Cap 05 by Cristian Manrique

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Chapter 5, Solution 1. \

Then

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

200 × 150 = 30000

−100

250

− 30 000000

6 750 000

400 × 300 = 120000

200

150

24 000 000

18000000

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

10 × 8 = 80

400

320

1 × 9 × 12 = 54 2

702

216

134

1102

536

Then

X =

ΣxA 1102 = ΣA 134

and

Y =

ΣyA 1102 = ΣA 134

X = 8.22 in.

or Y = 4.00 in.

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Chapter 5, Solution 3.

Then

A, mm 2

x , mm

xA, mm3

1 × 90 × 270 = 12 150 2

2 ( 90 ) = 60 3

729 000

1 × 135 × 270 = 18 225 2

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

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Chapter 5, Solution 4.

A,in 2

x ,in.

1 ( 21)( 24 ) = 252 2

(13)( 40 ) = 520

2 ( 21) = 14 3 21 +

1 (13) = 27.5 2

xA,in 3

y ,in. 40 −

1 ( 24 ) = 32 3

772

Then

yA,in 3

3528

8064

14 300

10 400

17 828

18 464

X =

ΣxA 17828 = in. ΣA 772

Y =

ΣyA 18464 = in. ΣA 772

or Y = 23.9 in.

X = 23.1 in.

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Chapter 5, Solution 5.

A, mm 2

x , mm

π ( 225 )

1 ( 375)( 225) = 42 188 2

−

= 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

81 949

Then

X =

ΣxA 1476600 mm = ΣA 81 949

or X = 18.02 mm

Y =

ΣyA 6961 000 mm = ΣA 81 949

Y = 84.9 mm

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Chapter 5, Solution 6.

1 2

−

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

× ( 4.5 ) = −15.9043 8 −

π 4

( 6 )2 = − 28.274

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

X = 8.32 in.

or Y = 3.61 in.

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Chapter 5, Solution 7.

A,in 2 1

π (16 ) 4

= 201.06

− ( 8 )( 8 ) = − 64

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

X = 8.09 in.

or Y = 8.09 in.

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Chapter 5, Solution 8.

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

35 343

63.662

2 250 006

− 4417.9

31.831

− 31.831

−140 626

140 626.2

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

X = 68.2 mm

or Y = 4.55 mm

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Chapter 5, Solution 9.

A −

Therefore, for X =

π2

( 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

( (

)

4 2r23 − r13 4r1 = 3π 3π 2r22 − r12

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:

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

)

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

)

( 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 − ∆) =

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).

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

8.1962

221.30

15.5885

3.4641

54.000

−18.8495

3.8197

−71.999

23.739

ΣxA 203.30 = ΣA 23.739

203.30

X = 8.56 in.

and by symmetry

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

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

Then

X =

ΣxA 720000 = mm ΣA 6000

Y =

ΣyA 360000 = mm ΣA 6000

or X = 120.0 mm or

Y = 60.0 mm

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Chapter 5, Solution 13.

A,in 2

x ,in.

y ,in.

(18)(8) = 144

−3

− 432

576

1 ( 6 )( 9 ) = 27 2

−3

−81

−5.0930

−3.8197

− 432.00

− 324.00

−810.00

171.00

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

yA,in 3

X = − 3.17 in.

or Y = 0.668 in.

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Chapter 5, Solution 14.

X = 90 mm

First, by symmetry

1 2

−

A, mm 2

y , mm

yA, mm3

(180 )(120 ) = 21 600

1 296 000

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

Y =

ΣyA 124 260 = 4635.4 ΣA

124 260

or Y = 26.8 mm

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Chapter 5, Solution 15.

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

18 240

−4

72 960

218 880

−1920

− 56

107520

−103 680

− 4071.5

− 41.441

168 731

186 731

12 248.5

−134171

−53 531.1

Then and

X =

ΣxA −134171 = ΣA 12 248.5

Y =

ΣyA −53 531 = ΣA 12 248.5

X = −10.95 mm or Y = − 43.7 mm

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Chapter 5, Solution 16. \

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

−

2 (100 )( 50 ) = − 3333.3 3

23 334

xA, mm3

yA, mm3

x , mm

y , mm

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

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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

8 × 15 = 120

7.5

900

− 4 × 10 = − 40

− 320

Then

y , in.

580

ΣyA 580 = = 7.2500 in. ΣA 80

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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.

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Chapter 5, Solution 18.

A, mm 2

Y =

yA, mm3

(80 )( 20 ) = 1600

144 000

( 20 )(80 ) = 1600

64 000

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 !

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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.

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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

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

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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

400

200

80 000

300

400

150

120 000

45 000

600

100

300

60 000

180 000

150

− 200

225

− 30 000

33 750

200

−100

150

− 20 000

30 000

150

11 250

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

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Chapter 5, Solution 22.

L, in.

x , in.

y , in.

xL, in 2

y , in 2

9.5

180.5

14.5

217.5

488

242

Then

X =

ΣxL 488 = ΣL 56

or X = 8.71 in.

and

Y =

ΣyA 242 = 56 ΣA

or Y = 4.32 in.

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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

45 000

187.5

112.5

81 998

49 199

− 50 625

50 625

76 373

99 824

1 2 3

3752 + 2252 = 437.32

π 2

( 225)

−

( 225)

1390.75

( 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

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Chapter 5, Solution 24.

L, mm

x , mm

y , mm

xL, mm 2

yL, mm 2

37.5

2812.5

150

11 250

95.492

45 000

−112.5

− 8437.5

47.746

− 47.746

5625.0

− 5625.0

53 437

− 2812.5

(150 )π

4 5

= 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

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Chapter 5, Solution 25.

O From Figure 5.8 b:

r =

( 20 in.) sin 30° 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:

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

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 )

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

cosθ =

5000 11250 or

θ = 63.6°

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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 )

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

l 2 + 300l − 197602 = 0. Solving for l :

l = 319.15, and l = − 619.15, and discarding the negative root l = 319 mm

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Chapter 5, Solution 28. The centroid coincides with the center of gravity because the wire is homogeneous.

L 1

2θ r

− −

−

r sin θ

θ l 2

X =

Then

r 2

−

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°

or l = 2.11 r

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Chapter 5, Solution 29.

y =

Then

ΣyA ΣA

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

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

(

)

2ζ ( c + kζ ) − c + kζ 2 = 0

(2)

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2cζ + 2ζ

Expanding (2)

ζ =

Then

− c − kζ − 2c ±

kζ

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

2 ( 0.6 )

or h = 0.387a !

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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

y =h

Q.E.D.

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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

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    

(

)

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:

x =

2 a! 3

y =

2 h! 3

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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 

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

a  h x 0  2

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:

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

k1 =

h a3

a = k 2 h3

k2 =

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 = ∫

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 

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

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

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 =

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

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 = ∫ 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 

(

)

3 2

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

= 0

 1    a 2 2 a (π − 2 )  3 

( )

3 2

1   1 +  − a3 + a3  3    2

or x =

2a ! 3 (π − 2 ) continued

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 

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 =

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

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,

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

b = k (0 − a)

Now

xEL = x, yEL =

and

dA = ydx =

and

b a2

b 2 x − a) 2( a

Then

k =

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

∫ yEL dA = ∫ 0 =

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  

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 =

∴ y =

b a2

b 2 x a2

x = a, −2b = k1a3

On line 2 at

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 0

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

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      

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

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

x =

39 a 50

or y = −

39 b 175

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 =

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

)

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

= a

continued

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 

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 =

148 b! 95

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 41.

For y2

x = a, y = b : a = kb2

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

continued

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

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

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 

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

and

Hence

1/2

−1/3

xEL = x

Now and

) ( −x )

 dy  dL = 1 +    dx 

  dx = 1 +  a 2/3 − x 2/3  

(

L = ∫ dL = ∫ 0

) ( −x ) 1/2

−1/3

 

1/2 2

  dx 

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

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

(

= 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

(

y = a 2/3 − x 2/3

)

3/2

2 a ! 5

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(

dy = a 2/3 − x 2/3 dx

Then

and

Hence

1/2

−1/3

xEL = x

Now and

) ( −x )

 dy  dL = 1 +    dx 

  dx = 1 +  a 2/3 − x 2/3  

(

L = ∫ dL = ∫ 0

) ( −x ) 1/2

−1/3

 

1/2 2

  dx 

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

x =

2 a ! 5

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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 = ∫ =

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

(

   0

)

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

∫ xEL dL = ∫



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

a 0

k =

(

)

x = 0.574a

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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

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

x sin

π2

−2

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 !

x = 0.363L !

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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

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

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

(

)

cosθ dθ =

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

(

)

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:

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

y =0

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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

V = 1.949 ft 3

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

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

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

A = 25.1 ft 2

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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

27.5

357.5

1360

800

578

1360

10.5

334.85

892.92

336

128

2966.4

3180.9

(a)

212 + 242 = 31.890

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

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(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 !

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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

(

A = 1.885 × 106 mm 2

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

V = 245 × 106 mm3

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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

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

V =

2 2 πa b 3

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Chapter 5, Solution 51. The area A and circumference C of the cross section of the bar are

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

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 

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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

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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

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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:

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

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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 !

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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

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A, mm 2

Seg . 1

1 ( 76.886 )( 37.5) = 1441.61 2

− α ( 75 ) = − 805.32

−

1 ( 73.951)(12.5) = − 462.19 2

− ( 2.9354 )(12.5 ) = − 36.693

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

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 !

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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

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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

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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

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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 )

a = 56.25 k

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

(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

Now

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

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

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

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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

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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

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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

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

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  

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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

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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

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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

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

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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

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

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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:

(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  

(8 − a )2 − 8 (8 − a ) + 8 = 0,

which gives continued

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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)

= 3531.4 N or A = 3.53 kN

B = 7.06 kN

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

COSMOS: Complete Online Solutions Manual Organization System

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:

(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 )

( ) 24 + ( 64 − 16a + a )

152 − 8a − a 2 88 − 16a + a 2

24 + 128 − 8a − a 2

continued

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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 

(

)

a − 20a + 72 = 0

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 )

= 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

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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:

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

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

(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

P = 1.527 kips !

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

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

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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.

COSMOS: Complete Online Solutions Manual Organization System

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

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:

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 

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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. !

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

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

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

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

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

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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 )

(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

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%

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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

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

or Finally

(3.7278d

)

+ 176.58 kN = 264.87 kN

d = 4.8667 m

4.8667 m = 20 × 10−3

m ×N year

or N = 243 years !

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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)

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

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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:

( 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

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

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

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

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

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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

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 =

(

)

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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:

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:

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:

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.

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Chapter 5, Solution 90.

Note, for the axes shown y

−R

−2π R 4

3 − r 8

1 4 πr 4

(π R ) ( 2R ) = 2π R

2 − π r3 3

 r3  2π  R 3 −  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 =

3 R: y = − 4

1−

1 3   3 4

1 3 1−   3 4

or y = −1.118R 1−

(b)

y = −1.2R : − 1.2R = −

1 r    8 R 

1 r  1−   3 R 

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

Solving numerically

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

(

)

1 2 πa h 2

h 2

1 2 2 πa h 4

1 2 πa h 6

h 4

1 π a 2h 2 24

2 2 πa h 3

7 π a 2h 2 24

or Y =

7 h 16

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

V 1 2

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

2 − a 3h 3 1 3 ah 6 1 − a 3h 2

) )

or Z = −

3a 4π

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Chapter 5, Solution 93.

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  

(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  

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

∴

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

( 60 )(105)(10 ) = 63000

−5

52.5

− 315 000

3 307 500

1 2 π ( 30 ) (10 ) = 14 137.2 2

−5

− 70 686

1 664 400

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

405 000

810 000

− π (19 ) (10 ) = −11 341.1

−5

105

56 706

−1 190 820

1 2 − π (19 ) (15 ) = − 8505.9 2

−186 585

−255 180

−110 565

4 335 900

Σ Then Y =

30 −

4 (19 ) 3π

= 21.936

105 +

4 ( 30 ) 3π

=117.732

84 290

ΣyV −110 565 mm = 84 290 ΣV

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

( 60 )(105)(10 ) = 63000

52.5

3 307 500

1 2 π ( 30 ) (10 ) = 14 137.2 2

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

105 +

4 ( 30 ) 3π

= 117.732

1 664 400

810 000

− π (19 ) (10 ) = −11 341.1

105

−1 190 820

1 2 − π (19 ) (15 ) = − 8505.9 2

− 255 180

84 290

ΣzV 4 335 900 = mm ΣV 84 290

4 335 900

or Z = 51.4 mm

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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

(100 )(88)(12 ) =105600

5 280 000

(100 )(12 )(88) = 105600

5 280 000

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

616 590

−

1 ( 66 )( 45)(12 ) = −17 820 2

Σ Then X =

34 +

2 ( 66 ) = 78 3

209 190

ΣxV 9 786 600 = mm ΣV 209190

−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

(100 )(12 )(88) = 105600

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

−

1 ( 66 )( 45)(12 ) = −17 820 2

Σ Then Z =

z , mm

zV , mm 4

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

−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

( 8)( 0.9)( 2.7) = 19.44

77.76

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

34.02

1 2 π (1.35) ( 0.9) = 2.5765 2 2

− π ( 0.8 ) ( 0.9 ) = −1.80956

37.217

ΣxV 119.392 = in. ΣV 37.217

1.8

22.088

π −14.4765 119.392

or X = 3.21 in.

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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

( 8)( 0.9)( 2.7) = 19.44

0.45

8.748

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

1.6

27.216

1 2 π (1.35) ( 0.9) = 2.5765 2

0.45

1.15943

− π ( 0.8 ) ( 0.9 ) = −1.80956

0.45

− 0.81430

37.217

ΣyV 36.309 = in. ΣV 37.217

36.309 or Y = 0.976 in.

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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

( 600 )( 400 ) = 240000

300

200

( 300 )( 400 ) = 120000

600

200

−150

−18

− (120 )( 280 ) = − 3360

600

140

− 240

− 20.160

− 4.7040

8.0640

( 600 )( 300 ) = 180000

300

400

−150

− 27

− 35 343

240

400

− 236.34

− 2.7000

− 4.5

2.6588

471 057

169.358

125.159

− 28.583 continued

COSMOS: Complete Online Solutions Manual Organization System

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

or Z = − 60.7 mm !

COSMOS: Complete Online Solutions Manual Organization System

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 ( 4.5)( 3) = 6.75 2

1.5

10.125

47.25

( 4.5)(10 ) = 45

2.25

101.25

135

180

− ( 2.25 )( 5 ) = −11.25 − 2.25

1.125

1.5

−12.6563

−16.875

− 67.5

2.25

17.8925

71.211

116.611

165.375

183.71 1

π 2

( 2.25)2 = 7.9522

48.452

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.

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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 )

= 241.1 mm

= 61.11 mm

For 3: Length DE =

(180 )2 + ( 96 )2

y = 90 mm,

= 204 mm

z = 48 mm

COSMOS: Complete Online Solutions Manual Organization System

y , mm

z , mm

yA, mm3

zA, mm3

(150 )2 = 35.34 × 103

339.7

12.005 × 106

( 96 )( 300 ) = 45.24 × 103

244.1

61.11

10.907 × 106

2.765 × 106

5.508 × 106

2.938 × 106

28.420 × 106

5.702 × 106

A, mm 2

π 2

( 204 )( 300 ) = 61.2 × 103 − 2.25

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 !

or Y = 40.2 mm !

COSMOS: Complete Online Solutions Manual Organization System

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

( 360 )( 270 ) = 97 200

135

13 122 000

( 339 )( 270 ) = 91 530

168

135

15 377 000

12 356 600

1 ( 339 )( 72 ) = 12 204 2

224

294

2 733 700

3 588 000

( 360 )( 343.63) = 123 707

168

306

20 783 000

37 854 000

1 ( 343.63)( 45) = 7731.73 2

224

318

1 731 900

2 458 700

7731.7

224

318

1 731 900

2 458 700

12 204

224

294

2 733 700

3 588 000

91 530

168

135

15 377 000

12 356 600

ÎŁ

443 838

60 468 200

87 782 600

COSMOS: Complete Online Solutions Manual Organization System

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 !

COSMOS: Complete Online Solutions Manual Organization System

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

16.2

1.8

29.16

32.4

16.2

1.8

29.16

32.4

97.2

2.7

262.44

1017.876

−15.2789

20.72113

−15552

21091.54

1017.876

−15.2789

20.72113

−15552

21091.54

− 706.858

−12.7324

23.2676

9000

−16446.9

− 706.858

−12.7324

23.2676

9000

−16446.9

1017.876

− 22.9183

13.08169

− 23328

13315.54

1769.511

− 36111.24

22669.6

COSMOS: Complete Online Solutions Manual Organization System

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. !

COSMOS: Complete Online Solutions Manual Organization System

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

576π

−128

−160π

− 42.667

96π

−

π 2

( 4 )2

= 8π

−

4 ( 4) 3π

=−

16 3π

(8)(12 ) = 96

576

1152

(8)(12 ) = 96

576

768

− 42.667

− 64π

Then

2 ( 4)

(8)( 4 ) = −16π

−

π 2

( 4 )2

= − 8π

4 ( 4) 3π

16 3π

( 4 )(12 ) = 48

288

480

( 4 )(12 ) = 48

288

480

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

COSMOS: Complete Online Solutions Manual Organization System

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 4

= 292.2 mm

= 212.2 mm = 398.3 mm

( 2 )( 500 )

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

(80)(500) = 40 000

250

1.6 × 106

10 × 106

196 350

292.2

212.2

57.4 × 106

41.67 × 106

III

(80)(680) = 54 400

500

0.2176 × 106

27.2 × 106

534 071

398.3

318.3

212.7 × 106

170 × 106

(80)(500) = 40 000

250

1.6 × 106

10 × 106

196 350

292.2

212.2

57.4 × 106

41.67 × 106

332.9 × 106

300.5 × 106

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

(

Z ΣA = Σ zA: Z 1.061 × 10 mm

) = 300.5 × 10

or Y = 314 mm 6

or Z = 283 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 107. Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

A,in 2

y ,in.

z , in.

yA, in 3

zA, in 3

(15)(14 ) = 120

1470

1.25

43.75

245

1.25

46.875

1.25

43.75

245

6.5

− 510.51

−21.848

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

− π ( 5 ) = − 78.540

1 2 − π (1.5 ) = −1.76715 4

( 4 )(12 ) = 48

288

480

( 4 )(12 ) = 48

288

480

235.43

134.375

1405.79

2 3

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

X = 7.50 in.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 108.

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

950

250

375

150

237.5 × 103

356.25 × 103

142.5 × 103

300

500

150

150 × 103

45 × 103

500

250

125 × 103

750

375

281.25 × 103

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

COSMOS: Complete Online Solutions Manual Organization System

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

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

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

COSMOS: Complete Online Solutions Manual Organization System

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

4cos 45° = 2.8284

28.284

4cos 45° = 2.8284

28.284

4π

12.5465

157.664

4π

2 ( 4)

125.664

2π

2 ( 4)

12.5465

78.832

51.416

104.568

462.16

π π

= =

π 8

Then

X =

ΣxL 104.568 = ΣL 51.416

Y =

ΣyL 462.16 = ΣL 51.416

or Y = 8.99 ft

and by symmetry:

X = 2.03 ft

COSMOS: Complete Online Solutions Manual Organization System

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)

3.5

2.5

(b)

6π = 18.8496

6.9099

168 37.699

120 130.249

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

Y =

ΣyL 340.25 = ΣL 84.850

or Y = 4.01 ft

X = 3.20 ft

COSMOS: Complete Online Solutions Manual Organization System

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 

(

)

(

)

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

(

Have

)

W , lb

x , in.

xW , in ⋅ lb

0.009791

0.3125

0.003060

0.093996

2.1

0.197393

III

−0.000846

2.7

−0.002284

0.017987

4.5

0.080942

0.000428

7.4

0.003169

0.12136

0.28228

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

or X = 2.33 in.

COSMOS: Complete Online Solutions Manual Organization System

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

0.0108372

0.075860

0.0142942

0.100059

0.0135717

0.040715

0.027489

0.137445

0.066192

0.35408

Σxm 0.35408 or = 0.066192 Σm

X = 5.35 mm

COSMOS: Complete Online Solutions Manual Organization System

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

continued

COSMOS: Complete Online Solutions Manual Organization System

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 !

COSMOS: Complete Online Solutions Manual Organization System

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.

( 2.5

)

− 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

(

)(

)

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

COSMOS: Complete Online Solutions Manual Organization System

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. !

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

21 a! 88

COSMOS: Complete Online Solutions Manual Organization System

5 π a3 24

and

a x ⎡π a/2 ⎣

∫2 xELdV = ∫

(

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

)

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

( )

Now

⎤⎫ ⎥ ⎪⎬ 4 ⎥⎪ ⎦⎥ ⎭

( a2 )

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

27 a! 40

COSMOS: Complete Online Solutions Manual Organization System

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

21 h! 88

COSMOS: Complete Online Solutions Manual Organization System

Component 2 h

V2 = ∫h/2 π

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

⎡ 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 )

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

27 h! 40

COSMOS: Complete Online Solutions Manual Organization System

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

2 h! 9

COSMOS: Complete Online Solutions Manual Organization System

Component 2 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 )

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

or x2 =

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

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

V =∫

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

continued

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 =

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

 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

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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 =

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 )   

(

)

= 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 =

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

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

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

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

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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 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=

1  4  a 1 − 2  ! 2  π 

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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

continued

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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=

(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

b (h − y) h

Then

dV = Aslicedy = ks 2dy = k

and

V = ∫0 k

= 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 

(

)

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

b 2 tan πN

  

1 h Q.E.D. 4

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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

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,

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

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 

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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

R2,

By observation

Abase =

Now

y ΣA = ΣyA

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

COSMOS: Complete Online Solutions Manual Organization System

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

1 3 y 3  −

3/2 R

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

and

∫ yELdV = ∫−

 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 = 350 mm

∴ y = −126.3 mm !

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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

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

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

= 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

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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



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   

z2   dx dz 

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

=∫

b y zx 0 0



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

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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

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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⎠

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

(

3/2 ⎪ ⎫

)

⎬ ⎭⎪ − 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

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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

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

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

or z = −

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

693

5082

−

1 ( 6 )( 9 ) = −27 2

−6

162

−54

−

1 ( 6 )(12 ) = −36 2

−288

−72

567

4956

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

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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

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

y , mm

xL, mm 2

yL, mm 2

80.50

40.25

192

224

1.5

31.50

462

−9

−144

224

− 4.5

− 48.67

32.45

111.32

982.7

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

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

Σ xi Li = 0

Hence

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

L2 = 144 in 2

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

h a3

k =

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

( )

= 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

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

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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 = ∫ 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  



 





 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

{

( )

COSMOS: Complete Online Solutions Manual Organization System

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

x =a y =

17 b 35

COSMOS: Complete Online Solutions Manual Organization System

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

COSMOS: Complete Online Solutions Manual Organization System

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

COSMOS: Complete Online Solutions Manual Organization System

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

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

M A = 6.33 kN ⋅ m

COSMOS: Complete Online Solutions Manual Organization System

Chapter 5, Solution 139.

Consider the free-body diagram of the side.

Have Now

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 

3 216 × 103 ft 3 = 374.4 d max

3 d max = 576.92 ft 3

(

)

d max = 8.32 ft

COSMOS: Complete Online Solutions Manual Organization System

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

− 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

or X = 2.44 in.

COSMOS: Complete Online Solutions Manual Organization System

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

( 25)( 60) = 1500

12.5

18 750

45 000

(12.5)( 60 ) = 750

− 6.25

18 750

− 4687.5

22 500

III

( 7.5)( 60 ) = 450

28.75

−12.5

12 937.5

− 5625

13 500

− (12.5 )( 30 ) = − 375

37.5

− 3750

−14 062.5

− 61.36

19.85

− 613.6

−1218.0

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

or Z = 29.0 mm