Chapter 5 Homework Locate the y coordinate of the center of gravity

PROBLEM 5.98 (Continued)

(b) ?when 0.4

hYa

a

Substituting into Eq. (1)

PROBLEM 5.99

Locate the centroid of the frustum of a right circular cone when r1 = 40 mm,

r2 = 50 mm, and h = 60 mm.

SOLUTION

PROBLEM 5.100

For the machine element shown, locate the x coordinate of

the center of gravity.

SOLUTION

Dimensions in mm

PROBLEM 5.100 (Continued)

34

33

848.18 10 mm

46.399 10 mm









XV xV

xV

XV 18.28 mmX 

PROBLEM 5.101

For the machine element shown, locate the z coordinate of

the center of gravity.

SOLUTION

Dimensions in mm

PROBLEM 5.101 (Continued)

zV

PROBLEM 5.102

For the machine element shown, locate the y coordinate of

the center of gravity.

SOLUTION

For half-cylindrical hole,

III

0.95 in.

4(0.95)

1.5 3

r

y







PROBLEM 5.103

For the machine element shown, locate the z coordinate of

the center of gravity.

SOLUTION

For half-cylindrical hole,

III

0.95 in.

4(0.95)

1.5 3

r

y







PROBLEM 5.104

For the machine element shown, locate the y coordinate of the

center of gravity.

SOLUTION

First assume that the machine element is homogeneous so that its center of gravity will coincide with the

centroid of the corresponding volume.

PROBLEM 5.105

For the machine element shown, locate the x coordinate of the

center of gravity.

SOLUTION

First assume that the machine element is homogeneous so that its center of gravity will coincide with the

centroid of the corresponding volume.

PROBLEM 5.106

Locate the center of gravity of the sheet-metal form shown.

SOLUTION

First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide

with the centroid of the corresponding area. Now note that symmetry implies

125.0 mmX



II

280

150

200.93 mm

280

y

z













PROBLEM 5.107

Locate the center of gravity of the sheet-metal form shown.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide

with the centroid of the corresponding area.

I

1(1.2) 0.4 m

3

1(3.6) 1.2 m

3

y

z

 



PROBLEM 5.108

A corner reflector for tracking by radar has two sides in the

shape of a quarter circle with a radius of 15 in. and one side in

the shape of a triangle. Locate the center of gravity of the

reflector, knowing that it is made of sheet metal of uniform

thickness.

SOLUTION

By symmetry, XZ

For I and II (Quarter-circle ),

22 2

44(15)

6.3662 in.

33

(15) 1.76715 in

44

r

xy

Ar



  

 

PROBLEM 5.109

A wastebasket, designed to fit in the corner of a room, is 16 in.

high and has a base in the shape of a quarter circle of radius 10

in. Locate the center of gravity of the wastebasket, knowing

that it is made of sheet metal of uniform thickness.

SOLUTION

By symmetry,

XZ

For III (Cylindrical surface),

22(10)

6.3662 in.

r

x



 

PROBLEM 5.110

An elbow for the duct of a ventilating system is made of sheet

metal of uniform thickness. Locate the center of gravity of the

elbow.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with

the centroid of the corresponding area. Also, note that the shape of the duct implies

38.0 mmY



II

2

Note that 400 (400) 145.352 mm

2

400 (200) 272.68 mm

2

300 (200) 172.676 mm

xz

x

z



  

 

PROBLEM 5.110 (Continued)

We have 23

: (227,723 mm ) 41,034, 760 mmXA xA X  or 180.2 mmX 

23

: (227, 723 mm ) 44,070, 260 mmZA zA Z  or 193.5 mmZ 

PROBLEM 5.111

A window awning is fabricated from sheet metal of uniform

thickness. Locate the center of gravity of the awning.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides

with the centroid of the corresponding area.

II VI

IV

(4)(25)

4 14.6103 in.

3

(4)(25) 100 in.

33

(2)(25)

4 19.9155 in.

(2)(25) 50 in.

yy

zz

y







 

 

 

PROBLEM 5.111 (Continued)

Now, symmetry implies 17.00 in.X 

and 23

: (2652.9 in ) 41,607 inYA yA Y  or

15.68 in.Y 

23

: (2652.9 in ) 37,567 inZA zA Z  or

14.16 in.Z 

PROBLEM 5.112

A mounting bracket for electronic components is formed

from sheet metal of uniform thickness. Locate the center of

gravity of the bracket.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides

with the centroid of the corresponding area. Then (see diagram)

V

4(0.625)

2.25 3

1.98474 in.

z







PROBLEM 5.113

A thin sheet of plastic of uniform thickness is bent to form

a desk organizer. Locate the center of gravity of the

organizer.

SOLUTION

First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide

with the centroid of the corresponding area. Now note that symmetry implies

30.0 mmZ

