Engineering Mechanics: statics, Instructor's Solutions Manual [4 ed.] 9781305885028

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An Instructor’s Solutions Manual to Accompany

ENGINEERING MECHANICS: STATICS, 4TH EDITION ANDREW PYTEL JAAN KIUSALAAS

ISBN: 978-1-305-88502-8

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Instructor's Solutions Manual to Accompany

Engineering Mechanics: Dynamics 4th EDITION

ANDREW PYTEL JAAN KIUSALAAS

Contents Chapter 1:............................................................................................................................ 1 Chapter 2:.......................................................................................................................... 23 Chapter 3:.......................................................................................................................... 64 Chapter 4:.......................................................................................................................... 93 Chapter 5:........................................................................................................................ 199 Chapter 6:........................................................................................................................ 237 Chapter 7:........................................................................................................................ 309 Chapter 8:........................................................................................................................ 361 Chapter 9:........................................................................................................................ 438 Chapter 10:...................................................................................................................... 486

Chapter 1 1.1

1.2 W

=

W

=

gV = (7850)(9:81) (0:042 )(0:110) = 42:58 N 0:2248 lb 42:58 N = 9:57 lb J 1:0 N

1.3 (a) 400 lb ft = 400 lb ft

(b) 6 m/s =

6m s

(c) 20 lb/in.2 = = 137:9 kPa J

4:448 N 1:0 lb

0:3048 ft 1:0 m

20 lb in.2

(d) 500 slug/in. =

1:0 mi 5280 ft

3600 s = 1: 247 mi/h J 1:0 h

1:0 in.2 = 1: 379 645:2 10 6 m2

4:448 N 1:0 lb

500 slug in.

0:3048 m = 542 N m J 1:0 ft

14:593 kg 1:0 slug

39:37 in. = 2: 87 1:0 m

105 N/m2

105 kg/m J

1.4 30 mi/gal

= =

30 mi 5280 ft 0:3048 m 1:0 gal gal 1:0 mi 1:0 ft 3:785 L 12 760 m/L = 12:76 km/L J

1.5 1 m 2 kg m2 (1000 kg) 6 = 18 000 = 18 000 2 s s2 = 18 000 N m = 18 kN m J

(a) E =

(b) E = 18 000 N m = 18 000 N m = 13 280 lb ft J

0:2248 lb 1:0 N

kg m s2

(m)

3:281 ft 1:0 m

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1.6

1.7

1.8 (a) 8 mm/ s =

8 mm s

(b) 8000 m/s =

8000 m s

1:0 m 1000 mm

1:0 s = 8000 m/s J 10 6 s

3:281 ft 1:0 m

1:0 mi 5280 ft

3600 s = 17 900 mi/h J 1:0 h

1.9

1.10 L T2

=

[A] L2 + [B] [L] [T ]

)

[A] =

1 LT 2

J

[B] =

1 T3

J

1.11 (a) The dimensions of x = At2

Bvt are

[L] = [A][T 2 ] [B][LT 1 ][T ] ) [A] = [LT 2 ] J [B] = [1] (dimensionless) J

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(b) The dimensions of x = Avte

Bt

are

[L] = [A][LT 1 ][T ]e[B][T ] [B][T ] = [1] ) [B] = [T 1 ] J [L] = [A][LT 1 ][T ] ) [A] = [1] J

1.12 d4 y dx4

=

!2 y D

=

Substituting [F ] = M LT

2

L = [L L4 [T

2]

[M L [F L2 ]

2

1

]

]

M T 2 F L2

[L] =

— see Eq. (1.2b)— we get

!2 M y = D T 2 L2 Substituting [F ] = M LT

3

T2 = [L ML

3

] Q.E.D.

— see Eq. (1.2b)— we get

!2 M y = D T 2 L2

T2 = [L ML

3

] Q.E.D.

1.13 The argument of the sine function must be dimensionless: Bx k

=

[F ]

[1]

[B][L]

L = [1] F

= [Akx2 ] = [A][F L

1

][L2 ]

[B] = [F L [A] = [L

1

2

] J

]J

1.14 (a) 110 hp = 110 hp

550 lb ft/s = 60 500 lb ft/s J 1:0 hp

(b) 110 hp = 110 hp

0:7457 kW = 82:0 kW J 1:0 hp

1.15 F W % of weight

(12)(12) mA mB = (6:67 10 11 ) = 6:003 10 8 N 2 R 0:42 = mg = (12)(9:81) = 117:7 N F 6:003 10 8 = 100% = 100% = 5:10 10 8 % J W 117:7

= G

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1.16

1.17 h = (28 000 ft)

On earth: We =

W = We

0:3048 m 1:0 ft

GMe m Re2

= 8534 m = 8:534 km

At elevation h: W =

GMe m (Re + h)2

63782 Re2 = 170 = 169:5 lb I 2 (Re + h) (6378 + 8:534)2

1.18 gm

=

gm ge

=

GMm GMe ge = 2 Rm Re2 2 0:07348(6378)2 Mm R e = = 0:1658 2 Me R m 5:974(1737)2

1 Q.E.D. 6

1.19

1.20 On earth: We =

GMe m Re2

At elevation h: W =

GMe m (Re + h)2

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W

=

We 10

GMe m GMe m = 2 (Re + h) 10Re2

(6378 + h)2

=

10(6378)2

(Re + h)2 = 10Re2

h = 13 790 km J

1.21 R

F

= G =

= Re + Rm + d = 6378 + 1737 + 384 = 392:1 103 km = 392:1 106 m

Me Mm = 6:67 R2

1:904

10

11

5:974

103

1024 (0:07348 (392:1

1024 )

2

106 )

1020 N J

1.22

90o

5m /s

3m /s o 40 α v

50o v=

p 52 + 32 = 5:83 m/s

= tan

1

3 = 31:0 5

|v1 + v2| = 5.83 m/s

31.0o 1.23

v1

90o v2

50o 40o 8 m/s

v1 = 8 sin 40 = 5:14 m/s J

v2 = 8 sin 50 = 6:13 m/s J

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1.24

240 lb

30o P 90o

Q Component parallel to AB Component pependicular to AB

: P = 240 cos 30 = 208 lb J : Q = 240 sin 30 = 120 lb J

1.25

v Pv sin 60 Pv Pu

u 50o 60o o Pv 70 20 kN 60o 50o Pu Pu 20 = sin 70 sin 50 sin 60 = 20 = 22:6 kN J sin 50 sin 70 = 20 = 24:5 kN J sin 50 =

1.26

140o v

α

3 mi/h

Law of cosines: Law of sines:

5 mi/h 40o

p v = 32 + 52 2(3)(5) cos 140 = 7:549 mi/h 5 7:549 = sin = 0:4257 = 25:2 sin sin 140

7.55 mi/h 25.2o

J

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1.27

R 35o

65o 8000 lb 80o 8000 P = sin 65 sin 35

P

P = 8000

sin 65 = 12 640 lb J sin 35

1.28

25o

θ R

α

8000 lb

80o

10 000 lb

Law of cosines: R

= =

p 80002 + 10 0002 11 671 lb J

2(8000)(10 000) cos 80

Law of sines: 10 000 sin

11 671 10 000 sin = sin 80 = 0:8438 sin 80 11 671 = sin 1 (0:8438) = 57:54 = 90 25 57:54 = 7:46 J =

7.46 o 11 670 lb

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1.29

1.30

1.31

60

o

360 lb

PAB

80 o

40 o P AC Law of sines: 360 sin 80

=

PAB

=

PAC

=

PAB = sin 40 360 sin 40 sin 80 360 sin 60 sin 80

PAC sin 60 = 235 lb J = 317 lb J

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1.32

α 360 lb

β

PAB = 190 lb

PAC = 210 lb

Law of cosines: 2102 1902

= 3602 + 1902 = 27:3 J = 3602 + 2102 = 24:5 J

2(360)(190) cos 2(360)(210) cos

1.33

1.34

Q = 500 lb α R = 800 lb 125o

P

β

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Law of sines: 500 sin

= =

800 = 30:8 sin 125 180 (125 + 30:8 ) = 24:2

65.8o

R=

800 lb 800 P = sin 125 sin 24:2

P = 400 lb J

1.35

1.36

C o

a

21.3o

116.2 b 42.5o 63.8o A 200 m B Law of sines: )a = b

=

200 a b = = sin 21:3 sin 116:2 sin 42:5 200 sin 116:2 = 494 m J sin 21:3 200 sin 42:5 = 372 m J sin 21:3

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1.37

12 lb 60o R

12 sin(90 sin(90 90

25 lb 90o−α

25 sin 60 12 sin 60 ) = = 0:4157 25 = 24:56 = 65:4

)

=

J

*1.38

300 N o 60 250 N α R First compute the resultant R of the two known forces. The smallest required F has the same direction as R and its magnitude is 500 N R. p Law of cosines: R = 2502 + 3002 2(250)(300) cos 60 = 278:4 N ) F = 500 278:4 = 222 N Law of sines: =

300 278:4 = sin sin 60 300 sin 60 = 68:9 sin 1 278:4

222 N 21.1o

J

1.39

β 30 lb γ 50 lb α 65 lb

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Law of cosines:

652 = 502 + 302 2(50)(30) cos 652 + 502 + 302 = cos 1 = 105:96 2(50)(30) = 180 = 180 105:96 = 74:0

Law of sines: =

30 65 = sin sin 105:96 30 sin 105:96 sin 1 = 26:3 65

J

J

1.40

1.41

1.42 ! AB = =

F=F

p ! 5i + 3j ft AB = 52 + 32 = 5:831 ft ! AB 5i + 3j ! = 5:831 = 0:8575i + 0:5145j AB

= 560( 0:8575i + 0:5145j) =

480i + 288j lb J

288 lb 480 lb

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1.43

1.44 (a)

=

AB

(b)

! AB ! = AB

2:2i + 7:5j + 3k = 8:372

0:2628i + 0:8958j + 0:3583k J

= 8 AB = 8( 0:2628i + 0:8958j + 0:3583k) = 2:10i + 7:17j + 2:87k m/s J

v

1.45 ! OA ! = OA

=

OA

3i + 4j + 2:5k = 5:590

0:5367i + 0:7156j + 0:4472k

F = F OA = 320( 0:5367i + 0:7156j + 0:4472k) = 172i + 229j + 143k N J

1.46 BA

=

! BA 14i 10j 18k = 0:5623i ! = 24:90 BA

F = F AB = 160(0:5623i 0:4016j = 90:0i 64:3j 115:7k lb J

0:4016j

0:7229k

0:7229k)

1.47 ! AB = = v

160i + 220j 70k ft ! AB 160i + 220j 70k ! = p1602 + 2202 + 702 = 0:5696i + 0:7832j AB

= v = 1400(0:5696i + 0:7832j = 797i + 1096j 349k ft/s J

0:2492k

0:2492k)

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1.48 (a) ! BA = =

20i + 60j 90k ft ! 20i + 60j BA ! = 110:0 BA

Fx Fy Fz

= F = F = F

x y z

p ! BA = 202 + 602 + 902 = 110:0 ft

90k

=

0:1818i + 0:5455j

0:8182k

= 600 ( 0:1818) = 109 lb J = 600(0:5455) = 327 lb J = 600( 0:8182) = 491 lb J

(b) = cos = cos = cos

x y z

1 1 1

x y z

= cos = cos = cos

( 0:1818) = 100:5 J (0:5455) = 56:9 J 1 ( 0:8182) = 144:9 J 1

1

1.49

z

C

5m

B 5m D O x

T

T 3m A

y

p ! 3j + 5k m AB = 32 + 52 = 5:831 m ! AB 3j + 5k = T ! = 35 = 18:01j + 30:01k kN 5:831 AB

! AB = TAB

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! AC TAC

=

5i

= T

3j + 5k m

p ! AC = 52 + 32 + 52 = 7:681 m

! 5i 3j + 5k AC = ! = 35 7:681 AC

22:78i

13:67j + 22:78k kN

= TAB + TAC = 22:78i + ( 18:01 13:67)j + (30:01 + 22:78)k = 22:8i 31:7j + 52:8k kN J

R

1.50 p ! 8j + 4k ft AB = 82 + 82 + 42 = 12:0 ft ! 8i 8j + 4k AB = F ! =F 12:0 AB

! AB = FAB

! AC

=

FAC

=

8i

p ! 8j + 4k ft AC = 42 + 82 + 42 = 9:798 ft ! AC 4i 8j + 4k 200 ! = 200 9:798 AC 4i

The resultant R lies in the yz-plane if Rx

=

(FAB )x + (FAC )x = 0

F

=

122:5 lb J

F

8 12:0

200

4 =0 9:798

1.51 (a) R

= (F1 + F2 sin 35 )i + (F2 cos 35 + F3 cos 65 )j + (F3 sin 65 )k = (1:6 + 1:2 sin 35 )i + (1:2 cos 35 + 1:0 cos 65 )j + (1:0 sin 65 )k = 2:288i + 1:4056j + 0:9063k kN J

(b) R

p 2:2882 + 1:40562 + 0:90632 = 2:834 kN R 2:288i + 1:4056j + 0:9063k = = R 2:834 = 0:807i + 0:496j + 0:320k =

) R = 2:83(0:807i + 0:496j + 0:320k) kN J

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1.52

1.53 4i + 3j 5i 12j = 72i + 54j lb Q=Q 5 13 Because R = P + Q lies in x-direction, we have P = 90

12 Q=0 Q = 58:5 lb J 13 5 = 94:5 lb J = Rx = 72 + 58:5 13

Ry R

=

0

54

1.54 Px + Qx Py + Qy

= Rx = Ry

P cos 30 P sin 30

P = 717 lb J

Solution is:

Q sin 30 = 360 cos 25 Q cos 30 = 360 sin 25

Q = 590 lb J

1.55 Px + Qx

= Rx

3 cos = 2 sin 55 2 = cos 1 sin 55 = 56:90 J 3 Py + Qy = Ry 3 sin Q = 2 sin 55 Q = 2 sin 55 + 3 sin 56:90 = 4:15 kN J

1.56 P

=

Q

=

F

=

6i + 8j

p

62

p

82

12k

+ + ( 12)2 6i + 6j 12k

= 0:3841i + 0:5121j

0:7682k

= 0:4082i + 0:4082j ( 6)2 + 62 + ( 12)2 8j 12k p = 0:5547j 0:8321k ( 8)2 + ( 12)2

0:8165k

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Px + Qx + Fx Py + Qy + Fy

= 0 = 0

0:3841P 0:4082Q + 0 = 0 0:5121P + 0:4082Q 0:5547(120) = 0

Solution is: P = 74:3 lb J

Q = 69:9 lb J

1.57 (a) A B (b) A B (c) A B

= 12( 2) + 8(3) = 0 J = 5(7) = 35 N m J = 3( 6) + 2(2) + ( 1)( 8) =

6 m2 J

1.58 (a) C =

i 0 4

j k 12 8 2 3

(b) C =

i 5 7

j 3 0

48k ft2 J

k 0 12

=

36i + 60j

k 1 8

=

14i + 30j + 18k m2 J

i j 3 2 6 2

(c) C =

= 52i + 32j

21k N m J

1.59 r

=

4 20 0

6 40 0:8

2 30 0:6

= 296 N m J

r F =

0 4 20

0:8 6 40

0:6 2 30

= 296 N m J

F

1.60 A = 2i + 1:2j m A

C

B=

B=

B = 2i + 1:2j + 1:5k m i 2 2

j 1:2 1:2

i j 0 0 2 1:2

k 0 1:5 k 1:5 1:5

= 1:8i

= 1:8i

C=

1:5k m

3j m2 J

3j m2 J

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1.61 2i + 1:2j m B = 2i + 1:2j + 1:5k m p p A = 22 + 1:22 = 2:332 m B = 22 + 1:22 + 1:52 = 2:773 m

A

=

cos

A B 2(2) + 1:2(1:2) = = 0:8412 AB (2:332) (2:773) = 32:7 J

= )

z

A B

θ C

Because the three vectors form a right triangle, we have in this case = cos

1

A = cos B

1

2:332 = 32:8 2:773

The di¤erence in the results is due to round-o¤ error.

1.62 p = Az = 142 + 92 csc 50 = 21:73 ft = 9i + 14j + 21:73k ft B = 6i + 21:73k ft

Bz A

A = B cos

p

92 + 142 + 21:732 = 27:37 ft

p

62 + 21:732 = 22:54 ft 9(6) + 21:73(21:73) A B = = 0:8529 = AB 27:37(22:54) = 31:5 J =

1.63

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1.64 =

3i + 4k in. Q = 2j + 4k in. p p 32 + 42 = 5 in. Q = 22 + 42 = 4:472 in. =

P P (a)

cos

= =

P Q 3( 2) + 4(4) = = 0:4472 PQ 5(4:472) 63:4 J

(b) P

Q = = =

i j k 3 0 4 = 8i 12j 6k in. 0 2 4 P Q 8i 12j 6k =p jP Qj 82 + 122 + 62 0:512i 0:768j 0:384k J

1.65 i j k 4 3 2 = 17i 8j 22k m2 2 4 3 17i 8j 22k A B = p jA Bj 172 + 82 + 222 ( 0:588i 0:277j 0:760k) J

B =

A

= =

1.66 ! CA = (0 3)i + ( 2 ! CB = ( 1 3)i + (4 ! CA ! CB ! CA

=

! CA ! CA

0)k =

3i

2j + 2k in.

0)j + (1

0)k =

4i + 4j + k in.

p ( 3)2 + ( 2)2 + 22 = 4:123 in. p = ( 4)2 + 42 + 12 = 5:745 in. =

! CB =

! CB ! = CB

0)j + (2

i 3 4

j k 2 2 4 1

=

10i 5j 20k = 4:123(5:745)

10i

5j

( 0:422i

20k in.2

0:211j

0:844k) J

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1.67 P = 3i + 4k m Q = 3i + 4j + 5k m ! 3i + 4j OA = 0:6i + 0:8j = ! =p 32 + 42 OA The component of P P

Q in direction of Q

=

3 3 0:6

is

0 4 4 5 0:8 0

=

12:0 m J

1.68

1.69 3(4)

A B = 0 a(1) 2(1) = 0

a = 10:0 J

*1.70

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1.71 By inspection, a unit vector perpendicular to the door is = sin 20 i + cos 20 j = 0:3420i + 0:9397j The component of F perpendicular to the plane of the door is F? = F

=

5(0:3420) + 12(0:9397) = 9:57 lb J

1.72

*1.73

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1.74

1.75

1.76

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Chapter 2 2.1 The resultant of each force system is 500N ". Each resultant force has the same line of action as the the force in (a), except (f) and (h) Therefore (b), (c), (d), (e) and (g) are equivalent to (a) J

2.2

2.3 Rx Ry

= = = = R

Fx = T1 cos 60 + T3 cos 40 110 cos 60 + 150 cos 40 = 59:91 lb Fy = T1 sin 60 + T2 + T3 sin 40 110 sin 60 + 40 + 150 sin 40 = 231:7 lb p 59:912 + 231:72 = 239 lb J 231:7 = 75:5 J = tan 1 59:91 =

R = 239 lb 75.5o x

2.4 Rx Rx

= =

Fx + ! 7:68 kN

Rx = 25 cos 45 + 40 cos 60

30

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Ry Ry

= =

Fy +" 16:96 kN R = 7:68i

Ry = 25 sin 45

40 sin 60

16:96k kN J

2.5 F1 F2

F3

120j + 80k = 80 p = 66:56j + 44:38k N ( 120)2 + 802 100i 120j + 80k = F2 AC = 60 p ( 100)2 + ( 120)2 + 802 = 34:19i 41:03j + 27:35k N 100i + 80k = F3 AD = 50 p = 39:04i + 31:24k N ( 100)2 + 802 = F1

R

AB

=

F = ( 34:19 39:04)i + ( 66:56 +(44:38 + 27:35 + 31:24)k = 73:2i 107:6j + 103:0k N J

41:03)j

2.6

(b)

24 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.7 = ( P2 cos 25 P3 cos 40 )i + (P1 + P2 sin 25 )j + P3 sin 40 k = 800i + 700j + 500k lb

R

Equating like coe¢ cients: P2 cos 25 P3 cos 40 P1 + P2 sin 25 P3 sin 40

= 800 = 700 = 500

Solution is P1 = 605 lb J

P2 = 225 lb J

P3 = 778 lb J

2.8 T1

=

T2

=

T3

=

R

i + 2j + 6k

90 p

( 1)2 + 22 + 62 2i 3j + 6k

=

14:06i + 28:11j + 84:33k kN

60 p

= 17:14i 25:71j + 51:43k kN ( 2)2 + ( 3)2 + 62 2i 3j + 6k 40 p = 11:43i 17:14j + 34:29k kN 22 + ( 3)2 + 62

= T1 + T2 + T3 = ( 14:06 17:14 + 11:43)i +(28:11 25:71 17:14)j + (84:33 + 51:43 + 34:29)k = 19:77i 14:74j + 170:05k kN J

2.9 T1 T2 T3

i + 2j + 6k

= T1 p

( 1)2 + 22 + 62 2i 3j + 6k

= T1 ( 0:15617i + 0:3123j + 0:9370k)

= T2 p

= T2 ( 0:2857i 0:4286j + 0:8571k) ( 2)2 + ( 3)2 + 62 2i 3j + 6k = T3 p = T3 (0:2857i 0:4286j + 0:8571k) 22 + ( 3)2 + 62 T1 + T2 + T3 = R

Equating like components, we get 0:15617T1 0:2857T2 + 0:2857T3 = 0 0:3123T1 0:4286T2 0:4286T3 = 0 0:9370T1 + 0:8571T2 + 0:8571T3 = 210 Solution is T1 = 134:5 kN J

T2 = 12:24 kN J

T3 = 85:8 kN J

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2.10

2.11 = 10 cos 20 i 10 sin 20 j = 9:397i 3:420j lb = F2 (sin 60 i + cos 60 j) = F2 (0:8660i + 0:5j) = F = ( 9:397 + 0:8660F2 )i + ( 3:420 + 0:5F2 )j ! AB = 4i + 6j in. ! Because R and AB are parallel, their components are proportional: F1 F2 R

9:397 + 0:8660F2 4 F2

= =

3:420 + 0:5F2 6 9:74 lb J

2.12 30 lb

35o

β

8.5"

Α P

8" − a a

R

First …nd the direction of R from geometry (the 3 forces must intersect at a common point). 8

a = =

Rx Ry Solution is

= =

Fx Fy

8:5 tan 35 ) a = 2:048 in. 2:048 1 a tan = tan 1 = 13:547 8:5 8:5 + ! R sin 13:547 = P sin 35 + 30 + # R cos 13:547 = P cos 35

P = 38:9 lb J

R = 32:8 lb J

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2.13 FAB

=

FAC

=

12i 6j + 9k 15 p = 11:142i 5:571j + 8:356k lb 122 + ( 6)2 + 92 11:142i 5:571j + 8:356k lb (by symmetry) Fy T

= =

0: 2( 5:571) + T = 0 11:14 lb J

2.14 P1 P2 P3

3i + 4k 100 p = 60i + 80k lb 32 + 42 3i + 3j + 4k = 61:74i + 61:74j + 82:32k lb = 120 p 32 + 32 + 42 = 60j lb

=

Q1

= Q1 i

Q2

= Q2 p

Q3

3i 3j = Q2 ( 0:7071i 0:7071j) 32 + 32 3j + 4k = Q3 p = Q3 (0:6j + 0:8k) 32 + 42

Equating similar components of

Q = P:

Q1 0:7071Q2 0:7071Q2 + 0:6Q3 0:8Q3

= 60 + 61:74 = 61:74 + 60 = 80 + 82:32

Solution is Q1 = 121:7 lb J

Q2 = 0

Q3 = 203 lb J

2.15 Rx Ry

= Fx + ! 8 = 40 sin 45 Q sin 30 Q = 40:57 lb W + 40:57 cos 30 = Fy + " 0 = 40 cos 45 ) W = 63:4 lb J

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2.16

P1

P2 40o

h

θ

50o

P3

b 4m

6m

The forces must be concurrent. From geometry: h

= (4 + b) tan 40 = (6 b) tan 50 ) b = 1:8682 m J ) h = (4 + 1:8682) tan 40 = 4:924 m 4:924 h = 69:22 J = tan 1 = tan 1 b 1:8682 R= F =

(25 cos 40 + 60 cos 69:22 80 cos 50 )i + (25 sin 40 + 60 sin 69:22 + 80 sin 50 )j = 10:99i + 133:45j kN J

2.17

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2.18 T1

=

T2

=

T3

=

R

= =

3i 2j 6k = 77:14i 51:43j 154:29k lb 32 + ( 2)2 + ( 6)2 3j 6k 250 p = 111:80j 223:61k lb 32 + ( 6)2 4i 6k 400 p = 221:88i 332:82k lb ( 4)2 + ( 6)2 T = (77:14 221:88)i + ( 51:43 + 111:80)j +( 154:29 223:61 332:82)k 144:7i + 60:4j 710:7k lb J acting through point A: 180 p

2.19 3i 12j + 10k = 120 p 32 + ( 12)2 + 102 = 22:63i 90:53j + 75:44k lb 8i 12j + 3k = TAC AC = 160 p ( 8)2 + ( 12)2 + 32 = 86:89i 130:34j + 32:59k lb = TAB

TAB

TAC

R

AB

= TAB + TAC W k = (22:63 86:89)i + ( 90:53 130:34)j + (75:44 + 32:59 = 64:3i 220:9j + 0:0k lb J

108)k

2.20 Choose the line of action of the middle force as the x-axis.

F

y

25o 40o F

x

F Rx Ry

= =

Fx = F (cos 25 + 1 + cos 40 ) = 2:672F Fy = F (sin 25 sin 40 ) = 0:2202F

p R = F 2:6722 + ( 0:2202)2 = 2:681F 400 = 2:681F ) F = 149:2 lb J

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*2.21

2.22 P2

800 N o

38

P1

0.6 m A

+

MA

0.5 m

= 0:6P1 + 0:5P2 = 0:6(800 cos 38 ) + 0:5(800 sin 38 ) = ) MA = 132:0 N m J

132:0 N m

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2.23 40 in.

A

24 in.

60 lb P2 P1 12 in. P1 P1 = 60 p

B

P2 C

40 = 57:47 lb + 122

402

With the force in the original position: J

MA = 24P1 = 24(57:47) = 1379 lb in. With the force moved to point C: MB = 36P1 = 36(57:47) = 2070 lb in:

J

2.24

P 5.5 m

2.5 m B

A

P cosθ

C P sinθ

Resolve the force at C into components as shown. Adding the moments of the forces about A yields + sin

MA = 5:5P 8P sin = 0 5:5 = 0:6875 = 43:4 J = 8

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2.25

y A P

O

3m 40 0.6

0.5 m

0.4 m

0.4 P 0.6403 B

x

0.5 P 0.6403 Since MA = MB = 0, the force P passes through A and B, as shown. + P

0:5 P (0:4) = 350 kN m P = 1120:5 N 0:6403 0:4 0:5 1120:5i 1120:5j = 700i 875j N J 0:6403 0:6403 MO =

=

2.26

2.27

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2.28

30.41 kN

16 m

20 m B

45o 30.41 kN

W = 38 kN

T = 43 kN (a) Moment of T:

MB = 30:41(20) = 608 kN m CCW J

+ (b) Moment of W : +

MB = 38(16) = 608 kN m CW J

(c) Combined moment: +

MB = 608

608 = 0 J

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2.29

2.30 (a)

F 2 B

F 45o F 2

y

d 65o

A

x Fd MA = p 2

J

(b) F = F cos 20 i + F sin 20 j ! r = AB = d cos 65 i + d sin 65 j

MA

= r =

F=

i cos 65 cos 20

(sin 20 cos 65

j sin 65 sin 20

k 0 0

Fd

cos 20 sin 65 ) F d k =

0:707F dk J

2.31 18 lb

24 lb 4 in. x A

34 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Because the resultant passes through point A, we have MA = 0

+

24(4)

18x = 0

x = 5:33 in. J

2.32 6' o 30

y x 8'

4'

6200 lb

7' 5.196' 4.804' 10' Wy

Wx W

7 = 0:8245W + 4:8042 Largest W occurs when the moment about the rear axle is zero. Wy = W p

+ )

72

Maxle = 6200(8) W = 6020 lb J

(0:8245W ) (10) = 0

2.33

B A

0.8 0.6 0.3 30o 0.7416 0.5196

0.15 0.3

Dimensions in meters + +

MA = 310 = MB = 120 =

310 120

Fx Fy

Fx (0:15) + Fy (0:5196 + 0:7416 + 0:3) 0:15Fx + 1:5612Fy Fx (0:3 + 0:15) + Fy (0:7416 + 0:3) 0:45Fx + 1:0416Fy

= =

(a) (b)

0:15Fx + 1:5612Fy 0:45Fx + 1:0416Fy

Solution of Eqs. (a) and (b) is Fx = 248:1 N and Fy = 222:4 N p ) F = 248:12 + 222:42 = 333 N J Fx 248:1 = tan 1 = tan 1 = 48:1 J Fy 222:4

35 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.34 70i 100k = Pp = ( 0:5735i ( 70)2 + ( 100)2 ! r = AB = 0:07i + 0:09j m

P

= r

MA

= MA

i 0:07 0:5735

P=

( 73:73i

j 0:09 0

0:8192k)P

k 0 P 0:8192

57:34j + 51:62k)

10

3

P

p = ( 73:73)2 + ( 57:34)2 + 51:622 (10 = 106:72 10 3 P

3

P)

Using MA = 15 N m, we get 15 = 106:72

10

3

P

P = 140:6 N J

2.35 P

= =

160

AB

93:02i

= 160 p

0:5i

0:6j + 0:36k

0:5)2

( + ( 0:6)2 + 0:362 111:63j + 66:98k N

(a) MO = rOB

P=

i 0 93:02

j 0 111:63

k 0:36 66:98

= 40:2i

33:5j N m J

(b) MC = rCB

P=

i 0 93:02

j 0:6 111:63

k 0 66:98

=

40:2i

55:8k N m J

2.36

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2.37 rOC

=

MO

= P

M0 P

2i + 4j

3k m

i 2 cos 25

P = P ( cos 25 i + sin 25 k) j k 4 3 = P (1:6905i + 1:8737j + 3:6252k) 0 sin 25

p = P 1:69052 + 1:87372 + 3:62522 = 4:417P = 350 kN m = 79:2 kN J

2.38

2.39 = P

BA

Q = Q

AC

P

2j + 4k

= 20 p

( 2)2 + 42 2i + 2j

=

8:944j + 17:889k kN

k = 20 p = 2 2 ( 2) + 2 + ( 1)2

13:333i + 13:333j

6:667k kN

! r = OA = 2i + 4k m

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P+Q = =

MO

13:333i + ( 8:944 + 13:333)j + (17:889 13:333i + 4:389j + 11:222k kN

= r =

6:667)k

i j k 2 0 4 13:333 4:389 11:222 75:78j + 8:78k kN m J

(P + Q) = 17:56i

2.40 Noting that both P and Q pass through A, we have MO = rOA

P

=

Q =

P+Q = )

60 p

4:2i

(P + Q)

rOA = 2k ft

2j + 2k

= 49:77i 23:70j + 23:70k lb + ( 2)2 + 22 2i 3j + 2k 80 p = 38:81i 58:21j + 38:81k lb ( 2)2 + ( 3)2 + 22 (

4:2)2

88:58i MO =

81:91j + 62:51k lb i j k 0 0 2 88:58 81:91 62:51

= 163:8i

177:2j lb ft J

2.41

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2.42

2.43 MO = r

F=

i x 50

j 0 100

k z 70

= 100zi + (70x + 50z)j

100xk

Equating the x- and z-components of MO to the given values yields 100z = 100x =

400 300

) z = 4 ft J ) x = 3 ft J

Check y-component: 70x + 50z = 70(3) + 50(4) = 410 lb ft O.K.

2.44 F = r = MO

=

MO

=

d

=

150 cos 60 j + 150 sin 60 k = 75j + 129:90k N ! OB = 50i 60j mm i j k 50 60 0 r F= = 7794i + 6495j 3750k N mm 0 75 129:90 p ( 7794)2 + 64952 + ( 3750)2 = 10 816 N mm = 10:82 N m J MO 10 816 = = 72:1 mm J F 150

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2.45

2.46 F =

2i 12j + 5k lb ! r = BA = ( x + 2)i + 3j zk i j k x+2 3 z MB = r F = 2 12 5 = ( 12z + 15)i + (5x

2z

10)j + (12x

30)k

Setting i and k components to zero: 12z + 15 12x 30

z = 1:25 ft J x = 2:5 ft J

= 0 = 0

Check j component: 5x

2z

10 = 5(2:5)

2(1:25)

10 = 0 Checks!

2.47 (a) Mx My Mz

= 75(0:85) = 63:75 kN m J = 75(0:5) = 37:5 kN m J = 160(0:5) 90(0:85) = 3:5 kN m J

40 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) MO = rOA

F=

i 0:5 90

j 0:85 160

k 0 75

=

63:75i + 37:5j + 3:5k kN m

The components of MO agree with those computed in part (a).

2.48 (a)

z B 250 mm

400 mm 20 kN 40 kN

O

C y

A

30 kN

x MOA = 20(400)

30(250) = 500 kN mm = 500 N m J

(b) F =

40i + 30j + 20k kN ! r = OC = 400j + 250k mm 0 400 250 40 30 20 MOA = r F i = 0 1 0 = 500 N m J

= 500 kN mm

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2.49

2.50 (a)

6.928'

z

Fy

30o 8'

4'

F = 55 lb Fx 2'

y

Only Fy has a moment about x-axis (since Fx intersects x-axis, it has no moment about that axis). Fy +

=

6:928 = 52:84 lb 6:9282 + 22 Mx = 6Fy = 6(52:84) = 317 lb ft J

55 p

42 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) 2i + 6:928k F = 55 p = 6:9282 + 22 Mx = r

F

0 0 1

=

15:26j + 52:84k lb 6 15:26 0

0 52:84 0

r = 6j ft

= 317 lb ft J

2.51

10 N 12 N

8m 0.4

0.16 m

18 N

0.4 0m

2m 0.1

x

a

(a)

y

1: 920j N m J

Ma = [ 10(0:48) + 18(0:16)] j = (b)

z

Mz = [ 12(0:48 + 0:12) + 18(0:4)] k = 0 J

2.52 (a)

O

y 1.2 m

1.2 cos 30o 3

A

D

30o

F1

C

30o F2

B x F = 160 N

We resolve F into components F1 and F2 , which are parallel and perpendicular to BC, respectively. Only F2 contributes to MBC : MBC = 1:8F2 = 1:8(160 cos 30 ) = 249 N m J

43 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) F =

160i N ! 1:2 cos 30 r = BA = 0:6i + j + 1:8k = 0:6i + 0:3464j + 1:8k m 3 = sin 30 i + cos 30 j = 0:5i + 0:8660j BC 0:6 0:3464 1:8 0 0 = 249 N m J MBC = r F BC = 160 0:5 0:8660 0

2.53 F = r = My

40i 8j + 5k N 350 sin 20 i 350 cos 20 k =119:7i 328:9k mm 119:7 0 328:9 40 8 5 = r F j= = 12 560 N mm 0 1 0 = 12:56 N m J

2.54

2.55

2.56 With T acting at A, only the component Tz has a moment about the y-axis: My = 4Tz .

44 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Tz

AB z 3 = 28:11 lb = 60 p 2 AB 4 + 42 + 32 My = 4(28:11) = 112:40 lb ft J

= T )

2.57 Only the x-component of each force has a moment about the z-axis. ) Mz

= (P cos 30 + Q cos 25 ) 15 = (32 cos 30 + 36 cos 25 ) 15 = 905 lb in. J

2.58 P

=

rCA

=

0:42i

360 p

(

0:81j + 0:54k

= 142:6i 275:0j + 183:4k N + ( 0:81)2 + 0:542 0:42i + 0:54k = 0:6139i + 0:7894k CD = p 0:422 + 0:542

0:42)2

0:42i m

0 0 275:0 183:4 0 0:7894

= rCA

MCD

= MCD CD = 91:18(0:6139i + 0:7894k) = 56:0i 72:0k N m J

P

CD

=

0:42 142:6 0:6139

MCD

=

91:18 N m

2.59

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2.60 MBC = rBA rBA

=

BC

=

4j 2k p = 0:8944j 42 + ( 2)2

= rBA

MBC

=

BC

3i + 3j 3k = 0:5774F ( i + j F=Fq 2 ( 3) + 32 + ( 3)2

5i

MBC

F

k)

0:4472k

5 0 0 1 1 1 = 1:2911F BC 0 0:8944 0:4472 1:2911F = 150 lb ft F = 116:2 lb J = 0:5774F

F

150 lb ft

2.61 The unit vector perpendicular to plane ABC is = ! AB = ! AB

! AC

=

! AB ! AB

! AC ! AC

! (0:3i 0:5k) AC = (0:4j 0:5k) m i j k 0:3 0 0:5 = 0:2i + 0:15j + 0:12k 0 0:4 0:5 0:2i + 0:15j + 0:12k = 200 p 0:22 + 0:152 + 0:122 144:24i + 108:18j + 86:55k N m

F = F =

Mx

! = OA

jMx j =

0 F i = 144:24 1

0 108:18 0

0:5 86:55 0

=

54:1 N m

54:1 N m J

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2.62

2.63 Equating moments about the x- and y- axis: 600(1:5) + 400(2) + 200(4) = 1200y 600(3) 200(3) = 1200x

y = 2:08 ft J x = 2:00 ft J

2.64

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2.65

2.66

48 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.67 (a) F = rBO

=

4i + 8j + 10k 180 p = 53:67i + 107:33j + 134:16k lb 42 + 82 + 102 ( 6 cot 40 ) i + 6k 6k ft = 0:7660i + 0:6428k AB = q 2 ( 6 cot 40 ) + 62

MAB = rBO

F

AB

0 0 53:67 107:33 0:7660 0

=

(b)

6 134:16 0:6428

=

493 lb ft J

z B

x

' 96 4.5

50o 6'

40o A

O

Note that only Fy = 107:33 lb has a moment about AB. From trigonometry, the moment arm is d = 6 sin 50 = 4:596 ft. ) MAB =

Fy d =

107:33(4:596) =

493 lb ft J

2.68

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2.69

2.70

F 2

F 2

F 2 F 2 15 in.

C F

F 15 p 2 p p 2 2 = C= (120) = 11:31 lb J 15 15

=

50 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.71

y

30 lb 50o

B

x

33 in.

12 in. A

50 o 30 lb

Choosing A as the moment center, we get + =

C = MA = (30 sin 50 ) (33) 527 lb in. J

(30 cos 50 )(12)

2.72 Choosing A as the moment center, we get C = MA = 60(3)i + 60(2)j 30(2)j = 180i + 60j 90k lb ft J

30(3)k

2.73 C = P

=

0:4i 0:3j + 0:4k = 37:48i 28:11j + 37:48k N m = 60 p 0:42 + ( 0:3)2 + 0:42 0:3i + 0:4k 300k N rAD = 0:4i m = 0:6j + 0:8k AB = 0:5

60

DB

Moment of the couple: (MAB )C = C

AB

=

28:11( 0:6) + 37:48(0:8) = 46:85 N m

Moment of the force: (MAB )P = rAD

P

AB

=

0:4 0 0

0 0 0:6

0 300 0:8

= 72:0 N m

Combined moment: MAB = (MAB )C + (MAB )P = 46:85 + 72:0 = 118:9 N m J

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*2.74

2.75 Moment of a couple is the same about any point. Choosing B as the moment center, we have F= C = MB = rBA

30i kN

F=

i 0 30

rBA = j 1:8 0

1:8j k 1:2 0

1:2k m

= 36:0j

54:0k kN m J

2.76 Moment of a couple is the same about any point. Choosing B as the moment center, we have rBA = 180i bj mm Cz

180 b 0 90 60 F k = 150 0 0 1 16 200 = 0 b = 108:0 mm J

=

(MB )z = rBA

)

150b

= 150b

16 200 kN mm

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2.77 C = MA = 20(24)i 80(16)j + 50(24)k = 480i 1280j + 1200k lb in. J

2.78

2.79

2.80

C0

z

30o

CP= 6P x

CP C0 CR

y CR = 2R

30o

= 6P i = 6(750)i = 4500i lb in. = C0 ( cos 30 i + sin 30 k) = C0 ( 0:8660i + 0:50k) = 2R( sin 30 i cos 30 k) = R(i + 1:7321k)

C = (4500

0:8660C0

R)i + (0:5C0

1:7321R)k = 0

Equating like components: 4500 0:8660C0 R 0:5C0 1:7321R

= 0 = 0

The solution is: R = 1125 lb J

C0 = 3900 lb in. J

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2.81 z

C

3 N.m 25o

θ 1.8 N.m

x y 1.8 N.m

The system consists of the four couples shown, where C = 0:36F (i cos + k sin ) N m C=

2(1:8)k + 3( i cos 25 + k sin 25 ) + 0:36F (i cos + k sin ) = 0

Equating like components: 3 cos 25 + 0:36F cos 3:6 + 3 sin 25 + 0:36F sin

F cos

=

F sin

=

tan F

2.82

= =

0 0

3 cos 25 = 7:553 0:36 3:6 3 sin 25 = 6:478 0:36

6:478 = 0:8577 = 40:6 J 7:553 p = 7:5532 + 6:4782 = 9:95 N J

=

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2.83

2.84 (a)

0.7 m

B + +

#

P

P

C

=

A

CR = C − 0.7P A

R = P = 140 N down J C R = MA = C 0:7P = 180

0:7(140) = 82:0 N m CCW J

(b)

P

CR = C − 0.7P

C R/0.15 − P

0.15 m

=

A

B

A

C R/0.15 FA

=

FB

=

CR 82 P = 140 = 407 N up J 0:15 0:15 CR 82 = = 547 N down J 0:15 0:15

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2.85

+

#

R = F = 15 20 + 20 = 15 kN J C R = MA = 15(3) 20(6) + 20(8) = 85 kN m J

2.86 R = +

90j + 50(i sin 30 j cos 30 ) = 25:0i R C = 90(9) 50(12) = 210 lb in.

133:3j lb J CR = 210k lb in. J

2.87 The resultant force R equals V . ) V = R = 1400 lb J CR 20 (1400)

=

MD = 0: 20V 10H 750 (12) = 0

10H C = 0 H = 1900 lb J

2.88 R CR

CR

= 250k N J = MO = 250(1:2)i + 250(0:8)j = 300i + 200j N m J p = ( 300)2 + 2002 = 361 N m

250 N

361 N. m

O 3

2

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2.89 2:2i + 2:0j 2:0k = 270 p ( 2:2)2 + 2:02 + ( 2:0)2 = 165:8i + 150:7j 150:7k kN J i j k 0 2 0 = = rOB F = 165:8 150:7 150:7

F =

CR

270

AB

301i + 332k kN m J

2.90 3i 2k 40 p = ( 3)2 + ( 2)2 3i 5j 90-lb ft couple: C = 90 p = ( 3)2 + ( 5)2 rOA = 3i + 5j ft 40-lb force: P

R CR

= P=

33:28i

= C + rOA =

=

157:3i

33:28i

22:19k lb

46:30i

77:17j lb ft

22:19k lb J

P=

46:30i

77:17j +

i j 3 5 33:28 0

k 0 22:19

10:6j + 166:4k lb ft J

*2.91 (a) R rOA CR

= F = 2800i + 1600j + 3000k lb J = 10i + 5j 4k in. i j k 10 5 4 = rOA F = 2800 1600 3000 = 21 400i 18 800j + 30 000k lb in. J

(b) Normal component of R

:

Shear component of R

:

P = jRy j = 1600 lb J p p V = Rx2 + Rz2 = ( 2800)2 + 30002 = 4100 lb J

(c) Torque: T Bending moment: M

CyR = 18 800 lb in. J q p (CxR )2 + (CzR )2 = 21 4002 + 30 0002 =

=

=

36 900 lb in. J

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2.92

2.93 F = C = rBA

=

1:2i + 0:8k = 499:2i + 332:8k N ( 1:2)2 + 0:82 1:2i + 1:8j = 665:6i + 998:5k N m 1200 p (1:2)2 + 1:82 1:2i 1:8j m

600 p

R

= F=

CR

= rBA

499:2i + 332:8k N J i j k 1:2 1:8 0 F+C= 499:2 0 332:8

+C

= ( 599:0i 399:4j 898:6k) + ( 665:6i + 998:5k) = 1265i 399j + 100k N m J

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2.94 MAB P MAB 6:128P

= rAO P AB = 850 lb ft rAO = 8j ft = P (cos 20 i + sin 20 k) cos 30 i + sin 30 k AB = 0 8 0 cos 20 0 sin 20 = 6:128P = P cos 30 0 sin 30 = 850 lb ft P = 138:7 lb J

2.95 Given force and couple: F = C =

3i 4j + 6k 32 p = 12:292i 16:389j + 24:58k kN ( 3)2 + ( 4)2 + 62 3i 4j 180 p = 108:0i 144:0j kN m 32 + ( 4)2

Equivalent force-couple ststem at A: R CR

= F=

12:29i

= C + rAB =

206i

16:39j + 24:6k kN J

F = 108:0i

144:0j +

i 3 12:292

j 4 16:389

k 0 24:58

70:3j + 98:3k kN m J

2.96 T1 T2 T3

R

= =

3i 7j 60 p = 23:64i 55:15j kN ( 3)2 + ( 7)2 6i 7j = 39:05i 45:56j kN = 60 p 62 + ( 7)2 3i 2j = 60 p = 49:92i 33:28j kN ( 3)2 + ( 2)2 =

T = ( 23:64 + 39:05 49:92)i + ( 55:15 34:51i 133:99j kN J

45:56

33:28)j

Noting that only the x-components of the tensions contribute to the moment about O: CR = MO = [7(23:64)

7(39:05) + 2(49:92)] k =

8:03k kN m J

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2.97 MO

My

= rOA = =

F=

i j b 0:25 10 20

k 0:3 5

7:25i + (3 + 5b)j + ( 2:5 + 20b)k kN m 3 + 5b = 8 ) b = 1:0 m J

MO =

7:25i + 8j + 17:5k kN m J

2.98

2.99 F =

160i ! r = BA =

120j + 90k N

0:36i + 0:52j 0:48k m i j k 0:36 0:52 0:48 C = MB = r F = 160 120 90 = 10:80i + 109:2j + 126:4k N m J

2.100 (a) MO P MO

= rOA P + C rOA = 4k ft 3i 4k = 480i 640k lb C = 1400k lb ft = 800 5 i j k 0 0 4 = + 1400k = 1920j + 1400k lb ft J 480 0 640

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(b) MOF

= MO =

OF

= (1920j + 1400k)

3i + 12j + 4k 13

1920(12) + 1400(4) = 2200 lb ft J 13

2.101 Rx Ry The solution is

= =

Fx = T1 sin 45 T3 sin 30 = 0 Fy = T1 cos 45 + T3 cos 30 + 250 = 750 T1 = 259 lb J

T3 = 366 lb J

2.102

y

F CT x A

C d

Transferring F to point A introduces the couple of transfer C T which is equal to the moment of the original F about point A: C T = Fy d = 300d The couples C and C T cancel out if C = CT

600 = 300d

d = 2 ft J

2.103

rOA

= F = 40i + 30k kN J = 0:8i + 1:2j m

CR

=

R

MO = rOA

R=

i 0:8 40

j k 1:2 0 0 30

= 36i

24j

48k kN m J

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2.104

2.105

2.106

62 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.107 F =

400i + 300j + 250k lb 3j + 4k C = C = C( 0:6j + 0:8k) 5 rDA = 3j ft 0:6i + 0:8k DE =

(MDE )P

= rDA

(MDE )C

= C

P DE

MDE 510 + 0:64C

DE

0 400 0:6

=

3 300 0

= C( 0:6j + 0:8k) (

0 250 0:8

= 510 lb ft

0:6i + 0:8k) = 0:64C

= (MDE )P + (MDE )C = 1200 lb ft = 1200 C = 1078 lb ft J

2.108 z B 300 O x

200

400 N 400 N A

400

D

300 N E

y 400 200 N

300 N

200 N Split the 500-N force at D into the 200-N and 300-N forces as shown. We now see that the force system consists of three couples. CR

= =

C= 120i

300(0:4)i 200(0:4)j 80j 80k N m J

400(0:2)k

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Chapter 3 3.1

3.2 (a) + +

"

R= F =0 J C R = MA = 94 + 20(7)

36(2) = 162:0 kN m J

(b) From part (a): R + CR

= =

0 J MB = 94 + 36(5)

16(7) = 162:0 kN m J

The answers make sense: since the resultant is a couple, its moment is the same about any point.

3.3 R = + =

F = 75i (20 + 60)j = 75i 80j lb J C R = MO = 75(6 cos 30 ) + 20(9 cos 30 ) 77:9 lb in. CCW J

60(9 cos 30 )

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3.4

3.5

400 N 374.53 N

8

3 140.45 N + + +

! " =

Rx = Fx = 200 140:45 = 59:6 N Ry = Fy = 374:53 + 300 = 674:5 N C R = MO = (374:53 + 300) (0:6) 200(0:4) 324:7 N m

The equivalent force-couple system with the force acing at O is R = 59:6i + 674:5j N J

C R = 324:7 N m CCW J

3.6 Fx Fy

= Rx + ! 200 + 120 cos 30 + Px = 80 ) Px = 176:1 lb = Ry +" 120 sin 30 + 80 + Py = 20 ) Py = 120 lb ) P = 176:1i 120j lb J MO

= 0 + 176:1(6 + b) + 200(6) + 80(3) = 0 ) b = 2:18 in. J

3.7 The equivalent force-couple system with the force acing at O is R CR

= = =

F = (400 + 300 500)k = 200k lb J MO = [8j ( 500k)] + (6i 400k) 400i 2400j lb ft J

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3.8 R=

CR

rDA rDB rDO

= = =

=

MD =

F = (240 + 200

400) k = 40k lb J

6(1 cos 40 )i 6 sin 40 j = 1:4037i 3:857j ft 6 cos 40 i + 6(1 sin 40 )j = 4:596i + 2:143j ft 6 cos 40 i 6 sin 40 j = 4:596i 3:857j ft

+

i 1:4037 0 i 4:596 0

j k 3:857 0 0 240 j 3:857 0

+

i 4:596 0

j k 2:143 0 0 200

k 0 400

= ( 925:7i 336:9j) + (428:6i + 919:2j) + (1542:8i = 1046i 1256j lb ft J

1838:4j)

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3.9

3.10

67 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.11 R = P + Q = ( 8i + 10j) + 8i = 10j lb

MO = CR i a 8

ai P + (8j i j k 0 0 + 0 8 10 0

O.K.

bk) Q = j k 8 b = 0 0

110j 110j

10ak + ( 8bj 64k) = 110j 10a 64 = 0 a = 6:4 in. J 8b = 110 b = 13:75 in. J

3.12 FAB

= F

FAC

= F

CR

2:4F CR R

3i

4k 5

3j

4k 5

= (0:6i

0:8k)F

= (0:6j

0:8k)F

=

MO = rOA (FAB + FAC ) 30(6)j i j k 0 0 4 = F 180j = 2:4F i + (2:4F 0:6 0:6 1:6 180 = 0 F = 75 kN J 2:4(75)i = 180i kN m J F = 75(0:6i + 0:6j 1:6k)

= =

30i = 15i + 45j

180)j

120k kN J

3.13 1:5i + 2j 120 p = 72:0i + 96:0j N ( 1:5)2 + 22 2j 2k PB = 100 p = 70:71j 70:71k N 22 + ( 2)2 1:5i + 2k C = 180 p = 108:0i + 144:0k N m ( 1:5)2 + 22 PA

R

=

= PA + PB = ( 72:0i + 96:0j) + (70:71j = 72:0i + 166:71j 70:71k N J

70:71k)

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CR

=

MC = rCD PA + rCD PB + C = rCD R + C i j k 0 0 2 + ( 108:0i + 144:0k) = 72:0 166:71 70:71 = 225i + 144j + 144:0k N m J

3.14

3.15 180i + 150j + 210k 20 p = 11:442i + 9:535j + 13:348k N ( 180)2 + 1502 + 2102 300i 200j + 210k = 4:314i 2:876j + 3:020k N m = 6p 3002 + ( 200)2 + 2102 = 0:48i 0:35j m

=

FAC CBC rBA CR

R

= rBA

FAC + CBC i j k 0:48 0:35 0 = + 4:314i 2:876j + 3:020k 11:442 9:535 13:348 = 0:358i 9:28j + 3:59k N m J = FAC = 11:44i + 9:54j + 13:35k N J

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3.16

3.17 + " + ! The solution is

F = 3200 F =0

T1 sin 45 + T2 sin 25 = 3200 T1 cos 45 + T2 cos 25 = 0

T1 = 3090 lb J

T2 = 2410 lb J

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3.18 Ry Rx P

3 + " 600 = 800 sin + (400) = 26:74 5 4 = Fx + ! 0 = 800 cos + (400) + P 5 = 800 cos 26:74 320 = 394 lb J

=

Fy

J

3.19

3.20 Rx Ry

= =

Fx = 50 cos 60 = 25 lb Fy = 150 50 sin 60 =

193:3 lb

Let A be the point where the line of action of R crosses the x-axis. + xA =

MO = Ry xA 0:1454 ft

Resultant force R = 25i

875

150(4)

(50 sin 60 ) (7) =

193:3xA

193:3j lb intersects the x-axis at x = 0:1454 ft. J

3.21 Rx Ry

= =

Fx = 8 6 = 2 kN Fy = 6 + 15 = 21 kN

Let the resultant intersect the x-axis at x = xA . + xA =

MO = Ry xA 1:429 m

12

8(1:5) + 15(2) = 21xA

Resultant force R = 2i + 21j kN intersects the x-axis at x = 1:429 m. J

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3.22 (a)

= 30 Rx Ry CR

= = =

Fx = 400 + 2(400) sin 30 = 0 Fy = 0 by inspection MO = 3(400)(2) = 2400 lb in. CCW

) Resultant is the couple C R = 2400 lb in. CCW J (b) = 45 Rx Ry MO

= Fx = 400 + 2(400) sin 45 = 165:69 lb = Fy = 0 by inspection = Rx d 3(400)(2) = 165:69d d = 14:49 in.

y O

x

d = 14.49 in. 165.7 lb Resultant force R = 165:7i lb intersects y-axis at y =

14:49 in. J

3.23 +

Rx MA

= =

Fx = 0 Ry = Fy = 1200 + 600 = 600 lb 1000(4:5) + 1200(8:2) 600(13:7) = 6120 lb ft 6120 MA = = 10:20 ft jRy j 600 600j lb intersecting the x-axis at x = 10:2 ft J x=

) Resultant is R =

3.24 Rx

=

Fx = 300

Ry

=

Fy = 300

4 5 3 5

240 = 0 120

60 = 0

The resultant is not a force. +

CR =

MO =

60(0:12) =

7:2 N m

The resultant is the couple C R = 7:2 N m CW J

72 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.25

3.26 MO C R

= 0 + C + (160 40)(12) + (80 200)(15) = 0 = 360 lb in. = 30 lb ft J = F = 200 + 80 160 40 = 80 lb ! J

3.27 1 10 P1

y 4P 5 3

P2 O

B 3 ft

4 ft

3 10 P1

2 ft x

3 5 P3

+

!

Rx =

Fx = 0

+

"

Ry =

Fy = 0

+

MB = 120

Solution is P1 = 34:5 lb J

1 p P1 10 3 p P1 10 3 3 p P1 10

P2 = 32:7 lb J

4 P2 + P3 = 0 5 3 P3 = 0 5 1 + 2 p P1 10

= 120

P3 = 54:5 lb J

73 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.28

Solution is

Fx

=

Fy

=

MB

=

0

+ ! 220

0

P1 = 134:1 kN J

+#

3 P2 + P3 = 0 5 4 P1 + P3 = 220 5

4 P1 (2) 5

+

P2 = 64:4 kN J

2P3 = 0 P3 = 107:3 kN J

*3.29

3.30 All three cases represent parallel force systems. (a) Mx My

= Ry = Rx

250 = 50y y = 5:0 ft 200 = ( 50)x x = 4:0 ft

z

50 lb y

x

4 ft

5 ft

J

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(b) Mx My

= Ry = Rx

0 = 50y y=0 250 = 50x

x = 5:0 m

z 50 kN y 5m J

x (c) Mx My

= Ry = Rx

320 = 40y 400 =

y = 8:0 m 40x x = 10:0 m

z 40 N

y

10 m x

8m

J

3.31

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3.32 T1

= T1

1

T2

= T2

2

T3

= T3

3

8j + 4k p = 254:4i 678:4j + 339:2k lb 89 4i 8j + 4k p = 600 = 244:9i 489:9j + 244:9k lb 96 8j 6k = 400 = 320:0j 240:0k lb 10

= 800

3i

(254:4i 678:4j + 339:2k) + ( 244:9i + ( 320:0j 240:0k) R=

T = 9:5i

489:9j + 244:9k)

1488j + 344k lb J

3.33 T1

= T1

1

T2

= T2

2

T3

= T3

3

8j + 4k p = T1 (0:3180i 0:8480j + 0:4240k) 89 4i 8j + 4k p = 620 = 253:1i 506:2j + 253:1k lb 96 8j 6k = T3 = T3 ( 0:8j 0:6k) 10

= T1

3i

R = Rj = T Equating like components: 0 = 0:3180T1 253:1 R = 0:8480T1 506:2 0:8T3 0 = 0:4240T1 + 253:1 0:6T3 Solution is R = 1969 lb

T1 = 796 lb J

T3 = 984 lb J

3.34 Due to symmetry P2 = P3 . P1 P2

20i + 32k = ( 0:5300i + 0:8480k)P1 = P1 p ( 20)2 + 322 20i + 15j = P2 p = (0:8000i + 0:6000j)P2 202 + 152 Fx Fz

= 0 = 200 lb

Solution is: P1 = 236 lb J

0:5300P1 + 2(0:8000)P2 = 0 0:8480P1 = 200 P2 = P3 = 78:1 lb J

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3.35

3.36 Let the resultant intersect the plate at (x; y). R

The resultant is R =

=

Fz = 6P My 2P a x = = R 6P Mx 3P b y = = = R 6P 6P k intersecting the plate

=

a 3

b 2 at (a=3; b=2).

3.37 Let the resultant R intersect the plate at (x; y). R

=

Fz = 60 + 70 10 = 120 lb My ( 60)(6) x = = = 3:0 ft R 120 Mx 60(6) + 70(0:75) 10(5:25) y = = = 3:0 ft R 120 ) Resultant is R = 120k lb acting at the center of the plate J

3.38

z R = 8 lb

x

Solution is

R

=

Fx

20R

=

Mx

15R

=

My

C = 360:0 lb in. J

15 in.

3

4 20 in.

y

A

8 = P1 + P2 3 20(8) = P2 (5) + C 5 15(8) =

P2 (15)

P1 = 19: 2 lb J

4 C 5 P2 =

11: 2 lb J

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3.39 Let x and y be the coordinates of the point where the resultant intersects the xy-plane

z 0.8C

4

C 3

y

0.6C

:x R Mx

= Fz = 35 + 20 = 55 lb = Ry P2 (5) + 0:6C = Ry 5P2 + 0:6C 5(20) + 0:6(80) y = = = 2:69 in. R 55 My = Rx P2 (15) 0:8C = Rx 15P2 + 0:8C 15(20) + 0:8(80) x = = = 6:62 in. R 55

The resultant force R = 55k lb passes through the point (6:62; 2:69; 0).J

3.40 Let the resultant R intersect the plate at (x; y). R

=

Fz = 300 + 120 200 = 380 lb 500 120(3) + 200(3 sin 30 ) My = = R 380 Mx 300(3) 200(3 cos 30 ) = = 1:001 ft R 380

x = y

=

) Resultant is R =

1:474 ft

380k lb acting at ( 1:474 ft; 1:001 ft; 0) J

3.41 R = F = 6i + 8j + 5k kN J C = 18i + 24j + 15k kN m

CR

= C + rOA =

F=

18i + 24j + 15k+

i 3 6

j 4 8

k 2 5

14i + 27j + 15k kN m J

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3.42 Note symmetry about the y-axis. ) My = 0. R MO

= F = 60j + 80k lb = 20(4)i = 80i lb ft

Because R and MO are perpendicular, the resultant is a force. Due to symmetry, the resultant must intersect the y-axis. Let the coordinate of the crossing point be y. Equating moments about the x-axis of the original force system and the resultant, we get Mx = R z y

80 = 80y

y = 1:0 ft

Resultant force R = 60j + 80k lb crosses the y-axis at y = 1:0 ft. J

3.43 The couple of the wrench is C = = CR

600i + 1400j + 700k R = 1200 p jRj 6002 + 14002 + 7002 429:5i + 1002:2j + 501:1k lb ft

1200

= C + rOA

R

i 3 600 1098j + 3501k lb ft J

=

429:5i + 1002:2j + 501:1k +

=

1830i

j 2 1400

k 0 700

79 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.44 Unit vector in the direction of R is =

R 250i + 360j 400k = 0:4213i + 0:6067j =p jRj 2502 + 3602 + ( 400)2

The component of CR in the direction of CtR CR t

0:6741k

is

= CR = 1200(0:4213) + 750(0:6067) + 560( 0:6741) = 583:1 N m R = Ct =583:1(0:4213i + 0:6067j 0:6741k) = 245:7i + 353:8j 393:1k N m

The component of CR that is normal to CR n

is

= CR CR t = (1200i + 750j + 560k) (245:7i + 353:8j = 954:3i + 396:2j + 953:1k N m

393:1k)

Let A(x; y; 0) be the point where the wrench intersects the xy-plane. rOA

R=

i x 250

j y 360

k 0 400

=

Equating x- and y-components of rOA 400y = 954:3 400x = 396:2

400yi + 400xj + (360x

250y)k

R = CR n yields y = 2:39 m x = 0:991 m

) The wrench consists of: Force R Couple CR t

= 250i + 360j 400k N J = 245:7i + 353:8j 393:1k N m J

The axis of the wrench passes through the point (0:991 m,

2:39 m, 0) J

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3.45 (a) R CR

F = 8k kN J MO = 6(1:5)i + 6(1:2)k =

= =

9:0i + 7:2k kN m J

(b) Note that = k is the unit vector in the direction of R. The component of CR in the direction of k is CR t = 7:2k kN m. ) The equivalent wrench consists of Couple = CR t = 7:2k kN m J

Force = R = 8k kN

The component of CR that is normal to R CR n =C

is

CR t = ( 9:0i + 7:2k)

7:2k = 9:0i kN m

Let A(x; y; 0) be the point where the wrench intersects the xy-plane. rOA

R

i j k x y 0 = 9:0i 0 0 8 x=0 J y = 1:125 m J

= CR n )

8:0yi

8:0xj = 9:0i

3.46 Area of sign: Resultant:

2502 = 515 103 mm2 = 0:515 m2 2 R = pA = 110(0:515) = 56:7 N A = 8002

4

Resultant is a 56.7 N normal force acting at the center of the sign. J

3.47

81 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.48 y

x3 x1

A

P1

=

P2

=

P3

=

R MA

P2

x2 P1

500 lb/ft

B

6 ft

P3

1200 lb/ft x C

8 ft

1 (1200)(6) = 3600 lb 2

2 (6) = 4 ft 3 1 1200(8) = 9600 lb x2 = 6 + (8) = 10 ft 2 2 1 (500)(8) = 2000 lb x3 = 6 + (8) = 11:333 ft 2 3 x1 =

= P = 3600 + 9600 + 2000 = 15 200 lb # J = Rx: 3600(4) + 9600(10) + 2000(11:333) = 15 200x ) x = 8:75 ft J

3.49 y x 1 P1

w0

L/2 L/2

+

P1

= P2 =

x1

=

1 3

L 2

R

=

P =0

CR

=

MA =

1 w0 2

w0

x2

L 2 L = 6

x P2

1 w0 L 4 L 2 x2 = + 2 3

=

L 2

) Resultant is a couple 1 L 1 5 w0 L + w0 L L 4 6 4 6

=

5 L 6

=

1 w0 L2 6

J

82 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.50 y

3000 lb 900 lb 2 ft

2 ft 2000 lb

3 ft O

x

x R

R = 3900i + x =

MO = Ry x 7:40 ft J

2000j lb J

900(2)

3000(3)

2000(2) =

2000x

3.51

83 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.52

z z

w0a2

y a/2

w0a2/8 y

+

a/2 a/3

a/2 x

x Loading is directed down

Loading is directed up

Use the superposition shown in the …gure. R=

1 w0 a2 + w0 a2 k = 8

7 w0 a2 k J 8

Let (x; y) be the point where R crosses the xy-plane. Due to symmetry x = y. My

=

w0 a2

Rx

x = y=

a 2

1 a a w0 a2 + = 8 2 3

7 w0 a2 x 8

19 aJ 42

3.53 y B

12 ft 7.5 ft

P1 y1

y2

P2 x

A 749 lb/ft2

468 lb/ft2

P1

=

P2

=

1 (468)(7:5)(22) = 38 610 lb 2 1 (749)(12)(22) = 98 870 lb 2

1 (7:5) = 2:50 ft 3 1 y2 = (12) = 4:0 ft 3 y1 =

84 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

+

R = P2 P1 = 98 870 38 610 = 60 260 lb J + MA = Ry P2 y2 P1 y1 = Ry 98 870(4) 38 610(2:5) = 60 260y y = 4:96 ft J

3.54 y

4

6

5 1 R1 2

8

12

Dimensions in feet

750 lb/ft2 R2

7.5 O

R1 R2 R3

R3

{

3.75 12

467 lb/ft

2.5

x

2

p 1 (750)(20) 62 + 122 = 100:6 103 lb 2 = 750(20)(7:5) = 112:5 103 lb 1 (7:5)(467)(20) = 35:0 103 lb = 2

=

85 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.55 y x1

4m

P1 6 MN/m 4 MN/m

Ry y

P2 Rx 2.5 m

y3

x1 =

x 4 MN/m

O

P1

=

1 (6)(2:5) = 7:5 MN 2

P2

=

4(4) = 16 MN

P3

=

1 (4)(4) = 8 MN 2

P3 y 2

2 (2:5) = 1:6667 m 3

1 (4) = 2 m 2 1 y3 = (4) = 1:3333 m 3

y2 =

R = (P2 + P3 )i P1 j = 24i 7:5j MN J MO = Rx y + P2 y2 + P3 y3 P1 x1 = Rx y 16(2) + 8(1:3333) 7:5(1:6667) = ( 24y) y = 1:257 m J

3.56 Rx Ry R

= Fx = 520 cos 50 + 340 cos 20 140 = 513:7 lb = Fy = 520 sin 50 340 sin 20 = 282:1 lb = 514i + 282j lb J

3.57

86 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.58 +"

R=

Fz =

200

50

150 =

400 kN

)R=

400k kN J

The coordinates of the point where R crosses the xy-plane are x = y

=

My 200(2) = = 1:0 m J R ( 400) 150(3) Mx = = 1:125 m J R ( 400)

3.59

3.60

z 270 lb

x

1.5 ' 4.5 '

5' 2.2

135 lb 3'

y

87 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

R = ( 270 My x Mx y

135) k =

405k lb J

= Rx 270(2:25) + 135(3) = ( 405)x = 2:50 ft J = Ry 270(1:5) 135(4:5) = 405y = 2:50 ft J

3.61 y

2 ft 240 lb

O

x 9.2 ft 240 lb

The resultant is the couple C R = 240(7:2) = 1728 lb ft CCW J

3.62 R CR

= 15i + 18j + 15k kN J = MO = 25(1:2)i + 30(0:9)j + 15(1:2)k kN m = 30i + 27j + 18k kN m J

3.63

3.64 Rx = 140 kN

Ry = 10 kN

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Find the y-coordinate of the point where the resultant crosses the y-axis: + y =

MO = 1:029 m

Rx y

140

40(0:4) + 60(0:2) =

140y

Resultant force R = 140i + 10j kN intersects the y-axis at y = 1:029 m. J

3.65 R Mx y My x

= Fz = 20 60 40 = 120 kN = Ry 60(4) + 40(4 sin 40 ) 20(4 sin 30 ) = 120y = 2:52 m J = Rx 40(4 cos 40 ) 20(4 cos 30 ) = ( 120x) = 0:444 m J

3.66 BE

=p

= = FA = C = =

FE

R CR

50i + 80j 502

+

802

30k + ( 30)2

= 0:5051i + 0:8081j

0:3030k

FE BE = 180(0:5051i + 0:8081j 0:3030k) 90:92i + 145:46j 54:54k lb 250k lb C( BE ) = 620(0:5051i + 0:8081j 0:3030k) 313:2i 501:0j + 187:9k lb in.

= FA + FE = 250k+ (90:92i + 145:46j 54:54k) = 90:9i + 145:5j + 195:5k lb J = MD = C + rDE FE i j k 0 80 0 = 313:2i 501j + 187:9k + 90:92 145:46 54:54 = 4676i 501j 7086k lb in. J

89 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.67 (a) + ! +

"

+

3 4 (200) (120) = 44 lb 5 5 4 3 Ry = Fy = (200) (120) = 88 lb 5 5 C R = MO = 600 + 200(2:5) = 1100 lb ft Rx = Fx = 260

The equivalent force-couple system is C R = 1100 lb ft CCW J

R = 44i + 88j lb acting at O (b) R = +

44i + 88j lb as in Part (a) C R = MA = 600 + 260(3)

200(2:5) = 880 lb ft

The equivalent force-couple system is R = 44i + 88j lb acting at A 96:0zD i

72:0zD j + (96:0xD

C R = 880 lb ft CCW J

144:0)k =

69:12i

51:84j + 144:0k

Equating i and k components:

(96:0xD

96:0zD = 144:0) =

69:12 144:0

zD = 0:720 m xD = 3:00 m

Wrench consists of the force P = 72:0i + 96:0j N and the couple CtR = 38:88i + 51:84j N m; the axis of the wrench passes through the point (3.00, 0, 0.720) m. J

90 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.68 + ! Rx = Fx = 300 cos 60 + 100 + 200 cos 30 = 423:2 lb + " Ry = Fy = 300 sin 60 200 sin 30 = 159:8 lb p R = 423:22 + 159:82 = 452 lb J

3.69

P

=

120

AC

CR

=

180

AB

1:5i + 2j = 72:0i + 96:0j N 2:5 1:5i + 2k = 180 = 108:0i + 144:0k N m 2:5

= 120

Component of CR in direction of CtR CR t

= CR = CtR

is

= ( 108:0i + 144:0k) ( 0:6i + 0:8j) = 64:80 N m = 64:80( 0:6i + 0:8j) = 38:88i + 51:84j N m

AC AC

Component of CR normal to CR n

AC

AC

is

= CR CR t = ( 108:0i + 144:0k ) = 69:12i 51:84j + 144:0k N m

( 38:88i + 51:84j)

Let D be the point where the axis of the wrench crosses the xz-plane. rAD rAC

P

=

(xD

= CR n

96:0zD i

1:5)i + zD k m i j k 0 zD xD 1:5 72:0 96:0 0

72:0zD j + (96:0xD

144:0)k =

=

69:12i

69:12i

51:84j + 144:0k

51:84j + 144:0k

Equating i and k components:

(96:0xD

96:0zD = 144:0) =

69:12 144:0

zD = 0:720 m xD = 3:00 m

Wrench consists of the force P = 72:0i + 96:0j N and the couple CtR = 38:88i + 51:84j N m; the axis of the wrench passes through the point (3.00, 0, 0.720) m. J

91 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3.70 T1

= T1

1

T2

= T2

2

T3

= T3

3

2j 8k p = T1 (0:3419i 0:2279j 77 3j 8k = T2 p = T2 (0:3511j 0:9363k) 73 4i 8k = 500 p = 223:6i 447:2k N 80

= T1

3i

0:9117k)

R = Rk = Ti Equating like components: 0 = 0:3419T1 223:6 0 = 0:2279T1 + 0:3511T2 0:9117T1 0:9363T2 R =

447:2

Solution is T1 = 654 N J

T2 = 425 N J

R=

1441 N J

92 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 4 4.1

T

1.5 m

40o

B 1.2 m

0.75 m Ax

A

W

Ay

W = 30(9:81) = 294:3 N Three unknowns: T; Ax ; Ay J

4.2

0.3 m

0.5 m

A

0.4 m

0.5 m RA W2

W1 W1

=

W2

=

B

Bx

By

0:5 (30) (9:81) = 113:19 N 1:3 0:8 (30)(9:81) = 181:11 N 1:3

Three unknowns: RA , Bx , By J

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4.3

0.9 m 0.8 m

FA 30o

P

W 0.9 m

A

NA

T B

60o W = 30(9.81) = 294.3 N P = 0.5(1.2)(900) =540 N

4.4 in. 7 . 12

B

NB 12 lb

FA

FB 24 in.

in. 40 in. 5 3 . 26

A NA 4 unknowns (NA ; FA ; NB and FB ) J

4.5

94 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.6 Weight of the plate is W = 12(9:81) = 117:72 N FA

117.72 N 140 70

Ax

NA

120

Ay (a) 3 unknowns

117.72 N 140 70

30o

120

30o

(b) 4 unknowns FB

NB

30o NB

30o

Dimensions in mm 117.72 N 140 70 N 4 3

117.72 N 140 70

y x

120

120 FB

Bx (c) 3 unknowns

T 45o

(d) 3 unknowns

By

NB

4.7 70 lb. ft 10" 20"

Ax Ay

70 lb. ft 10"

60o NB

20"

Ax Ay

(a)

NB

(b)

95 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(a) 3 unknowns (Ax ; Ay and NB ) J (b) 3 unknowns (Ax ; Ay and NB ) J

4.8

Cy Cx RB

C B

2 in.

20o 6 in. 12 lb A Three unknowns: RB , Cx , Cy J

96 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.9

4.10

4.11

80 lb D

2 ft C

E

Ex

2 ft

3 ft ND

Ey

3 unknowns ( ND , Ex , Ey ) J

97 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.12

4.13 Refer to the given FBD. MA P

= 0 + 4P 120(4 sin 15 ) = 0 = 120 sin 15 = 31:1 lb J

4.14 Refer to the given FBD. MO Fy Fx

= 0 = 0 = 0

+ 18TC 18TB = 0 + " TB cos 25 + TC 60 = 0 + ! F TB sin 25 = 0

Solution is TB = TC = 31:5 lb

F = 13:30 lb J

4.15 Refer to the given FBD. Weight of boom = Weight of load = MA T Fx Fy

180(9:81) = 1766 N = 1:766 kN 320(9:81) = 3139 N = 3:139 kN

= 0 + T (2 sin 30 ) 1:766(2 cos 30 ) 3:139(4 cos 30 ) = 0 = 13:933 kN J = 0 + ! Ax T = 0 Ax = T = 13:933 kN = 0 + " Ay 1:766 3:139 = 0 Ay = 4:905 kN p A = 13:9332 + 4:9052 = 14:77 kN J

98 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.16 Refer to the given FBD. MA = 0

+

CA + 0:6(10 sin 15 ) + 0:6(20 sin 15 +1:8(20 sin 15 24) = 0 CA = 35:2 lb in. J

10)

4.17 Refer to the given FBD. MA

=

0 + 54 sin

0:6(10 sin ) + 0:6(20 sin 49:2 = 0 = 65:7 J

10) + 1:8(20 sin

24) = 0

4.18 Refer to the given FBD. MB

=

0

Fx

=

0

Fy

=

0

+

1:2(0:25)

NC (0:25) = 0 3 +! Bx + NC = 0 5 4 + " By 1:2 NC = 0 5

Solution is Bx RB

4.19

0:72 kN By = 2:16 kN NC = 1:2 kN J p 2 2 = 0:72 + 2:16 = 2:28 kN J =

Refer to the given FBD. MA

=

0

12 T (18) 182 + 122 9:985T

p

+

CA = 2880 Fx

=

0

+ !

Fy

=

0

+"

p

Ax Ay

300

300(9:6) + CA = 0

12 T =0 Ax = 0:5547T + 122 18 p T =0 182 + 122

182

Ay = 300 + 0:8321T (a) With T = 490 lb: CA Ax

= 2880 9:985(490) = 2013 lb ft J = 0:5547(490) = 272 lb J Ay = 300 + 0:8321(490) = 708 lb J

99 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) With T = 0: CA = 2880 lb ft J

Ax = 0 J

Ay = 300 lb J

4.20 Refer to the given FBD. MA = 0

+

400(2)

35(9:81)(3 cos ) = 0

= 39:0

J

4.21 Refer to the given FBD. Fx Fy MA

= 0 = 0 = 0

+ % P 1200 sin 10 = 0 + - NA + NB 1200 cos 10 = 0 + 5NB 3:6P + 1200( 1:5 cos 10 + 1:8 sin 10 ) = 0

Solution is P = 208 lb J

NA = 752 lb J

NB = 430 lb J

4.22 Refer to the given FBD. MA T Fx Fy Ay

= 0 + T (3 + 2:5 sin 70 ) 400(1:5) = 0 = 112:17 N J = 0 + ! Ax + 112:17 cos 70 = 0 Ax = = 0 + " Ay + 112:17(1 + sin 70 ) 400 = 0 = 182:4 N J

38:4 N J

4.23 Refer to the given FBD. MD b

= 0 + = 0:437 m J

850b + (388 sin 30 ) (0:6

b)

(388 cos 30 ) (1:2) = 0

4.24 Refer to the given FBD. MB

=

0

+

MA

=

0

+

Fx

=

0

+

8T2 340(4) = 0 T2 = 170 lb J 3 T3 (8) 340(4) = 0 T3 = 283:3 lb J 5 4 T1 (283:3) = 0 T1 = 226:6 lb J 5

100 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.25

4.26

NB

B

4 ft

y

4 ft

x 25 lb o

60

Ax A Ay

MA

4.27

FBD

=

0 + NB (8 sin 60 ) 25(4 cos 60 ) = 0 NB = 7:217 lb J Fx = 0 + ! Ax 7:217 = 0 Ax = 7:217 lb Fy = 0 + " Ay 25 = 0 Ay = 25 lb p A = 7:2172 + 252 = 26:0 lb J

W P

θ A

N

FBD

101 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Fx Fy

= =

0 0

N=

W cos

Solution is

+ ! P N sin = 0 + " N cos W =0 P = W tan

J

4.28

960 lb

800 lb

160 lb/ft B FBD

Ax A Ay

6 ft 8 ft

NB

12 ft

1 (160)(12) = 960 lb 2 + 12NB 800(6) 960(8) = 0 Ax = 0 J + " Ay + 1040 800 960 = 0

Resultant of distributed load is MA Fx Fy

= 0 = 0 = 0

NB = 1040 lb J Ay = 720 lb J

4.29 Weight of the bar is 40(9:81) = 392: 4 N Fx = 0 MA = 0

+!

T

+ T (3:25 sin 30 ) 1:625T 2:0NB + 552:2 1:625T 4:0T + 552:2

NB sin 30 = 0

NB = 2T

NB (2:0) + 392:4(1:625 cos 30 ) = 0 = 0 = 0 T = 233 N J

4.30

102 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.31

C

P

a

TAC A

TBC

a

a a 3

a 2

B

W 3 2W 3 FBD MC

=

0

+

P

=

2 W J 9

2W a 3 2

Wa 3 3

Pa = 0

4.32

600 lb

900 lb 7200.lb ft

A Ay Fx MB Fy

= 0 = 0 = 0

6 ft

6 ft

8 ft

+ Bx = 0 + 24Ay + 600(18) By + 600 900 = 0

The reactions are: RA = 0 J

4 ft

B Bx By

900(4) 7200 = 0 By = 300 lb

Ay = 0

RB = 300 lb " J

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4.33

4.34

4.35

104 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.36 120 mm 75 mm Ox

200 N . m

O Oy

2 kN

MO = 0

+

FBD

T

0:075T + 200

2000(0:12) = 0

T = 533 N J

4.37

40(9.81) N 0.5 m G 30o B D FA

0.3 m

P A

NA

(a) MA = 0 DA = BD = 40(9:81)(0:25)

+

40(9:81) DA

P BD = 0

0:5 sin 30 = 0:25 m 0:5 cos 30 0:3 = 0:133 01 m P (0:133 01) = 0

P = 738 N J

(b) If the inclined surface is smooth (FA = 0), the equation be satis…ed. Hence equilbrium is not possible. J

MG = 0 cannot

105 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.38

y 40(9.81) N

x

30o

C G

0.3 m

P A

NA

= 0 + % P cos 30 40(9:81) sin 30 = 0 P = 226:6 N J = 0 + P (0:3) C = 0 226:6(0:3) C = 0 = 68:0 N m J

Fx MG C

4.39 NA

80(9.81) = 784.8 N NA

B A

2m

2m 20o

784.8 N

NB

(b)

2m

(a)

C

2m

C

A

54o B 20o NB

(a) MB = 0

+

784:8(2)

C=0

C = 1570 N m

(b) Fy Fx MB

= =

0 + " NB cos 20 784:8 = 0 NB = 835:2 N 0 + ! NA NB sin 20 = 0 NA 835:2 sin 20 = 0 NA = 285:7 N = 0 + C + NA (4 sin 54 ) 784:8(2 cos 54 ) = 0 C + 285:7(4 sin 54 ) 784:8(2 cos 54 ) = 0 C = 1:96 N m J

106 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.40

4.41

107 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.42

P

θ

20 o

40 lb

30 lb

(a)

30o

30 lb

N

(b)

θ

N

(a) + ! P cos 20

N sin 30 = 0

N=

P cos 20 = 1:8794P sin 30

Fx

=

0

Fy

=

0 + " N cos 30 P sin 20 30 = 0 (1:8794P ) cos 30 P sin 20 30 = 0 P = 23:3 lb J

(b) Fx

=

0

+ ! 40 cos

Fy

=

0 + " N cos 30 40 sin 30 = 0 80 cos cos 30 40 sin 30 = 0 69:282 cos 40 sin 30 = 0

Solving numerically:

N sin 30 = 0

N=

40 cos sin 30

= 80 cos

= 38:0 J

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4.43 A

1200 N. m

1.0 m B 1.0 m

C 45o

T1 T1

2m T2

D

2m E

25o

Fy

MD

Fx

=

NE

0

+ " NE cos 25 T1 cos 45 = 0 T1 cos 45 = 0:7802T1 NE = cos 25 = 0 + 1200 (T1 sin 45 ) (3) T1 (2) (NE sin 25 ) (2) = 0 1200 4:121T1 0:8452NE = 0 1200 4:121T1 0:8452(0:7802T1 ) = 0 T1 = 251:0 N J = 0 + ! T1 (1 + sin 45 ) T2 NE sin 25 = 0 1:7071T1 T2 0:4226NE = 0 1:7071(251:0) T2 0:4226(0:7802 251:0) = 0 T2 = 346 N J

109 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.44

4.45

L/2

120 lb θ B 30 lb

NA

MA

=

0

2L /3

P A

+

(120 sin ) 160 cos

2L cos 3 15 = 0

+ (120 cos ) = cos

1

2L sin 3

15 = 84:6 160

30

L sin 2

=0

J

110 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.46

4.47 D 5 in.

T

4 in.

6 in.

Ax

C

B 12 in.

A

20 lb

Ay MA

=

0

MD MB

= 0 = 0 6Ay

6 p T (9) 20(18) = 0 T = 52:1 lb J 61 + 9Ax 20(18) = 0 Ax = 40:0 lb + 6Ay 4Ax + 20(12) = 0 4(40:0) + 20(12) = 0 Ay = 13:33 lb p A = 40:02 + 13:332 = 42:2 lb J

+

111 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.48

4.49

4.50

20 in.

20 in.

40 in. A Ax

C

B 40 lb

2400 lb . in.

30 in .

20o D

Ay

ND MA = 0

+ 40(80) + 2400 + (ND sin 20 )(30 sin 20 ) (ND cos 20 )(40 30 cos 20 ) = 0 5600 7:588ND = 0 ND = 738:0 lb J

112 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Fy

=

0 + " ND cos 20 + Ay 40 = 0 738:0 cos 20 + Ay 40 = 0 Ay = 653:5 lb = 0 + ! ND sin 20 + Ax = 0 738:0 sin 20 + Ax = 0 Ax = 252:4 lb p A = 252:42 + 653:52 = 701 lb J

Fx

4.51

y

22' + 8' = 30' 11' W = 420 lb 1

W2 = 168 lb

Ax

x 19'

19'

Ay Fx MA Fy

NC

380 lb

= 0 + ! Ax = 0 = 0 + 38NC 420(11) 168(30) 380(19) = 0 NC = 444:2 lb = 0 + " Ay 420 168 380 + NC = 0 Ay 420 168 380 + 444:2 = 0 Ay = 523:8 lb A = 524j lb J

NC = 444j lb J

4.52 C

20o 800 lb

2 ft B T Ax

18 ft

A Ay FBD

113 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

MA Fx Fy RA

4.53

= 0 + 2T 18(800 cos 20 ) = 0 T = 6766 lb J = 0 + ! 800 cos 20 Ax = 0 Ax = 752 lb = 0 + " Ay T 800 sin 20 = 0 Ay = 7040 lb p 2 2 = 752 + 7040 = 7080 lb J

114 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.54

4.55

115 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.56

y

AD DC

5.389 D 2.611 C θ = 54.31o 34o 6.5 F Dimensions T in meters B x FBD W = 1962 N 3.635

A

= 6:5 cos 34 = 5:389 m BD = 6:5 sin 34 = 3:635 m = AC AD = 8 5:389 = 2:611 m BD 3:635 = tan 1 = tan 1 = 54:31 2:611 DC W = 200(9:81) = 1962 N

Fx Fy

= 0 + ! F cos 54:31 T cos 34 = 0 T = 0:7037F = 0 + " T sin 34 + F sin 54:31 1962 = 0 0:7037F sin 34 + F sin 54:31 1962 = 0 F = 1627 N J

4.57

W

L/2 C

A

NA

60o Fx Fy

= 0 = 0

L/2 B o

FBD

NB

30

+ ! NA sin 60 NB sin 30 = 0 + " NA cos 60 + NB cos 30 W =0

Solution is NA = 0:5W +

NB = 0:8660W

MA

=

0

(NB cos 30 ) L

C

=

0:8660W L cos 30

W

W

L 2

C=0

L = 0:250 W L J 2

116 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.58

4.59

117 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.60

4.61

118 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.62

119 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.63

20o

CA Ax FBD's CA Ax Ay

A

A Ay

1.8

B 0.4

T

D

D 0.8 25 kN By T 20o

Dimensions in meters Bx Bx B 0.4 B 1.8 m 0.8 25 kN By

D

6 unknowns, 6 independent equations J

4.64 NC

NC

Dimensions in mm ND

ND 150 300

By

100

Bx

Bx

250

400 N

400 N

150 Ax

50

50

By Ax

200

100

50

50 Ay

Ay 6 unknowns, 6 independent equations J

120 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.65

Dy Dy

D

D T3 F

Dx T3

B

F T4

B

T1 A

T4

Dx

T2 T2

T1 A

200 lb FBD's

200 lb

7 unknowns, 7 independent equations J

4.66

121 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.67

4.68

Cy

Cy CC

CC

Cx

Cx

0.9 m Dy

0.3 m Dx Dx

2.1 m

0.6 m 0.3 m 0.9 m Dy

T1 0.6 m 300 N. m

T2

T2

0.9 m

Bx

T1 0.6 m

0.9 m

300 N. m

Bx

By

By

9 unknowns, 9 independent equations. J

122 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.69 A

A B

o

20

20o N1

W N2

W

20o

o

20

N3

N1

20o

N4

B

N4

20o

W N2

W

20o N3

4 unknowns, 4 independent equations.

123 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.70

TA

TB 45o 6.93 ft

o

30 B 5.07 ft

A

C NC

480 lb (a) Bar ABC with pin A in bar ABC Ay A

TA 30o Ax

TB B 45o 5.07 ft 6.93 ft

Ax A

C NC

Ay (b) Bar ABC with pin A removed

480 lb (c) Pin A

4.71 1000 lb A

2.5 ft

4 ft

2.5 ft B

NA

C Cx MC Cy

1000 lb A

2.5 ft

Bx

2.5 ft B By

NA

4 ft

Bx B By

C Cx MC Cy

6 unknowns, 6 independent equations.

124 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.72 4 ft 6 ft Ax

3 ft

D

MA 3 ft A

E

1200 lb

Ay 1200 lb

2

Bx

B

4

3

6 ft

9

D

By

T

3 ft 1200 lb

E

6 unknowns, 6 independent equilibrium equations.

125 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.73

3 kN 0.75 m 0.75 m

3.6 kN 0.9 m 0.9 m Ax A

Bx

Bx

Ay

C

B By

B By

CA

NC

From FBD of BC: MB Fy Fx

= 0 = 0 = 0

+ 1:5NC 3(0:75) = 0 NC = 1:5 kN + " By + NC 3 = 0 By = 1:5 kN + ! Bx = 0

From FBD of AB: Fx MA Fy

+ ! Ax Bx = 0 Ax = 0 J + CA 3:6(0:9) 1:8By = 0 CA = 5:94 kN m J + " Ay 3:6 By = 0 Ay = 5:1 kN J

= 0 = 0 = 0

4.74 By 6 ft

Bx Bx

B

B By P = 3150 lb

9 ft

3 ft

Ax

5.4 ft

Q = 2430 lb 1.8 ft Cx C 900 lb/ft Cy

700 lb/ft A Ay P =

1 (700)(9) = 3150 lb 2

Q=

1 (900)(5:4) = 2430 lb 2

FBD of BC: MC = 0

+

5:4Bx

2430(1:8) = 0

Bx = 810 lb

126 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of AB: MA

=

0 + 6By + 9Bx 3150(3) = 0 6By + 9(810) 3150(3) = 0 By = 360 lb p B = 8102 + 3602 = 886 lb J

4.75

By B 800 N

Bx 600 N

250

250

125 125

600 N

200

Ax A Ay

B

C

200 Cx 125 125

Cy

C

200 200 Cx

Cy

From FBD of ABC: MA MC Fx

= 0 = 0 = 0

+ 750Cy 800(250) + 750Ay 800(500) + ! Ax Cx = 0

600(625) = 0 600(125) = 0

From FBD of BC: MB = 0

+

250Cy

400Cx

600(125) = 0

Solution is Ax = 291:7 N

Ay = 633:3 N RA RC

Cx = 291:7 N

Cy = 766:7 N

p 291:72 + 633:32 = 697 N J p = 291:72 + 766:72 = 820 N J =

127 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.76 200(9.81) N 1.444 m A 30o 60o 5m

Ay Ax A

4.330 m 30o 1.5 m 3m T

Bx

C

5.774 m

B By

NC

2m

C NC

FBD of entire structure: MB = 0

+

5:774NC

200(9:81)(1:444) = 0

NC = 490:7 N

FBD of bar AC (with the pin at A removed): MA = 0

+

490:7(4:330)

1:5T = 0

T = 1416 N J

4.77

128 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.78 Cx

C Cy

160 lb

6 ft

300 lb

B

Ay Ax

A 1.75 ft 1.75 ft 1.85 ft B 300 lb B y 3 ft

Bx

Ax A Ay From FBD of ABC: MA Fx Fy

= 0 = 0 = 0

+ 6Cx 460(1:75) = 0 + ! Ax Cx = 0 + " Ay + Cy 460 = 0

From FBD of AB: MB = 0

+

3:60Ay

3Ax

300(1:85) = 0

Solution of these equations is Ax = 134:17 lb

Ay = 266:0 lb RA RC

Cx = 134:17 lb

Cy = 194:0 lb

p 134:172 + 266:02 = 298 lb J p = 134:172 + 194:02 = 236 lb J =

129 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.79

4.80 1.65 m

675 N

A

675 N B

Ax

Bx

0.75 m A Ay

B

2.4 m

By Ay

Ax D

E 0.9 m A

2.4 m NE

ND

Cy C Cx 1.2 m

1.2 m

E NE

From FBD of entire table: MD = 0

+

2:4NE

675(1:65) = 0

NE = 464:1 N

130 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

From FBD of member AB: MB = 0

+

675(0:75)

2:4Ay = 0

Ay = 210:9 N

From FBD of member ACE: MC

=

0 + 1:2NE 0:9Ax + 1:2Ay = 0 1:2(464:1) 0:9Ax + 1:2(210:9) = 0 Ax = 900 N Fy = 0 + " Cy + NE Ay = 0 Cy + 464:1 210:9 = 0 Cy = 253:2 N Fx = 0 + ! Ax Cx = 0 Cx = Ax = 900 N p p A = 9002 + 210:92 = 924 N J C = 9002 + 253:22 = 935 N J

4.81

131 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.82

6

7500 lb Fx Fx

F

7.2

3

Fy F

Fy

A NA

1.8 4.5

Dimensions in feet

11 000 lb B

C

NB

NC

From FBD of the cab: MF Fx Fy

= 0 + 7500(10:2) 7:2NA = 0 NA = 10 625 lb J = 0 + ! Fx = 0 = 0 + " NA Fy 7500 = 0 10 625 Fy 7500 = 0 Fy = 3125 lb

From FBD of the trailer: MB MF

=

0 + 6:3NC 6Fy 11 000(1:8) = 0 6:3NC 6(3125) 11 000(1:8) = 0 NC = 6119 lb J = 0 + 6NB + 12:3NC 11 000(7:8) = 0 6NB + 12:3(6119) 11 000(7:8) = 0 NB = 1756 lb J

4.83

132 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.84

TC

TC

TA

TC

A

D TA

3 ft C B x 150 lb

From FBD of pulley D: F =0

+ " 2TC

TA = 0

TA = 2TC

From FBD of bar ABC: F MA

= =

0 + " TA + TC 150 = 0 TC = 50 lb J 0 + 3TC 150x = 0 x = 1:0 ft J

TA = 100 lb J

4.85

TD

2.4 m

TD D T2

A

Ax

882.9 N

2.7 m

Ay

T1 T1 B NB 882.9 N

W = 90(9:81) = 882:9 N From FBD of the assembly: MA Fx Fy

= 0 + 2:4TD 882:9(2:7) = 0 TD = 993:3 N = 0 + ! Ax = 0 = 0 + " Ay + TD 882:9 = 0 Ay + 993:3 882:9 = 0 Ay = 110:4 N ) A = 110:4 N J

From FBD of the pulley: MD Fy

= 0 T1 = T2 = 0 + " TD T1 T2 = 0 T1 = T2 = 0:5TD = 0:5(993:3) = 496:7 N

133 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

From FBD of the ball: Fy NB

= 0 + " T1 + NB = 386 N J

882:9 = 0

496:7 + NB

882:9 = 0

4.86

4.87 240 lb Ax

6 ft 200 lb 2 ft

A Ay

Bx B

Bx

2 ft B

By

By

Cx

2 ft C Cy

From FBD of BC : MB Fy

= =

0 0

+ 4Cy 200(2) = 0 + " By + Cy 200 = 0

Cy = 100 lb By = 100 lb

From FBD of AB: MA RB

=

0 + ( 240 + By ) (6) 2Bx = 0 p = 10202 + 1002 = 1025 lb J

Bx = 1020 lb

134 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.88

4.89 C

3 kN

3 kN By

0.3 m

Bx B

B

Bx

D 3 kN

By 1.0 m

1.0 m Ax

A Ay

0.3 m

0.5 m

Ex

0.5 m E (b)

(a)

Ey

From FBD (a): MA = 0

+

1:0Bx

3(1:3) = 0

Bx = 3:9 kN

135 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

From FBD (b): ME = 0

+

1:0Bx + 1:0By 3(1:3) 3(0:2) = 0 p RB = 3:92 + 0:62 = 3:95 kN J

4.90

By = 0:6 kN

1800 lb

By B

A

Bx

B

6 ft 6 ft T Fx Ex

F Fy

4 ft

E

D

Fx

2 ft 8 ft

20 ft

F Fy

1800 lb

Ey

Ax

A

B 6 ft

6 ft

Ay

Bx

By

From FBD of member AB: MA = 0

+

12By

1800(6) = 0

By = 900 lb

From FBD of member BDF : Fy = 0

+ " Fy

By = 0

Fy = 900 lb

From FBD of entire frame: ME

=

0 + 20(900)

20Fy 4Fx 1800(6) = 0 4Fx 1800(6) = 0 Fx = 1800 lb

From FBD of member BDF : MB

=

0 + 8Fx + 8Fy T = 3600 lb J

6T = 0

8(1800 + 900)

6T = 0

136 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.91 15 lb

E

24 Cy Cx Cx

B A

5.5

T

C D

C 12

Cy

P

3

3.75 Dx

Dy

FBD of ABC: Fx = 0

+ !

Cx = 0

FBD of DCE: MD = 0

+

3Cy

15(27:75) = 0

Cy = 138:75 lb

FBD of ABC: MB = 0

+

5:5P

138:75 (12) = 0

P = 303 lb J

4.92

137 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.93

A

800 N

By Bx

B

0.45 m B

1.05 m

0.3 m

900 N CC Cx

0.3 m

C 0.45 m Cy

900 N

D

D

ND

ND

From FBD of member BD: MB = 0

+

0:45ND

900(0:3) = 0

ND = 600 N

From FBD of entitre frame: MC Fx Fy

=

0 + CC + 0:45ND + 900(0:3) 800(1:05) = 0 CC + 0:45(600) + 900(0:3) 800(1:05) = 0 CC = 300 N J = 0 + ! Cx + 800 900 = 0 Cx = 100 N J = 0 + " ND Cy = 0 Cy = ND = 600 N J

138 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.94

NC B

240 N

40

35

C

NC

By Bx

40

35

B

C

90

90 140 D

ND D

A

(b)

Ax

Ay

(a)

MA = 0

+

From FBD (a): 75NC

240(140) = 0

NC = 448 N J

From FBD (b): MB = 0

+

90ND

75NC = 0

ND = 373 N J

4.95

139 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.96

TA TA

TB TB

180 lb

C 12'

40 lb

200 lb 3' 3'

D

6'

TA

TB

A

B 200 lb

180 lb NA

NB

(a) FBD of entire system: Fy

=

0 + " 2TA + 2TB (40 + 180 + 200) = 0 TA + TB = 210 MD = 0 + 40(12) + 180(9) + 200(6) TA (24 + 9) 33TA + 6TB = 3300 TB = 134:4 lb J Solving : TA = 75:6 lb J

TB (6) = 0

(b) FBD of man A: Fy NA

= 0 + " N A + TA = 104:4 lb J

180 = 0

NA + 75:6

200 = 0

NB + 134:4

180 = 0

FBD of man B: Fy NB

= 0 + " NB + TB = 65:6 lb J

200 = 0

140 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.97 75 N By A P

C

B 27

80

Cy

C

Cx Cx

E D Dy

Cy 24

72

FBD of CDE : Fx MD

= 0 = 0

+ ! Cx = 0 + 24Cy 75(72) = 0

Cy = 225:0 N

FBD of ABC: MB = 0

+

27P

225:0(80) = 0

P = 667 N J

4.98

141 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.99 A

A 1.6 m

B

T

T 2m

B Ex

C D

E

400(9.81) N

1.6 m Cy

Cx

C

1.1 m

Dy

1.4 m

1.4 m E D

Ey

T 2m

Dx

Dx 2.7 m

D Dy

400(9.81) N

FBD of the drum: ME = 0

+

1:4Dy

400(9:81)(0:7) = 0

Dy = 1962 N

FBD of tongs with drum: Fy = 0

2T p

+"

1:6 + 1:62

400(9:81) = 0

22

T = 3141 N

FBD of BCD: MC = 0 + 1:4Dx

Tp

1:4Dx (3141) p

2 (1:1) 0:7Dy = 0 22 + 1:62 2 (3141) p (1:1) 0:7(1962) = 0 2 2 + 1:62 Dx = 5711 N

1:6 (2) + 1:62

22

D=

4.100

1:6 (2) 22 + 1:62

p

57112 + 19622 = 6040 N J

C 8 kips

Tp

PBC

B

B

4 ft PBD

C 3 ft 3 ft

D

D

6 ft

6 ft

9 ft

A PAB

4 ft

E

Ex

Ey

E

Ex = 8 kips

Ey

142 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of entire frame (AB is a 2-force member): Fx = 0

+ !

Ex

8=0

Ex = 8:0 kips

FBD of member CDE (BC and BD are 2-force members): MD MC Fy E

4.101

4 PBC (6) 8:0(6) = 0 5 4 PBD (6) 8:0(12) = 0 = 0 + 5 3 3 = 0 + # Ey (20) (10) = 0 5 5 p = 8:02 + 18:02 = 19:70 kips J =

0

+

PBC = 10:0 kips J PBD = 20:0 kips J Ey = 18:0 kips

143 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.102 D

45 in.

C 50o

PBE (a)

70o

A Ax

CA

Ay Dy E

25 in D

50o

32 lb

B

Dx

45 in.

C

70o

40 in .

32 lb

E

25 in

70o

PBE

(b)

From FBD (b) (BE is a two-force member): MD

=

0 + (PBE sin 70 )(25) PBE = 46:96 lb

(32 sin 50 ) (45) = 0

From FBD (a): MA = 0

+

CA + (PBE sin 70 ) (25) (32 cos 50 )(40 sin 70 ) (32 sin 50 )(45 40 cos 70 ) = 0 CA + 23: 492PBE 1540: 9 = 0:0 CA + 23:49(46:96) 1540:9 = 0 CA = 438 lb in. J

4.103 70o

PAB B

200 lb

10 in. C

B

A

3 in.

70o PAB = 292.4 lb

PCD

3 in. E

Ex

Ey F PAF

144 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of links AB and AF (both are 2-force members): Fy Fx

= =

0 yields PAF = PAB 0 + ! 2PAB cos 70

200 = 0

PAB = 292:4 lb

FBD of BCE (CD is a 2-force member): Fx MC

= 0 = 0

+ ! Ex 292:4 cos 70 = 0 Ex = 100 lb + 3Ey + 100(3) (292:4 sin 70 ) (10) = 0

Ey = 816 lb J

4.104

B

FBC

C

150

C

ND

D

60o D 110 Ax

Ex E A 16 N m Ay 70 Ey (a)

Ex

60o E

16 N m Ey (b)

From FBD (a): MA = 0

+

0:07Ey

16 = 0

Ey = 228:6 N

From FBD (b) (BC is a two-force member): Fy

=

0

ME

=

0

+ # Ey + ND cos 60 = 0 ND = 457:2 N 0:11 + 0:26FBC + ND =0 FBC = 223 N J sin 60

145 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.105

4.106

FBC 80 kN m

B

30o 3m

Ax A Ay Bar BC is a two-force member. MA Fx Fy

=

0

= 0 = 0

RA RC

+

FBC

+ ! Ax + # Ay

3 80 = 0 cos 30 FBC sin 30 = 0 FBC cos 30 = 0

FBC = 23:09 kN Ax = 11:55 kN Ay = 20:00 kN

p 11:552 + 20:002 = 23:1 kN J = = FBC = 23:1 kN J

146 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.107 PAC G 600 lb

Fy

A 15 5 B By

F

Bx

Fx 25

C

PDE

E

D

30 G

B A 15 5 20

50

600 lb Both cylinders are 2-force members. FBD of bucket: Fx MB

= 0 = 0

+ ! Bx = 0 + 5PAC 600(20) = 0

PAC = 2400 lb J

From FBD of mechanism: MF = 0

+

25PDE

600(90) = 0 PDE = 2160 lb J

4.108

400 kN B 0.35 m FAB 45o

B FAB 45o

0.6 m Cx

(a)

Dx D Dy

0.6 m E

C

m 0.4

C Cy

400 kN 0.35 m

NE

0.6 m (b)

From FBD (a) (bar AB is a two-force member): MC = 0

+

(FAB sin 45 ) (0:6)

400(0:95) = 0

FAB = 895:7 kN

147 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

From FBD (b): MD = 0 + 0:4NE 400(1:35) + 1:0 (FAB sin 45 ) NE = 400 kN J

0:4 (FAB cos 45 ) = 0

4.109

4.110

12 lb

25o PAD A Cy

By B

Cx C (a)

F

8" 2" 6"

Bx

Bx

B (b)

By

AD is a two-force member.

148 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

From FBD (b): MB Fy

=

0 + (PAD cos 25 )(2) (PAD sin 25 )(6) + 12(14) = 0 PAD = 232:3 lb = 0 + " By PAD cos 25 = 0 By 232:3 cos 25 = 0 By = 210:5 lb

From FBD (a): MC = 0

+

F

By = 0

F = By = 210:5 lb J

4.111

149 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.112

60o

0.8

A

260 N. m m O C G

E The three bars are two-force members. MG

=

0 + (A cos 60 )(1:6) (A sin 60 )(0:8) 260 = 0 A = 2426 N J = 0 E = A cos 60 = 2426 cos 60 = 1213 N J = 0 C = A sin 60 = 2426 sin 60 = 2101 N J

Fx Fy

4.113 0.6 P 1.2

3.8

8 lb

By

2

Bx

Ay A

I C

B

Ax Ax

II

A Ay

Cy

CD is 2 two-force member. From FBD I: Fx MA

= =

0 0

+ ! Ax = 0 + 2Cy 3:2P = 0

Cy = 1:6P

From FBD II: MB = 0

+

0:6Ay

8(3:8) = 0

Ay = 50:67 lb

From FBD I: Fy = 0

+"

P

Cy +Ay = 0

P

1:6P +50:67 = 0

P = 84:5 lb J

150 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.114

120 kN 0.75 m

0.75 m

Ey E

NB Ex

0.8 m

B

FAC 4

C 0.7 m 3

Dy D

Dx

B NB From FBD of bar BE: ME Fy Fx

= 0 = 0 = 0

+ 1:5NB 120(0:75) = 0 NB = 60 kN + " Ey + NB 120 = 0 Ey = 60 kN + ! Ex = 0

From FBD of bar BCD (bar AC is a two-force member): MD

=

0

+

Fx

=

0

+!

Fy

=

0

+ # Dy + NB RA RD RE

1:5NB 4 FAC 5

3 FAC (0:7) = 0 5 Dx = 0

FAC = 214:3 kN

Dx = 171:4 kN

3 FAC = 0 5

Dy = 68:6 kN

= FAC = 214 kN J p = 171:42 + 68:62 = 184:6 kN J = 60 kN J

4.115

180 lb

FDE Cy

1.5 ft

C Cx A

Bx 3 ft

B By

NA

A

1.5 ft

E

1.5 ft

1.5 ft

NA

From FBD of entire frame: MB = 0

+

3NA

180(3) = 0

NA = 180 lb

151 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

From FBD of ACE (bar DE is a two-force member): MC Fx Fy

= 0 = 0 = 0

+ 1:5FDE 1:5NA = 0 FDE = 180 lb + ! Cx FDE = 0 Cx = 180 lb + # Cy NA = 0 Cy = 180 lb p RC = 180 2 = 255 lb J

4.116

152 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.117 FAB

3

8

B

FAB

P

150

FAB

(a)

E

120 70 (b)

20o

25 Ex

D

Dimensions in mm

P/2

Ey

P/2 P

1.2 kN

P/2 (c)

The two bars connected to pin A are two-force members. From FBD (a): Fx = 0

8 p FAB 73

2

P =0

FAB = 0:5340P

From FBD (b): ME = 0

8 p FAB (150) + 73

+

P (25) 2

(1:2 cos 20 ) (120)

3 p FAB (120) 73

(1:2 sin 20 )(70) = 0

182:58FAB 12: 5P 164: 05 = 0 182:58(0:5340P ) 12: 5P 164: 05 = 0 P = 1:930 kN J

4.118

ND D 30o

Ey 2 3.5 F EE x FBF

3.5

Ax

FBF

40 lb 10

C

B

A Ay

From FBD of ABC (link BF is a two-force member): MA = 0

+

3:5FBF

40(13:5) = 0

FBF = 154:29 lb

From FBD of DEF : ME = 0

+

2ND

3:5FBF = 0

ND = 270 lb J

153 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

*4.119

4.120 0.6 ft B

30o A Cy 0.75 ft PAB C Cx 0.75 ft

θ = 47.99

o

5 ft 3.7 ft 95 .64 0 d= F

D

200 lb 0.6 ft 30o

200 lb B

PAB Dx D

0.75 ft Dy

I

2 ft

II

E

PEF

FBD I (AB is a 2-force member): MD = 0

+

(PAB cos 30 )(0:75)

200(0:6) = 0

PAB = 184:75 lb

154 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Geometry (see FBD II): d

0:75 cos 30 = 0:6495 ft d = 30 + tan 1 = 30 + tan 2 FBD II (EF is a 2-force member): MC PEF

=

= + =

1

0:6495 = 47:99 2

0 + (PEF cos 47:99 ) (0:75) (184:75 cos 30 ) (0:75) 200(3:75 cos 30 + 0:6) = 0 1294 lb J

4.121

RA

FA

FB

4 3

0.15 m 0.15 m

P A 0.4 m

4

B

3

G 117.72 N 0.3 m

The two links are two-force bodies. Let RA be the resultant of P and FA . Because the plate is a three-force body, RA and FB intersect at G (the mass center of the plate). Fx

=

Fy

=

P

=

0

R A = FB 4 0 (RA + FB ) = 117:72 5 3 3 RA = (73:58) = 44:1 N J 5 5

8 RA = 117:72 5

RA = 73:58 N

4.122 120 lb

6.5 in.

4.5 in. NB

A B

o

45 6.5 in.

RC C

155 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

The bracket is a 3-force member. Hence the forces intersect at A. Fy Fx

= 0 + # RC sin 45 120 = 0 RC = 169:71 lb J = 0 + NB RC cos 45 = 0 NB 169:71 cos 45 = 0 NB = 120:0 lb J

4.123

4.124

156 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.125

R/2 P

B

2R

45o FBC Ax A Ay From FBD of AB (segment BC is a two-force member): MA

=

0

+

FBC

Fx

=

0

+ ! Ax

p

2R

P

FBC p =0 2

R =0 2 Ax =

P FBC = p 2 2 P J 4

157 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.126

4.127

158 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.128

A PBC

80 mm

PA

E

C

40 mm

B

d

40 mm

P

D

BC is a two-force member. Since ABD is a three-force member, all three forces intersect at point E. From similar triangles: 40 d = 80 40

d = 20 mm J

4.129

40o B

O

NB

6'

RA

6'

α

T 240 lb NA F

A

E

55o

RA 59.21o RAx

Bar AB is a three-force body. All three forces acting on it intersect at point O. From geometry: OF AF

= BE = 12 sin 40 = 7:714 ft = 6 cos 40 = 4:596 ft OF 7:714 = tan 1 = tan 1 = 59:21 4:596 AF

159 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

From FBD of bar AB: Fy

=

0 + " RA sin RA = 279:4 lb

240 = 0

RA sin 59:21

240 = 0

RA is the resultant of NA (the normal reaction at A) and cable tension T . Referring to the …gure on the right: RAx T

= RA cos 59:21 = 279:4 cos 59:21 = 143:02 lb 143:02 RAx = = 249 lb J = cos 55 cos 55

4.130

4.131

160 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.132

4.133

C

1C 2 1

1200 lb

1

PAC

6 ft 3 ft

6 ft

PBC PBC

Ax A Ay

1200 lb

B

PAB

NB

1 1

B

NB

FBD of entire truss: MA = 0

+

9NB

1200(6) = 0

NB = 800 lb

FBD of joint B: Fy

=

0 PBC

Fx

=

0 PAB

1 1 p PBC + 800 = 0 + " p PBC + NB = 0 2 2 = 1131:4 lb = 1131:4 lb (C) J 1 1 + PAB + p PBC = 0 PAB + p ( 1131:4) = 0 2 2 = 800 lb (T) J

FBD of joint C: Fy

=

0 PAC

2 1 + # p PAC + p PBC = 0 5 2 = 894 lb (T) J

2 1 p PAC + p ( 1131:4) = 0 5 2

161 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.134

5 2 1 C B Ax 6 ft D 8 ft A 6 ft Ay ND 2400 lb 1800 lb 4

PAE

Ax = 0 A

PAB Ay

PBE

PAB B 2400 lb

PCE

PBC

PDE D

C PCD PCD

PBC 1800 lb

ND

FBD of entire truss: MD Fx

= 0 = 0

+ +!

20Ay 2400(14) Ax = 0

1800(6) = 0

Ay = 2220 lb

FBD of joint A: Fy

=

0 PAE

Fx

=

4 4 p PAE + Ay = 0 p PAE + 2220 = 0 41 41 = 3554 lb = 3554 lb (C) J 5 5 + ! PAB + p PAE = 0 PAB + p ( 3554) = 0 41 41 = 2775 lb (T) J +"

0 PAB

FBD of joint B: Fy

=

0

+"

Fx

=

0

+!

PBC

2 p PBE 5 PBC

2400 = 0

PBE = 2683 lb (T) J

1 PAB + p PBE = 0 5

1 2775 + p 2683 = 0 5

PBC = 1575 lb (T) J

FBD of joint C: Fy

=

0

+"

Fx

=

0

+!

PCD

1575

2 p PCE 5 PCD

1800 = 0 PBC

1 p 2012 = 0 5

PCE = 2012 lb (T) J

1 p PCE = 0 5 PCD = 2475 lb (T) J

162 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of joint D: Fx

=

0

5 p PDE + PCD = 0 41 3170 lb = 3170 lb (C) J

+

PDE =

5 p PDE + 2475 = 0 41

4.135 D

D

P PAD

PBD PCD

L

PBD

Ax A Ay

B

L

P

C

L

NC

2P

PAB

PCD

PBC PBC

B

C NC

2P

FBD of entire truss: MA = 0

+

NC (2L)

2P L

PL = 0

NC = 1:5P

FBD of joint C: Fy

=

0 PCD

Fx

=

0 PBC

1 1 + " NC + p PCD = 0 1:5P + p PCD = 0 2 2 = 2:121P = 2:121P (C) J 1 1 PBC + p ( 2:121P ) = 0 + PBC + p PCD = 0 2 2 = 1:500P (T) J

FBD of joint B: Fx Fy

+ PAB PBC = 0 PAB = PBC = 1:500P (T) J + " PBD 2P = 0 PBD = 2P (T) J

= 0 = 0

FBD of joint D: Fx

=

0

+

1 p PAD 2 PAD =

1 1 p PAD p PCD P = 0 2 2 1 p ( 2:121P ) P = 0 2 0:7068P = 0:7068P (C) J

163 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.136 PBD = 160 kN

2 PBC 2 2 1 1 1 1 C NC A PAD PAD D PDE = PAD P B PBE P AB BC PBD PCE 160 kN 160 kN 2

PAB

FBD of joint A: 1 p PAB 5

Fy

=

0

+"

Fx

=

0

+!

PAB = 357:8 kN (T) J

160 = 0

2 PAD + p PAB = 0 5 320 kN = 320 kN (C) J

PAD =

2 PAD + p 357:8 = 0 5

FBD of joint D: PDE = PAD = 320 kN (C) J

PBD = 160 kN (T) J

FBD of joint B: Fy

=

0

+"

1 p (PBC 5 Fx

=

0

+!

PBC + PBE

1 p (PBC 5 PBE

PBE

357:8)

PAB )

PBD = 0

160 = 0

2 p (PBC + PBE 5 357:8 = 0

(1)

PAB ) = 0 (2)

Solving Eqs. (1) and (2): PBC = 536:7 kN (T) J

PBE =

178:9 kN = 178:9 kN (C) J

FBD of joint C: Fy

=

0

1 PCE + p PBC = 0 5 240 kN = 240 kN (C) J

+#

PCE =

1 PCE + p 536:7 = 0 5

4.137

164 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.138

1200 lb

C PAC

PCD 4

4

3

PCD

PAD

3 4

PAC D PBD

4 3

3

A

PAD PAB

NA

165 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of joint C: Fx

=

0

Fy

=

0 PAC

+!

4 PCD 5

1200 = 0

PCD = 1500 lb (T) J

3 3 + # PAC + PCD = 0 PAC + 1500 = 0 5 5 = 900 lb = 900 lb (C) J

FBD of joint D: Fx

Fy

4 4 PAD + PCD = 0 PAD + 1500 = 0 5 5 PAD = 1500 lb = 1500 lb (C) J 3 = 0 + # PBD + (PAD PCD ) = 0 5 3 PBD + ( 1500 1500) = 0 PBD = 1800 lb (T) J 5

=

0

+

FBD of joint A: Fx

=

0 PAB

4 + ! PAB + PAD = 0 5 = 1200 lb (T) J

PAB +

4 ( 1500) = 0 5

4.139

166 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.140

C

300 kN

C

300 kN 24.91o

PBC

6

PBC B

400 kN PAB

24.91o PCD

4 50o 40.74o A 4.644 3.356

D Dx

400 kN 50o PAB

B

40.74o PBD

Dimensions in meters Dy

FBD of entire truss: MD

=

0 + (PAB sin 50 ) (8) PAB = 750:6 kN (T) J

400(4)

300(10) = 0

FBD of joint B: Fx

Fy

=

0 + ! PBD cos 40:74 PAB cos 50 + 400 = 0 PBD cos 40:74 750:6 cos 50 + 400 = 0 PBD = 108:9 kN (T) J = 0 + " PBC PAB sin 50 PBD sin 40:74 = 0 PBC 750:6 sin 50 108:9 sin 40:74 = 0 PBC = 646:1 kN (T) J

FFD of joint C: Fx

=

0 + ! 300 + PCD sin 24:91 = 0 PCD = 712 kN = 712 kN (C) J

4.141 (a) BD and CD; (b) AE, BE and CE; (c) BD; (d) DF

167 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.142

160 lb B

PAB

PBD 420 lb

PBC 420 lb L

PBD

6

6 D

L

FBD of joint B: Fy = 0

+ " PBD

160 = 0

PBD = 160 lb

FBD of joint D: Fy

=

+ " 2(420) p

0

L + 62

PBD = 0

L2

L 160 = 0 L2 + 36 p 27:56L2 = L2 + 36 5:25L = L2 + 36 r 36 = 1:164 ft J L= 26:56 840 p

4.143 D 42 kN

PAD 45o PDE

PDE

PAD A o

PAE 45

PAB E

PAE PBE PEF

45o

FBD of joint D: Fy Fx

= 0 + # PDE sin 45 + 42 = 0 PDE = 59:40 kN = 0 + ! PAD + PDE cos 45 = 0 PAD + ( 59:40) cos 45 = 0 PAD = 42:0 kN

FBD of joint A: Fx Fy

=

0 PAB = 0 PAE

+ ! PAB sin 45 PAD = 0 = 59:40 kN + # PAE + PAB cos 45 = 0 = 42:0 kN

PAB sin 45

42:0 = 0

PAE + 59:40 cos 45 = 0

FBD of joint E: Fy

=

0 + " (PDE PEF ) sin 45 + PAE = 0 ( 59:40 PEF ) sin 45 + ( 42:0) = 0 PEF = 118:8 kN = 118:8 kN (C) J

168 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.144

12 kN 12 kN

E

9 kN

E

D 30o

A

3m

B

30o 30o 3m

NA

Cx C Cy

PAE PAE A

PBE

PED

PAB

NA

FBD of entire truss: MC = 0

+

6NA

12(3)

9(1:5) = 0

NA = 8:25 kN

FBD of joint A: Fy

=

0 + " PAE sin 30 + NA = 0 PAE sin 30 + 8:25 = 0 PAE = 16:5 kN = 16:5 kN (C) J

FBD of joint E: Fx Fy

=

0 + ! (PED PAE ) cos 30 = 0 PED ( 16:5) = 0 PED = 16:5 kN = 16:5 kN (C) J = 0 + # PBE + (PED + PAE ) sin 30 + 12 = 0 PBE + 2( 16:5) sin 30 + 12 = 0 PBE = 4:5 kN (T) J

169 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.145 8 kN

10 kN

16 kN

8

20 kN

8 A Ay

60o

60o

Ax F

9.238

Ex

G

E Ey

18.475 27.713 Cy Cx C

8

16 kN D

8

8 30o

Dimensions in meters G

Ex E

FBD of entire truss: MA = 0

+ 27:713Ey 20(8) 10(16) + (8 sin 60 ) (9:238) +(16 sin 60 )(18:475) = 0 Ey = 0

FBD of right half of trusss: MC = 0

+

16(8)

8Ex = 0

Ex = 16 kN

y' PCD

16 kN D

PCG 60o G

PDG PDG 60o PEG

PDE x' PDE 30o PEG

16 kN E

FBD of joint E: Fy Fx

= 0 = 0

+ " PDE = 0 J + PEG + 16 = 0

PEG =

16 kN = 16 kN (C) J

170 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of joint D: Fy 0 Fx0

= 0 = 0

+ . PDG + & PDE

16 = 0 PCD = 0

PDG = 16 kN (T) J 0 PCD = 0 PCD = 0 J

FBD of joint G: Fy

=

0 + " PCG + PDG = 0 PCG + 16 = 0 PCG = 16 kN = 16 kN (C) J

4.146

171 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.147

4.148

172 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.149 4 @3 m

P

P A AB B 4m P

AG

PFG G MA MG

= 0 = 0

+ +

PF G = 3P = 3P (C) J PAB = 2:25P (T) J

4PF G + 12P = 0 4PAB 9P = 0

4.150 C

PBC

B Ax

A

G

F

5000 lb

H 6 ft

6 ft

6 ft Ay

B

D

4000 lb

E A

6 ft 6000 lb

NE

7250 lb

PBG

3 ft PFG F 6 ft 6 ft

C 1 2 2

6 ft 1

G

5000 lb

FBD of entire truss: MA Fx Fy

=

0 + 24NE 5000(6) 4000(12) 6000(18) = 0 NE = 7750 lb = 0 + Ax = 0 = 0 + " Ay + NE 5000 4000 6000 = 0 Ay + 7750 5000 4000 6000 = 0 Ay = 7250 lb

FBD of part ABF : MB

=

0

MG

=

0 PBC

MA

=

0 PBG

3PF G 7250(6) = 0 PF G = 14 500 lb (T) J 2 p PBC (6) + 7250(12) 5000(6) = 0 + 5 = 10 620 lb = 10 620 lb (C) J 1 p PBG (12) + 5000(6) = 0 + 5 = 5590 lb = 5590 lb (C) J +

173 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.151

a

E

E

F

12 ft PEF

9 ft Ax

12 ft

9 ft

9 ft

B a 3000 lb

A Ay

C

F

9 ft PBF 3 4 9 ft PBC D A B ND 2100 lb 3000 lb

FBD of entire truss: MD Fx

= =

0 0

+ +

30Ay 3000(21) = 0 Ax = 0

FBD of truss left of section a MB

Ay = 2100 lb

a:

=

0 + 9PEF + 2100(9) = 0 PEF = 2100 lb = 2100 lb (C) J 3 = 0 + " PBF + 2100 3000 = 0 PBF = 1500 lb (T) J 5 = 0 + 9PBC 2100(21) + 3000(12) = 0 PBC = 900 lb (T) J

Fy MF

4.152

Jx 36 m J 12 m Jy 400 kN PEF

24

F

PNF 1

1

N

NR

PNO 6m

6m

E

6.25 m

1

R

O

18 m

R 100 kN

FBD of entire truss: MJ = 0

+

48NR

400(12) = 0

NR = 100 kN

174 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of right portion of truss: MF MN

Fy

6PN O 100(18) = 0 PN O = 300 kN (T) J 24 p = 0 + PEF (6:25) + 100(24) = 0 577 PEF = 384:3 kN = 384:3 kN (C) J 1 1 p = 0 + # p PN F PEF 100 = 0 2 577 1 1 p PN F p 384:3 100 = 0 PN F = 164:1 kN (T) J 2 577 =

0

+

24

E PEF F PNF

6.25 m

1

1

1

N

PNO

6m

4.153

O

6m

18 m

R 250 kN

400 kN From FBD of entire truss (not shown): MJ = 0

+

48NR

400(30) = 0

NR = 250 kN

From FBD of right portion of truss: MF MN

Fy

6PN O 250(18) = 0 PN O = 750 kN (T) J 24 p PEF (6:25) 400(6) + 250(24) = 0 = 0 + 577 PEF = 576:5 kN = 576:5 kN (C) J 1 1 = 0 +" p PEF p PN F + 250 400 = 0 577 2 1 1 p ( 576:5) p PN F 150 = 0 577 2 PN F = 246:1 kN = 246:1 kN (C) J =

0

+

175 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.154

4.155

176 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.156

D

2m A 2m

42 kN

B

2m E y

2m

x' 45ox

PBC PCF

F

PFG MF = 0 Fy

=

+ 0 PF G

Fx0

=

0

p PBC ( 2)

p 42(4 2) = 0

PBC = 168:0 kN (T) J

1 + # p (PF G + PBC + 42) = 0 PF G + 168:0 + 42 = 0 2 = 210:0 kN = 210:0 kN (C) J 1 + % p PCF 42 = 0 PCF = 59:4 kN (T) J 2

4.157 720 lb 1440 lb 1440 lb 8 ft 8 ft 8 ft PCD D C B A 6 ft

PDH F

G1

4

H

10 ft PHI

177 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

MH

=

0 + 10PCD 720(24) 1440(16 + 8) = 0 PCD = 5184 lb (T) J 4 p PHI (10) + 720(24) + 1440(16 + 8) = 0 = 0 + 17 PHI = 5344 lb = 5344 lb (C) J 1 = 0 + " PDH p PHI 720 2(1440) = 0 17 1 PDH p ( 5344) 720 2(1440) = 0 17 PDH = 2300 lb (T) J

MD

Fy

4.158

18 m PCD

A

6m D

PDF

9 4

8m

3 kN

5 kN

E

PEF

F MF

=

0

ME

=

0 PDF

+

8PCD 5(18) 3(24) = 0 4 p PDF (6) + 5(6) = 0 + 97 = 12:31 kN = 12:31 kN (C) J

PCD = 20:25 kN (T) J

178 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.159

4.160

179 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.161

F a

1200 lb E

G B

4.154 ft NA 6 ft

a 3 ft

E

6 ft D

A

1 2

3

PBF

2

Dx A 4.154 ft 4.154 ft 6 ft 6 ft Dy 1269.2 lb

C 3 ft

F

1200 lb PEF

1500 lb

B

6 ft

PBC

9 ft FBD of entire truss: MD

=

0 + 18NA NA = 1269:2 lb

1200(18

FBD of truss to left of section a MA

=

0 PBF

MB

=

0

MF

PEF = 0 PBC

4:154)

1500(4:154) = 0

a:

2 p PBF (6) 1200(4:154) = 0 5 = 929 lb (T) J 2 p PEF (6) 1200(6 4:154) + 1269:2(6) = 0 + 13 = 1622 lb = 1622 lb (C) J + 6PBC + 1200(9 4:154) 1269:2(9) = 0 = 935 lb (T) J +

4.162 4m A PAD

5 @ 3 m = 15 m

PAC C

D PDE

F

G

H

I

J

K L 30 kips

180 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Due to symmetry, the reaction at L is 30 kips. MA MC Fx

= 0 = 0 =

3PDE 30(4) = 0 PDE = 40 kips (T) J PAD = 0 J 4 4 + ! PDE + PAC = 0 40 + PAC = 0 5 5 = 50 kips = 50 kips (C) J + +

0 PAC

4.163 3 @3 m= 9 m

PGH

F 3m G PFH P

GI

H

I 2m

J

K L 30 kips

Due to symmetry, the reaction at L is 30 kips. MH MG ML

= 0 = 0 = 0

PGI = 0 J + 3(PF H + 30) = 0 PGH = 0 J

PF H =

30 kips = 30 kips (C) J

4.164

181 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.165

a

E F

D 1.5 m

8 kN

6

3m

3m

D

A

C 4m

NA

Cy

12 kN

A

Cx

4m

a B

2

1

5

PDE

PBD

4m PAB

B

9 kN

FBD of entire truss: MC = 0

+

8NA

FBD of truss left of section a MD

=

0

MB

=

0

+

PDE =

12(4)

8(3) = 0

NA = 9 kN

a:

3PAB 9(4) = 0 PAB = 12 kN (T) J 2 1 p PDE (3) + p PDE (2:5) + 9(4) = 0 5 5 9:470 kN = 9:47 kN (C) J

4.166 9 ft 3

E

4

PCF

PCE

I F

G 9 ft

H 9 ft

PDF 1500 lb 1500 lb

182 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(a) Fx ME MF

= =

0 0 PDF = 0

+ PCF = 0 J + 9PDF + 1500(18) + 1500(27) = 0 = 7500 lb = 7500 lb (C) J + 9PCE 1500(9 + 18) = 0 PCE = 4500 lb (T) J

(b) The zero-force members are CF , CD, BC and AB J

4.167 10 000 lb 8 ft PBC

B

8 ft

5

C 5 ft

8

PBE E 9 ft Gx

G Gy = 14 500 lb

From FBD of entire truss (not shown): Gy = 14 500 lb " MB

=

0

Fy

=

0 PBE

Fx

=

0 PBC

+

14Gx (14 500 10 000)(16) = 0 Gx = 5143 lb 5 + " p PBE + 14 500 10 000 = 0 89 = 8491 lb = 8491 lb (C) J 8 + PBC + p PBE + Gx = 0 89 8 + p ( 8491) + 5143 = 0 PBC = 2057 lb (T) J 89

183 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.168

4.169

4.170

2 ft 1.5 ft

P O 25o

Ax

320 lb A Ay MA = 0

+

P (1:5 + 2 sin 25 )

320(2 sin 25 ) = 0

P = 115:3 lb J

184 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.171

6 kN 4m

4m

B

C 65o

NC

5m A

Ax Ay

MA Fy Fx

4.172

=

0 + (NC cos 65 ) (5) + (NC sin 65 ) (8) NC = 2:563 kN = 0 + " Ay + NC sin 65 6=0 Ay + 2:563 sin 65 6=0 Ay = 3:677 kN = 0 + ! Ax NC cos 65 = 0 Ax 2:563 cos 65 = 0 Ax = 1:083 kN p A = 1:0832 + 3:6772 = 3:83 kN J

W

6(4) = 0

B W

L Ax C A

30o Ay

MA

=

0 + C + W (L sin 30 ) C = 0:366W L J

W (L cos 30 ) = 0

185 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.173 800 lb

8 ft

E E D

Ex

4 ft

Ey

1

D

4 ft

1

FCD 12 ft

C 4 ft A Ax Ay

8 ft

12 ft

B

B

NB

NB

FBD of entire structure: MA = 0

+

8NB

800(16) = 0

NB = 1600 lb

FBD of member BDE: ME

=

0

+

8NB

1 p FCD (4) 2

1 p FCD (8) = 0 2

12 p FCD = 0 FCD = 1508:5 lb 2 1 = 0 + ! Ex p FCD = 0 2 1 Ex p (1508:5) = 0 Ex = 1066:7 lb 2 1 = 0 + # Ey + p FCD NB = 0 2 1 Ey + p (1508:5) 1600 = 0 Ey = 533:3 lb 2 p E = 1066:72 + 533:32 = 1193 lb J 8(1600)

Fx

Fy

186 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.174

4.175 6000 lb

6000 lb

C

9 ft B

B

A

Ax Fx

F

18 ft

O

6 ft

9 ft Ay

PBE

2000 lb

E

6 ft Ax

9 ft

D 3 ft

12 ft

2000 lb

C P CD

A Ay

Fy BE and CD are two-force members.

187 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of entire frame: MF

=

0 + 3Ax + 12Ay + 2000(9) Ax + 4Ay = 30 000

6000(18) = 0 (a)

FBD of member ABC: MO 3Ay

Ax

= 0 + = 9000

6Ax

18Ay + 6000(9) = 0 (b)

Solving Eqs. (a) and (b): Ax = 7714 lb

Ay = 5571 lb

FBD of entire frame: Fx Fy

=

0 + ! Fx + Ax + 2000 6000 = 0 Fx + 7714 + 2000 6000 = 0 Fx = 3714 lb = 0 + " Ay + Fy = 0 Fy = Ay = 5571 lb p 77142 + 55712 = 9520 lb J p = 37142 + 55712 = 6700 lb J

A = F

4.176

By Bx

B 14 kN. m

14 kN. m 14 kN. m

Ay A 1.0 m

B 2m

Ay Ax

Ax

A

1.5 m

Dx

3m D Dy FBD of entire frame: MD = 0

+

1:0Ax + 3Ay = 0

Ax =

+

14 = 0

3Ay

(a)

FBD of member AB: MB = 0

2Ax

1:5Ay

(b)

188 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solving Eqs. (a) and (b): Ax = 5:60 kN Ay = 1:8667 kN p 2 ) A = 5:60 + 1:86672 = 5:90 kN J

From FBD of entire frame: Fx Fy

= 0 + ! Dx + Ax = 0 Dx = Ax = 0 + " Dy + Ay = 0 Dy = Ay ) D = A = 5:90 kN J

4.177 A

5 2

PAC 12 ft

B

=

Fy

=

D

4 kips

6 5

C 10 ft

NB

ME

PAD

8 ft

F

15 ft E

5 p PAD (8) 4(5) = 0 PAD = 2:693 kips (T) J 29 2 6 0 + " p PAD + p PAC 4 = 0 29 61 6 2 p (2:693) + p PAC 4 = 0 PAC = 3:90 kips (T) J 29 61

0

+

189 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.178

4.179 By Bx B

900 lb D

600 lb

3 ft 3 ft

4 ft

3 ft

E

3 ft

3 ft

C

NA

E Ey

C

NA

A

Ex

4 ft FCD

A

FBD of entire frame: MB = 0

+

6NA

900(3)

600(7) = 0

NA = 1150 lb

FBD of member ACE: ME Fx Fy

=

0 FCD = 0 = 0

+ 4FCD 3NA = 0 4FCD 3(1150) = 0 = 862:5 lb + Ex N A = 0 Ex 1150 = 0 Ex = 1150 lb + # Ey FCD = 0 Ey 862:5 = 0 Ey = 862:5 lb p E = 11502 + 862:52 = 1438 lb J

190 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.180

4.181

191 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.182

T

T

T 30o C 1.5 ft 1.5 ft 3 ft Ay 14 lb 30 lb

C

A

Ax

B

30 lb N

FBD of entire assembly: MA

=

0 + ( T sin 30 ) (6) + 1:5T T = 19:333 lb

30(1:5)

14(3) = 0

FBD of block C: Fy N

= 0 +" N +T = 10:67 lb J

30 = 0

N + 19:333

30 = 0

4.183 30 lb

60 lb B

A o

30

6 ft

NA

x 12 ft

Fx Fy The solution is MA

=

= 0 = 0

60o NB

+ ! NA cos 30 NB cos 60 = 0 + " NA sin 30 + NB sin 60 90 = 0

NA = 45:0 lb

NB = 77:94 lb

0 + (NB sin 60 ) (12) 30(6) (77:94 sin 60 ) (12) 30(6) 60x = 0

60x = 0 x = 10:50 ft J

192 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.184 16 000 lb

18 ft

D E NE

Ax A Ay

24 ft 16 000 lb

C

D

PBC

PDF

PCF

PDE 8 ft

4 3

PDE

4

PEF

3

12 ft

B

F

6 ft

3

6 ft

E

4

E

12000 lb

FBD of entire truss: MA = 0

+

24NE

16 000(18) = 0

NE = 12 000 lb

(a) FBD of joint E: Fy

=

0

Fx

=

0

4 PDE + 12 000 = 0 5 3 + PEF + PDE = 0 5 = 9000 lb (T) J +"

PEF

PDE =

15 000 lb

3 PEF + ( 15 000) = 0 5

(b) FBD of member CD: MF

=

PBC

=

4 4 PBC (12) PDE (12) 16 000(6) = 0 5 5 4 4 PBC (12) ( 15 000) (12) 16 000(6) = 0 5 5 5000 lb = 5000 lb (C) J

0

+

193 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.185 Bx

B

2 ft

Ax A Ay

3 ft

3.5 ft

80 lb 3.5 ft

80 lb 1.25 ft

2 ft

0.75 ft

C Cx Cy

By B

2.75 ft Ax A Ay

3 ft

FBD of entire frame: MC = 0

+

10Ay

80(2:75) = 0

Ay = 22 lb

FBD of member AB: MB Fx Fy

=

0 + 4Ax + 3Ay 80(1:25) = 0 4Ax + 3(22) 80(1:25) = 0 Ax = 8:5 lb = 0 + Bx + Ax 80 = 0 Bx + 8:5 80 = 0 Bx = 71:5 lb = 0 + # By Ay = 0 By 22 = 0 By = 22 lb p B = 71:52 + 222 = 74:8 lb J

194 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.186

W 6 in.

Dx

C

18 in.

D

NC NC

Dy 9 in. By B 4 in.

9 in.

C

Bx T = 400 lb

A NA FBD of entire stool (not shown): ME = 0

+

12W

18NA = 0

NA =

2 W 3

6W = 0

NC =

1 W 3

FBD of seat CD: MD = 0

+

18NC

FBD of member ABC:

1600

6W

MB 3W

= =

0 0

+ 400(4) 9NA W = 177:8 lb J

9NC = 0

4.187

40 kN

a 4m H 4m

F 4m 3m A NA

G

4m I

6m

B a

40 kN PGH

40 kN J Ex D

E Ey

C

F 3m A

4m G

4

B NA= 40 kN

H

3 PBH

PBC

FBD of entire truss: NA = Ey = 40 kN by symmetry

195 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of truss left of section a MB MH

MF

a:

=

0 + 3PGH + 40(4) PGH = 53:3 kN = 53:3 kN (C) J 3 = 0 + PBC (8) + 40(4) 40(8) = 0 5 PBC = 33:3 kN (T) J 3 = 0 + PBH (8) 40(4) = 0 PBH = 33:3 kN (T) J 5

196 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.188

80 N

15o

130

NB

E B

40 Dimensions in mm

Cx

C Cy 15o NB B 40

Ax

A

130

50

Ay

D ND

FBD of member CBE: MC = 0

+

80(130)

(NB sin 15 ) (40) = 0

NB = 1004:6 N

FBD of member AD: MA ND

=

0 + 180ND (NB cos 15 )(130) + (NB sin 15 )(40) = 0 180ND (1004:6 cos 15 )(130) + (1004:6 sin 15 )(40) = 0 = 643 N J

197 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4.189

W 12 in.

A

15 in. By 8 in. B

7 in. N

Bx

C W/2

MB = 0

+

W (8) 2

W (12) + N (7) = 0

N=

8 W J 7

198 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 5 5.1

z C

D

1.2

1.2

0.6 TC TD A 0.7 2

x

2.1

1800 N

B Dimensions in meters

By y

3 unknowns J

5.2 z C

x

0.6

1.2

D 1.2

TC TD Dimensions in meters A 0.7 2 1.0 5 Bz 1800 N 1.0 5 50(9.81) N B By y Bx 5 unknowns J

199 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.3

TA

TB B 1.2 m

m 1.2

TC C

A 2m

D 15 kN

40 kN

3 unknowns J

5.4

1.2 m

m 1.2

TB B

PAB

TC PBC

C

PAC

A 2m PAD D 15 kN

40 kN 6 unknowns J

200 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

z Dz x Dy Dx F

E

W D ft 2.5T TBF BG 3 ft 5 ft

A 3 ft

y

2 ft

TAG

G

2 ft

5.5

C

B 6 unknowns J

5.6

Az

Ax

22 in.

A

Dz

T B

z x

12 in. y

Dx D

Dy

in. 28

C 60 lb 6 unknowns J

201 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.7

5.8

z 14.5

Bz By

B

Bx

14.5

Ax

A

D TCD

23.5 60 lb 39

36 Ay

35

35 45o C

Dimensions in inches

40 lb 30

y

x 6 unknowns J

202 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.9

5.10 TD TA

D

4 W1

A

4.7 5 TC

W2

3 Dimensions in feet

3.5 1.5

B

1

1

1

C 1

360 W 2 = = 14:40 lb/ft A (4)(8) (2)(3:5) Weight of plate without cutout is W1 = A1 = 14:40(4)(8) = 460:8 lb Weight of cutout is W2 = A2 = 14:40(2)(3:5) = 100:8 lb

Weight of plate per unit area is

=

3 unknowns J

203 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.11

5.12

NB B

z C 1 ft T 3 ft

6 ft 240 lb x

2 ft

ft 1.5 Fx

A NA

Fy y

5 unknowns J

204 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.13

5.14

5.15

205 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.16

5.17

206 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.18 Fy

=

TA = TB 20 p (TA + TB ) (30) 80(15) = 0 = 0 202 + 222 20 p (2TA ) (30) 1200 = 0 2 20 + 222 TA = TB = 29:73 lb J

MCD

Mx

0

=

0

TC (32) + p

20 TB (44) 202 + 222

20 (29:73)(44) + 222 TC = 27:50 lb J

TC (32) + p

Fz

=

202

80(22) = 0 1760 = 0

20 (TA + TB ) 80 = 0 202 + 222 20 TD + 27:50 + p (29:73 + 29:73) 80 = 0 2 20 + 222 TD = 12:50 lb J TD + TC + p

0

5.19 Fx Fy Fz Mx My Mz

= = = = = =

0 0 0 0 0 0

Bx Cx = 0 By Ay = 0 Az + Cz = 0 12Cz 12By = 0 12Bx 10Az 120 = 0 12Cx 10Ay = 0

(a) (b) (c) (d) (e) (f)

Equations (a)-(d) yield Bx = Cx

Ay = By = Cz =

Az

Equations (e) and (f) can now be written as 12Bx 10Az 12Bx + 10Az

= =

120 0

The solution is Az =

6 lb

Bx = 5 lb

Therefore, Ay = By = Cz = Az = 6 lb Bx = Cx = 5 lb p p A = 62 + 62 = 8:49 lb J B = C = 62 + 52 = 7:81 lb J

207 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.20 Mx

Fx

MBC (MC )z

5.21 Mx Mz Fy My Fz Fx

2:8 T (3:5) 270(3:5) = 0 + 3:52 + 2:82 T = 473:3 lb J 2 = 0 Ax + p T =0 2 2 + 3:52 + 2:82 2 Ax + p (473:3) = 0 Ax = 192:86 lb 22 + 3:52 + 2:82 = 0 5:5Az 270(2) = 0 Az = 98:18 lb = 0 5:5Ay + 3:5Ax = 0 5:5Ay + 3:5( 192:86) = 0 Ay = 122:73 lb p A = ( 192:86)2 + 122:732 + 98:182 = 249 lb J =

0

p

22

= 0 1200PBC 196:2(600) = 0 PBC = 98:10 N J = 0 Ay = 0 J = 0 Ay + Oy = 0 Oy = 0 J = 0 1200Az + 196:2(600) 450PBC = 0 196:2(600) 1200Az 450(98:10) = 0 Az = 61:31 N J = 0 Oz + Az 196:2 + PBC = 0 Oz + 61:31 196:2 + 98:10 = 0 Oz = 36:79 N J = 0 Ox = 0 J

5.22 The force system is concurrent. Therefore, F = TBC + TBD + RA + P = 0

TBC TBD RA P

4i 12j + 12k = TBC (0:2294i 0:6883j + 0:6883k = TBC p 42 + 122 + 122 8i 12j + 4k = TBD ( 0:5345i 0:8018j + 0:2673k) = TBD p 82 + 122 + 42 4i + 12j + 6k = RA p = RA ( 0:2857i + 0:8571j + 0:4286k) 42 + 122 + 62 = 2200i 2800k lb

208 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

F = 0 results in the following equations: 0:2294TBC 0:5345TBD 0:2857RA + 2200 = 0 0:6883TBC 0:8018TBD + 0:8571RA = 0 0:6883TBC + 0:2673TBD + 0:4286RA 2800 = 0 The solution is RA = 3320 lb J

TBC = 936 lb J

TBD = 2740 lb J

5.23 TAD TBC rOA

1:8i 1:2j + 1:2k = TAD ( 0:7276i 0:4851j + 0:4851k) 1:82 + 1:22 + 1:22 1:8i 2:4j = TBC p = TBC (0:6i 0:8j) 1:82 + 2:42 = 1:2j m rOB = 2:4j m W = 784:8k N = TAD p

MO = rOA = TAD

i 0 0:7276

TAD + rOB j 1:2 0:4851

i + 0 0

k 0 0:4851

j 1:2 0

(0:5821i + 0:8731k) TAD

TBC + rOA + TBC

k 0 784:8

i 0 0:6

W j k 2:4 0 0:8 0

=0

(1:440k) TBC

941:8i = 0

Equating like components, we get 0:5821TAD 0:8731TAD

941:8

=

0

1:440TBC

=

0

TAD = 1618N J 0:8731(1618) TBC = = 981 N J 1:440

5.24 The 800-lb applied force in vector form is 3i 6j + 1:5k F = 800 p = ( 3)2 + ( 6)2 + 1:52 MA

= rAB

= =

F

i 0 349:2 (747:6

10Cx k j 6 698:3

349:2i

10Cz i k 0 124:6

10Cz ) i + (2095

698:3j + 124:6k lb

10Cx k

10Cz i

10Cx ) k = 0

209 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Cx C

5.25 Mx My Mz

= 0 = 0 = 0 Fx Fy

5.26

Mx

600(5) 6PBG = 0 PBG = 500 lb (T) J PAF = 0 J 600(2) 6PAE = 0 PAE = 200 lb = 200 lb (C) J

2i + 4j 5k = ( 0:2981i + 0:5963j 0:7454k) PBD 22 + 42 + 52 rCB PBD i + 600(5) 1 0 0 0 6 0 PBD + 3000 0:2981 0:5963 0:7454 4:472PBD + 3000 = 0 671 lb (T) J

= PBD p = =

PBD

209:5 lb Cz = 74:76 lb p 2 209:5 + 74:762 = 222 lb J =

= 0 Cx = 0 = 0 Cy 600 PAE = 0 Cy 600 ( 200) = 0 Cy = 400 lb = 0 Cz PBG PAF = 0 Cz 500 0 = 0 = 500 lb p C = 4002 + 5002 = 640 lb J

Fz Cz

PBD

=

= =

210 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.27

z C

D

1.2

1.2

0.6 TC TD A 0.7 2

x

2.1

1800 N

B

By

Dimensions in meters

TC TD

y

0:6i 0:72j + 1:2k = TC p = TC (0:3941i 0:4729j + 0:7881k 0:62 + 0:722 + 1:22 1:2i 0:72j + 1:2k = TD ( 0:6509i 0:3906j + 0:6509k = TD p 1:22 + 0:722 + 1:22 Fx Fz

= 0 0:3941TC 0:6509TD = 0 = 0 0:7881TC + 0:6509TD 1800 = 0 Solution is: TC = 1522:6 N J TD = 921:9 N J

MCD = 0

1:2By

1800(0:72) = 0

By = 1080 N J

5.28

B 1.2 m

C

m 1.2 A 2m PAD

PBD PCD D

15 kN

x

z y

40 kN

211 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

PAD

=

PCD

=

Fx Fy Fz

1:2i + 2k p PAD = (0:5145i + 0:8575k) PAD 1:22 + 22 (0:5145j + 0:8575k) PCD

=

0 0:5145PAD + 15 = 0 PAD = 29:15 kN PAD = 29:2 kN (C) J = 0 PCD = 0 J = 0 0:8575PAD + 0:8575PCD + PBD 40 = 0 0:8575( 29:15) + 0:8575(0) + PBD 40 = 0 PBD = 65:0 kN (T) J

5.29 Az

Ax

22 in.

A

Dz

T B

z x

12 in. y

Dx D

Dy

in. 28

C 60 lb

AD

=

rDC

=

28i + 22j 12k p = ( 28)2 + 222 + ( 12)2 28i in. rAB = 22j in.

MAD = 0 0 22 T 0 0:7451 0:5855

0:7451i + 0:5855j

0:3193k

rAB ( T i) AD + rDC ( 60k) AD 0 28 0 0 0 60 0 0 + 0:3193 0:7451 0:5855 0:3193

7:025T + 983:6 = 0

=

0

=

0

T = 140:0 lb J

212 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.30

5.31

213 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.32

PCF PBF

m 0.4

PAD D 0.3 m x PAE = PAE p MF

PAF

0.4 m

F

12 kN. m z C

A

PBE E

B

PAE

y

0:4i + 0:4j + 0:3k = PAE ( 0:6247i + 0:6247j + 0:4685k) 0:42 + 0:42 + 0:32

= rF E

PAE + 0:4PBE i 0:4PAD j + 12k i j k 0 0:4 0 PAE + 0:4PBE i 0:6247 0:6247 0:4685

= =

(0:4PBE + 0:1874PAE ) i PAE

=

PAD

=

PBE

=

0:4PAD j + 12k

0:4PAD j + (12 + 0:2499PAE ) k = 0

12 = 48:02 kN = 48:0 kN (C) J 0:2499 0 J 0:1874( 48:02) = 22:5 kN (T) J 0:4

5.33 TAB TAC TAD

6i 6j + 10:5k = TAB (0:4444i 0:4444j + 0:7778k) = TAB p 62 + 62 + 10:52 6i 3j + 10:5k = TAC p = TAC ( 0:4815i 0:2408j + 0:8427k 62 + 32 + 10:52 2:4i + 6j + 10:5k = TAD p = TAD (0:1947i + 0:4867j + 0:8516k 2:42 + 62 + 10:52 Fx Fy Fz

= = =

0 0 0

0:4444TAB 0:4815TAC + 0:1947TAD = 0 0:4444TAB 0:2408TAC + 0:4867TAD = 0 0:7778TAB + 0:8427TAC + 0:8516TAD 800 = 0

214 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution is TAB = 222 lb J

TAC = 359 lb J

TAD = 380 lb J

5.34 MBC = rBA

BC

+ rBA

W

BC

=0

2:4i + 6j + 10:5k = TAD p = TAD (0:1947i + 0:4867j + 0:8516k 2:42 + 62 + 10:52 = 800k lb rBA = 6i + 6j 10:5k ft 12i + 3j = 0:9701i + 0:2425j = p 122 + 32

TAD W BC

MBC

TAD

= =

6 6 10:5 6 0:1947 0:4867 0:8516 TAD + 0 0:9701 0:9701 0:2425 0 9:171TAD + 3492 = 0 TAD = 381 lb J

6 0 0:2425

10:5 800 0

5.35 NB

P

B

m 0.8

1.6 m G

z C x

245.3 N y 1.8 m

D A

Ax

MAC MAD

= 0 = 0

Ay Az

1:6P 245:3(0:4) = 0 1:6NB 245:3(0:9) = 0

P = 61:3 N J NB = 138:0 N J

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5.36

216 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.37 6 kN C 450

500

2.4 kN. m

B

D

z

270 A

x

9 kN y

RA

CA

F = 0 RA + 9i 6k = 0 RA = 9i + 6k kN J MA = 0 CA 2:4j 6(0:5)j + 9(0:27)j 9(0:45)k = 0 CA = 2:97j + 4:05k N m J

5.38

z

a 6 ft PA

PO

x PB

W1

W2 y

PA

= PB = PO =

W1

=

Mx b

=

b

2700 = 900 lb 3

12 (2700) = 1542:9 lb 21 My

ft 5 . 4

W2 =

9 (2700) = 1157:1 lb 21

0

6W1 aPA = 0 6W1 6(1542:9) a= = = 10:29 ft J PA 900 = 0 bPB 4:5W2 = 0 4:5W2 4:5(1157:1) = = = 5:79 ft J PB 900

217 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.39

z 100 TA

Bz

By

Bx

A B

150 x

TC

E 400 Dimensions in mm

C y

300 D 6 kN

Mx

=

My

=

Fx

=

Fy

=

Fz

=

6(400) = 0 TC = 6 kN J 1 3 0 6(300) p TA (150) p TA (300) = 0 10 10 TA = 7:590 kN J 3 3 Bx p (7:590) = 0 0 B x p TA = 0 10 10 Bx = 7:20 kN 0 By = 0 1 0 B z + TC + p T A 6 = 0 10 1 Bz + 6 + p 7:590 6 = 0 Bz = 2:40 kN 10 p B = 7:202 + 2:402 = 7:590 kN J 0

400TC

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5.40

5.41

219 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.42

z

4 ft

4 ft D By 7 ft Ay Az

Ax A

2.5 ft

Bz B 350 lb

5 ft

x

T C y

4i 7:5j + 7k p T = (0:3633i 0:6811j + 0:6357k) T 42 + 7:52 + 72 = 0 (0:6357T ) (7:5) 350(2:5) = 0 T = 183:52 lb J ) T = (0:3633i 0:6811j + 0:6357k) (183:52) = 66:67i 125:0j + 116:66k lb

T = Mx

Fx MAD Fy My Fz

A=

= 0 Ax + Tx = 0 Ax = Tx = 66:67 lb = 0 By = 0 = 0 Ay + By + Ty = 0 Ay + 0 125:0 = 0 Ay = 125:0 lb = 0 Az = Bz = 0 Az + Bz + Tz 350 = 0 2Az + 116:66 Az = Bz = 116:67 lb

p 66:672 + 125:02 + 116:672 = 183:5 lb J

350 = 0

B = 116:7 lb J

220 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.43

z Az

Ay

x

A

3 kN

300

Dimensions in mm 360

Ax

500

B

D

MA Dz

T

400 Dz

C y Dy

= (500T + 500Dz + 300Dy ) i 360Dz j + [3(400) + 360Dy ] k = 0 = 0 J Dy = 3:33 kN J T = 2:00 kN J

5.44

A z 480 N PBD D

3m

6m

4m

B PBE

4m O 4 m

Cy

Cz E

x

Cx

C

y

221 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Mx MCy

1 p PBE (4) = 0 PBD = 1:179PBE 2 1 4 = 0 480(10) 4 PBD + p PBE = 0 5 2 4800 3:20PBD 2:828PBE = 0 4800 3:20(1:179PBE ) 2:828PBE = 0 PBE = 727 N J PBD = 1:179(727) = 857 N J

=

0

4 PBD (3) 5

5.45

222 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.46 z

x D

Ay

Az

Ax

A n. i 18 10 in.

4 in .

20 in. 120 lb

80 lb

. 9 in T 3 C Bx

2

y

4 in . B Bz

2 p T (24) + 28Bz 80(14) 120(4) = 0 13 13:313T + 28Bz 1600 = 0 2 p T (18) 18Bz + 80(9) + 120(18) = 0 My = 0 13 9:985T 18Bz + 2880 = 0 Solution is: T = 1297:8 lb J Bz = 559:9 lb Mx = 0

Mz

=

0

3 p T (24) 13

28Bx = 0

3 p (1297:8)(24) 28Bx = 0 Bx = 925:6 lb 13 p B = 559:92 + 925:62 = 1082 lb J

223 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.47

224 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.48

z

150

Ay

B A Az

80

Bx 100

Dimensions 300 N in mm 40 x Cx

Fx Fy Fz Mx My Mz

= = = = = =

0 0 0 0 0 0

500 N By C y Cz

Bx + Cx + 300 = 0 Ay + By + 500 = 0 Az Cz = 0 180Ay 100By 40Cz 100Bx + 150Az = 0 150Ay 40Cx = 0

500(180) = 0

The solution is Ay Bx Cx

= = =

1600 N 6300N 6000 N

Az = 4200 N By = 2100 N Cz = 4200N

p 16002 + 42002 = 4490 N J p = 63002 + 21002 = 6640 N J p = 6002 + 42002 = 4240 N J

A = B C

225 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.49 Fz = 0

Fy

F

Fx = 0

F

Fx

Ey Dz Dy

Fz

Fy

C

Ez E

Ey

4.5 ft

Ex = 0

6 ft A

B

z

ft 4.5

C

Ez PEC E PEB

Ex

B

D Dx = 0

PFC

x

y

6000 lb FBD of joint F : Fx Fz

= =

0 0

Fx = 0 Fz = 0

FBD of joint E: Fx = 0

Ex = 0

FBD of entire truss: Fx MDE MAD (MD )z MEF Fz

= 0 Dx + Ex + Fx = 0 Dx + 0 + 0 = 0 Dx = 0 = 0 4:5Fy + 6000(6) = 0 Fy = 8000 lb = 0 4:5(Ez + Fz ) = 0 4:5(Ez + 0) = 0 Ez = 0 = 0 4:5(Ey + Fy ) = 0 4:5(Ey 8000) = 0 Ey = 8000 lb = 0 Dy = 0 = 0 Dz + Ez + Fz 6000 = 0 Dz + 0 + 0 6000 = 0 Dz = 6000 lb

) D = 6000k lb J

E = 8000j lb J

F=

8000j lb J

226 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.50

z Az

Ay x Ax

A

3

5 ft

4

180 lb

5 ft TBC

Mx = 0

5 5

TBD B y

5 5 p TBC + p TBD (10) 180(5) = 0 41 34 7:809TBC + 8:575TBD 900 = 0

3 4 p TBC + p TBD (10) = 0 41 34 6:247TBC + 5:145TBD = 0 Solution is: TBC = 49:4 lb J TBD = 60:0 lb J Mz = 0

5.51

227 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.52 PBA PCA PDA P

2:4j + 3:2k = PBA p = PBA ( 0:6j + 0:8k) 2:42 + 3:22 2:4i + 1:2j + 3:2k = PCA p = PCA (0:5747i + 0:2874j + 0:7663k) 2:42 + 1:22 + 3:22 2:4i + 1:2j + 3:2k = PDA p = PDA ( 0:5747i + 0:2874j + 0:7663k) 2:42 + 1:22 + 3:22 = Pk Fx Fy Fz

= 0 = 0 = 0

0:5747PCA 0:5747PDA = 0 0:6PBA + 0:2874PCA + 0:2874PDA = 0 0:8PBA + 0:7663PCA + 0:7663PDA P = 0

Solution is PBA = 0:4167P

PCA = PDA = 0:4350P

The limiting condition is 0:4350P = 8 kN

P = 18:39 kN J

228 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.53

229 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.54 z C T NB

30o

B

2.5 ft x

2 ft

3f t

G

A A y

50 lb Ax

My MAx

Az

y

=

0 (T sin 30 ) (2) (T cos 30 )(2:5) + 50(1:0) = 0 T = 15:80 lb J = 0 50(1:5) NB (2:5) (T sin 30 ) (3) = 0 50(1:5) NB (2:5) (15:80 sin 30 ) (3) = 0 NB = 20:5 lb J

5.55

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5.56

231 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.57

2000 lb

2 ft

D

z 4 ft C

T

6 ft

B

Ez E

8 ft

Ex

2 ft Ax x

A

F

Ay Az

3 ft y

1:0i + 2j 8k T = Tp = (0:1204i + 0:2408j 1:02 + 22 + 82 6j + 8k = p = 0:6j + 0:8k AE 62 + 82 MAE

T AE + rAB 2000j AE 3 2 0 2 0:1204 0:2408 0:9631 T + 0 = 0 0:6 0:8 0 = 2:504T + 3200 = 0 T = 1278 lb J

0:9631k)T

= rAF

0 8 2000 0 0:6 0:8

232 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.58 B

TCB W z

TDB 4 m G

Ay Az

3m A C x A 2.6 m 3m x

MAD = rAC TCB W rAC AD

MAD

TCB

m 4.5 y .2 m D 2

AD

+ rAG

W

7:5i 2:6j 4k = TCB q = (0:8438i 2 2 7:52 + ( 2:6) + ( 4)

AD

=0

0:2925j

0:4500k)TCB

= =

860(9:81)k = 8437k N 3i 2:6j m rAG = 2:25i + 2k m 2:2i + 3j = p = 0:5914i + 0:8064j 2:22 + 32

=

3 2:6 0 2:25 0:8438 0:2925 0:4500 TCB + 0 0:5914 0:8064 0 0:5914 1:7806TCB 15308 = 0 TCB = 8600 N J

0 0 0:8064

2 8437 0

=0

5.59

z PB

PC 4 ft

PA

4 ft

y

t 6f lb. ft 8000

x

233 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Mx

=

0

My

=

0

Fz PB

= 0 = PA

3 (8000) = 0 PC = 1200 lb (T) J 5 4 6PA + (8000) = 0 PA = 1067 lb (T) J 5 PA + PB + PC = 0 PC = 1067 1200 = 2267 lb = 2270 lb (C) J

4PC

5.60

234 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5.61 D

z

3m

2m C

3m

Az Ay Ax

x

A

3m

TBC

E 4 kN

3m

TBD B y

3 kN. m

TBD AC

MAC

3i 6j + 3k = ( 0:4083i 32 + 62 + 32 2i + 3k p = 0:5547i + 0:8321k 22 + 32

= TBD p =

= rCD =

=

TBD 5 0:4083 0:5547

+ 3k AC 0 3 0 0 0 0 0 4 0:8165 0:4083 TBD + 0:5547 0 0:8321 0 0:8321 AC

+ rAE

( 4k)

0:8165j + 0:4083k) TBD

+3(0:8321) 3:397TBD 6:656 + 2:496 = 0

AC

TBD = 1:225 kN J

5.62 Az

Ay

A 600 lb . in

Ax

z Bz T

21 in. x

B

Bx

C

12 in.

D

16 in.

in. 18

y

235 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

My

=

0

T = Tp )

MA rAB MA

12Tx 600 = 0 18i 16j 12k

Tx = 50 lb

= (0:6690i 182 + ( 16)2 + ( 12)2 0:6690T = 50 T = 74:74 lb J

= 0 rAB B + rAD T 600j = 0 = 21j in. rAD = 18i + 21j in. i j i j k 18 21 0 21 0 + 74:74 = 0:6690 0:5946 Bx 0 Bz 21(Bz i

Bx k) + ( 700i + 600j

1850k)

0:5946j

0:4460k)T

k 0 0:4460

600j = 0

600j = 0

Equating like components: (i-component) 21Bz 700 = 0 Bz = 33:33 lb (k-component) 21Bx 1850 = 0 Bx = 88:10 lb p ) B = 33:332 + ( 88:10)2 = 94:2 lb J

236 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6 6.1

3.75 m 60 kN

1 D

A NA

7.5 m

B

Bx

3m By

1.5 m 3.75 m 60 kN

M1 V1

M1 A

D 7.5 m

P1 P1 V1

D 3m

1.5 m

NA = 41.25 kN

B

Bx = 0

By = 18.75 kN

FBD of entire beam: MA Fx Fy

= 0 = 0 = 0

By (12) 60(3:75) = 0 Bx = 0 NA + 18:75 60 = 0

By = 18:75 kN NA = 41:25 kN

FBD of segment AD: Fx Fy MD

= 0 P1 = 0 J = 0 41:25 V1 60 = 0 V1 = 18:75 kN J = 0 M1 41:25(9) + 60(5:25) = 0 M1 = 56:25 kN m J

FBD of segment DB Fx Fy MD

= 0 = 0 = 0

P1 = 0 J V1 + 18:75 = 0 V1 = 18:75 kN J M1 18:75(3) = 0 M1 = 56:25 kN m J

237 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.2 12 kN

1 D

A 1.5 m

1.5 m

Bx

1.5 m

NA 12 kN A

1.5 m

1.5 m

M1 D V1

NA = 8 kN

P1

B By

P1

M1

V1 Bx = 0 D 1.5 m B By = 4 kN

FBD of entire beam: MA Fx Fy

= 0 = 0 = 0

4:5By 12(1:5) = 0 By = 4 kN Bx = 0 NA + 4 12 = 0 NA = 8 kN

FBD of segment AD: Fx Fy MD

= 0 = 0 = 0

P1 = 0 J 8 12 V1 = 0 V1 = M1 8(3) + 12(1:5) = 0

4 kN J M1 = 6 kN m J

FBD of segment DB: Fx Fy MD

= 0 = 0 = 0

P1 = 0 J V1 + 4 = 0 V1 = 4 kN J M1 4(1:5) = 0 M1 = 6 kN m J

238 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.3

6.4 240 lb 2.5 ft

3 ft B

1

A

Ax

5 ft

Ay

2 C

240 lb 2.5 ft

5 ft NB

M2

Ax = 0 A

5 ft

Ay = 60 lb

3 ft

P2 V1

C V2

C

5 ft M1 P1

239 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of entire structure: MB Fx

= 0 = 0

Ay (10) Ax = 0

240(2:5) = 0

Ay = 60 lb

FBD of segment above section 1: Fx Fy MC

= 0 = 0 = 0

V1 = 0 J P1 240 = 0 P1 = 240 lb J M1 240(2:5) = 0 M1 = 600 lb ft J

FBD of segment to the left of section 2: Fx Fy MC

= 0 = 0 = 0

P2 = 0 J V2 = 60 lb J M2 60(5) = 0

M2 = 300 lb ft J

6.5

240 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.6

6.7

241 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.8

V1

M1

6 in .

P1

6 in. 48 lb O

60o

O

B

(a)

V2

M2

P2

(b)

48 lb B

FBD (a): Fx Fy MO

P1 = 48 lb J V1 = 0 J 6P1 M1 = 0

= 0 = 0 = 0

M1 = 6(48) = 288 lb in. J

FBD (b): FP 2 FV1 MO

P2 48 sin 60 = 0 P2 = 41:6 lb J V2 + 48 cos 60 = 0 V2 = 24:0 lb J 6P2 M2 = 0 M2 = 6(41:6) = 250 lb in. J

= 0 = 0 = 0

6.9

400 lb A

D

C 12 150 lb

NA 20 20

A

C

2

By = 110 lb M2 12

P1 P2

D

Bx = 250 lb 20

B

V2

150 lb

NA = 110 lb

20

400 lb

V1

Bx

B

20

M1

12

By

12

1

FBD of entire stucture: MA Fx Fy

= 0 = 0 = 0

By (60) (150 + 400) (12) = 0 By = 110 lb Bx 400 + 150 = 0 Bx = 250 lb NA By = 0 NA = By = 110 lb

242 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of portion AC: MC Fx Fy

=

0 M1 + 150(12) 110(20) = 0 M1 = 400 lb in. J = 0 P1 = 150 lb J = 0 V1 = 110 lb J

FBD of portion DB: MD Fx Fy

=

0 M2 400(12) + 110(20) = 0 M2 = 2600 lb in. J = 0 P2 400 + 250 = 0 P2 = 150 lb J = 0 V2 = 110 lb J

6.10

0.6 m M1

3.6(2.4) = 8.64 kN Ax

C A Ay

3.6 m

3.6(1.2) =4.32 kN

B B P1 C 1.2 m NB V1 1.2 m oNB = 7.482 kN 30o 30

FBD of entire beam: MA = 0

(NB cos 30 ) (4:8)

8:64(3:6) = 0

NB = 7:482 kN

FBD of segment CB: MC Fx Fy

=

0 M1 + 4:32(0:6) (7:842 cos 30 ) (1:2) = 0 M1 = 5:56 kN m J = 0 P1 + 7:482 sin 30 = 0 P1 = 3:74 kN J = 0 V1 + 7:482 cos 30 4:32 = 0 V1 = 2:16 kN J

243 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.11 C 1200 N 1.6 m 2

B

M1 D

D

1 800 N Ax

P1 V1

1.6 m

1.6 m

1.8 m

1.8 m

E

A Ay

NE

0.9 m NE = 1466.7 N

E

FBD of entire frame: MA = 0

NE (3:6)

800(1:8)

1200(3:2) = 0

NE = 1466:7 N

FBD of segment DE: MD

=

0

M1

FV1

=

0

V1

F P1

=

0

P1

1466:7(0:9) = 0 M1 = 1320 N m J 0:9 V1 = 719 N J 1466:7 p 0:92 + 1:62 1:6 P1 = 1278 N J 1466:7 p 2 0:9 + 1:62

244 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.12 C

Cx 1.8 m

Cy By 0.9 m Bx

1.6 m

Dy Dx

0.9 m

D

Dx Dy = 400 N Dy = 400 N

D

B 800 N M2 P2

0.9 m D F V2

1.6 m

Dx = 1425 N

E

NE = 1466.7 N

FBD of member BD: MB = 0

Dy (1:8)

800(0:9) = 0

Dy = 400 N

FBD of member CE (NE was computed in the solution of Prob. 6.11): MC = 0

Dx (1:6) + 400(0:9)

1466:7(1:8) = 0

Dx = 1425 N

FBD of segment F D: MF Fx Fy

= 0 = 0 = 0

M2 400(0:9) = 0 P2 = 1425 N J V2 = 400 N J

M2 = 360 N m J

6.13

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6.14

246 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.15 T D 5.6 ft 1 A

2 B

6 ft

9 ft Cy

(a)

2800 lb

T = 4667 lb D

D T = 4667 lb M1

A

6 ft

5.6 ft M2

5.6 ft P1

B V1

(b)

2800 lb

Cx

C

A 2800 lb

6 ft

P2

B

(c)

V2

FBD (a): MC = 0

2800(15)

9T = 0

T = 4667 lb

FBD (b): MB

Fx Fy

5:6 (4667) + M1 = 0 5:62 + 62 M1 = 13 629 lb ft J 6 p = 0 (4667) + P1 = 0 P1 = 3410 lb J 2 5:6 + 62 5:6 p = 0 (4667) V1 = 0 V1 = 3180 lb J 5:62 + 62

=

0

2800(6)

p

FBD (c): MB Fx Fy

= 0 = 0 = 0

2800(6) + M2 = 0 M2 = 16 800 lb ft J P2 = 0 J 4667 2800 V2 = 0 V2 = 1867 lb J

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6.16 4

D

3

PCD 24 in. PBE 1080 lb. in. Fx 1080 lb. in.

18 in. Ay

Fy

E 12 in.

P1 M1

B

12 in.

1080 lb. in.

F 60 lb

V1

E

P2

G M2

F 60 lb

6 in.

V2 45 lb

12 in. A 60 lb

FBD of entire frame: MA Fx Fy

= 0 = 0 = 0

1080 18Ay = 0 Ay = 60 lb Fx = 0 Fy Ay = 0 Fy = Ay = 60 lb

FBD of member F ED: MD = 0

24PBE

1080 = 0

PBE = 45 lb

FBD of segment DEF below section 1: Fx Fy ME

= 0 = 0 = 0

V1 = 0 J P1 60 = 0 P1 = 60 lb J M1 1080 = 0 M1 = 1080 lb in. J

FBD of segment ABC below section 2: Fx Fy MG

= 0 = 0 = 0

V2 45 = 0 V2 = 45 lb J P2 + 60 = 0 P2 = 60 lb J M2 45(6) = 0 M2 = 270 lb in. J

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6.17

6.18

249 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.19

6000 lb B

6 ft

M2

C 3

PCE 4

P2

V2 6 ft B

3 4

PCE = 5000 lb

8 ft A

C

Ax

Ay Note that CDE is a two-force body. FBD of ABC: MA = 0

4 3 PCE (8) + PCE (6) 5 5

6000(8) = 0

PCE = 5000 lb

FBD of BD: MB Fx Fy

= 0 = 0 = 0

4000(6) M2 = 0 M2 = 24 000 lb ft J P2 3000 = 0 P2 = 3000 lb J V2 + 4000 = 0 V2 = 4000 lb J

250 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

*6.20

6.21 w0

A

B

x

L

w0L/2

w0L/2

w0L/2 x

V

_ w L/2 0

2

w0L /8 M

x

w0 x A w0L/2

x/2 x FBD

M V

251 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

V =

w0 L 2

w0 x J

M=

w0 Lx 2

w0 x2 J 2

6.22

6.23

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253 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.24

6.25 y 600 lb/ft 14 400 lb . ft

A

B 6 ft

6 ft

x (ft) 600 lb/ft

V (lb) -1800 14 400 7200 M (lb. ft)

254 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

x/2 x/3 50x2 100x 100(6 - x)

100(6x - x2)

A x 14 400 lb . ft

V

=

50x2

M

=

14 400

=

100(6x 50x2

300x2 +

M V 600x + 50x2 lb J 1 100(6x x2 ) x 2

x2 ) = 2 x 3

50 3 x + 14400 lb ft J 3

6.26

255 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.27 y A

120 kN/m

B

C x

3m

3m 90 kN

0.75

270 kN

90 V (kN)

270

-270

304

M (kN. m)

A

M

x 90 kN

V 120(x -3) B

3m

A

x 90 kN

M x -3 2 V

Segment AB: V M

= =

90 kN J 90x kN m J

256 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Segment BC: V

=

90

M

=

90x

3) = 450 120x kN J x 3 120(x 3) = 60x2 + 450x 2

120(x

Mmax = M j3:75 m =

60(3:752 ) + 450(3:75)

540 kN m J

540 = 304 kN m

6.28

257 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.29

y A

40 kN 4.5 m

3m B

C

30 kN. m

x

20 kN

20 kN 20 V (kN)

-20 90 60 M (kN. m)

M

M

A

7.5 - x

x 20 kN V =

M=

V

20 kN

V

20 kN if 20 kN if

20x kN m 20(7:5 x) kN m

C

x x if if

3m 3m

J

x x

3m 3m

J

258 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.30

259 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.31

1800 lb

y

C

A 3 ft 900 lb

3 ft 900 lb

B

x

900 V (lb) -900

1800

M (lb. ft) (6 - x)/3 200x 200(6 - x)

P1 x/3 A

M

x 900 lb

M

V

V

P2 6-x 900 lb

C

Segment AB: P1 V

= =

M

=

100x2 lb 900 P1 = 900 100x2 lb J x 100 3 900x P1 = 900x x lb ft J 3 3

Segment BC: P2 V

= =

M

=

100(6 x)2 lb 900 + P2 = 900 + 100(6 x)2 lb J 6 x 100 900(6 x) P2 = 900(6 x) (6 3 3

x)3 lb ft J

260 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6.32 12 kN

12 kN B

A

D

C

x (m)

2m 12 kN

4m

2m 12 kN

12 V (kN) -12 M (kN. m) 24

12 kN A

12 kN

V M

x

M

8-x

D

V 12 kN x A

12 kN V =

M=

8 < :

M

B

2m

V

8
FA =NA ( same at C and A). ) FC = 0:25NC

NC =

s

being the

9:50 FC = = 38:0 lb 0:25 2:5

From FBD of bar BC: MB

=

0 + FC (24 cos ) NC (24 sin ) 10(12 cos ) = 0 9:50(24 cos ) 38:0(24 sin ) 10(12 cos ) = 0 108 108 cos 912 sin = 0 = tan 1 = 6:75 J 912

311 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.5

7.6

W

0.5NA

ft 1.5

ft 1.5 y

θ

0.5NB B

NB

x

A NA

312 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Fx Fy

= =

0 0

0:5NA NB = 0 NA + 0:5NB W = 0

Solution is NA = 0:8W MA

=

NB = 0:4W

0 NB (3 sin ) + 0:5NB (3 cos ) 0:4W (3 sin ) + 0:5(0:4W )(3 cos ) 1:2 sin 0:9 cos = 0 0:9 = tan 1 = 36:9 J 1:2

W (1:5 cos ) = 0 W (1:5 cos ) = 0

7.7 0.3 m

490.5 N 294.3 N

2

G C0 0.5 m

T

1 1 2

T B

FC

C

FB

NB

NC

Assume impending sliding of the block. ) FB = 0:2NB FBD of block: Fx

=

0

+!

Fy

=

0

+"

0:2NB

2 p T =0 5

1 NB + p T 5

294:3 = 0

The solution is T = 59:83 N

NB = 267:5 N

FBD of spool: Fx

MG

2 2 FC = p T = 0 FC p (59:83) = 0 5 5 FC = 53:51 N = 0 + C0 FC (0:5) T (0:3) = 0 C0 (53:51)(0:5) (59:83)(0:3) = 0 C0 = 44:7 N m J =

0

+

313 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Check for sliding at C: Fy

=

0

+"

NC FC NC

1 p T 5

NC

1 p (59:83) 5

490:5 = 0

53:51 = 0:1034 < 0:15 517:3

=

490:5 = 0 NC = 517:3 N

Spool does not slide

B

P

12 in.

7.8

0.15N1

12 in.

Oy

N1 N1

0.15N2 N2 N2 Ox

O 5i n.

0.15N1

8 in. A

.

2200 lb in.

Ax

4 in.

0.15N2

8 in.

4 in.

F

P

Ex

E

Ey

Ay FBD of EF : ME = 0

20P

8N2 + 4(0:15N2 ) = 0

N2 = 2:703P

20P

8N1

N1 = 2:326P

FBD of AB: MA = 0

4(0:15N1 ) = 0

FBD of cylinder: MO

=

0 5(0:15N1 ) + 5(0:15N2 ) 2200 = 0 N1 + N2 = 2933 lb 2:326P + 2:703P = 2933 P = 583 lb J

314 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.9 y

y 200 lb

NA

R

O

200 lb

x NA 15o

FA

O

R 15o

0.3NA

0.12NB

x

FB

NB Sliding at B

NB Sliding at A

Assume sliding at B. Fx Fy MO

= 0 = 0 = 0

NA NB FA

0:12NB 200 sin 15 = 0 FA 200 cos 15 = 0 0:12NB = 0

Solution is: FA = 26: 34 lb

NA = P = 78: 11 lb

NB = 219:5 lb

Assume sliding at A. Fx Fy MO

= 0 = 0 = 0

NA FB 200 sin 15 = 0 NB 0:3NA 200 cos 15 = 0 0:3NA FB = 0

Solution is: FB = 22: 18 lb

NA = P = 73: 95 lb

NB = 215:4 lb

The smallest force that initiates motion is P = 74:0 lb J

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7.10

316 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.11

7.12

2 lb Ax

A

9 in. 12 in.

B FB NB

Ay

NB B

y FB 4 lb

x

8 in. P

D FC

2 in. C

NC

317 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of bar: MA = 0

12NB

2(9) = 0

NB = 1:50 lb

FBD of spool: Fy MB MC

= 0 = 0 = 0

NC 8P 2P

4 NB = 0 NC 4 1:50 = 0 10FC = 0 P = 1:25FC 10FB = 0 P = 5FB

NC = 5:50 lb

Assume impending sliding at B: FB =

s NB

= 0:3(1:50) = 0:45 lb

P = 5(0:45) = 2:25 lb

Assume impending sliding at C: FC =

s NC

= 0:3(5:50) = 1:65 lb

P = 1:25(1:65) = 2:06 lb

Largest P that does not cause sliding is P = 2:06 lb J

318 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.13 P G

R 30o

F = µsΝ

W A N Consider impending slipping at A: MG Fx

= 0 = 0

+ ( sN ) R P R = 0 P = sN +! N sin 30 + P = 0 s N cos 30 N sin 30 + s N = 0 s N cos 30 s = 0:268 J

7.14

319 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.15 Assume equilibrium.

Cy

6' C

Cx

C

170 lb

y

1.848'

x

A NA

60o FA

8'

2' 12 lb

20 lb 12 lb 3.2'

6.928'

2'

60o FB

B

60o 4' FB

B NB

NB

FBD of entire ladder: MA Fy Fx

= =

0 8NB 12(6) 20(2) 170(1:848) = 0 NB = 53:27 lb 0 NA + NB 12 20 170 = 0 NA + 53:27 202 = 0 NA = 148:73 lb = 0 FA FB = 0 FA = FB

FBD of BC: MC

FA NA FB NB

=

= =

0 4NB 12(2) 6:928FB = 0 4(53:27) 12(2) 6:928FB = 0

FB = FA = 27:29 lb

27:29 = 0:1835 < 0:4 Ladder will not slide at A 148:73 27:29 = 0:152 > 0:4 Ladder will slide at B J 53:27

320 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.16

7.17 B NB

NB L/2

L/2

FB

30o B L/2

W

L/2

W

60o FA

A

FB 60o

60o FC NC = W

NA = W

C

Due to symmetry of the assembly NA = NC = W . FBD of sheet AB: MA

=

0

+

Fy

=

0 +# FB sin 60

L cos 60 = 0 NB = 0:25W 2 FB sin 60 NB cos 60 NA + W = 0 (0:25W ) cos 60 W +W =0 FB = 0:14434W NB (L)

W

FB 0:14434W = = 0:5774 > 0:5 NB 0:25W

) Equilibrium is impossible

321 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.18

W/2 C

Bx

By B

0.3(W/2) h/2

h/2 A

0.5(W/2)

b W/2

Equilibrium is lost when sliding impends at A and C. MB

=

0 0:2h

h W h + 0:3 2 2 2 b 0:5b = 0 = 0:4 J h 0:5

W 2

W b=0 2

7.19 4500 lb G 3600 lb A NA 4.3 ft 5.7 ft

T 2 ft y B FB NB

T C x

NC

0.2NC

Assume impending sliding of the crate. FBD of the crate: Fy Fx

= =

0 0

NC 4500 = 0 0:2NC T = 0

NC = 4500 lb T = 0:2NC = 0:2(4500) = 900 lb

FBD of car: MA Fx

=

0 10NB 3600(4:3) 2T = 0 10NB 3600(4:3) 2(900) = 0 NB = 1728:0 lb = 0 T FB = 0 FB = T = 900 lb

Check for slipping at B: FB NB

=

900 = 0:521 < 0:6 Tires will not slip. 1728:0 Assumption was O.K. Crate will slide. J

322 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.20

4500 lb P P 2 ft y

G B

FB

A 3600 lb 5.7 ft 4.3 ft NA

NB

C 0.2NC x

NC

Assume impending sliding of the crate. FBD of the crate: Fy Fx

= =

0 0

NC 4500 = 0 P 0:2NC = 0

NC = 4500 lb P = 0:2NC = 0:2(4500) = 900 lb

FBD of car: MA Fx

=

0 3600(4:3) + 2P 10NB = 0 3600(4:3) 2(900) 10NB = 0 NB = 1368:0 lb = 0 FB P = 0 FB = P = 900 lb

Check for slipping at B: FB NB

=

900 = 0:658 > 0:6 Tires will slip. 1368:0 Assumption was incorrect. Crate will not slide. J

7.21 NB

B

FB = 0.3NB 196.2 N 5 m 1.0 m P 1.5 m FA = 0.3NA A

2m NA

323 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Consider impending slipping at A and B. MA Fx Fy

=

0 + NB (5) 0:3NB (2) 196:2(1:0) 4:4NB 1:5P 196:2 = 0 = 0 + ! 0:3NA + NB P = 0 = 0 + " NA 0:3NB 196:2 = 0

P (1:5) = 0

Solution is NA = 227 N

NB = 102:9 N

P = 171:0 N J

7.22 z'

10.791 N 30o

1

5.5 N 0.6 m

3

F2

y'

F1

N1

m 1.8

x' Let the x0 and y 0 axes lie on the inclined plane, and let z 0 be perpendicular to the plane. W = 1:1(9:81) = 10:791 N Assume equilibrium. Fx0

=

0

Fy 0

=

0

Fz 0

=

0

F = N1

p

1 F1 + p (10:791) 5:5 cos 30 = 0 F1 = 1:3507 N 10 F2 + 5:5 sin 30 = 0 F2 = 2:75 N 3 N1 = 10:237 N N1 p (10:791) = 0 10

1:35072 + 2:752 = 0:299 < 0:35 10:237

Disk is in equilibrium J

324 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.23 40 lb

P

45o

T 30o n. 6.4 i G 9 in .

The spool will slip against the wall. ) F = MG Fx Fy

= 0 = 0 = 0

N F = 0.25N

sN

= 0:25N .

+ 0:25N (9) P (6:4) = 0 + ! T cos 30 N P sin 45 = 0 + " T sin 30 0:25N P cos 45 40 = 0

Solution is N = 180:0 lb

T = 259 lb

P = 63:3 lb J

7.24

325 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.25

7.26

7.27 Cy Cx

C

b FA= µsΝA 2W NA A b/2 b/2 W

FA= µsΝA NA A b/2 B

By Bx b/2

B

W

Consider impending slipping at A. FBD of entire assembly: MC = 0

+

NA b

p ( 2 + 1)W

b 2

=0

NA = 1:2071W

326 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

FBD of bar AB: MB

=

0

+ s

s NA b

W

= 0:414 J

b =0 2

1:2071

s

1 =0 2

7.28 3 ft

P

120 lb 6 ft

h F = 0.4N A N Consider simultaneous impending sliding and tipping about A. Fy Fy MA

= 0 = 0 =

0

+ " N 120 = 0 N = 120 lb + ! P 0:4N = 0 P = 0:4(120) = 48 lb 120(1:5) = 3:75 ft J + P h 120(1:5) = 0 h= 48

7.29

y 784.8 N

35o P

0.6 m

1.8 m 0.75 m

B FB

x NB

x

W = 80(0:81) = 784:8 N

327 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Assume impending sliding (FB = 0:3NB ): Fx Fy

= 0 P cos 35 0:3NB = 0 = 0 P sin 35 + NB 784:8 = 0 Solution is NB = 993:5 N P = 363:8 N

Assume impending tipping (x = 0): MB

=

0 (P cos 35 ) (1:8) P = 319 N

784:8(0:6) = 0

P = 319 N determined by tipping J

7.30

y P 784.8 N 0.6 m

1.8 m 0.75 m

B x NB

FB

x

Assume impending sliding (FB = 0:3NB ): Fy Fx

= 0 = 0

NB 784:8 = 0 P 0:3NB = 0

NB = 784:8 N P 0:3(784:8) = 0

P = 235 N

Assume impending tipping (x = 0): MB = 0

1:8P

784:8(0:6) = 0

P = 262 N

P = 235 N determined by sliding J

328 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.31 N F = µsΝ

A

B 84 in.

36 in. P 120 lb Consider simultaneous impending sliding and tipping (door lifting o¤ the track at B). Fy MA Fx

= 0 = 0 = 0 s

+ " N 120 = 0 N = 120 lb + P (84) 120(36) = 0 P = 51:43 lb J +! P 51:43 sN = 0 s (120) = 0 = 0:429 J

7.32

y

156.96 N A FA

2.25 m x x o

50 NA P

W = 16(9:81) = 156:96 N Assume impending sliding (FA = 0:3NA ): Fx Fy

= 0 0:3NA P cos 50 = 0 = 0 NA P sin 50 156:96 = 0 Solution is NA = 244:3 N P = 114:02 N

Assume impending tipping about corner A (x = 1:5 m): MA

=

0 156:96(2:25 P = 102:4 N

1:5)

(P sin 50 )(1:5) = 0

P = 102:4 N determined by tipping J

329 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.33 2 ft

5 ft

t 1.0 f

t 2.5 f

W

θ

A F = 0.4N x

N

Assume impending sliding. Fx

=

0

+ ! 0:4N cos = 21:80 J

N sin = 0

tan = 0:4

Check for tipping: x = 1:0 Since x

2:5 tan = 1:0

2:5 tan 21:80 = 7:134

10

5

ft

0, tipping and sliding impend simultaneously.

7.34 T

3 in.

T

C

5 in. 3 lb

3 lb B

A

FA

x

NA

FB NB

Assume impending tipping of the block. From FBD of block using x = 0: Fy MA Fx

= 0 = 0 = 0

+ " NB 3 = 0 NB = 3 lb + T (5) 3(1:5) = 0 T = 0:9 lb + ! FB T = 0 FB = T = 0:9 lb

Assume impending sliding of the block. From FBD of block using FB = 0:35NB : Fy Fx

= 0 = 0

+" +

NB T

3=0 NB = 3 lb 0:35NB = 0 T = 1:05 lb

330 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Assume impending sliding of the cylinder. From FBD of cylinder with FA = 0:35NA : Fy Fx

= =

0 0

+ " NA +! T

3=0 NA = 3 lb 0:35NA = 0 T = 1:05 lb

Equilibrium is lost when T = 0:9 lb due to tipping of the block. From FBD of cylinder: MA = 0

+

C

C = 5T = 5(0:9) = 4:5 lb in J

5T = 0

7.35

7.36 Assume simultaneous impending tipping and sliding.

y

L 2

0.45NA A

W

P

L 2

θ

x

B NA

331 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

L 2

MB

=

0

W

Fx

=

0

P cos

MA

=

0

(P sin ) L

tan

=

P

=

NA L = 0

NA =

0:45NA = 0 W

W 2

P cos = 0:45

L =0 2

P sin =

W 2

W 2

1 = 2:222 = tan 1 (2:222) = 65:8 0:45 W W = 0:548W J = 2 sin 2 sin 65:8

J

7.37 Assume tipping

P

P

0.5 m W1

0.75 m FA

0.25 m A 1.5 m

0.5 m

W2

NA

0.25 m FB

W1 = 20(9:81) = 196:2 N

B NB

W2 = (20 + 45)(9:81) = 637:7 N

Top box: MA

=

0 0:25W1 P = 65:4 N

0:75P = 0

0:25(196:2)

0:75P = 0

1:5P = 0

0:25(637:7)

1:5P = 0

Both boxes together: MB P

= 0 0:25W2 = 106:3 N

332 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Assume slipping P

P W1 W2

A

0.35NA NA y

x

B

0.12NB NB

Top box: Fy Fx

= =

0 0

NA W1 = 0 NA = W1 = 196:2 N P 0:35NA = 0 P = 0:35NA = 0:35(196:2) = 68:7 N

= 0 = 0

NB W2 = 0 NB = W2 = 637:7 N P 0:12NB = 0 P = 0:12NB = 0:12(637:7) = 76:5 N

Both boxes: Fy Fx

Smallest P that results in impending motion (tipping of top box) is P = 65:4 N J

7.38

333 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.39 T y h/2

W

θ

x

h/2

b/2

F = µsN

A N

Consider simultaneous impending sliding and impending tipping about A. Fy Fx MA

= 0 N W cos = 0 N = W cos = 0 W sin = 0 T = W sin sN + T h b = 0 (W sin ) + (W cos ) Th = 0 2 2 1 b T = W sin + W cos 2 2h

sW

cos

(a)

(b)

Equating the right-hand sides of Eqs. (a) and (b): W sin

sW

cos

=

1 sin 2

=

b 1 W sin + W cos 2 2h b cos = tan s+ 2h

1

2

s

+

b h

J

334 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.40

y

60 lb 2 ft

6 ft

4 ft P

A

B

8 ft

C

FB

NA

FB

x

NB B

20 lb

NB

2.5 ft D

FD ND

1.2 ft

FBD of plank ABC: MA Fx

= 0 = 0

8NB 60(6) = 0 NB = 45 lb FB P = 0 FB = P

FBD of support BD: Fx Fy

= 0 = 0

FD ND

FB = 0 FD = FB = P NB 20 = 0 ND = NB + 20 = 65 lb

Assume impending tipping of support BD: MD

=

0 2:5FB (NB + 20)(0:6) = 0 2:5P (45 + 20)(0:6) = 0 P = 15:60 lb

Assume impending sliding at B: P =

BN

= 0:4(45) = 18:0 lb

Assume impending sliding at D : P = FD =

D ND

= 0:3(65) = 19:5 lb

The largest P that can be applied without causing impending motion (tipping of the support) is P = 15:60 lb J

335 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.41

7.42 FD

ND= P D 200 N. m

y x

0.0 5m

O

30o 30o

NA Fx Fy MO

= 0 = 0 = 0

A

B

NB

(NA NB ) sin 30 (NA + NB ) cos 30 0:05FD 200 = 0

FD = 0 P =0 FD = 4000 N

(a) (b) (c)

Assume impending slipping at D (FD = 1:5P ): From Eq. (c): 1:5P = 4000

P = 2667 N

Assume impending rolling about D (NB = 0): From Eq. (a): From Eq. (b):

NA sin 30 8000 cos 30

4000 = 0 P =0

NA = 8000 N P = 6930 N

Smallest force that prevents motion is P = 6930 N J

336 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.43 y W

x

C o

25

µNA

R A

NA

Assume simultaneous slipping and tipping about point A: Fy Fx

= 0 = 0

NA W cos 25 = 0 NA W sin 25 = 0

= tan 25 = 0:466 J MA C

= 0 W R sin 25 C=0 = W R sin 25 = 0:423W R J

7.44

W 30 + φs o

R1

60o− φs 30o L/2

60o L/2

R2

Assume impeding slipping (left end down, right end up). Because the bar is a 3-force body, the forces intersect at a common point. 30 +

s

= 60

s

s

= 15

s

= tan 15 = 0:268 J

337 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.45

338 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.46

7.47

R a

B

b φs A R Assume impending sliding at A (note that AB is a two-force body). s

= tan

s

=

a J b

339 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.48 A

RA

φs d

NB

C

B 7.5 in.

P The hook is a three-force body with the forces intersecting at C. Assuming impending sliding at A: tan

s

=

7:5 d

d=

7:5 7:5 7:5 = = = 15 in. J tan s 0:5 s

7.49

7.50

P

W

R W φs

θ

θP

φs R

The block is a three-force body (the forces must be concurrent). From the force triangle we see that P is minimized when it is perpendicular to R: Hence

P

= s J = W sin

s

J

340 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.51 P 200 lb

R1

27o

15o 15o

27o

27

o

15o

R1

R1

12o R2

Consider impending slipping. FBD of wedge: Fy = 0

+"

2R1 sin 27

P =0

FBD of block: Fy Fx

= =

0 0

+" +

R2 cos 12 R2 sin 12

R1 sin 27 200 = 0 R1 cos 27 = 0

Solution is R1 = 53:5 lb

P = 48:6 lb J

R2 = 229 lb

7.52 WA + WB

WB B

50o

R 15o

P

B

A

R W 35o P

R'

30o

P

R' o W + WB 30 A P

Assume impending sliding of block B on block A. From the force triangle W 300(9:81) = = 4203 N tan 35 tan 35 Assume impeding sliding of both blocks on ground. From the force triangle P =

P = (WA + WB ) tan 30 = 800(9:81) tan 30 = 4531 N Smallest P causing impending motion is P = 4200 N J

341 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.53

180 N

C R

φk

θ

96"

20 "

D O

960 N

θ R 180 N

960 N

The collar is a three-force body with the force intersecting at C. From the force triangle: = tan

1

180 = 10:620 960

From triangle ODC in the FBD: sin(

k)

96 k

= =

sin 20 tan

k

sin k sin 10:620 = 96 20 = tan 62:20 = 1:897 J

k

= 62:20

7.54 p = 2 r tan

= tan

1

p = tan 2 r

1

0:5 = 2:604 2 (1:75)

(a) Cup

= W r tan( s + ) = 4000(1:75) tan(8:5 + 2:604 ) = 1373:9 lb in. = 114:5 lb ft J

(b) Cdown

= W r tan( s ) = 4000(1:75) tan(8:5 = 722:9 lb in. = 60:2 lb ft J

2:604 )

342 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.55

18o R' 12o

6000 lb P

12o R"

30o

y R 12o

x

R' 18o

Assume impending sliding (wedge slides to the left). From FBD of the block: Fx Fy

= =

0 R0 sin 18 R00 cos 12 = 0 0 R0 cos 18 + R00 sin 12 6000 = 0 0 00 R = 5901 lb R = 1864 lb

From FBD of wedge: Fy Fx

0 R cos 12 R0 cos 18 = 0 R cos 12 5901 cos 18 = 0 R = 5738 lb 0 = 0 P R sin 18 + R sin 12 = 0 P 5901 sin 18 + 5738 sin 12 = 0 P = 631 lb J =

7.56 Given: r = 0:004 m Lead angle:

p = 0:0016 m 1

= tan s

=

tan

1 s

p = tan 2 r

= tan

1

1

s

= 0:2

0:0016 = 3:643 2 (0:004)

(0:2) = 11:310

(a) Clamping force from Eq. (7.7a): F =

C 1:50 = = 1404 N J r tan( s + ) 0:004 tan(11:310 + 3:643 )

(b) Unclamping torque from Eq. (7.7b): C 0 = F r tan(

s

) = 1404(0:004) tan(11:310

3:643 ) = 0:756 N m J

343 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.57

7.58

344 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.59 T2 T1 n

= e

=

s

1 s

ln

T2 1 8000 = ln = 21:19 rad T1 0:25 40

21:19 = = = 3:37 turns 2 2 4 turns are required to hold the ship J

7.60

7.61

7.62

θC 3

4 3

4

B

= 0:927 rad

C

θB

The angles of contact are B

= tan1

4 3

=

2

+ tan

1

4 3

= 2:498 rad

345 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Total angle of contact is =

B

+

C

= 0:927 + 2:498 = 3:425 rad

Assume impending motion with A moving up: P = 120(9:81)e0:25(3:425) = 2770 N Assume impending motion with A moving down: 120(9:81) = P e0:25(3:425) Equilibrium range is 500 N

P

P = 500 N

2770 N J

7.63

T2 = 34 lb 30o

θ = 360o + 30o = 390o

T1 T1 = T2 e

s

= 34e

0:6 390 180

= 0:5725 lb

But T1 is the weight of the free end of the rein: T1 =

3:5 L 16

0:5725 =

3:5 L 16

L = 2:617 ft = 31:4 in. J

7.64 Assume impending sliding with the weight about to move up: P = 30e0:3 = 77:0 lb Assume impending sliding with the weight about to move down: 30 = P e0:3 System is at rest if

P = 11:69 lb

11:69 lb < P < 77:0 lb J

346 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.65

7.66

y

T

35o

x

WA A N

0.2N

Assume impending sliding with block A moving up. Fy Fx

= 0 = 0

N T

WA cos 35 = 0 0:2N WA sin 35 = 0

Solution is WA = 1:3561T

N = 1:1109T

The contact angle between the rope and the peg is = 90 + 35 = 125 = 2:182 rad WB

0

= T e = T e0:25(2:182) = 1:7255T 1:7255 WB = = 1:272 ) WA 1:3561

Assume impending sliding with block A moving down. The friction force on the FBD must be reversed, yielding Fy Fx

= 0 = 0

N WA cos 35 = 0 T + 0:2N WA sin 35 = 0

347 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution is WA = 2:441T T

0

0

= WB e WB = T e WB 0:5796 ) = = 0:237 WA 2:441

N = 1:999T 0:25(2:182)

= Te

= 0:5796T

System is in equilibrium if WB WA

0:237

1:272 J

7.67 θA A 4

B

+

β

T1

T2

T2

T1

8

8

4 β

θB

+

150 lb

P

From geometry: =

cos

A

=

2

B

=

1

4 = rad 8 3

2 =2 =

2

2

2 3

3

=

= 6

4 3

rad

rad

Consider impending slipping: T2 T1 P

= 150e = T2 e = T1 e

= 150e 0:3( =6) = 128:20 lb = 128:20e 0:3(4 =3) = 36:49 lb = 36:49e 0:3( =6) = 31:2 lb J

s B s A s B

348 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.68 α

C

α

3

4

6 ft

2

ΤΑ

4 ft A

θ

ΤΑ

3

ΤΒ

C Detail of rope at C

ΤΒ

8 ft

P B

6 ft 50 lb

From FBD of bar: MA

=

0

+

MB

=

0

+

3 TB (12) 50(6) = 0 TB = 41:67 lb 5 3 p TA (12) 50(6) = 0 TA = 30:05 lb 13

Geometry: =

1

tan

=

4 2 + tan 1 = 1:5153 rad 3 3 = 1:5153 = 1:6263 rad

Condition for impending sliding of rope on peg: TB = TA e

s

s

=

1

ln

TB 1 = ln TA 1:6263

41:67 30:05

= 0:201 J

7.69

7.70 P

=

Z

p dA =

A

=

2 p0

Z

0

p0

=

2 P R2

Z

R

p(2 r dr) =

0

R

r

Z

R

r2 R2

p0 1

0

r3 R2

2 P p= R2

dr = 2 p0 1

r2 2

r4 4

(2 r)dr

R

= 0

1 p0 R 2 2

r2 R2

349 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Consider impending slipping: C

=

Z

s pr dA =

A

=

P s 2 R

4

Z

Z

R s

0

R

2 P R2

r4 R2

r2

0

r2 R2

1

r(2 r dr)

P r3 s 2 R 3

dr = 4

r5 5R2

R

= 0

8 15

sP R

J

7.71 C y Oy

x

Ox

ft 2.5 O

1.75 ft 160 lb

MO = 0

C

160(1:75) = 0

C = 280 lb ft

Assume unworn surfaces. From Eq. (7.12): C=

2 3

sP R

s

=

3C 3(280) = = 0:280 J 2P R 2(600)(2:5)

7.72

dA = r dr dθ R

F

=

Z

0

=

2

R

Z

kp

r

dF = µk p dA

θ

=2

dF cos =

Z

kp

=2

0

R

rdr =

Z

k pR

2

0

R

Z

=2

r cos dr d =2

J

350 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.73

7.74 From Eqs. (7.11) and (7.12): C

R03 Ri3 3 3 R02 Ri2 2(0:2)(290)(0:055) 2(0:08)(290 + 24 + = 3 3 = 2:127 + 5:009 = 7:14 N m J =

2

A P R0

+

2

B (P

+ W)

9:81)

0:1653 0:1652

0:0553 0:0552

7.75

7.76

351 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.77

7.78

P F/2

N Applying Eq. (7.16) to the top roller: F = 2

rP

F =2

rP

= 2(0:016)(80) = 2:56 kN J

352 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.79 y

y W P 30o R

W

x

R

φ

a (a)

= sin

1

P 30o x

φ

a N

N

a = sin R

1 d

= sin

(b) 1

0:1 = 5:739

(a) Fx

=

0 P cos(30 ) W sin = 0 P cos(30 5:739 ) 30(9:81)(0:1) = 0

P = 32:3 N J

0 W sin P cos(30 + ) = 0 30(9:81)(0:1) P cos(30 + 5:739 ) = 0

P = 36:3 N J

(b) Fx

=

7.80

C W

R O

φ W From the FBD of cylinder (the weight of the cylinder is not shown since it is irrelevant): C = W R sin = W R r J

353 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.81 P

C 13

10.8

y x

G 85 lb

5.4 A

10.8 B

FA NA

FA = ( MC

r )A NA

29

FB

Dimensions in inches

NB = 0:12NA

FB = (

r )B NB

= 0:18NB

=

0 23:8NA 29FA 29FB 85(10:8) = 0 23:8NA 29(0:12NA ) 29(0:18NB ) 85(10:8) = 0 20:32NA 5:22NB 918 = 0 = 0 NA + NB 85 = 0

Fy

(a) (b)

Solution of Eqs. (a) and (b) is NA = 53:32 lb Fx

=

NB = 31:68 lb

0 FA + FB P = 0 0:12(53:32) + 0:18(31:68)

P =0

P = 12:10 lb J

7.82 30o T

C L/2

y

L/2

x

W

θ

B

0.35NB

NB Assume impending slipping. Fx Fy Solution is T

= 0 0:35NB T cos 30 = 0 = 0 NB W T sin 30 = 0 = 0:5065W NB = 1:2532W

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MC

=

0 W (0:5L cos ) + 0:35NB (L sin ) NB (L cos ) = 0 0:5W cos + (1:2532W ) (0:35 sin cos ) = 0 (0:5 1:2532) cos + 1:2532(0:35) sin = 0 1:2532 0:5 tan = = 1:7172 = 59:8 J 1:2532(0:35)

7.83 P W

L/2

L/2 30o

A

F = µsN

x B

N

Consider impending sliding. MA Fy Fx

L cos 30 P = 0:4330W 2 = 0 + " P cos 30 W +N =0 0:4330W cos 30 W +N =0 N = 0:6250W = 0 +! P sin 30 = 0 sN 0:4330W sin 30 = 0 s (0:6250W ) s = 0:346 J =

0

+

PL = W

7.84

180 lb 2 ft

β P y

4 ft x FA

A NA

355 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Assume simultaneous impending sliding and tipping. MA Fx

= =

0 0

180(2) P cos

Fy

=

0

NA

tan

=

4P cos = 0 0:3NA = 0

P sin

4 120 = 90 3

P =

P cos = 90 lb 90 0:3NA = 0 NA = 300 lb 90 180 = 0 300 sin 180 = 0 cos 4 = tan 1 = 53:13 J 3

90 90 = 150:0 lb J = cos cos 53:13

7.85

7.86

356 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.87

7.88 C W φs h RA

RB

G

φs

B

2m

6m D

A

The panel is a 3-force body with the forces intersecting at C. Assuming impending sliding at A and B, we get from geometry: CD h

= 6 tan = 4 tan

s s

= 2 tan s + h = 4 s = 4(0:5) = 2:00 m J

7.89

357 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.90

7.5 in.

5 in.

28 lb p

= =

s

=

C

B

A

W

RC

2 r tan (pitch of screw) p 0:16 tan 1 = tan 1 = 4:852 (lead angle) 2 r 2 (0:3) 1

tanh

s

= tan

1

0:4 = 21:80

From FBD of member ABC: MC = 0 (a) C0 (b) C0

5W

= W r tan( = W r tan(

28(12:5) = 0

W = 70 lb (axial thrust on screw)

+ ) = 70(0:3) tan(21:80 + 4:85 ) = 10:54 lb in. J ) = 70(0:3) tan(21:80 4:85 ) = 6:40 lb in. J s s

7.91

358 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7.92

θR θ A R

θ

The wedge is a two-force body. There is no slipping at A if s > . ) Smallest coe¢ cient of static friction that prevents slipping is s

= tan

s

= tan = tan 18 = 0:325 J

7.93

P

y x 5(9.81) N

0.32N 35o

N

Assume impending sliding down the incline: Fx Fy

= 0 = 0

P cos 35 + 0:32N 5(9:81) sin 35 = 0 P sin 35 + N 5(9:81) cos 35 = 0

359 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Solution is N = 48:9N

P = 15:24 N

Assume impending sliding up the incline (the direction of the friction force is reversed): Fx Fy

= 0 = 0

P cos 35 0:32N 5(9:81) sin 35 = 0 P sin 35 + N 5(9:81) cos 35 = 0

Solution is: N = 77:2 N

P = 64:5 N

Block will remain at rest if 15:24 N

64:5 N J

P

7.94 Bx

B

t 8f

t 8f

ft 10

B B By

6 ft

6 ft 15 lb

15 lb

20 lb A

C

C 8 ft

FA

5.292 ft

5.292 ft

FC

FC NC

NC

NA

Assume equilibrium. FBD of assembly: MA

Fy Fx

=

0

+

NC (8 + 5:292)

20(4)

NC = 18:033 lb = 0 + " NA + NC 20 NA + 18:033 20 15 = 0 = 0 + ! FA FC = 0

15 8 +

5:292 2

=0

15 = 0 NA = 16:967 lb FC = FA

FBD of bar BC: MB

=

0

+

FC (6)

FC (6)

NC (5:292) + 15

18:033(5:292) + 15

5:292 2

5:292 2 =0

=0 FC = FA = 9:290 lb

Check for sliding: FA NA FC NC

= =

9:290 = 0:5475 < 16:967 9:290 = 0:5152 < 18:033

s

) No sliding at A

s

) No sliding at C

) Bars are in equilibrium J

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Chapter 8 8.1

8.2

361 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.3 y

y = 30x dx

x

y = 0.5x

60 mm

x

120 mm 1 p 0:5x dx yel = 30x + 0:5x 2 Z Z 120 p 2p A = dA = 30x 0:5x dx = 30x3=2 3 A 0

dA

=

p

30x

1200 mm2 Z Z = yel dA =

120

0:25x2 0

=

Qx

= Qy

A

0

=

x =

1 30x 2

0:25x2 dx =

30 2 x 4

0:25 3 x 6

120 0

3

36 000 mm Z Z = x dA = A

120

120

p

30x3=2

0:5x2 dx =

0

2p 30x5=2 5

0:5 3 x 3

120 0

3

57 600 mm Qy 57 600 = = 48:0 mm J A 1200

y=

Qx 36 000 = = 30:0 mm J A 1200

8.4

362 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.5

y 4 in. dy 8 in. y x = 2y x p p 2 2ydy = 2:828 y dy Z Z 8 p A = dA = 2:828 y dy = 42:67 in2

dA

=

A

Qx

=

Z

A

x =

0

y dA = 2:828

Z

8

y 3=2 dy = 204:8 in3

0

0 due to symmetry

y=

Qx 204:8 = = 4:80 in J A 42:67

363 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.6 y

xy = 5

y 0

1 x

5

x

dx

5 = y dx = dx x Z Z 5 5 A = dA = dx = 5 ln 5 = 8:047 in2 x A 1 Z 5 Z 5 Z y 5 5 25 dA = dx = Qx = dx = 10:0 in3 2 2 2x x 2x A 1 1 Z Z 5 Qy = x dA = 5dx = 20:0 in3 dA

A

x=

1

Qy 20:0 = = 2:49 in. J A 8:047

y=

Qx 10:0 = = 1:243 in. J A 8:047

8.7

364 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.8 y

x2 + y2 = 1 a 2 b2 y

0

a

x

r

dx

a

x

x2 bp 2 y=b 1 a x2 = 2 a a Ra Ra Use symmetry in integration: :::dx = 2 0 :::dx. a

bp 2 a x2 dx dA = y dx = a Z Z Z b ap 2 b ap 2 A = dA = a x2 dx = 2 a x2 dx a a a 0 A a x b 1 p 2 ab x2 + a2 sin 1 = 2 x a = a 2 a 0 2 Z a Z y b p 2 bp 2 dA = a x2 a x2 dx Qx = a a 2a A 2 Z b2 a 2 2ab2 2 = (a x )dx = a2 0 3 y=

Qx = A

2ab2 3 ab 2

=

4b J 3

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8.9

8.10

366 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.11 Consider half of the parabola

y ds y = x2

1.0 ft dx 1.0 ft

x

p p = x2 dy = 2x dx ds = dx2 + dy 2 = 1 + 4x2 dx Z Z 1p p 1 1 p L = ds = 1 + 4x2 dx = ln 2x + 4x2 + 1 + x 4x2 + 1 4 2 L 0 = 1:4789 ft Z Z 1 p Qx = y ds = x2 1 + 4x2 dx y

L

= = x =

1 0

0

1 x 1 + 4x2 16

3=2

0:6063 ft2 0 (by symmetry) J

1 p x 1 + 4x2 32 y=

p 1 ln 2x + 1 + 4x2 64

1 0

0:6063 Qx = = 0:410 ft J L 1:4789

8.12

367 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

*8.13

8.14

368 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.15

y

y

10

6 8 2

1

A1

=

A2

=

A3

=

x =

3

x

x

3

10(16) = 160 in2

x1 = 5 in.

4 (8) = 13:395 in. 2 3 (32 ) = 28:27 in2 x3 = 6 in. Ai xi 160(5) + 100:53(13:395) 28:27(6) = = 8:51 in. J A 160 + 100:53 28:27 (82 ) = 100:53 in2

x2 = 10 +

8.16

y

2

1 a

a

a

x

a2 + a2 = 1:7854a2 4 a2 4a a = A1 y1 + A2 y2 = + a2 = 0:8333a3 4 3 2 a2 4a a = A1 x1 + A2 x2 = + a2 = 0:166 67a3 4 3 2

A = A1 + A2 = Qx Qy

x = y

=

Qy 0:166 67a3 = = 0:0934a J A 1:7854a2 3 Qx 0:8333a = = 0:467a J A 1:7854a2

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8.17

150 1

60

2

240 y 60

60 3 300

= 150(60) = 9000 mm2 = 360(60) = 21 600 mm2 = 300(60) = 18 000 mm2 = Ai = (9 + 21:6 + 18)

A1 A2 A3 A

x = y

=

x x1 = 225 mm x2 = 330 mm x3 = 150 mm 3 10 mm2 = 48:6

y1 = 330 mm y2 = 180 mm y3 = 30 mm 103 mm2

9(225) + 21:6(330) + 18(150) Ai xi = = 244 mm J A 48:6 Ai y i 9(330) + 21:6(180) + 18(30) = = 152:2 mm J A 48:6

8.18

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8.19 y 20

60 3

30

4

30

150 2

1

30

x

A (mm2 ) x (mm) Ax (mm3 ) y (mm) Ay (mm3 ) 1800 50 90 000 15 27 000 3000 10 30 000 75 225 000 1800 50 90 000 135 243 000 900 40 36 000 110 99 000 7500 246 000 594 000 246 000 594 000 x= = 32:8 mm J y= = 79:2 mm J 7500 7500

Part 1 2 3 4 Sum

8.20

4.5

A1 A2 A3 A x = y

=

9

3

1

10.5

2

3

3 10.5

1.5

x

= 3(13:5) = 40:5 in2 x1 = 2:25 in. y1 = 15 in. 2 = 10:5(1:5) = 15:75 in x2 = 0 y2 = 8:25 in. 2 = 3(10:5) = 31:5 in x3 = 0:75 in. y3 = 1:5 in. = Ai = 40:5 + 15:75 + 31:5 = 87:75 in2 Ai xi 40:5(2:25) + 0 + 31:5(0:75) = = 1:308 in. J A 87:75 Ai y i 40:5(15) + 15:75(8:25) + 31:5(1:5) = = 8:94 in. J A 87:75

371 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.21

8.22 Semicircle: A1 = Parabola: A2 =

R2 2

y1 =

2 (2R)h 3

4 R 3

A1 y1 =

2 h 5

A2 y2 =

y2 =

8 2 3 R + Rh2 = 0 3 15

A1 y1 + A2 y2 = 0

2 3 R 3 8 Rh2 15 p h=

5 R J 2

8.23 Qx

=

Qx

= 0

Ai yi = h=

p

R2 2

4R 3

+ (Rh)

h 3

=

2R3 3

Rh2 3

2R J

372 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.24

8.25

y

y

R

1

2 x

R A = A1

A2 =

= A1 y1

A2 y2 =

Qy

= A1 x1

A2 x2 =

x = y

=

1 (3 12 Qy = A Qx = A

R R2 R2 = 1 2 2 2 R2 4R R2 R 4 3 2 3 R2 4R R2 R 4 3 2

R2 4

Qx

=

R

x

R3 6 2R 3

=

8)R3 1 12 (3 R2 2 2 R3 6 R2 2 2

8)R3 1 1

=

=

3 3

8 R = 0:416R J 6

2R 3(

2)

= 0:584R J

373 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.26 A =

4ab

y

Qx 2ab2 = A 4ab

=

2ah

Qx = 4ab ah2 2b2 = 2ah 4b

b 2 h2 2h

2ah

h 2

= 2ab2

ah2

y is maximized when dy dh

=

0

(4b

2h) ( 2h) (4b

4b2 8bh + 2h2 =0 (4b 2h)2

(2b2 2h)

h2 )( 2)

2

=0

h = 0:586b J

8.27

374 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.28

y 40 mm 2 3 120 mm 1 4 6 80 mm

5 40 mm x

L = Li = 2(120) + 2(80) = 400 mm Qx = Li yi = 120(60) + 40(120) + 80(80) + 40(40) + 40(20) + 0 = 20 800 mm2 Qy = Li xi = 0 + 40(20) + 80(40) + 40(60) + 40(80) + 80(40) = 12 800 mm2

x = y

=

12 800 Qy = = 32 mm J L 400 Qx 20 800 = = 52 mm J L 400

8.29

y 45o 1

.

40

r Dimensions in mm 2 70 x

375 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

r L1 x1 y1

x = y

R sin

40 sin 135 = 12:004 mm 0:75 = 0:75(2 R) = 0:75(2 )(40) = 188:50 mm = r sin 45 = 12:004 sin 45 = 8:488 mm = 110 + r cos 45 = 110 + 12:004 cos 45 = 118:49 mm =

=

=

Li xi 188:50( 8:488) + 0 = = 6:19 mm J L 188:50 + 70 Li yi 188:50(118:49) + 70(35) = = 95:9 mm J L 188:50 + 70

8.30

8.31

36 mm 1

y 36 mm 3

x

24 mm 2 L = Qx

=

Li = 36 + 24 + 36 = 147:40 mm 2(24) Li yi = 36(18) + 24 +0=

Qy

=

Li xi = 36( 24) + 0 + 36(42) = 648 mm2 x = y

=

504 mm2

Qy 648 = = 4:40 mm J L 147:40 Qx 504 = = 3:42 mm J L 147:40

376 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.32

y 6i n.

1

x 3

y

=

10 in.

2

8 in.

6 ( 2 6) + 10( 4) + 10( 4) Li yi = = Li 6 + 10 + 10 Due to symmetry x = 0 J

0:206 in. J

8.33

y

90 4

120 5

3 60 2

x

150 1

L1

=

L = x = = y

= =

2

(150) = 75 mm Li =

2

x1 =

x2 =

2

(150) =

300

mm

(150) + 60 + 120 + 90 + 270 = 775:6 mm

Li xi 75(300) + 60(120) + 120(90) + 90(45) + 0 = L 775:6 57:4 mm J Li yi 75( 300) + 0 + 120(60) + 90(120) + 270( 15) = L 775:6 11:02 mm J

377 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.34

8.35

378 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.36 y

y x

x

dx

Use single integration: y = y dx xel = x yel = 2 Z Z A = dA = y dx A A Z Z Z Z 1 y 2 dA Qy = xel dA = xy dA Qx = yel dA = 2 A A A A dA

Evaluate integrals with Simpson’s rule: 2 [8:5 + 4(8:0) + 2(7:2) + 4(5:4) + 0] = 51:0 in2 3 1 = 8:52 + 4(8:0)2 2 + 2(7:2)2 + 4(5:4)2 + 0 = 182:86 in3 3 2 = [(0) (8:5) + 4 (2) (8:0) + 2 (4) (7:2) + 4 (6) (5:4) + (8) (0)] 3 = 167:47 in3

A = Qx Qy

x=

Qy 167:47 = = 3:28 in. J A 51:0

y=

Qx 182:86 = = 3:59 in. J A 51:0

8.37 y ds dx

=

L =

1+

Z

0

x (m) 0 0:25 0:5 0:75 1:0

dy = dx

2

= e x s

y (m) 1:0 0:9394 0:7788 0:5698 0:3679

1

dy dx

2

ds dx dx

dy=dx = 0:0 0:4697 0:7788 0:8547 0:7358

=

2

2xex =

p 1 + (2xy)2

Qx =

Z

1

0

2xy

2xy

ds=dx 1:0 1:1048 1:2675 1:3155 1:2415

y

ds dx dx y(ds=dx) (m) 1:0 1:0379 0:9871 0:7496 0:4567

379 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

0:25 = 1:2048 m 3 0:25 = [1:0 + 4(1:0379) + 2(0:9871) + 4(0:7496) + 0:4567] = 0:8817 m2 3 Qx 0:8817 = = = 0:732 m J L 1:2048

L = Qx y

[1:0 + 4(1:1048) + 2(1:2675) + 4(1:3155) + 1:2415]

8.38

380 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.39

8.40

8.41

y

x

dx x/2 240 mm

120 mm 60 mm x

381 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Volume element is a thin disk obtained by rotating the shaded are about x-axis. xel

= x

dV

=

V

=

60 + Z

x 4

dV =

V

Qxy

Z

=

2

Z

0

xel dV =

V

(60) dx =

240

60 + Z

60 +

x x dx = 9:048 4

240

x x dx 4 106 mm2

x 2 x dx = 1:520 4

60 +

0

Qxy 1:520 = V 9:048

x =

2

dx

109 = 168:0 mm J 106

109 mm3

y=z=0 J

8.42 Volume element is a thin shell obtained by rotating the shaded area about y-axis. xel

= x

dV

=

V

yel = 60 +

x x 2 xel dx = 2 x dx = x2 dx 2 2 Z Z 240 = dV = x2 dx = 14:476 106 mm2 V

Qzx

Z

=

0

yel dV =

V

y

x 4

Z

240

60 +

0

x 2 x dx = 1:520 4

1:520 109 Qzx = = 105:0 mm J V 14:476 106

=

109 mm3

x=z=0 J

8.43 y

y = h(1 − x2 /b2)

h

y x

x

dx b

Volume element is a thin disk obtained by rotating the shaded area about x-axis. dV V

y 2 dx = h2 1

= =

Z

V

dV = h

2

Z

x2 b2

b

0

1

2

dx x2 b2

xel = x

2

dx =

8 bh2 15

382 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Qyz

=

Z

xel dV = h2

V

)

Z

b

0

Qyz 5 x= = b J V 16

2

x2 b2

x 1

dx =

1 2 2 b h 6

By symmetry y = z = 0 J

8.44 Volume element is a thin-walled cylinder obtained by rotating the shaded area about the y-axis (see …gure in solution of Prob. 8.43). dV

=

2 xy dx = 2 xh 1

V

=

Z

dV = 2 h

V

Qzx

=

b

x2 b2

x 1

0

Z

V

)

Z

x2 b2

yel dV = h

2

Z

dx dx =

b

x 1

0

Qzx 1 y= = h J V 3

yel =

x2 b2

1 h y= 2 2

1

x2 b2

1 2 b h 2 2

dx =

1 2 2 b h 6

By symmetry x = z = 0 J

8.45

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8.46

8.47

8.48

384 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.49 We choose double integration using the volume element shown in the solution of Prob. 8.48. dV zel V

x y cos dx dy 2a 2b 1 1 x y = z = h cos cos 2 2 2a 2b Z aZ b Z Z aZ b y x y x cos dx dy = 4h cos cos dx dy = dV = h cos 2a 2b 2a 2b 0 0 V a b 2a 2b 16 = 4h = 2 abh = z dx dy = h cos

Qxy

Z

Z Z 1 2 a b x y = zel dV = h cos2 cos2 dx dy 2 2a 2b V a b Z aZ b abh2 y a b x cos2 dx dy = 2h2 = = 2h2 cos2 2a 2b 2 2 2 0 0 2 Qxy ) z= = h J By symmetry x = y = 0 J V 32

8.50

385 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.51

8.52 Block : V1

=

Hole : V2

=

x =

160(340)(70) = 3:808 106 mm3 x1 = 170 mm (1302 ) (70) = 0:9291 106 mm3 x2 = 80 mm 4

V1 xi 3:808(170) 0:9291(80) = = 199:1 mm J Vi 3:808 0:9291 By symmetry y = 80 mm J z = 35 mm J

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8.53

9 in.

2

3

5 in.

5 in.

9 in. 10 in. 6 in.

1 5 in. V1

=

V2

=

V3

=

V Qxy

2

(2:5) (6) = 117:81 in3 3

(4:5)2 (9) = 190:85 in3 3

= = =

(2:5)2 (5) =

32:73 in3

z1 = 3 in. 9 z2 = 6 + = 8:25 in. 4 5 z3 = 10 + = 11:25 in. 4

Vi = 117:81 + 190:85 32:73 = 275:9 in3 Vi zi = 117:81(3) + 190:85(8:25) 32:73(11:25) 1559:7 in4

z

=

1559:7 Qxy = = 5:65 in. J V 275:9 By symmetry x = y = 0 J

8.54

z 32 in.

1 2 x

in. 24

in. 24

26 in. y

387 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

V1

=

V2

=

V

=

x = z

=

24(26)(32) = 19:968 103 in3 x1 = 12 in. z1 = 13 in. 1 26 V1 = 9:984 103 in3 x2 = 32 in. z2 = = 8:667 in. 2 3 Vi = (19:986 + 9:984) 103 = 29:97 103 in3 Vi xi 19:968(12) + 9:984(32) = = 18:66 in. J V 29:97 V i zi 19:968(13) + 9:984(8:667) = = 11:55 in. J V 29:97 By symmetry y = 16 in. J

8.55

5

z in.

16 in.

2

1

12 in.

6 in. 12 in.

x

V1 x1 y1

in. 15

y

= (15)(12)(5) = 900 in3 V2 = (5)(16)(12) = 960 in3 = 12:5 in. x2 = 2:5 in. = 6 in. y2 = 8 in. z1 = 3 in. z2 = 6 in. V Qyz Qzx Qxy

Vi = 900 + 960 = 1860 in3 Vi xi = 900(12:5) + 960(2:5) = 13 650 in4 Vi yi = 900(6) + 960(8) = 13 080 in4 Vi zi = 900(3) + 960(8) = 10 380 in4

= = = =

x = y

=

z

=

Qyz 13 650 = = 7:34 in. J V 1860 Qzx 13 080 = = 7:03 in. J V 1860 Qxy 10 380 = = 5:58 in. J V 1860

388 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.56 z

2 20

1

20 x

V1 V2 x1 z1

V Qyz Qxy

= = =

100

y

100

(120)(100)(20) = 240 000 mm3 (502 ) (20) = 78 540 mm3 = 2 = 50 mm x2 = 10 mm 4 = 10 mm z2 = (50) = 21:22 mm 3 =

Vi = 240 000 + 78 540 = 318 500 mm3 Vi xi = 240 000(50) + 78 540( 10) = 11:215 106 mm4 Vi zi = 240 000( 10) + 78 540(21:22) = 0:7334 106 mm4 11:215 106 Qyz = = 35:2 mm J V 318 500 6 Qxy 0:7334 10 = = = 2:30 mm J V 318 500 By symmetry y = 0 J

x = z

389 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.57

8.58 25

y

z

2 y

20

37 z 30

1

390 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

V1

=

y1

=

z1

=

V Qzx Qxy

= = =

(302 ) (202 ) (37) = 52 308 mm3 V2 = (25) = 15 708 mm3 2 2 37 25 = 18:5 mm y2 = = 12:5 mm 2 2 4 4 (30) = 12:732 mm z2 = (20) = 8:488 mm 3 3 Vi = 52 308 15 708 = 36 600 mm2 Vi yi = 52 308(18:5) 15 708(12:5) = 0:7714 Vi zi = 52 308( 12:732) 15 708( 8:488) =

y z

106 mm4 0:5327 106 mm4

0:7714 106 Qzx = = 21:1 mm J V 36 600 6 Qxy 0:5327 10 = = = 14:56 mm J V 36 600 By symmetry x = 0 J =

8.59

391 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.60

16 in. 1

2 6 in.

8 in. 8 in.

12 in. A1

=

A2

=

p 16 (6) 62 + 162 = 322:1 in2 z1 = = 5:333 in. 3 p 8 z2 = 8 + = 10:667 in. (3) 32 + 82 = 80:53 in2 3

A = A1 + A2 = 322:1 80:53 = 241:6 in2 Qxy = A1 z1 + A2 z2 = 322:1(5:333) 80:53(10:667) = 858:8 in3 z

=

Qxy 858:8 = = 3:55 in. J A 241:6 By symmetry x = y = 0 J

8.61

392 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.62

z . 8 in

8 in

1 2 3

4

1

x

8 in .

8 in.

2 y

A1 = A4 = 128 in2 A2 = A3 = 64 in2 A = Ai = 2 (128 + 64) = 384 in2 Ai xi 128(4) + 64(8=3) + 64(8) + 0 = = 3:11 in. J A 384 Ai y i 128(4) + 64(8 + 8=3) + 64(4) + 128(8) = = = 6:44 in. J A 384 By symmetry z = 4 in. J

x = y

8.63

393 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.64

2

18 in.

z

in. 22

x1 x2 x3

=

y

(222 ) = 380:1 in2

A2 = A3 =

4 4 = (22) = 9:337 in. z1 = 0 3 22 = 7:333 in. z2 = 6 in. = 3 = 0 z3 = 6 in.

A = Qyz = Qxy =

in.

1

x

A1

3 22

1 (22)(18) = 198 in2 2

Ai = 380:1 + 2(198) = 776:1 in2 Ai xi = 380:1(9:337) + 198(7:333) + 0 = 5001 in3 Ai zi = 0 + 2(198)(6) = 2376 in3 Qyz 5001 = = 6:44 in. J A 776:1 Qxy 2376 = = = 3:06 in. J A 776:1 By symmetry y = 6:44 in. J

x = z

394 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.65 z 500 2 700

0 180

1

3

1200

4

1300

5

y

x

= A5 = 700(1200) = 840 103 mm2 1 A2 = A4 = (500)(1200) = 300 103 mm2 2 A3 = 1300(1800) = 2340 103 mm2 A = Ai = 2(840 + 300) + 2340 103 = 4620 A1

y

= =

z

=

103 mm2

2 (840) (350) + 2 (300) ( 500=3) + 2340( 250) Ai y i = A 4620 21:0 mm J 2 (840) (600) + 2(300)(800) + 2340(600) A i zi = = 626:0 mm J A 4620

8.66

395 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.67

z

1 2 x

5 80 m mm m 80 4

3 96 mm y

L = 3(96) + 2(80) = 448 mm 96(80) + 80(40) Li xi = = 24:3 mm J L 448 Li zi 3(96)(48) = = = 30:9 mm J L 448 By symmetry y = x = 24:3 mm J

x = z

396 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.68

397 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.69

z 8 in . 3

10 in. 2 . 8 in 1

y

x L = Qyz = Qzx = Qxy =

x = z

=

Li = 26 in. Li xi = 8(4) + 0 + 0 = 32 in2 Li yi = 0 + 0 + 8(4) = 32 in2 Li zi = 0 + 10(5) + 8(10) = 130 in2

Qyz 32 = = 1:231 in. J L 26 Qxy 130 = = 5:0 in. J L 26

Qzx 32 = = 1:231 in. J L 26

y=

8.70 y 200 60 dy x' = x − 60 y 0

x

Volume element is the thin shell generated by rotating the shaded area about the x-axis. dV V

1 2 yx0 dy xel = 60 + x0 2 Z Z 200 0 = dV = 2 yx dy

=

V

Qyz

=

Z

0

xel dV = 2

V

Z

200

xel yx0 dy

0

398 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Evaluating the integrals with Simpson’s rule: y (mm) 0 50 100 150 200

V

yx0 (mm2 ) 0 5700 11 400 13 800 10 200

xel (mm) 85.5 117.0 117.0 106.0 85.5

xel yx0 (106 mm3 ) 0 0.6669 1.3338 1.4628 0.8752

50 [0 + 4(5700) + 2(11 400) + 4(13 800) + 10 200] 3 = 11:624 106 mm3 50 = 2 [0 + 4(0:6669) + 2(1:3338) + 4(1:4629) + 0:8752] 3 = 1:2631 109 mm4 =

Qyz

x0 (mm) 51 114 114 92 51

2

x =

106

1:2631 109 Qyz = = 108:7 mm J V 11:624 106 By symmetry y = z = 0 J

8.71

200 60 dy x' = x − 60 y 0

x

The volume element is a thin annular disk obtained by rotating the shaded area about the y-axis. h i 2 dV = (60 + x0 ) 602 dy = x0 (x0 + 120)dy yel = y Z Z 200 V = dV = x0 (x0 + 120)dy V

Qzx

=

Z

V

0

yel dV =

Z

200

yx0 (x0 + 120)dy

0

399 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Evaluating the integrals with Simpson’s rule: x0 (mm) 51 114 114 92 51

y (mm) 0 50 100 150 200

V

= =

Qzx

= =

x0 (x0 + 120) (mm2 ) 8721 26 676 26 676 19 504 8721

yx0 (x0 + 120) (103 mm3 ) 0 1334 2668 2926 1744

50 [8721 + 4(26 676) + 2(26 676) + 4(19 504) + 8721] 3 13:379 106 mm3 50 [0 + 4(1334) + 2(2668) + 4(2926) + 1744] 103 3 1:2629 109 mm4

y

=

Qzx 1:2629 109 = = 94:4 mm J V 13:379 106 By symmetry x = z = 0 J

8.72

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8.73

401 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.74

8.75

402 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.76

z

z

3 4

Dimensions in inches 1 18

3 3 6 V =2 A = =

2

18

2

y

3

3

1

2 9

y

Ai yi = 2 [(54)(4:5) + (54)(8)] = 4240 in3 J

Li yi = 2

1921 in2 J

h p i 182 + 62 (9) + 9(7:5) + 18(3) + 3(4:5)

8.77 V = 2 (Qarea )AB = 2 ( b2 )a = 2 2 ab2 J A = 2 (Qcurve )AB = 2 (2 b)a = 4 2 ab J

403 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.78

8.79

404 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.80

8.81

8.82

z 160 2 200 1 300 y 40 V

=

2

Ai y i = 2

=

11:90

(40

500)(20) +

1 2

160

200

40 +

1 3

160

106 mm3 J

405 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.83 Properties of the generating area:

5 in.

6 in.

B

6 in. Rectangle 1

2 3

r

3 in.

8 in.

8 in. A

A1

=

A2

=

A3

=

V

=

11 = 13:5 in. 2 5 1 (5)(6) = 15 in2 r2 = 19 = 17:333 in. 2 3 1 3 (3)(6) = 9 in2 r3 = 8 + = 9:0 in. 2 3 2 Ai ri = 2 [66(13:5) 15(17:333) 9(9:0)] = 3460 in3 J

(11)(6) = 66 in2

r1 = 8 +

8.84 y

h

C x10 ft radius

V = 2 Ax

2000 = 2

2 10h 3

x

3 10 8

h = 12:73 ft J

406 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.85

8.86

407 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.87

8.88

ft 1 0 1

2 20 ft

25 ft r

408 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

L1 L2 A=2

2 (10) ft r1 = (10) ft 2 p = 102 + 252 = 26:93 ft r2 = 15 ft =

Li ri = 2 [100 + 26:93(15)] = 3170 ft2 J

8.89

8.90 Approximate the bowl as a thin shell. Bowl:

W1 = 2 R 2 t

1

= 2 (6:152 )(0:3)

6:15 1 R= = 3:075 in. 2 2 2 3 2 62:4 W2 = R 2= (63 ) 3 3 123 3 3 y2 = R = (6) = 2:25 in. 8 8

162 123

= 6:684 lb

y1 = Water:

y=

= 16:336 lb

W i yi 6:684(3:075) + 16:336(2:25) = = 2:49 in. J Wi 6:684 + 16:336

409 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.91

8.92

z 7.5 in. deep 6 in. 1

10.5 in.

1/20 in. 2

W1 W2 W3 W4 W y

z

3 in. 3 4 3 in y

= 0:283(6)(7:5)(1=20) = 0:6368 lb = 0:283(10:5)(7:5)(1=20) = 1:1143 lb = 0:283(3)(7:5)(1=20) = 0:3184lb = 0:029(3)(3)(7:5) = 1:9575 lb = Wi = 0:6368 + 1:1143 + 0:3184 + 1:9575 = 4:027 lb W i yi 0 + 1:1143(5:25) + 0:3184(10:5) + 1:9575(9) = W 4:027 = 6:66 in J W i zi 0:6368(3) + 0 + 0:3184(1:5) + 1:9575(1:5) = = W 4:027 = 1:322 in. J By symmetry x = 3:75 in. J

=

410 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.93

8.94 2 2 3 R st = (9)3 (0:283) = 432:1 lb 3 3 3 3 y1 = R= (9) = 3:375 in. 8 8 1 Cylinder : W2 = R2 h = (9)2 h = 254:5h lb y2 = h 2

Hemisphere

:

W1 =

W1 y1 + W2 y2

=

0

432:1( 3:375) + 254:5h

h

=

3:39 in. J

1 h 2

=0

8.95 Rod:

m1 =

Collar:

m2 =

4 4

0:0062 (0:4)(7850) = 0:08878 kg (0:042

0:0062 )(0:02)(8300) = 0:2039 kg

mi xi = 0:08878( 0:1) + 0:2039x = 0

x = 0:0435 m = 43:5 mm J

411 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.96

8.97

x

1

18 150

99 195 40

m1 m2 m3

x = =

2 18

= = =

3

40 15

(92 )(150)(7850 10 9 ) = 0:2996 kg (202 )(15)(7850 10 9 ) = 0:14797 kg (202 92 )(195)(2660 10 9 ) = 0:5198 kg

0:2996(174) + 0:14797(256:5) + 0:5198(97:5) mi xi = mi 0:2996 + 0:14797 + 0:5198 145:5 mm J

412 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.98

8.99

413 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.100

y y

Wwater = 62:4 (102 )h = 19 604h lb h ywater = 2

Wtan k

=

18 000 lb

ytan k

=

8 ft

W i yi 18 000(8) + 19 604h (h=2) 144 000 + 9802h2 = = Wi 18 000 + 19 604h 18 000 + 19 604h 144 000 + 9802h2 4901h2 + 9000h 72 000 = h =h =0 18 000 + 19 604h 9802h + 9000 Positive root is h = 3:02 ft J =

8.101 20 20

80

. 25. 10. .

1 .

. 2

x

Pulley: m1 m2 x1

= (0:0252 0:012 )(0:02)(2660) = 0:08775 kg = (0:082 0:012 )(0:02)(2660) = 1:0529 kg = 0:01 m x2 = 0:03 m

Shaft: m3 x3

x = =

= (0:012 )(0:14)(7850) = 0:3453 kg = 0:07 m

mi xi 0:08775(0:01) + 1:0529(0:03) + 0:3453(0:07) = mi 0:08775 + 1:0529 + 0:3453 0:0381 m = 38:1 mm J

414 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.102

8.103

415 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.104 With R

x = 2:5 ft, Simpson’s rule yields Z = w dx L

=

2:5 [42:5 + 4(37:5) + 2(28:3) + 4(32:4) + 2(42:3) + 4(52:1) + 58:6] 3 608:6 lb

Rx =

Z

wx dx

L

2:5 [0 + 4(37:5)(2:5) + 2(28:3)(5) + 4(32:4)(7:5) + 2(42:3)(10) 3 +4(52:1)(12:5) + 58:6(15)] = 4967 lb ft R = 609 lb J

x=

Z

Z

4967 = 8:16 ft J 608:6

8.105 R

=

p dA = p0

A

=

Rx =

1 p0 2

Z

A

=

Ry

=

Z

A

=

0

Z

b

b

Z

b

0

Z

b

Z

0

Z

b

0

b

Z

0

y(b + y)dy =

0

1 p0 b3 2 x= 2 = b J 3 3 p0 b2 4

3 p0 b 2 J 4 b

x

0

(b + y)dy = Z

x xy + 2 dx dy b b

0

0

py dA = p0

1 p0 2

Z

(b + y)dy =

px dA = p0

1 p0 b 3

b

x xy + 2 dx dy b b

1 p0 b 3 2 b

y

x xy + 2 dx dy b b

5 p0 b 3 12

5 p0 b 3 5 12 y= = b J 3 9 p0 b 2 4

416 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.106

8.107

8.108

8.109

6 lb/in. R1

z x-1

y-2

R2

240 in.

in. 0 3 3

y

x

417 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

R1

=

R2

=

2 (330)(6) = 1320 lb 3 1 (240)(6) = 720 lb 2

y2 =

3 (330) = 123:75 in. 8

1 (240) = 80 in. 3

Ri xi 1320(123:75) + 0 = = 80:1 in. J Ri 1320 + 720 R i yi 0 + 720(80) = = 28:2 in. J Ri 1320 + 720

x = y

x1 =

=

8.110 R

=

Z

=2

w ds =

0

Mx

Z

2w0

= =

[sin

2w0

cos ]0

=2

wx ds =

0

=

2w0 a

=2

2

2

0

(a sin )(a d )

0

2w0 a2

Z

(a d ) =

0

w0 a = 0:785w0 a = 4 Z =2 Z =2 = wy ds = 0

My

=2

Z

=2

=2

=

2w0

2w0 a2

= 0:6366w0 a2

(a cos )(a d )

0

2w0 a2

[cos + sin ]0

x = y

=

=2

=

2w0 a2 2

1 = 0:3634w0 a2

My 0:3634w0 a2 = = 0:463a R 0:785w0 a Mx 0:6366w0 a2 = = 0:811a R 0:785w0 a

The resultant is the force R = 0:785w0 a crossing the xy-plane at x = 0:463a, y = 0:811a: J

418 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.111

8.112

419 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.113

8.114

420 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.115 2b/3 + 4

Dimensions in feet

W1

2 W3

P

24

b/3 + 4 8

W2 b

Ax 4

A Ay

Consider 1 ft length of dam at impending tipping.

MA

P

=

W1

=

W2

=

W3

=

1 (62:4)(242 ) = 17 971 lb/ft 2 1 w hb = (62:4)(24)b = 748:8b lb/ft 2 1 c hb = (150)(24)b = 1800:0b lb/ft 2 (4h) = 150(4)(24) = 14 400 lb/ft c

1 2 1 2 1 2

wh

2

=

2 b + 4 (748:8b) 3

=

8(17 971)

=

1099:2b2 10195b + 0:1149 7 b = 6:59 ft J

1 b + 4 (1800b) 3

2(14 400)

106 = 0

421 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.116

8.117

y

4m

5.7 m

2.3 m 2

1

3m 5.2 m

Rx y-

3 Ry

p

x x-

422 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

W1

p

=

Rx

=

y

=

gh = 1000(9:81)(9:2) = 90 250 Pa 1 1 ph = (90 250)(9:2) = 415 000 N/m 2 2 1 1 h = (9:2) = 3:07 m 3 3

=

1000(9:81)(5:7)(4) = 223 700 N/m 1 W2 = (1000)(9:81)(2:3)(4) = 45 130 N/m 2 1 (1000)(9:81)(3)(5:2) = 76 520 N/m W3 = 2 Ry = Wi = 223 700 + 45 130 + 76 520 = 345 400 N/m Wi xi 223 700(2:85) + 45 130(6:467) + 76 520(1:0) x = = = 2:91 m R 345 400 The resultant force is R = 415i 345j kN/m acting through the point (2:91 m, 3:07 m). J

8.118

423 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.119

F

h = 9.4 in.

p

W 2.4 in. p F F

2

= =

h = 0:036(9:4) = 0:3384 lb/in 0 + " F pA W = 0 h i 3:5 = pA + W = 0:3384 (2:42 ) + = 1:750 lb J 4 16

8.120

b y

b(1 - y/h) h

dy

y R

=

Z

A

p dA =

Z

0

h

p0

y b 1 h

y dy h

Z bp0 h bhp0 = 2 J hy y 2 dy = h 0 6 Z Z bp0 h bh2 p0 py dA = hy 2 y 3 dy = 2 h 0 12 A R py dA h y = RA = J 2 p dA A

424 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.121

R2

RA

0.15 m

0.2 m

p A A

0.3 m

R1

p

B

B

F

2

pA

=

hA = 1000(9:81)(0:2) = 1962 N/m

pB R1

= hB = 1000(9:81)(0:5) = 4905 N/m = pA A = 1962(0:32 ) = 176:58 N 1 1 (pB pA )A = (4905 1962)(0:32 ) = 132:44 N = 2 2

2

R2

MA F

=

0 0:15R1 + 0:2R2 0:3F = 0 0:15(176:58) + 0:2(132:44) 0:3F = 0 = 176:6 N J

8.122

y

x-

Ay Ax Ry

x-1

W1

x-2

W2

10 ft

Rx y-

x NB

425 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1 (62:4)(102 )(12) = 37 440 lb 2

Rx = W1

=

W2

=

Ry

=

62:4(102 )(12) = 74 880 lb

y=

10 = 3:333 ft 3

x1 = 5 ft

(102 ) (12) = 58 810 lb x2 = 10 4 Wi = 74 880 58 810 = 16 070 lb Wi xi 74 880(5) 58 810(5:756) = = 2:233 ft Ry 16 070 62:4

x =

4(10) = 5:756 ft 3

From FBD: MA NB

=

0 10NB Rx (10 y) Ry (10 x) = 0 10NB 37 440(10 3:333) 16 070(10 2:233) = 0 = 37 400 lb J

8.123

y 1

R2

6 in .

Li x i = 0

R 2

a2 = 0 2

a=

y

x

2R p

2a =

. 6 in

4 a

2

x

3 b

+ R p

2R

+0+a

a =0 2

2(6) = 8:485 in. J

R 2R b +0+b 6+ + a ( 6 + b) = 0 2 2 b2 + 2b(a 6) + 2R2 12a = 0 b2 + 2b(8:485 6) + 2(62 ) 12(8:485) = 0 b2 + 4:970b 29:82 = 0 b = 3:51 in. J

Li yi = 0

426 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.124

427 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.125

z mm 75

60 mm

R1 4 kN/m x

R2 180 mm

mm 150

y R1 R Ri xi R i yi

=

4(0:15) = 0:60 kN

R2 =

1 (4)(0:18) = 0:36 kN 2

= Ri = 0:96 kN J = 0:6(0:075) + 0 = 0:045 kN m = 0 + 0:36(0:06) = 0:0216 kN m

Line of action of R crosses the xy-plane at x = y

=

0:045 Ri xi = = 0:0469 m = 46:9 mm J R 0:96 0:0216 R i yi = = 0:0225 m = 22:5 mm J R 0:96

8.126

428 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.127 z

40

200

3

100 1

x

60 2

y

180 80

(0:18)(0:2)(0:06) = 2:16 10 3 m3 1 = (0:08)(0:2)(0:06) = 0:48 10 3 m3 2 = (0:18)(0:1)(0:04) = 0:72 10 3 m3 = Vi = (2:16 + 0:48 + 0:72) 10 3 = 3:36

V1

=

V2 V3 V x1 x2 x3

=

0:09 m

y1 = 0:14 m 0:08 = 0:2067 m = 0:18 + 3 z2 = 0:03 m = 0:09 m y3 = 0:02 m

10

3

m3

z1 = 0:03 m 2 y2 = 0:04 + (0:2) = 0:17333 m 3 z3 = 0:05 m

Vi xi 2:16(0:09) + 0:48(0:20670) + 0:72(0:09) = V 3:36 = 106:67 10 3 m = 106:7 mm J V i yi 2:16(0:14) + 0:48(0:17333) + 0:72(0:02) = = V 3:36 = 119:1 10 3 m = 119:1 mm J V i zi 2:16(0:03) + 0:48(0:03) + 0:72(0:05) = = V 3:36 = 34:3 10 3 m = 34:3 mm J

x =

y

z

429 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.128

8.129 R=

Z

a

0

Z

b

p0 p dx dy = a b

Z

a

x

0

Z

! y cos dy dx 2b b

b

Noting that Z

b

y dy = 2 cos 2b b

Z

b

0

we get R = p0

cos

4b

Z

0

Rx =

Z

a

0

= p0

Z

4b

a

2b y y dy = 2 sin 2b 2b

px dx dy = p0 Z

0

Z

a

0

a

4b

= 0

x 2ab dx = p0 J a

b b

b

x2 a

Z

b

y cos dy 2b b

!

dx

x2 4a2 b dx = p0 a 3

2

p0 x=

4a b 2 3 = a J 2ab 3

By symmetry y = 0 J

p0

430 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.130

b = 24''

b = 24''

x

2 18'' = r

_ y

TA

W2

=

W

=

y1

=

TB B

b y

y =

C W b

x

1

W1

TC

x

y

A

tb2 = 0:284(0:5)(242 ) = 81:79 lb r2 (182 ) t = 0:284(0:5) = 36:13 lb 4 4 Wi = 81:79 36:13 = 45:66 lb 4r 4(18) 12 in. y2 = b = 24 = 16:361 in. 3 3

W y = Wi yi = 81:79(12)

36:13(16:361) = 390:4 lb in.

From FBD: Mx

Fz

=

0 TA b W y = 0 TA (24) 390:4 = 0 TA = 16:267 lb J Due to symmetry TC = TA = 16:267 lb J = 0 TA + TB + TC W = 0 TB = W TA TC = 45:66 2 (16:267) = 13:13 lb J

8.131

431 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.132

8.133

1

3.2 in. 1.0 in. 2 3 4 W1

=

W2 W3 W4

= = =

x1

=

1.2 in.

x

0.32 in. 7.2 in.

0.32 in.

2 3 2 R = (0:53 )(0:055) = 0:01440 lb 3 3 R2 L = (0:52 )(3:2)(0:055) = 0:138 23 lb (0:162 )(1:2)(0:055) = 0:005 308 lb (0:162 )(7:2)(0:283) = 0:163 87 lb 3 3:2 (0:5) = 3:3875 in. 8

432 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

x = = =

Wi xi Wi 0:01440( 3:3875) + 0:13823( 1:6) 0:005308( 0:6) + 0:16387(2:4) 0:01440 + 0:13823 0:005308 + 0:16387 0:12652 = 0:407 in. I 0:31119

8.134

a

2

.

1

.

3 a

a

.a

.2

2

.

.

1

.

A .

r

a

r

a

.

A .

B

B

Volume QAB

=

V

=

Ai ri = 2a2

3 a + a2 2 2

2a +

4 a 3

= 6:808a3

2 QAB = 2 (6:808a3 ) = 42:8a3 J

Surface area QAB

=

A =

Li ri = (2a)(a) + (2a)

3 2 a + ( a) 2a + a 2

= 13:283a2

2 QAB = 2 (13:283a2 ) = 83:5a2 J

8.135

433 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8.136

8.137

8.138

y 6.3 in. Parabola

2 10.2 in. 1 x

6.3 in.

434 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

A1

=

x1

=

y1

=

1 1 (6:3)(10:2) = 21:42 in2 A2 = (6:3)(10:2) = 32:13 in2 3 2 3 1 (6:3) = 1:89 in. x2 = (6:3) = 2:1 in. 10 3 1 2 (10:2) = 2:55 in. y2 = (10:2) = 6:8 in. 4 3 A = Ai = 21:42 + 32:13 = 53:55 in2

x = y

=

Ai xi 21:42( 1:89) + 32:13(2:1) = = 0:504 in. J A 53:55 Ai y i 21:42(2:55) + 32:13(6:8) = = 5:10 in. J A 53:55

8.139

8.140 z

1.5 ft r cosθ

dθ

θ

y

r dr

435 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

R

Z

=

p dA

A

R

=

62:4

=

Z

=2

pz dA =

A

=

=2

Z

r2 cos dr d = 62:4

0

62:4

z=

Z

Z

p ( r cos ) dA =

62:4

A

=2

dA = (r d ) dr

1:5

140:40 lb J

=

Rz

Z

p = h = 62:4 (r cos )

1:2656 cos2

d =

Z

=2

1:125 cos d =2

Z

=2 =2

Z

1:5

r3 cos2

dr d

0

124:05 lb ft

=2

124:05 = 140:40

0:884 ft J

By symmetry y = 0 J

8.141

8.142

y

y

1 50 3 2

100 x

20 100 x

436 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

A1

=

A2 A3

= =

y=

4 (50) = 121:22 mm 2 3 1002 = 10 000 mm2 y2 = 50 mm 2 2 (20 ) = 1256:6 mm y3 = 100 mm (502 ) = 3927 mm2

y1 = 100 +

Ai y i 3927(121:22) + 10 000(50) 1256:6(100) = = 67:1 mm J Ai 3927 + 10 000 1256:6

437 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 9 9.1 y = 10 + 4 x 9 y x

10 mm

dx 18 mm

18 mm x

4 10 + x dx 9 Z Z 18 4 x3 4 x4 = x2 dA = x2 10 + x dx = 10 + 9 3 9 4 A 0

dA

=

Iy

=

31:1

18 0

103 mm4 J

9.2 y (in.) x = 12

C

d=

20

y = 16 x (in.)

O = JC + Ad2 = Ix + Iy

JO

Ix + Iy JC 7000 + 4000 1000 == = 25:0 in2 J 2 d 202 = Ix Ay 2 = 7000 25:0(162 ) = 600 in4 J = Iy Ax2 = 4000 25:0(122 ) = 400 in4 J

A = Ix Iy

9.3 Ix Ay 2 0:4 1:5y 2

= Iu A(1:0 y)2 (= Ix ) = 0:6 1:5(1:0 y)2 y = 0:4333 ft J

Ix = 0:4

1:5 (0:4333) = 0:1184 ft4 J 2

438 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.4 Ix

= Ix + Ay

Iu

= Ix + A(1:0

r

y=

Ix

Ix

r

0:35 0:08 = 0:4243 ft J A 1:5 y)2 = 0:08 + 1:5(1 0:4243)2 = 0:577 ft4 J

2

=

9.5

9.6 y dy 2 in y

x = 10 − 2.5y2 x

10 in.

dA = Ix

=

(10 Z A

=

2:5y 2 )dy Z 2 y 2 dA = y 2 (10

2 10

2:5y 2 )dy = 2

2

y3 3

2:5

y5 5

Z

2

y 2 (10

2:5y 2 )dy

0

2

= 21:3 in4 J 0

439 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.7

y

Ix

=

2

Z

6

y 3 dx 2 = 3 3

0

=

2 3

Z

Iy

6

0

=

Z

6

2

0

2

Z

3

dx

144x2 + 1728 dx = 3160 in4 J

6 2

x y dx = 2

0

=

x2 3

12

1 6 x + 4x4 27 Z

x

6 in.

6 in.

x2 3

dx y = 12

x

Z

6

12

0

6

x4 3

12x2

0

x2 3

x2 dx

dx = 691:2 in4 J

9.8

9.9 y y = h(x/b)2 y x

x b

dx

440 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Ix = Iy =

Z

Z

dIx =

A

A

Z

b

0

2

x dA =

Z

1 3 1 y dx = 3 3

Z

b 2

x (y dx) =

0

b

x b

h3

0

Z

0

b

6

x x h b 2

dx = 2

h3 x7 3b6 7

h x5 dx = 2 b 5

b

= 0

bh3 J 21

b

= 0

b3 h J 5

Results check with Table 9.2.

9.10

9.11

441 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.12

9.13

9.14

442 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.15

9.16 y y' 0.606 in.

C

x

For one channel: Ix = 78:9 in4

2

Iy = Iy0 + Ad2 = 2:81 + 5:88 (0:606) = 4:969 in4

443 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

For the assembly in Fig. (b): Ix Iy

= 2Ix = 2(78:9) = 157:8 in4 J = 2Iy = 2(4:969) = 9:94 in4 J

9.17

9.18

y 1

1 3 17

2

2 10

6.5 6.5

10

x

444 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(Ix )1

(Ix )1 + A1 y12 = 0:05488(104 ) +

=

35:99 103 in4 bh3 10(173 ) = = 16:377 103 in4 3 3 bh3 13(273 ) = = 85:29 103 in4 3 3

(Ix )2

=

(Ix )3

=

Ix

(102 ) 4(10) 17 + 4 3

=

2

= 2(Ix )1 + 2(Ix )2 + (Ix )3 = [2(35:99) + 2(16:377) + 85:29] 103 = 190:0 103 in4 J

(Iy )1

=

(Iy )1 + A1 x21 = 0:05488(104 ) +

4 (102 ) 6:5 + (10) 4 3

2

9:615 103 in4 bh3 17(103 ) = + A2 x22 = + (10 17)(11:52 ) 12 12 = 23:90 103 in4 27(133 ) bh3 = = 4:943 103 in4 = 12 12 =

(Iy )2

(Iy )3 Iy

= 2(Iy )1 + 2(Iy )3 + (Iy )3 = [2(9:615) + 2(23:90) + 4:943] = 72:0 103 in4 J

103

9.19

y 180 1

40

20 140 C 2 40

y

= =

Ai y i (180 = Ai 130:57 mm

y = 130.57 mm 3 100

x

40)(200) + (140 20)(110) + (100 40)(20) (180 40) + (140 20) + (100 40)

445 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Iy Ix

40(1803 ) 140(203 ) 40(1003 ) + + = 22:9 106 mm4 J 12 12 12 180(403 ) 20(1403 ) = + (180 40)(200 130:57)2 + 12 12 +(20 140)(110 130:57)2 100(403 ) + + (40 100)(20 130:57)2 12 = 90:9 106 mm4 J

=

9.20

y 30 1

15

12.5 C

45

2 y = 34.27 3 40 Ix

10

x

30(153 ) + (30 15)(62:5 34:27)2 12 12:5(453 ) + + (12:5 45)(32:5 34:27)2 12 40(103 ) + (40 10)(5 34:27)2 + 12 = (367:1 + 96:7 + 346:0) 103 = 810 103 mm4 J =

446 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.21 y 30 1

15

12.5 2 C

45

x = 12.93 3 40

10

x

15(303 ) + (30 15)(15 12:93)2 12 45(12:53 ) + + (12:5 45)(6:25 12:93)2 12 10(403 ) + + (40 10)(20 12:93)2 12 = (35:68 + 32:42 + 73:33) 103 = 141:4 103 mm4 J

Iy

=

9.22

y y 2

10.5

2

1 x

9 Ix = (Ix )1

(Ix )2 =

9(10:53 ) 3

(24 ) 4

3 3

x

(22 )(32 ) = 3347 in4 J

1 (9)(10:5) (22 ) = 34:68 in2 2 1 (9)(10:5)(3:5) (22 )(3) Ai y i = = 2 = 3:681 in. A 34:68 = Ix Ay 2 = 3347 34:68(3:6812 ) = 2880 in4 J

A = y Ix

447 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.23 y y

mm 80

=

Iy

= Iy =

x

20 mm

2 Ix

mm 60

1

3 x

(804 ) 160(203 ) (604 ) + = 11:422 106 mm4 8 3 8 (604 ) (804 ) 20(1603 ) + = 17:82 106 mm4 J (Iy )i = 8 12 8

(Ix )i =

A = Ai y i

= =

y Ix = Ix

= Ay 2

(802 ) (602 ) + 160(20) = 7598 mm2 2 2 (802 ) 4(80) (602 ) 4(60) + (160 20)( 10) 2 3 2 3 0:16533 106 mm3 0:16533 106 Ai y i = = 21:76 mm A 7598 = 11:422 106 7598(21:762 ) = 7:82 106 mm4 J Ai =

3 in. 4.5 in.

9.24

y 1

2 3 6 in.

6 in.

x

y 4 12 in. Part 1 2 3 4 Sum

2

A (in ) 22.5 13.5 18.0 18:0 36.0

y (in.) 2.5 4.5 1.5 1.0

3 in. x Ay (in3 ) 56.25 60.75 27.00 18:00 126.00

448 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

y=

Ai yi 126:0 = = 3:50 in. Ai 36:0

6(4:53 ) 6(33 ) 6(7:53 ) + + 13:5(4:52 ) + 12 36 3

Ix

=

Ix

= Ix

Ay 2 = 526:5

12(33 ) = 526:5 in4 12

36(3:502 ) = 85:5 in4 J

9.25

9.26

449 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.27

y

105 mm

75 mm

105 mm

2

1

75 mm

y

30 mm x

x

Ix

A1

=

A2

=

y

=

= = =

1052 = 11 025 mm2 y1 = 52:5 mm 1 (752 ) = 2813 mm2 y2 = 80 mm 2 Ai y i 11 025 (52:5) 2813(80) = 43:08 mm = Ai 11 025 2813 Ix

i

+ Ai (y

yi )2

1054 + 11 025(52:5 12 6:39 106 mm4 J

43:08)2

754 36

2813(80

43:08)2

9.28

9.29 y

R/2

3 2R 2

R

α = 30o

1 x

450 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Consider a quarter of the cross section: (Ix )1

=

(Ix )2

=

R4 R4 (2 sin 2 ) = sin 60 = 0:011323R4 16 16 !3 ! p p 3 bh3 3 3 1 1 R + Ay 2 = R R + R2 36 36 2 2 8 3

2

0:02706R4

=

For the full cross section: Ix = 4 [(Ix )1 + (Ix )2 ] = 4(0:011323 + 0:02706)R4 = 0:1535R4 J Before sawing R4 = 0:7854 in4 4 0:7854 0:1535 Percent reduction in Ix is 100% = 80:5% J 0:7854 Ix =

9.30

2 7.5 in. (Ix )1 (Ix )2 Ix (Iy )1 (Iy )2 Iy

12.990 in.

30o 30o

15 in.

1 in. 15

12.990 in.

y

3 7.5 in

x

R4 154 (2 + sin 2 ) = 2 + sin 60 = 12 107 in4 8 8 6 7:5(12:990)3 = (Ix )3 = = 1370 in4 12 = (Ix )i = 12 107 + 2(1370) = 14 850 in4 J =

154 2 sin 60 = 1146:5 in4 8 6 12:990(7:5)3 1 = (Iy )3 = Iy 2 + A2 d2 = + (7:5) (12:990) (52 ) 36 2 = 1370 in4 = (Iy )i = 1146:5 + 2(1370) = 3890 in4 J =

R4 (2 8

sin 2 ) =

451 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.31

9.32

452 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.33 a

a

A

R=

Z

Rh

p= y Z y dA = Qa

p dA =

A

y h

dA C

Z

=

Z

py dA =

y 2 dA = Ia

A

A

h

Q.E.D.

A

Ia Ia Ia = = R Qa Qa

=

Q.E.D.

9.34 y

dx

x

y x

b y=h dIxy Ixy

x b

2

y 1 x = dIxy + xel yel dA = 0 + x (y dx) = xh2 2 2 b Z b Z b 2 2 h2 b h = dIxy = 4 x5 dx = J Checks 2b 0 12 0

4

9.35 y

x

dx

y

R

x

453 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(a) x2 + y 2 = R2 dIxy Ixy

y=

p R2

x2

1 y (y dx) = 0 + x(R2 = dIxy + xel yel dA = 0 + x 2 2 Z R 4 R 1 x R2 x2 dx = J Checks = 2 0 8

x2 )dx

(b) Ixy

= Ixy = R4

Axy = 1 8

4 9

R4 8 =

R2 4

4R 3

4R 3

0:01647R4 J Checks

9.36

9.37 The terms in the above equation have the same magnitudes for the four triangles, but di¤er in signs. The sign of Ixy is determined by the quadrant in which the triangle lies. The sign of Axy is determined by the signs of x and y. Triangle a b c d

Ixy neg. pos. pos. neg.

x neg. pos. neg. pos.

y pos. pos. neg. neg.

Ixy b2 h2 =72 b2 h2 =72 b2 h2 =72 b2 h2 =72

J

9.38

454 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.39

9.40 Outer rectangle: A1 = 120(140) = 16 800 mm2

x1 = 60 mm

y1 = 70 mm

x2 = 70 mm

y2 = 80 mm

Inner rectangle: A2 =

60(80) =

A = x = y

Ixy

= = =

=

4800 mm2

Ai = 16 800 4800 = 12 000 mm2 Ai xi 16 800(60) 4800(70) = = 56:0 mm A 12 000 Ai y i 16 800(70) 4800(80) = = 66:0 mm A 12 000

Ai (xi x)(yi y) 16 800(60 56)(70 66) 672 000 mm4 J

4800(70

56)(80

66)

9.41

455 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.42

3

y

4

Flange

x

C

2 8 Centroid C of the region can be found by inspection. Ixy

=

(Ixy )i + Ai xi yi = 2(Axy)‡ange

=

2(8

2)( 3)(4) =

384 in4 J

9.43

y 2 . 6 in x 3 in.

1 Ixy = A1 x1 y1 +

R24 = (6 8

3)(3)( 1:5) +

64 = 81 in4 J 8

456 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.44

9.45

y 30 1

15

2 45 12.5 3 40 Ixy = Part 1 2 3 Sum Ixy = Ixy

x

(Ixy )i + Ai xi yi = 0 + Ai xi yi 2

A (mm ) 450.0 562.5 400.0 1412.5

Axy = 576:2

10

103

x (mm) 15.00 6.25 20.00

y (mm) 62.5 32.5 5.0

Axy (mm4 ) 421.9 103 114.3 103 40.0 103 576.2 103

1412:5(12:93)(34:27) =

49:7

103 mm4 J

9.46 Due to symmetry Ixy = 0 J

457 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

y y 6

1

3 4

2 x

x

18 Ixy

18

=

Ai xi yi

=

(18

12)(9)(6) +

=

13 210 in4 J

(62 ) 4(6) 18 + (6) 2 3

(42 )(18)(6)

9.47 y

y

12 in.

12 in. 12 in. 18 in.

2

1 d = 6 in. x

x

Find the centroid …rst. From Table 8.1: 2 (12)2 = 96 in2 A2 = 3 3 3 x2 = (12) = 4:5 in. y2 = 6 + (12) = 13:2 in. 8 5 A (in2 ) x (in.) y (in.) Ax (in3 ) Ay (in3 ) 216 6:0 9:0 1296 1944:0 96 4:5 13:2 432 1267:2 120 864 676:8 864 676:8 x= = 7:200 in. y= = 5:640 in. 120 120

Part 1 2 Sum

2

(Ixy )1

=

(Ixy )2

= =

b 2 h2 (12) (18)2 = = 11 664 in4 4 4 b 2 h2 2bh 3h 3h Ixy 2 + A2 x2 y2 = + d+ 60 3 8 5 2 2 (12) (12) 2(12)(12) 3(12) 3(12) + 6+ = 6048 in4 60 3 8 5

458 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Ixy

=

(Ixy )1

Ixy

= Ixy

(Ixy )2 = 11 664

6048 = 5620 in4

(7:200) (5:640) (120) = 1175 in4 J

xyA = 6048

9.48

9.49

y 2 15 45

b

b x

1

45

15

459 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Only the two cutouts contribute to Ixy . Therefore, Ixy = 0 if (Ixy )1 +(Ixy )2 = 0. A1 =

1 (452 ) = 1012:5 mm2 2

(Ixy )1

(Ixy )2

x1 = y1 =

2 (45) = 3

45 mm

(452 )(452 ) + 1012:5( 45)2 72

=

(Ixy )1 + A1 x1 y1 =

=

1:9934

=

(Ixy )2 + A2 x2 y2 = 0 + b2

(Ixy )1 + (Ixy )2 = 0

15

106 mm4

1:9934

106

b 2

b4 =0 4

b 2

=

b4 4

b = 53:1 mm J

9.50 (a) Due to symmetry, the xy-axes are the principal axes at C. I1 I2

12(163 ) bh3 = = 4096 in4 J 12 12 b3 h (123 )(16) = Iy = = = 2304 in4 J 12 12 = Ix =

(b) b

=

R

=

Iu Iv Iuv

= = =

Ix + Iy 4096 + 2304 = = 3200 in4 2 2 Ix Iy 4096 2304 = = 896 in4 2 2 b + R cos 2 = 3200 + 896 cos 60 = 3650 in4 J b R cos 2 = 3200 896 cos 60 = 2750 in4 J R sin 2 = 896 sin 60 = 776 in4 J

460 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.51

9.52

y 2 160 mm

θ1

1

C 120 mm Ix

=

Iy

=

Ixy

=

x

bh3 120(1603 ) = = 13:653 106 mm4 36 36 b3 h (1203 )(160) = = 7:680 106 mm4 36 36 b2 h 2 (1202 )(1602 ) = = 5:120 106 mm4 72 72

461 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Ix + Iy 13:653 + 7:680 = 106 = 10:667 106 mm4 2 2 s s 2 2 Ix Iy 13:653 7:680 2 = + Ixy + ( 5:120)2 = 2 2

b

=

R

sin 2

106 mm4

=

5:927

I1 I2

= b + R = (10:667 + 5:927) = b R = (10:667 5:927)

1

1

106

106 = 16:59 106 mm4 J 106 = 4:74 106 mm4 J

Ixy ( 5:120) Ix Iy = = 0:8638 cos 2 1 = >0 R 5:927 2R ) 2 1 lies in the …rst quadrant 2 1 = sin 1 (0:8638) = 59:75 = 29:9 J J 2 = 1 + 90 = 119:9

=

9.53

y 2 in.

2 d 14 in.

1

.C

30o

2 in. 3

x u 2 in.

6 in. d=

Ix

=

Iy

=

Ixy

= =

Iu

= =

28(1:0) + 2 [8(4)] = 2:091 in. 28 + 2(8)

2(14)3 4(2)3 +2 + 8(6)2 = 1038:7 in4 12 12 14(2)3 2(4)3 + 28(1:0 2:091)2 + 2 + (2 12 12

4)(4

2:091)2

122:30 in4 0 due to symmetry Ix + Iy Ix Iy + cos 2 Ixy sin 2 2 2 1038:7 + 122:3 1038:7 122:3 + cos( 60 ) = 810 in4 J 2 2

462 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.54 Ix + Iy 2 Ix Iy 2

Iu;v Iu Iv

= =

3000 + 2000 = 2500 in4 2 3000 2000 = 500 in4 2

Ix Iy Ix + Iy cos 2 Ixy sin 2 2 2 = 2500 + 500 cos 240 ( 500) sin 240 = 1817 in4 J = 2500 500 cos 240 + ( 500) sin 240 = 3180 in4 J =

Iuv

= =

Ix

Iy sin 2 + Ixy cos 2 2 500 sin 240 + ( 500) cos 240 =

183:0 in4 J

9.55

9.56

463 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.57

9.58

9.59 y

v

u

45 o

x

O

Due to symmetry, u and v are the principal axes at O. Iu;v = With Ix

Ix + Iy 2

Ix

Iy 2

cos 2

Ixy sin 2

Iy = 0 and 2 = 90 , this becomes Iu;v = [16:023

( 1:1310)]

106

464 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Iu Iv

= I1 = (16:023 + 1:1310) = I2 = (16:023 1:1310)

106 = 17:15 106 = 14:89

106 mm4 J 106 mm4 J

9.60

y 4 in. u 1 16 in. 4 in.

Iu

d

=

Ix

=

Iy

=

= =

45o C 2 16 in.

x d = 7 in.

Ai y i 64(12) + 64(2) = 7:0 in. = Ai 2(64) 4(163 ) 16(43 ) + 64(52 ) + + 64(52 ) = 4651 in4 12 12 16(43 ) 4(163 ) + = 1451 in4 12 12

Ix + Iy Ix Iy + cos 2 Ixy sin 2 2 2 4651 + 1451 4651 1451 + cos 90 2 2

0 = 3050 in4 J

9.61

465 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.62

9.63

y

3

y

3 2

4

4 x

x

10

1

10 Ix

=

Iy

=

Ixy

=

Ix

+ A1 d21 1

(Ix )2

104 12 0 due to symmetry (Iy )1

(Iy )2 =

A2 d22 =

104 + 100(1:02 ) 12

6(4)3 36

12

2 4 3

2

= 837:3 in4

63 (4) = 815:3 in4 48

466 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Iu

= =

Iv

= =

Ixy

= =

Ix + Iy Ix Iy + cos 2 Ixy sin 2 2 2 837: 3 + 815:3 837: 3 815:3 + cos(90 ) = 826 in4 J 2 2 Ix Iy Ix + Iy cos 2 + Ixy sin 2 2 2 837: 3 + 815:3 837: 3 815:3 cos(90 ) = 826 in4 J 2 2 Ix Iy sin 2 + Ixy cos 2 2 837: 3 815:3 sin 90 = 11:0 in4 J 2

9.64

9.65 y

2

. 8 in x 6 in.

1

467 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

48 in2 x1 = 4 in. y1 = 3 in. 2 4(8) (8 ) A2 = = 50:27 in2 x2 = y2 = = 3:395 in 4 3 A = Ai = 48 + 50:27 = 98:27 in2 Ai xi 48(4) + 50:27(3:395) x = = = 3:691 in. A 98:27 Ai y i 48( 3) + 50:27(3:395) y = = = 0:2714 in. A 98:27

A1

Ix Iy Ixy

R

=

Ix

=

Iy

=

Ixy

=

(84 ) 8(63 ) + = 1380:2 in4 3 16 (84 ) 6(83 ) + = 1828:2in4 3 16 84 48(4)( 3) + = 64:0 in4 8

= Ix Ay 2 = 1380:2 98:27(0:2714)2 = 1373:0 in4 = Iy Ax2 = 1828:2 98:27(3:691)2 = 489:4 in4 = Ixy Axy = 64:0 98:27(3:961)(0:2714) = 169:64 in4

=

s

Ix

Iy 2

2 2 = + Ixy

s

1373:0

489:4 2

2

+ ( 169:64)2

473:25 in4 Ix + Iy 1373:0 + 489:4 = +R= + 473:25 = 1405 in4 J 2 2 Ix + Iy 1373:0 + 489:4 = R= 473:25 = 458 in4 J 2 2 =

I1 I2

9.66

y 24 in.

30 in.

x-1

y

1 y-1

O

x O

24 in. 2 x-2 -y 2 15 in.

x

468 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

A1 A2

(24)(30) = 720 in2 x1 = 12 in. y1 = 15 in. 2 (15 ) 4(15) = = 176:71 in2 x2 = 24 = 17:634 in. 4 3 4(15) = 6:366 in. y2 = 3

=

Ix

=

Iy

=

Ixy

=

24(303 ) (154 ) = 206:1 106 in4 3 16 30(243 ) 0:05488(154 ) + 176:71(17:6342 ) = 80:51 103 in4 3 720(12)(15) 0:01647(154 ) + 176:71(17:634)(6:366)

=

110:60

R

=

s

103 in4

Ix + Iy 2

2 2 = + Ixy

s

206:1 + 80:51 2

2

+ 110:602

103

181:02 103 in4 110:60 Ix Iy Ixy = = 0:6110 cos 2 1 = >0 = R 181:02 2R ) 2 1 lies in the 4th quadrant = sin 1 ( 0:6110) = 37:66 18:83 J 1 = = 1 + 90 = 18:83 + 90 = 71:2 J =

sin 2

1

2

1 2

y 2

x 1

O 18.83o

9.67

y

2

6 in.

6 in. C 6 in.

x

45o 1

6 in.

469 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

The principal axes can be located by inspection— they are the axis of symmetry of the region: 45 J 1 = 2 = 45 J Ix Ixy

I1

= =

I2

= =

= Iy = 2 =

2

64 4

(6)4 64 + = 1372:9 in4 3 16 64 = 324:0 in4 8

Ix Iy Ix + Iy + cos 2 1 Ixy sin 2 1 2 2 1372:9 + 0 324:0 sin( 90 ) = 1697 in4 J Ix Iy Ix + Iy + cos 2 2 Ixy sin 2 2 2 2 1372:9 + 0 324:0 sin 90 = 1049 in4 J

9.68

470 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.69

u 896

2304

R=

P.I.

y

60o

x M.I.

Units: in4 v

b = 3200 4096

Ix b Iu Iv Iuv

12(163 ) 16(123 ) = 4096 in4 Iy = = 2304 in4 Ixy = 0 12 12 4096 + 2304 4096 2304 = = 3200 in4 R= = 896 in4 2 2 = 3200 + 896 cos 60 = 3650 in4 J = 3200 896 cos 60 = 2750 in4 J = 896 sin 60 = 776 in4 J =

471 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.70 Ix

=

Iy

=

Ixy

=

8bh3 8(120)(903 ) = = 3:999 106 mm4 175 175 19b3 h 19(1203 )(90) = = 6:156 106 mm4 480 480 b2 h 2 (1202 )(902 ) = = 1:9440 106 mm4 60 60

P.I. 7.680

Units: 106 mm4

y R= 27 5.9

1

5.120

2θ1

2

M.I.

x b = 10.665 13.650

b

=

R

= =

Iu Iv Iuv

3:999 + 6:156 2 s 6:156

sin

1

106 = 5:078

3:999

106 mm4

2

2 1:944 = 60:99 2:223

+ 1:94402 = 90

106 = 2:223

106 mm4

= 29:01

= b R cos = (5:078 2:223 cos 29:01 ) 106 = 3:13 106 mm4 J = b + R cos = (5:078 + 2:223 cos 29:01 ) 106 = 7:02 106 mm4 J = R sin = (2:223 sin 29:01 ) 106 = 1:078 106 mm4 J

472 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.71 P.I. 7.680

Units: 106 mm

y

R=

5.120

27 5.9 2θ1

2

1

M.I.

x b = 10.665 13.650

Ix

=

Iy

=

Ixy

=

120(1603 ) = 13:650 106 mm4 36 160(1203 ) = 7:680 106 mm4 36 (1202 )(1602 ) = 5:120 106 mm4 72

13:650 + 7:680 2 s

b

=

R

=

I1 I2

= (10:665 + 5:927) = (10:665 5:927)

13:650

7:680

2

2

1

106 = 10:665

106 mm4

2

+ 5:1202

106 = 5:927

106 = 16:59 106 mm4 J 106 = 4:74 106 mm4 J

5:120 = 59:75 ) 1 = 29:9 5:927 = 29:9 + 90 = 119:9 J

=

sin

)

2

106 mm4

1

J

9.72 Ix

=

Iy

=

ab3 = 4 a3 b = 4

(8)(43 ) = 402:1 in4 4 (83 )(4) = 1608:5 in4 4

Ixy = 0

473 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

v R=6 03.2

P.I. (in4)

x 402.1

70

y

o

M.I. (in4 )

b = 1005.3 u 1608.5

b

=

R

=

Ix + Iy 402:1 + 1608:5 = = 1005:3 in4 2 2 1608:5 402:1 Iy Ix = = 603:2 in4 2 2

= b R cos 70 = 1005:3 603:2 cos 70 = 799 in4 J = b + R cos 70 = 1005:3 + 603:2 cos 70 = 1212 in4 J = R sin 70 = 603:2 sin 70 = 567 in4 J

Iu Iv Iuv

9.73 P.I. 2000 y 240o

α = 15o Units: in4

u

R = 45o 70 7.1

v M.I. 500 x

b = 2500 3000

Iu Iv Iuv

= b R cos = 2500 707:1 cos 15 = 1817 in4 J = b + R cos = 2500 + 707:1 cos 15 = 3180 in4 J = R sin = 707:1 sin 15 = 183:0 in4 J

474 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.74

9.75

475 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.76

9.77

9.78

476 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.79 Ix

= =

Ix + Iy 2 Ix Iy 2

= =

Iu Iv Iuv

18(123 ) 12(183 ) = 2592 in4 Iy = = 5832 in4 12 12 6 2 = 67:38 tan 1 = 33:69 9 2592 + 5832 = 4212 in4 2 2592 5832 = 1620 in4 2

Ixy = 0

= 4212 + ( 1620) cos( 67:38 ) = 3590 in4 J = 4212 ( 1620) cos( 67:38 ) = 4840 in4 J = 1620 sin( 67:38 ) = 1495 in4 J

9.80

477 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.81

9.82 Ix

=

(Ix )circle =

ab3 8

R4 8

(754 ) = 78:5 106 mm4 J 8 a3 b R4 = (Iy )ellipse (Iy )circle = 8 8 (754 ) (2003 )(105) = 317 106 mm4 J = 8 8 =

Iy

(Ix )ellipse

(200)(1053 ) 8

9.83 y x

dx

60 mm 30 mm y = 30 − x/3 xel = x

90 mm

yel = 60

x 3

x dA = 60 dx

478 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

dIx

2 = dIx + yel dA =

= Ix

=

Z

20 2 x 3

0

90

603 dx + 60 12

x 3

2

(60 dx)

2400x + 234 000 dx

20 2 x 3

2400x + 234 000 dx = 12:96

dIy = x2el dA = 60x2 dx

Iy =

Z

90

60x2 dx = 14:58

0

106 mm4 J 106 mm4 J

9.84

9.85

y 1 Dimensions in mm

C

150 80

_ y

2 10 3

x

479 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

A (mm2 ) 2400 2400 1200 6000

Part 1 2 3 Sum y Ix

Iy = Ix Iy

y (mm) 150 80 10

Ay (mm3 ) 360 000 192 000 12 000 564 000

564 000 = 94:0 mm 6000 120(203 ) = + 2400(1502 ) 12 20(1203 ) + + 2400(802 ) 12 60(203) + 1200(102 ) + 12 = 72:48 106 mm4 =

20(1203 ) 120(203 ) 20(603 ) + + = 3:32 12 12 12

106 mm4

= Ix Ay 2 = 72:48 106 6000(94:02 ) = 19:46 = Iy = 3:32 106 mm4 J

106 mm4 J

9.86

9.87

480 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.88

6 in.

y 6 in. C

2 in. 2 in.

x 6 in.

6 in. For one triangle: Ix

= Ix + Ay 2 =

Iy

=

Ixy

6(123 ) 6(12) 2 + (2 ) = 432:0 in4 36 2

12(63 ) = 216:0 in4 12 (62 )(122 ) 6(12) + (2)( 2) = = Ixy + Axy = 72 2

216:0 in4

For the two triangles: Ix Iy Ixy

= 2(432:0) = 864 in4 J = 2(216:0) = 432 in4 J = 2( 216:0) = 432 in4 J

481 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.89

y

y 3 2 in. 1

6 in.

6 in. x

6 in. 64 12

"

Ix

=

64 3

Ixy

=

64 4

=

144:0 in4 J

2 6 in.

6(23 ) 2 6 + 36 2

64 62 + (2)(2) 72 2

6

2 3

2

#

x

= 152:0 in4 J

(62 )(22 ) 6 2 + (4) 6 72 2

2 3

9.90

Due to symmetry Ixy = 0 J

482 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.91

9.92

483 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.93 y 2 in. 1 4 in. 1

C

45o

2

2

6 in. Ai xi 8(1:0) + 12(3) = = 2:20 in. Ai 8 + 12

x=

Ix Ixy

2 in. x By symmetry y = x = 2:20 in.

6(2)3 2(4)3 + 8(4)2 + = 154:67 in4 12 3 By symmetry Iy = Ix = 154:67 in4 = A1 x1 y1 + A2 x2 y2 = 8(1:0)(4) + 12(3)(1:0) = 68:0 in4 =

(Ix )1 + A1 y12 + (Ix )2 =

Ix Ixy

= Ix Ay 2 = 154:67 20(2:20)2 = 57:87 in4 By symmetry Iy = Ix = 57:87 in4 = Ixy Axy = 68:0 20(2:20)2 = 28:80 in4

Due to symmetry, "1" and "2" are the principal axes. s Ix Iy Ix + Iy Ix Iy =0 = 57:87 in4 R= 2 2 2

I1

=

I2

=

2 2 = 28:80 in4 + Ixy

Ix + Iy + R = 57:87 + 28:20 = 86:1 in4 J 2 Ix + Iy R = 57:87 28:20 = 29:7 in4 J 2

484 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.94 Equation (9.17) is Iu =

Ix + Iy Ix Iy + cos 2 2 2

Ixy sin 2

Substituting = 90 + 28:5 = 118:5

2 = 237

we get 21:3

=

21:3

=

80:9 + 38:8 80:9 38:8 + cos 237 Ixy sin 237 2 2 48:39 + 0:8387Ixy Ixy = 32:3 in4 J

485 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 10 10.1

10.2

486 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.3 xC C

B

P

yG L G

θθ

L

2 2

W

W A

U = 2W yG + P xC xC yG

U

=

P

=

=

2L sin

=

L cos 2 2

xC = L cos

2

yG =

L sin 4 2

2W

2 L sin 4 2

+ P L cos

2

=0

W tan J 2 2

10.4

θ

yA

A yB

W

yD B W D W

C

F

xC a sin 2

yA

=

yB

= a sin +

yD

=

xC

yA = a 2

a cos 2

yB = a cos

a a sin + a yD = cos 2 2 = a cos xC = a sin

487 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

U

= W ( yA + yB + yD ) + F xC = 0 (2W a cos

F a sin )

=0

F =

2W tan

J

10.5

10.6 B G A

T mm

120 θ C xD

mm E W 120 yE D

Note that BE remains vertical. xD yE U

= =

U

=

= =

120 cos 240 sin

xB = 120 sin yE = yG = 240 cos

T xD W yG = [T (120 sin ) (2:4)(9:81)(240 cos )] (120T sin 5651 cos ) = (120T sin 30 5651 cos 30 ) 5651 0 T = cot 30 = 81:6 N J 120

488 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4"

1.5 lb

7.5 "

10.7

yC

1.0 lb 5"

A

β

5"

θ

C

7.5 "

yA

B P

xB Geometry:

yA xB yC

sin

=

sin

=

8 = 0:8 10 12 = 0:8 15

= 5 sin = 10 cos = 7:5 sin

6 = 0:6 10 9 = = 0:6 15

cos = cos

yA = 5 cos = 5(0:6) = 3 xB = 10 sin = 10(0:8) = 8 yC = 7:5 cos = 7:5(0:6) = 4:5

Constraint: 15 sin 15(0:6)

10 sin 10(0:6)

= 4 = 0

15 cos 10 cos = 0:6667

=0

Virtual work: U P

= = =

1:0 yA + 1:5 yC + P xB [1:0(3) + 1:5(4:5)(0:6667) + P ( 8)] 0:938 lb J

=0

10.8 D a P yD

C b

b

θ

θ A

xB

B

R

489 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

U= yD xB

P yD

= (a + b) sin = 2b cos

U

R xB

yD = (a + b) cos xB = 2b sin

=

[ P (a + b) cos R( 2b sin )] P (a + b) = cot J 2b

R

=0

10.9 a

a

θ

yB a R

a A

xB B

W/2

W/2 W 2

U =2 xB yB

b

b

R

yb + R xb

= b sin xB = b cos = (2a + b) cos yB = U R

(2a + b) sin

=

[ W (2a + b) sin + 2Rb cos ] 2a + b = W tan J 2b

=0

10.10

6'

B

G T 6'

W 55 θ o

A

yG

xA

490 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Release the cable at A, so that the cable tension T becomes an active force. yG xA

= 6 sin = 12 cos U

= =

T

=

yG = 6 cos = 6 cos 40 = 4:596 xA = 12 sin = 12 sin 40 =

T cos 55 xA W yG = 0:5736T xA 0:5736T ( 7:713 ) 280(4:596 ) = 0 280(4:596) = 291 lb J 0:5736(7:713)

7:713

280 yG

10.11

10.12 P A L

L

yA

θ

F B

xB yA xB U P F P

= L sin = 2L cos

yA = L cos xB = 2L sin

= P yA + F xB = L(P cos = 2F tan

2F sin )

= k spring = 2kL(1 cos ) = 4kL tan (1 cos ) = 4(3000)(0:2) (tan 40 ) (1 = 471 N J

=0

cos 40 )

491 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.13

10.14

C

D 25"

45" yC

40"

50o

CA

xC A xC yC When

E

40"

32 lb

θ B

= 45 40 cos xC = 40 sin = 40 sin yC = 40 cos

= 70 : xC U

= 40 sin 70 = 37:59 yC = 40 cos 70 = 13:681 = 32 sin 50 yC 32 cos 50 xC + CA = [(32 sin 50 ) (13:681) (32 cos 50 ) (37:59) + CA ] = ( 437:8 + CA ) = 0 CA = 438 lb in J

492 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.15

10.16 W

a

a

h a

a

θ

h U C

C

= 2a sin h = 2a cos = W h+C = ( 2aW cos + C) = 2aW cos J

=0

493 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.17

70 sin45o lb B 70 cos45o lb

40 lb L/2

yB

L/2

G y G θ xB

A

U

= =

yG xB yB

U

40 yG + 70 cos 45 xB + 70 sin 45 40 yG + 49:50( xB + yB ) L sin 2 = L cos = L sin =

L cos 2 xB = L sin yB = L cos yG =

L cos + 49:50L( sin + cos ) 2 29:50 cos 49:50 sin = 0 1 29:50 = tan = 30:8 J 49:50 =

yB

40

=0

10.18

494 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.19

495 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.20

10.21 0.5δy 750 mm

D

δy B

WCD

δy C D

WBC

δy

mm 600 0.5δy δθ 480 mm

WAB A

B

Virtual kinematics WAB WBC WCD

= (18 = (18 = (18

Active forces

C

C0 A

9:81)(0:480) = 84:76 N 9:81)(0:600) = 105:95 N 9:81)(0:750) = 132:44 N

Bar AB rotates about A : Bar BC translates:

y = 0:480 m yB = yC = y

496 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

U

= C0 = fC0 = (C0

WAB (0:5 y) WBC y WCD (0:5 y) 0:480 [84:76(0:5) + 105:95 + 132:44(0:5)]g 102:98) = 0 C0 = 103:0 N m J

10.22

U P

= =

24δθ

θ 12 δθ

θ

24δθ

24δθ

24 in .

12 in .

12 in .

θ

δθ P (12 ) cos + 60(24 120 tan lb J

) sin = 0

10.23

10.24

497 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.25 2.4 in.

WB = 0.4 lb

x

δyA

WA = 0.8 lb

B

B

δyB

A

A Active forces

Virtual kinematics The scales form a parallelogram linkage. yA yB = 2:4 x U

= =

x =

yB = 0:4167x yA

WA y A + W B y B 0:8 yA + 0:4(0:4167x yA ) = ( 0:8 + 0:16668x) yA = 0 0:8 = 4:80 in. J 0:16668

10.26

δθ

0.6 m

A

C

0.5 m

δxB

B

0.3 m

D

δyD

δrD

0.5 m

δθ O The I.C. of bar BD is located at O, determined by the known directions of xB and rD . Bar AB rotates about A : Bar BD rotates about O :

xB = 0:5 m yD = 0:3 m

498 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

U P Q

=

P xB + Q yD = ( 0:5P + 0:3Q) 0:3 = 0:6 J 0:5

=

=0

10.27 A

D

B

WAB

E

0.3

Cx

C

Active forces

δθ

Dimensions in meters

WBC D

B 0.15 O

0.3

A

δrB

δyD

δθBC 0.075

E δyE

Kinematics

δrE

0.2

C δxC

Release the horizontal constraint at C so that Cx becomes an active force. The I.C. of bar BC is located at O, determined by the known directions of rB and xC . Bar AB rotates about A : Bar BC rotates about O : yE =

yD = 0:3 m rB = 0:6 m rB = 0:15 BC m xC = 0:2 0:075 BC m

Equating the expressions for rB : 0:6 xC = 0:2(4 U Cx

) = 0:8

m

= 0:15

BC

yE = 0:075(4

m ) = 0:3

= WAB yD + WBC yE Cx xC = (0:3WAB + 0:3WBC 0:3W 0:3(20 9:81) = = = 73:6 N J 0:8 0:8

BC

BC

=4

m

m

m 0:8Cx )

=0

499 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.28

δrF

F

δθAC

A 12"

δrB

32" B

δθDF

E

24"

16"

δrC

16" C

Bar ABC rotates about A Bar CD translates:

:

Bar DEF rotates about E

:

rF

=

U P

D

δrD

rB = 12 AC rC = 36 rD = rC = 36 AC 36 rD = AC = 2:25 DF = 16 16 32 DF = 32(2:25 AC ) = 72:0

= P rB 4000 rF = [P (12) = 24 000 lb J

4000(72)]

AC

AC

AC AC

=0

10.29

500 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.30

δrC C D

P

δθ 15 in.

15 in.

A δrA

B

δrD = δrC

δθDF

F

A

F

10 in. E 12 in. Kinematics

Active forces

δrF

NF

Remove the roller at F so that NF becomes an active force. Bar AC rotates about B Bar DF rotates about E rD = rC U

= P rA =

(15P

: : 10

rC = 15 rD = 10 DF

= 15

NF rF = P (15 18NF )

=0

in. DF

in. DF

rA = 15 in. rF = 12 DF in. = 1:5

)

NF (12)(1:5 ) 15 15 NF = P = (60) = 50:0 lb J 18 18

10.31

501 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.32

502 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.33

δθCD n. 45 i

C δrC

D

65o 45 in.

δθBC

B

30 in. A δθAB

30 = 45 tan 65 45 = 45 sec 65

30 = 66:50 in. 45 = 61:48 in.

δrB

E Point E is the I.C. of bar BC: BE CE

= AE = DE

Bar AB rotates about A

:

Bar BC rotates about E

:

rC

= C1 =

BC

:

CD

C2

AB

AB

30 AB rB = = 0:4511 AB = 66:50 BE BC = 61:48(0:4511 AB ) = 27:73 27:73 AB rC = = 0:6162 AB = 45 45

= CE

Bar CD rotates about D

U C1 C2

rB = 30

CD

= [C1

C2 (0:6162)]

AB

AB

=0

0:616 J

10.34

δrA

A y 12"

δrB δr'B

x

B 9"

δθ

15 "

12" C

δθBD D

503 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

rB 0 rB

= virtual displacement of point B on barAC = virtual displacement of point B on sliding collar

Bar AC rotates about C Bar BD rotates about D

0 ( rB )x = rB

Constraint: ) U CD

: :

BD

rA = 21 0 rB = 15

in. BD

3 0 r = rB 5 B

rB = 9

in.

in. 3 (15 5

BD )

=9

=

= P rA CD BD = (21P CD ) = 21P = 21(25) = 525 lb in. J

=0

10.35

504 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.36

10.37

505 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.38

506 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.39

10.40 Choose the

= 0 position as the datum for gravitational potential energy. V

dV d

W

1 = W L sin + k(L sec 2 = W L cos + kL2 (sec

1:5L)2 1:5) sec tan = 0

W cos 30 + kL(sec 30 1:5) sec 30 tan 30 = 0 0:8660W 0:2302kL = 0 0:2302 0:2302 = kL = (1:5)(20) = 7:97 lb J 0:8660 0:8660

507 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10.41

B Datum for V

θ

A

V dV d dV d

1.5 ft

= W h = W (3 sin

0:

d2 V d 2

1:5 =0 cos3

3

d2 V d 2

1:5 tan )

1:5 sec2 )

= W (3 cos =

1.5 tanθ

t 3f C

3 sinθ

h

cos3 = 0:5

J

3 sec2 tan )

= W ( 3 sin

= W ( 3 sin 37:5 =37:5

=

= 37:5

5:48W < 0

3 sec2 37:5 tan 37:5 ) ) Unstable J

10.42

1.8R

B hB

R

θ

R

hA A

θ θ B hB

hG hA

1.8R sinθ

C

G

Datum for V

A

508 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

hA hB hG

= AC sin = (2R cos ) sin = BD hA = 1:8R sin 2R sin cos = R sin (1:8 2 cos ) 1 1 (hA hB ) = [2R cos sin (R sin ) (1:8 2 cos )] = 2 2 = R sin (2 cos 0:9) V dV d dV d

=

W hG =

=

W R [(cos ) (2 cos

=

W R 4 cos2

=

W R sin (2 cos

0: 4 cos2 cos = 0:8285

d2 V d 2 d2 V d 2

=34:1

0:9)

0:9) + (sin ) ( 2 sin )]

0:9 cos 0:9 cos

2 2=0

= 34:1

J

=

W R( 8 cos sin + 0:9 sin )

=

W R( 8 cos 34:1 sin 34:1 + 0:9 sin 34:1 ) = 3:21W R > 0

)

Equilibrium is stable J

10.43

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10.44 R 3/8 R

θ h1 Hemisphere:

W1 =

2 3 R 3

= W 1 h1 + W 2 h2 3 2 3 R R R cos = 3 8 dV d d2 V d 2 At

3 R cos 8 h h2 = R + cos 2 h1 = R

W2 = R 2 h

Cylinder: V

h

h2

= = = 0:

4 4

+ R2 h

R4 sin R4 cos

2 2

d2 V = R4 4 d 2

R+

h cos 2

R2 h2 sin R2 h2 cos

2

R 2 h2

Stability requirement is d2 V >0 d 2

h 1