AppendixAnswers Dynamics 2e Gray

8/13/2019 AppendixAnswers Dynamics 2e Gray

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Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha

Answers to Selected Even-Numbered ProblemsPlease 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=Axy system

D .4:000 O{ 1:000O| / ft

ErB=Apq system

D .3:313Oup 2:455Ouq/ ft

ErB=A xy system

DErB=A pq system

D 4:123 ft

1.6 EvD .79:68 O{ C 7:190O| / ft=s

1.8 D 3:229 radD 0:087 31 rad

1.10 vrD 504:6 ft=s and vD 353:3 ft=s

1.12 D 101:1

1.14 xP2D 0:7679 ft and yP2D 5:330 ft

1.16 EvAD .20:02 O{ C 12:81O| / ft=s and EaAD .1:414 O{ C 4:243O| / ft=s2

1.18 rD 26:36 mm

1.20 Ixx D Iyy D I D ML2

Units ofIxx ,Iyy , andI in theSI system: kg m2;Units ofIxx ,Iyy , andI in theU.S.Customary system: slugft2 D lbs2 ft.

1.22 Concept problem.

1.24 the units ofE are kg=

ms2

.

Chapter 2

2.2 Concept Problem

2.4 Concept Problem

1 Last modified: August 16, 2012


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2.6 Er1D 8:747Our m and Er2D 13:73OurmEvavg1D 5:467Our m=s and Evavg2D 8:579Ourm=s2.8 EvD .154:0 O{ C 266:7O| / ft=s

2.10 EvD .145:4 O{ C 69:81O| / ft=s2.12 Ev1D .0:4623 O{ C 0:08155O| / m=s

Ev2D .0:002550 O{ C 0:0004499O| / m=s

2.14 1D 12:21 and 2D 17:30

1D 102:2 and 2D 72:70

2.16 .x/ D cos1

3 C 4xp10 C 24x C 16x2

2.18 EaavgD .0:04588 O{ C 3:886O| / ft=s2Eaavg Ea.5 s/ D .0:001154 O{ 0:003175O| / ft=s2

2.20 ErDE0EvavgDE0dD 1:394 ftvavgD 0:3485 ft=s

2.22 Computer Problem

2.24 vmaxD 2v0D 58:67 ft=s and vminD 0 ft=syvmin D 0 ft and yvmax D 2RD 2:300 ft

Eavmin D v20

R O|D .748:2 ft=s2/O| and Eavmax D

v20R

O|D .748:2 ft=s2/O|

2.26 RxD 8v20a

3y2 C 4a22

RyD 4v20a

2y

y2 C 4a2

2

2.28 EvD .203:2 O{ C 117:3O| / ft=sEaD .27:53 O{ C 47:69O| / ft=s2

2.30 vmaxD 80:82 ft=s

2.32 EvD .51:32 O{ 1:046O| / ft=sEaD .0:3873 O{ C 19:00O| / ft=s2



3/41


2.34 vD !p

d2 x2s

1 C h2

x

d2 x

3

d4

2

vD 0:5236 ft=s; vD 0:4554 ft=s; and vD 0

2.36 EvAD .72:00 O{ C 96:00O| / ft=sEaAD .460:8 O{ 345:6O| / ft=s2


2.40 tbrakingD 7:854 s

2.42 jajmaxD 3:600 m=s2tjamaxj

1D 3:927 s and tjamaxj2D 11:78 s

sjamaxj1D 12:84 m and sjamaxj2D 128:5 m

2.44 Largest distance traveled in1 s is dD 0:2667 m, corresponding toaD 1pt

2.46 v.0/ D 0:080 00 ft=s

2.48 v0D 10:10 m=s

2.50 acD 22:36g

2.52 tstopD 2:880 s

2.54 D 98:10 s1

2.56 PxD 7:000 sin.1:000 rad=s/t C 10:50 cos.0:5000 rad=s/t 10:50m=sxD 7:000 7:000 cos.1:000 rad=s/t C 21:00 sin.0:5000 rad=s/t .10:50 s1/tm=s2

2.58 tstopD 0:2233 s

2.60 vtermD 4:998 m=s

2.62 jvjmaxD 1:128 m=ssjvjmax D 0:1250 m and sjvjmax D 0:1250 m

2.64 v.t/ D mgCd

1 eCd t=m

vtermD mg

Cd

2.66 vfD 3:563 m=s




4/41


2.70 swetD 26:31 m.swet sdry/

sdry.100%/ D 65:21%

2.72 P./ D r

P20C 2g

L cos cos 0

2.74 PminD 4:930 rad=s

2.76 PxD s

v20C 2

g C k L0m

.x x0/

k

m

x2 x20

2.78 txmax D 0:2433 s

2.80 (a)PrD p

2G .mA C mB /r

r0 rrr0

(b) (i)PrD 5:980 105 ft=s and (ii)Pr! 1

2.82 Concept Problem

2.84 dD 186:4 ft

2.86 sDap

r C hv

2p

2r3

srDr1

D 1:442 rad=s2 and srDr2

D 3:631 rad=s2

2.88 vD v0 C ac.t t0/sD s0 C v0.t t0/ C 12ac.t t0/2

2.90 vminD 113:5 ft=s

2.92 tiD 3:497 sEvBD .33:33 O{ 34:31O| / m=s

2.94 RD 5:047 m

2.96 tflightD 6:263 sErlandD .281:8 O{ 4:500O| / m

2.98 dD 24:09 m

2.100 (a) Rmax D

4 rad D 45

(b)RmaxD 233:5% of the actual maximum range(c)tRmax D 119:5 s

2.102 1D 12 sin1.gR=v20/ and 2D 90 12 sin1.gR=v20/1D 24:86 and 2D 65:14



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2.104 .v0/minD 50:75 ft=s

2.106 EvinitialD .5:920 O{ C 7:786O| / m=s:dy

dx

xDxB

D 0:2244

2.108 hmaxD v20sin2

2g cos


2.112 tfRmaxD 5:211 s

RmaxD 437:1 ft and Rmax D 45:66

2.114 tmaxD 3:000 sHmaxD 49:21 m

percent increase in height with no air resistance D 2:363%


2.118 (a) Ea EbD 1 6 C 2 3 C 3 0 D 0(b)Ea

Ea Eb

D

84 O{ 42O|C 0Ok

(c)they are the same

2.120 (a)E!1D

104:7Ok rad=s; E!2D 104:7 O{ rad=s and E!3D 104:7O| rad=s(b)E!`D

100

3p

3 O{ C O|COk

rad=s D 60:46

O{ C O|COk

rad=s

2.122 Derivation, so no answer needed.

2.124 EvAD 10:00O| m=s; EvBD 10:00O|m=s; EvCD 10:00O| m=s; and EvDD 10:00O|m=s

2.126 E!wheelD 65:71Ok rad=s

2.128 jEaPj D 19;300 ft=s2

Orientation ofEaP fromx axis D 110:0 .ccw/

2.130 v0D 263:7 ft=s and rD 6875 ft

2.132 PrD 1:204 m=s and PD 0:8504 rad=s

2.134 EvD .0:037 70OurC 0:5341Ou / m=sEaD 1678OurC 236:9Ou m=s2


2.138 Pr D90

D 8:825 ft=s and P D90

D 0:7746 rad=s



6/41


2.140 EaPD .252:9OuC 399:8OuB / ft=s2



2.146

D49:69 ft

2.148 PvD 3:213 m=s2

2.150 SdkurveD 50:56 mNordkurveD 92:54 m

2.152 D 327:3 ft

2.154 dD 831:0 ft

2.156 EaDE0

2.158 EaD .39:40 O{ C 42:88O| / m=s2


2.162Ea D .33:69103 cos / m=s2

2.164 EaD .121:2 O{ C 230:0O| / ft=s2

2.166 EaD 18:96 ft=s2OutC 69:91 ft=s20:2917 s2sOun2.168 PvD 5:343 m=s2 and D 17:50 m

2.170 D 282:2 ft

2.172 minD 564:5 fttf t0D 4:742 s

2.174 jEaj D 1

v0

2v0

s

s4v202

C

v0 2v0

s

2

2.176 dD 304:4 ft and tfD 3:106 s



2.182 PrD 684:1 ft=s and PD 0:011 57 rad=sRrD 4:944 ft=s2 and RD 0:000 428 1 rad=s2

2.184 vD 1:816 m=sEa D 4:573 m=s2



7/41


2.186 RrD 1:599 m=s2

2.188 0D 0r0D 0:014 86 m

2.190 EvD .1:3 m=s/OurC 0:22 s1rOuEaD 0:048 40 s2rOur 0:5720 m=s2Ou

2.192 EvD .0:1500OurC 1:732Ou / ft=s and EaD .0:3638OurC 0:1125Ou / ft=s2

2.194 ErD 30:20OurftEvD .3OurC 7:550Ou / ft=sEaD .1:888OurC 1:500Ou / ft=s2

2.196

D5:602 m

!D 2630 rad=s D 25;120 rpm

2.198 vD 5490 ft=sEa D 1:005106 ft=s2

2.200 EvD v0p2 C .r0 C /2

OurC v0.r0 C /p

2 C .r0 C /2Ou

EaD v20.r0 C /3

r20C 2r0C .1 C 2/22OurC

v20

.r0 C /2 C 2

2 C .r0 C /2

2 C .r0 C /22

Ou

2.202 PAD 0:001 157 rad=s and PHD 4:287 rad=sRAD 1:684104 rad=s2 and RHD 0:075 47 rad=s2

2.204 PEaDr 3r PR 3Pr P2

OurC

hrP3C 3 Rr PC 3Pr RiOu

2.206 RrD 5:781 ft=s2

RD 1:421 rad=s2

2.208 r

D qh2

C.d

C/2

PrD v0hph2 C .dC /2

and PD v0.dC /h2 C .dC /2

RrD v20

. C d /d2 C h2 C dd2 C h2 C 2 C 2d3=2 and RD

v20

h

d2 C h2 2d2 C h2 C 2 C 2d2

2.210 EvD 1:758O|m=s and EaD 136:9gO|

2.212 vD 0:1185 ft=s and Ea D 0:086 42 ft=s2



8/41


2.214 arD K2a

b2

1

r2



2.220 vH=FD h5:630105 0:8091 cos x2 im=saH=FD 5:237 sin

x

2

m=s2

2.222 vA=BD 144:7 m=s

2.224 EvC=AD .6:857 O{ C 30:27O| / m=s:

EaC=AD .1:376 O{ 1:000O| / m=s2:

2.226 .vP/avg

D17:30 ft=s


2.230 vpmax D 330:7 ft=s D 225:4 mph

2.232 EvBD 2:928 O{ft=s

2.234 vrainD 16:37 ft=s

2.236 EaBD

L1Rcos L1 P2 sin C L2 Rcos L2 P2 sin

O{C L1 Rsin C L1 P2 cos C L2 Rsin C L2 P2 cos O|

2.238

D63:57

2.240 D 14:90

2.242 EvBD 3:088O|ft=s:

EaBD 0:3192O|ft=s2:

2.244 EvBD 2:000O| m=s:

2.246 vAyD 4:596 ft=s:

2.248 EaBD 11:10 O{m=s2: @tdD 0:1644 s

2.250 a0D 0:9374 ft=s2

2.252 P D dv0ph2 C d2

2.254 Component ofEvC=B alongB CD 0



9/41


2.256 rDq

.dC cos /2 C .h C sin /2

PD v0. C dcos C h sin /.dC cos /2 C .h C sin /2 and PrD

v0.h cos dsin /p.dC cos /2 C .h C sin /2

2.258 tD 0:1556 s

2.260 RyeD 9:81 m=s2

RyCtD0:4391 s

D 44:51 m=s2



2.266 tD 83:82 s

2.268 EvCD .6:500OuRC 5:520OuC 5:300Ou/ ft=s

EaCD .0:6624OuR 1:560Ou / ft=s2

2.270 rP.r Rsin C 2 Pr Psin C 2rP Pcos / r Psin .rR C 2 PrP r P2 sin cos / D 0;r Psin .Rr rP2 r P2 sin2 / Pr.r Rsin C 2 Pr Psin C 2rP Pcos / D 0;

Pr.rR C 2 PrP r P2 sin cos / rP. Rr rP2 r P2 sin2 / D 02.272 EvD .0:04869OuRC 7:961Ou 0:2999Ou/ m=s

EaD .79:79OuRC 0:9738Ou 0:8120Ou/ m=s2

2.274

Eamax

D 12:36 ft=s2

2.276 EvD dfd

P OurC K

f./OuC POu

EaD

R C P2 d2f

d2 K

2

f3./

OuRC ROu

2.278 arD 0:1598 m=s2; aD 0:1884 m=s2; and aD 0:2232 m=s2

2.280 EvP=BD .10:17 O{BC 10:33O|B / m=s and vP=BD vP=AD 14:50 m=s

EaP=BD .7:822 O{B 1:258O|B/ m=s2 andEaP=B D EaP=A D 7:923 m=s2



2.286 sD mgC2d

CdtC m

e

Cdm t 1

2.288 xD

L0 Cg m

k

1 cos

rk

mt

2.290 v0p

gR



10/41


2.292 D 240:5 ft

2.294 The car will lose contact with the ground.

acD 1:606 m=s2

2.296 Pr D180D 0P D180

D 2:941 rad=s

Rr D180

D 40:85 m=s2 and R D180

D 0

2.298 PD 0:053 91 rad=sRD 0:2380 rad=s2

2.300 vmaxD 6:020 ft=s

2.302 hmax

D0:8898 ft

2.304 jEaAj D 18:65 m=s2

2.306Ea D 172;700 ft=s2

2.308Eamin D 2:534 m=s2

Chapter 3

3.2 Concept Problem

3.4 EaAD 2:147O|ft=s2:

3.6 amaxD 17:75 ft=s2

3.8 aBmax D 0:1650 m=s2 D 0:5417 ft=s2

3.10 Rx C EAmL

xD 0

3.12 aBmax D 5:048 m=s2 D 16:57 ft=s2

3.14 dD 6:630 ft

3.16 dD 417:5 ftIf an entire train weighing30106 lb was skidding to a stop, instead of a locomotive, we would have the samestopping distance because the weight does not appear in our solution.

3.18 EaD

Cdm

v2 gO| :

3.20 .s/minD 0:3436

3.22 viD 5:297 ft=s



11/41



3.26 viD 32:68106 m=s

3.28 vmaxD gr

m

k

3.30 Spring compression D 0:7367 ft3.32 kD 5:084104 lb=ft

3.34 maxD 0:1675 m.Fs/maxD 586:1 N


3.38 xstopD 0:1597 m

3.40 kD 5:930 lb=ft3.42 kD 9034 kg=s2

3.44 v.0/ Ds

2k

m

x20C 2L0

L

qx20C L2




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 releaseD mg

cos ; .FOA/after releaseD mg cos ; .% change inFOA/D30D 25:00%

3.58 jEaj D 3:620 gjFLj D 43;060 N

3.60 Rx C km

x

1 ru

px2 Cy2

!D 0;

Ry C km

y

1 rup

x2 C y2

!D 0

3.62 Rr r P2 C km

.r ru/ g cos D 0 and r RC 2Pr PC g sin D 0

3.64 tsD 0:7466 s

3.66 yD mCd

tan 0

eCdx=m 1

m

2g

2C2d

v20cos2 0

eCdx=m 1

2



12/41


3.68

aD 0 D a;

aRD g.sC tan /

1 stan D 120:1 ft=s2;

FD 0;FD smg

cos ssin D .83:16 ft=s2/ m;

ND mg

cos ssin D .92:40 ft=s2

/ m;

vmaxDp

g

s sC tan 1 stan

D 363:4 ft=s D 247:8 mph

3.70 FD 5971 lb

3.72 .!c/maxD 2:733 rad=s

3.74 (a) EaD .318:8103 OurC 603:0Ou / m=s2(b) PD 5:898106 N and RD 11:33103 N

3.76 !

D9:939 rad=s

3.78 BD 41:81


3.82 Rr r P2 C G mer2

D 0 and r RC 2 Pr PD 0

3.84 minDr

5mgr

k

vBDp

2gr


3.88 hmaxD 0:3618 ft

3.90 RxL2 y2C Ryxy C x Px2

L2 y2C Py2L2 x2C 2xy Px PyL2 x2 y2 C gx

qL2 x2 y2 D 0;

RyL2 x2C Rxxy C y Px2

L2 y2C Py2L2 x2C 2xy Px PyL2 x2 y2 C gy

qL2 x2 y2 D 0

3.92 R P2 sin cos C .g=L/ sin D 0 and Rsin C 2 P Pcos D 0

3.94 Rr r P2 D 0 and r RC 2 Pr PD 0

3.96 M D 2m!20rq

r2 r20

jvr j D !0q

d2 C 2d r0

vD !0q

2d2 C 4d r0 C r20




13/41


3.100 D tan1 s sin1

sv20

g

q1 C 2s

D 15:59


3.104 RrB=AD G

mA C mBr2

3.106 EaAD 16:10O| ft=s2

3.108 EaAD mA C mB 4mP

mA C mBgO| :

3.110 EaD 4:414 O{m=s2:

3.112 NABD 675:7 N and jEaAj D j EaB j D 3:355 m=s2

3.114 EaD 226:1 O{ft=s2

and EaBDE0:3.116 jEaAj D 4:075 m=s2 up the incline; jEaB j D 2:037 m=s2 downward; TD 38:86 N

3.118 EaAD 4:204 O{m=s2; EaBD 5:606O|m=s2; and jFbj D 21:02 N in compression

3.120 WmaxD 2mg

3.122 vimpactD 3:691 m=s

3.124 d 2k.mA C mB /gk

3.126 NBD 5922 sin N.NB /maxD 12mAd!2 D 5922 N

3.128 RD g2R

2 d

2

2R2

sin d

R2

qR2 14d2 cos

3.130 TD 239:6 N

3.132 dBmax D 2 ftvBmax D 8:579 ft=s

3.134 dBmax D 1:840 ftvBmax D 4:787 ft=s

3.136 .mA C mB / RxBC mAL Rcos mAL P2 sin D 0;mB RxB cot mAL Rsin mAL P2 cos D mAg

3.138 PD 748:4 N

3.140 dD 3:162103 ft



14/41


3.142 tcontactD 96;140 s D 26:71 h

3.144 Number of rotations D 0:7776

3.146 FrestraintD 1158 N

3.148 v0D

4:428 ft=s

3.150 yD

tan C mgv0 cos

x C m

2g

2 ln

1 x

mv0cos

3.152 EaAD 0:2688O|m=s2 and TD 1209 N:

3.154 tssD v0kg

and dssD v202kg

3.156 EaAD .2:685 O{ 2:139O| / m=s2:


Chapter 4

4.2 Concept Problem

4.4 .U1-2/ND 1810 J and .U1-2/Fg D 1570 J;j.U1-2/Nj j.U1-2/Fg j becauseN mgdue to the fact that the man is accelerating.

4.6 U1-2D 360:7103 ftlb

4.8 v2D dp

k= m

4.10 U1-2D 117:2 kJ

4.12 U1-2D 288:5 ftlb

4.14 .U1-2/frictionD 1870 J

4.16 dD 25:22 m


4.20 v2D 35:62 ft=s


4.24 Energy lost to permanent deformation D 4:9841011 J

4.26 v1D 7:093 ft=s


4.30 D 1:293105 lb=ft3



15/41


4.32 kD 273:9103 N=m

4.34 v0D 6:400 ft=s

4.36 kD 44:81 lb=ft

4.38 .U1-2/engineD

464:4

103 ft

lb

4.40 .Fp/avgD 1471 N

4.42 RD 4425 ft

4.44 v2D 6:348 m=s

4.46 v2D 18:52 m=s

4.48 kD 4316 N=m

4.50 VD k qAqBr

4.52 vAD 2336 m=s

4.54 .vB/max D 0.vB /maxD 0:3822 m=s

4.56 minDr

5mgr

k and vBD

p2gr

4.58 (a) v2D 1626 ft=s(b) v2D 1522 ft=s(c) The form of the potential energy allows us to interpret the work done by the expanding gas as the

work of the forceP0A along an effective distances0log s2=s0. The forceP0Ais the same in both

cases. However, the decrease ins0 in Part (b) is such that the effective distance over which the force

acts is smaller, thus causing v2 to be smaller in Part (b).

4.60 hmaxD 0:1055 m

4.62 (a) VcD 12k2 144

EvbottomD 5:161O|ft=s:(b) jEajmaxD 1:599g

4.64 kD 0:8343

4.66 VD 4"

rij

12

rij

6#

4.68 vmaxD gr

m

k

4.70 L0D 1:060 m

4.72 vA2D 1:093 m=s and vB2D 3:278 m=s

4.74 dBmax D 2:500 ft



16/41


4.76 vB2D 5:648 m=s

4.78 vB2D 33:23 ft=s

4.80 vA2D 13:77 ft=s

4.82 WD

2mg

4.84 The work done by the tension in the cord on A is equal and opposite to the corresponding work done by the

tension in the cord onB . Thus the net work done by the tension in the cord on the system is equal to zero.

EvA2D 0:1424 O{m=s and EvB2D 0:4273 O{m=s @

4.86 v2Dp

Pl.1 sin 0/=m

4.88 distance ofB from the floor D 0 and vmaxD 9:422 ft=s

4.90 hmaxD 21:85 ft

4.92 vA2D 7:001 m=s and vB2D 14:00 m=s

4.94 vD sr

g

l


4.98 PavgD 375:0 ftlb=s D 0:6818 hp


4.102 D 6:321

4.104 v2D 13:62 mph

4.106 PD 162:4 hp

4.108 vBD 2:128 m=s

4.110 PiD 1:247 hp

4.112 vD 9:895 ft=s

4.114 vB2D 10:84 m=s

4.116 vA2D 3:631 m=s


4.120 .U1-2/dD 0:1585 ftlb

4.122 kD 9034 N=m

4.124 v2D 73:55 ft=s



17/41


4.126 vmaxD 9:674 ft=s and is achieved when the distance ofB from the ground is zero.

4.128 EbD 1290 C

4.130 PD(

4637 ftlb=s; for case (a);343:5 ftlb=s; for case (b):

Chapter 5

5.2 Concept Problem

5.4 Concept Problem

5.6 Concept Problem

5.8 dD 28:461024 mdD 4:1521015 m

dD 1:423106 m

5.10

Z t2t1

EF dtD 93:17O|lb s

5.12

Z t2t1

EF dt D 2:635 lbs

EFavg

D 2396 lb5.14 FavgD 2589 lb

5.16 vtD2 sD 4:293 ft=s

5.18 D 1:610 s

5.20EFb avgD 40;640 lb

5.22 vtD2:5 s

D 9:045 m=s

5.24 Impulse imparted to the ball by kicker D 2:531O|lb sEFkavgD 316:4O|lb

5.26 t2D 10:57 sdD 1279 ft

5.28EFcavgD .219:3 O{ C 192:4O| / N

5.30 (a)EFavg D 8081 N

(b) sD 0:5148



18/41


5.32 Impulse provided by floor D .14:93 Ns/O| :

EaavgD

5420 m=s2O|D 552:5gO| :

5.34 Impulse of the rope D 16:30 O{lb s

5.36 EvBD 5:094106 O{ft=s:

5.38 number of worker bees D 9:062106

slowdown D 48:11 ft=s D 32:80 mph

5.40 EvAD 8:455 O{m=s and EvBD 7:455 O{m=s

5.42 EvAtD1:5 s

D 0:7905 O{m=s and EvBtD1:5 s

D 2:371 O{m=s

5.44 vP2D 1:714 ft=s

5.46 vP3D 1:852 ft=s

5.48 dD 4:951 m

5.50 .EvP/finalD .1:516 ft=s/ O{:

5.52 dD 1:017 ft

5.54 EvAD 11:17 O{m=s and EvBD 3:624 O{m=s

5.56 (a) EvA2D .8:826 O{ 10:76O| / ft=s and EvB2D 13:24 O{ft=s:

5.58 EvA2D .10:29 ft=s/ O{ and EvB2D .51:46 ft=s/ O{:

5.60 EvBD mAcos

p2gL

pcos cos 0p

.mA mB /2 mA.mA 3mB / cos2 O{;

vADp

2gLp

cos cos 0s

.mA mB /2 mA.mA 2mB / cos2

.mA mB /2 mA.mA 3mB / cos2 5.62 (a) vARD 0:3750 ft=s

(b) EFAD .447:2 lb/ C .10:48 lb=s/tOuRA C .139:8 lb/OuA ;EFBD .174:7 lb/ C .10:48 lb=s/t OuRB C .139:8 lb/OuB

(c) EvGD .1:170 ft=s/OuG C .1:400 ft=s/Ok Ea5.64 Concept Problem

5.66 EvCAD EvCBD .38:09 mph/ O{:

5.68 D 0:041 75 ft

5.70 vBD 209:7 m=s



19/41


5.72 vCAD 2:346 ft=s and vCBD 4:154 ft=s

5.74 eD mA=mB

5.76 0:7280 e 0:7616

5.78 tAD

0:5184 s and tBD

0:6600 s

5.80 dD 2:887 m

5.82 WmaxD 818:6 lb

5.84 vCAx

D 0; vCBx

D 0; vCCx

D 6 ft=s

5.86 Treat each impact as only involving two balls. Because the COReD 1and the masses are identical we see fromthe solution to Problem 5.74 that ball 1 will come to a complete stop after impacting with ball 2. We also see

that ball2 will have a post impact velocity identical to the pre impact velocity of ball 1. Each ball in the train

is tangent to the next so it will not appear to move at all during its impact with the next ball. Ball4 impacts

ball 5 which is free to move. Ball 5 will have a post impact velocity equal to the pre impact velocity of ball 1.

The work-energy principle tells us that ball5 will stop moving when it has reached the initial height ball 1 wasreleased from. Finally, since the lengths of the pendulums are identical the maximum swing angle of ball5 is

equal to the initial release angle of ball1.

5.88 Balls3,4, and5 will swing up as a single unit until reaching the height that balls 1,2, and3 were released from

while balls1and 2 will hang motionless.

5.90 EvCAD 53:19O|mph D 78:01O|ft=s and EvC

BD .50 O{ C 53:19O| / mph D .73:33 O{ C 78:01O| / ft=s

ErAD 135:0O|ft and ErBD .174:2 O{ C 185:3O| / ft:

5.92 Given the assumption that the masses are not identical, it is not possible to have a moving ballA hit a stationary

ballB so thatA stops right after the impact.

5.94 EvC

BD 0:7071O|m=s: @ 45

5.96 vBD 0:6886 ft=s

5.98 D tan1.cot /

5.100 EvCAD .8:309 O{ C 18:53O| / m=s and EvCBD .8:309 O{ 6:202O| / m=s:

5.102 dD 5:266 ft

5.104 vCBD 3p

2gh

5.106 hiD e2i

h0

tiD

1 ei 1 C e1 e

s2h0

g

tstopD 13:27 s


5.110 EhOD 513:1103Ok slugft2=s:



20/41


5.112 EhO .t1/ D0:009705 O{ C 0:08734O| 0:05823Ok slugft2=s

EhO .t2/ D0:01941 O{ 0:01941O|C 0:01941Ok slugft2=s


5.116EhBAD 120:0 O{ 84:00O| 42:00Ok kgm=s

5.118 The angular impulse provided to the pendulum bob between and is equal to zero.

5.120 EvAnD .13:33 O{ C 6:667O| / m=s

5.122 EhOD

142:8103 kgm2=s3t2Ok:5.124 Since

PEhEC EvE mPEvPD mPgtvE vP.0/ cos Ok and EMED mPgtvE vP.0/ cos Ok, it is true thatEMEDPEhEC EvE mPEvP.

5.126 RD gL

sin

5.128 vimpactD 0:4714 m=s

5.130 Rr r P2 D 0 and mr2 RC 2m r Pr PD M

5.132 kD 26:40 N=m

5.134 v2D 18:39 m=s

5.136 Area.P1OP2/=Area.P3OP4/ D 1

5.138 e

D s1 C 2E2

.GmB /2

(a) E < 0 ) 1 C 2E2

.GmB /2

< 1 ) e < 1 ) elliptical orbit.

ED 0 ) 1 C 2E2

.GmB /2D 1 ) eD 1 ) parabolic trajectory.

E > 0 ) 1 C 2E2

.GmB /2

> 1 ) e > 1 ) hyperbolic orbit.

(b) vcDs

GmBrc

; which agrees with Eq. (5.82)

5.140 vD 3232 m=s

5.142 GmeD 3:9711014 m3=s2

5.144 rgD 1:386108 ft

hgD 1:177108 ft

vcD 1:008104 ft=s

5.146 vD 3240 m=s D 11;660 km=h



21/41


5.148 vPD 8064 ft=s D 5498 mph and vAD 4850 ft=s D 3307 mphtD 18;930 s D 5:258 h



5.154 (a) v1Ds

rPv2P 2GmB

rP

(b) rPv2P> 2GmB



5.160 QnzD 3:662 ft3=s

dD 2s

Qnzvw

D 0:2678 ft

5.162 RD 14pAd2AC4 Q2

g

1

d2A 1

d2B

D 2097 lb

5.164 voD 269:6 m=s

5.166 sD 0:07578

5.168 Pmf1D 12Pmf.1 C cos / and Pmf 2D 12Pmf.1 cos /

5.170 vxD 26:92 ft=s

5.172 FRD 270:4 kNMO

DFRh

D20:28

106 N

m

5.174 FD 1:099 lb


5.178 ymaxD 30;470 ft

5.180

EFavg

D 1459 lb

5.182 .EvP/finalD .1:672 m=s/ O{;

5.184 D 14:26

5.186 MD 12m!20r20

e2!0t e2!0t 5.188 vD 2:919103 ft=s D 1990 mph

5.190 TjD 4:1601010 JTeD 2:1611011 JVD 5:1531011 J

5.192 .L y/. Ry g/ 12Py2 D 0



22/41


Chapter 6

6.2 Concept Problem

6.4 Concept Problem

6.6

E!B

D532:2

Ok rad=s and

EB

D187:8

Ok rad=s2

6.8 !BD RARB

!AD 600:0 rad=s

6.10 tstopD 87:94 sptD 806:1 rev

6.12 .!OA/maxD 2:476 rad=s

6.14 j E!ramj D 0

6.16

EvC

D 71:88 O{ 287:6O|C 198:6Ok cm=sEaCD

7244 O{ C 4061O|C 3260Ok cm=s2

6.18 EvCD43:13 O{ C 172:6O| 119:2Ok cm=s

EaCD

2585 O{ C 1554O|C 1110Ok cm=s2

6.20 EvDD

498:0O|C 211:7Ok cm=sEaDD

1:331104 O{ 5341O|C 1:237104Ok cm=s2

6.22 E!sD 10:00Ok rad=s

EsD 1:333Ok rad=s2

6.24 EvGD .0:1495Our 3:379Ou / m=sEaGD .20:41Our 3:660Ou / m=s2

6.26 EvCD .0:2917 O{ C 0:7361O| / ft=sEaCD .0:2045 O{ C 0:08102O| / ft=s2

6.28 EvCD .0:1791 O{ C 0:4920O| / m=sEaCD .2:576 O{ C 0:9377O| / m=s2

6.30 EvHD rAL

rC!A. sin O{ C cos O| /

EaHD rAL

rC

Asin

rArC

!2Acos

O{ C

Acos rArC

!2Asin O|

:



23/41


6.32 !sD 72:72106 rad=s

6.34 E!CD 40:00Ok rad=s:

ECD 5:200Ok rad=s2:

6.36 Chain ring/sprocket combination: C1/S3.

!wD !sD RcRs

!cD 126:4 rpm

6.38 EvBD 146:0O|ft=s

6.40 EvBD .9:881 O{ C 3:666O| / m=s

6.42 E!PD 9:600Ok rad=sEvOD 2:000 O{ft=s


6.46 EvAD r!d.O{ C O| /;

EvBD 2r !d O| :

EvCD r!d.O{ O| /:

6.48 E!PD 1:600Ok rad=sEvCD 5:333 O{m=s

6.50 EvCD .50:00 O{ C 50:00O| / ft=s

E!ABD 20:00Ok rad=s

6.52 E!AD 7:000Ok rad=s

6.54 E!ADD 2:745Ok rad=s

6.56 E!ADD 1:373Ok rad=s

EvC

D.

0:2745

O{

1:775

O| / m=s

6.58 E!ABD 2:208Ok rad=sEvAD 2:274O|ft=s

6.60 EvAD 11:67O| ft=s

6.62 EvBD 23R!OAsin O{:



24/41


6.64 E!ABD 0:06805Ok rad=s

6.66 PD 74:76 rad=sEvDD .29:33 O{ C 73:73O| / ft=s

6.68 the cable isunwindingat1:364 m=s.

6.70 E!CDD 0:8824Ok rad=s

E!BCD 0:3000Ok rad=sEvCD .1:200 O{ C 0:3000O| / ft=s

6.72 EvBD 35:00 O{ft=sEvBD 35:00 O{ft=s

6.74

E!AB

D3:333Ok rad=s

EvCD 19:36O|ft=s

6.76 E!sD 3:333Ok rad=sEvOD 5:000 O{rad=s

6.78 E!ABDE0 and E!BCD 7:667Ok rad=s

6.80 E!ABD R Psin. C /Hsin. /

Ok

E!BCD R Psin.C /L sin. /

Ok

6.82 E!CDD 12:00Ok rad=sEvbarD 3:000 O{m=s

6.84 E!cD 877:1Ok rad=s

6.86 EvBD"

R Pcos RP.H R cos / sin p

L2 .H R cos /2

# O|

6.88 E!BCD R

Psin

pL2 R2 cos2 Ok

EvCD

R!ABcos CR2!ABsin cos p

L2 R2 cos2

!O|

6.90 EaCD .12:86 O{ C 89:53O| / ft=s2

6.92 PD 0 and RD 13:50 rad=s2



25/41


6.94 EADD 4:620Ok rad=s2

EBDD 5:650Ok rad=s2

6.96 EADD 1:590Ok rad=s2

EaB

D.0:9707

O{

1:674

O| / m=s2

6.98 E!CD 378:6Ok rad=s

ECD 307;400Ok rad=s2

6.100 EaCD 1024Our ft=s2

EaPD 3482Our ft=s2

6.102 E!ABD 0:4577Ok rad=s and EABD 0:2582Ok rad=s2

6.104 EaBD 23R

OAsin C !2OAcos

O{:

6.106 EABD 27:08Ok rad=s2 and EaBD 266:6 O{ft=s2

6.108 EaBD 2R

W! 2Wp

5

!O{

6.110 EaBD R

32

h84 C

p2

WC

32 9p

2

!2W

iO{

6.112 EABD 0:2057Ok rad=s2

6.114 E

aAD

RW

tan R2!2

W

L sin3 !O|

6.116 EBCD 1639Ok rad=s2

6.118 EBCD HR P2R2 H2 sin H2 C R2 2HR cos 2 Ok

6.120 EaQD 82:03Ourft=s2

6.122 EOCD 0:3714Ok rad=s2

EaPD .15:79OurC 6:539Ou / ft=s2

6.124 EgateD 4:500Ok rad=s2

6.126 EBCD 0:01749Ok rad=s2 and ECDD 0:004516Ok rad=s2

EaCD .0:003164 O{ 0:03876O| / ft=s2

6.132 EABD 87:67Ok rad=s2 and EBCD 9:974Ok rad=s2

6.134 EaPDRs s!20 O{ C s0 C 2Ps!0O|



26/41


6.136 EaPDRs s!2D

O{

2Ps!DC d!2D

O|

EaPCoriolisD 2Ps!DO|6.138 EvDD .Ps d!1/ O{ C s .!1 !2/O|

6.140 E!CDD 22:43Ok rad=s;

ECDD 202:0Ok rad=s2:

6.142 E!ABD 4:561Ok rad=s;

EABD 18:43Ok rad=s2:

6.144 EaCD .30:00 O{ C 51:80O| / ft=s2

6.146

EvP

D.

11:54

O{B

C33:30

O|B / ft=s

EaPD .64:43 O{B 16:43O|B / ft=s2

6.148 EvDD v0 d!1 R.!1 !2/ O{ C R.!1 !2/O|

6.150 (a)E!CDD 2:513Ok rad=s and EvbarD 0:3016OIm=s

(b)ECDD 37:90Ok rad=s2 and EabarD 4:548OIm=s2

6.152 (a)E!CDD 6:283Ok rad=s and EvbarD 0:7540OIm=s

(b)ECDDE0 and EabarDE0


6.156 EvCD

7:500 O{ C 7:506O| ft=sEaCD

105:1 O{ 161:0O| ft=s2

6.158 PdABD 1:118 ft=sRdABD 2:012 ft=s2

6.160 EvAD R!s O{ C `

!sCP

R

!O|

EaAD

`

R2

P C R!s2 C R P!s

O{ C

1

R

` R CP2

R!2sC ` P!s

O|

6.162 vCD 0:1601 ft=s

6.164 EaDDhRs d1 s .!1 !2/2

iO{ C

h2 .!1 !2/ Ps C s .1 2/ d!21

iO|



27/41


6.166 EvDD .1:340 O{ C 5:000O| / ft=s

6.168 EvAD 0:8750O|m=s;

EaAD 0:3750O| m=s2;

EvD

D 1:750

O| m=s

EaDD 0:7500O| m=s2:

6.170 E!barD 14:29Ok rad=s;

6.172 E!ACD 2:442Ok rad=s

E!CED 2:442Ok rad=sEvED .15:00 O{ 42:01O| / ft=s

6.174 EvDD .vCC `!OAsin / O{ C .d!OA `!Ccos /O| `!Csin Ok

EaDD

2`!C!OAcos C `OAsin d!2OA O{

Ch

dOA C 2vC!OA C `

!2CC !2OA

sin iO| `!2Ccos Ok

6.176 EvCD .21:50 O{ C 7:506O| / ft=s D

21:50OIC 7:506OJ

ft=s

EaCD .105:1 O{ C 165:7O| / ft=s2 D

105:1OIC 165:7OJ

ft=s2

Chapter 7

7.2 tstopD

1:863 s

dstopD 16:77 ft

7.4 hmaxD 4:054 ft and aGxD 9:800 ft=s2

7.6 a0D 15:70 ft=s2

7.8 Concept Problem

7.10 a0D 66:80 ft=s2

7.12 P

D20:70 lb

NAD 152:1 lbNBD 147:9 lb

7.14 EaCD 2:875O|m=s2:

7.16 L D 5:604105 N

7.18 aGxD 2:147 ft=s2 and PD 49:83 lb



28/41


7.20 D 17:00 and D 17:00

7.22 ND 1:492105 ND 1:909 m

7.24 NfD 783:7 lbNrD 783:2 lbNBD 2233 lbHD 673:7 lbFD 576:9 lb.s/minD 0:7362

7.26 D tan1

aAg

Dtan1 aA

g

7.28 RxD 0:1115t2 N=s2;RyD 0:010 12 N;M D 3:110105 Nm

7.30 tfD 1:411 sn D 7:485 rev.cw/

7.32 EdD 10:06Ok rad=s2:n

D16:01 rev.cw/

7.34 !fD 31:42 rad=s D 300:0 rpm

7.36 M D 3208 ftlb

7.38 EFAD 9:197 .O{ C O| / NEFBD 9:197 .O{ O| / N

tsD 4:096 s

7.40 kD

1:443

EFAD .18:83 O{ C 26:19O| / N and EFBD .18:83 O{ 11:48O| / N

7.42 OrD 16:85 lb and OD 5:000 lb

7.44 jmaxj Dp

3g

L for` D 16

3

p3

L

7.46 PD 1:503 rad=s



29/41


7.48 DxD3mcgh

2 .d `/ mcC .2d L/ mp

h

w2 C 4h2 C 12.d `/2i

mcC 4

3d2 3dL C L2mpDyD g

L2m2pC

4h2 C w2m2cC 4h2 C w2 C 4 L2 3`L C 3`2mcmp4

3d2 3dL C L2mpC h4h2 C w2 C 12.d `/2imcp

Dc

D6g 2 .` d / mcC .L 2 d / mp

4

3d2 3dL C L2mpC h4h2 C w2 C 12.d `/2imc7.50 FfD

3

2mg cos

3

2sin 1

ND 14

mg .1 3 sin /2

.s/minD 1

7.52 TD 6:925Ok rad=s2

EFOD .1323 O{ C 48:47O| / N

7.54 EFAD .35:07 O{ C 48:22O| / N and EbD 6:679Ok rad=s2

7.56 EFBD .560:1 O{ C 558:6O| / N and MBD 423:8Ok Nm

7.58 FD k2G

k2GC r2mg sin ; ND mg cos ; and bD

r g sin

k2GC r2

.s/minD k2G

k2GC r2 tan

7.60 EaGD kg O{ and EbD grk

k2G

Ok:

7.62

7.64 TCDD 147:2 N; and EaGD 4:905O| m=s2

7.66 EFsD

R2

R2 C k2G

!PO{ C .mg P /O| and EsD

PR

m.R2 C k2G/Ok:

7.68 Notice that the sign of the angular accelerationc contradicts our assumption that the crate tips and slips, since

the only physically meaningful tipping would be in the clockwise direction. Therefore, this solution shows that

the crate cannot tip and slip.

7.70 sepD

3D 60:00

7.72 FD

12:13 lbtf

newD 6:845 s

7.74 EABD 6mABg cos

L

4mABC 18mB sin2 Ok

EaBD 6mABg sin cos

4mABC 18mB sin2 O{

EwD 6mABg sin cos

R

4mABC 18mBsin2 Ok



30/41


7.76 EABD 6g.2mA C mAB / cos

2L

6mAcos2 C 2mABC 9mB sin2

OkEaBD

6g.2mA C mAB/ sin cos 2

6mAcos2 C 2mABC 9mB sin2

O{EwD

6g.2mA C mAB / sin cos 2R 6m

Acos2

C2m

ABC9m

Bsin2 Ok

7.78 IGD mR2

7.80 EbD 6g sin

L

1 C 3 sin2 Ok; EFAD mg

1 C 3 sin2 O| ; and EaAD 3g sin cos

1 C 3 sin2 O{:



7.86 m

1 C k

2G

r2

!Rx C kxD mg sin C kL0

7.88 D cos1 23 D 48:197.90 EFdue to bowlD mg cos OurC 27mg sin Ou

EbD 5g sin

7Ok

EaGD 57g sin Ou

7.927

5.R / R C g sin D 0

7.94 h D 75 r


7.98 EaAD 60g

181O{ and EbD

45g

181ROk


7.102 TCDD 2061 lb; and EaED 5:655O|ft=s2

7.104 EaED .13:19 O{ 25:90O| / ft=s2

AxD 436:6 lb and AyD 255:6 lbCx

D628:8 lb and Cy

D353:5 lb

7.106 AxD 228:7 lb; AyD 1330 lb; CxD 336:2 lb; and CyD 1940 lb

EaED .6:996 O{ C 7:059O| / ft=s2

7.108 PmaxD 1595 lb

7.110 EaGD kg O{ and EbD krg

k2G

Ok



31/41


7.112 D tan1

v2cgR

!

7.114 EACD 1:459Ok rad=s2 and ECED 1:459Ok rad=s2:


7.118

ND

2k2GC 2R2 C 2

cos 2 cos. 2/ R sin.2 / 3 sin 2

k2GC R2 C 2 2R sin. / W

C

R cos.2 2/ 3R C 2 k2GC R2 C 2 sin. /P22g

k2GC R2 C 2 2R sin. /

WFD k

2Gsin C cos. /.R cos C sin /


k2GC 2 R sin. /P2 cos. /

g


RD g .R sin cos / RP2 cos. /

k2GC R2 C 2 2R sin. /

7.120 L .mABC 2mC/P2 sin C .2dC h/ mCP2 sin C 2 .mABC mC/RxAD L .mABC 2mC/Rcos C .2dC h/ mCRcos

4 .mABC 2mC/ gL sin C .mABC 2mC/ L2 P2 sin 2C mC.2dC h/ L

h3 sin . / C sin .C /

iP2 C 8ID R 4mCL RxAcos

Ch

mABC 6mC .mABC 2mC/ cos 2i

L2 RC 2.2dC h/ mCL

cos cos C 2 sin sin RD 0

2.2dC h/ mCh

g sin L P2 sin . / RxAcos C L Rcos . /iCh

4IEC .2dC h/2 mCiRD 0

7.122 PmaxD sg"

mdC mc

1 C R2

k2G

!#; EaCD sg

1 C R

2

k2G

!O{; and EaGD sg O{

7.124 PD 7:588 N and .s/minD 0:04620

Chapter 8

8.2 Concept Problem

8.4 Concept Problem

8.6 TD 0:011 21 J

8.8 TD 1719 Jh D 17:52 m

8.10 !b2D 3:693 rad=s

8.12 minD 53:13

8.14 D 8781 rev

8.16 kD 472:0 lb=ft



32/41


8.18 1D 33:02

8.20 !d2Ds

.2Md=R/ kd2m.k2GC R2/

dsD 2M

kR

8.22 MD 653:8 Nm

8.24 vcD 3:040 m=s

8.26 PD 2:344 N

8.28 dD 0:4990 m and dD 1:102 m

8.30 LfD 34:13 ft

8.32 !s2D 2:640 rad=s

8.34 .U1-2/ncD 2042 ftlb

8.36 vG2D 3:951 ft=s

8.38 vpersonDp

2gH.1 C cos / and vpersonD0

D 10:85 m=s

8.40 TD 15:85 ftlb

8.42 vcD 6:291 ft=s

8.44 vA2D 2:709 ft=s and vB2D 5:417 ft=s

8.46 !s2D 1:094 rad=s

8.48 TD 28:88 ftlb

8.50 vQ2D 20:45 ft=s

8.52 nminD 26vpersonD 26:54 ft=s

8.54

E!d

D sd.2M=R/ C 2mg sin d k

m.R2

C k2G/

Ok: @

MD 11:25 ftlb

8.56 D 1:031 ft

8.58 mCD 26:96 kg

8.60 vS2D 16:00 ft=s



33/41


8.62 WPD 799:5 lb

8.64 vCD 5:019 m=s and E!DD .44:61 rad=s/Ok

8.66 vmaxD 6:389 ft=s

8.68 .vB /finalD 1:279 m=s8.70 (a) aGxD 12L. Rcos P2 sin / and aGyD 12L. Rsin CP2 cos /

FD 12mL. Rcos P2 sin / and ND mg

1 L2g

. Rsin CP2 cos /

(b)P2 D 3gL

.1 cos / and RD 3g2L

sin

(c) FD 34mg sin .3 cos 2/ and ND 14mg.1 3 cos /2.s/maxD 0:3706 and slideD 35:10





8.80 EpwD .83:85 lbs/ O{EhCD .310:3 ftlbs/Ok

8.82 j E!r2 j D 77:59 rpm

8.84 PD 25:00 N8.86 kGD 5:218 ft

8.88EhAABD 13 mABR2!ABOkD .1:725 kgm2=s/Ok;EhABCD mBC!ABR2OkD .7:200 kgm2=s/Ok;EhDCDD 13mCDHR!ABOkD .7:750 kgm2=s/Ok;

8.90

EhCW

D .91:88106 kgm2=s/Ok; @EhOWD .669:4106 kgm2=s/Ok; @

8.92 EpAB D WABvA

2g cos D 3:227 lbs

EhGD WABLvA12g cos

OkD .2:420 ftlbs/Ok;

8.94 tD 16:98 s



34/41


8.96 j E!s2j D 2:470 rad=s

8.98 (a) Collar modeled as a particle: vimpactD 0:9905 ft=s,(b) Collar modeled as a rigid body: vimpactD 0:9432 ft=s,

8.100 j!Ajafter slip stopsD j!B jafter slip stopsD 18:70 rad=s

8.102 vC2D 25:19 m=s

8.104 vfD 18:34 ft=s and trD 2:502 s


8.108 vGtD3 s

D 7:563 m=s.s/minD 0:3951

8.110 !fD 30:13 rpm

8.112 dD 4r2b.2r b/2

8.114 (a) EvAD R!s O{ C `

!sCP

R

O|

(b) 12

IOC 2mR2

!20D m

R2!2sC `2

!sC

PR

2C 12IO!2s

(c)

IOC 2mR2

!0D

IOC 2m.`2 C R2/

!sC 2m`2

RP

(d) P D R!0 and !sD IOC 2m.R2 `2/IOC 2m.R2 C `2/

!0

`!sD0

D rIO

2mCR2

(e) `.t/ D R!0t

!s.t / D IOC 2mR2.1 !20 t2/IOC 2mR2.1 C !20 t2/

!0

8.116 EvCPD .5:682 ft=s/ O{ and EvCQD .6:198 ft=s/ O{:

8.118 v0D 1:397 ft=s

8.120 v0D 1326 ft=s

8.122 dD 2:094 ft

vCb D 226:1 ft=s and E!BD .26:60 rad=s/Ok:

8.124 sweptD 220:0

8.126 EvCGD .1:310 m=s/ O{ and E!CAD .3:434 rad=s/Ok:

8.128 dD 5:525 m

8.130 E!CAD .1:796 rad=s/Ok and E!CBD .0:4047 rad=s/Ok: @ 12



35/41



8.134 TD 0:025 70 ftlb

8.136 vB2D 0:4069 ft=s

8.138

EhD

D.14:90 ft

lb

s/

Ok:

8.140 j!Ajafter slip stopsD j!B jafter slip stopsD j!Cjafter slip stopsD 16:41 rad=s

8.142 .v0/maxD 2:176 m=s

8.144 vCGyD 1:431 ft=s and E!Cp D .0:3211 rad=s/Ok:

Chapter 9

9.2 IOD mgL2

42 or IGD

mgL2

42 mL2

9.4 D 2s

7L

6g

9.6 !nDs

Gr4

LR4t

9.8 ` Dr

IGm

9.10 fD 1:771 Hz

9.12 fD !n2

D d4

rg

mD 0:5663 Hz

9.14 D 2d

hrm

k

9.16 D 2vuuut 13

h3 C d3C md2k 12g

h2

9.18 m Ry C kyD 0m RC kD 0

9.20 RxGC17k

6mxGD 0

9.22 moving the disk upn D 4:778 turns

9.24 RxGC 8k

3mxGD 0

D 2:221 s

9.26 !nDs

2g

3.R r/

D 2s

3.R r/2g



36/41


9.28 D 2s

L

2g

9.30 D 2s

.3 2/R3g

9.32 x.t/ D A sin !ntC Bcos !ntC F0=k

1

.!0=!n/

2cos !0t

9.34 muD 0:01 kg ) jymj Dq

2:190106 cos.4:369t/ C 1:888105 m;

muD 0:1 kg ) jymj Dq

1:100105 cos.4:369t/ C 1:112105 m;

muD 1 kg ) jymj Dq

9:798104 cos.4:369t/ C 1:014103 m9.36 ampD 0:001 758 rad


9.40 x.t/ D

0:000 501 3 sin 10tC 0:1000 cos 10t 2:506105 sin 200t

m

9.42 m Ry C 2k 1 L0L yD F0sin !0ty.t/ D F0=keq

1 .!0=!n/2

sin !0t !0!n

sin !nt

;

where keqD 2k

1 L0L

and !nD

s2k

m

1 L0

L

9.4423mBC 12mA

RxA C 2kxAD 12F0sin !0t

.xA/ampD F0

4k 43mBC mA

!20

9.46 !nDs

3gEI

.Wu C We/ d3D 505:7 rad=s

fD !n2

D 80:48 Hz

MFDp

.Wu C We/ dWuR

D

!p=!n2

1 !p=!n2D 0:20719.48 Concept Problem


9.52 no peak inMFfor p1=29.54 Ry C 2!n Py C !2nyD

mu"!2r

m sin !r t; wherec= m D 2!n and k= m D !2n

9.56 F0D 0:015 84 Njyj D 0:000 09410 m

9.58 y.t/ D

0:5te28:28t

m



37/41


9.60 D 110

D 0:1 forMFD 5

D 120

D 0:05 forMFD 10

9.62 mL RC 0:18cL PC .mg C 0:72kL/D 0

!dDs

g

LC 0:72k

m 0:0081c

2

m2

9.64 yAD !0E

2k

k mA!20

C c2!202!d

hk mA!20

2 C c2!20i e.c=2mA/t sin !dt

C mAc!30E

k mA!202 C c2!20 e

.c=2mA/t cos !dt

C mA!20E

k mA!202 C c2!20

hk mA!20

sin !0t c!0cos !0t

iC Esin !0t

D 0:0002665 cos.94:25t/ C e2:500t 0:0002665 cos.5:204t/ C 0:0002121 sin.5:204t/ 4:642106 sin.94:25t/

9.66 DD 4:373106 m

9.68 mRs C c Ps C ksD mA!20sin !0t; where s.t/ D y.t/ u.t/

y.t/ D"

mA!20

k m!20

k m!20

2 C c2!20 C A#

sin !0t"

mcA!30k m!20

2 C c2!20#

cos !0t

DT Dvuut k2 C c2!20

k m!202 C c2!20

9.70 m Rx C k1k2k1 C k2 xD 0

9.72 kD 5:910%

9.74 .yc/ampD 0:000 454 4 m D 0:4544 mm

9.76 F0D 200:1 Nj Rxj D 282:0 m=s2

9.78 muD 0:01 kg W FtD 14:77 sin.125:7tC 0:8422/ C 17:67 cos.125:7tC 0:8422/ N;muD 0:1 kg W FtD 147:7 sin.125:7tC 0:8422/ C 176:7 cos.125:7tC 0:8422/ N;muD 1 kg W FtD 1477 sin.125:7tC 0:8422/ C 1767 cos.125:7tC 0:8422/ N

9.80 m Ry C 2k

1 L0py2 C L2

!yD 0

9.82 m Ry C 2k

1 L0L

yD 0



38/41


Chapter 10

10.2 EvBD `!1sin O{ C !1cos O|CPcos Ok

EaBD ` P2 cos !21cos P!1sin O{ C ` P!1cos !21 sin O|C ` Rcos Ok

10.4 EvAD 2.` C d /!1cos2

EaAD 2.` C d /!21cos2 O{ C 2.` C d / P!1cos2 O| .` C d /!21cos

10.6 EvED R!armcos O{ C .dC ` cos /!armO| .dC ` cos /!armOkEaED !2arm.dC ` cos / O{ !2armsin .dC ` cos /O|

!2arm

R

hd2 C 2d ` cos C

`2 C R2

cos2

iOk10.8 EvED R!armcos O{ C .dC ` cos /!armO| .dC ` cos /!armOk

EaED hRarm !2arm.dC ` cos /i

cos O{ C .dC ` cos /

arm !2armsin O|

1R

hd

Rarm C d!2arm

C `

Rarm C 2d !2arm

cos C

`2 C R2

!2armcos2

iOk10.10 EvBD

P cos ` Psin O{ C `!1cos O|C ` Pcos CP sin OkEaBD

h`

P2 C !21 cos 2 P Psin i O{ C 2!1 P cos ` Psin O|CP2 P cos ` Psin Ok10.12 E!diskD !b O{ C !dOk

EdiskD P!b O{ !d!bO|C P!dOk

10.14

E!c

D !0

cos

sin O{

EcD cos

sin

0 O{ C !20Ok

10.16 EvBD L!0cos O| L!0cos2 Ok

EaBD L!20 cos2 O{ C L0cos O| L cos cot

!20C 0sin Ok

10.18 EaAD

231:3O|C 107:9Ok

ft=s2

EABD

18:29 O{ C 15:08O|C 34:90Ok

rad=s2

10.20 E!ABD !sO|P OkEABDP!s O{ sO|R Ok

EaGD

L

2R sin C

dC L

2 cos

!2sC

L

2P2 cos

O{ C L

2

P2 sin R cos O|C

dC L2

cos

s L P!ssin

Ok

10.22 E!ABD

0:3139 O{ C 10:37O|C 2:786Ok

rad=s

EABD35:35 O{ C 135:2O|C 161:9Ok

rad=s2



39/41


10.24 EaAD

101:8O|C 47:49Ok

m=s2

10.26 EvAD0:7354O| 0:3429Ok

m=s and EaAD

12:49O| 5:825Ok

m=s2

10.28 EABD6:040O| 8:054Ok

rad=s2

10.30

EaA

D 124:2

O{m=s2

10.32

EABD 11:84 cos O{ C4:187

Z3=2

h26:25 cos C 0:750019:53 cos 2

5:400 cos 3C 0:5625 cos 4 24:55iO| 101:2

Z3=2

h7:438 cos C 0:75001:200 3:600 cos 2C 0:7500 cos 3iOk rad=s2;

whereZD .7:438 C 3:600 cos 0:5625 cos 2 / m2

EaAD 29:08

Z3=2

h31:37 C 25:92 cos 2 C 1:688 cos 2

4:800 cos

1:688 cos 2 6:312 22:63

pZsin

C 0:75000:5625 cos 4C 2p2.0:7500 cos 2:400/pZsin 2i O{m=s2;whereZD .7:438 C 3:600 cos 0:5625 cos 2 / m2

10.34 EvBD .vtC h!a/ O{ C .` C rt / !bO| .` C rt / !aOk

10.36 E!barD s

24g .1 C sin /L sin .5 C 3 sin /O{

10.38 !10D 4554 rad=s

10.40 13LR C

13L cos C 12d

!2s sin 12g cos D 0

10.42 OxD mh CL2 !

2s ;

OyD 0;OD mg;

MOxD 0;MOyD mg

h C L2

C 12mr 2!d!s ;MOD 0

10.44 vA2D 6:294 m=s

10.46 vA2D 6:165 m=s

10.48 MAxD 1

2mR2 PPcos ;

MAyD 1

4 mR

2

P2

cos sin C1

2 mR

2

Pcos PPsin ;MAD

1

2mR2 PPsin 1

2mR2 P PPsin

10.50

MXD 0;AYD

m

8LR2!2s cos sin ;

AZD 0;BYD

m

8LR2!2s cos sin ;

BZD 0;whereX Y Z is attached to the shaft.



40/41


10.52 RxD 0:5236t2 N; RyD 0:021 93 N; RD 0:098 10 N;MxD 0:000 09291 Nm; MyD 0:002219t2 Nm; M D 0:000 06855 Nm

10.54 !sDr

3g

L

10.56 ND 8mg .1 C cos / C mR!2

0

4 cos C 4 cos2

16 sin 17 cos sin 4 .2 C 2 cos sin /

10.58 .!0/minDr

5g

7R

10.60 EdD h

R!2bO{ C P!bO|

h

RP!bOk

10.62 EaPD"

h P!bcos C

h2 C R2R

!!2bsin

#O{ C

"h P!bsin

h2

R

!!2bcos

#O|

Ch

R P!bsin h!2b.1 C 2 cos /iOk

10.64 OXD mL!20 ; OYD mg;OZD mL P!0; MXD 112 mL

2 P!0cos sin ;

MYD mL2 P!0

1 sin2

12

!; MZD mgL

1

12mL2!20cos sin

10.66 OXD mL!20 ; OYD mg;

OZD mL P!0; MXD 3

16mL2 P!0cos sin ;

MYD 1

16mL2 P!0

17 3 sin2

; MZD mgL

1

16mL2!20 cos sin

10.68 PD 675 151=3 csc2

4mL2 .1 C sin /

10.70 EABD 2dhR2!2dd2 C h22O{ hR

2 hd4

C h2

.` R/2

C d2

.h C R `/.h R C `/i!2dd2 C h22 .` R/3 O|

dR2

hd4 C d2h2 C 2h2 .` R/2

i!2d

d2 C h22 .` R/3 Ok;D26:32 O{ 0:9870O| 17:77Ok

rad=s2;

where` Dq

L2AB d2 h2 C R,

10.72 EaAD Rh

d2R C ` .` R/2i

!2d

.` R/3 O{D 124:2 O{m=s2; where` D

qL2AB d2 h2 C R,



41/41


10.74 EaAD R!2

d

4 .` R sin /3h

2d2R 6R`2 d

R2 4`2

cos C 2Rd2 C `2

cos 2

C dR2 cos 3C 5R2` sin C 4`3 sin 4dR` sin 2C R2` sin 3iO{;

D 20:56.2:750 0:7500 sin /3

h31:87 C 35:62 cos C 9:184 cos 2C 0:6750 cos 3

C90:92 sin

9:900 sin 2

C1:547 sin 3i rad=s2;

where` Dq

L2AB

d2 h2 C R,

EABDhR!2

d

4

d2 C h2C 5R2 cos R .8dC R cos 3 /4

d2 C h2 C R cos .R cos 2d /2 O{C hR!

2d

d2 C h2 C R cos .R cos 2d /2 .R sin `/3

R3` cos4 .2R sin `/

dR2 cos3 h3`2 C R sin .6` C R sin /

i R cos2

h3d2 C h2

`2 C R sin 23d2 C h2 `

CR sin `2

3d2

CR sin .R sin

2`/i

C dcos h

d2 C h2

`2 C R sin 2d2 C h2 `C R sin

3d2 h2 C 2R sin .R sin 2`/

iC R sin

hsin

d4 C h2`2 C d2

h2 `2

C

d2 h2

R sin .2` R sin /C dR`2 sin 2iO|C R!

2d

.R cos d /d2 C h2 C R cos .R cos 2d /2 .R sin `/3

2h2R2 sin3 .2` R sin /

C `2 cos .R cos d /h

d2 C h2 C R cos .R cos 2d /i

C2R` cos .d

R cos /d2 C h2 C R cos .R cos 2d / sin

C Rhd2

d2 C h2

2h2`2

C dR cos

3d2 C h2 C R cos .R cos 3d /i

sin2

Ok;

D 133:2 C 218:6 cos 10:41 cos 32:250 C .1:800 C 0:5625 cos / cos 2O{

C .2:750 C 0:7500 sin /3

2:250 C .1:800 C 0:5625 cos / cos 2h1229 1142 cos 2C 149:8 cos 3

18:22 cos 4C 5:270 cos 5 0:5489 cos 6C cos .2490 1340 sin / C 434:1 sin C 289:8 sin 3 25:76 sin 4

iO|

C .2:750 C 0:7500 sin /32:250 C .1:800 C 0:5625 cos / cos 2

h1889 3302 cos 2

C 1170 cos 3 199:7 cos 4C 19:03 cos 5 0:7319 cos 6C cos .4329 1880 sin / C 278:2 sin C 815:8 sin 3

277:5 sin 4C 42:94 sin 5 3:355 sin 6iOk rad=s2

AppendixAnswers Dynamics 2e Gray

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Transcript of AppendixAnswers Dynamics 2e Gray