Solucionario

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 1.
\

A, mm 2
1 2

x , mm
−100 200

y , mm
250 150

xA, mm3 − 30 000000 24 000 000 21000 000

yA, mm3 6 750 000 18000000 24 750000

200 × 150 = 30000 400 × 300 = 120000 150 000

Σ

Then

X =
Y =

ΣxA 21 000000 = mm ΣA 150000
ΣyA 24 750000 = mm ΣA 150 000

or X = 140.0 mm

or Y = 165.0mm

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 2.

A,in 2
1 2
10 × 8 = 80

x ,in.
5
13

y ,in.
4

xA,in 3
400
702

yA,in3
320
216

1 × 9 × 12 = 54 2
134

4

Σ

1102

536

Then and

X = Y =

ΣxA 1102 = ΣA 134
ΣyA 1102 = ΣA 134

or

X = 8.22 in.

or Y = 4.00 in.

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 3.

A, mm 2
1

x , mm

xA, mm3 729 000 2 460 375 3 189 375

1 × 90 × 270 = 12 150 2 1 × 135 × 270 = 18 225 2

2 ( 90 ) = 60 3 90 + 1 (135) = 135 3

2

Σ

30375
ΣxA 3189375 mm = ΣA 30375

Then

X =

or X = 105.0 mm

For the whole triangular area by observation:

Y =

1 ( 270 mm ) 3

or Y = 90.0mm

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 4.

A,in 2
1

x ,in.
2 ( 21) = 14 3 21 + 1 (13) = 27.5 2 40 −

y ,in.
1 ( 24 ) = 323

xA,in 3
3528

yA,in 3
8064

1 ( 21)( 24 ) = 252 2

2

(13)( 40 ) = 520
772

20

14 300

10 400

Σ

17 828

18 464

Then

X = Y =

ΣxA 17828 = in. ΣA 772 ΣyA 18464 = in. ΣA 772

or

X = 23.1 in.

or Y = 23.9 in.

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 5.

A, mm 2

x , mm

y , mm

xA, mm3
− 3 796 900

yA, mm3 3 796 900

1

π ( 225 )
4

2

= 39 761

−

4 ( 225 ) 3π

= − 95.493

95.493

2

1 ( 375)( 225) = 42 188 2

125

75

5 273 500

3 164 100

Σ81 949

1 476 600

6 961 000

Then

X = Y =

ΣxA 1476600 mm = ΣA 81 949 ΣyA 6961 000 mm = ΣA 81 949

or X = 18.02 mm
or Y = 84.9 mm

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 SolutionsManual Organization System

Chapter 5, Solution 6.

A,in 2
1 2
17 × 9 = 153

x ,in.
8.5

y ,in.
4.5

xA,in 3
1300.5

yA,in 3
688.5

−

π
4
−

× ( 4.5 ) = −15.9043 8 −

2

4 × 4.5 4 × 4.5 = 6.0901 9 − = 7.0901 − 96.857 3π 3π
10.5465 6.4535

−112.761 −182.466
393.27

3

π
4

( 6 )2 = − 28.274
108.822

− 298.19
905.45

Σ

Then and

X =

ΣxA 905.45 =ΣA 108.822

or

X = 8.32 in.

Y =

ΣyA 393.27 = ΣA 108.22

or Y = 3.61 in.

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 7....