1 D . sin t O{1 2 O|1/ m=s2

.EaP=O /2 D Œ. cos Â sin t 2 sin Â / O{2 C .sin Â sin t 2 cos Â/ O|2� m=s2

jEvP=O.t D 2 s/j1 D 0:584 m=s

jEvP=O.t D 2 s/j2 D 0:584 m=s

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

vP=A D 24:5 ft=s

vP=B D 24:5 ft=s

EaP=A D .1:78 O{A C 5:96 O|A/ ft=s2

aP=A D 6:22 ft=s

2

aP=B D 6:22 ft=s2

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

EaA D

p 2 O{ C 3

p 2 O|Á

ft=s2 D .1:41 O{ C 4:24 O| / ft=s2



2.32 Rx D 8a3v204a2 C y2

2

Ry

D

4a2v20y

4a2 C y2


2.36 Ev D v0p 3a2 C b2

p 3a O{ C b O|

Á

2.38 tbraking D 5

2s D 7:85 s

2.40 tjajmaxD 3:93 s and tjajmax

D 11:8 s where jajmax D 3:60 m=s2

samaxD 12:8 m and samax

D 129 m

2.42 v0 D 10:1 m=s

2.44 Px.t/ Da0

!

sin 2! t C ˇ cos !v.t/ D 7:00

sin t C 1:50 cos.0:500t/ 1:50

m=sx.t/ D 7:00Œ1:00 1:50 cos t C 3:00 sin.0:500t/� m

2.46 vterm Dr

mg

C d D 5:00 m=s

2.48 v.t/ D mg

C d

Â1 e

C dt

m

Ã

vterm D mg

C d

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

2.56 PÂ.Â/ D ˙r PÂ 20 C 2g

L cos Â cos Â 02.58 . PÂ 0/min D ˙4:93 rad=s

2.60 Px D ˙s

v20 C 2

Âg C kL0

m

Ã.x x0/ k

m

x2 x20

2.62 txmax

D 0:243 s

2.64 (a) t D r3=20p

2G .mA C mB/

"sin1

r r

r0s

r

r0

Â1 r

r0

Ã

2

#

(b) t D 106 h

t D 106 h

2.66 tstop D v0gk

D 2:88 s

2.68 k D .vT /20aT

g

.vT /20 2daT

D 0:301

2.70 t D 8:63 s

Py D 77:7 m=s

2.72 v D v0 C ac.t t0/

s D s0 C v0.t t0/ C 12ac.t t0/2

2.74 ˛s D ap

rC hv2p

2 r3

˛sˇˇrDr1

D 1:44 rad=s2 and ˛sˇˇrDr2

D 3:63 rad=s2

2.76 Â D tan1Â

w hd

Ã

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

2.80 y D h C tan ˇ.x w/ g

2v20 cos2 ˇ.x w/2

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 Ä Â Ä 56:8ı and 69:4ı Ä Â Ä 74:5ı

The sizes of these ranges are Â 1 D 26:4ı and Â 2 D 5:09ı

The angle subtended by the target as seen from an observer at point O is ˇD

21:3ı

Unlike Example 2.11, the difference between the angle subtended by the target and Â 1 or Â 2 is significant. In

addition, we see that the value of Â 1 is much closer to ˇ than Â 2.

2.88 d AB D 82:0 m

2.90 v0 D 62:5 ft=s and ˇ D 20:6ı


d D 4:70 m

Evinitial 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.





2.94 v0 Dr

gR

2

cos Â p sin ˇ cos.ˇ Â/

2.96 Â max D

4D 45ı

R D v20g

D 447 ft

t D 2v0 sin Â maxg D 5:27 s

2.98 y Dh

59:9103 ln

1 2:84104xÁ

C 17:7xi

m


tI D 6:19 s

xI D 268 m

2.102 y D v0 sin ˇ

Äx cos ˛ y sin ˛

v0 cos.˛ C ˇ/

1

2g cos ˛

Äx cos ˛ y sin ˛

v0 cos.˛ C ˇ/

2

2.104 (a) Ea Eb D 1 6 C 2 3 C 3 0 D 0

(b)Ea

Ea EbÁ D 84

O{

42O

|C

0O

kÁ(c)

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

(b) E!` D 100

3p

3

O{ C O| C Ok

Árad=s ) 60:5

O{ C O| C Ok

Árad=s

2.108

EvA D 5 Ok 2 O{ D 10 O| m=s; EvB D 5 Ok

2 O{ C 1 OkÁ

D 10 O| m=s;

EvC D 5 Ok

2 O{ C 2 OkÁ

D 10 O| m=s; EvD D 5 Ok

2 O{ C 3 OkÁ

D 10 O| m=s

2.110 E! D v0R

Ok D .65:7 rad=s/ Ok

2.112 jEvj 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 70:0ı

2.114 Pr D v0 cos.Â C / and PÂ D v0r

sin.Â C /

2.116 EvB D .4:00 Our C 0:800 OuÂ / ft=sEa D .0:320 Our C 3:20 OuÂ / ft=s

2.118 Pr D 0 and PÂ D v0d

D 1:50 rad=s

Rr D v20d

D 9:00 ft=s2 and RÂ D 0 rad=s2

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

2.122 2! Pr C P!r D 0

2.124 Ea D .262 OuC 200 OuB/ ft=s









2.134Ea.s/

D f19:0

Out C

Œ69:9

.0:292/s�Oung

ft=s2


2.138ˇEaˇ D .33:7103 cos / m=s2

2.140 v0 D 19:2 ft=s

2.142 D v20g

D 282 ft

2.144 min D v202g

D 565 ft

tf Dv20

2g vf Cv0

D 4:74 s

2.146 Pv D 13:8 ft=s2

2.148 s D 304 ft



2.154 Pr D 466 mph

PÂ D 41:6 rad=h

Rr D 12;100 mi=h2

RÂ D 5550 rad=h2

2.156

Pr.0/

D 438 ft=s and

PÂ.0/

D347

106 rad=s

Rr.0/ D 14:9 ft=s2 and RÂ.0/ D 10:6106 rad=s2

2.158 Pr D 0:265 ft=s

2.160ˇEvˇ Dq Pr2 C .L PÂ /2 D 5490 ft=s

ˇEaˇ Dv uut v2s

2L L PÂ 2

!2C

2vsPÂ Á2 D 1:00106 ft=s2

2.162 Ev D .1:3 Our C 0:22r OuÂ / m=s

.0:0484r Our 0:572 OuÂ / m=s2

2.164 PÂ A D vAL

.d=2/2

CL2

D 1:16103 rad=s

PÂ H D1

L

s v2A C v2B

2D 4:29 rad=s

RÂ A D 1

r2

"v2A

v2B

L

2d v2

AdL

r2

#D 1:68104 rad=s2

RÂ D

v2A v2B

2dL

D 0:0755 rad=s2

2.166 Rr D 5:78 ft=s2 and RÂ D 1:42 rad=s2

2.168 PEa D«r 3r PÂ RÂ 3Pr PÂ 2

ÁOur C

hr«Â PÂ 3

ÁC 3 Rr PÂ C 3Pr RÂ

iOuÂ





2.170 r Dq

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

Pr D v0 .h cos d sin /q .d C cos /2 C .h C sin /2

and PÂ D v0 . C d cos C h sin /

d 2 C h2 C 2 C 2d cos C 2h sin :

Rr D v20. C d cos C h sin /

d 2 C d cos C h2 C h sin

d 2 C 2d cos C h2 C 2 C 2h sin 3=2

RÂ D v20

d 2

C h2

2

.d sin h cos /

d 2 C 2d cos C h2 C 2 C 2h sin 2

2.172 P D 0

R D R!2ABp

h2 R2D 15:8 rad=s2

2.174 Ev D .37:7103 Our C 0:534 OuÂ / m=s

Ea D .1680 Our C 237 OuÂ / m=s2

2.176 Evfollower D .1:76 m=s/ O|Eafollower D .137g/ O|

2.178 v

D0:119 ft=s

jEaj D 0:0864 ft=s2



2.184 ROS D EvB EvA

ErB=AjErB=Aj

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

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


For vB=W

D10 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 EaB D

L1RÂ cos Â L1

PÂ 2 sin Â C L2R cos L2

P2 sin Á

O{C

L1PÂ 2 cos Â C L1

RÂ sin Â C L2P2 cos C L2

R sin Á

O|

2.194 Â D C sin1Ä

.vA` vBd / cos

vP=Ad

D 63:6ı:

2.196 Â D 14:9ı

2.198 vpmaxD vs C `

2r g

2hD 331 ft=s D 225 mph

2.200 EvB D 3:09 ft=s

EaB D 0:319 ft=s2

2.202 EvB D .2:00 m=s/ O|

2.204 EaB D .11:1 m=s2/ O{ @Â

t Ds

d

2a0D 0:164 s





2.206 a0 D 26d

3t2D 0:937 ft=s2

2.208 EvA D v0p

w2 C h2

w C p w2 C h2

O{ D .3:08 ft=s/ O{

EaA Dv20

w

p h2 C w2

Á a0

hh2 C w

w C

p h2 C w2

Áiw

Cp

h2

Cw2Á2

O{ D .0:893 ft=s2/ O{


2.212 EvC D .6:61 ft=s/ sin Â

Â1 C 0:292 cos Â p

0:195 0:0851 sin2 Â

ÃO|

EaC D1:28104 ft=s2

Äcos Â

Â1 C 0:292 cos Â p

0:195 0:0851 sin2 Â

Ã

C sin Â

Â0:0248 cos2 Â sin Â

.0:195 0:0851 sin2/3=2 0:292 sin Â p

0:195 0:0851 sin2 Â

ÃO|

2.214 r Dq

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

PÂ

Dv0. C d cos C h sin /

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

Pr

Dv0.h cos d sin /p

.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 . OuR; OuÂ ; Ou´/ is right-handed

we have

EvC D .6:5 Our C 5:52 OuÂ C 5:3 Ou´/ ft=s

EaC D .0:662 Our 1:56 OuÂ / ft=s2

2.222 Ev D P tan ˇ OuR C K

´ tan ˇOuÂ C P Ou´

Ea DÂ

R tan ˇ K 2

´3 tan3 ˇ

ÃOuR C R Ou´

2.224

r P.r RÂ sin C 2Pr PÂ sin C 2r P PÂ cos / r PÂ sin .r R C 2Pr P r PÂ 2 sin cos / D 0

r PÂ sin . Rr r P2 r PÂ 2 sin2 / Pr.r RÂ sin C 2 Pr PÂ sin C 2r P PÂ cos / D 0

Pr.r R C 2Pr P r PÂ 2 sin cos / r P.Rr r P2 r PÂ 2 sin2 / D 0

2.226 Pr D 0

P D 0

PÂ D v0r sin

D 0:0177 rad=s

Rr D v20r

D 11:3 ft=s2





R D v20 cos

r2 sin D 78:1106 rad=s2

RÂ D 0


2.230 xland D 3:35 m; yland D 1:42 m; and ´land D 0:


2.234 EvP=B .10:2 O{2 C 10:3 O|2/ m=s

EaP=B D .7:82 O{2 1:26 O|2/ m=s2


h D 29:4106 m

2.238 s.t/ D mg

C 2d

ÄC d t C m

ÂeC dm

t 1

Ã

2.240 x DÂ

L0 C gm

k

ÃÄ1 cos

Âr k

mt

Ã

2.242 v0 p gR

2.244 D 240 ft

2.246 f D 629 m

tf D 5:24 s

2.248 Pr ˇÂD180ı

D 0

P ˇÂD180ı

D 2:94 rad=s

Rr ˇÂD180ı D 40:8 m=s2 and R ˇÂD180ı D 0

2.250 PÂ D30ı

D 0:0539 rad=s

RÂ ˇD30ı D 0:243 rad=s2

2.252 jEaj D 2L!2ˇ0 D 1:24 ft=s2

2.254 at D 3:33 ft=s2

2.256 jEaAj D 2:58 m=s2

2.258ˇEaˇ D

s Âd!2

2

Ã2CÂ

d˛

2

Ã2C a20 D 173;000 ft=s2

2.260ˇEaˇ

minD L

q 12

PÂ 4 C !421 !2 PÂ 21 D 2:53 m=s2

Chapter 3


3.4 ay D g .2P W /

W

3.6 Rx C EA

mLx D 0

3.8 ay D C d m

v2 g

3.10 d D v202kg

L D 50:8 ft





3.12 Py D mg

C d

Â1 e

C dt

m

ÃD 0:701 m=s

3.14 T D m

2

8<:g C v20 .` h/2

2h

x2A C .` h/2i3=2

9>=>; D 251 lb

3.16 .F s

/max D

2mg


3.20 t D v v0kg

D 7:59 s

3.22 xstop D v0

s W

gkD 0:737 ft

3.24 k D 50;800 lb=ft

3.26 ımax D 0:167 mF sp

Ámax

D kımax D 586 N


ımax D 0:568 ft; when t D 0:177 s

3.30 xstop D 4

s 2mv2i

ˇD 0:160 m

3.32 k D 5:93 lb=ft

3.34 k D 9:03 kN=m

3.36 v.0/ D 2

v uut k

m

"x202

C L0

ÂL

q x20 C L2

Ã#




3.44 D W v2

g .N W /D 1:40104 ft

3.46 an D v2

D 35:51 m=s2 D 3:62 g

F L D mg C mv2

D 43;100 N

3.48

Rx C k

mx

1 rup

x2 C y2

!D 0;

Ry C k

my

1 rup

x2 C y2

!D 0

3.50

Rr r PÂ 2 C k

m.r ru/ g cos Â D 0;

r RÂ C 2Pr PÂ C g sin Â D 0





3.52 y D m

C d tan Â 0

eC dx=m 1

Á m2g

2v20 cos2 Â 0C 2d

eC dx=m 1

Á2

3.54 Â s D cos1

2

3C v20

3gR

!D 47:7ı

3.56 vmax D p

gs s C tan Â

1 s tan Â D55:36 m=s

D199 km=h

vmin D p g

s tan Â s

1 C s tan Â D 35:33 m=s D 127 km=h

127 km=h Ä v Ä 199 km=h

3.58 (a) Ea D .319103 Our C 603 OuÂ / m=s2

(b) P D mar C mg sin Â D 5:90106 N

(c) R D maÂ C mg cos Â D 11:3103 N

3.60 ! D 9:94 rad=s


3.64 R

RÂ

C2

PR

PÂ

D0

RR.1 C cot2 / R PÂ 2 D g cot

3.66 Â B D sin1Â

2

3

ÃD 41:8ı

3.68

Rr r PÂ 2 C Gme

r2D 0;

r RÂ C 2 Pr PÂ D 0

3.70 k D 3v0m

.d max L0/2 .d max=2 L0/2D 4:74 lb=ft

3.72 sD

cos1 Â2

3Ã D

132ı

3.74RrL2

.L2 R2/ R PÂ 2 C

PR2L2R

.L2 R2/2C gRp

L2 R2D 0

R RÂ C 2 PR PÂ D 0


3.78 Pr D 0:471 m=s





3.88 aAy D F

2mA

C g D 16:1 ft=s2

3.90 RrA D GmB

r2; and RrB D G

mA

r2

RrB=A D G

ÂmA C mB

r2

Ã

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 k1g D 4:41 m=s2





3.96 s D tan Â D 0:424

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

EaB D .145 O{/ ft=s2

3.100 vmax D mgs 1

k m Cmp

3.102 vimpact D 3:69 m=s

3.104 gd p

4R2 d 2 cos Â g.4R2 d 2/ sin Â D 4R3 RÂ

3.106 tcontact D 78:5103 s D 21:8 h

3.108 j.yB/maxj Dˇˇ 2mBg

k

ˇˇ

j.yB/maxj D 2 ft

.vB/max D vB Œ.yB/max� D 8:58 ft=s

3.110 Defining a Cartesian coordinate system with the y axis opposite tot he direction of gravity and with origin at the

initial position of B , we havej.yB

/maxj D ˇ

2

k.m

Asin ˛

CmB

/g ˇ D 1:84 ft

.vB/max D vB j.yBD0:9202/ D 4:79 ft=s

3.112 EvA.tf / D .3:16 O{ C 3:13 O| / m=s and EvB.tf / D .3:26 O{/ m=s


3.116 EaG D 9mGg 39P

mB C mC C 25mD C 9mG

O| D .9:46 ft=s/ O|

T 4 D 3g.mC C 10mD/ mG C .4mB 9mC 30mD C 36mG/ P

mB C mC C 25mD C 9mG

D 1190 lb


C d D

0:0888 lb

s=ft

3.120 vc Ds

2G .mA C mB/

Â2

d A C d B 1

r0

ÃD 276106 ft=s

3.122 ab D g .1 cos Â/ D 1:94 ft=s2

an D g sin Â D 11 ft=s2

D v2

g sin Â D 101103 ft D 19:2 mi

3.124 Defining a Cartesian coordinate system with origin at O, y axis pointing opposite to gravity and x axis pointing

to the right, we have y D x

v0 cos ˇ

Ämg

ÁC v0 sin ˇ

C m2g

Á2ln

Ä1 Áx

mv0 cos ˇ

3.126 Verification problem

3.128 W A D W B .g aB/

2g C 4aBD 161 lb

3.130 m1L1RÂ D m2g cos sin m2 sin

hL1

RÂ sin L1PÂ 2 cos L2

P2i m1g sin Â

L2R D g sin L1

RÂ cos L1PÂ 2 sin





Chapter 4


4.4 9:02104 ftlb

4.6 U 1-2 D 12m.v22 v21/ D 180 kJ

4.8 U 1-2 D1

2m

v2

2 v2

1Á D 4:3010

5

J

4.10 (a) 283 m

(b) 283 m

(c) 14:8 km

4.12 d D 15:8 m

4.14 4:981011 J

4.16 43:5106 ft lb

23:3106 ft lb

50:8103 lb



4.22 1:29105 lb=ft3

4.24 k D 2:74105 N=m

4.26 v0 Dp

2gL.1 cos Â/ D 6:40 ft=s

4.28 k D mv2

ı2D 44:8 lb=ft

4.30 .U 1-2/Engine D 4:64105 ftlb

4.32 F avg D mg C m

2h.v21 v22/ D 1:47 kN

4.34 Â D sin123

Á D 41:8ı

4.36 V B D 14ˇx4

d D 0:420 ft

4.38ˇEvˇA D

s ˇEvˇ2P 2GmE

Â1

RP 1

RA

ÃD 2:34103 m=s

4.40 V D P 0s0A ln

Âs

s0

Ã

4.42 1:92 ftlb

4.44 D cos1

23 Á

D 132ı

4.46 v2 D 24:1 ft=s

4.48 k D 276 lb=ft

4.50 .vy/max D g

r m

k

4.52 L0 D 5R

3 mv2

B

4kRD 1:06 m

4.54 vA2 D

v uut2gd

mB 13mA

ÁmA C 9mB

D 1:09 m=s and vB2 D 3

v uut2gd

mB 13mA

ÁmA C 9mB

D 3:28 m=s





4.56 mB D mAv2B2 C kı2

2gı v2B2

D 0:685 kg

4.58 v2 Dr

2gR dg

q 1 d 2=.2R/2

N A D mg

Â3 d

R

q 1 d 2=.2R/2

Ã

4.60 vA Ds

2hg ŒmA.1 0:35k/ mB �

mA C mB

D 3:69 m=s

4.62 vB Ds

8gmA`OA mB`OB

mA C 4mB.sin 35ı C sin 75ı/ D 26:6 ft=s

4.64 yB D 1

k.mA sin ˛ C mB/g

.vB/max D vB.yB D 2/ D 9:42 ft=s

d B D 0 ft

4.66 W D 2mg

4.68 Part (a): T D 2m.v2

O C R2

!2

/ D 580 JPart (b): T D 1

2 .4m/v2O C 12m.4R2!2/ D 2m.v2O C R2!2/ D 580 J

4.70 U D 5760 kJ


4.74 P 1 D P 2 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 P i D mgv .k cos Â C sin Â/

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

4.82 ˇ D 1:07109 kg=.m2s2/

4.84 U 1-2 D 12m

Â0:99 mg

C d

Ã2D 0:0361 J

4.86 k D mv2 32mgL

ı2u2 C ı2l2

ı2u1 ı2l1

D 9030 N=m

4.88 V D 3EI cs

2L3ı2

4.90 .v2/max Ds

2gı.mA sin ˛ C mB kmA cos ˛/ kı2

mA

CmB

D 9:67 ft=s

The distance between B and the ground is 0 ft

4.92 Eb D mgvt sin Â

D 5:395103 kJ D 1290 C

Chapter 5

5.2 xe D 28:51024 m

xe D 4:151012 m

xe D 1:42106 m





5.4

ˇˇZ t2t1

EF dt

ˇˇ D mv2 D 2:64 lbs

ˇEF avg

ˇD mv2

t2 t1D 2400 lb


5.8 t2 Dmvx2

F T D 10:6 s

d D 1280 ft


5.12 (a)ˇ EF avg

ˇ D 8080 N

(b) s ˇ EF avg

ˇmg

D 0:515

5.14 (a) Impulse D .3:17 O{ C 1:20 O| / lbs(b) EF b

avgD .3170 O{ C 1200 O| / lb

(c) Within the accuracy of our calculation (4 significant figures), ˛ remains equal to 31ı.



5.20 n D 9:170106 bees

Slowdown D 48:5 ft=s D 33:0 mph

5.22 .xfp/2 D Lfpmp

mp C mfp

5.24 vP4 D 1:86 ft=s

5.26 EvG D .2580 O{ 3030 O| / m=s;

5.28 d D mALfp

mA C mC D 4:95 m

5.30 .EvP /final D .1:52 ft=s/ O{

5.32 (a) Amplitude D mAr

mA C mB

D 2:14106 m

(b) .vB/max D mAr!

mA C mB

D 0:0135 m=s

5.34 EvA2 D .5500 O{ C 9520 O| / m=s and EvB2 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


5.40 EvCA D EvCB D .55:9 ft=s/ O{

5.42 vB D mA C mB

mB

p 2gL.1 cos Â m/ D 210 m=s

5.44 vB D mA C mB

mB

p 2kgd D 199 m=s

5.46 0:817 Ä e Ä 0:882

5.48 d D m2B.vB/2

2kg.mA C mB/2D 2:89 m





5.50 W max D mAg Cq

.mA C mB/m2A

gk2Œ2hk C .mA C mB/g�

.mA C mB/kD 819 lb

5.52 EvCA D EvCB D .8:33 O{ C 78:0 O| / ft=s

Er D .14:5 O{ C 136 O| / ft

5.54 EvCA D 77:0 O| ft=s and EvCB D .73:3 O{ C 85:8 O| / ft=s

ErA D 132 O| ft and ErB 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.

5.58 ˇ D tan1 .cot ˛/

5.60 The answer is an explanation.


5.64 EvCA D .23:6 O{ C 18:5 O| / m=s and EvCB D .0:717 O{ 6:20 O| / m=s

5.66

d Ds

2

gh1.1 C e/h1 cos ˛ sin ˛

r gh1

2Œ.e 1/ C .1 C e/ cos 2˛�

Cq

2gh2 C 12gh1Œ.e 1/ C .1 C e/ cos 2˛�2

D 6:24 ft

5.68 d D 3:50 ft

5.70 vCN D

2N 1Áp

2gh

5.72EhOA D 593103 Ok slugft2=sEhOA D 79:9103 Ok slugft2=s

5.74 EhQ D 1:44106 Ok slugft2=s


5.78 ! D .L sin Â 0 C d /2

.L sin Â C d /2!0

5.80 SincePEhO D mP gvP .0/t cos Â and EM O D mP gvP .0/t cos Â Ok, it is indeed true that EM O D PEhO .

5.82 EhO D ˙W s 2L3

g .cos Â cos 33ı/ Ok D .3:99 lbfts/p cos Â 0:839 Ok @ 57ı

5.84

v1 D sin 2

s 2gL.cos 2 cos 1/

sin2 1 sin2 2D 6:85 ft=s;

v2 D sin 1

s 2gL.cos 2 cos 1/

sin2 1 sin2 2D 2:32 ft=s

5.86 M A D 2m!20r

q r2 r20






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.

5.94 b

D

p rP rA

5.96 v Ds

gr 2e

rc.p

2 1/ D 3:23103 m=s

5.98 vesc D .1:631010 m3=2=s/p r

vesc

Earth

D 4:21104 m=s D 151;000 km=h

5.100 Gme D 42a3

2D 3:971014 m3=s2

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

5.104 vP Ds gr2eÂ 2

rP 1

aÃ D 7:7110

3m=s D 27;800 km=h

vA Ds

gr 2e

Â2

rA 1

a

ÃD 7:70103 m=s D 27;700 km=h

e D rAa

1 D 7:45104

D 2

s a3

gr 2eD 5470 s D 91:2 min

5.106 vA D 4710 ft=s D 3210 mph

vA D 6830 ft=s D 4660 mph

t D s a3e

gr 2

e

D 10;600 s D 2:95 h


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) Â D 180ı ˇ D 5:30ı

5.112 (a) v1 Ds

rP v2P

2GmB

rP

(b) rP v2P > 2GmB



5.118 R D 14pAd 2A C 4Q2

g

Â1

d 2A 1

d 2B

ÃD 2120 lb

5.120 vB D 29:2 ft=s

5.122 vmax D vo ln

Â1 C mf

mb

Ã

5.124 k D d 2v2w4gımax

Œ1 cos.Â=2/� D 370 lb=ft





5.126 v D vo ln

ÂM

M Pmot

Ã gt

5.128

N D 14P Ad 2 C 4Q2

d 2D 53:4 N;

V D

4Q2

d 2 D0:354 N;

M D 4Q2`

d 2D 0:0707 Nm

5.130 F D

gL C v20

ÁD 1:10 lb


5.134 Pmo D 0:0150 slug=s

vo D 4530 ft=s

vmax D vo ln

Â1 C mf

mb

Ã gtbo D 1320 ft=s

5.136 Â max power

D0ı

v0 D 12vw

5.138 EF avg D m

t

q 2ghf C

p 2ghi

O| D .654 N/ O|

j EpC Epjjt.mg O| /j D j EF avgj

mgDq

2ghf Cp

2ghi

gtD 111

5.140 EvCA D .13:6 m=s/ O{ and EvCB D .21:3 m=s/ O{T T C

T 100% D 45:3%

5.142

EvCA

D.4:61

O{

C75:3

O| / ft=s and

EvCB

D.37:4

O{

C98:8

O| /ft=s

ErA D d AEvCAvCA

D .7:71 O{ C 126 O| / ft

ErB D d BEvCBvCB

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

D 8:64107 s D 1000 days

5.148 vw D 8<:1

2K 1

2q

K 2

4v

2

0 D 0:219 m=s;12K C 1

2

q K 2 4v20 D 18:3 m=s;

5.150 EvA D .0:813 m=s/ O{ and EvB D .3:25 m=s/ O{

Chapter 6

6.2 E!B D 1800 Ok rpm and E!C D 1160 Ok rpm

6.4 !B D RARB

!A D 600 rad=s








6.12 .!OA/max D 2:48 rad=s

6.14j E!

ramj D0

6.16 EvG D .0:150 Our 3:38 OuÂ / m=s

EaG D .20:4 Our 3:66 OuÂ / m=s2

6.18 E!s D 10:0 Ok rad=s

Es D 1:33 Ok rad=s2

6.20 EvC D .0:179 O{ C 0:492 O| / m=s

EaC D .2:58 O{ C 0:938 O| / m=s2

6.22 !s D 2 rad

24 h

1 h

3600 sD 72:7106 rad=s

6.24 E!B D 8:00 Ok rad=s; EB D 2:00 Ok rad=s2; E!D D 1:60 Ok rad=s; ED D 0:400 Ok rad=s2

6.26 E!C D 40:0 Ok rad=s

EC D 5:20 Ok rad=s2

6.28 Letting E!OB and EOB denote the angular velocity and acceleration of the turbine,

E!OB D 1:60 Ok rad=s and EOB D 0:461 Ok rad=s2

EaA D .93:0 O{ 110 O| / m=s2

6.30 Chain ring/sprocket combination: C1/S3

!W D !S D RC

RS

!C D52:6 mm

28:3 mm68 rpm D 126 rpm

6.32

EvB D

.9:88O{C

3:66O

| / m=s

6.34 E!P D 9:60 Ok rad=s

EvO D 2:00 O{ ft=s


6.38 E!P D 1:60 Ok rad=s

EvC D 5:33 O{ m=s

6.40 Â 1 D 67:7ı

Â 2 D 231ı

6.42 E!A D 7:00 Ok rad=s

6.44 E!C D 60:0 Ok rad=s

6.46 EvB D 10:6 O{ ft=s

EvC D .5:31 O{ 4:00 O| / ft=s

6.48 E!P D 8:80 Ok rad=s

ErIC=Q D 0:307 O| rad=s

6.50 EvC D 82:2 O| ft=s

6.52 E!OC D 2:86 Ok rad=s

EvP D .0:0286 Our 0:119 OuÂ / m=s





6.54 EvC D 16:3 O{ ft=s

the cable is unwinding at 12:2 ft=s

6.56 E!CD D 42:4 Ok rad=s

6.58 E!AB D 5:09 Ok rad=s

6.60

E!spool

D 3:33

Ok rad=s

EvO D 5:00 O{ rad=s

6.62 E!gate D 0:0467 Ok rad=s


E!BC D 8:58 Ok rad=s

6.66 E!R D 35:0 Ok rpm

EvD D 17:2 O| in.=s


EvB D 1:92 O| m=s

6.70 EvB D "R PÂ cos Â R

PÂ.H

R cos Â/ sin Â p

L2 .H R cos Â/2# O|


6.74 EaB D .17:0 O{ 5:38 O| / ft=s2

6.76 EaC D .16:8 O{ C 87:7 O| / ft=s2

6.78 EaQ D 9:16 Our m=s2

6.80 EAB D 48:6 Ok rad=s2 and EBC D E06.82 Letting Q denote the point on the ball in contact with the bowl, at the instant shown

EaA D .1020 Our C 24:0 OuÂ / ft=s2 and EaQ D 3480 Our ft=s2

6.84 EAB D 27:1 Ok rad=s2

EaB D 267 O{ ft=s2

6.86 E!AB D 0:458 Ok rad=s and EAB D 0:258 Ok rad=s2

6.88 EW D 7:97 Ok rad=s2

6.90 EBC D 1640 Ok rad=s2

6.92 EBC D HR PÂ 2.R2 H 2/ sin Â

.H 2 C R2 2HR cos Â /2Ok

6.94 Es D 0:400 Ok rad=s2 and EaO D 2:00 O{ ft=s2

6.96 Egate D 4:50 Ok rad=s2

6.98 EAB D 92:1 Ok rad=s2 and EBC D 191 Ok rad=s2

6.100 Letting Q denote the point of contact between the wheel W and the cylinder S , EaQ D 82:0 Our ft=s2

6.102 For C accelerating downward and to the right: EAB D 551 Ok rad=s2 and EBC D 320 Ok rad=s2

For C accelerating downward and to the left: EAB D 729 Ok rad=s2 and EBC D 320 Ok rad=s2

6.104 For D 10ı and Â D 56:11ı EaD D .18;300 O{ C 34;900 O| / ft=s2

For D 10ı and Â D 303:9ı EaD D .18;300 O{ C 32;900 O| / ft=s2





6.106 EaC D(

R˛AB cos Â

Â1 C R sin Â p

L2 R2 cos2 Â

Ã

C R!2AB

"R.L2 R2/ cos2 Â

.L2 R2 cos2 Â/3=2 R sin2 Â p

L2 R2 cos2 Â sin Â

#)O|


6.110 EaP D Rs s!20 O{ C s˛0 C 2Ps!0

O|

6.112 (a) E!CD D .R=h/!AB1 C R= h

D 3:14 Ok rad=s and Evbar D .R=h/d!AB1 C R= h

OI D 0:377 OI m=s

(b) ECD D E0 and Eabar D d˛CDOI D E0

6.114 (a) E!CD D ı!AB.ı C sin Â/

1 C ı2 C 2ı sin Â Ok

Evbar D ı!AB.ı C sin Â/

1 C ı2 C 2ı sin Â d OI

(b) ECD D ı

1 ı2

!2AB cos Â

1 C ı2 C 2ı sin Â 2

Ok

Eabar D d ı

1 ı2

!2AB cos Â

1 C ı2 C 2ı sin Â 2 Ok

6.116 (a) Evbar D ı

1 C ı!ABd OI

(b) Evbar D ı

1 ı!ABd OI

(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

6.118 EvP D .23:6 O{A C 26:1 O|A/ ft=s

EaP D .52:9 O{A 40:3 O|A/ ft=s

2

6.120 EaC D .30:0 O{ C 51:8 O| / ft=s2

6.122 EvC D .7:50 O{ C 7:51 O| / ft=s

EaC D .105 O{ 161 O| / ft=s2


6.126 Pd AB D 1:12 ft=s

Rd AB D 2:01 ft=s2

6.128 E!bar D 14:3 Ok rad=s

6.130

E!A

D 481 Ok rad=s

6.132 EaE D .1:50 O{ 0:320 O| / ft=s2

6.134 Pd EC D 0:214 ft=s and Rd EC D 0:0192 ft=s2

6.136 EvD D .vC C !OA` sin / O{ C .d!OA `!C cos / O| `!C sin OkEaD D .d!2

OA C 2!OA`!C cos C `˛OA sin / O{ C .d˛OA C .!2C C !2OA/` sin C 2!OAvC / O| !2C cos Ok:

6.138 EvC D .21:5 O{ C 7:51 O| / ft=s

EaC D .107 O{ 166 O| / ft=s2





Chapter 7

7.2 tstop D v0=.kg/ D 1:86 s and d stop D v20=.2gk/ D 16:8 ft

7.4 F D 179 N and aGx D 4:71 m=s2

7.6 Â 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

D577 lb; N f

D784 lb; N r

D783 lb; and N B

D2230 lb

EH D H x O{ C H y O| D .653 O{ C 166 O| / lb andˇ EH D 674 lb

.s/min D 0:736

7.12 m1 D 14m; m2 D 3

4m; xP D 23L

m1 D 12m; m2 D 1

2m; .I G/extra D 16mL2

7.14

Rx D .0:112 N=s2/t2;

Ry D 0:0101 N;

M D 3:11105 Nm


Or D 16:8 lb and OÂ D 5:00 lb

7.18 j˛barjmax Dp

3g

Lfor d D 1

6

3

p 3

L or d D 16

3 C

p 3

L


EO D Ox O{ C Oy O| D .1320 O{ C 48:5 O| / N and ET-bar D .6:92 rad=s2/ Ok






7.24 .vG/final Dp

2gL.sin Â k cos Â C sin Â/ D 18:4 ft=s

tfinal D .vG/final=aGx Dp

2gL.sin Â k cos Â C sin Â/

g.sin Â k cos Â/D 1:08 s

7.26 tr D 2:41 s and d D 50:7 ft

7.28 aGx D kg and ˛b D grk

k2G


T CD D 101 N: and EaG 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 coun-

terclockwise. 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 Â separation D

3D 60:0ı


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 Â , and angular acceleration of the sphere is as follows:

N D mg cos Â m.R C / PÂ 2; RÂ D g sin Â

R C ; and ˛s D 0:

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 I Gvf

r2tf D 12:1 lb

(b) .tf /new D mcr2tf

2I G C r2mc

D 6:85 s

7.40 ˛AB D 3mABg cos Â

L.2mAB C 9mB sin2 Â/; EaB D 3mABg sin Â cos Â

2mAB C 9mB sin2 Â O{; ˛w D 3mABg sin Â cos Â

R.2mAB C 9mB sin2 Â /





7.42 ˛AB D 3.2mA C mAB/g cos Â

L.6mA cos2 Â C 2mAB C 9mB sin2 Â/

EaB D 3.2mA C mAB/g sin Â cos Â

6mA cos2 Â C 2mAB C 9mB sin2 Â O{;

˛w D 3.2mA C mAB/g sin Â cos Â

R.6mA cos2 Â C 2mAB C 9mB sin2 Â/

7.44 I G D

mR2

7.46 ac D 1:35 m=s2; T D 3130 N; and ˛s D 1:68 rad=s

7.48 Rx D .3g cos Â C 2L PÂ 2/ sin Â

1 C 3 sin2 Â and RÂ D 3.2g C L PÂ 2 cos Â/ sin Â

L.1 C 3 sin2 Â/

7.50 (a)

N r D 1

4W c

Â1 C 2hac

g`

ÃD 736 lb;

N f D1

4W c

Â1 2hac

g`

ÃD 549 lb;

F r D W cac2g

D 503 lb

(b)

Ax D acW wg

F r D 485 lb

Ay D W r N r D 689 lb

M A D F rd

2C 2acI B

d D 522 ftlb


m

1 C k2G

r2

!Rx C kx D mg sin Â C kL0

7.54 ˛AB Dg

2L

cos Â 13 C cos2 Â.cos = sin /2 C sin Â cos Â.cos = sin / D 3:74 rad=s

2

7.56 aAx D 60

181g and ˛b D 45g

181R

7.58 EaG D 57g sin Ou ; ˛b D 5g sin

7; and EF due to bowl D mg cos Our C 2

7mg sin Ou

7.60 R C 5g

7.R /sin D 0

7.62 (a) EaG D .1 k/PR2

m.R2 C k2G/O{; and .s/min D kR2 C k2G

R2 C k2G

P

kP C mg

(b)

EaG

D ÄP

m

.1

2

k

/

kg O

{; and ˛d D

kR

k2GÄP

m

.1

k/

g

7.64

R.2mAR2 C I O/ P!s C 2mA`. P C R!s/2 D 0;

` R C P2 C `R P!s R2!2s D 0

7.66 (a) Ay D 34mg and By D 5

4mg

(b) Â D 2 tan1Â

P

mg

Ã; Â 0 D 0; and Â 1 D

7.68 T CD D 2060 lb; and EaE D .5:65 ft=s2/ O|





7.70 EaE D .13:2 O{ 25:9 O| / ft=s2

Ax D 3g.mb C mr /Œ.d C h/mb C 2d mr � sin Â cos Â C g.h d /mbmr tan Â

2.d C h/.2mb C 3mr /D 437 lb;

Ay D gŒmbh C d.mb C mr /�Œmb C 3.mb C 2mr / cos2 Â � C 3ghmbmr sin2 Â

2.d C h/.2mb C 3mr /D 256 lb;

C x D 3g.mb C mr /Œ.d C h/mb C 2hmr � sin Â cos Â g.h d /mbmr tan Â 2.d C h/.2mb C 3mr /

D 629 lb;

C y D gŒmbd C h.mb C mr /�Œmb C 3.mb C 2mr / cos2 Â � C 3gdmbmr sin2 Â

2.d C h/.2mb C 3mr /D 353 lb

7.72 EaE D .7:00 O{ C 7:06 O| / ft=s2

Ax D 229 lb; Ay D 1330 lb; C x D 336 lb; and C y D 1330 lb

7.74 P max D 1600 lb

7.76 Â D tan1

v2cgR

!

7.78 ˛AC D RÂ D 1:46 rad=s2 and ˛CE D RÂ D 1:46 rad=s2


tclosure D 1:29 s

.vE /at closure D 1:87 m=s

7.82

R D gR sin Â g cos R cos. Â/ P2R2 C 2 C k2

GC 2R sin. Â /

;

F D mg sin Â C maGx

D mg sin Â m

ŒRC sin.Â/�ŒgR sin Âg cosR cos.Â/ P2�

R2C2Ck2GC2R sin.Â/

C P2 cos. Â /

;

N D mg cos Â C maGy

Dmg cos Â

Cm cos.Â/ŒgR sin Âg cosR cos.Â/ P2�

R2C2Ck2GC2R sin.Â/

P2 sin.

Â/

7.84

.mAB C mC / RxA C 12 .mAB C 2mC /L cos Â RÂ C 1

2 .2d C h/mC cos R12 .mAB C 2mC /L sin Â PÂ 2 1

2 .2d C h/mC sin P2 D 0;

12LmC cos Â RxA C Œ 112mAB C 1

2mC cos2 Â C . 14mAB C mC / sin2 Â�L2 RÂ

C14 .2d C h/LmC .2 sin Â sin C cos Â cos / R

C14 .mAB C 2mC /L2 sin Â cos Â PÂ 2

C14 .2d C h/LmC .2 sin Â cos sin cos Â / P2

C12 .mAB C 2mC /gL sin Â D 0;

12 .2d

Ch/mC cos

RxA

C12 .2d

Ch/LmC cos.Â

/ RÂ

CŒ 112 .h2 C w2/ C 14 .2d C h/2�mC R

12 .2d C h/LmC .sin Â cos sin cos Â/ PÂ 2

C12g.2d C h/mC sin D 0:

7.86 (a) F D 34mg.3 cos Â 2/ sin Â and N D 1

4mg.1 3 cos Â/2

(b) Given any (finite) value of s , there always is a value of Â for which the rod will slip.

7.88 P max D mcgs

1 C md

mc

C R2

k2G

!; aCx D g

1 C R2

k2G

!; and aGx D sg

7.90 P D 7:59 N and .s/min D 0:0462





Chapter 8

8.2 T A D T B D 518 J

8.4 T R1 D 7760 ftlb and T R2 D 15;000 ftlb

8.6 T D 324;000 ftlb

8.8 T D1

2H 2 ŒI DR

2

C H

2

.I A C mBC R

2

/�!

2

AB D 25:3 J

8.10 T D 28:9 ftlb


8.14 Lf D 34:1 ft


8.18 !2 Ds

.2Md=R/ kd 2

m.k2G C R2/

d s D 2M

kR

8.20 M D 2:06106

ftlb

8.22 vperson D vG Dp

2gH.1 C cos Â /

vperson

ˇÂD0ı

D 10:8 m=s

8.24 vO D 10:9 ft=s

8.26 nmin D 26

(a) vperson D 26:5 ft=s

8.28 U fracture D 12gŒLA.mA C 2mS / C 2mSrS �.cos Â f cos Â i / D 318 J

8.30 U friction D 364 ftlb

8.32 k

D W

2g

v2max C g.L C d 2H /

.H d/.H C 2ı0 d / D11:3 lb=ft

8.34 ı D 1:03 ft

8.36 W P D3gH v2A2

3gH C 6v2A2W D 800 lb

8.38 vmax D 6:39 ft=s

8.40 vC D 4:96 m=s and E!D D .44:1 rad=s/ Ok8.42 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 NmP max D 132 W

8.48 .s/max D 0:371 and Â slide D 35:1ı


8.52EhAAB D 1

3mABR2!AB

Ok D .1:72 kgm2=s/ Ok

(a)EhABC D mBC !ABR2 Ok D .7:20 kgm2=s/ Ok

(b)EhDCD D 1

3mCDHR!ABOk D .7:75 kgm2=s/ Ok





8.54EhC W D .91:9106 kgm2=s/ Ok @Â EhOW D .669106 kgm2=s/ Ok @Â

8.56ˇ EpAB ˇ D W ABvA

2g cos Â D 3:23 lbs

EhG D W ABLvA12g cos Â

Ok D .2:42 ftlbs/ Ok

8.58 EF avg D W vfinal

gtO{ D .878 lb/ O{

EM avg D 2I Gvfinal

dtOk D .10:7 lbft/ Ok


8.62 tr D v02kg

8.64 !Aˇ

end of slipD 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

D0:943 ft=s,

8.68 E!.t/ D"

g.1 C w/g

.1 C gw/rt !0

#Ok

8.70 vGˇtD2 s

D 21:7 ft=s

.s/min D 0:0588


8.74 The system rotates clockwise.

Â D mBa0e

2I Ot2f ; (clockwise)

8.76 d

D4r2d

.2r b/2



8.82 d D 23`

8.84 d D 2:09 ft

E!B D .26:6 rad=s/ Ok and vCb

D 226 ft=s

8.86 E!Cb

D .10:7 rad=s/ Ok

8.88

EvCG

D.

1:31 m=s/

O{ and

E!CA

D.3:43 rad=s/ Ok

8.90 v0 Ds

4g` sin ˇ

3 cos2 ˇD 7:48 ft=s

8.92 E!CAB D .0:851 rad=s/ Ok and E!CBD D .2:55 rad=s/ Ok

8.94 E!CA

D .0:000 715 rad=s/ Ok and E!CB

D .0:0000241 rad=s/ Ok8.96 T D 0:0257 ftlb

8.98 d D 1:98 m







8.104 EvCP

D 2v0 O{ D .12 ft=s/ O{

8.106 E!CA D .1:03 rad=s/ OkÂ max D 14:8

ı

Chapter 9

9.2 I O D mgL 2

42

9.4 D 2

s 7L

6g

9.6 ` Dr

I Gm

9.8 !n Ds Gr4LR4t

9.10 D 2d

h

r m

k

9.12 D 2

s 13.h3 C d 3/ C md 2

kh2

9.14 W payload D 90;700 lb

9.16 f D !n2

D d

4

r g

mD 0:566 Hz

9.18 (a) m

Ry

Cky

D0

(b) m RÂ C kÂ D 0

9.20 mn D 35:6 g

9.22 RxG C 17k

6mxG D 0

9.24 RxG C 8k

3mxG D 0

D 2

!nD 2

r 3m

8kD 2:22 s

9.26 !n Ds

2g

3.R r/

D 2

s 3.R r/

2g

9.28 D 2

!nD 2

s L

2g

9.30 D 2

s .3 2/R

3g

9.32 x.t/ D A sin !nt C B cos !nt C F 0=k

1 .!0=!n/2cos !0t





9.34

mu D 0:01 kg ) jymj Dq

2:19106 cos.4:37t/ C 1:89105 m;

mu D 0:1 kg ) jymj Dq

1:10105 cos.4:37t/ C 1:11105 m;

mu D 1 kg ) jymj Dq

9:80104 cos.4:37t/ C 1:01103 m

9.36 Â amp D 0:001 76 rad=s

9.38 !n Ds

3EI

.mu C me/d 3 C 43mwh2d

D 312 rad=s

f D !n2

D 49:7 Hz

MF D jÂ pj.mu C me/d

muRD .!p=!n/2

1 .!p=!n/2D 0:820


9.42 x.t/ D 0:000 501 sin 10t C 0:100 cos 10t 2:51105 sin 200t

m

9.44 m Ry C 2kÂ1 L0

L Ãy D F 0 sin !0t

y.t/ D F 0=keq

1 .!0=!n/2

Ä!0!n

sin !nt C sin !0t

;

where keq D 2k

Â1 L0

L

Ãand !n D

s 2k

m

Â1 L0

L

Ã

9.4623mB C 1

2mA Rx C 2kx D 1

2F 0 sin !0t

xamp D F 0

4k 43mB C mA

!20



9.52 no peak in MF for p

1=2

9.54 Ry C 2!n Py C !2ny D mu"!2rm

sin !r t

9.56 jymj D 0:002 19 m

9.58 I GRÂ C cr2o

PÂ C k1r2o C k2r2i

Â D 0

Â.t/ Dh

0:0504e3:80t 0:000386e496ti

rad

9.60 x.t/ D e0:5t .A sin 3:12t C B cos 3:12t/

9.62 m Rx C c Px C 2k

Äq L2 C .L C x/2

p 2L

L C x

p L2 C .L C x/2

D F 0 sin !0t

mRx

CcPx

Ckx

DF 0

sin !0

t

D D F 0=kp Œ1 .!0=!n/2�2 C .2!0=!n/2

; where !n Dr

k

mand D c

2p

km

9.64 k > 1:35109 N=m and c D 0

9.66 (a) mRs C c Ps C ks D mA!20 sin !0t

(b) y.t/ D"

mA!20

k m!20

k m!20

2 C c2!20

C A

#sin !0t

"mcA!3

0k m!2

0

2 C c2!20

#cos !0t

(c) DT Dv uut k2 C c2!20

k m!20

2 C c2!20





9.68 m Rx C k1k2k1 C k2

x D 0

9.70 k D 5:91%

9.72 .mm C mp/ Rym C keqym D mu"!2r 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

9.76 m Ry C 2k

1 L0p

y2 C L2

!y D 0

9.78 m Ry C 2k

Â1 L0

L

Ãy D 0

Chapter 10

10.2 EvE D R!arm cos O{ C .d C ` cos /!arm O| .d C ` cos /!arm OkEaE D !2arm cos .d C ` cos / O{ !2arm sin .d C ` cos / O| !2arm

R

hd 2 C 2d ` cos C `2 C R2

cos2

i Ok

10.4 EvP D 2.d C ` cos /!armOk

EaE D !2arm cos .d C ` cos / O{ !2arm sin .d C ` cos / O|(

.d C ` cos /˛arm C !2arm

R

hd 2 C 2d ` cos C `2 C R2

cos2

i) Ok

10.6 EaA D

231 O| C 108 OkÁ

ft=s2

EAB D

18:3 O{ C 15:1 O| C 34:9 OkÁ

rad=s2

10.8

EvP

D R!d cos ˇ

O{

.h!b

CR!d sin ˇ/

O|

CR!b cos ˇ Ok

EaP D R

!2d sin ˇ P!d cos ˇÁ O{ R

h!2b C !2d

Ácos ˇ C P!d sin ˇ

i O|C

R P!b cos ˇ h!2b 2R!b!d sin ˇ

Á Ok

10.10 EvA D 2L!0 cos2 ˇ OkEaA D 2L!2

0 cos2 ˇ O{ L!20 cot ˇ O| 2L˛0 cos2 ˇ Ok

10.12 E!AB D !s O| P OkEAB D P!s O{ R OkEaG D

ÄL

2R sin ˇ C

Âd C L

2cos ˇ

Ã!2s C L

2P2 cos ˇ

O{ C L

2

P2 sin ˇ R cos ˇÁ

O| L P!s sin ˇ Ok

10.14 EvA D 1:432 O| 0:6677 Okm=s

10.16 EaA D 103 O| C 47:8 Okm=s2

10.18 E!AB D 0:0161 O{ C 2:69 O| C 0:917 Ok rad=s

EAB D 1:73 O{ 4:59 O| C 15:8 Ok rad=s2

10.20 E!bit D !d O{ C !a O| C !bOk

Ebit D . P!d !a!b/ O{ C . P!a C !d !b/ O| C . P!b !d !a/ Ok

10.22 EaB Dh

at C h P!a .` C rt /

!2a C !2b

ÁiO{ C Œ2.vt C h!a/ !b C .` C rt / P!b � O|

Œ.2v C h! / ! C .` C r / P! � Ok

Documents

Dynamics Answers 2009-11-2