Vector mechanics for engineers statics 7th chapter 5

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PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION 2 , in A , in. x , in. y 3 , in xA 3 , in yA 1 8 6 48 × = 4 9 192 432 2 16 12 192 × = 8 6 1536 1152 Σ 240 1344 1584 Then 3 2 1344 in 240 in xA X A Σ = = Σ or 5.60 in. X = and 3 2 1584 in 240 in yA Y A Σ = = Σ or 6.60 in. Y =

Transcript of Vector mechanics for engineers statics 7th chapter 5

PROBLEM 5.1 Locate the centroid of the plane area shown.

SOLUTION

2, inA , in.x , in.y 3, inxA 3, inyA

1 8 6 48× = 4− 9 192− 432

2 16 12 192× = 8 6 1536 1152

Σ 240 1344 1584

Then 3

21344 in240 in

xAXA

Σ= =

Σ or 5.60 in.X =

and 3

21584 in240 in

yAYA

Σ= =

Σ or 6.60 in.Y =

PROBLEM 5.2 Locate the centroid of the plane area shown.

SOLUTION

2, mmA , mmx , mmy 3, mmxA 3, mmyA

1 1 60 75 22502× × = 40 25 90 000 56 250

2 105 75 7875× = 112.5 37.5 885 900 295 300

Σ 10 125 975 900 351 600

Then 3

2975 900 mm10 125 mm

xAXA

Σ= =

Σ or 96.4 mmX =

and 3

2351 600 mm10 125 mm

yAYA

Σ= =

Σ or 34.7 mmY =

PROBLEM 5.3 Locate the centroid of the plane area shown.

SOLUTION

For the area as a whole, it can be concluded by observation that

( )2 24 in.3

Y = or 16.00 in.Y =

2, inA , in.x 3, inxA

1 1 24 10 1202× × = ( )2 10 6.667

3= 800

2 1 24 16 1922× × = ( )110 16 15.333

3+ = 2944

Σ 312 3744

Then 3

23744 in312 in

xAXA

Σ= =

Σ or 12.00 in.X =

PROBLEM 5.4 Locate the centroid of the plane area shown.

SOLUTION

2, mmA , mmx , mmy 3, mmxA 3, mmyA

1 21 22 462× = 1.5 11 693 5082

2 ( )( )1 6 9 272

− = − 6− 2 162 54−

3 ( )( )1 6 12 362

− = − 8 2 288− 72−

Σ 399 567 4956

Then 3

2567 mm399 mm

xAXA

Σ= =

Σ or 1.421 mmX =

and 3

24956 mm399 mm

yAYA

Σ= =

Σ or 12.42 mmY =

PROBLEM 5.5 Locate the centroid of the plane area shown.

SOLUTION

2, mmA , mmx , mmy 3, mmxA 3, mmyA

1 120 200 24 000× = 60 120 1 440 000 2 880 000

2 ( )260

5654.92

π− = − 94.5 120 534 600− 678 600−

Σ 18 345 905 400 2 201 400

Then 3

2905 400 mm18 345 mm

xAXA

Σ= =

Σ or 49.4 mmX =

and 3

22 201 400 mm

18 345 mmyAYA

Σ= =

Σ or 93.8 mmY =

PROBLEM 5.6 Locate the centroid of the plane area shown.

SOLUTION

2, inA , in.x , in.y 3, inxA 3, inyA

1 ( )29

63.6174

π=

( )( )4 9

3.89173π

−= − 3.8917 243− 243

2 ( )( )1 15 9 67.52

= 5 3 337.5 202.5

Σ 131.1 94.5 445.5

Then 3

294.5 in

131.1 inxAXA

Σ= =

Σ or 0.721 in.X =

and 3

2445.5 in131.1 in

yAYA

Σ= =

Σ or 3.40 in.Y =

PROBLEM 5.7 Locate the centroid of the plane area shown.

SOLUTION

First note that symmetry implies X Y=

2, mmA , mmx 3, mmxA

1 40 40 1600× = 20 32 000

2 2(40) 1257

− = − 16.98 21 330−

Σ 343 10 667

Then 3

210 667 mm

343 mmxAXA

Σ= =

Σ or 31.1 mmX =

and 31.1 mmY X= =

PROBLEM 5.8 Locate the centroid of the plane area shown.

SOLUTION

First note that symmetry implies 0X =

2, inA , in.y 3, inyA

1 ( )24

25.132

π− = − 1.6977 42.67−

2 ( )26

56.552

π= 2.546 144

Σ 31.42 101.33

Then 3

2101.33 in31.42 in

yAYA

Σ= =

Σ or 3.23 in.Y =

PROBLEM 5.9 For the area of Problem 5.8, determine the ratio 2 1/r r so that 13 /4.y r=

SOLUTION

A y yA

1 2

12rπ−

143rπ

31

23r−

2 2

22rπ

243rπ

32

23r

Σ ( )2 22 12r rπ

− ( )3 32 1

23r r−

Then Y A y AΣ = Σ

or ( ) ( )2 2 3 31 2 1 2 1

3 24 2 3r r r r rπ× − = −

2 32 2

1 1

9 1 116

r rr r

π − = −

Let 2

1

rpr

=

[ ] 29 ( 1)( 1) ( 1)( 1)16

p p p p pπ+ − = − + +

or 216 (16 9 ) (16 9 ) 0p pπ π+ − + − =

PROBLEM 5.9 CONTINUED

Then 2(16 9 ) (16 9 ) 4(16)(16 9 )

2(16)p

π π π− − ± − − −=

or 0.5726 1.3397p p= − =

Taking the positive root 2

11.340r

r=

PROBLEM 5.10 Show that as 1r approaches 2,r the location of the centroid approaches that of a circular arc of radius ( )1 2 / 2.r r+

SOLUTION

First, determine the location of the centroid.

From Fig. 5.8A: ( )

( ) ( )2 22 2 2 22

2

sin2 3

y r A rπ

ππ

αα

α

−= = −

( )2

2

2 cos3r

παα

=−

Similarly ( ) ( ) 2

1 1 1 122

2 cos 3

y r A rππ

α αα

= = −−

( ) ( ) ( ) ( )

( )

2 22 2 1 12 2

2 2

3 32 1

2 cos 2 cosThen3 32 cos3

yA r r r r

r r

π ππ π

α αα αα α

α

Σ = − − − − −

= −

( )

2 22 1

2 22 1

and 2 2

2

A r r

r r

π πα α

π α

Σ = − − − = − −

( ) ( )2 2 3 32 1 2 1

3 32 12 2

2 1 2

Now2 cos

2 32 cos 3

Y A yA

Y r r r r

r rYr r π

π α α

αα

Σ = Σ − − = −

−=

−−

PROBLEM 5.10 CONTINUED

Using Figure 5.8B,Y of an arc of radius ( )1 21 is2r r+

( ) ( )( )

21 2

2

sin12

Y r rπ

π

α

α

−= +

( )1 22

1 cos( )2r r

παα

= +−

(1)

( )( )( )( )

2 23 3 2 1 2 1 2 12 12 2

2 1 2 12 12 2

2 1 2 1

2 1

Now r r r r r rr rr r r rr r

r r r rr r

− + +−=

− +−+ +

=+

2

1

Let r rr r= + ∆= − ∆

Then ( )1 212

r r r= +

( ) ( )( ) ( )( ) ( )

2 23 32 12 2

2 12 2

and

32

r r r rr rr rr r

rr

+ ∆ + + ∆ − ∆ + − ∆−=

+ ∆ + − ∆−+ ∆

=

1 2In the limit as 0 (i.e., ), thenr r∆ → = 3 32 12 2

2 1

1 2

3 23 1 ( )2 2

r r rr r

r r

−=

= × +

so that ( )1 22

2 3 cos3 4

Y r r παα

= × +−

or ( )1 22

1 cos2

Y r r παα

= +−

Which agrees with Eq. (1).

PROBLEM 5.11 Locate the centroid of the plane area shown.

SOLUTION

First note that symmetry implies 0X =

2 2 2 in., 45r α= = °

( ) ( )( )

42

4

2 2 2 sin2 sin 1.6977 in.3 3ry

π

πα

α= = =′

2, inA , in.y 3, inyA

1 ( ) ( )1 4 3 62

= 1 6

2 ( )22 2 6.2834π

= 2 0.3024y− =′ 1.8997

3 ( ) ( )1 4 2 42

− = − 0.6667 2.667−

Σ 8.283 5.2330

Then Y A yAΣ = Σ

( )2 38.283 in 5.2330 inY = or 0.632 in.Y =

PROBLEM 5.12 Locate the centroid of the plane area shown.

SOLUTION

2, mmA , mmx , mmy 3, mmxA 3, mmyA

1 (40)(90) 3600= 15− 20 54 000− 72 000

2 ( ) ( )40 60

21214

π= 10 15− 6750 10 125−

3 ( ) ( )1 30 45 6752

= 25.47− 19.099− 54 000− 40 500−

Σ 6396 101 250− 21 375

Then XA xA= Σ

( )2 36396 mm 101 250 mmX = − or 15.83 mmX = −

and YA yA= Σ

( )2 36396 mm 21 375 mmY = or 3.34 mmY =

PROBLEM 5.13 Locate the centroid of the plane area shown.

SOLUTION

2, mmA , mmx , mmy 3, mmxA 3, mmyA

1 ( ) ( )2 40 80 21333

= 48 15 102 400 32 000

2 ( ) ( )1 40 80 16002

− = − 53.33 13.333 85 330− 21 330−

Σ 533.3 17 067 10 667

Then X A XAΣ = Σ

( )2 3533.3 mm 17 067 mmX = or 32.0 mmX =

and Y A yAΣ = Σ

( )2 3533.3 mm 10 667 mmY = or 20.0 mmY =

PROBLEM 5.14 Locate the centroid of the plane area shown.

SOLUTION

2, mmA , mmx , mmy 3, mmxA 3, mmyA

1 ( )( )2 150 240 24 0003

= 56.25 96 1 350 000 2 304 000

2 ( ) ( )1 150 120 90002

− = − 50 40 450 000− 360 000−

Σ 15 000 900 000 1 944 000

Then X A xAΣ = Σ

( )2 315 000 mm 900 000 mmX = or 60.0 mmX =

and Y A yAΣ = Σ

( )215 000 mm 1 944 000Y = or 129.6 mmY =

PROBLEM 5.15 Locate the centroid of the plane area shown.

SOLUTION

2, inA , in.x , in.y 3, inxA 3, inyA

1 ( ) ( )1 10 15 503

= 4.5 7.5 225 375

2 ( )215 176.714π

= 6.366 16.366 1125 2892

Σ 226.71 1350 3267

Then X A x AΣ = Σ

( )2 3226.71 in 1350 inX = or 5.95 in.X =

and Y A y AΣ = Σ

( )2 3226.71 in 3267 inY = or 14.41 in.Y =

PROBLEM 5.16 Locate the centroid of the plane area shown.

SOLUTION

2, inA , in.x , in.y 3, inxA 3, inyA

1 ( ) ( )2 8 8 42.673

= 3 2.8 128 119.47

2 ( ) ( )2 4 2 5.3333

− = − 1.5 0.8− 8− 4.267

Σ 37.33 120 123.73

Then X A x AΣ = Σ

( )2 337.33 in 120 inX = or 3.21 in.X =

and Y A y AΣ = Σ

( )2 337.33 in 123.73 inY = or 3.31 in.Y =

PROBLEM 5.17 The horizontal x axis is drawn through the centroid C of the area shown and divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

SOLUTION

Note that xQ yA= Σ

Then ( ) 21

5 1m 6 5 m3 2xQ

= × × or ( ) 3 3

125.0 10 mmxQ = ×

and ( ) 2 22

2 1 1 12.5 m 9 2.5 m 2.5 m 6 2.5 m3 2 3 2xQ

= − × × × + − × × ×

or ( ) 3 32

25.0 10 mmxQ = − ×

Now ( ) ( )1 20x x xQ Q Q= + =

This result is expected since x is a centroidal axis ( )thus 0y =

and ( )0 0x xQ y A Y A y Q= Σ = Σ = ⇒ =

PROBLEM 5.18 The horizontal x axis is drawn through the centroid C of the area shown and divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

SOLUTION

First, locate the position y of the figure.

2, mmA , mmy 3, mmyA

1 160 300 48 000× = 150 7 200 000

2 150 80 16 000− × = − 160 2 560 000−

Σ 32 000 4 640 000

Then Y A y AΣ = Σ

( )2 332 000 mm 4 640 000 mmY =

or 145.0 mmY =

PROBLEM 5.18 CONTINUED

( )I I

6 3

:155 115 (160 155) 80 115

2 2 1.393 10 mm

A Q yA= Σ

= × + − × = ×

( ) ( )II II

6 3

:145 85 160 145 80 85

2 2 1.393 10 mm

A Q yA= Σ = − × − − ×

= − ×

( )area I II Q 0x Q Q∴ = + =

Which is expected since xQ yA yA= Σ = and 0y = , since x is a centroidal axis.

PROBLEM 5.19 The first moment of the shaded area with respect to the x axis is denoted by .xQ (a) Express xQ in terms of r and .θ (b) For what value of θ is xQ maximum, and what is the maximum value?

SOLUTION

( ) With and using Fig. 5.8 A,xa Q yA= Σ

( ) ( ) ( )

( )

23 2 2 2

2 32

3 2

sin 1sin 2 cos sin2

2 cos cos sin3

xr

Q r r r r

r

ππ

π

θθ θ θ θ

θ

θ θ θ

− = − − × × −

= −

or 3 32 cos3xQ r θ=

( )b By observation, is maximum when xQ 0θ =

and then 323xQ r=

PROBLEM 5.20 A composite beam is constructed by bolting four plates to four 2 2 3/8-in.× × angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal x axis of the red shaded areas shown, respectively, in parts a and b of the figure. Knowing that the force exerted on the bolt at A is 70 lb, determine the force exerted on the bolt at B.

SOLUTION

From the problem statement: xF Q∝

so that ( ) ( )A B

x xA B

F FQ Q

=

and ( )( )x B

B Ax A

QF F

Q=

Now xQ yA= ∑

So ( ) ( ) 30.3757.5 in. in. 10 in. 0.375 in. 28.82 in2x AQ = + × =

( ) ( ) ( )( )

( ) ( )( )

0.375and 2 7.5 in. in. 1.625 in. 0.375 in.2

2 7.5 in. 1 in. 2 in. 0.375 in.

x xB AQ Q = + − + −

3 3 328.82 in 8.921 in 9.75 in= + +

347.49 in=

Then ( )3

347.49 in 70 lb 115.3 lb28.82 inBF = =

PROBLEM 5.21 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION

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

, in.L , in.x , in.y 2, inxL 2, inyL

1 16 8 0 128 0 2 12 16 6 102 72 3 24 4 12 96 288 4 6 8− 9 48− 54 5 8 4− 6 32− 48 6 6 0 3 0 18 Σ 72 336 480

Then X L x LΣ = Σ

( ) 272 in. 336 inX = or 4.67 in.X =

and Y L yLΣ = Σ

2(72 in.) 480 inY = or 6.67 in.Y =

PROBLEM 5.22 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION

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

, mmL , mmx , mmy 2, mmxL 2, mmyL

1 165 82.5 0 13 612 0 2 75 165 37.5 12 375 2812 3 105 112.5 75 11 812 7875

4 2 260 75 96.05+ = 30 37.5 2881 3602 Σ 441.05 40 680 14 289

Then X L x LΣ = Σ

2(441.05 mm) 40 680 mmX = or 92.2 mmX =

and Y L yLΣ = Σ

2(441.05 mm) 14 289 mmY = 32.4 mmY =

PROBLEM 5.23 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION

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

, mmL , mmx , mmy 2, mmxL 2, mmyL

1 2 212 6 13.416+ = 6 3 80.50 40.25

2 16 12 14 192 224

3 21 1.5 22 31.50 462

4 16 9− 14 144− 224

5 2 26 9 10.817+ = 4.5− 3 48.67− 32.45

Σ 77.233 111.32 982.7

Then X L x LΣ = Σ

2(77.233 mm) 111.32 mmX = or 1.441 mmX =

and Y L yLΣ = Σ

2(77.233 mm) 982.7 mmY = or 12.72 mmY =

PROBLEM 5.24 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION

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

By symmetry 0X =

, in.L , in.y 2, inyL

1 2 0 0

2 ( )6π ( )2 63.820

π= 72

3 2 0 0

4 ( )4π ( )2 42.546

π= 32

Σ 35.416 104

Then Y L yLΣ = Σ

2(35.416 in.) 104 inY = or 2.94 in.Y =

PROBLEM 5.25 A 750 g= uniform steel rod is bent into a circular arc of radius 500 mm as shown. The rod is supported by a pin at A and the cord BC. Determine (a) the tension in the cord, (b) the reaction at A.

SOLUTION

( )0.5 m sin 30First note, from Figure 5.8B:

/6X

°=

π

1.5 mπ

=

( )( )2Then mg

0.75 kg 9.81 m/s7.358 N

W ===

Also note that ∆ ABD is an equilateral triangle. Equilibrium then requires

( ) ( )

( ) 0:

1.5 0.5 m m cos30 7.358 N 0.5 m sin 60 0

A

BC

a M

Σ =

− ° − ° =

or 1.4698 NBCT = or 1.470 NBCT =

( )( ) 0: 1.4698 N cos60 0x xb F AΣ = + ° =

or 0.7349 NxA = −

( )0: 7.358 N 1.4698 N sin 60 0y yF AΣ = − + ° =

or 6.085 NyA = thus 6.13 N=A 83.1°

PROBLEM 5.26 The homogeneous wire ABCD is bent as shown and is supported by a pin at B. Knowing that 8 in.,l = determine the angle θ for which portion BC of the wire is horizontal.

SOLUTION

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

Thus 0, which implies that 0BM xΣ = = or 0xLΣ =

Hence

( ) ( )2(6 in.) 8 in.6 in. 8 in.2

ππ

− × +

( )6 in.8 in. cos 6 in. 02

θ + − =

Then 4cos9

θ = or 63.6θ = °

PROBLEM 5.27 The homogeneous wire ABCD is bent as shown and is supported by a pin at B. Knowing that 30 ,θ = ° determine the length l for which portion CD of the wire is horizontal.

SOLUTION

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

Thus 0, which implies that 0BM xΣ = = or 0i ix LΣ =

Hence ( ) ( ) ( )2 6 in.cos30 6 in. sin 30 6 in.π

π

− ° + ° ×

( ) ( ) in.cos30 in.

2l

l

+ °

( ) ( )6 in.in. cos30 6 in. 02

l + ° − =

or 2 12.0 316.16 0l l+ − =

1with roots 12.77 and 24.77.l = − Taking the positive root

12.77 in.l =

PROBLEM 5.28 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which the portion BCD of the wire is horizontal.

SOLUTION

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 0, which implies that 0CM xΣ = =

or 0i ix LΣ =

( ) ( ) ( ) ( ) ( )Hence 4 in. 8 in. 4 in. 10 in. 02L L + − + − =

or 2 2144 inL = or 12.00 in.L =

PROBLEM 5.29 Determine the distance h so that the centroid of the shaded area is as close to line BB′ as possible when (a) 0.2,k = (b) 0.6.k =

SOLUTION

Then yAyA

Σ=

Σ

( ) ( ) ( )

( )2 2

or

a ha ab kb a hy

ba kb a h

+ − −

=− −

( )2 2112 (1 )a k kha k kh

− +=

− +

Let 1 and hc ka

ζ= − =

Then 2

2a c kyc k

ζζ

+=

+ (1)

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

( ) ( )

( )

2

2

20

2

k c k k c kdy ad c k

ζ ζ ζ

ζ ζ

+ − += =

+ or ( ) ( )22 0c k c kζ ζ ζ+ − + = (2)

PROBLEM 5.29 CONTINUED

Expanding (2) 2 22 2 0c c kζ ζ ζ+ − − = or 2 2 0k c cζ ζ+ − =

Then ( ) ( ) ( )22 2 4

2c c k c

− ± −=

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

( ) ( ) ( )22 1 4 1 4 1

2k k k k

h ak

− − + − + −=

(a) 0.2: k = ( ) ( ) ( )( )

( )

22 1 0.2 4 1 0.2 4 0.2 1 0.22 0.2

h a− − + − + −

= or 0.472h a=

(b) 0.6: k = ( ) ( ) ( )( )

( )

22 1 0.6 4 1 0.6 4 0.6 1 0.62 0.6

h a− − + − + −

= or 0.387h a=

PROBLEM 5.30 Show when the distance h is selected to minimize the distance y from line BB′ to the centroid of the shaded area that .y h=

SOLUTION

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

Then from Eq. (2)

We see 2

2 c kc k

ζζζ

+=

+ (3)

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

We obtain ( )22ay ζ=

But ha

ζ =

So y h= Q.E.D.

PROBLEM 5.31 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION

For the element of area (EL) shown hy xa

=

( )and

1

dA h y dx

xh dxa

= −

= −

( )

Then 12

12

EL

EL

x x

y h y

h xa

=

= +

= +

2

00

1Then area 12 2

aa x xA dA h dx h x ah

a a = = − = − =

∫ ∫

2 32

00

2 2

20 0

2 32

20

1and 12 3 6

1 1 12 2

12 33

aa

EL

a aEL

a

x x xx dA x h dx h a ha a

h x x h xy dA h dx dxa a a

h xx aha

= − = − =

= + − = −

= − =

∫ ∫

∫ ∫ ∫

2

Hence

1 12 6

ELxA x dA

x ah a h

=

=

13

x a=

21 12 3

ELyA y dA

y ah ah

=

=

23

y h=

PROBLEM 5.32 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION

For the element (EL) shown

At 3, :x a y h h ka= = = or 3hka

=

Then 1/31/3ax yh

=

1/31/3

Now dA xdya y dyh

=

=

1/31/ 3

1 1 , 2 2EL EL

ax x y y yh

= = =

Then ( )1/3 4/31/3 1/30

0

3 34 4

hh a aA dA y dy y ahh h

= = = =∫ ∫

1/3 1/3 5/3 21/3 1/3 2/30

0

1 1 3 3and 2 2 5 10

hh

ELa a ax dA y y dy y a hh h h

= = =

∫ ∫

1/3 7/3 21/3 1/30

0

3 37 7

hh

ELa ay dA y y dy y ahh h

= = =

∫ ∫

Hence 23 3: 4 10ELxA x dA x ah a h = =

∫ 2

5x a=

23 3: 4 7ELyA y dA y ah ah = =

∫ 4

7y h=

PROBLEM 5.33 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION

For the element (EL) shown

At 31, :x a y h h k a= = = or 1 3

hka

=

32a k h= or 2 3

akh

=

Hence, on line 1

33hy xa

=

and on line 2

1/31/3hy xa

=

Then

1/3 3 1/3 31/3 3 1/3 3

1 and 2EL

h h h hdA x x dx y x xa a a a

= − = +

1/3 3 4/3 41/3 3 1/3 30

0

3 1 1 24 4

aa h hA dA x x dx h x x aha a a a

∴ = = − = − =

∫ ∫

1/3 3 7/3 5 21/3 3 1/3 30

0

3 1 8357 5

aa

ELh hx dA x x x dx h x x a ha a a a

= − = − =

∫ ∫

1/3 3 1/3 31/3 3 1/3 30

12

aEL

h h h hy dA x x x x dxa a a a

= + −

∫ ∫

2 2/3 6 2 5/3 6

22/3 6 5/3 60

0

3 1 82 2 5 7 35

aah x x h x xdx aha a a a

= − = − =

From 28:2 35ELahxA x dA x a h = =

∫ or 16

35x a=

and 28:2 35ELahyA y dA y ah = =

∫ or 16

35y h=

PROBLEM 5.34 Determine by direct integration the centroid of the area shown.

SOLUTION

First note that symmetry implies 0x =

For the element (EL) shown

2 (Figure 5.8B)ELry

dA rd rππ

=

=

Then ( )2

2

11

22 2

2 12 2

rrr

r

rA dA rd r r rππ π

= = = = −

∫ ∫

and ( ) ( )2

2

11

3 3 32 1

2 1 223 3

rr

EL rr

ry dA rd r r r rππ

= = = − ∫ ∫

So ( ) ( )2 2 3 32 1 2 1

2:2 3ELyA y dA y r r r rπ = − = −

or 3 32 12 2

2 1

43r ryr rπ

−=

PROBLEM 5.35 Determine by direct integration the centroid of the area shown.

SOLUTION

First note that symmetry implies 0x =

For the element (EL) shown

cos , siny R x Rθ θ= =

cos dx R dθ θ=

2 2cosdA ydx R dθ θ= =

Hence

( )2 2 2 20

0

sin 2 12 cos 2 2 sin 22 4 2

A dA R d R Rα

α θ θθ θ α α = = = + =

∫ ∫

( )

( )

2 2 3 20

03

2

1 22 cos cos cos sin sin2 3 3

cos sin 2sin3

ELRy dA R d R

R

αα θ θ θ θ θ θ

α α α

= = +

= +

∫ ∫

But so ELyA y dA= ∫ ( )

( )

32

2

cos sin 2sin3

2 sin 22

R

yR

α α α

α α

+=

+

or ( )( )

2cos 22 sin3 2 sin 2

y Rα

αα α

+=

+

Alternatively, 22 3 sinsin

3 2 sin 2y R αα

α α−

=+

PROBLEM 5.36 Determine by direct integration the centroid of the area shown.

SOLUTION

For the element (EL) shown

2 2by a xa

= −

( )

( )2 2

and dA b y dx

b a a x dxa

= −

= − −

( ) ( )2 21;2 2EL EL

bx x y y b a a xa

= = + = + −

Then ( )2 20a bA dA a a x dxa

= = − −∫ ∫

To integrate, let 2 2sin : a cos , cosx a x a dx a dθ θ θ θ= − = =

( )( )/20

/22 2

0

Then cos cos

2sin sin 12 4 4

bA a a a da

b a a aba

π

π

θ θ θ

θ θ πθ

= −

= − + = −

( ) ( )/2

3/22 2 2 2 20

0

3

1and 2 3

16

aEL

b b ax dA x a a x dx x a xa a

a b

π = − − = + −

=

∫ ∫

( ) ( )

( )

2 2 2 20

2 2 32 2

2 200

2

1 3 62 2

aEL

aa

b by dA a a x a a x dxa a

b b xx dx aba a

= + − − −

= = =

∫ ∫

21: 14 6ELxA x dA x ab a bπ = − =

∫ or ( )

23 4

axπ

=−

21: 14 6ELyA y dA y ab abπ = − =

∫ or ( )

23 4

byπ

=−

PROBLEM 5.37 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION

For the element (EL) shown on line 1 at

22, x a b k a= = or 2 2

bka

=

22 by xa

∴ =

On line 2 at 31, 2x a b k a= − = or 2 3

2bka−

=

33

2 by xa−

∴ =

2 32 3

2b bdA x x dxa a

= +

3 3 42

2 200

2 2Then 3 4

1 1 53 2 6

aa b x b x xA dA x dx

x aa a

ab ab

= = + = +

= + =

∫ ∫

4 52 3 2

2 3 200

2

2 2 1 2and 4 5 4 5

1320

aa

ELb b b x xx dA x x x dx a b

aa a a

a b

= + = + = +

=

∫ ∫

2 3 2 32 3 2 30

2 2 2 52 3 7

2 3 4 200

2 5 2

1 2 22

1 2 22 52 7

1 2 1310 7 70

aEL

aa

b b b by dA x x x x dxa a a a

b b b xx x dx xa a a a

b a ab

= − +

= − = −

= − = −

∫ ∫

Then 25 13 :6 20ELxA x dA x ab a b = =

∫ or 39

50x a=

25 13:6 70ELyA y dA y ab ab = −

∫ or 39

175y b= −

PROBLEM 5.38 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION

At 0,x y b= =

( )20b k a= − or 2bka

=

Then ( )22 by x aa

= −

Now ( )22, 2 2EL ELy bx x y x a

a= = = −

and ( )22bdA ydx x a dxa

= = −

( ) ( )2 32 20 0

1Then 33

aa b bA dA x a dx x a aba a

= = − = − = ∫ ∫

( ) ( )

( ) ( ) ( )

2 3 2 22 20 0

4 23 2 2

2

22 2 5

2 2 400

2

and 2

2 14 3 2 12

152 2

110

a aEL

aa

EL

b bx dA x x a dx x ax a x dxa a

b x aax x a ba

b b by dA x a x a dx x aa a a

ab

= − = − +

= − + =

= − − = −

=

∫ ∫ ∫

∫ ∫

Hence 21 1:3 12ELxA x dA x ab a b = =

∫ 1

4x a=

21 1:3 10ELyA y dA y ab ab = =

∫ 3

10y b=

PROBLEM 5.39 Determine by direct integration the centroid of the area shown.

SOLUTION

2

2

2

2

Have

1 12 2

1

EL

EL

x x

a x xy yL L

x xdA ydx a dxL L

=

= = − +

= = − +

Then 22 2 3

22 20

0

812 33

LL x x x xA dA a dx a x aL

L LL L

= = − + = − + = ∫ ∫

22 2 3 42

2 200

2

2 22

2 20

2 2 3 4

2 3 40

2 2 3 4

2 3

and 12 3 4

103

1 12

1 2 3 22

2 2

LL

EL

LEL

EL

x x x x xx dA x a dx aL LL L

aL

a x x x xy dA a dxL LL L

a x x x x dxL L L L

a x x xxL L L

= − + = − +

=

= − + − +

= − + − +

= − + − +

∫ ∫

∫ ∫

252

40

1155

Lx a LL

=

Hence 28 10:3 3ELxA x dA x aL aL = =

∫ 5

4x L=

21 11:8 5ELyA y dA y a a = =

∫ 33

40y a=

PROBLEM 5.40 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION

For 1 at , 2y x a y b= = 22b ka= or 22bka

=

Then 21 2

2by xa

=

By observation ( )2 2 2b xy x b ba a

= − + = −

Now and for 0 :

ELx xx a

=

≤ ≤

2 21 12 2

1 2 and 2EL

b by y x dA y dx x dxa a

= = = =

For 2 :a x a≤ ≤

2 21 2 and 22 2EL

b x xy y dA y dx b dxa a

= = − = = −

2220

223

20 0

2Then 2

2 723 2 6

a aa

aa

b xA dA x dx b dxaa

b x a xb abaa

= = + −

= + − − =

∫ ∫ ∫

( ) ( ){ ( ) ( )

2220

24 32

20 0

2 2 32 2

2

2and 2

24 3

1 12 22 376

a aEL a

a a

b xx dA x x dx x b dxaa

b x xb xaa

a b b a a a aa

a b

= + −

= + −

= + − + −

=

∫ ∫ ∫

PROBLEM 5.40 CONTINUED

22 22 20 0

232 5 2

40

2

2 2 22

2 25 2 3

1730

a aEL

aa

a

b b b x xy dA x x dx b dxa aa a

b x b a xaa

ab

= + − −

= + − −

=

∫ ∫ ∫

Hence 27 7:6 6ELxA x dA x ab a b = =

∫ x a=

27 17:6 30ELyA y dA y ab ab = =

∫ 17

35y b=

PROBLEM 5.41 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION

2For y at , :x a y b= = 2a kb= or 2akb

=

Then 1/22

by xa

=

1/2 1/22

2

Now

and for 0 : , 2 2 2

EL

EL

x x

a y b x xx y dA y dx b dxa a

=

≤ ≤ = = = =

For ( )1/2

1 21 1:

2 2 2 2ELa b x xx a y y y

a a

≤ ≤ = + = − +

( )1/2

2 112

x xdA y y dx b dxaa

= − = − +

( )

( ) ( )

1/2 1/2/2

0 /2

/2 3/2 23/2

0 /2

3/2 3/23/2

22

1Then 2

2 2 13 3 2 2

23 2 2

1 1 2 2 2 2

a aa

aa

a

x x xA dA b dx b dxaa a

b x xx b xaa a

b a aaa

a ab a aa

= = + − +

= + − +

= + −

+ − − + −

∫ ∫ ∫

1324ab

=

PROBLEM 5.41 CONTINUED

( )

( ) ( )

1/2 1/2/2

0 /2

/2 5/2 3 45/2

0 /2

5/2 5/25/2

33 2

1and 2

2 25 5 3 4

25 2 2

1 1 3 2 4

a aEL a

aa

a

x x xx dA x b dx x b dxaa a

b x x xx baa a

b a aaa

ab a aa

= + − +

= + − +

= + −

+ − − + −

∫ ∫ ∫

2

2

1/2 1/2/2

0

1/2 1/2

/2

/2 32 2 22

0 /2

2

71240

2

1 1 2 2 2

1 1 12 2 2 2 3 2

4

aEL

aa

aa

a

a

a b

b x xy dA b dxa a

b x x x xb dxa aa a

b b x xxa a a a

b

=

=

+ − + − +

= + − −

=

∫ ∫

( )2 2 32

2

2

12 2 6 2 2

1148

a a b aaa a

ab

+ − − −

=

Hence 213 71:24 240ELxA x dA x ab a b = =

∫ 17 0.546

130x a a= =

213 11:24 48ELyA y dA y ab ab = =

∫ 11 0.423

26y b b= =

PROBLEM 5.42 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a.

SOLUTION

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

Have at 2, :x a y a a ka= = = or 1ka

=

Thus 21 2 and y x dy xdxa a

= =

( )

2 2

2 22

2 2 20

0

2

2

2Then 1 1

4 4 2 4 1 1 ln 12 4

5 ln 2 5 1.47892 4

4 1

aa

EL

dydL dx x dxdx a

x x a xL dL x dx xaa a a

a a a

xx dL x dxa

= + = +

∴ = = + = + + + +

= + + =

= +

∫ ∫

( )

3/222

200

23/2 2

2 413 8

5 1 0.848412

aa a x

a

a a

= +

= − =

Then ( ) 2: 1.4789 0.8484ELxL x dL x a a= =∫ 0.574x a=

PROBLEM 5.43 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid.

SOLUTION

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

Now cosELx r θ= and dL rdθ=

Then [ ]7 /47 /4/4 /4

32

L dL rd r rπππ πθ θ π= = = =∫ ∫

and ( )7 /4/4 cosELx dL r rdπ

π θ θ=∫ ∫

[ ]7 /42 2 2/4

1 1sin 22 2

r r rππθ = = − − = −

23Thus : 22

xL xdL x r rπ = = −

∫ 2 23

x rπ

= −

PROBLEM 5.44 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid.

SOLUTION

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

Now 3cosELx a θ= and 2 2dL dx dy= +

Where 3 2cos : 3 cos sinx a dx a dθ θ θ θ= = −

3 2sin : 3 sin cosy a dy a dθ θ θ θ= =

( ) ( )

( )

1/22 22 2

1/22 2

/ 20

Then 3 cos sin 3 sin cos

3 cos sin cos sin

3 cos sin

1 3 cos sin 3 sin2

dL a d a d

a d

a d

L dL a d aπ

θ θ θ θ θ θ

θ θ θ θ θ

θ θ θ

θ θ θ θ

= − +

= +

=

∴ = = = ∫

/ 22

0

3 2a

π

=

( )/ 2 30

/22 5 2

0

and cos 3 cos sin

1 33 cos5 5

ELx dL a a d

a a

π

π

θ θ θ θ

θ

=

= − =

∫ ∫

Hence 23 3 :2 5ELxL x dL x a a = =

∫ 2

5x a=

PROBLEM 5.44 CONTINUED

Alternative solution

2/33 2

2/33 2

cos cos

sin sin

xx aa

yy aa

θ θ

θ θ

= ⇒ =

= ⇒ =

( )

( ) ( )

2/3 2/3 3/22/3 2/3

1/22/3 2/3 1/3

1 or

Then

x y y a xa a

dy a x xdx

∴ + = = −

= − −

( ) ( )1/22 21/22/3 2/3 1/3

Now

and 1 1

ELx x

dydL dx a x x dxdx

=

= + = + − −

1/31/3 2/3

1/ 300

3 3Then 2 2

aa aL dL dx a x ax

= = = = ∫ ∫

and 1/3

1/3 5/3 21/30

0

3 35 5

aa

ELax dL x dx a x ax

= = = ∫ ∫

Hence 23 3 :2 5ELxL x dL x a a = =

∫ 2

5x a=

PROBLEM 5.45 Determine by direct integration the centroid of the area shown.

SOLUTION

2 2Have cos cos3 32 2sin sin3 3

EL

EL

x r ae

y r ae

θ

θ

θ θ

θ θ

= =

= =

( )( ) 2 21 1and 2 2

dA r rd a e dθθ θ= =

Then

( )2 2 2 2 2 2 20

0

1 1 1 1 1 133.6232 2 2 4

A dA a e d a e a e aπ

π θ θ πθ = = = = − = ∫ ∫

and 2 2 3 30 0

2 1 1cos cos3 2 3ELx dA ae a e d a e dπ πθ θ θθ θ θ θ = =

∫ ∫ ∫

To proceed, use integration by parts, with 3 3 and 3 cos and sin

u e du e ddv d v

θ θ θθ θ θ

= =

= =

Then ( )3 3 3cos sin sin 3e d e e dθ θ θθ θ θ θ θ= −∫ ∫

Now let 3 3 then 3u e du e dθ θ θ= =

sin , then cosdv d vθ θ θ= = −

Then ( )( )3 3 3 3sin sin 3 cos cos 3e d e e e dθ θ θ θθ θ θ θ θ θ− = − − − − ∫ ∫

So that ( )3

3 cos sin 3cos10ee dθ

θ θ θ θ θ= +∫

( ) ( )3 3

3 3 3

0

1 sin 3cos 3 3 1239.263 10 30EL

e ax dA a e aπθ

πθ θ

∴ = + = − − = −

Also 2 2 3 30 0

2 1 1sin sin3 2 3ELy dA ae a e d a e dπ πθ θ θθ θ θ θ = =

∫ ∫ ∫

PROBLEM 5.45 CONTINUED

Using integration by parts, as above, with

3 3 and 3u e du e dθ θ θ= =

sin and cosdv d vθ θ θ= = −∫

Then ( ) ( )3 3 3sin cos cos 3e d e e dθ θ θθ θ θ θ θ= − − −∫ ∫

So that ( )3

3 sin cos 3sin10ee d

θθ θ θ θ θ= − +∫

( ) ( )3 3

3 3 3

0

1 cos 3sin 1 413.093 10 30EL

e ay dA a e aπθ

πθ θ

∴ = − + = + =

Hence ( )2 3: 133.623 1239.26ELxA x dA x a a= = −∫ or 9.27x a= −

( )2 3: 133.623 413.09ELyA y dA y a a= =∫ or 3.09y a=

PROBLEM 5.46 Determine by direct integration the centroid of the area shown.

SOLUTION

Have 1, sin2EL EL

xx x y xLπ

= =

and dA ydx= /22 2

/22 20

0

sin sin cosL

L x L x L x LA dA x dx xL L Lπ π π

ππ π

= = = − =

∫ ∫

and /2

0 sinLEL

xx x dA x x dxLπ = =

∫ ∫

/22 3 3 32

2 3 2 30

2 2sin cos sin 2L

L x L x L x L Lx xL L Lπ π π

ππ π π π = + − = −

Also /2

01 sin sin2

LEL

x xy y dA x x dxL Lπ π = =

∫ ∫

/22 3

2 30

1 2 2sin cos2

LL x L L xx x

L Lπ π

ππ π

= − −

( ) ( )3 2 3

22 2

1 1 1 62 6 8 24 96

L L L L ππ π

= − − = +

PROBLEM 5.46 CONTINUED

Hence 2

32 2 3

1:ELL zxA x dA x Lπ π π = = −

or 0.363x L=

2 3

2 2 2 31 2:

96ELL LyA y dA yπ π π π = = −

or 0.1653y L=

PROBLEM 5.47 Determine the volume and the surface area of the solid obtained by rotating the area of Problem. 5.2 about (a) the x axis, (b) the line

165x = mm.

SOLUTION

From the solution to Problem 5.2:

( )2area area10 125 mm , 96.4 mm, 34.7 mm AreaA X Y= = =

From the solution to Problem 5.22:

( )line line441.05 mm 92.2 mm, 32.4 mm LineL X Y= = =

Applying the theorems of Pappus-Guldinus, we have

(a) Rotation about the x axis:

( )( ) 3 2lineArea 2 2 32.4 mm 441.05 mm 89.786 10 mmY Lπ π= = = ×

3 289.8 10 mmA = ×

( )( ) 6 3areaVolume 2 2 34.7 mm 10125 mm 2.2075 10 mmY Aπ π= = = ×

6 32.21 10 mmV = ×

(b) Rotation about 165 mm:x =

( ) ( ) ( ) 5 2lineArea 2 165 2 165 92.2 mm 441.05 mm 2.01774 10 mmX Lπ π = − = − = ×

6 20.202 10 mmA = ×

( ) ( ) ( ) 6 3areaVolume 2 165 2 165 96.4 mm 10 125 mm 4.3641 10 mmX Aπ π = − = − = ×

6 34.36 10 mmV = ×

PROBLEM 5.48 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.4 about (a) the line 22y = mm, (b) the line 12x = mm.

SOLUTION

From the solution to Problem 5.4: 2

area area399 mm , 1.421 mm, 12.42 mm (Area)A X Y= = = From the solution to Problem 5.23:

line line77.233 mm, 1.441 mm, 12.72 mm (Line)L X Y= = = Applying the theorems of Pappus-Guldinus, we have

(a) Rotation about the line 22 mm:y =

( ) ( ) ( ) 2lineArea 2 22 2 22 12.72 mm 77.233 mm 4503 mmY Lπ π = − = − =

3 24.50 10 mmA = ×

( ) ( ) ( )2 3areaVolume 2 22 2 22 12.42 mm 399 mm 24 016.97 mmY Aπ π = − = − =

3 324.0 10 mmV = ×

(b) Rotation about line 12 mm:x =

( ) ( ) ( ) 2lineArea 2 12 2 12 1.441 mm 77.233 mm 5124.45 mmX Lπ π = − = − =

3 25.12 10 mmA = ×

( ) ( ) ( )2 3Volume 2 12 1.421 2 12 1.421 mm 399 mm 26 521.46 mmAπ π = − = − =

3 326.5 10 mmV = ×

PROBLEM 5.49 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.1 about (a) the x axis, (b) the line

16 in.x =

SOLUTION

From the solution to Problem 5.1: 2

area area240 in , 5.60 in., 6.60 in. (Area)A X Y= = = From the solution to Problem 5.21:

line line72 in., 4.67 in., 6.67 in.L X Y= = = Applying the theorems of Pappus-Guldinus, we have

( ) Rotation about the axis:a x

( )( ) 2line2 2 6.67 in. 72 in. 3017.4 inxA Y Lπ π= = =

23020 inA =

( )( )2 3area2 2 6.60 in. 240 in 9952.6 inxV Y Aπ π= = =

39950 inV =

( ) Rotation about 16 in.:b x =

( ) ( ) ( ) 216 line2 16 2 16 4.67 in. 72 in. 5125.6 inxA X Lπ π= = − = − =

2

16 5130 inxA = =

( ) ( ) ( )2 316 area2 16 2 16 5.60 in. 240 in 15 682.8 inxV X Aπ π= = − = − =

3 3

16 15.68 10 inxV = = ×

PROBLEM 5.50 Determine the volume of the solid generated by rotating the semielliptical area shown about (a) the axis ,AA′ (b) the axis ,BB′ (c) the y axis.

SOLUTION

Applying the second theorem of Pappus-Guldinus, we have

(a) Rotation about axis :AA′

( ) 2 2Volume 2 22abyA a a bππ π π = = =

2 2V a bπ=

(b) Rotation about axis :BB′

( ) 2 2Volume 2 2 2 22abyA a a bππ π π = = =

2 22V a bπ=

(c) Rotation about y-axis:

24 2Volume 2 2

3 2 3a abyA a bππ π ππ

= = =

223

V a bπ=

PROBLEM 5.51 Determine the volume and the surface area of the chain link shown, which is made from a 2-in.-diameter bar, if 3R = in. and 10L = in.

SOLUTION

First note that the area A and the circumference C of the cross section of the bar are

2 and4

A d C dπ π= =

Observe that the semicircular ends of the link can be obtained by rotating the cross section through a horizontal semicircular arc of radius R. Then, applying the theorems of Pappus-Guldinus, we have

( ) ( ) ( ) ( ) ( )side endVolume 2 2 2 2 2V V AL RA L R Aπ π= + = + = +

( ) ( )2

3

2 10 in. 3 in. 2 in.4

122.049 in

ππ = +

=

3122.0 inV =

( ) ( ) ( ) ( ) ( )side endArea 2 2 2 2 2A A CL RC L R Cπ π= + = + = +

( ) ( )2 10 in. 3 in. 4 in.π π = +

2488.198 in=

2488 inA =

PROBLEM 5.52 Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on page 261 are correct.

SOLUTION

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 242 2

3 4aV yA aππ ππ

= =

32or

3V aπ=

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

Have 42 23 4

aV yA haππ ππ

= =

22or

3V a hπ=

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

Have 3 22 28 3

V yA a ahπ π = =

21or

2V a hπ=

(4) Cone: the generating area is a triangle

Have 12 2

3 2aV yA haπ π = =

21or

3V a hπ=

PROBLEM 5.53 A 15-mm-diameter hole is drilled in a piece of 20-mm-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process.

SOLUTION

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 12 2 7.5 mm 5 mm 5 mm3 2

V xAπ π = = + × × ×

3or 720 mmV =

PROBLEM 5.54 Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design.

SOLUTION

Applying the first theorem of Pappus-Guldinus, the contact area CA of a belt is given by

CA yL yLπ π= = Σ Where the individual lengths are the “Lengths” of the belt cross section that are in contact with the pulley

( )

( ) ( )

1 1 2 2Have 2

2.5 2.5 mm2 60 mm2 cos 20

60 2.5 mm 12.5 mm

CA y L y Lπ

π

= +

= − °

+ −

3 2or 3.24 10 mmCA = ×

Have ( )1 12CA y Lπ =

7.5 7.5 mm2 60 1.6 mm2 cos 20

π = − − ×

3 2or 2.74 10 mmCA = ×

Have ( ) ( )1 12 560 mm 5 mmCA y Lπ π ππ

× = = − ×

3 2or 2.80 10 mmCA = ×

PROBLEM 5.55 Determine the capacity, in gallons, of the punch bowl shown if 12 in.R =

SOLUTION

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

( )

( )

1 1 2 2

2

3 33

3 3

2 2 2

1 1 1 1 3 2 sin 302 cos303 2 2 2 2 63

6

3 32816 3 2 3

3 3 12 in. 3526.03 in8

V xA xA x A x A

RR R R R

R R R

π π π

ππ π

π π

π

= = Σ = +

= × × × + ×

= + =

= =

3Since 1 gal 231 in= 3

33526.03 in 15.26 gal231 in /gal

V = =

15.26 galV =

PROBLEM 5.56 The aluminum shade for a small high-intensity lamp has a uniform thickness of 3/32 in. Knowing that the specific weight of aluminum is

30.101 lb/in , determine the weight of the shade.

SOLUTION

The weight of the lamp shade is given by

W V Atγ γ= = where A is the surface area of the shade. This area can be generated by rotating the line shown about the x axis. Applying the first theorem of Pappus-Guldinus, we have

( )1 1 2 2 3 3 4 42 2 2A yL yL y L y L y L y Lπ π π= = Σ = + + +

( ) ( ) ( )2 20.6 mm 0.60 0.752 0.6 mm mm 0.15 mm 1.5 mm2 2

π + = + × +

( ) ( )2 20.75 1.25 mm 0.50 mm 0.40 mm2+ + × +

( ) ( )2 2

2

1.25 1.5 mm 0.25 mm 1.25 mm2

22.5607 in

+ + × +

=

Then 3 2 3lb/in0.101 22.5607 in in. 0.21362 lb

32W ×= × =

0.214 lbW =

PROBLEM 5.57 The top of a round wooden table has the edge profile shown. Knowing that the diameter of the top is 1100 mm before shaping and that the density of the wood is 3690 kg/m , determine the weight of the waste wood resulting from the production of 5000 tops.

SOLUTION

All dimensions are in mm

( ) ( )waste blank top

2 6 3blank

top 1 2 3 4

Have

550 mm 30 mm 9.075 10 mm

V V V

V

V V V V V

π π

= −

= × = ×

= + + +

Applying the second theorem of Pappus-Guldinus to parts 3 and 4

( ) ( ) ( ) ( )

( )

( )

( )

2 2top

2

2

6 3

6 3

529 mm 18 mm 535 mm 12 mm

4 122 535 mm 12 mm3 4

4 182 529 mm 18 mm3 4

5.0371 3.347 0.1222 0.2731 10 mm

8.8671 10 mm

V π π

πππ

πππ

π

π

= × + ×

× + + ×

× + + ×

= + + + ×

= ×

( ) 6 3waste

3 3

9.0750 8.8671 10 mm

0.2079 10 m

V π

π −

∴ = − ×

= ×

( ) ( )waste wood waste tops

3 3 3 2

Finally

690 kg/m 0.2079 10 m 9.81 m/s 5000 tops

W V g Nρ

π −

=

= × × × ×

wasteor 2.21 kNW =

PROBLEM 5.58 The top of a round wooden table has the shape shown. Determine how many liters of lacquer are required to finish 5000 tops knowing that each top is given three coats of lacquer and that 1 liter of lacquer covers 12 m2.

SOLUTION

Referring to the figure in solution of Problem 5.57 and using the first theorem of Pappus-Guldinus, we have

( ) ( )

( )

( )

surface top circle bottom circle edge

2 2

3 2

535 mm 529 mm

2 122 535 mm 12 mm2

2 182 529 mm 18 mm2

617.115 10 mm

A A A A

π π

πππ

πππ

π

= + +

= +

× + + ×

× + + ×

= ×

surface tops coats

3 22

Then # liters Coverage

1 liter617.115 10 m 5000 312 m

A N N

π −

= × × ×

= × × × ×

or # liters 2424 L=

PROBLEM 5.59 The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from yellow brass. Knowing that the specific weight of yellow brass is 30.306 lb/in . determine the weight of the escutcheon.

SOLUTION The weight of the escutcheon is given by (specific weight)W V= where V is the volume of the plate. V can be generated by rotating the area A about the x axis.

Have 3.0755 in. 2.958 in. 0.1175 in.a = − =

and 0.5sin 0.16745 R 9.59413

φ φ= ⇒ = = °

Then 2 26 9.5941 16.4059 or 8.20295 0.143169 radα α= − = = = The area A can be obtained by combining the following four areas, as indicated.

Applying the second theorem of Pappus-Guldinus and then using Figure 5.8A, we have

2 2V yA yAπ π= = Σ

PROBLEM 5.59 CONTINUED

2, inA , in.y 3, inyA

1 ( )( )1 3.0755 1.5 2.30662

= ( )1 1.5 0.53

= 1.1533

2 ( )23 1.28851α− = − ( ) ( )2 3 sin

sin 0.609213

αα φ

α× + = –0.78497

3 ( )( )1 2.958 0.5 0.73952

− = − ( )1 0.5 0.166673

= –0.12325

4 ( )( )0.1755 0.5 0.05875− = − ( )1 0.5 0.252

= –0.14688

30.44296 inyAΣ = Then ( )3 32 0.44296 in 1.4476 inV π= =

so that ( )3 31.4476 in 0.306 lb/in 0.44296 lbW = = 0.443 lbW =

PROBLEM 5.60 The reflector of a small flashlight has the parabolic shape shown. Determine the surface area of the inside of the reflector.

SOLUTION

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

2A yLπ= ( )22Now, since , at : 7.5x ky x a a k= = =

or 56.25a k= (1)

At ( ) ( )215 mm: 15 12.5x a a k= + + =

or 15 156.25a k+ = (2)

Then Eq. (2) 15 156.25: or 8.4375 mmEq. (1) 56.25

a k aa k+

= =

1Eq. (1) 0.15

mmk⇒ =

20.15 and 0.3dxx y ydy

∴ = =

22Now 1 1 0.09dxdL dy y dy

dy

= + = +

So 2 andA yL yL ydLπ= = ∫

( )

12.5 27.5

12.53/22

7.5

2 1 0.09

2 1 2 1 0.093 0.18

A y y dy

y

π

π

∴ = +

= +

21013 mm=

2or 1013 mmA =

PROBLEM 5.61 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLUTION

( )( )

( )( )

1 2

1

2

Resultant

( ) Have 40 lb/ft 18 ft 720 lb

1 120 lb/ft 18 ft 1080 lb2

R R R

a R

R

= +

= =

= =

or 1800 lbR = The resultant is located at the centroid C of the distributed load x

Have ( ) ( )( )( ) ( )( )( )1: 1800 lb 40 lb/ft 18 ft 9 ft 120 lb/ft 18 ft 12 ft2AM xΣ = +

or 10.80 ftx =

1800 lb10.80 ft

Rx=

=

( )b 0: 0x xF AΣ = =

0: 1800 lb 0, 1800 lby y yF A AΣ = − = = 1800 lb∴ =A

( )( )0: 1800 lb 10.8 ft 0A AM MΣ = − =

19.444 lb ftAM = ⋅ or 19.44 kip ftA = ⋅M

PROBLEM 5.62 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLUTION

( )( )

( )( )

I

II

( ) Have 300 N/m 6 m 1800 N

1 6 m 900 N/m 1800 N3

a R

R

= =

= =

Then I II:yF R R RΣ − = − −

or 1800 N 1800 N 3600 NR = + =

( ) ( )( ) ( )( ): 3600 N 3 m 1800 N 4.5 m 1800 NAM xΣ − = − −

or 3.75 mx =

3600 NR = 3.75 mx =

(b) Reactions

0: 0x xF AΣ = =

( ) ( )( )0: 6 m 3600 N 3.75 m 0A yM BΣ = − =

or 2250 NyB = 2250 N=B

0: 2250 N 3600 Ny yF AΣ = + =

or 1350 NyA = 1350 N=A

PROBLEM 5.63 Determine the reactions at the beam supports for the given loading.

SOLUTION

( )( )

( )( )

( )( )

I

II

III

Have 100 lb/ft 4 ft 400 lb

1 200 lb/ft 6 ft 600 lb2200 lb/ft 4 ft 800 lb

R

R

R

= =

= =

= =

Then 0: 0x xF AΣ = =

( )( ) ( )( ) ( )( ) ( )0: 2 ft 400 lb 4 ft 600 lb 12 ft 800 lb 10 ft 0A yM BΣ = − − + = or 800 lbyB = 800 lb=B

0: 800 lb 400 lb 600 lb 800 0y yF AΣ = + − − − =

or 1000 lbyA = 1000 lb=A

PROBLEM 5.64 Determine the reactions at the beam supports for the given loading.

SOLUTION

( )( )

( )( )

I

II

Have 9 ft 200 lb/ft 1800 lb

1 3 ft 200 lb/ft 300 lb2

R

R

= =

= =

Then 0: 0x xF AΣ = =

( )( ) ( )( ) ( )0: 4.5 ft 1800 lb 10 ft 300 lb 9 ft 0A yM BΣ = − − + =

or 1233.3 lbyB = 1233 lb=B

0: 1800 lb 300 lb 1233.3 lb 0y yF AΣ = − − + =

or 866.7 lbyA = 867 lb=A

PROBLEM 5.65 Determine the reactions at the beam supports for the given loading.

SOLUTION

( )( )I1Have 200 N/m 0.12 m 12 N2

R = =

( )( )II 200 N/m 0.2 m 40 NR = =

Then 0: 0x xF AΣ = =

0: 18 N 12 N 40 N 0y yF AΣ = + − − =

or 34 NyA = 34.0 N=A

( )( ) ( )( ) ( )( )0: 0.8 m 12 N 0.22 m 40 N 0.38 m 18 NA AM MΣ = − − +

or 2.92 N mAM = ⋅ 2.92 N mA = ⋅M

PROBLEM 5.66 Determine the reactions at the beam supports for the given loading.

SOLUTION

First, replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a linear relation between load and distance, and the values at the end points are the same.

( )( )IHave 1.8 m 2000 3600 NN/mR = =

( )( )II1 1.8 m 4500 N/m 4050 N2

R = =

Then 0: 0x xF AΣ = =

( ) ( )( ) ( )( )0: 3 m 2.1 m 3600 N 2.4 m 4050 NB yM AΣ = − − +

or 270 NyA = 270 N=A

0: 270 N 3600 N 4050 N 0y yF BΣ = − + − =

or 720 NyB = 720 N=B

PROBLEM 5.67 Determine the reactions at the beam supports for the given loading.

SOLUTION

( )( )I1Have 4 m 2000 kN/m 2667 N3

R = =

( )( )II1 2 m 1000 kN/m 666.7 N3

R = =

Then 0: 0x xF AΣ = =

0: 2667 N 666.7 N 0y yF AΣ = − − =

or 3334 NyA = 3.33 kN=A

( )( ) ( )( )0: 1 m 2667 N 5.5 m 666.7 NA AM MΣ = − −

or 6334 N mAM = ⋅ 6.33 kN mA = ⋅M

PROBLEM 5.68 Determine the reactions at the beam supports for the given loading.

SOLUTION

First, replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance, and the values at end points are the same.

( )( )IHave 8 ft 100 lb/ft 800 lbR = =

( )( )II2 8 ft 600 lb/ft 3200 lb3

R = =

Then 0: 0x xF AΣ = =

( )( ) ( )( )0: 11 5 ft 800 lb 4 ft 3200 lb 0AM BΣ = + − =

or 800 lb=B

0: 3200 lb 800 lb 800 lb 0y yF AΣ = − + + =

or 1600 lb=A

PROBLEM 5.69 Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports.

SOLUTION

( )( ) ( )I1( ) Have ft 120 lb/ft 60 lb2

a R a a= =

( )( ) ( )II1 12 40 lb/ft 240 20 lb2

R a a= − = −

Then ( )0: 60 240 2 0y y yF A a a BΣ = − − − + =

or 240 40y yA B a+ = +

Now 120 20y y y yA B A B a= ⇒ = = + (1)

Also ( ) ( ) ( ) ( )10: 12 m 60 lb 12 ft 12 ft 240 20 lb 03 3B yaM A a a a Σ = − + − + − − =

or 2140 1080

3 9yA a a= − − (2)

Equating Eqs. (1) and (2)

2140 10120 20 803 9

a a a+ = − −

or 240 320 480 0

3a a− + =

Then 1.6077 ft, 22.392a a= =

Now 12 fta ≤ 1.608 fta =

( ) Haveb 0: 0x xF AΣ = =

Eq. (1) ( )120 20 1.61 152.2 lby yA B= = + =

152.2 lb= =A B

PROBLEM 5.70 Determine (a) the distance a so that the vertical reaction at support B is minimum, (b) the corresponding reactions at the supports.

SOLUTION

( )( )I1( ) Have ft 120 lb/ft 60 lb2

a R a a= =

( ) ( ) ( )II1 12 ft 40 lb/ft 240 20 lb2

R a a = − = −

Then ( ) ( ) ( )0: ft 60 lb 240 20 lb 8 ft 12 ft 03 3A ya aM a a B Σ = − − − + + =

or 210 20 160

9 3yB a a= − + (1)

Then 20 20 09 3

ydBa

da= − = or 3.00 fta =

( ) Eq. (1)b ( ) ( )210 203.00 3.00 1609 3yB = − +

150 lb= 150.0 lb=B

and 0: 0x xF AΣ = =

( ) ( )0: 60 3.00 lb 240 20 3.00 lb 150 lb 0y yF A Σ = − − − + =

or 210 lbyA = 210 lb=A

PROBLEM 5.71 Determine the reactions at the beam supports for the given loading when

0 1.5 kN/m.w =

SOLUTION

( )( )I1Have 9 m 2 kN/m 9 kN2

R = =

( )( )II 9 m 1.5 kN/m 13.5 kNR = =

Then 0: 0x xF CΣ = =

( )( ) ( )( ) ( )0: 50 kN m 1 m 9 kN 2.5 m 13.5 kN 6 m 0B yM CΣ = − ⋅ − − + =

or 15.4583 kNyC = 15.46 kN=C

0: 9 kN 13.5 kN 15.4583 0y yF BΣ = − − + =

or 7.0417 kNyB = 7.04 kN=B

PROBLEM 5.72 Determine (a) the distributed load 0w at the end D of the beam ABCD for which the reaction at B is zero, (b) the corresponding reactions at C.

SOLUTION

( ) ( ) ( )I 0 01Have 9 m 3.5 kN/m 4.5 3.5 kN2

R w w = − = −

( )( )II 0 09 m kN/m 9 kNR w w= =

( ) Thena ( ) ( ) ( )( )0 00: 50 kN m 5 m 4.5 3.5 kN 3.5 m 9 kN 0CM w w Σ = − ⋅ + − + =

or 09 28.75 0w + =

so 0 3.1944 kN/mw = − 0 3.19 kN/mw = Note: the negative sign means that the distributed force 0w is upward.

( )b 0: 0x xF CΣ = =

( ) ( )0: 4.5 3.5 3.19 kN 9 3.19 kN 0y yF CΣ = − + + + =

or 1.375 kNyC = 1.375 kN=C

PROBLEM 5.73 A grade beam AB supports three concentrated loads and rests on soil and the top of a large rock. The soil exerts an upward distributed load, and the rock exerts a concentrated load RR as shown. Knowing that P = 4 kN and 1

2 ,B Aw w= determine the values of wA and RR corresponding to equilibrium.

SOLUTION

( )( )IHave 1.2 m kN/m 1.2 kNA AR w w= =

( )II1 11.8 m kN/m 0.45 kN2 2 A AR w w = =

( )III11.8 m kN/m 0.9 kN2 A AR w w = =

Then ( ) ( ) ( ) ( )0: 0.6 m 1.2 kN 0.6 m 0.45 kN/mC A AM w w Σ = − +

( ) ( ) ( )( )0.9 m 0.9 kN/m 1.2 m 4 kN/mAw + −

( )( ) ( )( )0.8 m 18 kN/m 0.7 m 24 kN/m 0− + =

or 6.667 kN/mAw = 6.67 kN/mAw =

and ( )( ) ( )( )0: 1.2 m 6.67 kN/m 0.45 m 6.67 kN/my RF RΣ = + +

( )( )0.9 m 6.67 kN/m 24 kN 18 kN 4 kN+ − − −

or 29.0 kNRR = 29.0 kNRR =

PROBLEM 5.74 A grade beam AB supports three concentrated loads and rests on soil and the top of a large rock. The soil exerts an upward distributed load, and the rock exerts a concentrated load RR as shown. Knowing that wB = 0.4wA, determine (a) the largest value of P for which the beam is in equilibrium, (b) the corresponding value of wA.

In the following problems, use γ = 62.4 lb/ft3 for the specific weight of fresh water and γc = 150 lb/ft3 for the specific weight of concrete if U.S. customary units are used. With SI units, use ρ = 103 kg/m3 for the density of fresh water and ρc = 2.40 × 103 kg/m3 for the density of concrete. (See the footnote on page 222 for how to determine the specific weight of a material given its density.)

j

SOLUTION

( )( )IHave 1.2 m kN/m 1.2 kNA AR w w= =

( )( )II1 1.8 m 0.6 kN/m 0.54 kN2 A AR w w= =

( )( )III 1.8 m 0.4 kN/m 0.72 kNA AR w w= =

( ) Thena ( ) ( ) ( ) ( ) ( )0: 0.6 m 1.2 kN 1.2 m 1.8 m 0.54 kNA A R AM w R w Σ = + +

( ) ( ) ( )( )2.1 m 0.72 kN 0.5 m 24 kNAw + −

( )( ) ( )2.0 m 18 kN 2.4 m 0P− + =

or 3.204 1.2 2.4 48A Rw R P+ − = (1)

and 0: 1.2 0.54 0.72 24 18 0y R A A AF R W W W PΣ = + + + − − − =

or 2.46 42R AR W P+ − = (2) Now combine Eqs. (1) and (2) to eliminate :Aw

( ) ( )3.204 Eq. 2 2.46 Eq. 1 0.252 16.488 2.7RR P− ⇒ = −

Since RR must be 0,≥ the maximum acceptable value of P is that for which 0,R =

or 6.1067 kNP = 6.11 kNP =

( ) Then, from Eq. (2):b 2.46 6.1067 42AW − = or 19.56 kN/mAW =

PROBLEM 5.75

The cross section of a concrete dam is as shown. For a dam section of unit width, determine (a) the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of the reaction forces of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

In the following problems, use γ = 62.4 lb/ft3 for the specific weight of fresh water and γc = 150 lb/ft3 for the specific weight of concrete if U.S. customary units are used. With SI units, use ρ = 103 kg/m3 for the density of fresh water and ρc = 2.40 × 103 kg/m3 for the density of concrete. (See the footnote on page 222 for how to determine the specific weight of a material given its density.)

SOLUTION

The free body shown consists of a 1-m thick section of the dam and the triangular section BCD of the water behind the dam.

Note: 1 6 mX =

( )2 9 3 m 12 mX = + =

( )3 15 2 m 17 mX = + =

( )4 15 4 m 19 mX = + =

( ) Now so thata W gVρ=

( )( ) ( )( )( )3 21

12400 kg/m 9.81 m/s 9 m 15 m 1 m 1589 kN2

W = =

( )( ) ( )( )( )3 22 2400 kg/m 9.81 m/s 6 m 18 m 1 m 2543 kNW = =

( )( ) ( )( )( )3 23

12400 kg/m 9.81 m/s 6 m 18 m 1 m 1271 kN2

W = =

( )( ) ( )( )( )3 24

12400 kg/m 9.81 m/s 6 m 18 m 1 m 529.7 kN2

W = =

( )( ) ( )( )( )3 3 21 1Also 18 m 1 m 10 kg/m 9.81 m/s 18 m2 2

P Ap = =

1589 kN=

Then 0: 1589 kN 0xF HΣ = − =

or 1589 kNH = 1589 kN=H

0: 1589 kN 2543 kN 1271 kN 529.7 kNyF VΣ = − − − −

or 5933 kNV = 5.93 MN=V

PROBLEM 5.75 CONTINUED

( ) Haveb ( ) ( )( )0: 5933 kN 6 m 1589 kNAM XΣ = +

( )( ) ( )( )( )( ) ( )( )6 m 1589 kN 12 m 2543 kN

17 m 1271 kN 19 m 529.7 0

− −

− − =

or 10.48 mX = 10.48 mX = to the right of A

(c) Consider water section BCD as the free body.

Have 0Σ =F

Then 1675 kN− =R 18.43°

or 1675 kN=R 18.43°

Alternative solution to part (c)

Consider the face BC of the dam.

2 2Have 6 18 18.9737 mBC = + =

6tan 18.43

18θ θ= = °

( ) ( )( )( )3 2

2

and 1000 kg/m 9.81 m/s 18 m

176.6 kN/m

p g hρ= =

=

( )( ) ( )21 1Then 18.97 m 1 m 176.6 kN/m2 2

1675 kN

R Ap = =

=

1675 kN∴ =R 18.43°

PROBLEM 5.76 The cross section of a concrete dam is as shown. For a dam section of unit width, determine (a) the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of the reaction forces of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLUTION

The free body shown consists of a 1-ft thick section of the dam and the parabolic section of water above (and behind) the dam.

Note ( )15 16 ft 10 ft8

x = =

( )2116 6 ft 19 ft2

x = + =

( )3122 12 ft 25 ft4

x = + =

( )4522 12 ft 29.5 ft8

x = + =

PROBLEM 5.76 CONTINUED

Now W Vγ=

( ) ( )( ) ( )31

2150 lb/ft 16 ft 24 ft 1 ft 38,400 lb3

W = × =

( ) ( )( ) ( )32 150 lb/ft 6 ft 24 ft 1 ft 21,600 lbW = × =

( ) ( )( ) ( )33

1150 lb/ft 12 ft 18 ft 1 ft 10,800 lb3

W = × =

( ) ( )( ) ( )34

262.4 lb/ft 12 ft 18 ft 1 ft 8985.6 lb3

W = × =

( ) ( )2 31 1Also 18 1 ft 62.4 lb/ft 18 ft 10,108.8 lb2 2

P Ap = = × × × =

( ) Thena 0: 10,108.8 lb 0xF HΣ = − =

or 10.11 kips=H

0: 38,400 lb 21, 600 lb 10,800 lb 8995.6 lb 0yF VΣ = − − − − =

or 79,785.6V = 79.8 kips=V

( )b ( ) ( )( ) ( )( ) ( )( )0: 79,785.6 lb 6 ft 38,400 lb 19 ft 21,600 lb 25 ft 10,800 lbAM XΣ = − − −

( )( ) ( )( )29.5 ft 8985.6 lb 6 ft 10,108.8 lb 0− + =

or 15.90 ftX = The point of application of the resultant is 15.90 ft to the right of A

(c) Consider the water section BCD as the free body.

Have 0Σ =F

13.53 kips41.6

∴ =

= °

On the face BD of the dam

13.53 kips=R 41.6°

PROBLEM 5.77 The 9 12-ft× side AB of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 40 kips, and the design specifications require the force in the rod not exceed 20 percent of this value. If the tank is slowly filled with water, determine the maximum allowable depth of water d in the tank.

SOLUTION

Consider the free-body diagram of the side.

( )1 1Have2 2

P Ap A dγ= =

Now ( )0: 9 ft 03AdM T PΣ = − =

maxThen, for :d

( ) ( )( ) ( )( ) ( )3 3maxmax max

19 ft 0.2 40 10 lb 12 ft 62.4 lb/ft 03 2

d d d × − =

or 3 3 3

max216 10 ft 374.4 d× =

or 3 3max 576.92 ftd = max 8.32 ftd =

PROBLEM 5.78 The 9 12-ft× side of an open tank is hinged at its bottom A and is held in place by a thin rod. The tank is filled with glycerine, whose specific weight is 380 lb/ft . Determine the force T in the rod and the reactions at the hinge after the tank is filled to a depth of 8 ft.

SOLUTION

Consider the free-body diagram of the side.

( )1 1Have2 2

P Ap A dγ= =

( )( ) ( )( )31 8 ft 12 ft 80 lb/ft 8 ft 30,720 lb2 = =

Then 0: 0y yF AΣ = =

( ) ( )80: 9 ft ft 30,720 lb 03AM T Σ = − =

or 9102.22 lbT =

9.10 kips=T

0: 30,720 lb 9102.22 lb 0x xF AΣ = + − = or 21,618 lbA = −

21.6 kips=A

PROBLEM 5.79 The friction force between a 2 2-m× square sluice gate AB and its guides is equal to 10 percent of the resultant of the pressure forces exerted by the water on the face of the gate. Determine the initial force needed to lift the gate that its mass is 500 kg.

SOLUTION

Consider the free-body diagram of the gate.

( ) ( )( )( )2 3 3 2I I

1 1Now 2 2 m 10 kg/m 9.81 m/s 3 m2 2

P Ap = = ×

58.86 kN=

( ) ( )( )( )2 3 3 2II II

1 1 2 2 m 10 kg/m 9.81 m/s 5 m2 2

P Ap = = ×

98.10 kN=

( )I IIThen 0.1 0.1F P P P= = +

( )0.1 58.86 98.10 kN= +

15.696 kN=

Finally ( )( )20: 15.696 kN 500 kg 9.81 m/s 0yF TΣ = − − =

or 20.6 kN=T

PROBLEM 5.80 The dam for a lake is designed to withstand the additional force caused by silt which has settled on the lake bottom. Assuming that silt is equivalent to a liquid of density 3 31.76 10 kg/msρ = × and considering a 1-m-wide section of dam, determine the percentage increase in the force acting on the dam face for a silt accumulation of depth 1.5 m.

SOLUTION

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

Have ( )1 12 2w wP Ap A ghρ= =

( )( ) ( )( )( )3 3 21 6 m 1 m 10 kg/m 9.81 m/s 6 m2

=

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

( )( ) ( )( )( )3 3 21Have 4.5 m 1m 10 kg/m 9.81 m/s 4.5 m2wP ′ =

99.326 kN=

( ) ( )( ) ( )( )( )3 3 2I 1.5 m 1 m 10 kg/m 9.81 m/s 4.5 msP =

66.218 kN=

( ) ( )( ) ( )( )( )3 3 2II

1 1.5 m 1 m 1.76 10 kg/m 9.81 m/s 1.5 m2sP = ×

19.424 kN=

Then ( ) ( )I II 184.97 kNw s sP P P P′ ′= + + =

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

( )184.97 176.58% inc. 100% 100% 4.7503%

176.58w

w

P PP

−′ −= × = × =

% inc. 4.75%=

PROBLEM 5.81 The base of a dam for a lake is designed to resist up to 150 percent of the horizontal force of the water. After construction, it is found that silt (which is equivalent to a liquid of density 3 31.76 10 kg/msρ = × ) is settling on the lake bottom at a rate of 20 mm/y. Considering a 1-m-wide section of dam, determine the number of years until the dam becomes unsafe.

SOLUTION

From Problem 5.80, the force on the dam face before the silt is deposited, is 176.58 kN.wP = The maximum allowable force allowP on the dam is then:

( )( )allow 1.5 1.5 176.58 kN 264.87 kNwP P= = =

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

Have ( ) ( ) ( )( )( )3 3 21 6 m 1 m 10 kg/m 9.81 m/s 6 m2wP d d ′ = − × −

( )24.905 6 kNd= −

( ) ( ) ( )( )( )3 3 2I 1 m 10 kg/m 9.81 m/s 6 msP d d = −

( )29.81 6 kNd d= −

( ) ( ) ( )( )( )3 3 2II

1 1 m 1.76 10 kg/m 9.81 m/s m2sP d d = ×

28.6328 kNd=

( ) ( ) ( ) ( )2 2 2I II

2

4.905 36 12 9.81 6 8.6328 kN

3.7278 176.58 kN

w s sP P P P d d d d d

d

′ ′= + + = − + + − +

= +

PROBLEM 5.81 CONTINUED

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

( )23.7278 176.58 kN 264.87 kNd∴ + =

or 4.8667 md =

Finally 3 m4.8667 m 20 10

yearN−= × ×

or 243 yearsN =

PROBLEM 5.82 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water 3.5d = m, determine the force exerted on the gate by the shear pin.

SOLUTION

First consider the force of the water on the gate.

Have ( )1 12 2

P Ap A ghρ= =

Then ( ) ( )( )( )2 3 3 2I

1 18 m 10 kg/m 9.81 m/s 1.7 m2

P =

26.99 kN=

( ) ( )( )( )2 3 3 2II

1 18 m 10 kg/m 9.81 m/s 1.7 1.8cos30 m2

P = × °

51.74 kN=

Now ( ) ( )I II1 20: 03 3A AB AB AB BM L P L P L FΣ = + − =

( ) ( )1 2or 26.99 kN 51.74 kN 03 3 BF+ − =

or 43.49 kNBF =

4.35 kNB =F 30.0°

PROBLEMS 5.83 AND 5.84 Problem 5.83: A temporary dam is constructed in a 5-ft-wide fresh water channel by nailing two boards to pilings located at the sides of the channel and propping a third board AB against the pilings and the floor of the channel. Neglecting friction, determine the reactions at A and B when rope BC is slack.

Problem 5.84: A temporary dam is constructed in a 5-ft-wide fresh water channel by nailing two boards to pilings located at the sides of the channel and propping a third board AB against the pilings and the floor of the channel. Neglecting friction, determine the magnitude and direction of the minimum tension required in rope BC to move board AB.

SOLUTION

First, consider the force of the water on the gate.

Have ( )1 12 2

P Ap A hγ= =

( )( ) ( )( )3I

1So that 1.5 ft 5 ft 62.4 lb/ft 1.8 ft2

P =

421.2 lb=

( )( ) ( )( )3II

1 1.5 ft 5 ft 62.4 lb/ft 3 ft2

P =

702 lb= 5.83 Find the reactions at A and B when rope is slack.

( ) ( )( ) ( )( )0: 0.9 ft 0.5 ft 421.2 lb 1.0 ft 702 lb 0AM BΣ = − + + =

or 1014 lbB =

1014 lb=B

( ) ( )4 40: 2 421.2 lb 702 lb 05 5x xF AΣ = + + =

or 449.28 lbxA = − Note that the factor 2 (2 )xA is included since xA is the horizontal force exerted by the board on each piling.

( ) ( )3 30: 1014 lb 421.2 lb 702 lb 05 5y yF AΣ = − − + =

or 340.08 lbyA = −

563 lb∴ =A 37.1°

PROBLEMS 5.83 AND 5.84 CONTINUED

5.84 Note that there are two ways to move the board:

1. Pull upward on the rope fastened at B so that the board rotates about A. For this case 0→B and BCT is perpendicular to AB for minimum tension.

2. Pull horizontally at B so that the edge B of the board moves to the left. For this case 0yA → and the board remains against the pilings because of the force of the water.

Case (1) ( )( )0: 1.5 0.5 ft 421.2 lbA BCM TΣ = − +

( )( )1.0 ft 702 lb 0+ =

or 608.4 lbBCT =

Case (2) ( )( ) ( )( )0: 1.2 ft 2 0.5 ft 702 lbB xM AΣ = − −

( )( )1.0 ft 421.2 lb 0− =

or 2 643.5 lbxA = −

( )40: 643.5 lb 421.2 lb5x BCF TΣ = − − +

( )4 702 lb 05

+ =

or 255.06 lbBCT =

( )min 255 lbBC∴ =T

PROBLEMS 5.85 AND 5.86 Problem 5.85: A 2 3-m× gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 12 kN/m. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor.

Problem 5.86: Solve Problem 5.85 if the mass of the gate is 500 kg.

SOLUTION

First, determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor.

Have 1sin 302

θ θ= ∴ = °

Then ( )1.5 m tan 30spx = °

and sp spF kx=

( )( )12 kN/m 1.5 m tan 30= °

10.39 kN= Assume 2 md ≥

Have ( )1 12 2

P Ap A g hρ= =

Then ( )( ) ( )( )( )3 3 2I

1 2 m 3 m 10 kg/m 9.81 m/s 2 m2

P d = −

( )29.43 2 kNd= −

( )( ) ( )( )( )3 3 2II

1 2 m 3 m 10 kg/m 9.81 m/s 2 2cos30 m2

P d = − + °

( )29.43 0.2679 kNd= −

PROBLEMS 5.85 AND 5.86 CONTINUED

5.85 Find mind so that gate opens, 0.W =

Using the above free-body diagrams of the gate, we have

( )20: m 29.43 2 kN3AM d Σ = −

( )4 m 29.43 0.2679 kN3

d + −

( )( )1.5 m 10.39 kN 0− =

or ( ) ( )19.62 2 39.24 0.2679 15.585d d− + − =

58.86 65.3374d = or 1.1105 md = 1.110 md = 5.86 Find mind so that the gate opens.

( )( )29.81 m/s 500 kg 4.905 kNW = =

Using the above free-body diagrams of the gate, we have

( )20: m 29.43 2 kN3AM d Σ = −

( )4 m 29.43 0.2679 kN3

d + −

( )( )1.5 m 10.39 kN− +

( )( )0.5 m 4.905 kN 0− =

or ( ) ( )19.62 2 39.24 0.2679 18.0375d d− + − =

or 1.15171 md = 1.152 md =

PROBLEMS 5.87 AND 5.88 Problem 5.87: The gate at the end of a 3-ft-wide fresh water channel is fabricated from three 240-lb, rectangular steel plates. The gate is hinged at A and rests against a frictionless support at D. Knowing that d = 2.5 ft, determine the reactions at A and D.

Problem 5.88: The gate at the end of a 3-ft-wide fresh water channel is fabricated from three 240-lb, rectangular steel plates. The gate is hinged at A and rests against a frictionless support at D. Determine the depth of water d for which the gate will open.

SOLUTION

5.87 Note that in addition to the weights of the gate segments, the water exerts pressure on all submerged surfaces ( ).p hγ=

( )( )3 20.5Thus, at 0.5 ft, 62.4 lb/ft 0.5 ft 31.2 lb/fth p= = =

( )( )3 22.52.5 ft, 6.24 lb/ft 2.5 ft 156.0 lb/fth p= = =

( )( ) ( )21

1Then 0.5 ft 3 ft 31.2 lb/ft 23.4 lb2

P = =

( )( ) ( )22 2 ft 3 ft 31.2 lb/ft 187.2 lbP = =

( )( ) ( )23

1 2 ft 3 ft 31.2 lb/ft 93.6 lb2

P = =

( )( ) ( )24

1 2 ft 3 ft 156 lb/ft 468 lb2

P = =

and ( )( ) ( )( ) ( ) ( )( )10: 4 2 ft 240 lb 1 ft 240 lb 2 0.5 ft 23.4 lb 1 ft 187.2 lb3AM D Σ = − + + − + × −

( )( ) ( )( )2 12 ft 93.6 lb 2 ft 468 lb 03 3

− − =

or 11.325 lbD = 11.33 lb∴ =D

PROBLEMS 5.87 AND 5.88 CONTINUED

0: 11.32 23.4 93.6 468 0x xF AΣ = + + + + =

or 596.32 lbxA = −

0: 240 240 240 187.2 0y yF AΣ = − − − + =

or 532.8 lbyA = 800 lb∴ =A 41.8°

5.88 At ( ) ( ) 2 322 ft, 2 lb/ft where 62.4 lb/ftdh d p dγ γ−= − = − =

( ) 2ft, lb/ftdh d p dγ= =

( ) ( ) ( ) ( )231 1 2

1 1 3Then 2 ft 3 ft lb/ft 2 ft 2 lb2 2 2dP A p d d dγ γ− = = − × − = −

(Note: For simplicity, the numerical value of the densityγ will be substituted into the equilibrium equations below, rather than at this level of the calculations.)

( )( ) ( ) ( )2 2 2 2 ft 3 ft 2 ft 6 2 lbdP A p d dγ γ− = = − = −

( )( ) ( ) ( )3 3 21 1 2 ft 3 ft 2 ft 3 2 lb2 2dP A p d dγ γ− = = − = −

( )( ) ( ) ( ) ( )4 41 1 2 ft 3 ft ft 3 lb 3 2 6 lb2 2dP A p d d dγ γ γ γ = = = = − +

As the gate begins to open, 0→D

∴ ( )( ) ( )( ) ( ) ( )21 30: 2 ft 240 lb 1 ft 240 lb 2 ft 2 ft 2 lb3 2AM d dγ Σ = + − + − − +

( ) ( ) ( ) ( )21 ft 6 2 lb 2 ft 3 2 lb3

d dγ γ − − − −

( ) ( )1 2 ft 3 2 lb 6 lb 03

dγ γ − − + =

( ) ( ) ( )3 21 720or 2 3 2 12 2 42d d d

γ− + − + − = −

720 462.4

= −

7.53846= Solving numerically yields 2.55 ftd =

PROBLEM 5.89 A rain gutter is supported from the roof of a house by hangers that are spaced 0.6 m apart. After leaves clog the gutter’s drain, the gutter slowly fills with rainwater. When the gutter is completely filled with water, determine (a) the resultant of the pressure force exerted by the water on the 0.6-m section of the curved surface of the gutter, (b) the force-couple system exerted on a hanger where it is attached to the gutter.

SOLUTION

(a) Consider a 0.6 m long parabolic section of water.

Then ( )1 12 2

P Ap A ghρ= =

( )( ) ( )( )( )3 3 21 0.08 m 0.6 m 10 kg/m 9.81 m/s 0.08 m2

=

18.84 N=

wW gVρ=

( )( ) ( )( )( )3 3 2 210 kg/m 9.81 m/s 0.12 m 0.08 m 0.6 m3 =

37.67 N=

Now ( )0: 0wΣ = − + + =F R P W

2 2So that , tan w

wWR P WP

θ= + =

42.12 N, 63.4θ= = ° 42.1 N=R 63.4° (b) Consider the free-body diagram of a 0.6 m long section of water and gutter.

Then 0: 0x xF BΣ = =

0: 37.67 N 0y yF BΣ = − =

or 37.67 NyB =

( ) ( )0: 0.06 0.048 m 37.67 N 0B BM M Σ = + − =

or 0.4520 N mBM = − ⋅ The force-couple system exerted on the hanger is then

37.7 N , 0.452 N m⋅

PROBLEM 5.90 The composite body shown is formed by removing a hemisphere of radius r from a cylinder of radius R and height 2R. Determine (a) the y coordinate of the centroid when r = 3R/4, (b) the ratio r/R for which

1.2 .y R= −

SOLUTION

Note, for the axes shown

V y yV

1 ( )( )2 32 2R R Rπ π= R− 42 Rπ−

2 323rπ− 3

8r− 41

4rπ

Σ 3

323rRπ

442

8rRπ

− −

Then

4 4

3 3

1813

R ryVYV R r

−Σ= = −

Σ −

4

3

118113

rRrR

− = −

( )a

4

3

1 313 43 :

4 1 313 4

r R y R

− = = − −

or 1.118y R= −

( )b

4

3

1181.2 : 1.2113

rRy R R RrR

− = − − = − −

or 4 3

3.2 1.6 0r rR R

− + =

Solving numerically 0.884rR=

PROBLEM 5.91 Determine the y coordinate of the centroid of the body shown.

SOLUTION

First note that the values of Y will be the same for the given body and the body shown below. Then

V y yV

Cone 213a hπ

14h−

2 2112

a hπ−

Cylinder 2

212 4a b a bπ π − = −

12b−

2 218a bπ

Σ ( )2 4 312a h bπ

− ( )2 2 22 324a h bπ

− −

Have Y V yVΣ = Σ

( ) ( )2 2 2 2Then 4 3 2 312 24

Y a h b a h bπ π − = − −

( )2 22 3or

2 4 3h bYh b−

= −−

PROBLEM 5.92 Determine the z coordinate of the centroid of the body shown. (Hint: Use the result of Sample Problem 5.13.)

SOLUTION

First note that the body can be formed by removing a “half-cylinder” from a “half-cone,” as shown.

V z zV

Half-Cone 216a hπ *

− 31

6a h−

Half-Cylinder 2

2

2 2 8a b a bπ π − = −

4 2

3 2 3a a

π π − = −

3112a b

Σ ( )2 4 324a h bπ

− ( )31 212a h b− −

From Sample Problem 5.13

Have Z V zVΣ = Σ

( ) ( )2 31Then 4 3 224 12

Z a h b a h bπ − = − −

2 2or

4 3a h bZh bπ−

= −−

PROBLEM 5.93 Consider the composite body shown. Determine (a) the value of x when

/2h L= , (b) the ratio /h L for which x L= .

SOLUTION V x xV

Rectangular prism Lab 12L

212L ab

Pyramid 13 2

ba h

14

L h+ 1 16 4abh L h +

21 1 1Then 36 6 4

V ab L h xV ab L h L h Σ = + Σ = + +

Now so thatX V xVΣ = Σ

2 21 1 136 6 4

X ab L h ab L hL h + = + +

or 2

21 1 11 36 6 4h h hX LL L L

+ = + + (1)

1( ) ? when2

a X h L= =

1Substituting into Eq. (1)2

hL=

21 1 1 1 1 11 36 2 6 2 4 2

X L + = + +

or 57

104X L= 0.548X L=

( ) ? whenhb X LL= =

Substituting into Eq. (1) 2

21 1 11 36 6 4h h hL LL L L

+ = + +

or 2

21 1 1 116 2 6 24h h hL L L

+ = + +

or 2

2 12hL

= 2 3hL

∴ =

PROBLEMS 5.94 AND 5.95 Problem 5.94: For the machine element shown, determine the x coordinate of the center of gravity.

Problem 5.95: For the machine element shown, determine the y coordinate of the center of gravity.

SOLUTIONS

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

3, inV , in.x , in.y 4, inxV

4, inyV I (4)(3.6)(0.75) 10.8= 2.0 0.375 21.6 4.05 II (2.4)(2.0)(0.6) 2.88= 3.7 1.95 10.656 5.616 III 2(0.45) (0.4) 0.2545π = 4.2 2.15 1.0688 0.54711

IV 2(0.5) (0.75) 0.5890π− = − 1.2 0.375 0.7068− 0.22089− Σ 13.3454 32.618 9.9922

5.94

Have X V x VΣ = Σ

( )3 413.3454 in 32.618 inX = or 2.44 in. X =

5.95

Have Y V yVΣ = Σ

( )3 413.3454 in 9.9922 inY = or 0.749 in. Y =

PROBLEMS 5.96 AND 5.97 Problem 5.96: For the machine element shown, locate the x coordinate of the center of gravity.

Problem 5.97: For the machine element shown, locate the y coordinate of the center of gravity.

SOLUTIONS

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

3, mmV , mmx , mmy 4, mmxV 4, mmyV

1 ( )( )( )160 54 18 155 520= 80 9 12 441 600 1 399 680

2 ( )( )( )1 120 42 54 136 0802

= 40 32 5 443 200 4 354 560

3 2(27) (18) 65612π π=

36160π

+ 9 3 534 114 185 508

4 2(16) (18) 4608π π− = − 160 9 2 316 233− 130 288−

Σ 297 736 19 102 681 5 809 460

5.96

( )3 4

Have

297 736 mm 19 102 681 mm

X V x V

X

Σ = Σ

= or 64.2 mmX =

5.97

( )3 4

Have

297 736 mm 5 809 460 mm

Y V yV

Y

Σ = Σ

= or 19.51 mmY =

PROBLEMS 5.98 AND 5.99 Problem 5.98: For the stop bracket shown, locate the x coordinate of the center of gravity.

Problem 5.99: For the stop bracket shown, locate the z coordinate of the center of gravity.

SOLUTIONS

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

( )

( )

( )

( )

( )

II

III

III

IV

IV

1Have.. 24 mm 90 86 mm 112 mm2

124 mm 102 mm 58 mm3

168 mm 20 mm 78 mm2

2110 mm 90 mm 170 mm3

260 mm 132 mm 156 mm3

Z

Z

X

Z

X

= + + =

= + =

= + =

= + =

= + =

3, mmV , mmx , mmz 4, mmxV 4, mmzV

I ( )( )( )200 176 24 844 800= 100 12 84 480 000 1 013 760

II ( )( )( )200 24 176 844 800= 100 112 84 480 000 94 617 600

III ( )( )( )1 20 124 102 126 4802

= 78 58 9 865 440 733 840

IV ( )( )( )1 90 132 24 142 5602

− = − 156 170 22 239 360− 24 235 200−

Σ 1 673 520 156 586 080 8 785 584

5.98

Have X V xVΣ = Σ

( )3 41 673 520 mm 156 586 080 mmX = or 93.6 mmX =

5.99 Have Z V zVΣ = Σ

( )3 41 673 520 mm 8 785 584 mmZ = or 52.5 mmZ =

PROBLEM 5.100 Locate the center of gravity of the sheet-metal form shown.

SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity coincides with the centroid of the corresponding area.

2, mmA , mmx , mmy , mmz 3, mmxA

3, mmyA 3, mmzA

1 ( )( )1 90 602 2700=

30 120 20 140

+

= 0 81 000 378 000 0

2 ( )( )90 200

18 000= 45 60 80 810 000 1 080 000 1 440 000

3 ( )( )45 100

4500

= − 22.5 30 120 101 250− 135 000− 540 000−

4 ( )245

21012.5

π

π= 45 0

( )( )4 45160

3 179.1

π+

= 143 139 0 569 688

Σ 19 380.9 932 889 1 323 000 1 469 688

( )2 3

Have :

19 380.9 mm 932 889 mm

or 48.1 mm

X A xA

X

Σ = Σ

=

=

48.1 mmX =

( )2 3

19 380.9 mm 1 323 000 mm

or 68.3 mm

Y A yA

Y

Y

Σ = Σ

=

=

68.3 mmY =

( )2 3

19 380.9 mm 1 469 688 mm

or 75.8 mm

Z A zA

Z

Z

Σ = Σ

=

=

75.8 mmZ =

PROBLEM 5.101 A mounting bracket for electronic components is formed from sheet metal of uniform thickness. Locate the center of gravity of the bracket.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with the centroid of the corresponding area. Then (see diagram)

( )

( )

V

2V

2

4 6.2522.5

3

19.85 mm

6.252

61.36 mm

z

A

π

π

= −

=

= −

= −

2, mmA , mmx , mmy , mmz 3, mmxA 3, mmyA

3, mmzA I ( ) ( )25 60 1500= 12.5 0 30 18 750 0 45 000

II ( )( )12.5 60 750= 25 6.25− 30 18 750 4687.5− 22 500

III ( )( )7.5 60 450= 28.75 12.5− 30 12 937.5 5625− 13 500 IV ( )( )12.5 30 375− = − 10 0 37.5 3750− 0 14 062.5− V 61.36− 10 0 19.85 613.6− 0 1218.0− Σ 2263.64 46 074 10 313− 65 720

Have X A xAΣ = Σ

( )2 32263.64 mm 46 074 mmX = or 20.4 mmX =

( )2 3

2263.64 mm 10 313 mm

Y A yA

Y

Σ = Σ

= − or 4.55 mmY = −

Z A zAΣ = Σ

( )2 32263.64 mm 65 720 mmZ = or 29.0 mmZ =

PROBLEM 5.102 Locate the center of gravity of the sheet-metal form shown.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the form coincides with the centroid of the corresponding area.

( )

( ) ( )

( ) ( )

I

I

II II

IV

46 7.333 in.3

1 6 2 in.3

1 2 6 3.8197 in.

111 4 1.5 10.363 in.3

y

z

x y

x

π

π

= + =

= =

= = =

= − =

2, inA , in.x , in.y , in.z 3, inx A

3, iny A 3, inz A

I 12 0 7.333 2 0 88 24 II 56.55 3.8197 3.8197 3 216 216 169.65 III 30 8.5 0 3 255 0 90 IV 3.534 − 10.363 0 3 36.62− 0 10.603− Σ 95.01 434.4 304 273.0

( )2 3

Have

95.01 in 434 in

X A xA

X

Σ = Σ

= or 4.57 in.X =

Y A yAΣ = Σ

( )2 395.01in 304.0 inY = or 3.20 in.Y =

( )2 3

95.01 in 273.0 in

Z A zA

Z

Σ = Σ

= or 2.87 in.Z =

PROBLEM 5.103 An enclosure for an electronic device is formed from sheet metal of uniform thickness. Locate the center of gravity of the enclosure.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the form coincides with the centroid of the corresponding area.

Consider the division of the back, sides, and top into eight segments according to the sketch.

Note that symmetry implies 6.00 in.Z = and

8 2

7 3

6 5

A A

A A

A A

=

=

=

Thus

2, inA , in.x , in.y 3, inxA

3, inyA 1 ( )( )12 9 108= 0 4.5 0 486

2 ( )( )11.3 9 101.7= 5.6 4.5 569.5 457.6

3 ( )( )1 11.3 2.4 13.562

= 7.467 9.8 101.25 132.89

4 ( )( )12 11.454 137.45= 5.6 10.2 769.72 1402.0

5 ( )( )1 1.5 11.454 8.5912

= 7.467 10.6 64.15 91.06

6 8.591 7.467 10.6 64.15 91.06 7 13.56 7.467 9.8 101.25 132.89 8 101.7 5.6 4.5 569.5 457.6 Σ 493.2 2239.5 3251.1

( )2 3Have : 493.2 in 2239.5 inX A xA XΣ = Σ = or 4.54 in.X =

( )2 3 : 493.2 in 3251.1 inY A yA YΣ = Σ = or 6.59 in.Y =

PROBLEM 5.104 A 200-mm-diameter cylindrical duct and a 100 200-mm× rectangular duct are to be joined as indicated. Knowing that the ducts are fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also note that symmetry implies 0Z =

2, mA , mx , my 3, mxA

3, myA 1 ( )( )0.2 0.3 0.1885π = 0 0.15 0 0.028274

2 ( )( )0.2 0.1 0.03142π

− = −

( )2 0.10.06366

π= 0.25 0.02000− 0.007854−

3 ( )20.1 0.015712π

= ( )4 0.1

0.042443π

−= −

0.30 0.000667− 0.004712

4 ( )( )0.3 0.2 0.060= 0.15 0.30 0.00900 0.001800

5 ( )( )0.3 0.2 0.060= 0.15 0.20 0.00900 0.001200

6 ( )20.1 0.15712π

− = − ( )4 0.13π 0.20 0.000667− 0.003142−

7 ( )( )0.3 0.1 0.030= 0.15 0.25 0.004500 0.007500

8 ( )( )0.3 0.1 0.030= 0.15 0.25 0.004500 0.007500 Σ 0.337080 0.023667 0.066991

Have 2 3: (0.337080 mm ) 0.023667 mmX A xA XΣ = Σ =

or 0.0702 mX = 70.2 mmX =

( )2 3: 0.337080 mm 0.066991 mmY A yA YΣ = Σ =

or 0.19874 mY = 198.7 mmY =

PROBLEM 5.105 An elbow for the duct of a ventilating system is made of sheet metal of uniform thickness. Locate the center of gravity of the elbow.

SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also, note that the shape of the duct implies 1.5 in.Y =

( )

( )

( )

( )

( )

( )

I I

II

II

IV IV

V

V

2Note that 16 in. 16 in. 5.81408 in.

216 in. 8 in. 10.9070 in.

212 in. 8 in. 6.9070 in.

416 in. 16 in. 9.2094 in.3

416 in. 8 in. 12.6047 in.3412 in. 8 in. 8.6047 in.

3

x z

x

z

x z

x

z

π

π

π

π

π

π

= = − =

= − =

= − =

= = − =

= − =

= − =

Also note that the corresponding top and bottom areas will contribute equally when determining and .x z

Thus 2, inA , in.x , in.z 3, inxA

3, inzA

I ( )( )16 3 75.39822π

= 5.81408 5.81408 438.37 438.37

II ( )( )8 3 37.6991

= 10.9070 6.9070 411.18 260.39

III ( )4 3 12= 8 14 96.0 168.0

IV ( )22 16 402.1239 4π = 9.2094 9.2094 3703.32 3703.32

V ( )22 8 100.5309

4π − = −

12.6047 8.6047 1267.16− 865.04−

VI ( )( )2 4 8 64− = − 12 14 768.0− 896.0− Σ 362.69 2613.71 2809.04

Have ( )2 3: 362.69 in 2613.71 inX A xA XΣ = Σ =

or 7.21 in.X =

( )2 3: 362.69 in 2809.04 inZ A zA ZΣ = Σ =

or 7.74 in.Z =

PROBLEM 5.106 A window awning is fabricated from sheet metal of uniform thickness. Locate the center of gravity of the awning.

SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with the centroid of the corresponding area.

( )( )

( )( )

( )( )

( )( )

( )

( )( )

II VI

II VI

IV

IV

2 2II VI

2IV

4 50080 292.2 mm

34 500

212.2 mm3

2 50080 398.3 mm

2 500318.3 mm

500 196 350 mm4

500 680 534 071 mm2

y y

z z

y

z

A A

A

π

π

π

ππ

π

= = + =

= = =

= + =

= =

= = =

= =

2, mmA , mmy , mmz 3, mmyA 3, mmzA

I (80)(500) 40 000= 40 250 61.6 10× 610 10× II 196 350 292.2 212.2 657.4 10× 641.67 10× III (80)(680) 54 400= 40 500 60.2176 10× 627.2 10× IV 534 071 398.3 318.3 6212.7 10×

6170 10× V (80)(500) 40 000= 40 250 61.6 10×

610 10× VI 196 350 292.2 212.2 657.4 10×

641.67 10× Σ 61.061 10× 6332.9 10×

6300.5 10× Now, symmetry implies 340 mmX = and ( )6 2 6 3: 1.061 10 mm 332.9 10 mmY A yA YΣ = Σ × = ×

or 314 mmY =

( )6 2 6 3: 1.061 10 mm 300.5 10 mmZ A zA ZΣ = Σ × = ×

or 283 mmZ =

PROBLEM 5.107 The thin, plastic front cover of a wall clock is of uniform thickness. Locate the center of gravity of the cover.

SOLUTION First, assume that the plastic is homogeneous so that the center of gravity of the cover coincides with the centroid of the corresponding area.

Next, note that symmetry implies 150.0 mmX =

2, mmA , mmy , mmz 3, mmy A 3, mmz

1 ( )( )300 280

84 000= 0 140 0 11 760 000

2 ( )( )280 50

14 000= 25 140 350 000 1 960 000

3 ( )( )300 50

15 000= 25 0 375 000 0

4 ( )( )280 50

14 000= 25 140 350 000 1 960 000

5 ( )2100

31 416

π−

= − 0 130 0 4 084 070−

6 ( )2304

706.86

π−

= − 0 ( )( )4 30

260 247.293π

− = 0 174 783−

7 ( )2304

706.86

π−

= − 0 ( )( )4 30

260 247.293π

− = 0 174 783−

Σ 94 170 1 075 000 11 246 363

Have ( )2 3: 94 170 mm 1 075 000 mmY A yA YΣ = Σ =

or 11.42 mmY =

( )2 3: 94 170 mm 11 246 363 mmZ A zA ZΣ = Σ =

or 119.4 mmZ =

PROBLEM 5.108 A thin steel wire of uniform cross section is bent into the shape shown, where arc BC is a quarter circle of radius R. Locate its center of gravity.

SOLUTION

First, assume that the wire is homogeneous so that its center of gravity coincides with the centroid of the corresponding line.

2, inL , in.x , in.y , in.z 2, inx L

2, inyL 2, inz L

1 15 0 7.5 0 0 112.5 0

2 14 7 0 0 98 0 0

3 13 59

1211.5

= 0 6 149.5 0 78

4 ( )152

23.56

π

=

3 2 155

5.73π×

=

30

9.549π=

24

7.639π=

135.0 225.0 180.0

Σ 65.56 382.5 337.5 258.0

( ) 2Have : 65.56 in. 382.5 inX L x L XΣ = Σ = or 5.83 in.X =

( ) 2: 65.56 in. 337.5 inY L yL YΣ = Σ = or 5.15 in.Y =

( ) 2: 65.56 in. 258.0 inZ L z L ZΣ = Σ = or 3.94 in.Z =

PROBLEM 5.109 A thin steel wire of uniform cross section is bent into the shape shown, where arc BC is a quarter circle of radius R. Locate its center of gravity.

SOLUTION

First, assume that the wire is homogeneous so that its center of gravity coincides with the centroid of the corresponding line

4

( ) ( )2 2

2 21

2

2 8 16Have ft

8 3 8.5440 ft8 4 ft2

x z

L

L

π π

ππ

= = =

= + =

= =

, ftL , ftx , fty , ftz 2, ftx L

2, fty L 2, ftz L

1 8.5440 4 1.5 0 34.176 12.816 0

2 4π 16π 0 16π 64.0 0 64.0

3 8 0 0 4 0 0 32 4 3 0 1.5 0 0 4.5 0 32.110 98.176 17.316 96.0

Have ( ) 2: 32.110 ft 98.176 ftX L x L XΣ = Σ = or 3.06 ftX =

( ) 2: 32.110 ft 17.316 ftY L yL YΣ = Σ = or 0.539 ftY =

( ) 2: 32.110 ft 96.0 ftZ L z L ZΣ = Σ = or 2.99 ftZ =

PROBLEM 5.110

The frame of a greenhouse is constructed from uniform aluminum channels. Locate the center of gravity of the portion of the frame shown.

SOLUTION

First, assume that the channels are homogeneous so that the center of gravity of the frame coincides with the centroid of the corresponding line.

( )( )

( )( )

( )

8 9

8 9

7 8

2 0.9Note 0.57296 m

2 0.91.5 2.073 m

0.9 1.4137 m2

x x

y y

L L

π

ππ

= = =

= = + =

= = =

, mL , mx , my , mz 2, mxL 2, myL

2, mzL 1 0.6 0.9 0 0.3 0.540 0 0.18 2 0.9 0.45 0 0.6 0.4050 0 0.54 3 1.5 0.9 0.75 0 1.350 1.125 0 4 1.5 0.9 0.75 0.6 1.350 1.125 0.9 5 2.4 0 1.2 0.6 0 2.880 1.44 6 0.6 0.9 1.5 0.3 0.540 0.9 0.18 7 0.9 0.45 1.5 0.6 0.4050 1.350 0.54 8 1.4137 0.573 2.073 0 0.8100 2.9306 0 9 1.4137 0.573 2.073 0.6 0.8100 2.9306 0.8482

10 0.6 0 2.4 0.3 0 1.440 0.18 Σ 11.827 6.210 14.681 4.8082

( ) 2Have : 11.827 m 6.210 mX L xL XΣ = Σ = or 0.525 mX =

( ) 2: 11.827 m 14.681 mY L yL YΣ = Σ = or 1.241 mY =

( ) 2: 11.827 m 4.8082 mZ L zL ZΣ = Σ = or 0.406 mZ =

*

PROBLEM 5.111 The decorative metalwork at the entrance of a store is fabricated from uniform steel structural tubing. Knowing that R = 1.2 m, locate the center of gravity of the metalwork.

SOLUTION

First, assume that the tubes are homogeneous so that the center of gravity of the metalwork coincides with the centroid of the corresponding line.

Note that symmetry implies 0Z =

, mL , mx , my 2, mxL 2, myL

1 3 ( )1.2 cos 45 0.8485° = 1.5 2.5456 4.5

2 3 ( )1.2 cos 45 0.8485° = 1.5 2.5456 4.5 3 1.2π 0 3.7639 0 14.1897

4 1.2π ( )( )2 1.2

0.7639π

= 3 2.88 11.3097

5 0.6π ( )( )2 1.2

0.7639π

= 3.7639 1.44 7.0949

Σ 15.425 9.4112 41.594

( ) 2Have : 15.425 m 9.4112 mX L xL XΣ = Σ = or 0.610 mX =

( ) 2: 15.425 m 41.594 mY L yL YΣ = Σ = or 2.70 mY =

PROBLEM 5.112 A scratch awl has a plastic handle and a steel blade and shank. Knowing that the specific weight of plastic is 30.0374 lb/in and of steel is

30.284 lb/in , locate the center of gravity of the awl.

SOLUTION

First, note that symmetry implies 0Y Z= =

( ) ( ) ( )

( )( )( ) ( )

( ) ( ) ( )

33I I

23II II

23III III

5 20.5 in. 0.3125 in., 0.0374 lb/in 0.5 in. 0.009791 lb8 3

1.6 in. 0.5 in. 2.1 in. 0.0374 lb/in 0.5 in. 3.2 in. 0.093996 lb

3.7 in. 1 in. 2.7 in., 0.0374 lb/in 0.12 in. 2 in. 0.004

x W

x W

x W

π

π

π

= = = =

= + = = =

= − = = − = −

( ) ( ) ( )

( ) ( ) ( ) ( )

2 23IV IV

23V V

0846 lb

7.3 in. 2.8 in. 4.5 in., 0.284 lb/in 0.12 in. 5.6 in. 0.017987 lb4

17.3 in. 0.4 in. 7.4 in., 0.284 lb/in 0.06 in. 0.4 in. 0.000428 lb4 3

x W

x W

π

π

= − = = =

= + = = =

, lbW , in.x , in lbxW ⋅ I 0.009791 0.3125 0.003060 II 0.093996 2.1 0.197393 III 0.000846− 2.7 0.002284− IV 0.017987 4.5 0.080942 V 0.000428 7.4 0.003169 Σ 0.12136 0.28228

( )Have : 0.12136 lb 0.28228 in. lbX W xW XΣ = Σ = ⋅ or 2.33 in.X =

PROBLEM 5.113 A bronze bushing is mounted inside a steel sleeve. Knowing that the density of bronze is 38800 kg/m and of steel is 37860 kg/m , determine the center of gravity of the assembly.

SOLUTION

First, note that symmetry implies 0X Z= =

( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3 2 2 2 2I I

3 2 2 2 2II II

Now

4 mm , 7860 kg/m 9.81 m/s 0.036 0.015 m 0.008 m4

0.51887 N

18 mm, 7860 kg/m 9.81 m/s 0.0225 0.05 m 0.02 m4

W g V

y W

y W

ρ

π

π

=

= = − =

= = −

( )( ) ( ) ( )3 2 2 2 2III III

0.34065 N

14 mm, 8800 kg/m 9.81 m/s 0.15 0.10 m 0.028 m4

0.23731 N

y W π

=

= = − =

Have Y W yWΣ = Σ

( ) ( ) ( ) ( ) ( ) ( )4 mm 0.5189 N 18 mm 0.3406 N 14 mm 0.2373 N0.5189 N 0.3406 N 0.2373 N

Y+ +

=+ +

( )or 10.51 mm

above base

Y =

PROBLEM 5.114 A marker for a garden path consists of a truncated regular pyramid carved from stone of specific weight 160 lb/ft3. The pyramid is mounted on a steel base of thickness h. Knowing that the specific weight of steel is 490 lb/ft3 and that steel plate is available in 1

4 in. increments, specify the minimum thickness h for which the center of gravity of the marker is approximately 12 in. above the top of the base.

SOLUTION

First, locate the center of gravity of the stone. Assume that the stone is homogeneous so that the center of gravity coincides with the centroid of the corresponding volume.

( ) ( )( )( )

( ) ( )( )( )

1 1

3

2 2

3

3 1Have 56 in. 42 in., 12 in. 12 in. 56 in.4 3

2688 in3 128 in. 21 in., 6 in. 6 in. 28 in.4 3

366 in

y V

y V

= = =

=

= = = −

= −

3 3stone

3

Then 2688 in 366 in

2352 in

V = −

=

( )( ) ( )( )3 3

3

and

42 in. 2688 in 21 in. 366 in

2352 in45 in.

yVYV

Σ=

Σ

+ −=

=

Therefore, the center of gravity of the stone is ( )45 28 in.− = 17 in. above the base.

( )( )

( ) ( )( )

( )

33 3

stone stone stone

steel steel steel3

3

1 ftNow 160 lb/ft 2352 in12 in.

217.78 lb

1 ft490 lb/ft 12 in. 12 in. 12 in.

1b40.833

W V

W V

h

h

γ

γ

= =

=

=

=

=

PROBLEM 5.114 CONTINUED

Then marker 12 in.yWYW

Σ= =

Σ

( )( ) ( )

( )

17 in. 217.78 lb in. 40.833 h lb2

217.78 40.833 lb

h

h

+ − =

+

or 2 24 53.334 0h h+ − =

With positive solution 2.0476 in.h =

specify 2 in.h∴ =

PROBLEM 5.115 The ends of the park bench shown are made of concrete, while the seat and back are wooden boards. Each piece of wood is 36 120× × 1180 mm. Knowing that the density of concrete is 32320 kg/m and of wood is 3470 kg/m , determine the x and y coordinates of the center of gravity of the bench.

SOLUTION

First, note that we will account for the two concrete ends by counting twice the weights of components 1, 2, and 3.

( ) ( )( ) ( )( )( )

( ) ( )( ) ( )( )( )

( ) ( )( ) ( )( )( )

3 21 1

3 22 2

3 23 3

2320 kg/m 9.81 m/s 0.480 m 0.408 m 0.072 m

320.9 N

2320 kg/m 9.81 m/s 0.096 m 0.048 m 0.072 m

7.551 N

2320 kg/m 9.81 m/s 0.096 m 0.384 m 0.072 m

c

c

c

W g V

W g V

W g V

ρ

ρ

ρ

= =

=

= − = −

= −

= =

( )( ) ( )( )( )

4 5 6 7 board

3 2

60.41 N

470 kg/m 9.81 m/s 0.120 m 0.036 m 1.180 m 23.504 N

wW W W W Vρ=

= = = =

=

=

PROBLEM 5.115 CONTINUED

( )Have : 865.06 N 236 590 mm NX W xW XΣ = Σ = ⋅

or 274 mmX =

( ): 865.06 N 89 671 mm NY W yW YΣ = Σ = − ⋅

or 103.6 mmY = −

, NW , mmx , mmy , mm Nx W ⋅ , mm NyW ⋅ 1 ( )2 320.4 641.83= 312 204− 200 251.4 130 933.6− 2 ( )2 7.551 15.10− = − 312 384− 4711.8− 5799.1

3 ( )2 60.41 120.82= 84 192 10 148.5 23 196.5 4 23.504 228 18 5358.8 423.1 5 23.504 360 18 8461.3 423.1 6 23.504 442 18 10 388.5 423.1 7 23.504 124.7 328.3 2930.9 7716.2 8 23.504 160.1 139.6 3762.9 3281.1 Σ 865.06 236 590 89 671−

PROBLEM 5.116

Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height.

A hemisphere.

SOLUTION

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

2 , ELdV r dx x xπ= =

The equation of the generating curve is 2 2 2x y a+ = so that 2 2 2r a x= − and then

( )2 2dV a x dxπ= −

Component 1

( )/23

/2 2 2 21 0

0

3

3

1124

aa xV a x dx a x

a

π π

π

= − = −

=

( )/2 2 21 0

/22 42

0

4

and

2 4

764

aEL

a

x dV x a x dx

x xa

a

π

π

π

= −

= −

=

∫ ∫

Now 3 4

1 1 1111 7:24 64ELx V x dV x a aπ π = =

121or88

x a=

Component 2

( )

( ) ( )

32 2 2

2 /2/2

33

22 2

3

3

3 2 3

524

aaa

a

a

xV a x dx a x

a aa a a

a

π π

π

π

= − = −

= − − −

=

PROBLEM 5.116 CONTINUED

( )

( ) ( ) ( ) ( )

2 42 2 2

EL2 /2/2

2 42 42 22 2

4

and 2 4

2 4 2 4

964

aaa

a

a a

x xx dV x a x dx a

a aa a

a

π π

π

π

= − = −

= − − −

=

∫ ∫

Now 3 42 2 EL 22

5 9:24 64

x V x dV x a aπ π = =

227or40

x a=

PROBLEM 5.117

Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height.

A semiellipsoid of revolution.

SOLUTION

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

2EL,dV r dx x xπ= =

The equation of the generating curve is 2 2

2 2 1x yh a

+ = so that

( )2

2 2 22

ar h xh

= − and then

( )2

2 22

adV h x dxh

π= −

Component 1

( )/22 2 3

/2 2 2 21 2 20

0

2

3

1124

hh a a xV h x dx h x

h h

a h

π π

π

= − = −

=

( )2

/2 2 221 0

/22 2 42

20

2 2

and

2 4

764

hEL

h

ax dV x h x dxh

a x xhh

a h

π

π

π

= −

= −

=

∫ ∫

Now 2 2 21 1 11

11 7:24 64ELx V x dV x a h a hπ π = =

121or88

x h=

PROBLEM 5.117 CONTINUED

Component 2

( )

( ) ( ) ( )

2 2 32 2 2

2 2 2/2/2

33222 2

2

2

3

3 2 3

524

hhh

h

h

a a xV h x dx h xh h

ha hh h hh

a h

π π

π

π

= − = −

= − − −

=

( )

( ) ( ) ( ) ( )

22 2

22 /2

2 2 42

2/2

2 42 422 22 2

2

2 2

and

2 4

2 4 2 4

964

hEL h

h

h

h h

ax dV x h x dxh

a x xhh

h ha h hh

a h

π

π

π

π

= −

= −

= − − −

=

∫ ∫

Now 2 2 2

2 2 225 9:24 64ELx V x dV x a h a hπ π = =

227or40

x h=

PROBLEM 5.118

Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height.

A paraboloid of revolution.

SOLUTION

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

2EL,dV r dx x xπ= =

The equation of the generating curve is 22

hx h ya

= − so that

( )2

2 ar h xh

= − and then

( )2adV h x dx

hπ= −

Component 1

( )2

/21 0

/22 2

0

2

2

38

h

h

aV h x dxh

a xhxh

a h

π

π

π

= −

= −

=

( )2

/21 0

/22 2 3

0

2 2

and

2 3

112

hEL

h

ax dV x h x dxh

a x xhh

a h

π

π

π

= −

= −

=

∫ ∫

Now 2 2 21 1 11

3 1:8 12ELx V x dV x a h a hπ π = =

12or9

x h=

PROBLEM 5.118 CONTINUED

Component 2

( )

( ) ( ) ( )

2 2 2

2 /2/2

2222

2

2

2 2 2

18

hhh

h

h

a a xV h x dx hxh h

ha hh h hh

a h

π π

π

π

= − = −

= − − −

=

( )

( ) ( ) ( ) ( )

2 2 2 3

2 /2/2

2 32 322 2

2 2

and 2 3

2 3 2 3

112

hh

EL hh

h h

a a x xx dV x h x dx hh h

h ha h hh

a h

π π

π

π

= − = −

= − − −

=

∫ ∫

Now 2 2 2

2 2 221 1:8 12ELx V x dV x a h a hπ π = =

22or3

x h=

PROBLEM 5.119 Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

SOLUTION

First note that symmetry implies 0y =

0z =

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

2 , ELdV r dx x xπ= =

Now 2

21 xr ba

= −

so that

222

21 xdV b dxa

π

= −

22 2 42 2

2 2 40 0

3 52

2 40

2

2

2Then 1 1

23 5

2 113 5

815

a a

a

x x xV b dx b dxa a a

x xb xa a

ab

ab

π π

π

π

π

= − = − +

= − +

= − +

=

∫ ∫

and 2 4

22 40

21aEL

x xx dV b x dxa a

π

= − +

∫ ∫

2 4 6

22 4

0

22 4 6

ax x xb

a aπ

= − +

2 2 1 1 12 2 6

a bπ = − +

2 216a bπ=

PROBLEM 5.119 CONTINUED

Then 2 2 28 1:15 16ELxV x dV x ab a bπ π = =

15or 6

x a=

PROBLEM 5.120 Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

SOLUTION

First, note that symmetry implies 0y = 0z = Choose as the element of volume a disk of radius r and thickness dx. Then

2 , ELdV r dx x xπ= =

Now 11rx

= − so that 2

2

11

2 11

dV dxx

dxxx

π

π

= −

= − +

( )

33

211

3

2 1 1Then 1 2 ln

1 13 2 ln3 1 2 ln 13 1

0.46944 m

V dx x xx xx

π π

π

π

= − + = − −

= − − − − −

=

( ) ( )

( )

323

211

2 3

2 1and 1 2 ln2

3 12 3 ln 3 2 1 ln12 2

1.09861 m

ELxx dV x dx x x

x xπ π

π

π

= − + = − +

= − + − − +

=

∫ ∫

( )3 4Now : 0.46944 m 1.09861 mELxV x dV X π π= =∫

or 2.34 mx =

PROBLEM 5.121 Locate the centroid of the volume obtained by rotating the shaded area about the line .x h=

SOLUTION

First, note that symmetry implies x h= 0z = Choose as the element of volume a disk of radius r and thickness dx. Then

2 , ELdV r dy y yπ= =

Now ( )2

2 2 22hx a ya

= − so that 2 2hr h a ya

= − −

Then ( )2 22 2

2hdV a a y dya

π= − −

and ( )2 22 2

20a hV a a y dya

π= − −∫

Let sin cosy a dy a dθ θ θ= ⇒ =

( )( ) ( )

( )

2 2/2 2 2 22 0

2/2 2 2 2 2

2 0

/22 2 20

/22 3

0

2 2

Then sin cos

2 cos sin cos

2cos 2cos sin cos

sin 2 12sin 2 sin2 4 3

12 22 3

0.095870

hV a a a a dah a a a a a a da

ah d

ah

ah

ah

π

π

π

π

π

π θ θ θ

π θ θ θ θ

π θ θ θ θ θ

θ θπ θ θ

π

π

= − −

= − + −

= − −

= − + −

= − −

=

2

PROBLEM 5.121 CONTINUED

( )( )

( )

( ) ( )

2 22 2

20

22 2 2 3

2 0

2 3/22 2 2 2 42

02 3/222 4 22

2 2

and

2 2

2 13 4

1 24 3

112

aEL

a

a

hy dV y a a y dya

h a y ay a y y dya

h a y a a y ya

h a a a a aa

a h

π

π

π

π

π

= − −

= − − −

= + − −

= − −

=

∫ ∫

( )2 2 21Now : 0.09587012ELyV y dV y ah a hπ π= =∫

or 0.869y a=

PROBLEM 5.122 Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x axis.

SOLUTION

First, note that symmetry implies 0y = 0z = Choose as the element of volume a disk of radius r and thickness dx. Then

2EL,dV r dx x xπ= =

Now sin2

xr baπ

=

so that 2 2sin2

xdV b dxaππ=

( ) ( )

2 2 2

22 2

2 22 2

2

Then sin2

sin2 2

12

aa

axa

a a

a a

xV b dxa

xb

b

ab

π

π

ππ

π

π

π

=

= −

= −

=

and 2 2 2sin2

aEL a

xx dV x b dxa

ππ =

∫ ∫

Use integration by parts with

2

2

sin2

sin2

xa

a

xu x dVa

xdu dx Vπ

π

π= =

= = −

PROBLEM 5.122 CONTINUED

( ) ( )

222

2 2

222 2

2

2 22 22 2

2 2

sin sinThen

2 2

2 12 cos2 2 4 2

3 1 122 4 42 2

ax x

aa aEL a

a aa

a

a

x xx dV b x dx

a a a xb a a xa

a ab a a a

π π

π ππ

πππ

ππ π

= − − −

= − − +

= − + − +

∫ ∫

2 22

2 2

3 14

0.64868

a b

a b

ππ

π

= −

=

2 2 21Now : 0.648682ELxV x dV x ab a bπ π = =

or 1.297x a=

PROBLEM 5.123 Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y axis. (Hint: Use a thin cylindrical shell of radius r and thickness dr as the element of volume.)

SOLUTION

First note that symmetry implies 0x = 0z = Choose as the element of volume a cylindrical shell of radius r and thickness dr.

Then ( ) ( ) ( ) 12 ,2ELdV r y dr y yπ= =

Now sin2

ry baπ

=

so that 2 sin2

rdV br draππ=

Then 2 2 sin2

aa

rV br draππ= ∫

Use integration by parts with

sin2

2 cos2

ru r dv dra

a rdu dr va

π

ππ

= =

= = −

( )

( ) ( )

22

22

2

2 2Then 2 cos cos2 2

2 42 2 1 sin2

aa

aa

a

a

a r a rV b r dra a

a a rb aa

π πππ π

πππ π

= − −

= − − +

2 2

2

2

2

4 42

18 1

5.4535

a aV b

a b

a b

ππ π

π

= −

= −

=

2 12

22 2

Also sin 2 sin2 2

sin2

aEL a

aa

r ry dV b br dra a

rb r dra

π ππ

ππ

=

=

∫ ∫

PROBLEM 5.123 CONTINUED

Use integration by parts with

2

2

sin2

sin2

ra

a

ru r dv dra

rdu dr vπ

π

π= =

= = −

( )

( ) ( )

( ) ( )

222

2 2

22 22

2

2 22 22 2

2 2

2

sin sinThen

2 2

22 cos2 2 4 2

232 4 42 2

ar r

aa aEL a

a aa

a

a

r ry dV b r dr

a a r a rb a aa

a aa ab a

a

π π

π ππ

πππ

ππ π

π

= − − −

= − − +

= − + − +

=

∫ ∫

22

2 2

3 14

2.0379

b

a b

π −

=

Now ( )2 2 2: 5.4535 2.0379ELyV y dV y a b a b= =∫

or 0.374y b=

PROBLEM 5.124

Show that for a regular pyramid of height h and n sides ( 3, 4,n = … ) the centroid of the volume of the pyramid is located at a distance / 4h above the base.

SOLUTION

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

2baseA kb=

where ( ) ;k k N= see note below. Using similar triangles, have s h yb h

−=

or ( )bs h yh

= −

( )2

22slice 2Then bdV A dy ks dy k h y dy

h= = = −

( ) ( )2 2

2 32 20

0

2

1and3

13

hh b bV k h y dy k h y

h h

kb h

= − = − −

=

Also ELy y=

( ) ( )2 2

2 2 2 32 20 0

22 2 3 4 2 2

20

so then 2

1 2 1 12 3 4 12

h hEL

h

b by dV y k h y dy k h y hy y dyh h

bk h y hy y kb hh

= − = − +

= − + =

∫ ∫ ∫

Now 2 2 21 1:

3 12ELyV y dV y kb h kb h = =

1or Q.E.D.4

y h=

Note

( )

2base tan

2

2

12

4 tan

N

b

N

A N b

N b

k N b

π

π

= × ×

=

=

PROBLEM 5.125 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R.

SOLUTION

First note that symmetry implies 0x = The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now

( )( )2

EL

dA r Rd

ry

π θ

π

=

= −

where sinr R θ= 2

EL

so that sin2 sin

dA R dRy

π θ θ

θπ

=

= −

[ ] 22 2 20 0

2

Then sin cosA R d R

R

πππ θ θ π θ

π

= = −

=

( )2

2

20

3

0

3

2and sin sin

sin 222 4

2

ELRy dA R d

R

R

π

π

θ π θ θπ

θ θ

π

= −

= − −

= −

∫ ∫

( )2 3Now :2ELyA y dA y R Rππ= = −∫

1or2

y R= −

Symmetry implies 12

z y z R= ∴ = −

PROBLEM 5.126 The sides and the base of a punch bowl are of uniform thickness t. If t R<< and 350R = mm, determine the location of the center of gravity of (a) the bowl, (b) the punch.

SOLUTION

(a) Bowl First note that symmetry implies 0x = 0z = 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

( )( )wall 2 sindA R Rdπ θ θ=

and ( )wall cosELy R θ= −

[ ] /2/2 2 2wall /6 /6

2

Then 2 sin 2 cos

3

A R d R

R

πππ ππ θ θ π θ

π

= = −

=

( )

( )( )wall wall wall

/2 2/6

/23 2/6

3

and

cos 2 sin

cos

34

ELy A y dA

R R d

R

R

ππ

π

π

θ π θ θ

π θ

π

=

= −

=

= −

2base base

3By observation , 4 2

A R y Rπ= = −

Now y A yAΣ = Σ

2 2 3 23 3or 34 4 4 2

y R R R R Rπ ππ π + = − + −

or 0.48763 350 mmy R R= − =

170.7 mmy∴ = −

(b) Punch First note that symmetry implies 0x = 0z = 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

2 , ELdV x dy y yπ= =

PROBLEM 5.126 CONTINUED

Now 2 2 2x y R+ = so that ( )2 2dV R y dyπ= −

( )0

0 2 2 2 33/2

3/2

32 3

1Then3

3 1 3 3 32 3 2 8

RR

V R y dy R y y

R R R R

π π

π π

−−

= − = −

= − − − − =

( ) ( )0

0 2 2 2 2 43/2

3/2

2 42 4

1 1and 2 4

1 3 1 3 152 2 4 2 64

EL RR

y dV y R y dy R y y

R R R R

π π

π π

−−

= − = −

= − − − − = −

∫ ∫

3 43 15Now : 38 64ELyV y dV y R Rπ π = = −

5or 350 mm8 3

y R R= − =

126.3 mmy∴ = −

PROBLEM 5.127 After grading a lot, a builder places four stakes to designate the corners of the slab for a house. To provide a firm, level base for the slab, the builder places a minimum of 60 mm of gravel beneath the slab. Determine the volume of gravel needed and the x coordinate of the centroid of the volume of the gravel. (Hint: The bottom of the gravel is an oblique plane, which can be represented by the equation .)y a bx cz= + +

SOLUTION

First, determine the constants a, b, and c. At 0, 0 : 60 mmx z y= = = − 60 mm ; 60 mma a∴ − = = −

( )At 7200 mm, 0 : 100 mm

100 mm 60 7200

1180

x z y

b

b

= = = −

∴ − = − +

= −

( )At 0 , 12 000 mm: 120 mm

120 mm 60 mm 12000

1200

x z y

c

c

= = = −

∴ − = − +

= −

It follows that 1 160180 120

y x z= − − − where all dimensions

are in mm. Choose as the element of volume a filament of base dx dz× and height | |.y Then

,

1 1or 60180 200

ELdV y dxdz x x

dV x z dxdz

= =

= − − −

( )( ) ( ) ( )

12000 72000 0

720012000 20

0

2120000

120002

09 3 3

1 1Then 60180 200

1 160360 200

7200 720060 7200

360 200

365760002

9.504 10 mm 9.50 m

V x z dxdz

x x xz dz

z dz

z z

= + +

= + +

= + +

= +

= × =

∫ ∫

39.50 mV =

PROBLEM 5.127 CONTINUED

( )

12000 72000 0

720012000 2 3 20

012000 60

120006 2

013 13

13 4 4

1 1and 60180 200

60 1 12 540 400

2246.4 10 129 600

129 6002246.4 102

2.695 10 0.933 10

3.63 10 mm 36.3 m

ELx dV x x z dxdz

x x x z dz

z dz

z z

= + +

= + +

= × +

= × −

= × + ×

= × =

∫ ∫ ∫

Now ( )3 4: 9.50 m 36.3 mELxV x dV x= =∫

or 3.82 mx =

PROBLEM 5.128 Determine by direct integration the location of the centroid of the volume between the xz plane and the portion shown of the surface

( )( )2 2

2 2

16.

h ax x bz zy

a b

− −=

SOLUTION

First note that symmetry implies 2ax =

2bz =

Choose as the element of volume a filament of base dx dz× and height y. Then

( ) ( )EL

2 22 2

1,2

16or

dV ydxdz y y

hdV ax x bz z dxdza b

= =

= − −

Then ( ) ( )2 22 20 0

16b a hV ax x bz z dxdza b

= − −∫ ∫

( )

( ) ( ) ( ) ( )

2 2 32 2 0

0

3 2 32 2 32 2 2

0

16 13

16 1 1 8 1 42 3 2 3 2 3 93

ab

b

h aV bz z x x dzza b

h a b ah ba a z z b b abha b b

= − −

= − − = − =

( )( ) ( )( )

( )( )

( )

( ) ( ) ( )

2 2 2 22 2 2 20 0

22 2 3 4 2 2 3 4

4 4 0 0

2 22 2 3 4 3 4 5

2 4 00

2 2 23 4 5 3

4 4

1 16 16and2

128 2 2

128 123 2 5

128 13 2 5 3

b aEL

b a

ab

h hy dV ax x bz z ax x bz z dxdza b a b

h a x ax x b z bz z dxdza b

h a ab z bz z x x x dza b

h a a b ba a a Za b

= − − − −

= − + − +

= − + − +

= − + −

∫ ∫ ∫

∫ ∫

( ) ( ) ( )

4 5

0

2 33 4 5 2

4

15

64 1 323 2 5 22515

b

Z ZZ

ah b bb b b abhb

+

= − + =

PROBLEM 5.128 CONTINUED

Now 24 32:

9 225ELyV y dV y abh abh = =

8or25

y h=

PROBLEM 5.129 Locate the centroid of the section shown, which was cut from a circular cylinder by an inclined plane.

SOLUTION

First note that symmetry implies 0x =

Choose as the element of volume a vertical slice of width 2x, thickness dz, and height y. Then 12 , ,2EL ELdV xy dz y y z z= = =

Now 2 2x a z= − and 12 2 2h h h zy z

a a = − = −

So 2 2 1 zdV h a z dza

= − −

( )

( ) ( )

3/22 2 2 2 2 1 2 20

2 1 1

2

1 1Then 1 sin2 3

1 sin 1 sin 12

2

aa

a

z zV h a z dz h z a z a a za a a

a h

a hπ

− −

= − − = − + + −

= − −

=

PROBLEM 5.129 CONTINUED

Then 2 21 1 12 2

aEL a

h z zy dV h a z dza a−

= × − − −

∫ ∫

2 2

2 221 2

4aa

h z za z dza a−

= − − +

( )32

22 2 2 1 2 21 2sin

4 2 3h zz a z a a z

a a− = − + + −

( )32

2 42 2 2 2 1

21 sin

4 8 8

a

a

z a z a za z a zaa

+ − − + − +

( ) ( )2 2

1 15 sin 1 sin 132h a − − = − −

Then ( )2 2 25:

2 32ELa h ayV y dV y hπ π

= =

5or

16y h=

and 2 2 1a

EL azz dV z h a z dza−

= − −

∫ ∫

( ) ( )3 32 2

2 42 2 2 2 2 2 11 1 sin

3 4 8 8

a

a

z a z a zh a z a z a za a

= − − − − − + − +

( ) ( )3

1 1sin 1 sin 18

a h − − = − − −

2 3

:2 8ELa h a hzV z dV z π π

= = −

or 4az = −

PROBLEM 5.130 Locate the centroid of the plane area shown.

SOLUTION

2, inA , in.x , in.y 3, inxA 3, inyA

1 14 20 280× = 7 10 1960 2800

2 ( )24 16π π− = − 6 12 301.59− 603.19−

Σ 229.73 1658.41 2196.8

( )2 3

Then

229.73 in 1658.41 in

X A xA

X

Σ = Σ

=

or 7.22 in.X =

( )2 3

and

229.73 in 2196.8 in

Y A yA

Y

Σ = Σ

=

or 9.56 in.Y =

PROBLEM 5.131 For the area shown, determine the ratio a/b for which .x y=

SOLUTION

A x y xA yA

1 23

ab 38

a 35

b 2

4a b

225ab

2 12

ab− 13

a 23

b 2

6a b

− 2

3ab

Σ 16

ab 2

12a b

2

15ab

2

Then

16 12

X A xA

a bX ab

Σ = Σ

=

or 12

X a=

216 15

Y A yA

abY ab

Σ = Σ

=

or 25

Y b=

Now 1 22 5

X Y a b= ⇒ =

4or5

ab=

PROBLEM 5.132 Locate the centroid of the plane area shown.

SOLUTION

Dimensions in mm

2, mmA , mmx , mmy 3, mmxA 3, mmyA 1 126 54 6804× = 9 27 61 236 183 708

2 1 126 30 18902× × = 30 64 56 700 120 960

3 1 72 17282× × = 48 16− 82 944 27 648−

Σ 10 422 200 880 277 020

( )2 2

Then

10 422 m 200 880 mm

X A xA

X

Σ = Σ

=

or 19.27 mmX =

( )2 3

and

10 422 m 270 020 mm

Y A yA

Y

Σ = Σ

=

or 26.6 mmY =

PROBLEM 5.133 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION

By observation

1

1

hy x ha

xha

= − +

= −

For 2:y At ( )0, : 1 0 orx y h h k k h= = = − =

( )22

1At , 0: 0 1 or x a y h ca Ca

= = = − =

2

2 2Then 1 xy ha

= −

( )2

2 1 2

2

2

Now 1 1x xdA y y dx h dxaa

x xh dxa a

= − = − − −

= −

( )2

1 2 2

2

2

1 1 12 2

22

EL ELh x xx x y y y

a a

h x xa a

= = − = − + −

= − −

2 2 3

2 200

Then2 3

16

aa x x x xA dA h dx h

a aa a

ah

= = − = −

=

∫ ∫

2 3 4

2 200

2

and 3 4

112

aa

ELx x x xx dA x h dx ha aa a

a h

= − = −

=

∫ ∫

PROBLEM 5.133 CONTINUED

2 2

2 20

2 2 4

2 40

2 2 3 52

2 40

22

2 32

12 105

aEL

a

a

h x x x xy dA h dxa aa a

h x x x dxa a a

h x x x aha a a

= − − −

= − +

= − + =

∫ ∫

21 1:

6 12ELxA x dA x ah a h = = ∫ 12

x a=

21 1:

6 10ELyA y dA y ah a h = = ∫ 35

y h=

PROBLEM 5.134 Member ABCDE is a component of a mobile and is formed from a single piece of aluminum tubing. Knowing that the member is supported at C and that l = 2 m, determine the distance d so that portion BCD of the member is horizontal.

SOLUTION

First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus, 0X =

So that 0xLΣ =

( )

( ) ( )

( ) ( )

0.75Then cos55 m 0.75 m2

0.75 m 1.5 m

11.5 m 2 m cos55 2 m 02

d

d

d

− − ° ×

+ − ×

+ − − × × ° × =

( ) ( ) ( )( )21or 0.75 1.5 2 0.75 2 cos 55 0.75 1.5 32

d + + = − ° + +

or 0.739 md =

PROBLEM 5.135 A cylindrical hole is drilled through the center of a steel ball bearing shown here in cross section. Denoting the length of the hole by L, show that the volume of the steel remaining is equal to the volume of a sphere of diameter L.

SOLUTION Calculate volumes by rotating cross sections about a line and using Theorem II of Pappus-Guldinus

For the sector: 22 sin

3AARy A Rα αα

= =

22 2 2 2 21 2 1For the triangle: 4 4 ,

2 2 3 3AALh R R L y h R L = − = − = = −

( )( )12

A L h=

2 21 4R4

L L= −

Using Theorem II of Pappus-Guldinus

( ) ( )

( )

( )

ball 1 21 2

2 2 2 2 2

3 2 2

2 2

2 sin 1 12 4 43 3 4

22 sin 43 12

AA AAV y A y A

R R R L L R L

LR R L

π π

απ αα

π α

= −

= − − −

= − −

Now sin2LR α =

2 2 3

3

2 1 1Then 23 2 3 12

6

LV R LR L

L

π

π

= − +

=

Note sphere43

V rπ= where r is the radius

If , then2Lr =

33

sphere43 2 6

LV Lππ = =

Therefore, 3

ball sphere Q.E.D.6

V V Lπ= =

PROBLEM 5.136 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLUTION

( )( )

( )( )

I

II

( ) Have 4 m 200 N/m 800 N

2 4 m 600 N/m 1600 N3

a R

R

= =

= =

I IIThen :

or 800 1600 2400 NyF R R R

R

Σ − = − −

= + =

( ) ( ) ( )and : 2400 2 800 2.5 1600AM XΣ − = − −

or 7 m3

X =

2400 N∴ =R , 2.33 mX = (b) Reactions

0: 0x xF AΣ = =

( ) ( )70: 4 m m 2400 N 03A yM B Σ = − =

or 1400 NyB =

0: 1400 N 2400 N 0y yF AΣ = + − =

or 1000 NyA =

1000 N∴ =A , 1400 N=B

PROBLEM 5.137 Determine the reactions at the beam supports for the given loading.

SOLUTION

( )( )

( )( )

( )( )

I

II

III

1Have 3 ft 480 lb/ft 720 lb2

1 6 ft 600 lb/ft 1800 lb2

lb2 ft 600 lb/ft 1200

R

R

R

= =

= =

= =

Then 0: 0x xF BΣ = =

( )( ) ( )( )0: 2 ft 720 lb 4 ft 1800 lbBMΣ = −

( ) ( )( )6 ft 7 ft 1200 lb 0Cy+ − =

or C 23601by = 2360 lb=C

0: 720 lb 1800 lb 2360 lb 1200 lb 0y yF BΣ = − + − + − =

or 1360 lbyB = 1360 lb=B

PROBLEM 5.138 Locate the center of gravity of the sheet-metal form shown.

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area.

( )

( )

( )

I

I

III

1 1.2 0.4 m3

1 3.6 1.2 m3

4 1.8 2.4 m3

y

z

xπ π

= − = −

= =

= − = −

2, mA , mx , my , mz 3, mxA 3, myA 3, mzA

I ( ) ( )1 3.6 1.2 2.162

= 1.5 −0.4 1.2 3.24 −0.864 2.592

II ( ) ( )3.6 1.7 6.12= 0.75 0.4 1.8 4.59 2.448 11.016

III ( )21.8 5.08942π

= 2.4π

− 0.8 1.8 −3.888 4.0715 9.1609

Σ 13.3694 3.942 5.6555 22.769

( )2 3Have : 13.3694 m 3.942 mX V xV XΣ = Σ =

or 0.295 mX =

( )2 3: 13.3694 m 5.6555 mY V yV YΣ = Σ =

or 0.423 mY =

( )2 3: 13.3694 m 22.769 mZ V zV ZΣ = Σ =

or 1.703 mZ =

PROBLEM 5.139 The composite body shown is formed by removing a semiellipsoid of revolution of semimajor axis h and semiminor axis 2

a from a hemisphere of radius a. Determine (a) the y coordinate of the centroid when h = a/2, (b) the ratio h/a for which 0.4 .y a= −

SOLUTION

V y yV

Hemisphere 323

aπ 38

a− 414

aπ−

Semiellipsoid 2

22 13 2 6

a h a hπ π − = −

38

h− 2 2116

a hπ+

( ) ( )2 2 2 2Then 4 46 16

V a a h yV a a hπ πΣ = − Σ = − −

Now Y V yVΣ = Σ

so that ( ) ( )2 2 2 24 46 16

Y a a h a a hπ π − = − −

or 234 4 (1)

8h hY aa a

− = − −

(a) ?Y = when 2ah =

Substituting 12

ha= into Eq. (1)

21 3 14 4

2 8 2Y a

− = − −

or 45112

Y a= − 0.402Y a= −

PROBLEM 5.139 CONTINUED

( ) ?hba= when 0.4Y a= −

Substituting into Eq. (1)

( )230.4 4 4

8h ha aa a

− − = − −

or 2

3 3.2 0.8 0h ha a

− + =

Then ( ) ( )( )

( )

23.2 3.2 4 3 0.82 3

ha

± − −=

3.2 0.86±

=

2 2or and5 3

h ha a= =

PROBLEM 5.140 A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity.

SOLUTION

First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the corresponding line.

( ) ( )

1

1

26

26

2

0.3sin 60 0.15 3 m0.3cos60 0.15 m

0.6sin 30 sin 30

0.9 m

0.6sin 30 cos30

0.9 3 m

0.6 0.2 m3

xz

x

z

L

π

π

π

ππ π

= ° =

= ° =

°= °

=

°= °

=

= =

, mL , mx , my , mz 2, mxL 2, myL 2, mzL

1 1.0 0.15 3 0.4 0.15 0.25981 0.4 0.15

2 0 .2π 0.9π

0 0.9 3π

0.18 0 0.31177

3 0.8 0 0.4 0.6 0 0.32 0.48 4 0.6 0 0.8 0.3 0 0.48 0.18 ∑ 3.0283 0.43981 1.20 1.12177

Have ( ) 2: 3.0283 m 0.43981 mX L x L XΣ = Σ = or 0.1452 mX =

( ) 2: 3.0283 m 1.20 mY L yL YΣ = Σ = or 0.396 mY =

( ) 2: 3.0283 m 1.12177 mZ L z L ZΣ = Σ = or 0.370 mZ =

PROBLEM 5.141 Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

SOLUTION

First note that symmetry implies 0y = and 0z =

Have ( )2y k X h= −

( )2At 0, :x y a a k h= = = −

2or akh

=

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

2 , ELdV r dx X xπ= =

Now ( )22ar x hh

= −

so that ( )2

44

adV x h dxh

π= −

( ) ( )2 2

4 54 40 0

2

Then5

15

hh a aV x h dx x hh h

a h

ππ

π

= − = −

=

( )

( )

24

40

25 4 2 3 3 2 4

4 0

26 5 2 4 3 3 4 2

40

2 2

and

4 6 4

1 4 3 4 16 5 2 3 2

130

hEL

h

h

ax dV x x h dxh

a x hx h x h x h x dxh

a x hx h x h x h xh

a h

π

π

π

π

= −

= − + − +

= − + − +

=

∫ ∫

2 2 2Now :5 30ELxV x dV x a h a hπ π = =

1or6

x h=