Cap 05

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Chapter 5, Solution 1. \

Then

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

200 × 150 = 30000

−100

250

− 30 000000

6 750 000

2

400 × 300 = 120000

200

150

24 000 000

18000000

Σ

150 000

21000 000

24 750000

X =

ΣxA 21 000000 = mm ΣA 150000

or X = 140.0 mm

Y =

ΣyA 24 750000 = mm ΣA 150 000

or Y = 165.0 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.

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Chapter 5, Solution 2.

A,in 2

x ,in.

y ,in.

xA,in 3

yA,in 3

1

10 × 8 = 80

5

4

400

320

2

1 × 9 × 12 = 54 2

13

4

702

216

Σ

134

1102

536

Then

X =

ΣxA 1102 = ΣA 134

and

Y =

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.

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Chapter 5, Solution 3.

Then

A, mm 2

x , mm

xA, mm3

1

1 × 90 × 270 = 12 150 2

2 ( 90 ) = 60 3

729 000

2

1 × 135 × 270 = 18 225 2

Σ

30375

X =

90 +

ΣxA 3189375 mm = ΣA 30375

1 (135) = 135 3

2 460 375 3 189 375

or X = 105.0 mm

For the whole triangular area by observation:

Y =

1 ( 270 mm ) 3

or Y = 90.0 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.

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Chapter 5, Solution 4.

A,in 2

x ,in.

1

1 ( 21)( 24 ) = 252 2

2

(13)( 40 ) = 520

Σ

2 ( 21) = 14 3 21 +

1 (13) = 27.5 2

xA,in 3

y ,in.

40 −

1 ( 24 ) = 32 3

20

772

Then

yA,in 3

3528

8064

14 300

10 400

17 828

18 464

X =

ΣxA 17828 = in. ΣA 772

or

Y =

ΣyA 18464 = in. ΣA 772

or Y = 23.9 in.

X = 23.1 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.

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Chapter 5, Solution 5.

A, mm 2

x , mm

2

1

π ( 225 )

2

1 ( 375)( 225) = 42 188 2

4

−

= 39 761

4 ( 225 ) 3π

= − 95.493

125

y , mm

xA, mm3

yA, mm3

95.493

− 3 796 900

3 796 900

5 273 500

3 164 100

1 476 600

6 961 000

75

81 949

Σ

Then

X =

ΣxA 1476600 mm = ΣA 81 949

or X = 18.02 mm

Y =

ΣyA 6961 000 mm = ΣA 81 949

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.

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Chapter 5, Solution 6.

1 2

3

−

π 4 −

A,in 2

x ,in.

y ,in.

xA,in 3

yA,in 3

17 × 9 = 153

8.5

4.5

1300.5

688.5

2

× ( 4.5 ) = −15.9043 8 −

π 4

( 6 )2 = − 28.274

Σ

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

−112.761

− 298.19

−182.466

905.45

393.27

10.5465

6.4535

108.822

Then

X =

ΣxA 905.45 = ΣA 108.822

and

Y =

ΣyA 393.27 = ΣA 108.22

or

X = 8.32 in.

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.

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Chapter 5, Solution 7.

A,in 2 1

π (16 ) 4

2

= 201.06

2

− ( 8 )( 8 ) = − 64

Σ

137.06

ΣxA 1109.32 = in. ΣA 137.06

Then

X =

and

Y = X by symmetry

x ,in.

4 (16 ) 3π

= 6.7906 4

xA,in 3 1365.32

− 256 1109.32

or

X = 8.09 in.

or Y = 8.09 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.

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Chapter 5, Solution 8.

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

35 343

63.662

0

2 250 006

0

2

− 4417.9

31.831

− 31.831

−140 626

140 626.2

Σ

30925.1

2 109 380

140 626.2

Then

X =

ΣxA 2109 380 = ΣA 30 925.1

and

Y =

ΣyA 140 625 = ΣA 30 925.1

or

X = 68.2 mm

or Y = 4.55 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.

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Chapter 5, Solution 9.

A −

1

Therefore, for X =

4

π

2

Σ

π2

2

π

( 2r 4

2 2

r12

r22 − r12

4r1 3π

r13  π 2  4r1   − r1  =− 3  4  3π 

4r2 3π

 π 2  4r2  2r23  r2  = 3  2  3π 

1 2r23 − r13 3

(

)

ΣxA 4r1 = : ΣΑ 3π

( (

or

xA

)

)

4 2r23 − r13 4r1 = 3π 3π 2r22 − r12

or

x

π =

)

  r 3  r13  2  2  − 1   r1   4   = 2  r   3π r12  2  2  − 1   r1    

r 2ρ 3 − 1 , where ρ = 2 2 r1 2ρ − 1

2 ρ 3 − 2πρ 2 + (π − 1) = 0.

Solving numerically for ρ and noting that ρ > 1:

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.

r2 = 3.02 r1

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Chapter 5, Solution 10.

First, determine the location of the centroid. y2 =

From Fig. 5.8A:

= y1 =

Similarly Then

Σ yA =

(π

2 sin 2 − α r2 π 3 −α 2

(

2 cos α r2 π 3 −α 2

)

(

)

(

)

2 cos α r1 π 3 −α 2

2 cosα  r2 π 3 −α  2

(

)

(

)

A2 =

A1 =

( π2 − α ) r12

( π2 − α ) r22  − 23 r1

(

cosα  −α  2

π

)

2 3 r2 − r13 cosα 3 π  π  Σ A =  − α  r22 −  − α  r12 2  2  =

and

π  =  − α  r22 − r12 2   Y Σ A = Σ yA

(

Now

( π2 − α ) r22

)

 π  2 3  Y  − α  r22 − r12  = r2 − r13 cos α   2  3

(

)

Y =

(

)

2 r23 − r13 cos α 3 r22 − r12 π2 − α

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.

)

( π2 − α ) r12 

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Using Figure 5.8B, Y of an arc of radius

1 ( r1 + r2 ) is 2 Y =

=

(π

sin − α 1 ( r1 + r2 ) π 2 2 −α 2

(

)

)

1 cos α (r1 + r2 ) π 2 −α 2

(

(

( r2 − r1 ) r22 + r1 r2 + r12 r23 − r13 = r22 − r12 ( r2 − r1 )( r2 + r1 )

Now

=

(1)

)

)

r22 + r1 r2 + r12 r2 + r1

r2 = r + ∆

Let

r1 = r − ∆

r =

Then

1 ( r1 + r2 ) 2 2

and

( r + ∆ ) + ( r + ∆ )( r − ∆ ) + ( r − ∆ ) r23 − r13 = 2 2 r2 − r1 (r + ∆) + (r − ∆) =

2

3r 2 + ∆ 2 2r

In the limit as ∆ → 0 (i.e., r1 = r2 ), then

r23 − r13 3 = r 2 2 2 r2 − r1 =

so that

Y =

3 1 × (r1 + r2 ) 2 2

2 3 cos α × ( r1 + r2 ) π 3 4 −α 2

Which agrees with Eq. (1).

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.

or Y =

1 cos α ! ( r1 + r2 ) π 2 −α 2

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Chapter 5, Solution 11.

Then

X =

A,in 2

x ,in.

xA,in 3

1

27

8.1962

221.30

2

15.5885

3.4641

54.000

3

−18.8495

3.8197

−71.999

Σ

23.739

ΣxA 203.30 = ΣA 23.739

203.30

or

X = 8.56 in.

and by symmetry

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.

Y =0

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Chapter 5, Solution 12.

1 2

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1 ( 240 )(150 ) = 18 000 2

160

50

2 880 000

900 000

3 ( 240 ) = 180 4

3 (150 ) = 45 10

−2160000

−540 000

720 000

360 000

−

1 ( 240 )(150 ) = 12 000 3 6000

Σ

Then

X =

ΣxA 720000 = mm ΣA 6000

Y =

ΣyA 360000 = mm ΣA 6000

or X = 120.0 mm or

Y = 60.0 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.

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Chapter 5, Solution 13.

A,in 2

x ,in.

y ,in.

1

(18)(8) = 144

−3

4

− 432

576

2

1 ( 6 )( 9 ) = 27 2

2

−3

54

−81

−5.0930

−3.8197

− 432.00

− 324.00

−810.00

171.00

3

Σ

Then

π 4

(12 )( 9 ) = 84.823 255.82

X =

ΣxA −810.00 = in.

255.82 ΣA

Y =

ΣyA 171.00 = in. ΣA 255.82

xA,in 3

or

yA,in 3

X = − 3.17 in.

or Y = 0.668 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.

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Chapter 5, Solution 14.

X = 90 mm

First, by symmetry

1 2

−

3

−

A, mm 2

y , mm

yA, mm3

(180 )(120 ) = 21 600

60

1 296 000

π 4

π 4

( 90 )(120 ) = − 8482.3

120 −

4 × 120 = 69.070 3π

−585 870

( 90 )(120 ) = − 8482.3

120 −

4 × 120 = 69.070 3π

−585 870

4635.4

Σ

Y =

ΣyA 124 260 = 4635.4 ΣA

124 260

or Y = 26.8 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.

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Chapter 5, Solution 15.

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

18 240

−4

12

72 960

218 880

2

−1920

− 56

54

107520

−103 680

3

− 4071.5

− 41.441

− 41.441

168 731

186 731

Σ

12 248.5

−134171

−53 531.1

Then and

X =

ΣxA −134171 = ΣA 12 248.5

Y =

ΣyA −53 531 = ΣA 12 248.5

or

X = −10.95 mm or Y = − 43.7 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.

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Chapter 5, Solution 16. \

A, mm 2 2 ( 200 )( 200 ) = 26 667 3

1

−

2

2 (100 )( 50 ) = − 3333.3 3

23 334

Σ

xA, mm3

yA, mm3

x , mm

y , mm

75

70

2 000 000

1866 690

37.5

− 20

−125 000

66 666

1875 000

1 933 360

Then X =

ΣxA 1875 000 = mm ΣA 23 334

or X = 80.4 mm

Y =

ΣyA 1 933 360 = mm ΣA 23 334

or Y = 82.9 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.

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Chapter 5, Solution 17.

Locate first Y :

Note that the origin of the X axis is at the bottom of the whole area.

A, in 2

Y =

yA, in 3

1

8 × 15 = 120

7.5

900

2

− 4 × 10 = − 40

8

− 320

Σ

Then

y , in.

80

580

ΣyA 580 = = 7.2500 in. ΣA 80

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.

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Now, to find the first moment of each area about the x-axis: Area I:

QI = ΣyA =

7.75 5.75  − ( 4 × 5.75 )  , (8 × 7.75) + 2 2 

or QI = 174.125 in 3 !

Area II:

QII = ΣyA = −

7.75 4.25  − ( 4 × 4.25 )  , (8 × 7.25) − 2 2 

or QII = −174.125 in 3 !

Note that Q( area ) = QI + QII = 0 which is expected as y = 0 and Q( area ) = yA since x is a centroidal axis.

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.

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Chapter 5, Solution 18.

A, mm 2

Y =

yA, mm3

1

(80 )( 20 ) = 1600

90

144 000

2

( 20 )(80 ) = 1600

40

64 000

Σ

Then

y , mm

3200

208 000

ΣyA 208 000 = = 65.000 mm ΣA 3200

Now, for the first moments about the x-axis: Area I

QI = ΣyA = 25 ( 80 × 20 ) + 7.5 ( 20 × 15 ) = 42 250 mm3 ,

or QI = 42.3 × 103 mm3 !

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.

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Area II

QII = ΣyA = − 32.5 ( 20 × 65 ) = 42 250 mm3 ,

or QII = 42.3 × 103 mm3 !

Note that Q( area ) = QI + QII = 0 which is expected as y = 0 and Q( area ) = yA since x is a centroidal axis.

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.

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Chapter 5, Solution 19.

(a) With Qx = Σ yA and using Fig. 5.8 A,

(

)

 2 r sin π − θ    r 2 π2 − θ  − Qx =  3 π 2  − θ   2   2 = r 3 cos θ − cos θ sin 2 θ 3

(

) ( 32 r sin θ )  12 × 2r cos θ × r sin θ 

(

)

or Qx =

(b) By observation, Qx is maximum when and then

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.

2 3 r cos3 θ 3

θ =0 Qx =

2 3 r 3

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Chapter 5, Solution 20.

From the problem statement: F is proportional to Qx . Therefore:

FA FB = , or ( Qx ) A ( Qx )B

FB =

( Qx )B F ( Qx ) A A

For the first moments:

Then

( Qx ) A

12   =  225 +  ( 300 × 12 ) = 831 600 mm3 2 

( Qx )B

12   = ( Qx ) A + 2  225 −  ( 48 × 12 ) + 2 ( 225 − 30 )(12 × 60 ) = 1 364 688 mm 3 2 

FB =

1364688 ( 280 N ) , 831600

or FB = 459 N

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.

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Chapter 5, Solution 21.

Because the wire is homogeneous, its center of gravity will coincide with the centroid for the corresponding line.

L, mm

x , mm

y , mm

xL, mm 2

yL, mm 2

1

400

200

0

80 000

0

2

300

400

150

120 000

45 000

3

600

100

300

60 000

180 000

4

150

− 200

225

− 30 000

33 750

5

200

−100

150

− 20 000

30 000

6

150

0

75

0

11 250

Σ

1800

210 000

300 000

Then

X =

ΣxL 210 000 = = 116.667 mm ΣL 1800

or X = 116.7 mm

and

Y =

ΣyL 300 000 = = 166.667 mm ΣL 1800

or Y = 166.7 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.

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Chapter 5, Solution 22.

L, in.

x , in.

y , in.

xL, in 2

y , in 2

1

19

9.5

0

180.5

0

2

15

14.5

6

217.5

90

3

4

10

10

40

40

4

10

5

8

50

80

5

8

0

4

0

32

Σ

56

488

242

Then

X =

ΣxL 488 = ΣL 56

or X = 8.71 in.

and

Y =

ΣyA 242 = 56 ΣA

or Y = 4.32 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.

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Chapter 5, Solution 23.

Because the wire is homogeneous, its center of gravity will coincide with the centroid for the corresponding line.

L, mm

x , mm

y , mm

xL, mm 2

yL, mm 2

600

75

0

45 000

0

187.5

112.5

81 998

49 199

2

2

− 50 625

50 625

76 373

99 824

1 2 3

3752 + 2252 = 437.32

π 2

Σ

( 225)

−

π

( 225)

1390.75

π

( 225)

Then

X =

ΣxL 76 373 = ΣL 1390.75

or X = 54.9 mm

and

Y =

ΣyL 99 824 = ΣL 1390.75

or Y = 71.8 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.

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Chapter 5, Solution 24.

L, mm

x , mm

y , mm

xL, mm 2

yL, mm 2

1

75

37.5

0

2812.5

0

2

150

0

75

0

11 250

95.492

0

45 000

0

0

−112.5

0

− 8437.5

47.746

− 47.746

5625.0

− 5625.0

53 437

− 2812.5

3

(150 )π

4 5

Σ

= 471.24 75

( 75)

π 2

= 117.81

889.05

Then

X =

ΣxL 53 437 = , ΣL 889.05

or X = 60.1 mm

and

Y =

ΣyA − 2812.5 = ΣA 889.05

or Y = − 3.16 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.

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Chapter 5, Solution 25.

O From Figure 5.8 b:

r =

( 20 in.) sin 30° π

=

60

π

in.

6 Note also that triangle ABO is equilateral, where O is the origin of the coordinate system in the figure.

For equilibrium: (a) ΣM A = 0:

   60   20 in. −  in.  cos 30° (1.75 lb ) − ( 20 in.) sin 60° TBC = 0 π   

Solving for TBC :

TBC = 0.34960 lb (b) ΣFx = 0:

or

TBC = 0.350 lb

Ax + ( 0.34960 lb ) cos 60° = 0 Ax = − 0.174800 lb

ΣFx = 0:

Ay − 1.75 lb + ( 0.34960 lb ) sin 60° = 0

Ay = 1.44724 lb Therefore:

A = 1.458 lb

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.

83.1°

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Chapter 5, Solution 26.

The wire supported only by the pin at B is a two-force body. For equilibrium the center of gravity of the wire must lie directly under B. Also, because the wire is homogeneous the center of gravity will coincide with the centroid. In other words, x = 0, or ΣxL = 0.

ΣxL = −

2 (150 mm )

π

150 mm  200 mm    π (150 mm )  +  cosθ  (150 mm )  ( 200 mm ) +  200 mm − 2 2    

or

cosθ =

5000 11250 or

θ = 63.6°

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 27.

The wire supported only by the pin at B is a two-force body. For equilibrium the center of gravity of the wire must lie directly under B. Also, because the wire is homogeneous the center of gravity will coincide with the centroid. In other words, x = 0, or ΣxL = 0.

ΣxL = −

2 (150 mm )

π

150 mm  200 mm    π (150 mm )  +  cosθ  (150 mm )  ( 200 mm ) +  200 mm − 2 2    

or

l 2 + 300l − 197602 = 0.

Solving for l :

l = 319.15, and l = − 619.15, and discarding the negative root l = 319 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 Solutions Manual Organization System

Chapter 5, Solution 28.

The centroid coincides with the center of gravity because the wire is homogeneous.

L 1

r

2

2θ r

x

− −

l

3

−

r sin θ

θ l 2

X =

Then

r 2

xL

−

r2 2

− 2r 2 sin θ

l2 2

ΣxL = 0 ⇒ ΣxL = 0 and ΣL

r2 l2 − 2r 2 sin θ + = 0, or l = r 1 + 4sin θ 2 2

(a) θ = 15° :

l = r 1 + 4sin15°

or l = 1.427 r

(b) θ = 60° :

l = r 1 + 4sin 60°

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.

or l = 2.11 r

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 29.

y =

Then

ΣyA ΣA

 (a + h)  a ( ab ) −    kb ( a − h )  2 2   y = ba − kb ( a − h )

or

=

2 2 1 a (1 − k ) + kh 2 a(1 − k ) + kh

Let

c =1− k

Then

y =

and

ζ =

h a

a c + kζ 2 2 c + kζ

(1)

Now find a value of ζ (or h) for which y is minimum:

(

)

2 dy a 2kζ ( c + kζ ) − k c + kζ = =0 dζ 2 ( c + kζ ) 2

or

(

)

2ζ ( c + kζ ) − c + kζ 2 = 0

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.

(2)

COSMOS: Complete Online Solutions Manual Organization System

2cζ + 2ζ

Expanding (2)

2

ζ =

Then

− c − kζ − 2c ±

2

=0

or

kζ

2

+ 2cζ − c = 0

( 2c )2 − 4 ( k ) ( c ) 2k

Taking the positive root, since h > 0 (hence ζ > 0 ) 2

h=a

− 2 (1 − k ) + 4 (1 − k ) + 4k (1 − k )

2k 2

(a) k = 0.2:

h=a

(b) k = 0.6:

h=a

− 2 (1 − 0.2 ) + 4 (1 − 0.2 ) + 4 ( 0.2 )(1 − 0.2 )

2 ( 0.2 )

or h = 0.472a !

2

− 2 (1 − 0.6 ) + 4 (1 − 0.6 ) + 4 ( 0.6 )(1 − 0.6 )

2 ( 0.6 )

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.

or h = 0.387a !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 30.

From Problem 5.29, note that Eq. (2) yields the value of ζ that minimizes h.

Then from Eq. (2) We see

2ζ =

c + kζ 2 c + kζ

(3)

Then, replacing the right-hand side of (1) by 2ζ , from Eq. (3) We obtain

y =

a ( 2ζ) 2

But

ζ=

h a

So

y =h

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.

Q.E.D.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 31. \

Note that y1 = −

=

h x+h a

h (a − x) a

Choose the area element (EL) as

dA = ( h − y1 ) dx =

h xdx a a

Then

A=

h a h 1  1 xdx =  x 2  = ah ∫ 0 a a  2 0 2

Now, noting that xEL = x, and yEL =

1 ( h + y1 ) 2 a

1 2 a h 2 1  2  x = ∫ xdA = x xdx  = 2  x3  = a ∫ 0  A ah 3 a  a  3 0

y =

1 1 2 h 1 2 1 a 2  2 ∫ ( h + y1 ) dA = ah ∫ 0  2 ( h + y1 )  ( h − y1 ) dx  = ah  2  ∫ 0 h − y1 dx A 2    

(

)

A

1 a  2 h2 h h 1  2 2 3  1  1  = h − 2 ( a − x )  dx =  x +   2  ( a − x )  =  a − a  = h ∫ 0  ah  a a 3  3 a  3  a  0  Therefore:

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.

x =

2 a!

3

y =

2 h!

3

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 32.

First determine k: For x = a, y = 0 and therefore

(

)

0 = h 1 − ka3 or k = a −3 , and therefore  x3  y = h 1 − 3  a   Choosing an area element as in the figure:

xEL = x,

yEL = a 0

y , 2

A = ∫ dA = ∫ ydx = ∫

∫ xEL dA = ∫

a xydx 0

=∫

and dA = ydx a

 x3  x4  3 − 3 dx = h  x − 3  = ah 4 a  4a  0  

a  h 1 0 

a

 x2 x4  x5  3 2 − 3  dx = b  − 3  = ab a  5a  0 10  2

a  h x 0  2

a

1 a 2 x3  b2 a  2 x3 x 6  b2  x4 x7  9 a y  1 y dA ydx h x dx dx x ab 2 = = − = − + = − + =  ∫ EL ∫0  2  ∫ 0  ∫ 0  3 3 6 3 6   2 2 2 28 2 7 a a a a a        0 Now

x =

1 4 3a 2b 2 x dA = = a ∫ EL A 3ab 10 5

y =

1 4 9ab 2 3 yEL dA = = b ∫ A 3ab 28 7

and

Therefore:

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.

2 a 5 3 y = b 7

x =

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 33.

For the element (EL) shown

x = a, y = h: h = k1a3

At

k1 =

or

h a3

a = k 2 h3

k2 =

or

a h3

Hence, on line 1

y =

h 3 x a3

and on line 2

y =

h 1/3 x a1/3

Then

h   h dA =  1/3 x1/3 − 3 x3  dx a  a

and

yEL =

1  h 1/3 h 3  1/3 x + 3 x  2 a a a

h 1/3 h  1 1  3  ∴ A = ∫ dA = ∫ x − 3 x3  dx = h  1/3 x 4/3 − 3 x 4  = ah 1/3 2 a 4a a   4a 0 a 0

∫ xEL dA = ∫

a

h 1/3 h  1 8 2  3  x − 3 x3  dx = h  1/3 x 7/3 − 3 x5  = a h 1/3 a 5a a   7a  0 35

a  x 0 

1

a  1/3 3  1/3 3 ∫ yEL dA = ∫ 0 2  a1/3 x + a3 x  a1/3 x − a3 x  dx   

h

h

h

h

a

h 2 a  x 2/3 x 6  h 2  3 x5/3 1 x 6  8 2 = − dx = − ah    = ∫ 2/3 6 0 2  a 2  5 a5/3 7 a 6  35 a  0

From

8 2  ah  xA = ∫ xEL dA: x   = a h  2  35

or x =

16 a 35

and

8 2  ah  yA = ∫ yEL dA: y   = ah  2  35

or y =

16 h 35

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 34.

Choose as an area element (EL) the shaded area shown:

π  dA =  r  dr 2 

xEL =

2r

π

and r

π 1  2 π 2 r π  A = ∫r 2  r  dr =  r 2  = r2 − r12 1  2  2  2  r1 4

(

)

Then r

x =

1 4 4 r2  2r   π  1 3 2 x dA rdr = = ∫ EL ∫     r  A π r22 − r12 r1  π   2 π r22 − r12  3  r1

(

)

(

)

or x =

4 r23 − r13

3π r22 − r12

y =

4 r23 − r13

3π r22 − r12

and by symmetry

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 35.

Note that y1 = −

y2 =

b b x + b = ( a − x ) , and a a

b 2 a − x2 a

Then for the shaded area element: dA = ( y2 − y1 ) dx =

b 2 a − x 2 − ( a − x )  dx and  a  a

a

A = ∫ dA = ∫ 0

=

b 2 b  1  2   x  1 a − x 2 − ( a − x )  dx =   x a 2 − x 2 + a 2 sin −1    + ( a − x )      a a  2   a  2  0

b  1 π 1 2  ab (π − 2 )  × − a = 4 a2 2 2 

Noting that xEL = x, and that yEL = x =

1 ( y1 + y2 ): 2

1 4 ab 2 2 ∫ xELdA = ab π − 2 ∫ 0 a  x x − a − x ( a − x )dx A ( )

 21 = −   a2 − x2 ab (π − 2 )  3  2 

4

(

)

3 2

1   1 +  − ax 2 + x3   3    2

a

= 0

 1    a 2 2 a (π − 2 )  3 

4

( )

3 2

1   1 +  − a3 + a3  3    2

or x =

2a !

3 (π − 2 ) continued

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

y =

=

1 4 a 1  yEL dA = y + y1 )  ( y2 − y1 ) dx  ∫ ∫ 0  ( 2 A ab (π − 2 )  2 

2

ab (π − 2 )

(

)

2 2 ∫ y2 − y1 =

2 a 2 (π − 2 )

∫

2 ab 0  2

a

(

)

a2 − x2 −

b2 2 a − x )  dx 2( a  a

2b 4b 4b a a 1 2 1 3 = 3 2 ax − x 2 dx = 3 ax − x 2 dx = 3 ax − x  ∫ ∫ 0 0 3 0 a (π − 2 ) a (π − 2 ) a (π − 2 )  2

(

)

(

)

or y =

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.

2b !

3 (π − 2 )

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 36.

x =0

First note that symmetry implies For the element (EL) shown

y = R cos θ, x = R sin θ dx = R cos θ d θ

dA = ydx = R 2 cos 2θ dθ Hence α

1 2 α  θ sin 2θ  A = ∫ dA = 2∫ 0 R 2 cos 2 θ dθ = 2R 2  +  = R ( 2α sin 2α ) 4 0 2 2 α

∫ yEL dA = 2∫ 0 =

(

)

R3 cos 2 α sin α + 2sin α 3

(

)

R3 cos 2 α sin α + 2sin α 3 y = R2 ( 2α + sin 2α ) 2

(

But yA = ∫ yEL dA so

or

α

R 2 1  cosθ R 2 cos 2 θ dθ = R3  cos 2 θ sin θ + sin θ  2 3 3  0

(

)

)

cos 2 α + 2 2 y = R sin α 3 ( 2α + sin 2α )

Alternatively,

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.

y =

2 3 − sin 2 α R sin α

3 2α + sin 2α

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 37.

x = 0, y = b

At

b = k (0 − a)

2

y=

Now

xEL = x, yEL =

and

dA = ydx =

and

b a2

b 2 x − a) 2( a

Then

Then

k =

or

a

A = ∫ dA = ∫ 0

y b 2 = x − a) 2( 2 2a

b ( x − a )2 dx a2

a b b  1 2 3 x − a dx = x − a = ab ( ) ( ) 2 2     3 0 a 3a

2 a  a 3  2 2 ∫ xEL dA = ∫ 0 x  a 2 ( x − a ) dx  = a 2 ∫ 0 ( x − 2ax + a x )dx  

b

=

b  x4 2 3 a2 2  1 2 x  = ab − ax +  3 2 12 a 2  4  a

∫ yEL dA = ∫ 0 =

b

a

b b2  1 2 b 2 5  x − a )  2 ( x − a ) dx  = x − a)  2( 4  ( 2a a  2a  5 0

1 2 ab 10

1 2 1  ab Hence xA = ∫ xEL dA: x  ab  =  3  12 1 2 1  yA = ∫ yEL dA: y  ab  = ab 3 10  

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.

x = y =

1 a 4

3 b 10

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 38.

For the element (EL) shown on line 1 at

x = a, b = k 2 a 2 ∴ y =

or

∴ y =

b a2

b 2 x a2

x = a, −2b = k1a3

On line 2 at

k2 =

or

k2 =

−2b a3

−2b 3 x a3

2b   b dA =  2 x 2 + 3 x3  dx a  a Then

b  2 x3  b  x3 2 x 4  A = ∫ dA = ∫ 2  x 2 + +  dx = 2   x  4a  a  a  3

a

a 0

0

1 1 5 = ab  +  = ab 3 2 6 and ∫ xEL dA = ∫

a

b 2 2b 3  b  x 4 2 x5  2 2 1 + x + 3 x  dx = 2   = a b  +  2 a 4 5 4 5 a a  a    0

a  x 0 

13 2 ab 20 2b 3   b 2 2b 3   a1 b 2 ∫ yEL dA = ∫ 0 2  a 2 x − a3 x   a 2 x + a3 x  dx      

=

2

1  b 2   2b  =∫  2 x  −  3 x 3  2  a  a  a 0

a

2

 b 2  x5 2 − 2 x 7   dx = 4  2a  5 7a  0

2 13  1 = b 2a5  −  = − ab 2 10 7 70   Then

xA = ∫ xEL dA:

yA = ∫ yEL dA:

 5  13 2 x  ab  = ab  6  20

 5  13 2 y  ab  − ab  6  70

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.

x =

39 a 50

or y = −

39 b 175

or

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 39.

Using the area element shown:

xEL = x,

yEL =

A = ∫ dA = ∫

L  h 1 0 

y , 2

and dA = ydx L

 x x2  x 2 2 x3  5 + − 2 2  dx = h  x + − = hL 2 L 2L 3 L  0 6 L    L

  x 2 1 x3 2 x 4  x x2  x2 x3  1 2 L L ∫ xEL dA = ∫ 0 xh 1 + L − 2 L2  dx = h∫ 0  x + L − 2 L2  dx = h  2 + 3 L − 4 L2  = 3 hL      0 2

1 2 h2 L  x x2  h2 L  x2 x4 x x2 x3  ∫ yEL dA = 2 ∫ y dx = 2 ∫ 0 1 + L − 2 L2  dx = 2 ∫ 0 1 + L2 + 4 L4 + 2 L − 4 L2 − 4 L3  dx     L

h2  x3 4 x5 x 2 4 x3 x 4  4 2 h L = − 2 − 3 = x + 2 + 4 + L 2  L  0 10 3L 5L 3L Now

x =

1 6 1 2 2 ∫ xEL dA = 5hL  3 hL  = 5 L and A  

y =

1 6  4 2  12 ∫ yEL dA = 5hL  10 h L  = 25 h A  

Therefore:

x = y =

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.

2 L 5 12 h 25

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 40.

Note that y1 = 0 at x = a, or

(

)

0 = 2b 1 − ka 2 , i.e. k =

1 a2

Also, note that the slope of y2 is

y2 =

− 3b and y2 = 0 at x = 2a. Therefore a

3b ( 2a − x ) . a

Pick the area element dA ( EL ) such that: for 0 ≤ x ≤ a dA = ( 3b − y1 ) dx,

and xEL = x,

yEL =

1 ( 3b + y1 ) 2

and for a ≤ x ≤ 2a dA = y2 dx,

and xEL = x,

yEL =

1 y2 2

Then: 2b a 2a a 2 a 3b  A = ∫ dA = ∫ 0 ( 3b − y1 ) dx + ∫a y2dx = ∫ 0 3b − 2 a 2 − x 2  dx + ∫a ( 2a − x ) dx = a a  

(

a b 0

)

a

2b  2 3b 1 2 a 3b + 2 x 2 dx + ∫a ( 2a − x ) dx = b  x + 2 x3  +  −  ( 2a − x )2 ∫ a a  2 a 3a    0

2a

= a

continued

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

2  3b  19 2  ab 1 +  − ab − ( 2a − a )  =    3 2 a 6   Now for the centroid: x =

 1 6  a  2 2 3b 2a ∫ xdA = 19ab b∫ 0 x 1 + a 2 x  dx + a ∫a x ( 2a − x ) dx  = A     6  1 2 1 3 1  x + 2 x 4  +  ax 2 − x3  19a  2 3 a 2a 0 a 

2a 

a

=

6  1 1  8 1    =  +  + 3  4 − − 1 +   3 3   19  2 2  

or x =

y =

18 a!

19

1 6  a1 1 yEL dA = ( 3b + y1 )( 3b − y1 ) dx + ∫ a2a y2 y2dx  ∫ ∫  0 A 19ab  2 2  =

1 6  a1 2a 1 2  y2 dx  = 9b 2 − y12 dx + ∫ a ∫ 2 19ab  0 2 2 

=

2  2 1 6  a 1  2 4b 2 2 2 2a 9b 2   ∫ 0 9b − 4 a − x  dx + ∫a 2 ( 2a − x ) dx  2 19ab  2  a a  

=

 3b  a  8 4 2 2a 9  5 + 2 x 2 − 4 x 4  dx + ∫ a 2 ( 2a − x ) dx  ∫  0 19a   a a a  

=

a 2a  3b  8 4 9 1 3  5 x + 2 x3 − 4 x5  + 2  −  ( 2a − x )  19a  a  3a 5a 0 a  3  

=

3b  8 4 3  5 + −  + 3 (1)   19  3 5 

(

)

(

)

y =

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.

148 b!

95

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 41.

For y2

x = a, y = b : a = kb2

at

or

a b2

k =

b 1/2 x a

Then

y2 =

Now

xEL = x a y b x1/2 x1/2 : yEL = 2 = , dA = y2dx = b dx 2 2 2 a a

and for

0≤ x≤

For

a 1 b  x 1 x1/2  ≤ x ≤ a : yEL = ( y1 + y2 ) =  − +  2 2 2a 2 a 

 x1/2 x 1  dA = ( y2 − y1 ) dx = b  − +  dx  a a 2

Then

a/2

A = ∫ dA = ∫ 0 b

 x1/2 x 1  x1/2 a dx + ∫a/2 b  − +  dx a  a a 2 a

=

a/2  2 x3/2 b  2 3/2  x2 1  x + b − + x   2a 2  a/2 a 3 0 3 a

3/2 3/2 2 b  a  3/2 a  =   + ( a ) −    3 a  2   2   2   1  a  1  a   + b  −  a 2 −    + ( a ) −      2   2   2    2a  

( )

13 ab 24 1/2   x1/2 x 1    a/2  x a x dA x b dx x b = + − +   dx    ∫ EL ∫0  ∫ a  a/2   a a 2    =

and

b = a

a

a/2  2 x5/2 x3 x 4   2 5/2  x + b − +   5  3a 4  a/2  0 5 a

continued

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

=

5/2 5/2 2 b  a  5/2 a    + ( a ) −    5 a  2   2  

3 2   1  3  a   1  2  a    + b  − ( a ) −    + ( a ) −      2   4   2     3a   71 2 = ab 240

b x1/2  x1/2



a/2 dx  b ∫ yEL dA = ∫ 0 2 a a 

1 x1/2    x1/2 x 1   a b x + ∫ a/2  − + − +  dx   b  2a 2 a    a a 2   a

a/2 3 b2  1 2  b 2  x 2 1  x 1     = + − − x    2a  2  0 2  2a 3a  a 2      a/2

=

2 2 3 b  a  2  a   b2  a 1  −   + ( a ) −    −   4a  2   2   6a  2 2 

=

11 2 ab 48

Hence xA = ∫ xEL dA:

yA = ∫ yEL dA:

71 2  13  x  ab  = ab  24  240  13  11 2 y  ab  = ab  24  48

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.

x =

17 a = 0.546a !

130

y =

11 b = 0.423b !

26

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 42.

First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line Now Then and

xEL = r cos θ

and

dL = rd θ

7π /4

7π /4

L = ∫ dL = ∫π /4 rdθ = r [θ ]π /4 =

3 πr 2

7π /4 ∫ xEL dL = ∫π /4 r cosθ ( rdθ )

1   1 7π /4 2 = r 2 [sin θ ]π /4 = r 2  − −  = −r 2 2 2 

Thus

3  xL = ∫ xdL : x  π r  = −r 2 2 2 

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.

x =−

2 2 r 3π

COSMOS: Complete Online Solutions Manual Organization System

SOLUTION 5.43 CONTINUED

(

dy = a 2/3 − x 2/3 dx

Then

Then

and

Hence

1/2

−1/3

xEL = x

Now and

) ( −x )

 dy  dL = 1 +    dx 

2

  dx = 1 +  a 2/3 − x 2/3  

(

a

L = ∫ dL = ∫ 0

) ( −x ) 1/2

−1/3

 

1/2 2

  dx 

a

a1/3 3 3  dx = a1/3  x 2/3  = a 1/ 3 2 x 2 0

a 1/3  3 2 a a 1/3  3 5/3  x dL x dx a x = =   ∫ EL ∫ 0  x1/3  5  = 5a  0  

3  3 xL = ∫ xEL dL : x  a  = a 2 2  5

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.

x =

2 a 5

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 43.

First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line Now

xEL = a cos3 θ

dx 2 + dy 2

dL =

and

x = a cos3 θ : dx = −3a cos2 θ sin θ dθ

Where

y = a sin 3 θ : dy = 3a sin 2 θ cosθ dθ

Then

(

 dL =  −3a cos 2 θ sin θ dθ 

1/2

) + (3a sin θ cosθ dθ )  2

2

2

(

= 3a cosθ sin θ cos 2 θ + sin 2 θ

)

1/2

dθ

= 3a cosθ sin θ dθ π /2

∴ L = ∫ dL = ∫0 = and

π /2

1  3a cosθ sin θ dθ = 3a  sin 2 θ  2 0

3 a 2

π /2 3 ∫ xELdL = ∫0 a cos θ ( 3a cosθ sin θ dθ ) π /2

 1  = 3a  − cos5 θ   5 0 2

=

3 2 a 5

3  3 xL = ∫ xEL dL : x  a  = a 2 2  5

Hence

x =

Alternative solution  x x = a cos3 θ ⇒ cos 2 θ =    a  y y = a sin 3 θ ⇒ sin 2 θ =    a

 x ∴   a

2/3

 y +  a

2/3

=1

or

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.

2/3

2/3

(

y = a 2/3 − x 2/3

)

3/2

2 a !

5

COSMOS: Complete Online Solutions Manual Organization System

(

dy = a 2/3 − x 2/3 dx

Then

Then

and

Hence

1/2

−1/3

xEL = x

Now and

) ( −x )

 dy  dL = 1 +    dx 

2

  dx = 1 +  a 2/3 − x 2/3  

(

a

L = ∫ dL = ∫ 0

) ( −x ) 1/2

−1/3

 

1/2 2

  dx 

a

a1/3 3 3  dx = a1/3  x 2/3  = a 1/ 3 2 x 2 0

a 1/3  3 2 a a 1/3  3 5/3  x dL = x dx = a x   ∫ EL ∫ 0  x1/3  5  = 5a  0  

3  3 xL = ∫ xEL dL : x  a  = a 2 2  5

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.

x =

2 a !

5

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 44.

First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line Have at

x = a, y = a : a = ka 2

Thus

y =

Then

 dy  2  dL = 1 +   dx = 1 +  x  dx  dx  a 

1 2 x a

dy =

and

∴ L = ∫ dL = ∫ =

2

x 4 4x2 a 2 4x2 1 + 2 x 2 dx =  1 + 2 + ln  x + 1 + 2 4 a 2 a a a  

(

a

   0

)

a a 5 + ln 2 + 5 = 1.4789a 2 4

∫ xEL dL = ∫

a



3/2 4 4x2   2  a2     1 + 2 dx  =    1 + 2 x 2     3  8  a a      0

a x 0 

a 2 3/2 5 − 1 = 0.8484a 2 12 xL = ∫ xEL dL: x (1.4789a ) = 0.8484a 2

=

Then

1 a

2 xdx a

2

a 0

k =

or

(

)

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.

x = 0.574a

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 45.

xEL = x,

Have

1 πx x sin L 2

dA = ydx

and L/2

A = ∫ dA = ∫0 x sin

and

yEL =

L/2

 L2 πx L πx − x cos  dx =  2 sin π L L L 0 π

πx

=

L2

π2

πx  L/2  x = ∫ xEL dA = ∫0 x  x sin dx  L   L/2

 2 L2  π x  2 L3  π x  L 2  π x  =  2 x sin   + 3 cos   − x sin    L  π  L  π  L  0 π

Also

L/2 1

y = ∫ yEL dA = ∫0

2

x sin

=

L3

π2

−2

L3

π3

πx

πx  dx   x sin L  L  L/2

1  2 L2 πx  L 2 L3  πx =  2 x sin −  x − 3  cos  2  π L π L  π  0

=

Hence

 L2  L  L3 1  1  L3  1 6 + π2 − = ( )    −    2 2 2  6  8  4π  2   96π

(

)

 L2  z   1 xA = ∫ xEL dA: x  2  = L3  2 − 3  π  π π 

or  L2  L3  1 2  − 3 yA = ∫ yEL dA: y  2  = 2 2 π   π  96π  π

or y = 0.1653L !

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.

x = 0.363L !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 46.

First note that by symmetry y = 0.

Using the area element shown in the figure,

xEL =

dA =

2 2 r cosθ = R cos 2θ cosθ 3 3

1 2 1 r dθ = R 2 cos 2 2θ dθ 2 2 π

π

π

1 1 A = ∫ dA = R 2 ∫ 4π cos 2 θ dθ = R 2 ∫ 04 cos 2 θ dθ = R 2 ∫ 04 (1 + cos 4θ ) dθ − 2 2 4

π

1  1 4 1 = R 2 θ + sin 4θ  = π R 2 2  4 0 8 π

2

1

2

π

   2 2 2 3 ∫ xEL dA = ∫−4π  3 R cos 2θ cosθ  2 R cos 2θ dθ  = 3 R ∫ 04 cos 2θ cosθ dθ    ρ π

=

2 3 4 R ∫ 0 1 − 2sin 2 θ 3

(

)

3

cosθ dθ =

π

2 3 4 R ∫ 0 1 − 6sin 2 θ + 12sin 4 θ − 8sin 6 θ cosθ dθ 3

(

)

π

2  12 8 4 = R3 sin θ − 2sin 3 θ + sin 5 θ − sin 7 θ  3  5 7 0 =

2 3 2 2 12 1 8 1  16 2 3 1− + R − = R  3 2  2 5 4 7 8  105

Now:

x=

1 8  16 2 3  128 2 xEL dA = R  = R ∫ 2  A 105π π R  105  or x = 0.549 R

y =0

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 47.

From the solution to problem 5.2:

A = 134 in 2 ,

ΣxA = 1102 in 3,

ΣyA = 536 in 3

and from the solution of problem 5.22

L = 56 in., and ΣxL = 488 in 2 Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x-axis:

(

)

Volume = 2π yarea A = 2πΣyA = 2π 536 in 3 = 3367.8 in 3

or

V = 1.949 ft 3

Area = 2π ylength L = 2πΣyL = 2π 6 (15 ) + 10 ( 4 ) + 8 (10 ) + 4 (18 )  = 1520.53 in 2

or

A = 10.56 ft 2

(b) Rotation about x = 19 in.:

(

)

Volume = 2π (19 − xarea ) A = 2π (19 A − ΣxA ) = 2π (19 in ) 134 in 2 − 1102 in 3   

= 9072.9 in 3

or

V = 5.25 ft 3

Area = 2π (19 − xline ) L = 2π (19L − ΣxL ) = 2π (19 in.)( 56 in.) − 488 in 2  = 3619.1 in 2

or

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.

A = 25.1 ft 2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 48.

From the solution to problem 5.4:

A = 772 in 2 ,

ΣxA = 17828 in 3 ,

( Area )

ΣyA = 18464 in 3

and for the line

L

y

xL

yL

1

13

27.5

0

357.5

0

2

40

34

20

1360

800

3

34

17

40

578

1360

10.5

28

334.85

892.92

21

8

336

128

2966.4

3180.9

4

(a)

x

212 + 242 = 31.890

5

16

Σ

134.89

(

)

or V = 64.8 ft 3 !

(

)

or A = 129.4 ft 2 !

V = 2π xarea A = 2πΣxA = 2π 17828 in 3 = 112 017 in 3 A = 2π xline L = 2πΣxL = 2π 2966.4 in 2 = 18 638.1 in 2

continued

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

(b) V = 2π ( 40 − yarea ) A = 2π ( 40 A − ΣyA )

(

)

= 2π ( 40 in.) 772 in 2 − 18 464 in 3  = 78 012 in 3   or V = 45.1 ft 3 !

A = 2π ( yline ) y = 40 L = − 2π ΣL ( y − 40 )  = − 2π ( ΣLy − 40ΣL )

= − 2π ( 3180.9 − 40 × 134.89 ) = 13 915.3 in 2 or A = 96.6 ft 2 !

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 49.

From the solution of Problem 5.1:

A = 150000 mm 2 ,

x A = 140 mm,

y A = 165 mm

From the solution of Problem 5.21:

L = 1800 mm,

xL = 116.667 mm,

yL = 166.667 mm

Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x-axis:

Ax = 2π yL L = 2π (166.667 mm )(1800 mm ) = 1884 960 mm 2

(

or

A = 1.885 × 106 mm 2

or

V = 155.5 × 106 mm3

)

Vx = 2π y A = 2π (165 mm ) 150 000 mm 2 = 155 509 000 mm 3

(b) Rotation about x = 400 mm:

Ax = 400 mm = 2π ( 400 mm − xL ) L = 2π ( 400 − 116.667 ) mm  (1800 mm ) = 3 204 420 mm 2 or

(

A = 3.20 × 106 mm 2

)

Vx = 2π ( 400 mm − x A ) A = 2π ( 400 − 140 ) mm  150 000 mm 2 = 245 040 000 mm 3 or

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.

V = 245 × 106 mm3

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 50.

Applying the second theorem of Pappus-Guldinus, we have (a) Rotation about axis AA′:

 π ab  2 2 Volume = 2π yA = 2π ( a )  =π a b  2 

V = π 2a 2b

(b) Rotation about axis BB′:

 π ab  2 2 Volume = 2π yA = 2π ( 2a )   = 2π a b  2 

V = 2π 2a 2b

(c) Rotation about y-axis:

 4a  π ab  2 2 Volume = 2π yA = 2π    = πa b  3π  2  3

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.

V =

2 2 πa b 3

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 51.

The area A and circumference C of the cross section of the bar are

A=

π 4

d 2 and C = π d .

Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V:

V = 2 (Vside ) + 2 (V

end

) = 2 ( AL ) + 2 (π RA) = 2 ( L + π R ) A

2 π or V = 2 3 in. + π ( 0.75 in.)   ( 0.5 in.)  = 2.1034 in 3 4  

or V = 2.10 in 3

For the area A:

A = 2 ( Aside ) + 2 ( Aend ) = 2 ( CL ) + 2 (π RC ) = 2 ( L + π R ) C or A = 2 3 in. + π ( 0.75 in.)  π ( 0.5 in.)  = 16.8270 in 2

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.

or A = 16.83 in 2

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Chapter 5, Solution 52.

Following the second theorem of Pappus-Guldinus, in each case a specific generating area A will be rotated about the x axis to produce the given shape. Values of y are from Fig. 5.8A. (1) Hemisphere: the generating area is a quarter circle

Have

 4a  π  V = 2π yA = 2π   a 2   3π  4 

or V =

2 3 πa !

3

(2) Semiellipsoid of revolution: the generating area is a quarter ellipse Have

 4a  π  V = 2π yA = 2π   ha   3π  4  or V =

2 2 πa h!

3

(3) Paraboloid of revolution: the generating area is a quarter parabola Have

 3  2  V = 2π yA = 2π  a  ah   8  3  or V =

1 2 πa h!

2

or V =

1 2 πa h!

3

(4) Cone: the generating area is a triangle

Have

 a  1  V = 2π yA = 2π   ha   3  2 

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.

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Chapter 5, Solution 53.

The required volume can be generated by rotating the area shown about the y axis. Applying the second theorem of Pappus-Guldinus, we have

 5  1   V = 2π xA = 2π  + 7.5  mm  ×  × 5 mm × 5 mm  3 2     

or V = 720 mm3

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.

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Chapter 5, Solution 54.

Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt is given by: AC = π yL = π ΣyL

where the individual lengths are the lengths of the belt cross section that are in contact with the pulley. (a)

 0.125    0.125 in.    AC = π  2 ( y1L1 ) + y2 L2  = π  2  3 − + ( 3 − 0.125 ) in. ( 0.625 in.)   in.   2    cos 20°     or AC = 8.10 in 2 (b)

 0.375    0.375 in.  AC = π  2 ( y1L1 )  = 2π  3 − 0.08 −  in.   2    cos 20°   or AC = 6.85 in 2 (c)

 2 ( 0.25 )   AC = π  2 ( y1L1 )  = π  3 −  in. π ( 0.25 in.)  π     or AC = 7.01 in 2

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.

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Chapter 5, Solution 55.

Volume: The volume can be obtained by rotating the triangular area shown through π radians about the y axis.

The area of the triangle is:

A=

1 ( 52 )( 60 ) = 1560 mm2 2

Applying the theorems of Pappus-Guldinus, we have

(

V = π xA = π ( 52 mm ) 1560 mm 2

)

or V = 255 × 103 mm3 !

The surface area can be obtained by rotating the triangle shown through an angle of π radians about the y axis.

Considering each line BD, DE, and BE separately: 22 = 31 mm 2

Line BD : L1 = 222 + 602 = 63.906 mm

x1 = 20 +

Line DE : L2 = 52 mm

x2 = 20 + 22 + 26 = 68 mm

Line BE : L3 = 742 + 602 = 95.268 mm

x1 = 20 +

74 = 57 mm 2 continued

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.

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Then applying the theorems of Pappus-Guldinus for the part of the surface area generated by the lines: AL = πΣxA = π ( 31)( 63.906 ) + ( 68 )( 52 ) + ( 57 )( 95.268 )  = π [10947.6] = 34.392 × 103 mm 2

The area of the “end triangles”: 1  AE = 2  ( 52 )( 60 )  = 3.12 × 103 mm 2 2  Total surface area is therefore:

A = AL + AE = ( 34.392 + 3.12 ) × 103 mm 2

or A = 37.5 × 103 mm 2 !

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.

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Chapter 5, Solution 56.

The mass of the escutcheon is given by m = ( density )V , where V is the volume. V can be generated by rotating the area A about the x-axis.

From the figure: L1 = 752 − 12.52 = 73.9510 m

L2 =

37.5 = 76.8864 mm tan 26°

a = L2 − L1 = 2.9354 mm

φ = sin −1 α=

12.5 = 9.5941° 75

26° − 9.5941° = 8.2030° = 0.143168 rad 2

Area A can be obtained by combining the following four areas:

Applying the second theorem of Pappus-Guldinus and using Figure 5.8 a, we have V = 2π yA = 2π ΣyA continued

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.

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A, mm 2

Seg .

1

1 ( 76.886 )( 37.5) = 1441.61 2

2

− α ( 75 ) = − 805.32

2

3

−

1 ( 73.951)(12.5) = − 462.19 2

− ( 2.9354 )(12.5 ) = − 36.693

4

y A , mm3

y , mm

1 ( 37.5) = 12.5 3

18 020.1

2 ( 75) sin α sin (α + φ ) = 15.2303 3α

−12 265.3

1 (12.5) = 4.1667 3

−1925.81

1 (12.5) = 6.25 2

− 229.33

Σ

3599.7

Then

(

)

V = 2π ΣyA = 2π 3599.7 mm3 = 22618 mm3 m = ( density )V

(

)(

= 8470 kg/m3 22.618 × 10−6 m3

)

= 0.191574 kg or m = 191.6 g !

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.

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Chapter 5, Solution 57.

The volume of the waste wood is:

Vwaste = Vblank − Vtop , where 2

Vblank = π ( 22 in.) (1.25 in.) = 1900.664 in 3

Vtop = V1 + V2 + V3 + V4 The volumes Vi can be obtained through the use of the theorem of Pappus-Guldinus: 2 2 Vtop = π ( 21.15 in.) ( 0.75 in.)  + π ( 21.4 in.) ( 0.5 in.)     

  ( 4 )( 0.5)  in. × π 0.5 in. 2  + 2π   21.15 + ( 4 )( 0.75)  in. × π 0.75 in. 2  + 2π   21.4 + ( ) ( )       3π   4 3π   4       = (1053.979 + 719.362 + 26.663 + 59.592 ) in 3 = 1859.596 in 3 Therefore

Vwaste = 1900.664 in 3 − 1859.596 in 3 = 41.068 in 3 Then

Wwaste = γ woodVwaste N tops

(

)(

)

= 0.025 lb/in 3 41.068 in 3 ( 5000 tops ) = 5133.5 lb,or Vwaste = 5.13 kips

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.

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Chapter 5, Solution 58.

The total surface area can be divided up into the top circle, bottom circle, and the edge.

ATotal = ATop circle + ABottom circle + AEdge,or 2 2 ATotal = π ( 21.4 in.)  + π ( 21.15 in.)     

   2 ( 0.5 )   π +  2π  21.4 +  in. × ( 0.5 in.)  + π   2   

   2 ( 0.75 )   π  in. × ( 0.75 in.)   2π  21.15 + π   2   

= (1438.72 + 1405.31 + 107.176 + 160.091) = 3111.3 in 2 Now, knowing that 1 gallon of lacquer covers 500 ft2, the number of gallons needed, NGallons is

N Gallons = ASurface × coverage × ( number of tops ) × ( number of coats ) N Gallons = 3111.3 in 2 ×

1 Gallon

( 500 ) (144 in 2 )

× 5000 × 3

= 648.19 gal or N Gallons = 648

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.

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Chapter 5, Solution 59.

The mass of the lamp shade is given by

m = ρV = ρ At where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line shown about the x axis. Applying the first theorem of Pappus-Guldinus we have

A = 2π yL = 2πΣyL = 2π ( y1L1 + y2 L2 + y3L3 + y4 L4 )  13 mm 13 + 16  or A = 2π  (13 mm ) +   mm × 2  2    16 + 28  +  mm × 2  

( 32 mm )2 + ( 3 mm )2

(8 mm )2 + (12 mm )2

 28 + 33  +  mm × 2  



( 28 mm )2 + ( 5 mm )2  

= 2π ( 84.5 + 466.03 + 317.29 + 867.51) = 10903.4 mm 2 Then

(

)(

)

m = ρ At = 2800 kg/m 3 10.9034 × 10−3 m 2 ( 0.001 m ) or m = 30.5 g

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.

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Chapter 5, Solution 60.

Free-Body Diagram: First note that the required surface area A can be generated by rotating the parabolic cross section through 2π radians about the x axis. Applying the first theorem of Pappus-Guldinus, we have

A = 2π yL 2

Now, since

x = ky ,

x = a : a = k ( 7.5 )

at

2

or

a = 56.25 k

(1) x = ( a + 15 ) mm: a + 15 = k (12.5 )

At

2

or

(2)

a + 15 = 156.25k

Then

Eq. (2) a + 15 156.25k : = Eq. (1) a 56.25k

or a = 8.4375 mm

Eq. (1) ⇒ k = 0.15 ∴ x = 0.15 y 2

and

1 mm dx = 0.3 y dy

2

Now

 dx  dL = 1 +   dy = 1 + 0.09 y 2 dy  dy 

So

A = 2π yL

yL = ∫ ydL

and 12.5

∴ A = 2π ∫7.5 y 1 + 0.09 y 2 dy 2 1  2 = 2π    1 + 0.09 y  3  0.18 

(

= 1013 mm 2

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.

12.5 3/2 

)

  7.5

or A = 1013 mm 2

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Chapter 5, Solution 61.

(a) Note that in the free-body diagram: R1 =

1 ( 4.2 m )( 600 N/m ) = 1260 N, 2

and R2 =

1 ( 4.2 m )( 240 N/m ) = 504 N 2

Then for the equivalence of the systems of forces: ΣFy :

R = R1 + R2 = 1260 + 504 = 1764 N

ΣM A :

  1 2     − x (1764 N ) =  2 + 4.2  m  (1260 N ) +  2 + 4.2  m  ( 504 N ) = 3.8000 m 3 3      

R = 1764 N

or x = 3.80 m (b) Equilibrium: ΣFx = 0:

Ax = 0

ΣFy = 0:

Ay − 1764 = 0 A = 1764 N

ΣΜ Α = 0:

M A − ( 3.80 m )(1764 N ) = 0 M A = 6.70 kN ⋅ m

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.

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Chapter 5, Solution 62.

With R1 = ( 20 lb/ft )(18 ft ) = 360 lb, and R2 =

ΣFy : or

1 ( 60 lb/ft )(18 ft ) = 360 lb: 3

− R = − R1 − R2 R = 360 lb + 360 lb = 720 lb R = 720 lb

+ ΣM A :

− x ( 720 lb ) = − ( 9 ft )( 360 lb ) − (13.5 ft )( 360 ft ) x = 11.25 ft

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.

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Chapter 5, Solution 63.

R = (1800 N/m )( 3.2 m ) = 5.76 kN + ΣM A = 0:

− ( 5.76 kN )(1.2 m + 1.6 m ) + By ( 3.6 m ) = 0, or

By = 4.48 kN B = 4.48 kN ΣFy = 0:

Ay + 4.48 − 5.76 = 0, or Ay = 1.28 kN

+ ΣFx = 0:

Ax = 0

Therefore:

A = 1.28 kN

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.

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Chapter 5, Solution 64.

 kN  R1 =  1.5  (1.6 m ) = 2.4 kN m  

R2 =

1  kN  3  ( 2.4 m ) = 3.6 kN 2 m 

 kN  R3 =  3  (1.6 m ) = 4.8 kN  m  Equilibrium:

+ ΣFx = 0:

Ax = 0

+ ΣM B = 0:

( 4.8 m )( 2.4 kN ) + ( 2.4 m )( 3.6 kN ) + ( 0.8 m )( 4.8 kN ) − ( 4.0 m ) Ay Ay = 6.0000 kN

+ ΣFy = 0:

6 kN − 2.4 kN − 3.6 kN − 4.8 kN + By = 0 By = 4.8000 kN

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.

=0

A = 6.00 kN B = 4.80 kN

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Chapter 5, Solution 65.

lb   R1 =  240  ( 4.8 ft ) = 1152 lb ft   R2 =

1 lb  180  ( 3.6 ft ) = 324 lb 2 ft 

Equilibrium:

+ ΣFx = 0:

Ax = 0

+ ΣFy = 0:

Ay − 1152 lb + 324 lb = 0 Ay = 828.00 lb

+ ΣM A = 0:

A = 828 lb

M A − ( 2.4 ft )(1152 lb ) + ( 6 ft )( 324 lb ) = 0 M A = 820.80 lb ⋅ ft

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.

M A = 821 lb ⋅ ft

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Chapter 5, Solution 66.

The distributed load given can be simplified as in the diagram below with the resultants R1 and R2.

The resultants are:

R1 = ( 6 ft )( 30 lb/ft ) = 180 lb, and R2 =

1 ( 4.5 ft )(120 lb/ft ) = 270 lb 2

Now, for equilibrium:

ΣFx = 0:

Ax = 0

ΣFy = 0:

Ay + 180 − 270 = 0 Ay = 90.0 lb

Therefore:

ΣM A = 0:

A = 90.0 lb

or M = 675 lb ⋅ ft

2   M A + ( 3 ft )(180 lb ) − 1.5 + × 4.5  ft × ( 270 lb ) = 0 3  

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.

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Chapter 5, Solution 67.

 kN  R1 = 1.5  ( 2.4 m ) = 3.6 kN m  

R2 =

2  kN  9  ( 2.4 m ) = 14.4 kN 3 m 

Equilibrium:

+ ΣFx = 0: + ΣM B = 0:

Ax = 0

− ( 3.3 m ) Ay − (1.8 m )( 3.6 kN ) + ( 2.1 m )(14.4 kN ) = 0 Ay = 7.2000 kN

+ ΣFy = 0:

A = 7.20 kN

B = 3.60 kN

7.2 kN + 3.6 kN − 14.4 kN + By = 0 By = 3.6000 kN

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.

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Chapter 5, Solution 68.

The resultants:

R1 =

2 ( 3.2 ft )(120 lb/ft ) = 256 lb 3

R2 =

1 ( 2.4 ft )(120 lb/ft ) = 96 lb 3

R3 =

1 (1.6 ft )( 45 lb/ft ) = 24 lb 3

Then for equilibrium:

ΣFx = 0: ΣM B = 0:

Ax = 0

3   − ( 7.2 ft ) Ay +  4 + × 3.2  ft × ( 256 lb ) 8   3   1  +  1.6 + × 2.4  ft × ( 96 lb ) +  × 1.6 ft  ( 24 lb = 0 ) 4 4    

Ay = 231.56 lb

ΣFy = 0:

A = 232 lb

B = 144.4 lb

23.56 − 256 − 96 − 24 + By = 0

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.

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Chapter 5, Solution 69.

Have

RI =

1 ( 9 m )( 2 kN/m ) = 9 kN 2

RII = ( 9 m )(1.5 kN/m ) = 13.5 kN Then

ΣFx = 0: C x = 0 ΣM B = 0: − 50 kN ⋅ m − (1 m )( 9 kN ) − ( 2.5 m )(13.5 kN ) + ( 6 m ) C y = 0

or

C y = 15.4583 kN

C = 15.46 kN

B = 7.04 kN

ΣFy = 0: By − 9 kN − 13.5 kN + 15.4583 = 0 or

By = 7.0417 kN

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.

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Chapter 5, Solution70.

Have

RI =

1 ( 9 m ) ( 3.5 − w0 ) kN/m  = 4.5 ( 3.5 − w0 ) kN 2

RII = ( 9 m ) ( w0 kN/m ) = 9w0 kN (a) Then or

ΣM C = 0: − 50 kN ⋅ m + ( 5 m )  4.5 ( 3.5 − w0 ) kN  + ( 3.5 m ) ( 9w0 kN ) = 0 9w0 + 28.75 = 0

so

w0 = − 3.1944 kN/m

w0 = 3.19 kN/m

C = 1.375 kN

Note: the negative sign means that the distributed force w0 is upward.

(b)

ΣFx = 0: C x = 0 ΣFy = 0: − 4.5 ( 3.5 + 3.19 ) kN + 9 ( 3.19 ) kN + C y = 0 or

C y = 1.375 kN

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.

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Chapter 5, Solution 71.

The distributed load can be represented in terms of resultants: R1 = ( 8 m )( 300 N/m ) = 2400 N

R2 =

1 ( 8 − a ) m  ( 2400 N/m ) = 1200 ( 8 − a ) N 2

For equilibrium: ΣM B = 0:

1  − 8 Ay + 4 ( 2400 ) +  ( 8 − a )  1200 ( 8 − a )  = 0 3  Ay = 1200 + 50 ( 8 − a )

ΣM A = 0:

2

(1)

2   8By − 4 ( 2400 ) −  a + ( 8 − a )  1200 ( 8 − a )  = 0 3   By = 1200 + 50 (16 + a )( 8 − a )

(a) ΣFy = 0:

(2)

Ay + By − 2400 − 1200 ( 8 − a ) = 0

(3)

Using the requirement By = 2 Ay and (1) 2 3 1200 + 50 ( 8 − a )  − 2400 − 1200 ( 8 − a ) = 0  

or

(8 − a )2 − 8 (8 − a ) + 8 = 0,

which gives continued

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.

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(8 − a ) =

8±

( − 8 )2 − 4 ( 8 ) 2

= 6.82843 m or 1.17157 m

a = 1.17157 m or a = 6.82843 m, and therefore

amin = 1.17157 m or amin = 1.172 m ! (b) ΣFx = 0:

Ax = 0

Equation (1) gives: Ay = 1200 + 50 ( 6.82843)

2

= 3531.4 N or A = 3.53 kN

!

B = 7.06 kN

!

By = 2 Ay gives By = 2 ( 3531.4 N ) , and

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.

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Chapter 5, Solution 72.

The distributed load can be represented in terms of resultants: R1 = ( 8 m )( 300 N/m ) = 2400 N

R2 =

1 ( 8 − a ) m  ( 2400 N/m ) = 1200 ( 8 − a ) N 2

For equilibrium: ΣM B = 0:

1  − 8 Ay + 4 ( 2400 ) +  ( 8 − a )  1200 ( 8 − a )  = 0 3  Ay = 1200 + 50 ( 8 − a )

ΣM A = 0:

2

(1)

2   8By − 4 ( 2400 ) −  a + ( 8 − a )  1200 ( 8 − a )  = 0 3  

By = 1200 + 50 (16 + a )( 8 − a )

(2)

(a) Dividing Equation (1) by Equation (2): By Ay

=

1200 + 50 (16 + a )( 8 − a ) 1200 + 50 ( 8 − a )

2

=

( ) 24 + ( 64 − 16a + a )

=

152 − 8a − a 2 88 − 16a + a 2

24 + 128 − 8a − a 2

2

continued

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.

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Differentiating d  By  da  Ay

By Ay

:

(

) (

)

 ( − 8 − 2a ) 88 − 16a + a 2 − 152 − 8a − a 2 ( −16 + 2a ) =0 = 2  88 − 16a + a 2 

(

)

2

a − 20a + 72 = 0

or

a=

or

20 ±

( − 20 )2 − 4 ( 72 ) 2

Knowing that a ≤ 8 m: a = 4.7085 m or

a = 4.71 m !

(b) For equilibrium:

ΣFx = 0:

Ax = 0

and from (1): Ay = 1200 + 50 ( 8 − 4.7085 )

2

= 1741.70 N A = 1.742 kN

!

B = 4.61 kN

!

Also, ΣFy = 0:

1741.70 − 2400 − 1200 ( 8 − 4.7085 ) + By = 0 By = 4608.1 N

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.

or

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Chapter 5, Solution 73.

R1 = ( 3.6 ft )( wA kips/ft ) = 3.6wA kips R2 =

1 1 ( 5.4 ft )  wA kips/ft  = 1.35wA kips 2 2 

1  R3 = ( 5.4 ft )  wA kips/ft  = 2.7wA kips 2   Equilibrium:

− (1.8 ft ) ( 3.6wA ) kips  + (1.8 ft ) (1.35wA ) kips 

+ ΣM C = 0:

+ ( 2.7 ft ) ( 2.7 wA ) kips  + ( 2.1 ft )( 6 kips ) − ( 2.4 ft )( 4.5 kips ) − ( 3.6 ft )(1 kip ) = 0 wA = 0.55556 kips/ft

+ ΣFyA = 0:

or

wA = 556 lb/ft

RR − ( 3.6 )( 0.55556 ) kips + 1.35 ( 0.55556 ) kips

+ 2.7 ( 0.55556 ) kips − 6 kips − 4.5 kips − 1 kip = 0 Solving for RR :

RR = 7.2500 kips

or

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.

RR = 7.25 kips

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Chapter 5, Solution 74.

R1 = ( 3.6 ft )( wA kips/ft ) = 3.6wA kips

R2 =

1 ( 5.4 ft )( 0.6wA kips/ft ) = 1.62wA kips 2

R3 = ( 5.4 ft )( 0.4wA kips/ft ) = 2.16wA kips Equilibrium: + ΣM A = 0:

− (1.8 ft ) ( 3.6wA ) kips  + ( 3.6 ft ) RR + ( 5.4 ft ) (1.62wA ) kips  + ( 6.3 ft ) ( 2.16 wA ) kips  − (1.5 ft )( 6 kips )

− ( 6 ft )( 4.5 kips ) − ( 7.2 ft ) P = 0 or

+ ΣFy y = 0: or

28.836wA + 3.6RR − 7.2 P − 36 = 0

(1)

RR + 3.6wA + 1.62wA + 2.16wA − 6 − 4.5 − P = 0

7.38wA + RR − P − 10.5 = 0

( 28.836 ) Eq. ( 2 ) − ( 7.38) Eq. (1) = 0 gives 2.268RR − 37.098 + 24.3P = 0

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.

(2)

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Since RR ≥ 0, the maximum acceptable value of P is that for which RR = 0, and P = 1.52667 kips

or

P = 1.527 kips !

(b) Now, from (2): 7.38wA − 1.52667 − 10.5 = 0

or

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.

wA = 1.630 kips/ft !

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Chapter 5, Solution 75.

Noting that the weight of a section of the dam is Wi = γ Vi (Vi being the volume of that section ) :

lb   W1 = 150 3  (10.5 ft )( 9 ft )(1 ft )  = 14175 lb ft  

lb   1   W2 = 150 3   (10.5 ft )( 21 ft )(1 ft )  = 16537.5 lb ft   2   lb   W3 = 150 3  (18 ft )( 30 ft )(1 ft )  = 81000 lb ft   lb   1   W4 = 150 3   ( 3 ft )( 30 ft )(1 ft )  = 6750 lb 2 ft    From the free-body diagram: x1 = 5.25 ft, x2 =

2 (10.5 ft ) = 7 ft, x3 = 19.5 ft, and x4 = 29.5 ft 3

For the distance a: a 3 = , or a = 2.4 ft 24 30

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.

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Therefore: lb   1   Ws =  62.4 3   ( 2.4 ft )( 24 ft )(1 ft )  = 1797.12 lb, and ft   2  

xs = 31.5 −

1 ( 2.4 ) = 30.7 ft 3

Now, for the pressure force P: P= =

1 1 PB A = (γ W hB ) A 2 2 1 lb   62.4 3  ( 24 ft ) ( 24 ft )(1 ft )  2 ft 

= 17971.2 lb Then, for equilibrium: (a) ΣFx = 0:

H −P=0 H = 17971.2 lb or H = 17.97 kips

ΣFy = 0:

!

V − 14175 − 16537.5 − 81000 − 6750 − 1797.12 = 0 V = 120259.62 lb or V = 120.3 kips

!

(b) From moment equilibrium: ΣM A = 0:

1  x (120259.62 lb ) +  × 24 ft  (17971.2 lb ) − ( 5.25 ft )(14175 lb ) − ( 7 ft )(16537.5 lb ) 3  

(19.5 ft )(81000 lb ) − ( 29.5 ft )( 6750 lb ) − ( 30.7 ft )(1797.12 lb ) = 0 or x = 15.63 ft ! (c) free-body diagram for section of water:

continued 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.

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For equilibrium: ΣF = 0:

Ws + P + ( − R ) = 0

where R is the force of the water on the face BD of the dam, and

P = 17971.2 lb, and Ws = 1797.12 lb Then from the force triangle: R=

(17971.2 lb )2 + (1797.12 lb )2

= 18.06 kips

 1797.12  θ = tan −1   = 5.71°  17971.2 

Therefore: R = 18.06 kips

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.

5.71° !

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Chapter 5, Solution 76.

Free-Body Diagram:

Locations of centers of gravity:

x1 =

1   x2 = 5 + ( 2 )  m = 6 m 2  

5 ( 5 m ) = 3.125 m 8

1  25  x3 = 7 + ( 4 )  = m 3  3 

5   x4 = 7 + ( 4 )  = 9.5 m 8  

Weights: Wi = ρi gVi 2  W1 = 2400 kg/m3 9.81 m/s 2  ( 5 m )( 8 m )(1 m )  = 627 840 N 3  

(

)(

)

(

)(

)

W2 = 2400 kg/m3 9.81 m/s 2 ( 2 m )( 8 m )(1 m )  = 376 700 N

1  W3 = 2400 kg/m3 9.81 m/s 2  ( 4 m )( 6 m )(1 m )  = 188 352 N 3  

(

)(

)

2  W4 = 2400 kg/m3 9.81 m/s 2  ( 4 m )( 6 m )(1 m )  = 156 960 N 3  The pressure force P is: 1 1 P = Aρ gh = ( 6 m )(1 m )   1000 kg/m3 9.81 m/s 2 ( 6 m )  = 176 580 N   2 2

(

)(

)

(

)(

)

Equilibrium: (a) + ΣFx = 0:

H − 176.580 kN = 0 H = 176.580 kN

+ ΣFy = 0:

H = 176.6 kN

!

V − 627.84 kN − 376.70 kN − 188.352 kN − 156.960 kN = 0 V = 1349.85 kN

(b) + ΣM A = 0:

or

or

V = 1350 kN

x (1349.85 kN ) − ( 3.125 m )( 627.84 kN ) − ( 6 m )( 376.70 kN )  25  − m  (188.352 kN ) − ( 9.5 m )(156.960 kN ) + ( 2 m )(176.580 kN ) = 0  3 

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.

!

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x = 5.1337 m Thus the point of application of the resultant is:

5.13 m to the right of A. !

(c) Free-body diagram and force triangle for the water section BCD

From the force triangle: R=

(176.580 )2 + (156.960 )2

= 236.26 kN

 156.960 

θ = tan −1   = 41.634°  176.580  or on the face BD of the dam

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.

R = 236 kN

41.6° !

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Chapter 5, Solution 77.

Free-Body Diagram:

Note that valve opens when B = 0.

Pressures p1 and p2 at top and bottom of valve:

( = (10

)( ) kg/m )( 9.81 m/s ) ( d ) = ( 9810d ) N/m

p1 = 103 kg/m 3 9.81 m/s 2 ( d − 0.225 m ) = ( 9810d − 2207.3) N/m 2 p2

3

3

2

2

Force P1 and P2:

P1 =

1 1 p1 A = ( 9810d − 2207.3) N/m 2  ( 0.225 m )( 0.225 m )  2 2

= ( 248.32d − 55.872 ) N P2 =

1 1 p2 A = ( 9810d ) N/m 2  ( 0.225 m )( 0.225 m )  2 2

= ( 248.32d ) N + ΣM A = 0:

− ( 0.15 − 0.09 ) m  ( 248.32d − 55.872 ) N  + ( 0.09 − 0.075 ) m  ( 248.32d ) N  = 0

Thus d = 0.30000 m, or

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.

d = 300 mm

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Chapter 5, Solution 78.

Free-Body Diagram:

Note that valve opens when B = 0.

Pressures p1 and p2 at top and bottom of valve:

( = (10

)( ) kg/m )( 9.81 m/s ) ( 0.450 m ) = 4414.5 N/m

p1 = 103 kg/m 3 9.81 m/s 2 ( 0.225 m ) = 2207.3 N/m 2 p2

3

3

2

2

Force P1 and P2:

P1 =

1 1 p1 A = 2207.3 N/m 2 ( 0.225 m )( 0.225 m )  2 2

(

)

= 55.872 N P2 =

1 1 p2 A = 4414.5 N/m 2 ( 0.225 m )( 0.225 m )  2 2

(

)

= 111.742 N + ΣM A = 0:

− ( 0.15 − h ) m  ( 55.872 N ) + ( h − 0.075 ) m  (111.742 N ) = 0

Solving for h: h = 0.100 000 m, or h = 100.0 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.

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Chapter 5, Solution 79.

Since gate is 4 ft wide: p1 = 4γ ( h − 3)

p2 = 4γ h p1′ = 4γ ′ ( d − 3)

p2′ = 4γ ′d

) )

( (

1 1 ( 3 ft ) p1′ − p1 = ( 3 ft ) 4γ ′ ( d − 3) − 4γ ( h − 3) = 6γ ′ ( d − 3) − 6γ ( h − 3) 2 2 1 1 P2′ − P2 = ( 3 ft ) p2′ − p2 = ( 3 ft ) [ 4γ ′d − 4γ h ] = 6γ ′d − 6γ h 2 2 P1′ − P1 =

This gives the free-body diagram:

+ ΣM A = 0: or

( 3 ft ) B − (1 ft ) ( P1′ − P1 ) − ( 2 ft ) ( P2′ − P2 ) = 0 B= =

) (

(

1 ′ 2 ′ P1 − P1 − P2 − P2 3 3

)

1 2 6γ ′ ( d − 3) − 6γ ( h − 3)  − [ 6γ ′d − 6γ h ] 3 3

= 2γ ′ ( d − 3) − 2γ ( h − 3) + 4γ ′d − 4γ h or + ΣFx = 0:

B = 6γ ′ ( d − 1) − 6γ ( h − 1)

(

) (

(1)

)

A + B − P1′ − P1 − P2′ − P2 = 0, or using (1) A + 6γ ′ ( d − 1) − 6γ ( h − 1)  − 6γ ′ ( d − 3) − 6γ ( h − 3)  − [ 6γ ′d − 6γ h ] = 0, or

A = 6γ ′ ( d − 2 ) − 6γ ( h − 2 ) Using the given data in (1) and (2): h = 6 ft, d = 9 ft, γ = 62.4 lb/ft 3 , γ ′ = 64 lb/ft 3 A = 6 ( 64 )( 9 − 2 ) − 6 ( 62.4 )( 6 − 2 )

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.

(2)

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= 2688 lb − 1497.6 lb = 1190.4 lb B = 6 ( 64 )( 9 − 1) − 6 ( 62.4 )( 6 − 1) = 3072 lb − 1872 lb = 1200 lb

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.

A = 1190 lb

!

B = 1200 lb

!

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Chapter 5, Solution 80.

First, determine the force on the dam face without the silt.

Pw =

Have

=

1 1 Apw = A ( ρ gh ) 2 2 1 ( 6 m )(1 m )   103 kg/m 3 9.81 m/s 2 ( 6 m )    2

(

)(

)

= 176.58 kN Next, determine the force on the dam face with silt.

Pw′ =

Have

1 ( 4.5 m )(1m )   103 kg/m 3 9.81 m/s 2 ( 4.5 m )    2

(

)(

)

= 99.326 kN

( Ps )I

(

)(

)

= (1.5 m )(1 m )   103 kg/m 3 9.81 m/s 2 ( 4.5 m )    = 66.218 kN

( Ps )II

=

1 (1.5 m )(1 m )   1.76 × 103 kg/m3 9.81 m/s 2 (1.5 m )    2

(

)(

)

= 19.424 kN Then

P′ = Pw′ + ( Ps )I + ( Ps )II = 184.97 kN

The percentage increase, % inc., is then given by

% inc. =

(184.97 − 176.58) × 100% = 4.7503% P′ − Pw × 100% = 176.58 Pw % inc. = 4.75%

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.

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Chapter 5, Solution 81.

From Problem 5.80, the force on the dam face before the silt is deposited, is Pw = 176.58 kN. The maximum allowable force Pallow on the dam is then: Pallow = 1.5Pw = (1.5 )(176.58 kN ) = 264.87 kN

Next determine the force P′ on the dam face after a depth d of silt has settled.

Have

Pw′ =

(

)(

)(

)

)

1 ( 6 − d ) m × (1 m )   103 kg/m3 9.81 m/s 2 ( 6 − d ) m    2 2

= 4.905 ( 6 − d ) kN

( Ps )I

(

=  d (1 m )   103 kg/m3 9.81 m/s 2 ( 6 − d ) m   

(

)

= 9.81 6d − d 2 kN

( Ps )II

=

(

)(

)

1  d (1 m )   1.76 × 103 kg/m3 9.81 m/s 2 ( d ) m    2

= 8.6328d 2 kN

(

)

(

)

P′ = Pw′ + ( Ps )I + ( Ps )II =  4.905 36 − 12d + d 2 + 9.81 6d − d 2 + 8.6328d 2  kN   = 3.7278d 2 + 176.58 kN

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.

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Now required that P′ = Pallow to determine the maximum value of d. ∴

or Finally

(3.7278d

2

)

+ 176.58 kN = 264.87 kN

d = 4.8667 m

4.8667 m = 20 × 10−3

m ×N year

or N = 243 years !

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.

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Chapter 5, Solution 82.

Pressure force from the water on board AB:

1 Api where p1 and p2 are the pressures at the top and bottom of the board: 2   1 kg  m P1 = ( 0.5 m )(1.5 m )  103 3  9.81 2  ( 0.6 m )  = 2207.3 N 2 m  s    Pi =

P2 =

  1 kg  m ( 0.5 m )(1.5 m )  103 3  9.81 2  (1 m )  = 3678.8 N  2 m  s   

Free-Body Diagram:

Ax denotes the force from one piling and is therefore multiplied by two in the free-body diagram. 1  2  + ΣM A = 0: − ( 0.3 m ) B +  ( 0.5 ) m  ( 2207.3 N ) +  ( 0.5) m  ( 3678.8 N ) = 0, or 3  3  B = 5313.8 N 4 4 + ΣFx = 0: 2 Ax + ( 2207.3 N ) + ( 3678.8 N ) = 0, or 5 5 Ax = − 2354.4 N 3 3 5318.8 N − ( 2207.3 N ) − ( 3678.8 N ) + Ay = 0, or + ΣFy = 0: 5 5 Ay = −1782.14 N

Therefore: (a) A x = 2.35 kN (b) (c)

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.

!

A y = 1.782 kN !

B = 5.31 kN !

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Chapter 5, Solution 83.

Pressure force from the water on board AB:

Pi =

1 Api where p1 and p2 are the pressures at the top and bottom of the board: 2

P1 =

  1 kg  m ( 0.5 m )(1.5 m )  103 3  9.81 2  ( 0.6 m )  = 2207.3 N 2 m  s   

P2 =

  1 kg  m ( 0.5 m )(1.5 m )  103 3  9.81 2  (1 m )  = 3678.8 N 2 m  s   

Note that the board can move in two ways: by rotating about A if the rope is pulled upward, and by sliding down at A if the rope is pulled sideways to the left.

Case 1 (rotation about A): For minimum tension the rope will be perpendicular to the board.

Free-Body Diagram:

+ ΣM A = 0:

1  − ( 0.5 m ) TBC +  ( 0.5 ) m  ( 2207.3 N ) + 3 

2   3 ( 0.5 ) m  ( 3678.8 N ) = 0, or  

TBC = 3188.3 N Case 2 (sliding down at A): When the board is just about to slide down at A, A y = 0. continued

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.

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Free-Body Diagram:

+ ΣM B = 0:

1  − ( 0.4 m ) ( 2 Ax ) −  ( 0.5 ) m  ( 3678.8 N ) − 3  

2   3 ( 0.5 ) m  ( 2207.3 N ) = 0, or  

2 Ax = − 3372.3 N + ΣFx = 0:

− TBC − 3372.3 +

4 4 ( 2207.3 N ) + ( 3678.8 N ) = 0, or 5 5

TBC = 1336.58 N Thus:

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.

( TBC )min

= 1.337 kN

!

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Chapter 5, Solution 84.

Free-Body Diagram:

Force from water pressure:

1 ApB where A is the rectangular cross sectional area through line BD, and pB is the pressure at 2 point B. Thus 1 1 P = A (γ h ) = (16 ft )(10 ft )   62.4 lb/ft 3 (10 ft )  = 18720.0 lb = 18.72 kips   2 2 1  W = γ V = 62.4 lb/ft 3  ( 3 ft )( 6 ft )( 6 ft )  = 3369.61 lb = 3.3696 kips 2  Equilibrium: 20 + ΣM A = 0: (18.72 kips )  ft  − B ( 3 ft ) + ( 3.3696 kips )( 2 ft ) = 0.  3  P−

(

(

)

)

Solving for B:

B = 43.846 kips, or + ΣFx = 0:

B = 43.8 kips

18.72 kips + Ax = 0, or Ax = −18.7200 kips

+ ΣFy = 0:

Ay − 3.3693 kips + 43.846 kips = 0, or Ay = − 40.476 kips A=

( −18.7200 )2 + ( −40.476 )2

= 44.595 kips

 40.476 

θ = tan −1   = 65.180°  18.7200 

Therefore: A = 44.6 kips

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.

65.2°

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Chapter 5, Solution 85.

Consider a 1-in. thick section of the gate and a triangular section BDE of water above the gate Free-Body Diagram:

Pressure force P:

1 1 1 ApB = ( d × 1 in.)(γ d ) = γ d 2 lb 2 2 2 Weight of water section above gate: P=

4 1 8  WW = γ VW = γ  × d × d × 1 in.  = γ d 2 lb  2 15  15 For impending motion of gate: B y = 0, and for equilibrium: + ΣM a = 0:

2 1  8   4 d  1 2  2  (16 ) −  d    γ d  −  − 6   γ d  = 0, and 3  15    15  3  2  3

d = 27.301 in., or

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.

d = 27.3 in.

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Chapter 5, Solution 86.

Consider a 1-in. thick section of the gate and a triangular section BDE of water above the gate Free-Body Diagram:

Pressure force P:

1 1 1 ApB = ( d × 1 in.)(γ d ) = γ d 2 lb 2 2 2 Weight of water section above gate: P=

4 1 8  WW = γ VW = γ  × d × d × 1 in.  = γ d 2 lb  2 15  15 For impending motion of gate: B y = 0, and for equilibrium: + ΣM a = 0:

2 1  8   4 d  1 2  2  (16 ) −  d    γ d  −  − (10 − h )   γ d  = 0, and 3 3 15 15 3       2  

with d = 30 in.

2 1  8   4 2  (16 ) −  30    γ 30  − 3 3 15 15     

d  1 2  3 − (10 − h )   2 γ 30  = 0, and   

h = 2.8444 in., or

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 = 2.84 in.

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Chapter 5, Solution 87.

Free-Body Diagram:

(

)

W = (125 kg ) 9.81 m/s 2 = 1226.25 N Denoting the water pressure at a depth h by ph, the forces due to the water pressure P1, P2, P3, P4 can be obtained as follows: 1 P1 = ADC p( 0.15 m ) , or with 2

(

)( ) 1 = ( 0.15 m )(1 m )  (1471.50 N/m ) = 110.363 N 2

p( 0.15 m ) = 1000 kg/m3 9.81 m/s 2 ( 0.15 m ) = 1471.50 N/m 2

P1

2

P2 = ACB p( 0.15 m) , or

(

)

P2 = ( 0.6 m )(1 m )  1471.50 N/m 2 = 882.90 N 1 ABA p( 0.15 m ) , or 2 1 P3 = ( 0.6 m )(1 m )  1471.50 N/m 2 = 441.45 N 2 1 P4 = ABA p( 0.75 m ) , or with 2

P3 =

(

(

)

)(

)

p( 0.75 m ) = 1000 kg/m3 9.81 m/s 2 ( 0.75 m ) = 7357.5 N/m 2

P4 =

(

)

1 ( 0.16 m )(1 m )  7357.5 N/m 2 = 2207.3 N 2

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.

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Now from the free-body diagram: − (1.2 m ) D + ( 0.6 m )(1226.25 N ) + ( 0.3 m )(1226.25 N )

+ ΣM A = 0:

− ( 0.6 + 0.05 m )  (110.363 N ) − ( 0.3 m )( 882.90 N ) − ( 0.4 m )( 441.45 N )

− ( 0.2 m )( 2207.3 N ) = 0, or

D = 124.149 N, and D = 124.1 N

+ ΣFx = 0:

!

Ax + 110.363 N + 441.45 N + 2207.3 N = 0, or Ax = −2759.1 N

+ ΣFy = 0:

Ay − 3 (1226.25 N ) + 882.90 N = 0, or Ay = 2795.9 N

Then, A=

( −2759.1)2 + ( 2795.9 )2

θ = tan −1

= 3930 N, and

2795.9 = 45.4° 2759.1 Therefore: A = 3930 N

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.

45.4° !

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Chapter 5, Solution 88.

Free-Body Diagram:

(

)

W = (125 kg ) 9.81 m/s 2 = 1226.25 N Denoting the water pressure at a depth h by ph, the forces due to the water pressure P1, P2, P3, P4 can be obtained as follows: 1 P1 = ADC p( d − 0.6 m ) , or w 2 1 P1 = ( d − 0.6 ) m × (1 m )  γ N/m3 ( d − 0.6 ) m 2 1 2 = γ ( d − 0.6 ) N 2 where γ denotes the specific weight of water. In the same way 1 P2 = ACB p( d − 0.6 m) , or 2

(

(

)

)

P2 = ( 0.6 m ) × (1 m )  γ N/m3 ( d − 0.6 ) m = 0.6γ ( d − 0.6 ) N

1 ABA p( d − 0.6 m ) , or 2 1 P3 = ( 0.6 m ) × (1 m )  γ N/m3 ( d − 0.6 ) m 2 = 0.3γ ( d − 0.6 ) N

P3 =

(

)

1 ABA p( d m ) , or 2 1 P4 = ( 0.6 m ) × (1 m )  γ N/m3 ( d m ) 2 = 0.3γ d N P4 =

(

)

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

Now from the free-body diagram: + ΣM A = 0:

( 0.6 m )(1226.25 N ) + ( 0.3 m )(1226.25 N )  1 1  2 −  0.6 m + ( d − 0.6) m   γ ( d − 0.6) N  − ( 0.3 m ) 0.6γ ( d − 0.6) N  2 3  

2  1  −  ( 0.6 m )  0.3γ ( d − 0.6 ) N  −  ( 0.6 m )  0.3γ ( d − 0.6 ) N + 0.18γ N  = 0, or 3 3     1 1103.63 − 0.036 ( d − 0.6 )3 + 0.3 ( d − 0.6 )2 + 0.36 ( d − 0.6 ) = γ 6

(

)(

)

With γ = 1000 kg/m3 9.81 m/s 2 = 9810 N/m3 , this gives 1 1103.63 − 0.036 = 0.076501 ( d − 0.6 )3 + 0.3 ( d − 0.6 )2 + 0.36 ( d − 0.6 ) = 6 9810 N/m3 Solving for d numerically:

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.

d = 0.782 m !

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Chapter 5, Solution 89. (a) Free-body diagram for a 24-in. long parabolic section of water: In the free body diagram force P is:

P=

1 1 1  3  24    3  AP = A (γ h ) =  ft  ft   62.4 lb/ft 3  ft   = 3.9000 lb 2 2 2  12  12    12   Ww = γ V

(

)

 2  4.5  3  24   = 62.4 lb/ft 3   ft  ft  ft    3  12  12  12   = 7.8000 lb

(

)

From the force triangle:

R=

P 2 + Ww 2 =

θ = tan −1

( 3.9 )2 + ( 7.8)2

= 8.7207 lb

Ww 7.8 = tan −1 = 63.435°, or P 3.9

R = 8.72 lb

(b)

63.4°

Free-body diagram for a 24-in. long section of the water: From (a) WW = 7.8000 lb From the free-body diagram:

By = 7.8000 lb +ΣM B = 0:

M B + ( 2.25 − 1.8 ) in. ( 7.8000 lb ) = 0, or

M B = −3.5100 lb ⋅ in.

Therefore, the force-couple system on the gutter is:

R = 7.8 lb ; M = 3.51 lb ⋅ 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 90.

Note, for the axes shown y

yV

−R

−2π R 4

3 − r 8

1 4 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−

1 r    8 R 

1−

1 r    3 R 

=

(a )

r =

4

3

3 R: y = − 4

1−

1 3   3 4

1 3 1−   3 4

4

3

R

or y = −1.118R 1−

(b)

y = −1.2R : − 1.2R = −

4

or

1 r    8 R 

1 r  1−   3 R 

4

3

R

3

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

Solving numerically

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.

r = 0.884 R

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Chapter 5, Solution 91.

Labeling the two parts of the body as follows:

ΣyV Then Y = = ΣV

(

7 π a 2h 2 24 2 π a 2h 3

(

)

)

V

y

yV

1

1 2 πa h 2

h 2

1 2 2 πa h 4

2

1 2 πa h 6

h 4

1 π a 2h 2 24

Σ

2 2 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:

z

V 1 2

Σ

Then Z =

( (

− 12 a3h ΣzV = 2 π a 2h ΣV 3

1 2 πa h 2 1 2 πa h 6 2 2 πa h 3

−

4a 3π a

π

zV

2 − a 3h 3 1 3 ah 6 1 − a 3h 2

) )

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.

or Z = −

3a

4π

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Chapter 5, Solution 93.

V

x

xV

Rectangular prism

Lab

1 L 2

1 2 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

h 1 = 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 1 +   = L  3 + + 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

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Chapter 5, Solution 94.

Assume that the machine element is homogeneous so that its center of gravity coincides with the centroid 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 ) = − 8505.9 2

30

−186 585

−255 180

−110 565

4 335 900

2

Σ Then Y =

30 −

4 (19 ) 3π

= 21.936

105 +

4 ( 30 ) 3π

=117.732

84 290

ΣyV −110 565 mm = 84 290 ΣV

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.

or Y = −1.312 mm

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Chapter 5, Solution 95.

Assume that the machine element is homogeneous so that its center of gravity coincides with the centroid of the volume.

Then Z =

V , mm3

z , mm

zV , mm 4

1

( 60 )(105)(10 ) = 63000

52.5

3 307 500

2

1 2 π ( 30 ) (10 ) = 14 137.2 2

3

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

4

105 +

4 ( 30 ) 3π

= 117.732

1 664 400

30

810 000

− π (19 ) (10 ) = −11 341.1

105

−1 190 820

5

1 2 − π (19 ) (15 ) = − 8505.9 2

30

− 255 180

Σ

84 290

2

ΣzV 4 335 900 = mm ΣV 84 290

4 335 900

or Z = 51.4 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 Solutions Manual Organization System

Chapter 5, Solution 96.

Assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the volume.

V , mm3

x , mm

xV , mm 4

1

(100 )(88)(12 ) =105600

50

5 280 000

2

(100 )(12 )(88) = 105600

50

5 280 000

3

1 ( 62 )( 51)(10 ) = 15 810 2

39

616 590

4

−

1 ( 66 )( 45)(12 ) = −17 820 2

Σ Then X =

34 +

2 ( 66 ) = 78 3

209 190

ΣxV 9 786 600 = mm ΣV 209190

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.

−1 389 960 9 786 600 or X = 46.8 mm

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Chapter 5, Solution 97.

Assume that the bracket is homogeneous so that it center of gravity coincides with the centroid of the volume.

V , mm 1

(100 )(88)(12 ) = 105600

2

(100 )(12 )(88) = 105600

3

1 ( 62 )( 51)(10 ) = 15 810 2

4

−

1 ( 66 )( 45)(12 ) = −17 820 2

Σ Then Z =

3

z , mm

zV , mm 4

6

633 600

12 +

1 (88) = 56 2

5 913 600

12 +

1 ( 51) = 29 3

458 490

55 +

2 ( 45) = 85 3

209 190

ΣzV 5 491 000 = mm ΣV 209 190

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.

−1 514 700 5 491 000

or Z = 26.2 mm

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Chapter 5, Solution 98.

Assume that the machine element is homogeneous so that its center of gravity coincides with the centroid of the volume.

Then X =

V , in 3

x , in.

xV , in 4

1

( 8)( 0.9)( 2.7) = 19.44

4

77.76

2

1 ( 2.1)( 6)( 2.7) = 17.01 2

2

34.02

3

1 2 π (1.35) ( 0.9) = 2.5765 2 2

4

− π ( 0.8 ) ( 0.9 ) = −1.80956

Σ

37.217

ΣxV 119.392 = in. ΣV 37.217

8+

1.8

8

22.088

π −14.4765 119.392

or X = 3.21 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 99.

Assume that the machine element is homogeneous so that its center of gravity coincides with the centroid of the volume.

Then Y =

V , in 3

y , in.

yV , in 4

1

( 8)( 0.9)( 2.7) = 19.44

0.45

8.748

2

1 ( 2.1)( 6)( 2.7) = 17.01 2

1.6

27.216

3

1 2 π (1.35) ( 0.9) = 2.5765 2

0.45

1.15943

4

− π ( 0.8 ) ( 0.9 ) = −1.80956

0.45

− 0.81430

Σ

37.217

ΣyV 36.309 = in. ΣV 37.217

2

36.309 or Y = 0.976 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 100.

Labeling the five parts of the body as follows, and noting that the center of gravity coincides with the centroid of the area due to the uniform thickness.

4 × 150   z5 = −  300 −  = − 236.34, 3π  

A5 = −

π 2

(150 )2 = −11 250π

= − 35 343

xA,106 mm3 yA, 106 mm3 zA, 106 mm3

A mm 2

x , mm

y , mm

z , mm

1

( 600 )( 400 ) = 240000

300

200

0

72

48

0

2

( 300 )( 400 ) = 120000

600

200

−150

72

24

−18

3

− (120 )( 280 ) = − 3360

600

140

− 240

− 20.160

− 4.7040

8.0640

4

( 600 )( 300 ) = 180000

300

400

−150

54

72

− 27

5

− 35 343

240

400

− 236.34

− 2.7000

− 4.5

2.6588

Σ

471 057

169.358

125.159

− 28.583 continued

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.

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Therefore: X =

ΣxA 169 358 000 = = 359.53 mm 471 057 ΣA

or X = 360 mm !

Y =

ΣyA 125159 000 = = 265.70 mm 471 057 ΣA

or Y = 266 mm !

Z =

ΣzA − 28 583 000 = = − 60.678 mm 471 057 ΣA

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.

or Z = − 60.7 mm !

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Chapter 5, Solution 101.

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

A, in 2

x , in.

y , in.

z , in.

xA, in 3

yA, in 3

zA, in 3

1

1 ( 4.5)( 3) = 6.75 2

1.5

7

0

10.125

47.25

0

2

( 4.5)(10 ) = 45

2.25

3

4

101.25

135

180

3

− ( 2.25 )( 5 ) = −11.25 − 2.25

1.125

1.5

6

−12.6563

−16.875

− 67.5

2.25

0

17.8925

0

71.211

116.611

165.375

183.71 1

4

π 2

( 2.25)2 = 7.9522

Σ

48.452

8+

4 ( 2.25 ) 3π

Then X =

ΣxA 116.611 = in. ΣA 48.452

or X = 2.41 in.

Y =

ΣyA 165.375 = in. ΣA 48.452

or Y = 3.41 in.

Z =

ΣzA 183.711 = in.

48.452 ΣA

or Z = 3.79 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.

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Chapter 5, Solution 102.

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

X = 150 mm !

First note that by symmetry: For 1: y = 180 + 96 +

4 (150 ) = 339.7 mm 3π

z =0 For 2: y = 180 + z =

2 ( 96 )

π

2 ( 96 )

π

= 241.1 mm

= 61.11 mm

For 3: Length DE =

(180 )2 + ( 96 )2

y = 90 mm,

= 204 mm

z = 48 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.

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y , mm

z , mm

yA, mm3

zA, mm3

(150 )2 = 35.34 × 103

339.7

0

12.005 × 106

0

( 96 )( 300 ) = 45.24 × 103

244.1

61.11

10.907 × 106

2.765 × 106

90

48

5.508 × 106

2.938 × 106

28.420 × 106

5.702 × 106

A, mm 2

π

1

2

2

π 2

3

( 204 )( 300 ) = 61.2 × 103 − 2.25

Σ

141.78 × 103

Then

Y =

ΣyA 28.420 × 106 = mm ΣA 141.78 × 103

Z =

ΣzA 5.702 × 106 = mm ΣA 141.78 × 103

or Y = 200 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.

or Y = 40.2 mm !

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Chapter 5, Solution 103.

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

( 360 )( 270 ) = 97 200

0

135

0

13 122 000

2

( 339 )( 270 ) = 91 530

168

135

15 377 000

12 356 600

3

1 ( 339 )( 72 ) = 12 204 2

224

294

2 733 700

3 588 000

4

( 360 )( 343.63) = 123 707

168

306

20 783 000

37 854 000

5

1 ( 343.63)( 45) = 7731.73 2

224

318

1 731 900

2 458 700

6

7731.7

224

318

1 731 900

2 458 700

7

12 204

224

294

2 733 700

3 588 000

8

91 530

168

135

15 377 000

12 356 600

ÎŁ

443 838

60 468 200

87 782 600

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 Š 2007 The McGraw-Hill Companies.

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Then X =

ΣxA 60 468 200 = mm ΣA 443 838

or X = 136.2 mm !

Y =

ΣyA 87 782 600 = mm ΣA 443 838

or Y = 197.8 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.

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Chapter 5, Solution 104.

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

Note that by symmetry X = 9 in.

A, in 2

y , in.

z , in.

yA, in 3

zA, in 3

1

16.2

1.8

2

29.16

32.4

2

16.2

1.8

2

29.16

32.4

3

97.2

2.7

0

262.44

0

4

1017.876

−15.2789

20.72113

−15552

21091.54

5

1017.876

−15.2789

20.72113

−15552

21091.54

6

− 706.858

−12.7324

23.2676

9000

−16446.9

7

− 706.858

−12.7324

23.2676

9000

−16446.9

8

1017.876

− 22.9183

13.08169

− 23328

13315.54

Σ

1769.511

− 36111.24

22669.6

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.

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Therefore X = 9 in. !

Y =

ΣyA − 36111.24 = ΣA 1769.511

or Y = − 20.4 in. !

Z =

ΣzA 22669.6 = ΣA 1769.511

or Z = 12.81 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.

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Chapter 5, Solution 105.

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

1 2

A, in 2

x , in.

y , in.

xA, in 3

yA, in 3

π ( 8 )(12 ) = 96π

0

6

0

576π

10

−128

−160π

12

− 42.667

96π

−

π 2

π

3

2

( 4 )2

= 8π

=

π

−

4 ( 4) 3π

8

π

=−

16 3π

4

(8)(12 ) = 96

6

12

576

1152

5

(8)(12 ) = 96

6

8

576

768

8

− 42.667

− 64π

6

Then

2 ( 4)

(8)( 4 ) = −16π

−

π 2

( 4 )2

= − 8π

4 ( 4) 3π

=

16 3π

7

( 4 )(12 ) = 48

6

10

288

480

8

( 4 )(12 ) = 48

6

10

288

480

Σ

539.33

1512.6

4287.4

X =

ΣxA 1514.67 = in. or X = 2.81 in. ΣA 539.33

Y =

ΣyA 4287.4 = in. or Y = 7.95 in. ΣA 539.33

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.

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Chapter 5, Solution 106.

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.

yII = yVI = 80 + zII = zVI = yIV = 80 + zIV =

3π ( 2 )( 500 )

π

2

π 4

= 292.2 mm

= 212.2 mm = 398.3 mm

π

( 2 )( 500 )

π

3π

( 4 )( 500 )

AII = AVI = AIV =

( 4 )( 500 )

= 318.3 mm

( 500 )2

= 196 350 mm 2

( 500 )( 680 ) = 534 071 mm 2

yA, mm3

zA, mm3

A, mm 2

y , mm

z , mm

I

(80)(500) = 40 000

40

250

1.6 × 106

10 × 106

II

196 350

292.2

212.2

57.4 × 106

41.67 × 106

III

(80)(680) = 54 400

40

500

0.2176 × 106

27.2 × 106

IV

534 071

398.3

318.3

212.7 × 106

170 × 106

V

(80)(500) = 40 000

40

250

1.6 × 106

10 × 106

VI

196 350

292.2

212.2

57.4 × 106

41.67 × 106

332.9 × 106

300.5 × 106

Σ

1.061 × 106

X = 340 mm

Now, symmetry implies and

(

)

Y ΣA = Σ yA: Y 1.061 × 106 mm 2 = 332.9 × 106 mm 3

(

6

Z ΣA = Σ zA: Z 1.061 × 10 mm

2

) = 300.5 × 10

or Y = 314 mm 6

mm

3

or Z = 283 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.

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Chapter 5, Solution 107.

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

1

A,in 2

y ,in.

z , in.

yA, in 3

zA, in 3

(15)(14 ) = 120

0

7

0

1470

1.25

7

43.75

245

1.25

0

46.875

0

1.25

7

43.75

245

6.5

0

− 510.51

0

−21.848

4

(14)( 2.5) = 35.0 (15)( 2.5) = 37.5 (14)( 2.5) = 35.0

5

− π ( 5 ) = − 78.540

0

6

1 2 − π (1.5 ) = −1.76715 4

0

7

( 4 )(12 ) = 48

6

10

288

480

8

( 4 )(12 ) = 48

6

10

288

480

Σ

235.43

134.375

1405.79

2 3

2

13 −

4 (1.5 ) 3π

= 12.36348

Then Y =

ΣyA 134.375 = in. ΣA 235.43

or Y = 0.571 in.

Z =

ΣzA 1405.79 = in. ΣA 235.43

or Z = 5.97 in.

and by symmetry

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.

X = 7.50 in.

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Chapter 5, Solution 108.

2

2

2

AB 2 = ( 500 mm ) + ( 750 mm ) + ( 300 mm ) , or

AB = 950 mm L, mm

x , mm

y , mm

z , mm

xL, mm 2

yL, mm 2

zL, mm 2

AB

950

250

375

150

237.5 × 103

356.25 × 103

142.5 × 103

BD

300

500

0

150

150 × 103

0

45 × 103

DO

500

250

0

0

125 × 103

0

0

OA

750

0

375

0

0

281.25 × 103

0

Σ

2500

512.5 × 103

637.5 × 103

187.5 × 103

Then

X =

ΣxL 512.5 × 103 = 2500 ΣL

or X = 205 mm

Y =

ΣyL 637.5 × 103 = 2500 ΣL

or Y = 255 mm

Z =

ΣzL 187.5 × 103 = ΣL 2500

or Z = 75 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.

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Chapter 5, Solution 109.

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the line

1 2 3 4

Σ

π 2

L, mm

x , mm

y , mm

z , mm

xL, mm 2

yL, mm 2

zL, mm 2

300 280 260

0 140 230

150 0 0

0 0 120

0 39 200 59 800

45 000 0 0

0 0 31 200

3  2 × 300  360  = 5 π π 

600

480

π

π

54 000

90 000

72 000

153 000

135 000

103 200

( 300 ) = 150π 1311.24

Then

X =

ΣxL 153 000 = ΣL 1311.24

or X = 116.7 mm

Y =

ΣyL 135 000 = ΣL 1311.24

or Y = 103.0 mm

Z =

ΣzL 103 200 = ΣL 1311.24

or Z = 78.7 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.

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Chapter 5, Solution 110.

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the line.

L,ft

x , ft

y ,ft

xL, ft 2

yL,ft 2

1

10

4cos 45° = 2.8284

5

28.284

50

2

10

4cos 45° = 2.8284

5

28.284

50

3

4π

0

12.5465

0

157.664

4

4π

2 ( 4)

10

32

125.664

5

2π

2 ( 4)

12.5465

16

78.832

Σ

51.416

104.568

462.16

π π

= =

8

π 8

π

Then

X =

ΣxL 104.568 = ΣL 51.416

or

Y =

ΣyL 462.16 = ΣL 51.416

or Y = 8.99 ft

and by symmetry:

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.

X = 2.03 ft

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Chapter 5, Solution 111.

First note by symmetry:

Z = 3.00 ft To simplify the calculations replace: (a) The two rectangular sides with an element of length

L(a) = 2  2 ( 7 ft ) + 2 ( 5 ft )  = 48 ft and center of gravity at (3.5 ft, 2.5 ft, 3 ft) (b) The two semicircular members with an element of length

Lb = 2 π ( 3 ft )  = 6π ft 2×3   ft, 3 ft  = ( 2 ft, 6.9099 ft, 3 ft ) and with center of gravity at  2 ft, 5 + π   (c) The cross members 1 and 2 with an element of length

Lc = 2 ( 6 ft ) = 12 ft and with center of gravity at ( 2 ft, 5 ft, 3 ft ) (d) This leaves a single straight piece of pipe, labeled (d) in the figure.

Now for the centroid of the frame:

L,ft

x , ft

y ,ft

xL, ft 2

yL,ft 2

(a)

48

3.5

2.5

(b)

6π = 18.8496

2

6.9099

168 37.699

120 130.249

(c) (d)

12 6 85.850

2 7

5 5

24 42 271.70

60 30 340.25

Σ Then

X =

ΣxL 271.70 = ΣL 84.850

or

Y =

ΣyL 340.25 = ΣL 84.850

or Y = 4.01 ft

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.

X = 3.20 ft

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Chapter 5, Solution 112.

Y = Z = 0

First, note that symmetry implies

xI =

5 2π ( 0.5 in.) = 0.3125 in., WI = 0.0374 lb/in 3   ( 0.5 in )3 = 0.009791 lb 8  3 

(

)

(

)

2

xII = 1.6 in. + 0.5 in. = 2.1 in. WII = 0.0374 lb/in 3 (π )( 0.5 in ) ( 3.2 in.) = 0.093996 lb 2 π  xIII = 3.7 in. − 1 in. = 2.7 in., WIII = − 0.0374 lb/in 3   ( 0.12 in ) ( 2 in.) = −0.000846 lb 4  

(

)

2 2 π  xIV = 7.3 in. − 2.8 in. = 4.5 in., WIV = 0.284 lb/in 3   ( 0.12 in ) ( 5.6 in ) = 0.017987 lb 4  

(

xV = 7.3 in. +

π 1 ( 0.4 in.) = 7.4 in., WV = 0.284 lb/in 3   ( 0.06 in )2 ( 0.4 in.) = 0.000428 lb 4 3

(

Σ

Have

)

)

W , lb

x , in.

xW , in ⋅ lb

I

0.009791

0.3125

0.003060

II

0.093996

2.1

0.197393

III

−0.000846

2.7

−0.002284

IV

0.017987

4.5

0.080942

V

0.000428

7.4

0.003169

0.12136

0.28228

X ΣW = ΣxW : X ( 0.12136 lb ) = 0.28228 in. ⋅ lb

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.

or X = 2.33 in.

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Chapter 5, Solution 113.

Determine first the masses of the component pieces:

π  m1 = 8800 kg/m3  0.0162 − 0.0122 m 2 × ( 0.014 m )  = 0.0108372 kg 4 

(

) (

)

π  m2 = 1250 kg/m 3  0.0362 − 0.0162 m 2 × ( 0.014 m )  = 0.0142942 kg 4  

(

) (

)

π  m3 = 1250 kg/m 3  0.0602 − 0.0362 m 2 × ( 0.006 m )  = 0.0135717 kg 4 

(

) (

)

π  m4 = 1250 kg/m3  0.0802 − 0.0602 m 2 × ( 0.010 m )  = 0.027489 kg 4 

(

) (

)

Now, for the center of mass:

Then X =

m, kg

x , mm

xm, kg ⋅ mm

1

0.0108372

7

0.075860

2

0.0142942

7

0.100059

3

0.0135717

3

0.040715

4

0.027489

5

0.137445

Σ

0.066192

0.35408

Σxm 0.35408 or = 0.066192 Σm

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.

X = 5.35 mm

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Chapter 5, Solution 114.

Assume that the stone is homogeneous so that its center of gravity coincides with the centroid of the volume and locate the center of gravity.

To determine the centroid of the truncated pyramid note that: 3 1 y1 = (1.4 m ) = 1.05 m, and V1 = ( 0.3 m )( 0.3 m )(1.4 m ) = 0.042 m3 4 3

y2 =

3 ( 0.7 m ) = 0.525 m, and 4

V2 = −

1 ( 0.15 m )( 0.15 m )( 0.7 m ) = − 0.00525 m3 3

Then Vstone = V1 + V2 = 0.042 m3 − 0.00525 m3 = 0.03675 m3, and

(

)

(

3 3 ΣyV (1.05 m ) 0.042 m + ( 0.525 m ) − 0.00525 m y= = ΣV 0.03675 m3

)

= 1.12500 m The center of gravity of the stone is therefore 0.425 m (i.e. 1.125 m – 0.7m) above the base.

Now to determine the center of gravity of the marker:

( )( )( ) = ( 7860 kg/m )( 9.81 m/s ) ⎡⎣( 0.3 m )( 0.3 m ) h ⎤⎦ = ( 6939.6 h ) N

Wstone = ( ρ gV ) stone = 2570 kg/m3 9.81 m/s 2 0.03675 m3 = 926.53 N Wsteel = ( ρ gV ) steel

3

2

continued

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.

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Then ymar ker =

0.3 m =

ΣyW , or ΣW

( 0.425 m )( 926.53 N ) + ( − h2 m ) ( 6939.6 h ) N , or ( 926.53 + 6939.6 h ) N

h 2 + 0.6 h − 0.033378 = 0.

Solving for h and discarding the negative root, this gives h = 0.051252 m, or

h = 50 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.

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Chapter 5, Solution 115.

Since the brass plates are equally spaced and by the symmetry of the cylinder: X =Y =0!

For the pipe: Specific weight of steel: γ s = 0.284 lb/in 3

y1 = 4 in. outside diameter: 2.5 in.

Inside diameter: 2.5 in. − 2 ( 0.25 in.) = 2.00 in.

π

( 2.5

)

2

− 2.02 8 = 14.137 in 3

Volume:

V1 =

Weight:

W1 = γ sV1 = 0.284 lb/in 3 14.137 in 3 = 4.015 lb

4

(

)(

)

For each brass plate: Specific weight for brass: γ B = 0.306 lb/in 3

8 2.667 in.

3 1 Volume: V2 = ( 8 )( 4 )( 0.2 ) = 3.2 in 3 2 y2 =

(

)(

)

Weight: W1 = γ sV1 = 0.306 lb/in 3 3.2 in 3 = 0.979 lb continued

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.

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For flagpole base: ΣW = ( 4.015 lb ) + 3 ( 0.979 lb ) = 6.952 lb ΣyW = ( 4 in.)( 4.015 lb ) + 3 ⎡⎣( 2.667 in.)( 0.979 lb ) ⎤⎦ = 23.892 in.⋅ lb, or

Y =

ΣyW 23.892 in.⋅ lb = = 3.437 in. ΣW 6.952 lb Y = 3.437 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 116.

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

Then dV = πr 2dx, xEL = x The equation of the generating curve is

r 2 = a 2 − x 2 and then

(

x 2 + y 2 = a 2 so that

)

dV = π a 2 − x 2 dx Component 1 a/2

⎡ x3 ⎤ a/2 V1 = ∫0 π a 2 − x 2 dx = π ⎢ a 2 x − ⎥ 3 ⎦0 ⎣

(

=

and

)

11 3 πa 24

a/2 2 2 ∫1 xEL dV = ∫0 x ⎡⎣π ( a − x ) dx ⎤⎦ a/2

⎡ x2 x4 ⎤ = π ⎢a 2 − ⎥ 4 ⎦0 ⎣ 2

= Now

7 π a4 64

7 ⎛ 11 ⎞ x1V1 = ∫1 xEL dV : x1 ⎜ π a3 ⎟ = π a4 ⎝ 24 ⎠ 64

or x1 = Component 2 a

⎡ x3 ⎤ a V2 = ∫a /2 π a 2 − x 2 dx = π ⎢ a 2 x − ⎥ 3 ⎦ a/2 ⎣

(

)

⎧ ⎡ a3 ⎤ ⎢ 2 ⎛ a ⎞ ⎪⎡ 2 = π ⎨⎢a ( a ) − ⎥ − a ⎜ ⎟ − 3 ⎦ ⎢ ⎝2⎠ ⎪⎣ ⎢⎣ ⎩

3 ⎤⎫

( a2 )

⎥ ⎪⎬ 3 ⎥⎪ ⎥⎦ ⎭ continued

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.

21 a!

88

COSMOS: Complete Online Solutions Manual Organization System

=

5 π a3 24

and

a x ⎡π a/2 ⎣

∫2 xELdV = ∫

(

a

⎡ x2 x4 ⎤ a − x dx ⎤ = π ⎢ a 2 − ⎥ ⎦ 4 ⎦ a/2 ⎣ 2 2

2

)

2 ⎧⎡ 2 4 ⎡ a ) ⎤ ⎢ 2 a2 ( ⎪ 2 (a) ⎥− a = π ⎨⎢a − − 2 4 ⎥ ⎢ 2 ⎪ ⎢⎣ ⎦ ⎣⎢ ⎩ 9 = π a4 64

( )

Now

⎤⎫ ⎥ ⎪⎬ 4 ⎥⎪ ⎦⎥ ⎭

( a2 )

4

9 ⎛ 5 ⎞ x2V2 = ∫2 xELdV : x2 ⎜ π a3 ⎟ = π a4 ⎝ 24 ⎠ 64 or x2 =

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.

27 a!

40

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Chapter 5, Solution 117.

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

Then dV = πr 2dx, xEL = x x2 y2 + = 1 so that h2 a 2

The equation of the generating curve is r2 =

a2 2 h − x 2 and then 2 h

(

)

dV = π

a2 2 h − x 2 dx 2 h

(

)

Component 1 h/2

V1 = ∫0 π =

and

a2 2 a2 2 − = h x dx π h2 h2

(

)

h/2

⎡ 2 x3 ⎤ ⎢h x − ⎥ 3 ⎦0 ⎣

11 2 πa h 24

2 ⎤ h/2 ⎡ a 2 2 x dV x π = ∫1 EL ∫0 ⎢ h 2 h − x dx ⎥ ⎣ ⎦

(

)

h/2

a2 ⎡ x2 x4 ⎤ = π 2 ⎢h2 − ⎥ 4 ⎦0 h ⎣ 2

7 π a 2h 2 64 7 ⎛ 11 ⎞ x1V1 = ∫1 xEL dV : x1 ⎜ π a 2h ⎟ = π a 2h 2 ⎝ 24 ⎠ 64 =

Now

or x1 =

continued

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.

21 h!

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Component 2 h

V2 = ∫h/2 π

h

a2 2 a2 ⎡ 2 x3 ⎤ 2 − = − π h x dx h x ⎢ ⎥ 3 ⎦h/2 h2 h2 ⎣

(

)

3 ⎧ 3 ⎡ h ⎤⎫ h) ⎤ ⎢ 2 ⎛ h ⎞ a 2 ⎪⎡ 2 ( 2 ⎥⎪ ⎥− h ⎜ ⎟ − = π 2 ⎨⎢h ( h ) − ⎬ 3 ⎥ ⎢ ⎝ 2⎠ 3 ⎥⎪ h ⎪⎣⎢ ⎦ ⎢ ⎥ ⎣ ⎦⎭ ⎩ 5 πa 2 h = 24

()

and

⎡ a2

⎤

h 2 2 ∫2 xELdV = ∫h/2 x ⎢π h2 ( h − x ) dx ⎥ ⎣ ⎦

=π

a2 h2

h

⎡ 2 x2 x4 ⎤ − ⎢h ⎥ 4 ⎦ h/2 ⎣ 2

2 ⎧ 2 4 ⎡ h ) ⎤ ⎢ 2 h2 ( a2 ⎪⎡ 2 ( h ) ⎥− h = π 2 ⎨⎢h − − 2 4 ⎥ ⎢ 2 h ⎪⎢ ⎣ ⎦ ⎢ ⎣ ⎩ 9 = π a 2h 2 64

( )

Now

⎤⎫ ⎥ ⎪⎬ 4 ⎥⎪ ⎥⎦ ⎭

( h2 )

4

9 ⎛ 5 ⎞ x2V2 = ∫2 xEL dV : x2 ⎜ π a 2h ⎟ = π a 2h 2 24 64 ⎝ ⎠ or x2 =

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.

27 h!

40

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Chapter 5, Solution 118.

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

Then dV = πr 2dx, xEL = x

x=h−

The equation of the generating curve is

r2 =

h 2 y so that a2

a2 ( h − x ) and then h dV = π

a2 ( h − x ) dx h

Component 1 h/2

V1 = ∫0 π

a2 ( h − x ) dx h h/2

a2 ⎡ x2 ⎤ = π ⎢ hx − ⎥ h ⎣ 2 ⎦0 = and

3 2 πa h 8 ⎡ a2

a2 =π h =

Now

⎤

h/2 ∫1 xELdV = ∫0 x ⎢π h ( h − x ) dx ⎥ ⎣ ⎦ h/2

⎡ x 2 x3 ⎤ − ⎥ ⎢h 3 ⎦0 ⎣ 2

1 π a 2h 2 12

1 ⎛3 ⎞ x1V1 = ∫1 xEL dV : x1 ⎜ π a 2h ⎟ = π a 2h 2 8 12 ⎝ ⎠

or x1 = continued

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.

2 h!

9

COSMOS: Complete Online Solutions Manual Organization System

Component 2 h

h

V2 = ∫h/2 π

a2 a2 ⎡ x2 ⎤ ( h − x ) dx = π ⎢hx − ⎥ 2 ⎦ h/2 h h ⎣

⎧ 2 ⎡ h) ⎤ ⎢ ⎛ h ⎞ ( a2 ⎪⎡ ⎢ ⎥ = π ⎨ h (h) − − h⎜ ⎟ − 2 ⎥ ⎢ ⎝2⎠ h ⎪⎢ ⎣ ⎦ ⎢⎣ ⎩ 1 = πa 2 h 8

⎤⎫ ⎥ ⎪⎬ 2 ⎥⎪ ⎥⎦ ⎭

( h2 )

2

h

and

⎡ a2 ⎤ a 2 ⎡ x 2 x3 ⎤ h ∫2 xEL dV = ∫h/2 x ⎢π h ( h − x ) dx ⎥ = π h ⎢ h 2 − 3 ⎥ ⎣ ⎦ ⎣ ⎦ h/2 2 ⎧ 2 3 ⎡ h ) ⎤ ⎢ h2 ( a2 ⎪⎡ ( h ) ⎥− h =π − − ⎨⎢h 3 ⎥ ⎢ 2 h ⎪⎢ 2 ⎣ ⎦ ⎢ ⎣ ⎩ 1 = π a 2h 2 12

( )

Now

x2V2 =

3 ⎤⎫

( h2 )

⎥ ⎪⎬ 3 ⎥⎪ ⎥⎦ ⎭

⎛ 2 ⎞ 2 2 ∫2 xEL dV : x2 ⎜ 8 π a h ⎟ = 12 π a h ⎝ ⎠ 1

1

or x2 =

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.

2 h!

3

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Chapter 5, Solution 119.

y = 0!

First note that symmetry implies

z = 0!

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

Then dV = πr 2dx, xEL = x ⎛ x2 ⎞ Now r = b⎜⎜1 − 2 ⎟⎟ so that a ⎠ ⎝ 2

2⎛

x2 ⎞ dV = πb ⎜⎜1 − 2 ⎟⎟ dx a ⎠ ⎝

Then

⎛ a π b 2 ⎜⎜1 0

2

⎛ x2 ⎞ 2x2 x4 ⎞ a − 2 ⎟⎟ dx = ∫0 π b 2 ⎜⎜1 − 2 + 4 ⎟⎟ dx a ⎠ a a ⎠ ⎝ ⎝

V =∫

a

2⎛

2 x3 x5 ⎞ = π b ⎜⎜ x − 2 + 4 ⎟⎟ 3a 5a ⎠ ⎝ 0 2 1⎞ ⎛ = π ab 2 ⎜1 − + ⎟ 3 5⎠ ⎝ 8 = π ab 2 15 and

2x2 x4 ⎞ a 2 ⎛ 1 x dV π b x = − + 4 ⎟⎟ dx ⎜ ∫ EL ∫0 ⎜ a2 a ⎠ ⎝ 2⎛

x2 2x4 x6 ⎞ = π b ⎜⎜ − 2 + 4 ⎟⎟ 4a 6a ⎠ ⎝ 2

a

0

continued

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

⎛1 1 1⎞ = π a 2b 2 ⎜ − + ⎟ ⎝2 2 6⎠

= Then

1 2 2 πa b 6

1 ⎛ 8 ⎞ xV = ∫ xEL dV : x ⎜ π ab 2 ⎟ = π a 2b 2 ⎝ 15 ⎠ 16 or x =

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.

15 a!

6

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Chapter 5, Solution 120. y = 0

First, note that symmetry implies

z = 0 Choose as the element of volume a disk of radius r and thickness dx.

Then dV = πr 2dx, xEL = x Now r = 1 −

1 so that x 2

1  dV = π 1 −  dx x  2 1   = π 1 − + 2  dx x x   3

Then

2 1  1 3   V = ∫1 π 1 − + 2  dx = π  x − 2 ln x −  x x 1 x     1  1  = π  3 − 2 ln3 −  − 1 − 2 ln 1 −   3  1   = ( 0.46944π ) m 3

and

3

 x2  2 1    1 − + 2  dx  = π  − 2 x + ln x  x x     2 1

3  x π 1 

∫ x EL dV = ∫

  32  13   = π   − 2 ( 3) + ln 3 −  − 2 (1) + ln1   2     2

= (1.09861π ) m Now

(

)

xV = ∫ x EL dV : X 0.46944π m 3 = 1.09861π m 4 or x = 2.34 m

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 121.

First, by symmetry:

x =a! y =0!

Next determine the constants k in y = kx1/3 : x = a, b = ka1/3 or k =

At

b a1/3

b 1/3 a x , or x = 3 y 3 1/3 a b Choosing horizontal disks of thickness dy for volume elements ( dV in the figure above)

Therefore, y =

2 b V = ∫0 π  a 2 − ( a − x )   

b

(

)

= π ∫0 2ax − x 2 dy a a2  b = π ∫0  2a × 3 y 3 − 6 y 6  dy b b   b

a2  1 1 1  5 = π 3  2 × y 4 − 3 × y 7  = π a 2b 4 7  0 14 b  b 1 ∫ yELdV , or V 14 b   a 2 3 a 2 6   y= ∫ y π  2 y − b6 y  dy  5π a 2b 0   b3  

Now y =

b

14  y5 1 y8  = 4  2 × − 3  5 5b  b 8  0

or

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.

y=

77 b!

100

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Chapter 5, Solution 122.

First note that by symmetry:

y =0! z = 0!

Choose as a volume element a disk of radius y and thickness dx. Then:

xEL = x, and dV = π y 2dx, or dV = π h 2 cos 2

πx 2a

dx

Using the identity: cos 2 x = dV =

1 (1 + cos 2 x ) , this gives 2 1 2 πx π h 1 + cos  dx.

2 a  

Then: V = ∫ dV =

π h2

a

πx π h2  a πx 1 a 1 cos dx x + sin + = = π h 2a. ∫    0 2 a  2  π a 0 2  continued

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

Also,

∫ xELdV = =

=

=

π h2

πx

a x + x cos ∫ 0 2 a 

π h 2   x 2 

 dx. Integrating by parts, 

a a πx π x   sin sin + − x    ∫ 2   2  0 π  a a  0   

π h 2  a 2  2  2

+

a

a a πx a π x   sin + cos x  π  a π a  0 

π h 2  a 2

a 2a   1 2 2  4  + 0 − 0 −    = π a h 1 − 2  2  2 π π   4 π  

Now, x=

1 2 1 2 2  4  xEL dV = π a h 1 − 2   , or ∫ 2  V πh a 4 π   x=

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.

1  4  a 1 − 2  !

2  π 

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 123.

First note that by symmetry: x =0! y =0!

Choosing the volume element shown in the figure, i.e. a cylindrical shell of radius r, height h and thickness dr:

yEL =

1 y, 2

and

dV = 2π ry dr = 2π r cos

πr

dr , and 2a πr a V = ∫ dV = 2π h ∫0 r cos dr , or, integrating by parts 2a a

V = 2π h

2a  πr πr  r sin dr − ∫ sin π  2a 2a  0 a

π r 2a πr   cos  = 4ah  r sin + π 2 a 2a  0 

2a  2  2  = 4ah  a −  = 4a h 1 −  π  π   Also, a

2 2 ∫ yEL dV = π h ∫0 r cos

π h 2   1

πr 2a

dr =

π h2

πr

a   ∫ r 1 + cos a  dr 2 0  

a a πr π r   = + − ∫ sin r sin dr   r 2   2  0 π  a a  0  2

a

continued

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

=

=

π h 2  1

a a πr a π r   2 + + a r sin cos   π  π 2  2 a a  0 

π h2  1

2a   1 2 2  4  a 2  a + 0 + 0 −   = π a h 1 − 2  π π  4 2 2 π  

Now, y=

1 ∫ yEL dV = V

1 2 2  1 4   π a h 1 − 2   , or 2   4 π   4a 2 h  1 −   π  y=

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.

(π

+ 2) h!

16

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Chapter 5, Solution 124.

Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of the pyramid is given by

Abase = kb 2 where k = k ( N ) ; see note below. Using similar triangles, have

s h− y = b h

or

s=

b (h − y) h

Then

dV = Aslicedy = ks 2dy = k

and

V = ∫0 k

h

= Also

b2 2 h − y ) dy 2( h h

b2 b2  1 2 3 h y dy k − = − (h − y)  ( ) h2 h 2  3 0

1 2 kb h 3 yEL = y

2  b2 h h  b 2 so then ∫ y EL dV = ∫0 y k 2 ( h − y ) dy  = k 2 ∫0 h 2 y − 2hy 2 + y 3 dy h  h 

(

)

h

2 1  1 2 2 b2  1 kb h = k 2  h 2 y 2 − hy 3 + y 4  = 3 4 0 12 h 2 Now

1 2 2 1  yV = ∫ y EL dV : y  kb 2h  = kb h 3 12   or y =

Note:

1 Abase = N  × b × 2  N = b2 π 4 tan N = k ( N ) b2

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.

b 2 tan πN

  

1 h Q.E.D. 4

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 125.

Since the spherical cup is uniform, the center of gravity will coincide with the centroid. Also, because the cup is thin, it can be treated like an area in finding the centroid.

An element of area is obtained by rotating arc ds about the y axis. With the y axis pointing downwards,

dA = 2π rds = 2π ( R sin θ ) Rdθ = 2π R 2 sin θ dθ yEL = y = R cosθ φ

φ

A = ∫ dA = 2π R 2 ∫0 sin θ dθ = 2π R 2 [ − cosθ ]0 = 2π R 2 (1 − cos φ ) φ 2 3 φ ∫ yEL dA = ∫0 ( R cosθ ) ( 2π R sin θ dθ ) = 2π R ∫0 cosθ sin θ dθ φ

 1  = 2π R3  − cos 2 θ  = π R3 1 − cos 2 φ 2  0

(

)

Then,

y=

1 1 3 2 ∫ yEL dA = 2π R 2 1 − cos φ π R 1 − cos φ , or A ( )

y=

R (1 + cos φ ) 2

(

)

Using

cos φ = y=

R−h h =1− : R R R  h  h 1 + 1 −   = R − R  2  2

The center of gravity is therefore located at a distance of

h h  R − y = R −  R −  = , above the base.(Q.E.D) 2 2 

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 126. (a) Bowl First note that symmetry implies

x = 0!

z = 0! for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of area is obtained by rotating the arc ds about the y axis. Then dAwall = ( 2π R sin θ )( Rdθ ) and Then

and

( yEL ) wall

= − R cos θ π /2

π /2

Awall = ∫π /6 2π R 2 sin θ dθ = 2π R 2 [ − cosθ ]π /6 = π 3R 2 ywall Awall = ∫ ( yEL )wall dA

(

π /2

= ∫π /6 ( − R cosθ ) 2π R 2 sin θ dθ = π R3 cos 2 θ 

)

π /2 π /6

3 = − π R3 4

π

R2,

By observation

Abase =

Now

y ΣA = ΣyA

or

or

4

ybase = −

3 R 2

π 3 π  3    y  π 3R 2 + R 2  = − π R3 + R 2  − R 4  4 4  2   y = −0.48763R

R = 350 mm ∴ y = −170.7 mm !

(b) Punch First note that symmetry implies

x = 0! z = 0! and that because the punch is homogeneous, its center of gravity will coincide with the centroid of the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then dV = π x 2dy, yEL = y continued

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.

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x2 + y 2 = R2

Now

so that 0

V = ∫−

Then

(

)

dV = π R 2 − y 2 dy

(

)

 

π R 2 − y 2 dy = π  R 2 y − 3/2 R

0

1 3 y 3  −

3/2 R

3   3  1 3   3 = −π  R 2  − R − − R  = π 3R3   2  3  2   8  

and

0

∫ yELdV = ∫−

0

 y π R 2 − y 2 ) dy  = π  3/2 R ( )  ( 

1 2 2 1 4 R y − y  4 − 2

3/2 R

4

 1  3  1 3   15 = −π  R 2  − R − − R  = − π R4     2  2  4 2   64   2

Now or

15 3  yV = ∫ yEL dV : y  π 3 R3  = − π R 4 64 8  y =−

5 8 3

R

R = 350 mm

∴ y = −126.3 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.

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Chapter 5, Solution 127.

The centroid can be found by integration. The equation for the bottom of the gravel is: y = a + bx + cz, where the constants a, b, and c can be determined as follows: For x = 0, and z = 0: y = − 3 in., and therefore

−

3 1 ft = a, or a = − ft 12 4

For x = 30 ft, and z = 0: y = − 5 in., and therefore

−

5 1 1 ft = − ft + b ( 30 ft ) , or b = − 12 4 180

For x = 0, and z = 50 ft: y = − 6 in., and therefore

−

6 1 1 ft = − ft + c ( 50 ft ) , or c = − 12 4 200

Therefore:

1 1 1 y = − ft − x− z 4 180 200 Now x dV x = ∫ EL V

A volume element can be chosen as:

dV = y dxdz, or

dV =

1⎛ 1 1 ⎞ x+ z ⎟ dx dz, and ⎜1 + 4⎝ 45 50 ⎠

xEL = x continued

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.

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Then 50 30 ⎛ ⎞ ∫ xEL dV = ∫0 ∫0 4 ⎜1 + 45 x + 50 z ⎟ dx dz ⎝ ⎠

x

1

1

30

1 50 ⎡ x 2 1 3 z 2⎤ = ∫0 ⎢ + x + x ⎥ dz 4 ⎣ 2 135 100 ⎦ 0

=

1 50 ∫ ( 650 + 9 z ) dz 4 0

=

1⎡ 9 ⎤ 650 z + z 2 ⎥ ⎢ 4⎣ 2 ⎦0

50

= 10937.5 ft 4

The volume is: 50 30 1 ⎛

V ∫ dV = ∫0 ∫0

1 1 ⎞ x+ z ⎟ dx dz ⎜1 + 4⎝ 45 50 ⎠ 30

1 50 ⎡ 1 2 z ⎤ = ∫0 ⎢ x + x + x dz 4 ⎣ 90 50 ⎥⎦ 0

=

1 50 ⎛ 3 ⎞ 40 + z ⎟ dz ∫ 0 ⎜ 4 ⎝ 5 ⎠ 50

1⎡ 3 2⎤ = ⎢ 40 z + z 4⎣ 10 ⎥⎦ 0

= 687.50 ft 3 Then

x dV 10937.5ft 4 x = ∫ EL = = 15.9091 ft V 687.5 ft 3 Therefore:

V = 688 ft 3 W x = 15.91 ft W

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.

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Chapter 5, Solution 128.

Choosing the element of volume shown, i.e. a filament of sides, y, dx, and dz:

dV = y dx dy, and z EL = z

x z b a V = ∫ dV = ∫ 0 ∫ 0  y0 − y1 − y2  dx dz a b  =∫

b y x 0 0



a

x2 zx  1 a  b − y1 − y2  dz = ∫ 0  y0a − y1a − y2 z  dz b 0 b  2a 2  b

 1 a z2  1 1   =  y0az − y1az − y2  =  y0 − y1 − y2  ab b 2 0  2 2 2   

xz

z2   dx dz 

b a ∫ zEL dV = ∫ 0 ∫ 0  y0 z − y1 a − y2 b 

=∫

b y zx 0 0



a

x2 z z2x  za z 2a  b − y1 − y2 − y2  dz  dz = ∫ 0  y0 za − y1 b 0 b  2a 2  b

 z 2a z 2a z 3a  1 1  2 1 =  y0 − y1 − y2  =  y0 − y1 − y2  ab 2 4 3b  0  2 4 3   continued

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.

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Now,

z =

1 1 1 1  1 z EL dV = y0 − y1 − y2  ab 2 , or ∫  1 1  2  4 3  V  y0 − y1 − y2  ab 2 2   1 1 1 y0 − y1 − y2 2 4 3 b z = 1 1 y0 − y1 − y2 2 2

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.

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Chapter 5, Solution 129. x =0

First note that symmetry implies

Choose as the element of volume a vertical slice of width 2x, thickness dz, and height y. Then 1 dV = 2 xy dz, yEL = y, zEL = z 2 h h h⎛ z⎞ Now and y = − z = ⎜1 − ⎟ x = a2 − z2 2 2a 2⎝ a⎠ z⎞ ⎛ dV = h a 2 − z 2 ⎜1 − ⎟ dz ⎝ a⎠

So

Then

V =∫ = =

Then

a h 0

z⎞ 1 2 ⎪⎧ 1 ⎡ ⎛ ⎛ z ⎞⎤ a − z ⎜1 − ⎟ dz = h ⎨ ⎢ z a 2 − z 2 + a 2 sin −1 ⎜ ⎟ ⎥ + a − z2 a⎠ ⎝ ⎝ a ⎠ ⎦ 3a ⎪⎩ 2 ⎣ 2

(

2

3/2 ⎪ ⎫

)

a

⎬ ⎭⎪ − a

1 2 ⎡ −1 a h ⎣sin (1) − sin −1 ( −1) ⎤⎦ 2

π 2

a 2h

h⎛ z ⎞⎤ ⎡ a ⎡1 2 2⎛ ∫ yELdV = ∫ − a ⎢ 2 × 2 ⎜1 − a ⎟ ⎥ ⎢h a − z ⎜1 − ⎝ ⎠⎦ ⎣ ⎝ ⎣ =

z⎞ ⎤ ⎟ dz ⎥ a⎠ ⎦

h2 a z z2 ⎞ 2 2⎛ a z − − + 1 2 ⎜ ⎟ dz ∫ ⎜ a a 2 ⎟⎠ 4 −a ⎝ continued

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.

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=

h2 4

⎧⎪ 1 ⎡ ⎡2 2 2 2 2 −1 ⎛ z ⎞ ⎤ a − z2 ⎨ ⎢ z a − z + a sin ⎜ ⎟ ⎥ + ⎢ a a 2 3 ⎝ ⎠⎦ ⎣ ⎩⎪ ⎣

(

1 ⎡ z + 2 ⎢− a2 − z 2 a ⎣ 4

(

=

3 2

⎤ ⎥ ⎦ a

3 2

a2 z 2 a4 ⎛ z ⎞ ⎤ ⎪⎫ + a − z 2 + sin −1 ⎜ ⎟ ⎥ ⎬ 8 8 ⎝ a ⎠ ⎦ ⎭⎪ −a

5h 2a 2 ⎡ −1 sin (1) − sin −1 ( −1) ⎤⎦ ⎣ 32

⎛ π a2 yV = ∫ yEL dV : y ⎜⎜ ⎝ 2

Then

)

)

⎞ 5h 2a 2 h ⎟⎟ = (π ) 32 ⎠

or y =

and

5 h 16

⎡ z⎞ ⎤ a 2 2⎛ ∫ zELdV = ∫ − a z ⎢ h a − z ⎜1 − a ⎟ dz ⎥ ⎝

⎣

⎪⎧ 1 = h ⎨− a 2 − z 2 ⎩⎪ 3

(

=−

)

3 2

⎠

⎦

1⎡ z − ⎢− a 2 − z 2 a⎣ 4

(

)

3 2

a2z 2 a 4 −1 ⎛ z ⎞ ⎤ ⎪⎫ + a − z2 + sin ⎜ ⎟ ⎥ ⎬ 8 8 ⎝ a ⎠ ⎦ ⎭⎪

a

−a

a3h ⎡ −1 sin (1) − sin −1 ( −1) ⎤⎦ ⎣ 8

⎛ π a 2h ⎞ π a 3h zV = ∫ z EL dV : z ⎜⎜ ⎟⎟ = − 8 ⎝ 2 ⎠

or z = −

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.

a 4

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Chapter 5, Solution 130.

A, mm 2 1

xA, mm3

yA, mm3

x , mm

y , mm

21 × 22 = 462

1.5

11

693

5082

2

−

1 ( 6 )( 9 ) = −27 2

−6

2

162

−54

3

−

1 ( 6 )(12 ) = −36 2

8

2

−288

−72

567

4956

Σ

399

Then

X =

Σ xA 567 mm 3 = ΣA 399 mm 2

or X = 1.421 mm

and

Y =

Σ yA 4956 mm 3 = ΣA 399 mm 2

or Y = 12.42 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.

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Chapter 5, Solution 131.

A, in 2

1 2 Σ

1 (10)(15) = 50 3

π 4

(15)2

= 176.71

x , in.

y , in.

xA, in 3

yA, in 3

4.5

7.5

225

375

6.366

16.366

1125

2892

226.71

1350

3267

X Σ A = Σx A

Then

(

)

X 226.71 in 2 = 1350 in 3

or X = 5.95 in.

Y ΣA = Σy A

and

(

)

Y 226.71 in 2 = 3267 in 3

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.

or Y = 14.41 in.

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Chapter 5, Solution 132.

First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line.

L, mm

1

y , mm

xL, mm 2

yL, mm 2

6

3

80.50

40.25

2

16

12

14

192

224

3

21

1.5

22

31.50

462

4

16

−9

14

−144

224

− 4.5

3

− 48.67

32.45

111.32

982.7

5

Σ Then

122 + 62 = 13.416

x , mm

62 + 92 = 10.817 77.233

X ΣL = Σx L

X (77.233 mm) = 111.32 mm 2 and

or X = 1.441 mm

Y ΣL = Σ y L

Y (77.233 mm) = 982.7 mm 2

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.

or Y = 12.72 mm

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Chapter 5, Solution 133.

First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through C. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line.

Thus ΣM C = 0, which implies that x = 0

or

Σ xi Li = 0

Hence

L ( L ) + ( − 4 in.)(8 in.) + ( − 4 in.)(10 in.) = 0 2

or

L2 = 144 in 2

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.

or L = 12.00 in.

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Chapter 5, Solution 134.

For the element (EL) shown At

x = a, y = h : h = ka3

Then

x=

h a3

k =

or

a 1/3 y h1/3

dA = xdy

Now

= xEL = h

A = ∫ dA =∫ 0

Then

a 1/3 y dy h1/3 1 1 a 1/3 x= y , yEL = y 2 2 h1/ 3

a 1/3 3 a y dy = y 4/3 4 h1/3 h1/3

( )

h

= 0

3 ah 4 h

1 a 1/3  a 1/3  1 a  3 5/3  3 2 and ∫ xEL dA = ∫ y  1/3 y dy  = y  = a h 1/3 2/3  2h h  2 h 5  0 10 h 0

h

a  3 7/3  3 2  a 1/3  ∫ yEL dA = ∫ y  h1/3 y dy  = h1/3  7 y  = 7 ah    0 h 0

Hence

3 2 3  xA = ∫ xEL dA : x  ah  = a h 4 10  

x =

2 a 5

3  3 yA = ∫ yEL dA: y  ah  = ah 2 4  7

y =

4 h 7

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 135.

For

2b = ka 2

y1 at x = a, y = 2b

or k =

2b a2

2b 2 x a2

Then

y1 =

By observation

y2 = −

b x ( x + 2b) = b  2 −  a a

xEL = x

Now and for 0 ≤ x ≤ a :

1 b y1 = 2 x 2 2 a

and

dA = y1dx =

1 b x y2 =  2 −  2 2 a

and

x  dA = y2dx = b  2 −  dx a  

yEL =

2b 2 x dx a2

For a ≤ x ≤ 2a : yEL = Then

a

A = ∫ dA = ∫ 0

2b 2 x 2a  x dx + ∫ a b  2 −  dx 2 a a  2a

a 2  a 2b  x3  x  7 = 2   + b  −  2 −   = ab a 2 6 a  3 0     0

and

x  a  2b 2 2a    ∫ xEL dA = ∫ 0 x  a 2 x dx  + ∫ a x b  2 − a  dx  



 

a





2a

=

 2 x3  2b  x 4   + b x −  2  3a  0 a  4 0 

=

1 2 1  2 2 2 3  2a − ( a )   a b + b ( 2a ) − ( a )  +   3a   2

=

7 2 ab 6

{

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.

( )

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x   x  a b 2  2b 2  2a b  ∫ yEL dA = ∫ 0 a 2 x  a 2 x dx  + ∫ 0 2  2 − a  b  2 − a  dx         2a

a 3 2b 2  x5  b2  a  x  = 4   + −  2 −   2  3  a   a  5 0 a 17 2 = ab 30

Hence

7  7 xA = ∫ xEL dA: x  ab  = a 2b 6  6  7  17 2 yA = ∫ yEL dA: y  ab  = ab  6  30

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.

x =a y =

17 b 35

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Chapter 5, Solution 136.

The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the second theorem of Pappus-Guldinus and using Fig. 5.8A, we have

V = 2π xA = 2πΣxA = 2π ( x1 A1 + x2 A2 )      1 1   1 1   3   2R sin 30o o π 2 = 2π  × R   × R × R +  cos 30   R    2   3 × π 6   3 2   2 2     6     R3 R3  3 3 = 2π  + π R3  = 8  16 3 2 3  =

Since

3 3 3 π (12 in.) = 3526.03 in 3 8

1 gal = 231 in 3 V =

3526.03 in 3 = 15.26 gal 231 in 3/gal V = 15.26 gal

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.

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Chapter 5, Solution 137.

Have

RI = ( 9 ft )( 200 lb/ft ) = 1800 lb RII =

Then

1 ( 3 ft )( 200 lb/ft ) = 300 lb 2

ΣFx = 0: Ax = 0

ΣM A = 0: − ( 4.5 ft )(1800 lb ) − (10 ft )( 300 lb ) + ( 9 ft ) B y = 0 or

By = 1233.3 lb

B = 1233 lb

A = 867 lb

ΣFy = 0: Ay − 1800 lb − 300 lb + 1233.3 lb = 0 or

Ay = 866.7 lb

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.

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Chapter 5, Solution 138.

Have

Then

RI =

1 ( 4 m )( 2000 kN/m ) = 2667 N 3

RII =

1 ( 2 m )(1000 kN/m ) = 666.7 N 3

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 2667 N − 666.7 N = 0

or

Ay = 3334 N

A = 3.33 kN

ΣM A = 0: M A − (1 m )( 2667 N ) − ( 5.5 m )( 666.7 N ) or

M A = 6334 N ⋅ m

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.

M A = 6.33 kN ⋅ m

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Chapter 5, Solution 139.

Consider the free-body diagram of the side.

Have Now

P=

1 1 Ap = A (γ d ) 2 2

ΣM A = 0:

( 9 ft ) T

−

d P=0 3

Then, for d max:

( 9 ft ) ( 0.2 ) ( 40 × 103 lb ) −

d max  1  3  (12 ft ) ( d max )  62.4 lb/ft d max  = 0 3 2 

or

3 216 × 103 ft 3 = 374.4 d max

or

3 d max = 576.92 ft 3

(

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.

)

d max = 8.32 ft

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Chapter 5, Solution 140.

First, assume that the machine element is homogeneous so that its center of gravity coincides with the centroid of the corresponding volume.

x , in.

I II

V , in 3 (4)(3.6)(0.75) = 10.8 (2.4)(2.0)(0.6) = 2.88

2.0 3.7

0.375 1.95

21.6 10.656

4.05 5.616

III

π(0.45)2 (0.4) = 0.2545

4.2

2.15

1.0688

0.54711

1.2

0.375

− 0.7068

− 0.22089

32.618

9.9922

IV

Σ

2

− π (0.5) (0.75) = − 0.5890 13.3454

y , in.

xV , in 4

yV , in 4

X ΣV = Σ x V

Have

(

)

X 13.3454 in 3 = 32.618 in 4

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.

or X = 2.44 in.

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Chapter 5, Solution 141.

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)

zV = 22.5 −

4 ( 6.25 ) 3π

= 19.85 mm AV = −

π 2

( 6.25)2

= − 61.36 mm 2

A, mm 2

x , mm

y , mm

z , mm

xA, mm3

yA, mm3

zA, mm3

I

( 25)( 60) = 1500

12.5

0

30

18 750

0

45 000

II

(12.5)( 60 ) = 750

25

− 6.25

30

18 750

− 4687.5

22 500

III

( 7.5)( 60 ) = 450

28.75

−12.5

30

12 937.5

− 5625

13 500

IV

− (12.5 )( 30 ) = − 375

10

0

37.5

− 3750

0

−14 062.5

V

− 61.36

10

0

19.85

− 613.6

0

−1218.0

Σ

2263.64

46 074

−10 313

65 720

Have

X ΣA = ΣxA

(

)

X 2263.64 mm 2 = 46 074 mm 3

Y ΣA = Σ yA

(

)

Y 2263.64 mm 2 = −10 313 mm 3

or X = 20.4 mm or Y = − 4.55 mm

Z ΣA = Σ zA

(

)

Z 2263.64 mm 2 = 65 720 mm 3

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.

or Z = 29.0 mm