Juanfra

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 90.

Note, for the axes shown
y

yV

−R

−2π R 4

3
−r
8

14
πr
4

V

1

(π R ) ( 2R ) = 2π R

2

2
− π r3
3

Σ


r3 
2π  R 3 − 

3



2

3


r4 
−2π  R 4 − 

8



1
R4 − r 4
Σ yV
8
Y=
=−
1
ΣV
3
R − r3
3

Then

1 r 

8 R 4

1−

1 r 

3 R 

3

1−

=

(a )

r=

3
R: y = −
4

1 3

3 4

4

1 3
1−  
3 4

3

1−

R

or y = −1.118R ⊳
1 r 

8 R 

4

1 r 
1−  
3 R 

3

1−

(b)

y = −1.2R : − 1.2R = −

4

or

R

3

r
r
  − 3.2   + 1.6 = 0
R
R

Solving numerically

Vector Mechanics for Engineers: Statics andDynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

r
= 0.884 ⊳
R

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 91.
Labeling the two parts of the body as follows:

V

(

)

)

h
2

1 22
πa h
4

12
πa h
6

h
4

1π a 2h 2
24

Σ

7 π a 2h 2
24
2 π a 2h
3

12
πa h
2

2

(

yV

1

ΣyV
Then Y =
=
ΣV

y

22
πa h
3

7
π a 2h 2
24

or Y =

7
h⊳
16

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 92.
Labeling the two parts of the body as follows:

V
1
2

Σ

(
(

− 1 a3h
ΣzV
2
Then Z =
=
2 π a 2h
ΣV
3

12
πa h
2
12
πa h
6
22
πa h
3

z

−

4a
3π
a

π

zV

2
− a 3h
3
13
ah
6
1
− a 3h
2

)
)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, FerdinandP. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

or Z = −

3a
⊳
4π

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 93.

V

x

xV

Rectangular prism

Lab

1
L
2

12
L ab
2

Pyramid

1 b
a h
3 2

1

ΣV = ab  L + h 
6

Then

Now

L+

ΣxV =

X ΣV = ΣxV

1
h
4

1
1

abh  L + h 
6
4


1 2
1 

ab 3L + h  L + h  
6
4 


so that


1  1 
1
X  ab  L + h   = ab  3L2 + hL + h 2 
6  6 
4

1 h 1 
h 1 h2 

X 1 +
= L3 + +


6 L 6 
L 4 L2 




or

(a) X = ? when h =
Substituting

(1)

1
L
2

1
h
=
into Eq.(1)
L
2
2

1  1  1 
1 11 
X 1 +    = L 3 +   +   
6  2  6 
2 42 




or X =

(b)

57
L
104

X = 0.548L ⊳

h
= ? when X = L
L

Substituting into Eq. (1)

or

or

1 h 1 
h 1 h2 

= L3 + +
L 1 +


6 L 6 
L 4 L2 




1+

1h
1 1h
1 h2
=+
+
6L
2 6 L 24 L2
h2
= 12
L2

Vector Mechanics for Engineers:Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

∴

h
= 2 3⊳
L

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 94.
Assume that the machine element is homogeneous so that its center of gravity coincides with thecentroid of
the volume.

V , mm3

y , mm

z , mm

yV , mm 4

zV , mm 4

1

( 60 )(105)(10 ) = 63000

−5

52.5

− 315 000

3 307 500

2

1
2
π ( 30 ) (10 ) = 14 137.2
2

−5

− 70 686

1 664 400

3

(15)( 30 )( 60 ) = 27 000

15

30

405 000

810 000

4

− π (19 ) (10 ) = −11 341.1

−5

105

56 706

−1 190 820

5

1
2
− π (19 ) (15 ) = −...