capitulo 5 estatica

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

240 1344 1584

Then3

21344 in240 in

xAXA

or 5.60 in.X

and3

21584 in240 in

yAYA

or 6.60 in.Y


SOLUTION

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

11 60 75 22502

40 25 90 000 56 250

2 105 75 7875 112.5 37.5 885 900 295 300

10 125 975 900 351 600

Then3

2975 900 mm10 125 mm

xAXA

or 96.4 mmX

and3

2351 600 mm10 125 mm

yAYA

or 34.7 mmY


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

11 24 10 1202

2 10 6.6673

800

21 24 16 1922

110 16 15.3333

2944

312 3744

Then3

23744 in312 in

xAXA

or 12.00 in.X


SOLUTION


1 21 22 462 1.5 11 693 5082

21 6 9 272

6 2 162 54

31 6 12 362

8 2 288 72

399 567 4956

Then3

2567 mm399 mm

xAXA

or 1.421 mmX

and 3

24956 mm399 mm

yAYA

or 12.42 mmY


SOLUTION


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

2260

5654.92

94.5 120 534 600 678 600

18 345 905 400 2 201 400

Then3

2905 400 mm18 345 mm

xAXA

or 49.4 mmX

and3

22 201 400 mm

18 345 mmyAYA

or 93.8 mmY


SOLUTION


129

63.6174

4 93.8917

33.8917 243 243

21 15 9 67.52

5 3 337.5 202.5

131.1 94.5 445.5

Then3

294.5 in

131.1 inxAXA

or 0.721 in.X

and3

2445.5 in131.1 in

yAYA

or 3.40 in.Y


SOLUTION

First note that symmetry implies X Y

2, mmA , mmx 3, mmxA

1 40 40 1600 20 32 000

22(40) 1257

416.98 21 330

343 10 667

Then3

210 667 mm

343 mmxAXA

or 31.1 mmX

and 31.1 mmY X


SOLUTION

First note that symmetry implies 0X

2, inA , in.y 3, inyA

124

25.132

1.6977 42.67

226

56.552

2.546 144

31.42 101.33

Then3

2101.33 in31.42 in

yAYA

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

12

12r 14

3r 3

123

r

22

22r 24

3r 3

223

r

2 22 12

r r 3 32 1

23

r r

Then Y A y A

or 2 2 3 31 2 1 2 1

3 24 2 3

r r r r r

2 32 2

1 1

9 1 116

r rr r

Let 2

1

rpr

29 ( 1)( 1) ( 1)( 1)16

p p p p p

or 216 (16 9 ) (16 9 ) 0p p

PROBLEM 5.9 CONTINUED

Then2(16 9 ) (16 9 ) 4(16)(16 9 )

2(16)p

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 r

22

2 cos3

r

Similarly 21 1 1 12

2

2 cos 3

y r A r

2 22 2 1 12 2

2 2

3 32 1

2 cos 2 cosThen3 32 cos3

yA r r r r

r r

2 22 1

2 22 1

and 2 2

2

A r r

r r

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


Using Figure 5.8B,Y of an arc of radius 1 21 is2

r r

21 2

2

sin12

Y r r

1 22

1 cos( )2

r r (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 r

r r r rr rr r r r

r 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 r3 3

2 12 2

2 1

1 2

323 1 ( )2 2

r r rr r

r r

so that 1 22

2 3 cos3 4

Y r r or 1 22

1 cos2

Y r r

Which agrees with Eq. (1).


SOLUTION

First note that symmetry implies 0X

2 2 2 in., 45r

42

4

2 2 2 sin2 sin 1.6977 in.3 3

ry


11 4 3 62 1 6

22

2 2 6.2834

2 0.3024y 1.8997

31 4 2 42

0.6667 2.667

8.283 5.2330

Then Y A yA

2 38.283 in 5.2330 inY or 0.632 in.Y


SOLUTION


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

240 60

21214

10 15 6750 10 125

31 30 45 6752 25.47 19.099 54 000 40 500

6396 101 250 21 375

Then XA xA

2 36396 mm 101 250 mmX or 15.83 mmX

and YA yA

2 36396 mm 21 375 mmY or 3.34 mmY


SOLUTION


12 40 80 21333 48 15 102 400 32 000

21 40 80 16002

53.33 13.333 85 330 21 330

533.3 17 067 10 667

Then X A XA

2 3533.3 mm 17 067 mmX or 32.0 mmX

and Y A yA

2 3533.3 mm 10 667 mmY or 20.0 mmY


SOLUTION


12 150 240 24 0003

56.25 96 1 350 000 2 304 000

21 150 120 90002

50 40 450 000 360 000

15 000 900 000 1 944 000

Then X A xA

2 315 000 mm 900 000 mmX or 60.0 mmX

and Y A yA

215 000 mm 1 944 000Y or 129.6 mmY


SOLUTION


11

10 15 503

4.5 7.5 225 375

2215 176.71

46.366 16.366 1125 2892

226.71 1350 3267

Then X A x A

2 3226.71 in 1350 inX or 5.95 in.X

and Y A y A

2 3226.71 in 3267 inY or 14.41 in.Y


SOLUTION


12 8 8 42.673

3 2.8 128 119.47

22 4 2 5.3333

1.5 0.8 8 4.267

37.33 120 123.73

Then X A x A

2 337.33 in 120 inX or 3.21 in.X

and Y A y A

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

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 2

0x 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

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

32 000 4 640 000

Then Y A y A

2 332 000 mm 4 640 000 mmY

or 145.0 mmY


I I

6 3

:155 115 (160 155) 80 115

2 2 1.393 10 mm

A Q yA

II II

6 3

:145 85 160 145 80 85

2 2 1.393 10 mm

A Q yA

area I II Q 0x

Q Q

Which is expected since xQ yA yA 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 . (b) For what value of is

xQ maximum, and what is the maximum value?

SOLUTION

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

23 2 2 2

2 32

3 2

sin 1sin 2 cos sin2

2 cos cos sin3

x

rQ r r r r

r

or 3 32 cos3xQ r

( )b By observation, is maximum when xQ 0

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 BB A

x A

QF F

Q

Now xQ yA

So 30.3757.5 in. in. 10 in. 0.375 in. 28.82 in2x A

Q

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 02 12 16 6 102 723 24 4 12 96 2884 6 8 9 48 545 8 4 6 32 486 6 0 3 0 18

72 336 480

Then X L x L

272 in. 336 inX or 4.67 in.X

and Y L y L

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


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 441.05 40 680 14 289

Then X L x L

2(441.05 mm) 40 680 mmX or 92.2 mmX

and Y L y L

2(441.05 mm) 14 289 mmY 32.4 mmY


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

77.233 111.32 982.7

Then X L x L

2(77.233 mm) 111.32 mmX or 1.441 mmX

and Y L y L

2(77.233 mm) 982.7 mmY or 12.72 mmY


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 6 2 63.820 72

3 2 0 0

4 4 2 42.546 32

35.416 104

Then Y L y L

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 30

First note, from Figure 5.8B:/6

X

1.5 m

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

T

or 1.4698 NBCT or 1.470 NBCT

( ) 0: 1.4698 N cos60 0x xb F A

or 0.7349 NxA

0: 7.358 N 1.4698 N sin 60 0y yF A

or 6.085 NyA thus 6.13 NA 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 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 x or 0xL

Hence

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

6 in.8 in. cos 6 in. 02

Then 4cos9

or 63.6

PROBLEM 5.27 The homogeneous wire ABCD is bent as shown and is supported by a pin at B. Knowing that 30 , determine the length l for which portion CDof 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 x or 0i ix L

Hence 2 6 in.

cos30 6 in. sin 30 6 in.

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.lTaking 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 x

or 0i ix L

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

2 2or

a ha ab kb a hy

ba kb a h

2 2112 (1 )

a k kha k kh

Let 1 and hc ka

Then 2

2a c ky

c k (1)

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

2

2

20

2

k c k k c kdy ad c k

or 22 0c k c k (2)


Expanding (2) 2 22 2 0c c k or 2 2 0k c c

Then22 2 4

2c c k c

k

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

22 1 4 1 4 12

k k k kh a

k

(a) 0.2: k22 1 0.2 4 1 0.2 4 0.2 1 0.2

2 0.2h a or 0.472h a

(b) 0.6: k22 1 0.6 4 1 0.6 4 0.6 1 0.6

2 0.6h 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 that minimizes h.

Then from Eq. (2)

We see 2

2 c kc k

(3)

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

We obtain 22ay

Butha

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 x

a

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


SOLUTION For the element (EL) shown

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

Then 1/31/3ax y

h

1/31/3

Now dA xdya y dy

h

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 ah

h 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 h

h h h

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

0

3 37 7

hh

ELa ay dA y y dy y ah

h 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


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

33

hy xa

and on line 2

1/31/3hy x

a

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 1 124 4

aa h hA dA x x dx h x x ah

a 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 h

a 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 62

2/3 6 5/3 600

3 1 82 2 5 7 35

aah x x h x xdx ah

a a a a

From 28:2 35EL

ahxA x dA x a h or 1635

x a

and 28:2 35EL

ahyA y dA y ah or1635

y 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 r

Then2

2

11

22 2

2 12 2

rrr

r

rA dA rd r r r

and2

2

11

3 3 32 1

2 1 223 3

rr

EL rr

ry dA rd r r r r

So 2 2 3 32 1 2 1

2:2 3ELyA y dA y r r r r

or3 32 12 2

2 1

43

r ryr r


SOLUTION First note that symmetry implies 0x

For the element (EL) shown

cos , siny R x R

cos dx R d2 2cosdA ydx R d

Hence

2 2 2 20

0

sin 2 12 cos 2 2 sin 22 4 2

A dA R d R R

2 2 3 20

0

32

1 22 cos cos cos sin sin2 3 3

cos sin 2sin3

ELRy dA R d R

R

But soELyA y dA

32

2

cos sin 2sin3

2 sin 22

R

yR

or2cos 22 sin

3 2 sin 2y R

Alternatively, 22 3 sinsin

3 2 sin 2y R


SOLUTIONFor the element (EL) shown

2 2by a x

a

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 dx

a

To integrate, let 2 2sin : a cos , cosx a x a dx a d

/20

/22 2

0

Then cos cos

2sin sin 12 4 4

bA a a a da

b a a aba

/23/22 2 2 2 2

00

3

1and 2 3

16

aEL

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

a b

2 2 2 20

2 2 32 2

2 200

2

13 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 b or

23 4

ax

21: 14 6ELyA y dA y ab ab or 2

3 4by

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

SOLUTIONFor 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

2by 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

39175

y b


SOLUTIONAt 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 22

bdA 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


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

Then22 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

54

x L

21 11:8 5ELyA y dA y a a 33

40y a


SOLUTIONFor 1 at , 2y x a y b 22b ka or 2

2bka

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


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


SOLUTION2For y at , :x a y b 2a kb or 2

akb

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 12 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

1324

ab


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 13 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 12 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 2 322

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.

SOLUTIONFirst 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

a

a

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.


Now cosELx r and dL rd

Then 7 /47 /4/4 /4

32

L dL rd r r

and 7 /4/4 cosELx dL r rd

7 /42 2 2/4

1 1sin 22 2

r r r

23Thus : 22

xL xdL x r r2 23

x r

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


Now 3cosELx a and 2 2dL dx dy

Where 3 2cos : 3 cos sinx a dx a d

3 2sin : 3 sin cosy a dy a d

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 a/ 2

2

0

3 2

a

/ 2 30

/22 5 2

0

and cos 3 cos sin

1 33 cos5 5

ELx dL a a d

a a

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

5x a


Alternative solution 2/3

3 2

2/33 2

cos cos

sin sin

xx aa

yy aa

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 a

x

and1/3

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


SOLUTION2 2Have cos cos3 32 2sin sin3 3

EL

EL

x r ae

y r ae

2 21 1and 2 2

dA r rd a e d

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 a

and 2 2 3 30 0

2 1 1cos cos3 2 3ELx dA ae a e d a e d

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

u e du e ddv d v

Then 3 3 3cos sin sin 3e d e e d

Now let 3 3 then 3u e du e d

sin , then cosdv d v

Then 3 3 3 3sin sin 3 cos cos 3e d e e e d

So that 3

3 cos sin 3cos10ee d

3 33 3 3

0

1 sin 3cos 3 3 1239.263 10 30EL

e ax dA a e a

Also 2 2 3 30 0

2 1 1sin sin3 2 3ELy dA ae a e d a e d


Using integration by parts, as above, with 3 3 and 3u e du e d

sin and cosdv d v

Then 3 3 3sin cos cos 3e d e e d

So that 3

3 sin cos 3sin10ee d

3 33 3 3

0

1 cos 3sin 1 413.093 10 30EL

e ay dA a e a

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


SOLUTION

Have 1, sin2EL EL

xx x y xL

and dA ydx

/22 2/2

2 200

sin sin cosL

L x L x L x LA dA x dx xL L L

and /20 sinL

ELxx x dA x x dx

L

/22 3 3 32

2 3 2 30

2 2sin cos sin 2L

L x L x L x L Lx xL L L

Also /20

1 sin sin2

LEL

x xy y dA x x dxL L

/22 3

2 30

1 2 2sin cos2

LL x L L xx x

L L

3 2 32

2 21 1 1 62 6 8 24 96

L L L L


Hence2

32 2 3

1:ELL zxA x dA x L

or 0.363x L

2 3

2 2 2 31 2:

96ELL LyA y dA y

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: 2

area 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 2

lineArea 2 2 32.4 mm 441.05 mm 89.786 10 mmY L

3 289.8 10 mmA

6 3areaVolume 2 2 34.7 mm 10125 mm 2.2075 10 mmY A

6 32.21 10 mmV

(b) Rotation about 165 mm:x5 2

lineArea 2 165 2 165 92.2 mm 441.05 mm 2.01774 10 mmX L

6 20.202 10 mmA

6 3areaVolume 2 165 2 165 96.4 mm 10 125 mm 4.3641 10 mmX A

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


area area399 mm , 1.421 mm, 12.42 mm (Area)A X Y


line line77.233 mm, 1.441 mm, 12.72 mm (Line)L X Y


(a) Rotation about the line 22 mm:y2

lineArea 2 22 2 22 12.72 mm 77.233 mm 4503 mmY L

3 24.50 10 mmA

2 3areaVolume 2 22 2 22 12.42 mm 399 mm 24 016.97 mmY A

3 324.0 10 mmV

(b) Rotation about line 12 mm:x2

lineArea 2 12 2 12 1.441 mm 77.233 mm 5124.45 mmX L

3 25.12 10 mmA

2 3Volume 2 12 1.421 2 12 1.421 mm 399 mm 26 521.46 mmA

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


area area240 in , 5.60 in., 6.60 in. (Area)A X Y


line line72 in., 4.67 in., 6.67 in.L X Y


( ) Rotation about the axis:a x2

line2 2 6.67 in. 72 in. 3017.4 inxA Y L23020 inA

2 3area2 2 6.60 in. 240 in 9952.6 inxV Y A

39950 inV

( ) Rotation about 16 in.:b x2

16 line2 16 2 16 4.67 in. 72 in. 5125.6 inxA X L2

16 5130 inxA

2 316 area2 16 2 16 5.60 in. 240 in 15 682.8 inxV X A

3 316 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 b 2 2V a b

(b) Rotation about axis :BB

2 2Volume 2 2 2 22abyA a a b 2 22V a b

(c) Rotation about y-axis:

24 2Volume 2 23 2 3

a abyA a b 223

V a b

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 d

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 A

2

3

2 10 in. 3 in. 2 in.4

122.049 in

3122.0 inV

side endArea 2 2 2 2 2A A CL RC L R C

2 10 in. 3 in. 4 in.

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 23 4

aV yA a

32or3

V a

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

Have 42 23 4

aV yA ha

22or3

V a h

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

Have 3 22 28 3

V yA a ah

21or2

V a h

(4) Cone: the generating area is a triangle

Have 12 23 2aV yA ha

21or3

V a h

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 xA

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 yL

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 cos20

60 2.5 mm 12.5 mm

CA y L y L

3 2or 3.24 10 mmCA

Have 1 12CA y L

7.5 7.5 mm2 60 1.6 mm2 cos20

3 2or 2.74 10 mmCA

Have 1 12 560 mm 5 mmCA y L

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

3Since 1gal 231 in

3

33526.03 in 15.26 gal231in /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 At

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 L

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

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

Then3 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

Applying the second theorem of Pappus-Guldinus to parts 3 and 4 2 2

top

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

6 3waste

3 3

9.0750 8.8671 10 mm

0.2079 10 m

V

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 N

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

surface tops coats

3 22

Then # liters Coverage

1 liter617.115 10 m 5000 312 m

A N N

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 Vwhere 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

Then 2 26 9.5941 16.4059 or 8.20295 0.143169 rad

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 yA



11 3.0755 1.5 2.30662

1 1.5 0.53

1.1533

2 23 1.288512 3 sin

sin 0.609213

–0.78497

31 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 inyA

Then 3 32 0.44296 in 1.4476 inV

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 2 radians about the x axis. Applying the first theorem of Pappus-Guldinus, we have

2A yL22Now, 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) 0.15mm

k

20.15 and 0.3dxx y ydy

22Now 1 1 0.09dxdL dy y dy

dy

So 2 andA yL yL ydL

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

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 x

or 10.80 ftx

1800 lb10.80 ft

Rx

( )b 0: 0x xF A

0: 1800 lb 0, 1800 lby y yF A A 1800 lbA

0: 1800 lb 10.8 ft 0A AM M

19.444 lb ftAM or 19.44 kip ftAM


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 R

or 1800 N 1800 N 3600 NR

: 3600 N 3 m 1800 N 4.5 m 1800 NAM x

or 3.75 mx

3600 NR3.75 mx

(b) Reactions

0: 0x xF A

0: 6 m 3600 N 3.75 m 0A yM B

or 2250 NyB 2250 NB

0: 2250 N 3600 Ny yF A

or 1350 NyA 1350 NA

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 A

0: 2 ft 400 lb 4 ft 600 lb 12 ft 800 lb 10 ft 0A yM B

or 800 lbyB 800 lbB

0: 800 lb 400 lb 600 lb 800 0y yF A

or 1000 lbyA 1000 lbA


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 A

0: 4.5 ft 1800 lb 10 ft 300 lb 9 ft 0A yM B

or 1233.3 lbyB 1233 lbB

0: 1800 lb 300 lb 1233.3 lb 0y yF A

or 866.7 lbyA 867 lbA


SOLUTION

I1Have 200 N/m 0.12 m 12 N2

R

II 200 N/m 0.2 m 40 NR

Then 0: 0x xF A

0: 18 N 12 N 40 N 0y yF A

or 34 NyA 34.0 NA

0: 0.8 m 12 N 0.22 m 40 N 0.38 m 18 NA AM M

or 2.92 N mAM 2.92 N mAM


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 A

0: 3 m 2.1 m 3600 N 2.4 m 4050 NB yM A

or 270 NyA 270 NA

0: 270 N 3600 N 4050 N 0y yF B

or 720 NyB 720 NB


SOLUTION

I1Have 4 m 2000 kN/m 2667 N3

R

II1 2 m 1000 kN/m 666.7 N3

R

Then 0: 0x xF A

0: 2667 N 666.7 N 0y yF A

or 3334 NyA 3.33 kNA

0: 1 m 2667 N 5.5 m 666.7 NA AM M

or 6334 N mAM 6.33 kN mAM


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 A

0: 11 5 ft 800 lb 4 ft 3200 lb 0AM B

or 800 lbB

0: 3200 lb 800 lb 800 lb 0y yF A

or 1600 lbA

PROBLEM 5.69 Determine (a) the distance a so that the vertical reactions at supports Aand 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 B

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

or 2140 10803 9yA a a (2)

Equating Eqs. (1) and (2)

2140 10120 20 803 9

a a a

or 240 320 480 03

a a

Then 1.6077 ft, 22.392a a

Now 12 fta 1.608 fta

( ) Haveb 0: 0x xF A

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

152.2 lbA 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

or 210 20 1609 3yB a a (1)

Then 20 20 09 3

ydBa

daor 3.00 fta

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

150 lb 150.0 lbB

and 0: 0x xF A

0: 60 3.00 lb 240 20 3.00 lb 150 lb 0y yF A

or 210 lbyA 210 lbA

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 C

0: 50 kN m 1 m 9 kN 2.5 m 13.5 kN 6 m 0B yM C

or 15.4583 kNyC 15.46 kNC

0: 9 kN 13.5 kN 15.4583 0y yF B

or 7.0417 kNyB 7.04 kNB

PROBLEM 5.72 Determine (a) the distributed load 0w at the end D of the beam ABCDfor 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

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 C

0: 4.5 3.5 3.19 kN 9 3.19 kN 0y yF C

or 1.375 kNyC 1.375 kNC

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 Pã 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

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 R

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 ã

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 ã 62.4 lb/ft3 for the specific weight of fresh water and c ã 150 lb/ft3 for the specific weight of concrete if U.S. customary units are used. With SI units, use ã 103 kg/m3 for the density of fresh water and c ã 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

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 P

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 ã 62.4 lb/ft3 for the specific weight of fresh water and c ã 150 lb/ft3 for the specific weight of concrete if U.S. customary units are used. With SI units, use ã 103 kg/m3 for the density of fresh water and c ã 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 gV

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 H

or 1589 kNH 1589 kNH

0: 1589 kN 2543 kN 1271 kN 529.7 kNyF V

or 5933 kNV 5.93 MNV


( ) Haveb 0: 5933 kN 6 m 1589 kNAM X

6 m 1589 kN 12 m 2543 kN

17 m 1271 kN 19 m 529.7 0or 10.48 mX 10.48 mX

to the right of A

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

Have 0F

Then 1675 kNR 18.43

or 1675 kNR 18.43

Alternative solution to part (c)

Consider the face BC of the dam.

2 2Have 6 18 18.9737 mBC

6tan 18.4318

3 2

2

and 1000 kg/m 9.81 m/s 18 m

176.6 kN/m

p g h

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

1675 kN

R Ap

1675 kNR 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


Now W V

31

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

W

32 150 lb/ft 6 ft 24 ft 1ft 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 H

or 10.11 kipsH

0: 38,400 lb 21,600 lb 10,800 lb 8995.6 lb 0yF V

or 79,785.6V 79.8 kipsV

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

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 0F

13.53 kips41.6

R

On the face BD of the dam

13.53 kipsR 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 d

Now 0: 9 ft 03AdM T P

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 3max216 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 d

31 8 ft 12 ft 80 lb/ft 8 ft 30,720 lb2

Then 0: 0y yF A

80: 9 ft ft 30,720 lb 03AM T

or 9102.22 lbT

9.10 kipsT

0: 30,720 lb 9102.22 lb 0x xF A

or 21,618 lbA

21.6 kipsA

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 T

or 20.6 kNT

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/ms 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 gh

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 II184.97 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/ms ) 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


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 10year

N

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 gh

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 F

1 2or 26.99 kN 51.74 kN 03 3 BF

or 43.49 kNBF

4.35 kNBF 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 h

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 B

or 1014 lbB

1014 lbB

4 40: 2 421.2 lb 702 lb 05 5x xF A

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 A

or 340.08 lbyA

563 lbA 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 0B 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 T

1.0 ft 702 lb 0

or 608.4 lbBCT

Case (2) 0: 1.2 ft 2 0.5 ft 702 lbB xM A

1.0 ft 421.2 lb 0

or 2 643.5 lbxA

40: 643.5 lb 421.2 lb5x BCF T

4 702 lb 05

or 255.06 lbBCT

min255 lbBCT

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

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 h

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 2 cos30 m2

P d

29.43 0.2679 kNd


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

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

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 h

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 1ft 187.2 lb3AM D

2 12 ft 93.6 lb 2 ft 468 lb 03 3

or 11.325 lbD 11.33 lbD


0: 11.32 23.4 93.6 468 0x xF A

or 596.32 lbxA

0: 240 240 240 187.2 0y yF A

or 532.8 lbyA 800 lbA 41.8

5.88 At 2 322 ft, 2 lb/ft where 62.4 lb/ftdh d p d

2ft, lb/ftdh d p d

231 1 2

1 1 3Then 2 ft 3 ft lb/ft 2 ft 2 lb2 2 2dP A p d d d

(Note: For simplicity, the numerical value of the density 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 d

3 3 21 1 2 ft 3 ft 2 ft 3 2 lb2 2dP A p d d

4 41 1 2 ft 3 ft ft 3 lb 3 2 6 lb2 2dP A p d d d

As the gate begins to open, 0D

21 30: 2 ft 240 lb 1 ft 240 lb 2 ft 2 ft 2 lb3 2AM d d

21 ft 6 2 lb 2 ft 3 2 lb3

d d

1 2 ft 3 2 lb 6 lb 03

d

3 21 720or 2 3 2 12 2 42

d d d

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 gh

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

18.84 N

wW gV

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

37.67 N

Now 0: 0wF R P W

2 2So that , tan ww

WR P WP

42.12 N, 63.4 42.1 NR 63.4

(b) Consider the free-body diagram of a 0.6 m long section of water and gutter.

Then 0: 0x xF B

0: 37.67 N 0y yF B

or 37.67 NyB

0: 0.06 0.048 m 37.67 N 0B BM M

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 R R 42 R

2 323

r 38

r 414

r

332

3rR

442

8rR

Then

4 4

3 3

1813

R ryVYV R r

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

or4 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 213

a h 14

h 2 2112

a h

Cylinder 2

212 4a b a b

12

b 2 218

a b

2 4 312

a h b 2 2 22 324

a h b

Have Y V yV

2 2 2 2Then 4 3 2 312 24

Y a h b a h b

2 22 3or2 4 3

h 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 216

a h *a 31

6a h

Half-Cylinder2

2

2 2 8a b a b

4 23 2 3

a a 3112

a b

2 4 324

a h b 31 212

a h b

From Sample Problem 5.13

Have Z V zV

2 31Then 4 3 224 12

Z a h b a h b

2 2or4 3

a h bZh b

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 12

L 212

L ab

Pyramid 13 2

ba h 14

L h1 16 4

abh L h

21 1 1Then 36 6 4

V ab L h xV ab L h L h

Now so thatX V xV

2 21 1 136 6 4

X ab L h ab L hL h

or2

21 1 11 36 6 4

h 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 57104

X L 0.548X L

( ) ? whenhb X LL

Substituting into Eq. (1)2

21 1 11 36 6 4

h h hL LL L L

or2

21 1 1 116 2 6 24

h h hL L L

or2

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, inyVI (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) 0.2545 4.2 2.15 1.0688 0.54711

IV 2(0.5) (0.75) 0.5890 1.2 0.375 0.7068 0.2208913.3454 32.618 9.9922

5.94

Have X V x V3 413.3454 in 32.618 inX or 2.44 in. X

5.95

Have Y V yV

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

21 120 42 54 136 0802

40 32 5 443 200 4 354 560

3 2(27) (18) 65612

36160 9 3 534 114 185 508

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

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 or 64.2 mmX

5.97

3 4

Have

297 736 mm 5 809 460 mm

Y V yV

Y 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 mm

3

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

III1 20 124 102 126 4802

78 58 9 865 440 733 840

IV1 90 132 24 142 5602

156 170 22 239 360 24 235 200

1 673 520 156 586 080 8 785 584

5.98Have X V xV

3 41 673 520 mm 156 586 080 mmX or 93.6 mmX

5.99Have Z V zV

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

11 90 602 2700

30120 20 140

0 81 000 378 000 0

290 200

18 00045 60 80 810 000 1 080 000 1 440 000

345 100

450022.5 30 120 101 250 135 000 540 000

4245

21012.5

45 0 4 45

1603

179.1143 139 0 569 688

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

48.1 mmX

2 3

19 380.9 mm 1 323 000 mm

or 68.3 mm

Y A yA

Y

Y 68.3 mmY

2 3

19 380.9 mm 1 469 688 mm

or 75.8 mm

Z A zA

Z

Z 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

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 14 062.5V 61.36 10 0 19.85 613.6 0 1218.0

2263.64 46 074 10 313 65 720

Have X A xA

2 32263.64 mm 46 074 mmX or 20.4 mmX

2 3

2263.64 mm 10 313 mm

Y A yA

Y or 4.55 mmY

Z A zA

2 32263.64 mm 65 720 mmZ or 29.0 mmZ


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

12 6 3.8197 in.

111 4 1.5 10.363 in.3

y

z

x y

x

2, inA , in.x , in.y , in.z 3, inx A 3, iny A 3, inz AI 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

95.01 434.4 304 273.0

2 3

Have

95.01 in 434 in

X A x A

X or 4.57 in.X

Y A yA

2 395.01 in 304.0 inY or 3.20 in.Y

2 3

95.01 in 273.0 in

Z A z A

Z 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.Zand

8 2

7 3

6 5

A A

A A

A A

Thus2, 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

31 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

51 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

493.2 2239.5 3251.1

2 3Have : 493.2 in 2239.5 inX A xA X or 4.54 in.X

2 3 : 493.2 in 3251.1 inY A yA Y 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 0.1885 0 0.15 0 0.028274

2 0.2 0.1 0.03142

2 0.10.06366 0.25 0.02000 0.007854

3 20.1 0.015712

4 0.10.04244

3 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 0.15712

4 0.13

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 0.337080 0.023667 0.066991

Have 2 3: (0.337080 mm ) 0.023667 mmX A xA X

or 0.0702 mX 70.2 mmX

2 3: 0.337080 mm 0.066991 mmY A yA Y

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

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 75.39822

5.81408 5.81408 438.37 438.37

II 8 3 37.69912

10.9070 6.9070 411.18 260.39

III 4 3 12 8 14 96.0 168.0

IV22 16 402.1239

4 9.2094 9.2094 3703.32 3703.32

V22 8 100.5309

4 12.6047 8.6047 1267.16 865.04

VI 2 4 8 64 12 14 768.0 896.0

362.69 2613.71 2809.04

Have 2 3: 362.69 in 2613.71 inX A xA X

or 7.21 in.X

2 3: 362.69 in 2809.04 inZ A zA Z

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

2, mmA , mmy , mmz 3, mmyA 3, mmzAI (80)(500) 40 000 40 250 61.6 10 610 10II 196 350 292.2 212.2 657.4 10 641.67 10III (80)(680) 54 400 40 500 60.2176 10 627.2 10IV 534 071 398.3 318.3 6212.7 10 6170 10V (80)(500) 40 000 40 250 61.6 10 610 10VI 196 350 292.2 212.2 657.4 10

641.67 1061.061 10 6332.9 10

6300.5 10

Now, symmetry implies 340 mmXand 6 2 6 3: 1.061 10 mm 332.9 10 mmY A yA Y

or 314 mmY

6 2 6 3: 1.061 10 mm 300.5 10 mmZ A zA Z

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

1300 280

84 0000 140 0 11 760 000

2280 50

14 00025 140 350 000 1 960 000

3300 50

15 00025 0 375 000 0

4280 50

14 00025 140 350 000 1 960 000

52100

31 4160 130 0 4 084 070

6230

4 706.86

04 30

260 247.293

0 174 783

7230

4 706.86

0 4 30260 247.29

30 174 783

94 170 1 075 000 11 246 363

Have 2 3: 94 170 mm 1 075 000 mmY A yA Y

or 11.42 mmY2 3: 94 170 mm 11 246 363 mmZ A zA Z

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, iny L 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

415

223.56

3 2 155

5.73

30

9.549

24

7.639135.0 225.0 180.0

65.56 382.5 337.5 258.0

2Have : 65.56 in. 382.5 inX L x L X or 5.83 in.X

2: 65.56 in. 337.5 inY L y L Y or 5.15 in.Y

2: 65.56 in. 258.0 inZ L z L Z 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

, ftL , ftx , fty , ftz 2, ftx L 2, fty L 2, ftz L1 8.5440 4 1.5 0 34.176 12.816 0

2 4 16 016

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 X or 3.06 ftX2: 32.110 ft 17.316 ftY L y L Y or 0.539 ftY

2: 32.110 ft 96.0 ftZ L z L Z 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

, mL , mx , my , mz 2, mxL 2, myL 2, mzL1 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 11.827 6.210 14.681 4.8082

2Have : 11.827 m 6.210 mX L xL X or 0.525 mX

2: 11.827 m 14.681 mY L yL Y or 1.241 mY

2: 11.827 m 4.8082 mZ L zL Z 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 ã 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 4.53 1.2 0 3.7639 0 14.1897

4 1.22 1.2

0.7639 3 2.88 11.3097

5 0.6 2 1.20.7639 3.7639 1.44 7.0949

15.425 9.4112 41.594

2Have : 15.425 m 9.4112 mX L xL X or 0.610 mX2: 15.425 m 41.594 mY L yL Y 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. 2.1in. 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

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

, lbW , in.x , in lbxWI 0.009791 0.3125 0.003060 II 0.093996 2.1 0.197393 III 0.000846 2.7 0.002284IV 0.017987 4.5 0.080942 V 0.000428 7.4 0.003169

0.12136 0.28228

Have : 0.12136 lb 0.28228 in. lbX W xW X 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

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

Have Y W yW

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. 21in., 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

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


Then marker 12 in.yWYW

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 1201180 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

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 V


Have : 865.06 N 236 590 mm NX W xW X

or 274 mmX

: 865.06 N 89 671 mm NY W yW Y

or 103.6 mmY

, NW , mmx , mmy , mm Nx W , mm Ny W

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

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 x

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 dx

Component 1 /23

/2 2 2 21 0

0

3

3

1124

aa xV a x dx a x

a

/2 2 21 0

/22 42

0

4

and

2 4

764

aEL

a

x dV x a x dx

x xa

a

Now 3 41 1 11

11 7:24 64ELx V x dV x a a

121or88

x a

Component 2 3

2 2 22 /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


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

Now 3 42 2 EL 22

5 9:24 64

x V x dV x a a

227or40

x a

PROBLEM 5.117


A semiellipsoid of revolution.


2EL,dV r dx x x

The equation of the generating curve is 2 2

2 2 1x yh a

so that

22 2 2

2ar h xh

and then

22 2

2adV h x dxh

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

2/2 2 2

21 0

/22 2 42

20

2 2

and

2 4

764

hEL

h

ax dV x h x dxh

a x xhh

a h

Now 2 2 21 1 11

11 7:24 64ELx V x dV x a h a h

121or88

x h


Component 2 2 2 3

2 2 22 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

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

Now 2 2 22 2 22


227or40

x h

PROBLEM 5.118


A paraboloid of revolution.


2EL,dV r dx x x

The equation of the generating curve is 22

hx h ya

so that

22 ar h x

h and then

2adV h x dxh

Component 1 2

/21 0

/22 2

0

2

2

38

h

h

aV h x dxh

a xhxh

a h

2/2

1 0

/22 2 3

0

2 2

and

2 3

112

hEL

h

ax dV x h x dxh

a x xhh

a h

Now 2 2 21 1 11


12or9

x h


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

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

Now 2 2 22 2 22


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 x

Now 2

21 xr ba

so that

222

21 xdV b dxa

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

and 2 4

22 40

21aEL

x xx dV b x dxa a

2 4 62

2 40

22 4 6

ax x xb

a a

2 2 1 1 12 2 6

a b

2 216

a b


Then 2 2 28 1:15 16ELxV x dV x ab a b

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 0y0z


2 , ELdV r dx x x

Now 11rx

so that

2

2

11

2 11

dV dxx

dxxx

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

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 x

3 4Now : 0.46944 m 1.09861 mELxV x dV X

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

0zChoose as the element of volume a disk of radius r and thickness dx.Then

2 , ELdV r dy y y

Now 2

2 2 22

hx a ya

so that 2 2hr h a ya

Then2 2

2 22

hdV a a y dya

and2 2

2 220

a hV a a y dya

Let sin cosy a dy a d

2 2/2 2 2 2

2 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 2 cos 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

ah2


2 22 2

20

22 2 2 3

2 0

2 3/22 2 2 2 42

0

2 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

2 2 21Now : 0.09587012ELyV y dV y ah a h

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 0y0z


2EL,dV r dx x x

Now sin2

xr ba

so that 2 2sin2

xdV b dxa

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

and 2 2 2sin2

aEL a

xx dV x b dxa

Use integration by parts with

2

2

sin2

sin2

xa

a

xu x dVa

xdu dx V


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

2 22

2 2

3 14

0.64868

a b

a b2 2 21Now : 0.64868

2ELxV x dV x ab a b

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 0x0z

Choose as the element of volume a cylindrical shell of radius r and thickness dr.

Then 12 ,2ELdV r y dr y y

Now sin2

ry ba

so that 2 sin2

rdV br dra

Then 2 2 sin2

aa

rV br dra


sin2

2 cos2

ru r dv dra

a rdu dr va

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

2 2

2

2

2

4 42

18 1

5.4535

a aV b

a b

a b

2 12

22 2

Also sin 2 sin2 2

sin2

aEL a

aa

r ry dV b br dra a

rb r dra



2

2

sin2

sin2

ra

a

ru r dv dra

rdu dr v

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 22

2 2

3 14

2.0379

b

a b

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

222

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

2 22 3

2 200

2

1and3

13

hh b bV k h y dy k h y

h h

kb h

Also ELy y2 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

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 0xThe element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now

2EL

dA r Rd

ry

where sinr R2

EL

so that sin2 sin

dA R dRy

22 2 20 0

2

Then sin cosA R d R

R

2

2

20

3

0

3

2and sin sin

sin 222 4

2

ELRy dA R d

R

R

2 3Now :2ELyA y dA y R R

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

0zfor 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 Rd

and wall cosELy R/2/2 2 2

wall /6 /6

2

Then 2 sin 2 cos

3

A R d R

R

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

2base base

3By observation , 4 2

A R y R

Now y A yA

2 2 3 23 3or 34 4 4 2

y R R R R R

or 0.48763 350 mmy R R

170.7 mmy

(b) Punch First note that symmetry implies 0x

0zand 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 y


Now 2 2 2x y R so that 2 2dV R y dy0

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

00 2 2 2 2 4

3/23/2

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

3 43 15Now : 38 64ELyV y dV y R R

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 aAt 7200 mm, 0 : 100 mm

100 mm 60 7200

1180

x z yb

b

At 0 , 12 000 mm: 120 mm120 mm 60 mm 12000

1200

x z yc

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

0

9 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


12000 72000 0

720012000 2 3 20

0

12000 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 mEL

xV 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 dxdz

a 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


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 h


Then 2 21 1 12 2

aEL a

h z zy dV h a z dza a

2 22 2

21 24

aa

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 21 15 sin 1 sin 1

32h a

Then2 2 25:

2 32ELa h ayV y dV y h

5or 16

y h

and2 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

31 1sin 1 sin 1

8a h

2 3:

2 8ELa h a hzV z dV z

or 4az


SOLUTION


1 14 20 280 7 10 1960 2800

2 24 16 6 12 301.59 603.19

229.73 1658.41 2196.8

2 3

Then

229.73 in 1658.41 in

X A xA

X

or 7.22 in.X

2 3

and

229.73 in 2196.8 in

Y A yA

Y

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

123

ab 38

a 35

b2

4a b 22

5ab

212

ab 13

a 23

b2

6a b 2

3ab

16

ab2

12a b 2

15ab

2

Then

16 12

X A xA

a bX ab

or 12

X a

216 15

Y A yA

abY ab

or 25

Y b

Now 1 22 5

X Y a b

4or5

ab


SOLUTION

Dimensions in mm

2, mmA , mmx , mmy 3, mmxA 3, mmyA1 126 54 6804 9 27 61 236 183 708

21 126 30 18902

30 64 56 700 120 960

31 72 17282

48 16 82 944 27 648

10 422 200 880 277 020

2 2

Then

10 422 m 200 880 mm

X A xA

X

or 19.27 mmX

2 3

and

10 422 m 270 020 mm

Y A yA

Y

or 26.6 mmY


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


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 1

2x a

21 1:6 10ELyA y dA y ah a h 3

5y 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 Cand that l ã 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 0xL

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 sin3AA

Ry A R

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

Now sin2LR

2 2 3

3

2 1 1Then 23 2 3 12

6

LV R LR L

L

Note sphere43

V r where r is the radius

If , then2Lr

33

sphere43 2 6

LV L

Therefore, 3

ball sphere Q.E.D.6

V V L


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

Rand : 2400 2 800 2.5 1600AM X

or 7 m3

X

2400 NR , 2.33 mX(b) Reactions

0: 0x xF A

70: 4 m m 2400 N 0

3A yM B

or 1400 NyB

0: 1400 N 2400 N 0y yF A

or 1000 NyA

1000 NA , 1400 NB


SOLUTION

I

II

III

1Have 3 ft 480 lb/ft 720 lb21 6 ft 600 lb/ft 1800 lb2

lb2 ft 600 lb/ft 1200

R

R

R

Then 0: 0x xF B

0: 2 ft 720 lb 4 ft 1800 lbBM

6 ft 7 ft 1200 lb 0Cy

or C 23601by 2360 lbC

0: 720 lb 1800 lb 2360 lb 1200 lb 0y yF B

or 1360 lbyB 1360 lbB


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 2.4m

3

y

z

x

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

I1 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

III21.8 5.0894

22.4

0.8 1.8 3 .888 4.0715 9.1609

13.3694 3.942 5.6555 22.769

2 3Have : 13.3694 m 3.942 mX V xV X

or 0.295 mX2 3: 13.3694 m 5.6555 mY V yV Y

or 0.423 mY

2 3: 13.3694 m 22.769 mZ V zV Z

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

a 38

a 414

a

Semiellipsoid 2

22 13 2 6

a h a h38

h 2 2116

a h

2 2 2 2Then 4 46 16

V a a h yV a a h

Now Y V yV

so that 2 2 2 24 46 16

Y a a h a a h

or234 4 (1)

8h hY aa a

(a) ?Y when 2ah

Substituting 12

ha

into Eq. (1)

21 3 14 42 8 2

Y a

or 45112

Y a 0.402Y a


( ) ?hba

when 0.4Y a

Substituting into Eq. (1)

230.4 4 48

h ha aa a

or 2

3 3.2 0.8 0h ha a

Then23.2 3.2 4 3 0.8

2 3ha

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

, 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 0.20.9

0 0.9 3 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 X or 0.1452 mX

2: 3.0283 m 1.20 mY L yL Y or 0.396 mY

2: 3.0283 m 1.12177 mZ L z L Z or 0.370 mZ

capitulo 5 estatica

Education