Chapter 5 Homework Locate the center of gravity of the portion of the frame

PROBLEM 5.113 (Continued)

2

,mmA ,mm

x

,mm

y

3

,mmxA 3

,mmyA

1 (74)(60) 4440 0 43 0 190,920

2 565.49 2.1803 2.1803 1233 1233

3 (30)(60) 1800 21 0 37,800 0

4 565.49 39.820 2.1803 22,518 1233

PROBLEM 5.114

A thin steel wire of uniform cross section is bent into the shape

shown. Locate its center of gravity.

SOLUTION

First assume that the wire is homogeneous so that its center of gravity

will coincide with the centroid of the corresponding line.

22

22.4 4.8

mxz





 

PROBLEM 5.115

The frame of a greenhouse is constructed from uniform aluminum

channels. Locate the center of gravity of the portion of the frame shown.

SOLUTION

First assume that the channels are homogeneous so that the center of gravity

of the frame will coincide with the centroid of the corresponding line.

89

23 6

ft

xx





 

PROBLEM 5.116

Locate the center of gravity of the figure shown, knowing that it is

made of thin brass rods of uniform diameter.

SOLUTION

Uniform rod:

22 2 2

(1 m) (0.6 m ) (1.5 m)AB  

1.9 mAB 

PROBLEM 5.117

Locate the center of gravity of the figure shown, knowing that it is

made of thin brass rods of uniform diameter.

SOLUTION

By symmetry,

0X



PROBLEM 5.118

A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is

1030 kg/m

3

and of steel is 3

7860 kg/m ,

locate the center of gravity of the awl.

SOLUTION

First, note that symmetry implies

0YZ



I

33

I

5(12.5 mm) 7.8125 mm

8

2

(1030 kg/m ) (0.0125 m)

3

x

W











PROBLEM 5.118 (Continued)

, kgW , mm

x

, kg mmxW 

I 3

4.123 10



 7.8125 3

32.916 10





II 3

40.948 10



 52.5 3

2123.5 10





III 3

0.49549 10



 67.5 3

33.447 10





PROBLEM 5.119

A bronze bushing is mounted inside a steel sleeve. Knowing that the

specific weight of bronze is 0.318 lb/in

3

and of steel is 0.284 lb/in

3

,

determine the location of the center of gravity of the assembly.

SOLUTION

First, note that symmetry implies 0XZ



PROBLEM 5.120

A brass collar, of length 2.5 in., is mounted on an aluminum rod

of length 4 in. Locate the center of gravity of the composite body.

(Specific weights: brass  0.306 lb/in

3

, aluminum  0.101 lb/in

3

)

SOLUTION

PROBLEM 5.121

The three legs of a small glass-topped table are equally spaced and are

made of steel tubing, which has an outside diameter of 24 mm and a

cross-sectional area of

2

150 mm .

The diameter and the thickness of the

table top are 600 mm and 10 mm, respectively. Knowing that the

density of steel is

3

7860 kg/m and of glass is

3

2190 kg/m , locate the

center of gravity of the table.

SOLUTION

First note that symmetry implies

0XZ



Also, to account for the three legs, the masses of components I and II will each bex

multiplied by three

I

2180

12 180

y



 

362

II

7860 kg/m (150 10 m ) (0.180 m)

ST

mV







  

PROBLEM 5.122

Determine by direct integration the values of

x

for the two volumes obtained by passing a vertical

cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given

shape and divides the shape into two volumes of equal height.

A hemisphere

SOLUTION

Choose as the element of volume a disk of radius r and thickness dx. Then

2

,EL

dV r dx x x





PROBLEM 5.122 (Continued)

and

 

24

22 2

2/2 /2

24

24 22

22



() 24

() ()

24 24

a

EL aa

aa

xx

xdV x a xdx a

aa

























































PROBLEM 5.123

Determine by direct integration the values of

x

for the two volumes obtained by passing a vertical

cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given

shape and divides the shape into two volumes of equal height.

A semiellipsoid of revolution

SOLUTION

Choose as the element of volume a disk of radius r and thickness dx. Then

2

,EL

dV r dx x x





The equation of the generating curve is

22

1

xy

ha



so that

2

222

2

()

a

rhx

h



PROBLEM 5.123 (Continued)

and

2

22

2

2/2

224

2

()

24

h

EL h

h

a

x

dV x h x dx

h

axx

h



























PROBLEM 5.124

Determine by direct integration the values of

x

for the two volumes obtained by passing a vertical

cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given

shape and divides the shape into two volumes of equal height.

A paraboloid of revolution

SOLUTION

Choose as the element of volume a disk of radius r and thickness dx. Then

2

,

EL

dV r dx x x





The equation of the generating curve is

2

h

xh y

a

 so that

2

().

a

rhx

h



PROBLEM 5.124 (Continued)

and

 

2223

2/2 /2

23

22322

() 23

() ()

23 2 3

h

EL hh

hh

aaxx

xdV x h xdx h

hh

ahh

hh

h







 





 



 





























PROBLEM 5.125

Locate the centroid of the volume obtained by rotating the shaded area

about the x-axis.

SOLUTION

First, note that symmetry implies

0y



0z



Choose as the element of volume a disk of radius r and thickness dx.

PROBLEM 5.126

Locate the centroid of the volume obtained by rotating the shaded area

about the x-axis.

SOLUTION

First note that symmetry implies

0y



and

0z



We have

2

()ykXh

PROBLEM 5.127

Locate the centroid of the volume obtained by rotating the shaded

area about the line

.xh

SOLUTION

First, note that symmetry implies

xh



0z



Choose as the element of volume a disk of radius r and thickness dx. Then

2

,

EL

dV r dy y y





Now

2

222

2

()

h

xay

a



so that

22

.

h

rh a y

a

 

PROBLEM 5.127 (Continued)

and







22

2

0

2

2223

20

22

a

EL

a

h

ydV y a a y dy

a

hay ay a y y dy

a















