Solution Manual for Probability, Statistics, and Random Processes Chapter 5: Random Vector, Exams of Probability and Statistics

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• CHAPTERj

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Xj (11jin

2

j)j=j11.5jin

3

Yj (11jin

2

j)j=j39.5jin

3

A ,jin

2

xj ,jin. yj ,jin. xjA ,in

3

yjA ,in

3

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Areaj1:j Rectanglej 42 jmmjbyj 44 jmm.

Areaj2:j Trianglej bj =j 18 jmm,j hj =j 12 jmm.jA

reaj3:j Trianglej bj =j 24 jmm,j hj =j 12 jmm.

Then Xj j Aj =jj xjA

Xj (1596jmm

2

j)j=j 4536 jmm

3

and Yj j Aj =jj yA

Yj (1596jmm

2

j)j=j39,j 648 jmm

3

A ,jmm

2

xj ,jmm yj ,jmm xA ,jmm

3

yA ,jmm

3

1 42 jj 44 j=j 1848 3 22 5544 40,

−j

1 j

 18 j 12 j=j− 108 j

2

−j

1 j

j 24 j 12 j=j− 144 j 2 16 4 - 2304 - 576



Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Areaj1:j Rectanglej 6 jin.jbyj 4 jin.

Areaj2:j Trianglej bj =j 6 jin.,j hj =j 3 jin.

Xj (33jin

2

j)j=j 108 jin

3

Yj (33jin

2

j)j=j 93 jin

3

A ,jin

2

xj ,jin yj ,jin xjA ,in

3

yjA ,in

3

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Then XAj =j xjA

YAj =j yjA

Xj (7200jmm

2

j)j=j−72.000j 10

3

j mm

3

Xj =j−10.00jmmj ◀

Yj (7200jmm

2

j)j=j629.83 10

3

j mm

3

A ,jmm

2

xj ,jmm yj ,jmm xjA ,jmm

3

yjA ,jmm

3

(60)(120)j=j 7200

• 30 60 − 216 j 10

3

432 j 10

3

j

(60)

2

j =j2827.

4

72.000jj 10

3

269.83jj 10

3

−j

j

(60)

2

j =j−2827.4j 4

• 25.465 25. 72.000jj 10

3

−72.000j 10

3

−72.000j 10

3

629.83jj 10

3

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Then Xj =j

j xAj

3200 jin

3

j A 1948.23jin

2

Yj =j

j Aj

=j

1948.23jin

2

j yA 34,j 021 jin

3

A ,jin

2

xj ,jin. yj ,jin.

xjA ,jin

3

yjA ,jin

3

j

(38)

2

j =j2268.

2

− 20 j 16 j=j 320 − 10 8 3200 − 2560

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Areaj1:j Squarej 75 jmmjbyj 75 jmm.

Areaj2:j Quarterjcirclejradiusjofj 75 jmm.

A ,jmm

2

xj ,jmm yj ,jmm

xjA ,jmmj

3

yjA ,jmmj

3

1 75 jj 75 j=j 5625 37.5 37.5 210,938 210,

−j

j

j 75

2

j =j−4417.9j 4



Thenj byjsymmetry

Xj =j Yj =j

j xjA

j A

Xj (1207.14jmm

2

j)j=j20,j 223 jmm

3

Xj =j Yj =j16.75jmmj ◀

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

A ,jmm

2

xj ,jmm yj ,jmm

xjA ,jmm

3

yjA ,jmm

3

2 j

(75)(120)j=j 6000

3

−j

1 j

(75)(60)j=j− 2250

2

Then Xj j Aj =j xA

and

Xj (3750jmm

2

j)j=j112,500jmm

3

Yj  Aj =jj yA

or

Xj =j30.0jmm

Yj (3750jmm

2

j)j=j243,j 000 jmm

3

or

Yj =j64.8jmm

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Areaj1:j Outerjsemicirclejdiameterj 120 jmm.jArea

j2:j Innerjsemicirclejdiameterj 72 jmm.

A ,jmm

2

xj ,jmm yj ,jmm xA ,jmm

3

yA ,jmm

3

j

(120)

2

j =j22,j 620

2

• 50.93 0 −1152.04j 10

3

−j

j

(72)

2

j =j−8143.

2

248.80j 10

3

 14,477 −903.20j 10

3

Then XAj =j xA

Xj (14,j 477 jmm

2

j)j=j−903.20j 10

3

j mm

3

YAj =j yA

Xj =j−62.4jmmj ◀

Yj =j 0 j ◀

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Firstjnotejthatjsymmetryjimplies Xj =j 0 j ◀

3

Then

j A 12.9106jm

2

A ,jm

2

yj ,mj yA ,jm

3

4 j

j4.5jj 3 j=j 18

3

2 −j

j

2

j=j−5.0894j

Locatejthejcentroidjofjthejplanejareajshown.

SOLUTION

Then XAj =jj xjA

YAj =j yjA Yj (7454.9jmm

2

j)j=j141,j 290 jmmj

3

Xj =j 0 jmmj ◀

Yj =j18.95jmmj ◀

A ,jmm

2

xj ,jmm yj ,jmm xjA ,jmm

3

yjA ,jmm

3

1 j

(120)(90)j=j 5400

2

j

(60j j 2 j)

2

j =j5654.

4

−j

1 j

(120)(60)j=j− 3600

2

Showj thatj asj r 1

j approachesj r 2

,j thej locationj ofj thej centroidjappr

oachesjthatjforjanjarcjofjcirclejofjradiusj ( r 1

j+j r 2

j)/2.

3

2 1

 Aj = −j r − −j r

j

j

2 j

j

1

j−j 2

j

Yj  Aj =jj yA

j 2

= r −j r cosj

= −j r −j r



j

−j

j j

2

2 j

j

2

j j 2 1

2 1 j

r −j r 

3

j 2 jcos 

j

r

2

cos



3 3

2

1 j j 

2

2 1

Yj =

2 j j r −j r

2 2

2

Aj = −j



2

Similarly,

=j

2 j

r

cosj

−j

FromjFigurej5.8A:

SOLUTION

2 2

Now

2

3 j

2

j

y =

sinj

(j

j

−jj

−j r −j j r

Aj = −j r

and

First,jdeterminejthejlocationjofjthejcentroid.

rj

2

−j

2

2

 2 cosj



−j

−j r

cosj



Then

Y −j r −j r = r −j r cosj

j

−j



2

j j −j 2 

Sojthat

Injthejlimitjasj  0 j(i.e.,j r 1

j=j r 2

j),j then

1 2

whichjagreesjwithjEquationj(1).

and

1 2

2

1

or Yj =j( rj +j rj )j

cosj j

j

−j

1 2

Yj =j

2 j

j

3 j

( rj +j rj )j

cos

(j

2 j

−jj)

Then

=j

3 j

j

1 j

( rj +j rj )

= r

2 1

r

2

j −j r

2

3 3

r −j r

2

2

(j



j −jj)

1

=j

1 j

( rj +j rj )j

cos

2



r

2

j −j r

2

Let

sin(



j −jj) 1

Yj = ( r

1 j

+j r

2 j

1 2

UsingjFigurej5.8B,j Yj ofjanjarcjofjradiusj

1 j

( rj +j rj )jis

PROBLEMj5.17j (Continued)

j 2 1 j j

2

1 j 2 j 1 j

r

2 j

+j r

1

2

r +j rjrj +j r

2

( r

2 j

−j r

1 j

)( r

2 j

+j r

1 j

2 1

2 1 j

Now

2 1

2 2

( rj −j rj )j r +j rjrj +j r

r

3

j−j r

3

2 1 2 1

r

2 j

=j rj +j

r

1 j

=j rj −j