Chapter 8 Solutions: Random Sequences | Probability and Statistics for Engineers Solution, Exams of Probability and Statistics

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• CHAPTERj

Determinejwhetherjthejblockjshownjisjinjequilibriumjandjfindjthejmagnitudejan

djdirectionjofjthejfrictionjforcejwhen P =j 150 jN.

SOLUTION

Assumejequilibrium:

 F

xj

=j0:j j Fj +j(500j N)jsinj 20 j−j(150j N)jcosj 20 j=j 0

Fj =j−30.056j Nj Fj =j30.056j N

 F

yj

=j0:j j Nj −j(500jN)jcosj 20 j−j(150jN)jsinj 20 j=j 0

Nj =j+521.15jNj Nj =j521.15jN

Maximumjfrictionjforce:

F

mj

=j sjN

=j0.30(521.15jN)

=j156.345jN

Sincej Fj is andj Fj j F

mj

, blockjisjinjequilibrium◀

Fj =j30.1jN 20.0j◀

Determinej whetherj thej blockj shownj isj inj equilibriumj andj findj thej

magnitudejandjdirectionjofjthejfrictionjforcejwhenj Pj =j 120 jlb.

SOLUTION

Assumejequilibrium:

 F

xj

=j0:j j Fj +j(50jlb)jsinj 30 j−j(120j lb)jcosj 40 j=j 0

Fj =j+66.925j lb

 F

yj

=j0:j j Nj −j(50jlb)jcosj 30 j−j(120jlb)jsinj 40 j=j 0

Nj =j+120.436j lb

Maximumjfrictionjforce:

F

mj

=j

sj

N

=j0.40(120.436jlb)

=j48.174jlb

Wejnotejthatj Fj j F

mj

.j Thus,jActual

jfrictionjforce:

Fj =j F

kj

=j kjNj =j0.30(120.436j lb)j=j36.131jlb,

blockjmovesjup◀

Fj =j36.1jlb 30.0j◀

Determinej whetherj thej blockj shownj isj inj equilibriumj andj findj thejm

agnitudejandjdirectionjofjthejfrictionjforcejwhenj Pj =j 80 jlb.

SOLUTION

Assumejequilibrium:

 F

xj

=j0:j j Fj +j(50j lb)jsinj 30 j−j(80j lb)jcosj 40 j=j 0

Fj =j+36.284j lb

 F

yj

=j0:j j Nj −j(50jlb)jcosj 30 j−j(80jlb)jsinj 40 j=j 0

Nj =j+94.724j lb

Maximumjfrictionjforce:

F

mj

=j

sj

N

=j0.40(94.724jlb)

=j37.890jlb

Wejnotejthatj Fj j F

mj

.j Thus,

Thus

blockjisjinjequilibrium◀

Fj =j36.3jlb 30.0j◀

Note:

F

kj

=j

kj

Nj =j0.30(94.724j lb)j=j28.417j lb,j Fj j F

Kj

. Ifj blockj isj originallyj inj motion,j itj willjke

ep moving with F = 28.4 lb.

PROBLEMj8.5j (Continued)

Fromjforcejtriangle:

P 50 jlb

sinj46.70 sinj33.30°

= Pj =j66.3jlbj◀

SOLUTION

Free-bodyjdiagrams:

 sj =j0.

 sj =jtan

j0.35j=j19.29

Forcejtrianglejforjpulleyj C :

Lawjofjsines:

sinj

=j

sinj s

2 T T

sinjj=j 2 jsin sj =j 2 jsin19.

j=j41.

j=j sj +jj=j19.29j+j41.

j =j60.

∘

j◀

Note:j Answerjisjindependentjofj W

Aj

Thej 20 -

lbjblockjAjhangsjfromjajcablejasjshown.jPulleyjCjisjconnectedjbyjajshortjlinkj

tojblockjE,jwhichjrestsjonjajhorizontaljrail.jKnowingjthatjthejcoefficientjofjsta

ticjfrictionjbetweenjblockjEjandjthejrailjisjsj=j0.35jandjneglectingjthejweightj

ofjblockjEjandjthejfrictionjinjthejpulleys,jdeterminejthejmaximumjallowablejv

aluejofjθjifjthejsystemjisjtojremainjinjequilibrium.

Pj =j20.9j N ◀

F

ABj

=j25.5jNj C

Rj =j86.053j N

Free-bodyjdiagramjofjbelt:jF

xj

=j 0 j:j j Pj −j(86.053jN)jsin14.

j=j 0

98.1jN

sinj110.

R

sinj 55

F

AB

sinj14.04

Forcejtriangle:

( b ) Free-bodyjdiagramjofjblock:Free-bodyjdiagramjofjbelt:

PROBLEMj8.7j (Continued)

Consideringjonlyjvaluesjofjj lessjthanj 90 ,jdeterminejthejsmallestjvaluejofj

required to start the block moving to the right when ( a ) Wj =j 75 jlb,

( b )j Wj =j 100 jlb.

or sin(j −j14.036)j=

j =j51.4j◀

( b )

Wj =j 75 jlb: j =j14.036j+jsin

− 1

( a )

30 jlb

j =j68.0j◀

s

sin(j−jj )j=j

Wj sinj14.036

sinj s sin(j−j sj )

30 jlbj W

sj

=jtan

j

sj

=j14.036

SOLUTION

FBDjblockj(Motionjimpending):

SOLUTION

Free-bodyjdiagramjofjblockjandjforcejtriangle:

Forjmotionjimpendingjdownward, Wj =j(7.5jkg)(9.81jm/s

j)j=j73.575jN

 sj =jtan

j sj =jtan

j(0.45)j=j24.23

100 jN

sinj65.

sinj(j+j sj )

j

j+j24.

j=j42.

j =j17.

∘

Forjmotionjimpendingjupward,

73.575jN

Knowingjthatj Pj =j 100 jN,jdeterminejthejrangejofjvaluesjofj θj forjwhichjequilibriumjofjt

hej7.5-kgjblockjisjmaintained.

PROBLEMj8.10j (Continued)

100 jN

sinj114.

∘ sinj

(j−j sj )

j

j−j24.

j=j42.

j =j66.

∘

73.575jN

∘

jjj j66.

∘

j◀

Thej 50 - lbjblockj Aj andjthej 25 -

lbjblockj Bj arejsupportedjbyjanjinclinejthatjisjheldjinjthejpositionjshown.

jKnowingjthatjthejcoefficientjofjstaticjfrictionjisj0.15jbetweenjalljsurfa

cesjofjcontact,jdeterminejthejvaluejofjjforjwhichjmotionjisjimpending.

SOLUTION

Sincejmotionjimpends,j Fj =j

sj

Nj betweenj A + B

Freejbody:jBlockj A

Impendingjmotion:j F

yj

=j0:j j N

1 j

−j 50 jcosj=j 0

 F

xj

=j0:j j Tj −j 50 jsinj +j 1 N

1 j

=j 0

Tj =j 50 jsinj −j 1 (50)jcos (1)

Freejbody:jBlockj B

Impendingjmotion:  F

yj

=j0:j j N

2 j

−j N

1 j

−j25cosj =j 0

Nj 2 j =j 75 jcos

 F

xj

: Tj −j

1 j

N

1 j

−j

2 j

N

2 j

−j25sin

Tj =j 1 j( (^50) )cosj+j 2 j( (^75) )cosj+j25sin

0 j=j25sinj−j 1 j( (^100) )cosj−j 2 j( (^75) )cosj=j 0

Eq.j(1)-Eq.j(2):

Substitutingjinjforj  1 j=j 2 j =j0.15,j wejhave:

(^175) (0.15)cosj =j 25sinj : tanj =j

26.25j

; j =j46.4j◀

SOLUTION

Considerj Cj byjitself:j Assumejequilibrium

 F

yj

=j0:

 F

xj

=j0:

But

N

Cj

−j Wj cosj 15 j=j 0

N

Cj

=j Wj cosj 15 j=j0.966 Wj

F

Cj

−j Wj sinj 15 j=j 0

F

Cj

=j Wj sinj 15 j=j0.259 W

F

mj

=j sjN

C

=j0.30(0.966 Wj )

=j0.290 W

Thus,j F

Cj

j F

m

Packagej Cj doesjnotjmove◀

F

Cj

=j0.259 W

=j0.259(4jkg)(9.81jm/s

j)

=j10.16jN

Considerj Bj byjitself:j Assumejequilibrium.jWejfind,

F

Bj

=j0.259 Wj

N

Bj

=j0.966 W

But F

mj

=j sjN

B

=j0.10(0.966 Wj )

=j0.0966 W

Thus,j F

Bj

j F

mj

. Packagej Bj wouldjmovejifjalone◀

Threej 4 -

kgjpackages A ,j B ,jandj Cj arejplacedjonjajconveyorjbeltjthatj isj atj rest.

j Betweenj thej beltj andj bothj packagesj Aj andj Cj the

coefficientsj ofj frictionj arej 

sj

=j0.30j andj 

kj

=j0.20;j betweenjpackag

ej Bj andjthejbeltjthejcoefficientsjarej 

sj

=j0.10j andj 

kj

=j0.08.jThejpack

agesjarejplacedjonjthejbeltjsojthatjtheyjarejinjcontactjwith

eachjotherjandjatjrest.jDeterminejwhich,jifjany,jofjthejpackagesjwilljm

ovejandjthejfrictionjforcejactingjonjeachjpackage.

SOLUTION

Considerjpackagej Bj byjitself:j Assumejequilibrium

 F

yj

=j0:j j N

Bj

−j Wj cosj 15 j=j 0

N

Bj

=j Wj cosj 15 j=j0.966 W

 F

xj

=j0:j j F

Bj

−j Wj sinj 15 j=j 0

F

Bj

=j Wj sinj 15 j=j0.259 W

But F

mj

=j sjN

B

=j0.10(0.966 Wj )

=j0.0966 W

Thus,j F

Bj

. F

mj

.j Packagej Bj wouldjmovejifjalone.

Considerjalljpackagesjtogether:j Assumejequilibrium.jInjajmannerjsimilarjtojabove,jwejfind

N

Aj

=j N

Bj

=j N

Cj

=j0.966 WjF

Aj

=j F

Bj

=j F

Cj

=j0.259 W

F

Aj

+j F

Bj

+j F

Cj

=j3(0.259 Wj )

=j0.777 W

But

and

( F

Aj

mj

=j( F

Cj

mj

=j sjN

=j0.30(0.966 Wj )

=j0.290 W

( F

Bj

mj

=j0.10(0.966 Wj )

=j0.0966 W

SolvejProblemj8.13jassumingjthatjpackagej Bj isjplacedjtojthejrightjofjbo

thjpackagesj Aj andj C.

PROBLEMj8.13j Threej 4 -

kgjpackages A ,j B ,jandj Cj arejplacedjonjajconveyorjbeltjthatjisjatjrest.jBet

weenjthejbeltjandjbothjpackagesj Aj andj Cj thej coefficientsj ofj friction

j arej j 

sj

=j0.30j j andj j 

kj

=j0.20;

betweenj packagej Bj andj thej beltj thej coefficientsj arej 

sj

=j0.10j and

kj

=j0.08.j Thejpackagesjarejplacedjonjthejbeltjsojthatjtheyjarejinjcontac

tjwithjeachjotherjandjatjrest.jDeterminejwhich,jifjany,jofjthe

packagesjwilljmovejandjthejfrictionjforcejactingjonjeachjpackage.

PROBLEMj8.14j (Continued)

Thus,

andjwejnotejthat

( F

Aj

mj

+j( F

Cj

mj

+j( F

Bj

mj

=j2(0.290 Wj )j+j0.0966 W

=j0.677 W

Alljpackagesjmove◀

F

Aj

=j F

Cj

=j

kj

N

=j0.20(0.966)(4jkg)(9.81jm/s

j)

=j7.58jN

F

Bj

=j kjN

=j0.08(0.966)(4jkg)(9.81jm/s

j)

=j3.03jN

; F

Bj

=j3.03jN