An Instructor’s Solutions Manual to Accompany
ENGINEERING MECHANICS: STATICS, 4TH
EDITION
ANDREW PYTEL
JAAN KIUSALAAS
ISBN: 978-1-305-88502-8
© 2017, 2010 Cengage Learning
WCN: 01-100-101
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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
1
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.6
1.7
1.8
(a) 8 mm/ s =
8 mm
s
1:0 m
1000 mm
(b) 8000 m/s =
8000 m
s
3:281 ft
1:0 m
1:0 s
= 8000 m/s J
10 6 s
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
2
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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 3 ]
L4
[T
2]
[M L 1 ]
M
[L] =
[F L2 ]
T 2 F L2
— see Eq. (1.2b)— we get
!2
M
y =
D
T 2 L2
Substituting [F ] = M LT
2
T2
= [L 3 ] Q.E.D.
ML
— see Eq. (1.2b)— we get
!2
M
y =
D
T 2 L2
T2
= [L 3 ] Q.E.D.
ML
1.13
The argument of the sine function must be dimensionless:
L
= [1]
F
Bx
k
=
[F ]
= [Akx2 ] = [A][F L 1 ][L2 ]
[1]
[B][L]
[B] = [F L 2 ] J
[A] = [L 1 ] 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
3
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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
4
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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
5
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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
6
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
7
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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
PAC
=
sin 40
sin 60
360 sin 40
= 235 lb J
sin 80
360 sin 60
= 317 lb J
sin 80
8
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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
9
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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
10
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
11
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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
12
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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)
13
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1.48
(a)
!
BA =
=
20i + 60j 90k ft
!
20i + 60j
BA
! =
110:0
BA
Fx
Fy
Fz
= F
= F
= F
p
!
BA = 202 + 602 + 902 = 110:0 ft
90k
=
0:1818i + 0:5455j
0:8182k
0:1818) = 109 lb J
y = 600(0:5455) = 327 lb J
491 lb J
z = 600( 0:8182) =
x = 600 (
(b)
= cos 1
= cos 1
= cos 1
x
y
z
( 0:1818) = 100:5 J
=
cos
(0:5455)
= 56:9 J
y
1
( 0:8182) = 144:9 J
z = cos
x = cos
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
14
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
p
!
AC = 52 + 32 + 52 = 7:681 m
!
AC
=
5i
TAC
= T
!
5i 3j + 5k
AC
=
! = 35
7:681
AC
3j + 5k m
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
15
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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
12k
62 + 82 + (
p
6i + 6j
12)2
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
16
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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) =
(a) C =
i
0
4
j k
12 8
2 3
= 52i + 32j
(b) C =
i
5
7
j
3
0
k
0
12
=
36i + 60j
(c) C =
i
j
3 2
6 2
k
1
8
=
14i + 30j + 18k m2 J
6 m2 J
1.58
48k ft2 J
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
B = 2i + 1:2j + 1:5k m
C=
A
B=
i
2
2
j
1:2
1:2
k
0
1:5
= 1:8i
3j m2 J
C
B=
i
j
0 0
2 1:2
k
1:5
1:5
= 1:8i
3j m2 J
1:5k m
17
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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
2:332
A
= cos 1
= 32:8
B
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
18
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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)
5j
20k in.2
( 0:422i
0:211j
10i
0:844k) J
19
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
20
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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
21
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1.74
1.75
1.76
22
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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
23
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Ry
Ry
=
=
Fy
+"
16:96 kN
Ry = 25 sin 45
R = 7:68i
16:96k kN J
40 sin 60
2.5
F1
= F1
F2
= F2
F3
AB = 80 p
AC = 60 p
120j + 80k
= 66:56j + 44:38k N
( 120)2 + 802
100i 120j + 80k
( 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
R
=
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
25
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
26
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
27
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
28
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
= TAB
TAB
3i
12j + 10k
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
TAC
R
AB = 120 p
32 + (
= 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
29
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
*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
30
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2.23
40 in.
A
24 in.
60 lb
P2
P1
12 in.
P2
C
P1
P1 = 60 p
B
40
402 + 122
= 57:47 lb
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
31
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
32
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
33
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
Wy = W p
7'
5.196' 4.804'
10'
Wy
7
72 + 4:8042
Wx
W
= 0:8245W
Largest W occurs when the moment about the rear axle is zero.
+
)
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 3 P )
= 106:72 10 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 = 160 p
93:02i
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
P=
i
0
93:02
j
0:6
111:63
k
0
66:98
=
55:8k N m J
(b)
MC = rCB
40:2i
2.36
36
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2.37
rOC
=
2i + 4j
3k m
MO
= P
M0
P
p
= P 1:69052 + 1:87372 + 3:62522 = 4:417P = 350 kN m
= 79:2 kN J
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
2.38
2.39
P
= P
2j + 4k
BA = 20 p
( 2)2 + 42
2i + 2j
=
8:944j + 17:889k kN
k
Q = Q AC = 20 p
=
2
2
( 2) + 2 + ( 1)2
!
r = OA = 2i + 4k m
13:333i + 13:333j
6:667k kN
37
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 =
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 + (
P+Q =
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
38
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
39
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
41
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
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(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
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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
45
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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
0 0:8944
0:4472
1:2911F = 150 lb ft
F = 116:2 lb J
BC = 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
46
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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
47
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2.65
2.66
48
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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
0
0
53:67
107:33
0:7660
0
AB =
(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
49
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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 DB = 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
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
51
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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
52
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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
53
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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
=
54
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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
55
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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
=
250k N J
= MO = 250(1:2)i + 250(0:8)j
=
300i + 200j N m J
p
C R = ( 300)2 + 2002 = 361 N m
R
CR
250 N
361 N. m
O
3
2
56
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2.89
2:2i + 2:0j 2:0k
270 AB = 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
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
=
33:28i
22:19k lb
46:30i
77:17j lb ft
i
j
3
5
33:28 0
k
0
22:19
22:19k lb J
R
= P=
CR
= C + rOA
P=
=
10:6j + 166:4k lb ft J
157:3i
33:28i
46:30i
77:17j +
*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
57
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2.92
2.93
F =
C =
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
rBA
=
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
58
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
= F=
CR
= C + rAB
=
206i
12:29i
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
59
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2.97
MO
= rOA
My
=
=
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
60
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b)
MOF
= MO
OF = (1920j + 1400k)
3i + 12j + 4k
13
=
1920(12) + 1400(4)
= 2200 lb ft J
13
=
=
Fx = T1 sin 45
T3 sin 30 = 0
Fy = T1 cos 45 + T3 cos 30 + 250 = 750
2.101
Rx
Ry
The solution is
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
61
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2.104
2.105
2.106
62
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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 = C(
MDE
510 + 0:64C
0
400
0:6
DE =
3
300
0
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
63
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 )
64
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
65
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3.8
R=
CR
F = (240 + 200
rDA
rDB
rDO
=
=
=
=
MD =
i
1:4037
0
+
i
4:596
0
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
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)
66
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
=
3i
4k
5
3j
4k
5
= (0:6i
0:8k)F
= (0:6j
0:8k)F
2:4F
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
CR
R
=
=
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)
68
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
69
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
70
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
71
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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
60(0:12) =
7:2 N m
The resultant is not a force.
+
CR =
MO =
The resultant is the couple C R = 7:2 N m CW J
72
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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
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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
74
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(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
75
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3.32
T1
= T1
T2
= T2
T3
= T3
8j + 4k
p
= 254:4i 678:4j + 339:2k lb
89
4i 8j + 4k
p
= 244:9i 489:9j + 244:9k lb
2 = 600
96
8j 6k
= 320:0j 240:0k lb
3 = 400
10
1 = 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
T2
= T2
T3
= T3
8j + 4k
p
= T1 (0:3180i 0:8480j + 0:4240k)
89
4i 8j + 4k
p
= 253:1i 506:2j + 253:1k lb
2 = 620
96
8j 6k
= T3 ( 0:8j 0:6k)
3 = T3
10
1 = 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
76
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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
a
x =
=
=
R
6P
3
Mx
3P b
b
y =
=
=
R
6P
2
6P k intersecting the plate 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
15 in.
3
4
20 in.
A
R
=
Fx
8 = P1 + P2
20R
=
Mx
3
20(8) = P2 (5) + C
5
15R
=
My
C = 360:0 lb in. J
y
15(8) =
P2 (15)
P1 = 19: 2 lb J
4
C
5
P2 =
11: 2 lb J
77
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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
78
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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
=
400yi + 400xj + (360x
Equating x- and y-components of rOA
R = CR
n yields
400y = 954:3
400x = 396:2
y = 2:39 m
x = 0:991 m
250y)k
) 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
80
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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
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3.48
y
x3
x1
A
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
P1
=
P2
=
P3
=
R
MA
=
P = 3600 + 9600 + 2000 = 15 200 lb # J
= Rx: 3600(4) + 9600(10) + 2000(11:333) = 15 200x
) x = 8:75 ft J
x1 =
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
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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
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3.55
y
x1
4m
P1
6 MN/m
4 MN/m
Ry
y
P2
Rx
2.5 m
y3
x1 =
=
1
(6)(2:5) = 7:5 MN
2
P2
=
4(4) = 16 MN
P3
=
1
(4)(4) = 8 MN
2
2
x
4 MN/m
O
P1
P3 y
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
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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
88
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
30k
502 + 802 + (
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
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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
=
CR
=
1:5i + 2j
= 72:0i + 96:0j N
2:5
1:5i + 2k
180 AB = 180
= 108:0i + 144:0k N m
2:5
120 AC = 120
Component of CR in direction of
CtR
CR
t
AC is
= CR AC = ( 108:0i + 144:0k) ( 0:6i + 0:8j) = 64:80 N m
= CtR AC = 64:80( 0:6i + 0:8j) = 38:88i + 51:84j N m
Component of CR normal to
CR
n
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
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3.70
T1
= T1
T2
= T2
T3
= T3
2j 8k
p
= T1 (0:3419i 0:2279j
77
3j 8k
= T2 (0:3511j 0:9363k)
2 = T2 p
73
4i 8k
p
= 223:6i 447:2k N
3 = 500
80
1 = 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
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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
93
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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
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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
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(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
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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
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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
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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
12
182 + 122
p
300(9:6) + CA = 0
T =0
Ax = 0:5547T
18
T =0
182 + 122
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
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(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
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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
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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
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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
103
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4.33
4.34
4.35
104
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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
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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
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4.40
4.41
107
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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
108
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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
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4.44
4.45
L/2
120 lb θ
B
30 lb
NA
MA
=
0
2L
/3
P
A
+
(120 sin )
2L
cos
3
+ (120 cos )
160 cos
15 = 0
= cos 1
2L
sin
3
15
= 84:6
160
30
L
sin
2
=0
J
110
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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
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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
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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
A
C
B
60o
FBD
o
NA
Fx
Fy
L/2
= 0
= 0
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
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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
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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
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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
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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
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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
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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
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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
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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
2m
C
NC
NC
200(9:81)(1:444) = 0
NC = 490:7 N
FBD of entire structure:
MB = 0
+
5:774NC
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
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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
MB = 0
+
+
6Cx 460(1:75) = 0
+ ! Ax Cx = 0
+ " Ay + Cy 460 = 0
From FBD of AB:
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
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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
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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
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4.84
TC
TC
D
TA
TC
A
3 ft
C
B
x
150 lb
TA
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
20 ft
8 ft
F
Fy
B
Bx
1800 lb
Ey
Ax
A
6 ft
6 ft
Ay
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
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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
+
5:5P
138:75 (12) = 0
P = 303 lb J
FBD of ABC:
MB = 0
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
+
75NC
240(140) = 0
NC = 448 N J
+
90ND
75NC = 0
ND = 373 N J
From FBD (a):
From FBD (b):
MB = 0
4.95
139
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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
180 = 0
200 = 0
NB + 134:4
200 = 0
FBD of man B:
Fy
NB
= 0
+ " NB + TB
= 65:6 lb J
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
400(9:81) = 0
T = 3141 N
FBD of tongs with drum:
1:6
+"
2T p
MC = 0 +
1:4Dx
Tp
1:4Dx
(3141) p
Fy = 0
22 + 1:62
FBD of BCD:
1:6
22 + 1:62
D=
4.100
1:6
(2)
22 + 1:62
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
(2)
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
Ex
60o
E
16 N m
Ey
(b)
(a)
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
=
0
Fx
Fy
= 0
= 0
RA
RC
3
80 = 0
cos 30
+ ! Ax FBC sin 30 = 0
+ # Ay FBC cos 30 = 0
+
FBC
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
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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
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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
+
1:5NB
3
FAC (0:7) = 0
5
Fx
=
0
+!
4
FAC
5
Dx = 0
Fy
=
0
+ # Dy + NB
RA
RD
RE
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
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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
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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
2ND
3:5FBF = 0
ND = 270 lb J
From FBD of DEF :
ME = 0
+
153
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*4.119
4.120
200 lb
200 lb 0.6 ft
0.6 ft
o
B
30o A
Cy
0.75 ft P
AB
Cx
C
θ = 47.99
E
0.75 ft
o
5 ft
3.7
ft
95
.64
0
d= F
D
30
B
PAB
Dx D
0.75 ft
Dy
I
2 ft
II
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
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Geometry (see FBD II):
d
=
0:75 cos 30 = 0:6495 ft
d
0:6495
= 30 + tan 1 = 30 + tan 1
= 47:99
2
2
FBD II (EF is a 2-force member):
MC
=
+
=
PEF
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
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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
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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
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4.126
4.127
158
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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
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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
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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:
1
1
p PBC + 800 = 0
+ " p PBC + NB = 0
2
2
PBC = 1131:4 lb = 1131:4 lb (C) J
1
1
= 0
+
PAB + p PBC = 0
PAB + p ( 1131:4) = 0
2
2
PAB = 800 lb (T) J
Fy
=
Fx
0
FBD of joint C:
Fy
=
2
1
+ # p PAC + p PBC = 0
5
2
PAC = 894 lb (T) J
0
2
1
p PAC + p ( 1131:4) = 0
5
2
161
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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
Fx
4
4
p PAE + Ay = 0
p PAE + 2220 = 0
41
41
PAE = 3554 lb = 3554 lb (C) J
5
5
= 0
+ ! PAB + p PAE = 0
PAB + p ( 3554) = 0
41
41
PAB = 2775 lb (T) J
=
0
+"
FBD of joint B:
Fy
=
0
+"
2
p PBE
5
Fx
=
0
+!
PBC
PBC
1
2775 + p 2683 = 0
5
2400 = 0
PBE = 2683 lb (T) J
1
PAB + p PBE = 0
5
PBC = 1575 lb (T) J
FBD of joint C:
Fy
=
0
+"
2
p PCE
5
Fx
=
0
+!
PCD
PCD
1575
1800 = 0
PBC
1
p 2012 = 0
5
PCE = 2012 lb (T) J
1
p PCE = 0
5
PCD = 2475 lb (T) J
162
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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
L
C
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
Fx
1
1
+ " NC + p PCD = 0
1:5P + p PCD = 0
2
2
PCD = 2:121P = 2:121P (C) J
1
1
PBC + p ( 2:121P ) = 0
= 0
+
PBC + p PCD = 0
2
2
PBC = 1:500P (T) J
=
0
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
=
1
1
p PAD p PCD P = 0
2
2
1
1
p PAD p ( 2:121P ) P = 0
2
2
PAD = 0:7068P = 0:7068P (C) J
0
+
163
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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
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4.138
1200 lb
C4
PAC
PCD
4
3
PCD
PAD 4 3
PAC
D
PBD
4
3
3
A
PAD
PAB
NA
165
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FBD of joint C:
Fx
=
Fy
=
0
+!
4
PCD
5
1200 = 0
PCD = 1500 lb (T) J
3
3
+ # PAC + PCD = 0
PAC + 1500 = 0
5
5
PAC = 900 lb = 900 lb (C) J
0
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
=
4
+ ! PAB + PAD = 0
5
PAB = 1200 lb (T) J
0
PAB +
4
( 1500) = 0
5
4.139
166
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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
PBD = 0
L2 + 62
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
PAD A
45o
PDE
PAE 45
PDE
o
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 sin 45
PAD = 0
PAB = 59:40 kN
= 0
+ # PAE + PAB cos 45 = 0
PAE = 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
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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
MG
MA
3PF G 7250(6) = 0
PF G = 14 500 lb (T) J
2
p PBC (6) + 7250(12) 5000(6) = 0
= 0
+
5
PBC = 10 620 lb = 10 620 lb (C) J
1
p PBG (12) + 5000(6) = 0
= 0
+
5
PBG = 5590 lb = 5590 lb (C) J
=
0
+
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
24
E
PEF
PNF
N
NR
F
1
1
R
PNO
6m
6m
1
6.25 m
36 m
J 12 m
Jy
400 kN
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
Fx0
+
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
PF G = 210:0 kN = 210:0 kN (C) J
1
= 0
+ % p PCF 42 = 0
PCF = 59:4 kN (T) J
2
=
0
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
ME
=
0
+
8PCD 5(18) 3(24) = 0
4
p PDF (6) + 5(6) = 0
= 0
+
97
PDF = 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
MB
MF
4:154)
1500(4:154) = 0
a:
2
p PBF (6) 1200(4:154) = 0
5
PBF = 929 lb (T) J
2
p PEF (6) 1200(6 4:154) + 1269:2(6) = 0
= 0
+
13
PEF = 1622 lb = 1622 lb (C) J
= 0
+
6PBC + 1200(9 4:154) 1269:2(9) = 0
PBC = 935 lb (T) J
=
0
+
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
3PDE 30(4) = 0
PDE = 40 kips (T) J
PAD = 0 J
4
4
= 0
+ ! PDE + PAC = 0
40 + PAC = 0
5
5
PAC = 50 kips = 50 kips (C) J
= 0
= 0
+
+
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
+
PCF = 0 J
0
+
9PDF + 1500(18) + 1500(27) = 0
PDF = 7500 lb = 7500 lb (C) J
= 0
+
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
Fy
Fx
=
0
+
14Gx (14 500 10 000)(16) = 0
Gx = 5143 lb
5
= 0
+ " p PBE + 14 500 10 000 = 0
89
PBE = 8491 lb = 8491 lb (C) J
8
= 0
+
PBC + p PBE + Gx = 0
89
8
PBC + 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
30o
A
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
8 ft
12 ft
B
B
NB
NB
Ax
Ay
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
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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
CD
9 ft
D
3 ft
12 ft
2000 lb
C P
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
+
4FCD 3NA = 0
4FCD 3(1150) = 0
FCD = 862:5 lb
= 0
+
Ex N A = 0
Ex 1150 = 0
Ex = 1150 lb
= 0
+ # 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
N + 19:333
30 = 0
FBD of block C:
Fy
N
= 0
+" N +T
= 10:67 lb J
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
Fx
4
PDE + 12 000 = 0
5
3
= 0
+
PEF + PDE = 0
5
PEF = 9000 lb (T) J
=
0
+"
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
+
18NC
6W = 0
NC =
1
W
3
=
=
0
0
FBD of seat CD:
MD = 0
FBD of member ABC:
1600
6W
MB
3W
+
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
TC (32) + p
20
TB (44)
202 + 222
20
202 + 222
(29:73)(44)
80(22) = 0
1760 = 0
TC = 27:50 lb J
Fz
=
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
0
TD + TC + p
Fx
Fy
Fz
Mx
My
Mz
=
=
=
=
=
=
5.19
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
Az =
Bx = 5 lb
The solution is
6 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
=
p
0
2:8
22 + 3:52 + 2:82
T = 473:3 lb J
Fx
=
0
Ax + p
T
2
22 + 3:52 + 2:82
(3:5)
270(3:5) = 0
T =0
2
(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
Ax + p
MBC
(MC )z
5.21
Mx
Mz
Fy
My
Fz
Fx
= 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
209:5 lb
Cz = 74:76 lb
p
2
209:5 + 74:762 = 222 lb J
=
600(5) 6PBG = 0
PBG = 500 lb (T) J
PAF = 0 J
600(2) 6PAE = 0
PAE = 200 lb = 200 lb (C) 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
5.26
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
= PBD p
Mx
=
=
PBD
=
=
=
210
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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
PAE
B
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
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
BC + rBA
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
215
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
218
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 2
T 3
C
Bx
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
230
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
3
C
2.6
m
x
MAD = rAC
TCB
W
rAC
AD
MAD
m
Ax
m
4.5
A
3m
y
.2 m
D 2
AD + rAG
TCB
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
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Mx
My
Fz
PB
3
(8000) = 0
PC = 1200 lb (T) J
5
4
= 0
6PA + (8000) = 0
PA = 1067 lb (T) J
5
= 0
PA + PB + PC = 0
=
PA PC = 1067 1200 = 2267 lb = 2270 lb (C) J
=
0
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
x
Ax
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
AC + rAE
=
0
0
0:8165 0:4083 TBD +
0
0:8321
=
TBD
5
0:4083
0:5547
( 4k)
0:8165j + 0:4083k) TBD
+3(0:8321)
3:397TBD 6:656 + 2:496 = 0
AC + 3k
AC
0
3
0
0
0:5547 0
0
4
0:8321
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):
= 0
= 0
= 0
P1 = 48 lb J
V1 = 0 J
6P1 M1 = 0
= 0
= 0
= 0
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
Fx
Fy
MO
M1 = 6(48) = 288 lb in. J
FBD (b):
FP 2
FV1
MO
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
NE
0.9 m
NE = 1466.7 N
1.8 m
1.8 m
E
A
Ay
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
245
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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
6 ft
P2
B
2800 lb
(c)
9T = 0
T = 4667 lb
V2
FBD (a):
MC = 0
2800(15)
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
247
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.16
4
D
3
PCD
24 in.
PBE
E
12 in.
P1
M1
E
V1
12 in.
1080 lb. in.
1080 lb. in. F
Fx 1080 lb. in.
F
Fy
60 lb
60 lb
18 in.
Ay
P2
G
M2
B
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
248
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
252
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
100(6x
M
=
14 400
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)
A
P1 x/3
(6 - x)/3
200x 200(6 - x)
x
M
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
<
12 kN if
0
if
:
12 kN if
x 2m
2m x 6m
x 6m
12x kN m
24 kN m
12(8 x) kN m
if
if
if
J
x 2m
2m x 6m
x 6m
J
261
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.33
4 kN/m
y
A
B
2m
C
3m
x
32.7 kN. m
V (kN)
-4
-16
M (kN. m)
-2.67
-32.7
x-4
3
(4x -8) kN
4 kN
2x kN/m
x
3
x kN
2
A
M
x
x-2
2
A
M
B
2m
V
4 kN/m
V
x
Segment AB:
V
=
M
=
x2 kN J
1 3
x kN m J
3
Segment BC:
V
=
M
=
(4x 8) = 4(1 x) kN J
4
x 2
4(x
) (4x 8)
=
3
2
4
Mmax = M jx=5 m =
2(52 ) + 4(5)
2x2 + 4x
8
=
3
8
kN m J
3
32:7 kN m
262
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.34
263
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.35
264
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.36
10 kips
A
B
3 ft
RA = 10 kips
.
C 30 kip ft D
3 ft
3 ft
RD= 0
10
V (kips)
30
M (kip. ft)
10 kips
A
10 kips
x
M
A
B
3 ft
M
x
V 10 kips
M
V
9-x
D
V
8
x 3 ft
< 10 kips if
0
if 3 ft x 6 ft J
V =
:
0
if
x 6 ft
8
x 3 ft
< 10x kip ft if
30 kip ft if 3 ft x 6 ft kip ft J
M=
:
0
if
x 6 ft
265
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.37
y
16 kN/m
8m C
B
A
4m
32 kN
32
V (kN)
D
x
4m
160 kN
64
6 ft
-96
160
M (kN. m)
-128
16(x - 4)
A
M
x
A
4m
M M
x
V 32 kN
32 kN
V =
M=
8
<
V
8
<
32 kN
if
32 16(x 4) kN if
:
16(16 x) kN
if
16 - x
8(6
D
V
x 4m
4 m x 12 m
x 12 m
32x kN m
if
32x 8(x 4)2 kN m if
:
8(16 x)2 kN m
if
Mmax = M jx=6 m = 32(6)
16(16 - x)
J
x 4m
4 m x 12 m
x 12 m
J
4)2 = 160 kN m
266
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.38
y
A
B
750 lb
6 ft
300 lb
60 lb/ft
x
C
9 ft
378
V (lb)
-300
-372
M (lb . ft)
-1944
x
2x lb 3
2
A
2x2 lb
4x lb/ft
M
x
A
300 lb
V
300 lb
x
3
B
6 ft
750 lb
x
4x lb/ft
M
V
Segment AB:
V
=
M
=
2x2 lb J
2 3
300x
x lb ft J
3
300
Segment BC:
V
=
M
=
450
2x2 lb J
300x + 750(x
6)
2 3
x lb ft J
3
267
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6.39
6.40
268
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.41
y
24w0 lb 4 ft
b/2 w0b
w0 lb/ft
Ax
A
Ay
16 ft
B
NB
x
8 ft
A
b
Ay = 6w0 lb
MD
D
VD
FBD of entire beam:
Fx
MB
= 0
= 0
Ax = 0
16Ay 24w0 (4) = 0
Ay = 6w0
FBD of segment AD:
MD = 0
MD
6w0 (b) + w0 b
b
2
=0
MD =
w0 b
(12
2
b)
Note that MD = 0 when b = 12 ft. Therefore, the most advantageous position
for the joint is at b = 12 ft J
269
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.42
Case 1
P
y
L/2
A
L/2
P/2
M
A
B
L −x
x
P/2
V
P/2
x
P/2
V
M
B
The reactions were determined by symmetry (each support carries half the load).
From FBD of left segment (0 < x < L=2):
M=
P
x J
2
V =
P
J
2
From FBD of right segment (L=2 < x < L):
M=
P
(L
2
x) J
P
J
2
V =
M
PL/4
V
P/2
L/2
x
L/2
−P/2
x
L/2
Case 2
y
P
x
A
Ax
B
L/2
L/2
1
Ay
M
A
L/2
P
L (L − x)
x > L/2
NB
V
M
L/2
V
x
x < L/2
B
Px
L
270
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FBD of entire beam:
MA
=
0
NB L
Px = 0
Fx
=
0
Ax = 0
Fy
=
0
Ay + NB
NB =
P =0
P
x
L
Ay +
P
x
L
P =0
Ay =
P
(L
L
x)
When P is in the range 0 < x < L=2 (use FBD of right half):
M=
P
x
L
L
2
=
P
x J
2
V =
P
x
L
When P is in the range L=2 < x < L (use FBD of left half):
M=
P
(L
L
x)
L
2
=
x) J
V =
P
(L
L
x) J
M
PL/4
V
P/2
L/2
−P/2
P
(L
2
L/2
x
L/2
x
271
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6.43
400 lb
1200 lb
800 lb
4 ft
4 ft
4 ft
4 ft
1100 lb
1300 lb
1100
700
4400
2800
V (lb)
-2000
-500
-5200
-1300
7200
5200
4400
M (lb. ft)
Vmax =
1300 lb J
Mmax = 7200 lb ft J
272
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.44
273
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.45
274
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.46
2 kN/m
40
2m
2m
12
8
6m
4m
14
6
8
6
12
12
16
9
V (kN)
-9
3m
-6
12
M (kN.m)
-7
-28
-16
-16
Vmax = 8 kN J
Mmax =
28 kN m J
275
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.47
276
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.48
240 lb/ft
2880 lb
720 lb
A
6 ft
C
12 ft
B
2280 lb
1320 lb
5070
6.5 ft
-720
-3630
1560
-1440
V (lb)
-1320
3630
M (lb . ft)
-1440
Vmax = 1560 lb J
Mmax = 3630 lb ft J
277
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.49
278
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.50
40 kN/m
120 kN
B
A
2m
120 kN. m
C
3m
60 kN
60 kN
60
120
V (kN)
90
1.5 m
-90
-60
120
90
M (kN. m)
Vmax = 60 kN J
Mmax = 120 kN m J
279
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.51
2500
15 000
3 ft
4.5 ft
1000
3 ft
1500
2500
7500
1000
3000
4500
V (lb)
3000
M (lb. ft)
-7500
-12 000
Vmax = 2500 lb J
Mmax =
12 000 lb ft J
280
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.52
281
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.53
12 kN/m
24 kN
36 kN
A
B
2m
C
3m
60 kN
25 kN
D
3m
25 kN
25
V (kN)
1
75
0.042
-24
-51.05
-24
-35
0.083 m
M (kN. m)
-24 -23.96
-75
Vmax =
35 kN J
Mmax =
75 kN m J
282
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.54
11 kN/m
-44
2m
1m
4m
48
24
24 kN/m
24 kN/m
24 kN
4 kN
24
12
26.18
-8.333
V (kN)
4
-0.333
-18.18
-12
-24 1 m 1 m
2.182 m
1.818 m
-20
0.1667 m
26.18
8
M (kN .m)
-0.333
-12
Vmax =
24 kN J
Mmax = 26:2 kN m J
283
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.55
284
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.56
285
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.57
2500 lb
5500 lb
4.5 ft
3 ft
3 ft
1250 lb
3 ft
6250 lb
3000 lb
3000
9000
3750
500
V (lb)
1250
2250
-15 000
-5000
11 250
9000
M (lb. ft)
-3750
Vmax =
5000 lb J
Mmax = 11 250 lb ft J
286
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.58
287
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.59
..
288
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.60
900 lb/ft
4 ft
5 ft
5 ft
885 lb
5415 lb
3615
0.983 ft
7261
V (lb)
-2400
-1800
-435
4861
-4425
-885
4426
M (kN.m)
-2400
289
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.61
290
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.62
6.63
The shape of the cable is parabolic. From Eq. (6.10):
H = yjx=L=2 =
w0 (L=2)2
2T0
T0 =
1402
L2
w0 =
w0 = 122:50w0 N
8H
8(20)
From Eq. (6.9):
Tmax
= T jx=L=2 =
=
)
s
w0 L
2
T02 +
2
= w0
s
122:502 +
140:0
2
2
141:09w0 N
Tmax
4 106
w0 =
=
= 28 400 N/m J
141:09
141:09
6.64
The shape of the cable is parabolic. From Eq. (6.10):
w0 (L=2)2
2T0
L2
19902
w0 =
(444:7
8H
8(233)
H
= yjx=L=2 =
T0
=
103 ) = 944:8
106 N
From Eq. (6.9):
Tmax
= T jx=L=2 =
s
=
106 ) +
=
s
(944:8
1:043
T02 +
2
w0 L
2
2
(444:7
103 )1990
2
2
109 N = 1:043 GN J
291
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.65
Tmax
y
22.5 m
22.5 m
B
30 m
180 kN
To
x
A
The shape of the cable is parabolic.
(a)
W = w0 L = 4(45) = 180 kN
MB = 0
180(22:5)
30T0 = 0
T0 = 135 kN
From Eq. (6.12):
Tmax =
q
p
T02 + W 2 = 1352 + 1802 = 225 kN J
(b) From Eq. (6.12):
s
L
s(L) =
1+
2
w0 L
T0
=
4(45)
4
=
135
3
)
45
s(L) =
2
=
56:0 m J
w0 L
T0
s
1+
2
2
s
1 T0 4 w0 L
ln
+ 1+
+
2 w0
T0
4
3
2
w0 L
T0
0
s
1 135 @ 4
+
ln
+ 1+
2 4
3
4
3
2
3
5
2
1
A
6.66
292
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.67
1200 lb
B
30 ft
θA
50 ft
A
TA
MA
720 lb
θB
60 ft
= 0
(1200 cos B ) (50) + (1200 sin B ) (60) 720(30) = 0
72 sin B 60 cos B = 21:6
J
B = 53:13
Fx
Fy
=
0
1200 cos B TA cos A = 0
TA cos A = 1200 cos 53:13 = 720:0
= 0
1200 sin B TA sin A 720 = 0
TA sin A = 1200 sin 53:13
720 = 240:0
tan A
=
TA
=
240
= 0:3333
J
A = 18:43
720
cos B
cos 53:13
1200
= 1200
= 759 lb J
cos A
cos 18:43
293
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.68
The shape of the string is a catenary.
3.2 lb
B
θ
FBD
Η
2.8 lb O
w0s
L
w0 =
0:4
= 0:025 lb/ft
16
From FBD:
Fx
Fy
=
=
0
+ ! 3:2 cos
2:8 = 0
0
+ # w0 s 3:2 sin = 0
s = 61:98 ft J
= 28:96
0:025s 3:2 sin 28:96 = 0
From Eq. (6.15):
s =
)L =
T0
w0 L
sinh
w0
T0
61:98(0:025)
T0
2:8
1 sw0
sinh 1
= 59:19 ft
sinh
=
w0
T0
0:025
2:8
From Eq. (6.16):
H
T0
w0
16:01 ft J
= yjx=L =
=
cosh
w0 L
T0
1
=
2:8
0:025
cosh
0:025(59:19)
2:8
1
6.69
6.70
This is a catenary cable.
294
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(a) From Prob. 6.69:
T
= T0 + w0 y
) TB = T0 + w0 yB
400 = T0 + 16(6)
T0 = 304 N
) TA = T0 + w0 yA = 304 + 16(3) = 352 N J
(b) From Eq. (6.17):
w0 x
T0
)x=
w0 xB
T0
1800 = 1200 cosh
T0
T
cosh 1
w0
T0
TA
TB
T0
cosh 1
+ cosh 1
L = xA + xB =
w0
T0
T0
304
352
400
=
cosh 1
+ cosh 1
= 25:3 m J
16
304
304
T
= T0 cosh
6.71
From Eq. (6.17):
TB
= TO cosh
1:5
=
cosh
w0
53:33
w0 (22:5)
1200
w0 = 53:33 cosh 1 1:5 = 51:33 N/m
From Eq. (6.15):
sB
W
TO
51:33(22:5)
w0 xB
1200
sinh
= 26:14 m
sinh
=
w0
T0
51:33
1200
= w0 (2sB ) = 51:33(2)(26:14) = 2680 N J
=
6.72
Eq. (6.16):
Given:
H
Tmax
wO x
dy
= sinh
dx
TO
dy
24(22:5)
= tan 20
) tan 20 = sinh
dx x=22:5
TO
24(22:5)
sinh 1 (tan 20 ) =
TO
24(22:5)
TO =
= 1515:2 N
sinh 1 (tan 20 )
y=
TO
wO
cosh
wO x
TO
1
)
1515:2
24(22:5)
cosh
1
24
1515:2
TO
1515:2
= TB =
=
= 1612 N J
cos 20
cos 20
=
yjx=22:5 =
= 4:05 m J
295
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.73
w0 = 1:8(9:81) = 17:658 N/m
From Eq. (16.17):
Tmax
w0 (L=2)
17:658(9)
40(9:81) = T0 cosh
T0
T0
158:92 392:4
f (T0 ) = cosh
=0
T0
T0
= T0 cosh
f(T0) 4
2
0
100
200
300
400
500
T0
-2
The plot of f (T0 ) shows two roots which can be computed numerically. They
are T0 = 63:38 N and 356:4 N.
From Eq. (6.16):
H
) H1
=
H2
=
T0
w0
w0 L
1
2T0
17:658(18)
63:38
cosh
1 = 18:58 m J
17:658
2(63:38)
356:4
17:658(18)
cosh
1 = 2:04 m J
17:658
2(356:4)
= yjx=L=2 =
cosh
6.74
y
TB
L/2 = 9 m
B
H = 1.6 m
T0
W
x
O
The shape of the cable is caternary.
w0 = 1:2(9:81) = 11:772 N/m
296
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Equation (6.16):
y
=
1:6
=
T0
w0 x
cosh
1
w0
T0
T0
w0
cosh 9
1
w0
T0
T0
w0 (L=2)
cosh
1
w0
T0
w0
w0
1:6
= cosh 9
1
T0
T0
)H=
Solving numerically for ! 0 =T0 yields
w0
= 0:03910
T0
) T0 =
11:772
= 301:1 N
0:03910
Equation (6.17):
T
M
= T0 cosh
w0 x
T0
w0 L
T0 2
)
TB = T0 cosh
=
319:9
TB
=
= 32:6 kg J
g
9:81
= 301:1 cosh [0:03910(9)] = 319:9 N
6.75
40 lb
A
14o
FBD
s
T0 O
w0 s
L
The shape of the hose is a catenary. From FBD:
Fy
Fx
=
0
+ # w0 s = 40 sin 14
40 sin 14
40 sin 14
s=
=
= 19:35 ft J
w0
0:5
= 0
+
T0 40 cos 14 = 0
T0 = 38:81 lb
From Eq. (6.15):
s =
)L =
T0
w0 L
sinh
w0
T0
T0
sw0
38:81
19:35(0:5)
sinh 1
=
sinh 1
= 19:15 ft J
w0
T0
0:5
38:81
297
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.76
y
TO
1.2 lb
O
w0 =
7.5 lb
θ x
B
L/2
W
2:4
=
= 0:048 lb/ft
Length
50
From FBD:
Fy
=
0
Fx
TO
w0
=
0
7:5 cos
TO = 0
7:403
= 154:23 ft
0:048
=
7:5 sin
1:2
= 9:207
7:5
TO = 7:5 cos 9:207 = 7:403 lb
= sin 1
1:2 = 0
Eq. (6.15):
TO
w0 x
sinh
w0
TO
TO
w0 s
sinh 1
w0
TO
L
25
1
= xjs=25 = 154:23 sinh
= 24:892 ft
2
154:23
L = 2 (24:892) = 49:78 ft J
s =
)x=
6.77
By
30 N
B
5m
W/2
O
T0
L/2
From FBD:
Fx = 0
+
T0
30 = 0
T0 = 30 N
From Eq. (6.16):
yjx=L=2
=
w0 L
T0
cosh
w0
2T0
L = 28:5 m J
1
5=
30
5:2L
cosh
5:2
2(60)
1
298
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.78
V = 125 lb
A
y
TO
W = 125 lb
H
TO
O
x
20 ft
W = w0 sA = 5(25) = 125 lb
From Eq. (6.15):
TO
w0 xA
TO
20w0
TO
sinh
25 =
sinh
)
= 16:91 ft
w0
TO
w0
TO
w0
= 16:91w0 = 16:91(5) = 84:55 lb
q
p
= TA = TO2 + V 2 = 1252 + 84:552 = 150:9 lb J
sA
=
TO
Tmax
From Eq. (6.16):
H=
TO
w0
cosh
w0 xA
TO
1
= 16:91 cosh
20
16:91
1
= 13:27 ft I
6.79
y
V0
xB
50 ft
C
T0 = 170 lb
h
T0 = 170 lb B
O
55 lb
Each cable is a catenary.
TB =
Equation (6.17):
W
x
p
1702 + 552 = 178:68 lb
w0 xB
T0
T0
TB
170
178:68
cosh 1
=
cosh 1
= 67:62 ft
w0
T0
0:8
170
TB
= T0 cosh
xB
=
299
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Equation (6.15):
T0
w0 x
sinh
w0
T0
T0
w0 xC
w0 xB
= sC sB =
sinh
sinh
w0
T0
T0
170
0:8(117:62)
0:8(67:62)
=
sinh
sinh
0:8
170
170
s =
sBC
= 55:0 ft J
Equation (6.16):
T0
w0
w0 x
1
T0
w0 xB
w0 xC
T0
cosh
cosh
= y C yB =
w0
T0
T0
0:8(117:62)
0:8(67:62)
170
cosh
cosh
= 22:5 ft J
=
0:8
170
170
y
=
h
cosh
6.80
y
80 lb
B θ1
60 lb
y
TBC
x
θ1
TBC
TCD
θ2
C
x
60 lb
FBD of joint B:
Fx
Fy
1
= 0
= 0
=
FBD of joint C:
Fx
Fy
2
TBC cos 1 80 = 0
TBC sin 1 60 = 0
p
60
tan 1
= 36:87
TBC = 802 + 602 = 100 lb J
80
= 0
= 0
=
TCD cos 2 TBC cos 1 = 0
TCD cos 2 80 = 0
TCD sin 2 TBC sin 1 60 = 0
TCD sin 2 120 = 0
p
120
tan 1
= 56:31
TCD = 802 + 1202 = 144:2 lb J
80
h = sBC sin 1 + sCD sin 2 = 8(sin 36:87 + sin 56:31 ) = 11:46 ft J
300
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.81
4 ft D 3
6 ft
TCD
2
6 ft
h
C
P
B
40 lb
40 lb
From FBD:
Fy
Fx
MD
3
p TCD 80 = 0
TCD = 96:15 lb J
13
2
2
= 0
+
P p TCD = 0
P p (96:15) = 0
13
13
P = 53:33 lb J
= 0
+
P h 40(10) 40(4) = 0
55:33h 560 = 0
h = 10:121 ft J
=
0
+"
Horizontal component of force in the cable is P:
)
6
p
62 + (h
6)2
6
p
2
6 + (10:121
TBC = P
TBC = 64:7 lb J
6)2
TBC = 53:33
6.82
TAB
A
3 ft
4 ft
4 ft
β1 = _θ1
4 ft
B
TCD
D
β3= θ
β2= θ2
C
3
150 lb
200 lb
Geometry:
MD
=
)
1 = 45
J
0
+
(TAB sin 45 ) (11) 200(7)
TAB = 237:9 lb J
T0 = 237:9 cos 45 = 168:22 lb
150(3) = 0
301
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Using Eqs. (6.19):
T0 (tan 2
T0 (tan 3
tan 1 ) = W1
168:22(tan 2 + 1) = 200
tan 2 = 0:18892
J
2 = 10:698
tan 2 ) = W2
168:22(tan 3
tan 3 = 1:0806
3 = 47:22
0:18892) = 150
J
Using Eqs.(6.18):
TBC
=
TCD
=
T0
168:22
=
= 171:2 lb J
cos 2
cos 10:698
T0
168:22
=
= 247:7 lb J
cos 3
cos 47:22
6.83
Ay
Ey
T0
8 ft
8 ft
T0
12 ft
12 ft
E
A
14 ft
B
D
400 lb
C
400 lb
400 lb
520 lb
TAB 680 lb
T0
8 ft
8 ft
12 ft
12 ft
A
E
B
VCD
400 lb
C
14 ft
TBC
TCD
T0 480 lb
400 lb
14 ft
VBC
D
C
400 lb
TDE
480 lb
400 lb
FBD of entire cable:
ME
Fy
=
=
0
400(12 + 24 + 32) 40Ay = 0
Ay = 680 lb
0
Ay + Ey 3(400) = 0
680 + Ey 1200 = 0
Ey = 520 lb
302
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FBD of portion ABC:
Fy
MA
= 0
VCD + 680 2(400) = 0
VCD = 120 lb
= 0
16VCD + 14T0 400(8 + 16) = 0
16(120) + 14T0 400(8 + 16) = 0
T0 = 548:6 lb
FBD of portion CDE:
Fy = 0
TAB
TBC
TCD
TDE
6.84
VBC + 520
q
2(400) = 0
VBC = 280 lb
p
548:62 + 6802 = 874 lb J
q
p
2 =
T02 + VBC
548:62 + 2802 = 616 lb J
=
q
p
2 =
=
T02 + VCD
548:62 + 1202 = 562 lb J
q
p
=
T02 + Ey2 = 548:62 + 5202 = 756 lb J
=
T02 + A2y =
303
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6.85
6.86
A 2m
4m
4m
4m
TAB
D
TCD
B
C
12 kN
P
304
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MC
Fx
Fy
2
p TAB (4) 12(4) = 0
TAB = 26:83 kN J
20
2
4
= 0
+ ! p TCD p TAB = 0
32
20
4
2
p TCD p (26:83) = 0
TCD = 16:969 kN J
32
20
4
4
= 0
+ # P p TAB p TCD 12 = 0
20
32
4
4
P p (26:83) p (16:969) 12 = 0
P = 48:0 kN J
20
32
=
0
+
Horizontal component of force in the cable is constant.
2
2
) TBC = p TAB = p (26:83) = 12:00 kN J
20
20
6.87
TAB
A
3 ft
P
6 ft
B
TCD 3 ft
D
3 ft
4 ft
C
80 lb
MB
=
0
+
80(2)
4
TCD (3)
5
3
TCD (6) = 0
5
TCD = 26:67 lb
MA
=
0
+
3P
80(4)
3P
80(4)
4
TCD (6) = 0
5
4
(26:67) (6) = 0
5
P = 149:3 lb J
305
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6.88
6.89
306
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6.90
5m
2m
A
θ1
11 m
H
θ2
C
T
B
P θ
1
B
T
θ2
14 kN
Geometry:
sAB + sBC
1
2
p
p
20 m
52 + (H 2)2 + 112 + H 2 = 2
Solution is H = 6:912 m
6:912 2
H 2
= tan 1
= 44:49
= tan 1
5
5
6:912
H
= tan 1
= 32:14
= tan 1
11
11
=
FBD of joint B:
Fy
Fx
=
0
T (sin 1 + sin 2 ) 14 = 0
T (sin 44:49 + sin 32:14 ) 14 = 0
T = 11:356 kN
= 0
T (cos 2 cos 1 ) P = 0
11:356(cos 32:14
cos 44:49 ) P = 0
P = 1:515 kN J
6.91
T0 = 20 cos 32 = 16:961 kN
307
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(a) From Eqs. (6.19):
T0 (tan 2
T0 (tan 3
tan 1 )
= 12
16:961 [tan 2
=
4:72
J
2
tan 2 ) = 18
16:961(tan 3
= 48:84 J
3
tan( 32 )] = 12
tan 4:72 ) = 18
(b)
TAB
TBC
TCD
20 kN J
T0
16:961
= 17:02 kN J
=
=
cos 2
cos 4:72
T0
16:961
=
= 25:77 kN J
=
cos 3
cos 48:84
=
(c)
L = 6(cos 1 + cos 2 + cos 3 )
= 6 [cos( 32 ) + cos 4:72 + cos 48:84 ] = 15:02 m J
H = 6( sin 1 + sin 2 + sin 3 )
= 6 [ sin 32 + sin 4:72 + sin 48:84 ] = 1:83 m J
308
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Chapter 7
7.1
7.2
309
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7.3
12 in. 12 in
14 lb
NB 14 lb
FB FB B
P
B
NB
20 in.
FA
FC
A
FD
C
NA
D
ND
NC
Assume impending sliding at B, C and D.
) FB = 0:2NB
FC = 0:35NC
FD = 0:35ND
From FBD of left chair:
MA = 0
+
0:2NB (20)
NB (24) + 14(12) = 0
NB = 8:400 lb
From FBD of right chair:
Fy
=
0
+ " NC + ND NB 14 = 0
NC + ND 8:400 14 = 0
NC + ND = 22:40 lb
= 0
+ ! P 0:35(NC + ND ) 0:2NB = 0
P 0:35(22:40) 0:2(8:400) = 0
P = 9:52 lb J
Fx
Check for sliding at A:
From FBD of left chair:
Fy
Fx
=
0
+ " NA + NB 14 = 0
NA = 5:600 lb
= 0
+
FA 0:2NB = 0
FA = 1:6800 lb
1:6800
FA
=
= 0:3 <
NA
5:600
A
NA + 8:400
FA
14 = 0
0:2(8:400) = 0
) The assumption was correct
310
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7.4
FA
NA
.
12 in
A
.
12 in
B 1
2 in.
θ
θ
8 lb
10 lb
By
B
Bx
12 in
.
θ
10 lb
FC
12 in
.
NC
C
FC
12 in
.
C
NC
From FBD of assembly:
MA
Fy
Fx
=
0
+
FC (48 cos ) 8(12 cos ) 10(36 cos ) = 0
FC (48) 96 360 = 0
FC = 9:50 lb
= 0
+ " FA + FC 8 10 = 0
FA + 9:50 8 10 = 0
FA = 8:50 lb
= 0
NA = NC
Impending sliding will …rst occur at C since FC =NC > FA =NA ( s being the
same at C and A).
) FC = 0:25NC
NC =
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
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7.5
7.6
W
0.5NA
ft
1.5
ft
1.5
y
θ
0.5NB
B
NB
x
A
NA
312
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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:
2
p T =0
5
Fx
=
0
+!
0:2NB
Fy
=
0
+"
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
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Check for sliding at C:
Fy
=
FC
NC
1
p T
5
0
+"
NC
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
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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
315
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7.10
316
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7.11
7.12
2 lb
Ax
A
9 in.
12 in.
Ay
B FB
NB
y
NB
B
FB
4 lb
x
8 in.
P
D
FC
2 in.
C
NC
317
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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
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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
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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
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7.16
7.17
NB
B NB
L/2
FB
L/2
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
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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
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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
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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
s N = 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
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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 NA b
s = 0:414
W
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
= 0
= 0
MA
=
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
+ " N 120 = 0
N = 120 lb
+
P (84) 120(36) = 0
P = 51:43 lb J
+! P
51:43
sN = 0
s (120) = 0
s = 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
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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
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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
s W cos
(a)
(b)
Equating the right-hand sides of Eqs. (a) and (b):
W sin
s W cos
=
1
sin
2
=
b
1
W sin +
W cos
2
2h
b
b
cos
= tan 1 2 s +
s+
2h
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
FB
NA
FB
x
C
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 =
B N = 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
R1
o
30 + φs
60o− φs
60o
L/2
30o
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
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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
W φs
θ
R
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
0
0
R2 cos 12
R2 sin 12
R1 sin 27
200 = 0
R1 cos 27 = 0
R2 = 229 lb
P = 48:6 lb J
FBD of block:
Fy
Fx
=
=
+"
+
Solution is
R1 = 53:5 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
D
O
20
"
96"
θ
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
sin k
sin 10:620
=
20
96
20
tan k = tan 62:20 = 1:897 J
k = 62:20
7.54
p = 2 r tan
= tan 1
0:5
p
= tan 1
= 2:604
2 r
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
Cdown
= W r tan( s
) = 4000(1:75) tan(8:5
= 722:9 lb in. = 60:2 lb ft J
(b)
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
p = 0:0016 m
= tan 1
Lead angle:
s
=
tan 1
s = 0:2
0:0016
p
= tan 1
= 3:643
2 r
2 (0:004)
s = tan
1
(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
4
3
θB
B
The angles of contact are
1
B = tan
4
3
= 0:927 rad
C =
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
390
= 34e 0:6 180 = 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
= T e 0:25(2:182) = 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 1
A
=
2
B
=
2
4
= rad
8
3
2 =2
=
2
2
3
3
=
=
6
4
3
rad
rad
Consider impending slipping:
T2
T1
P
= 150e s B = 150e 0:3( =6) = 128:20 lb
= T2 e s A = 128:20e 0:3(4 =3) = 36:49 lb
= T1 e s B = 36:49e 0:3( =6) = 31:2 lb J
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
3
α
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:
=
tan 1
=
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 R
Z R
0
=
2 P
R2
Z R
p0 1
0
r
r3
R2
dr = 2 p0
p=
2 P
R2
1
0
p0
p(2 r dr) =
r2
2
r2
R2
(2 r)dr
R
r4
1
=
p0 R 2
4 0
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 =
4 s
P
R2
A
=
Z R
Z R
s
0
r2
0
2 P
R2
r4
R2
r2
R2
1
dr = 4 s
r(2 r dr)
R
P r3
R2 3
8
r5
=
2
5R 0
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 RZ
0
=
2 kp
r
dF = µk p dA
θ
=2
dF cos =
kp
=2
Z R
rdr =
Z RZ
0
k pR
2
0
=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
2 (P + W ) R03 Ri3
2 A P R0
+ B
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
=
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 r P = 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
W
x
P
30o
o
30
φ
a
x
φ
a
N
N
a
= sin 1
R
d = sin
(a)
= sin 1
R
R
(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
FB
Dimensions
in inches
NB
FA = ( r )A NA = 0:12NA
MC
29
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
x
L/2
C
L/2
y
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
354
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
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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
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7.87
7.88
C
W φs
h
RA
φs
B
RB
G
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 = 2 tan
s+h
s = 4 s = 4(0:5) = 2:00 m
J
7.89
357
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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)
tanh 1
s = tan
1
0:4 = 21:80
From FBD of member ABC:
MC = 0
(a) C0
(b) C0
5W
28(12:5) = 0
W = 70 lb (axial thrust on screw)
= W r tan( s + ) = 70(0:3) tan(21:80 + 4:85 ) = 10:54 lb in. J
= W r tan( s
) = 70(0:3) tan(21:80
4:85 ) = 6:40 lb in. J
7.91
358
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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
= tan 18 = 0:325 J
s = tan
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
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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
360
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Chapter 8
8.1
8.2
361
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8.3
y
x
y = 30x
dx
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 120
1
=
yel dA =
30x
2
A
0
120
0:25x2
0
=
Qx
0:25x2 dx =
36 000 mm3
Z
Z 120 p
30x3=2
=
x dA =
30 2
x
4
120
0:25 3
x
6
0
=
Qy
=
x =
A
0:5x2 dx =
0
2p
30x5=2
5
120
0:5 3
x
3
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
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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
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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
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8.8
y
x2 + y2 = 1
a 2 b2
y
0
a
x
dx
r
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
2
y=
2ab
Qx
4b
= 3ab =
J
A
3
2
365
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8.9
8.10
366
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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
1
p
1
1 p
L =
ds =
1 + 4x2 dx =
ln 2x + 4x2 + 1 + x 4x2 + 1
4
2
L
0
0
= 1:4789 ft
Z
Z 1 p
Qx =
y ds =
x2 1 + 4x2 dx
y
L
0
=
1
3=2
x 1 + 4x2
16
=
0:6063 ft2
x =
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
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*8.13
8.14
368
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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
369
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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
370
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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
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8.21
8.22
Semicircle:
A1 =
R2
2
A2 =
2
(2R)h
3
Parabola:
y1 =
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
R2
4
A2 =
Qx
= A1 y1
A2 y2 =
Qy
= A1 x1
A2 x2 =
=
x =
y
=
R
R
R2
R2
=
1
2
2 2
R2
4R
R2
R
4
3
2
3
R2
4R
R2
R
4
3
2
1
(3
8)R3
12
1
(3
8)R3
Qy
3
= 12R2
=
A
3
1
2
2
Qx
= R2
A
2
R3
6
2
1
=
R3
6
2R
3
=
8
R = 0:416R J
6
2R
3(
x
2)
= 0:584R J
373
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8.26
h
2
2h) ( 2h)
(2b2
h2 )( 2)
(4b
2
4ab
2ah
Qx = 4ab
y
Qx
2ab2
=
A
4ab
ah2
2b2
=
2ah
4b
=
b
2
h2
2h
2ah
A =
= 2ab2
ah2
y is maximized when
dy
dh
=
0
(4b
4b2 8bh + 2h2
=0
(4b 2h)2
2h)
=0
h = 0:586b J
8.27
374
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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
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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
3
5
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 1
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
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
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8.39
8.40
8.41
y
x
dx
x/2
240 mm
120 mm
60 mm
x
381
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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
(60) dx =
Z 240
0
xel dV =
V
60 +
Z 240
60 +
x
x dx = 9:048
4
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 b
0
x2
b2
1
2
dx
x2
b2
xel = x
2
dx =
8
bh2
15
382
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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
=
x2
b2
x 1
0
Z
V
)
Z b
x2
b2
yel dV = h
2
Z b
dx
dx =
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
383
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8.46
8.47
8.48
384
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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
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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
386
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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.
2
x
in.
24
1
26 in.
in.
24
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
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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
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8.57
8.58
25
y
z
2
y
20
37
z
30
1
390
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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
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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
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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
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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
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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
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8.68
397
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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
y=
Qzx
32
=
= 1:231 in. J
L
26
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
400
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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
8 in.
3
r
3 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 R2 t 1 = 2 (6:152 )(0:3)
162
123
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
= 16:336 lb
= 6:684 lb
y1 =
Water:
y=
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
4967
= 8:16 ft J
608:6
x=
8.105
R
=
Z
p dA = p0
A
=
Rx =
1
p0
2
Z
Ry
=
1
p0 b
3
Z
A
=
0
0
Z b
1
p0
2
(b + y)dy =
0
y(b + y)dy =
0
x xy
+ 2 dx dy
b
b
1
p0 b 3
2
Z bZ b
0
1
p0 b3
2
x= 2
= b J
3
3
p0 b2
4
x
0
0
Z b
3
p0 b 2 J
4
Z bZ b
0
py dA = p0
x xy
+ 2 dx dy
b
b
0
(b + y)dy =
px dA = p0
A
=
Z b
Z bZ 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
=2
wx ds =
0
=
2w0 a
2w0
=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
cos ]0
=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
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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
Dimensions
in feet
2b/3 + 4
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
1
h2 = (62:4)(242 ) = 17 971 lb/ft
2 w
2
1
1
hb = (62:4)(24)b = 748:8b lb/ft
2 w
2
1
1
hb = (150)(24)b = 1800:0b lb/ft
2 c
2
(4h)
=
150(4)(24)
= 14 400 lb/ft
c
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 h
0
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
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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
V3
V
(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
=
0:09 m
V1
=
V2
x1
x2
x3
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 aZ b
p0
p dx dy =
a
b
0
Z a
x
0
!
y
cos
dy dx
2b
b
Z b
Noting that
Z b
y
dy = 2
cos
2b
b
Z b
2b
y
4b
y
dy = 2
sin
=
2b
2b 0
Z a
x
2ab
dx = p0
J
a
0
we get
R = p0
b
cos
4b
0
Rx =
Z aZ b
0
= p0
px dx dy = p0
b
4b
Z a
0
Z a
0
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
.a
1
.
3
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
Rz
=
=
62:4
=2 Z 1:5
Z
140:40 lb J
Z
pz dA =
62:4
z=
Z
Z
p ( r cos ) dA =
62:4
A
=2
dA = (r d ) dr
r2 cos dr d = 62:4
0
=2
=
A
=
p = h = 62:4 (r cos )
1:2656 cos2
d =
Z
=2
1:125 cos d
=2
=2 Z 1:5
Z
r3 cos2
dr d
0
=2
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
20
100
100
x
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
18
4
x3
4 x4
=
x2 dA =
x2 10 + x dx = 10 +
9
3
9 4 0
A
0
dA
=
Iy
=
31:1
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
y 2 (10
2:5y 2 )dy
0
2
2:5
Z 2
y5
= 21:3 in4 J
5 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
0
=
2
3
Z 6
Iy
y 3 dx
2
=
3
3
0
=
Z 6
2
0
2
Z 6
3
dx
144x2 + 1728 dx = 3160 in4 J
2
x y dx = 2
0
=
x2
3
12
1 6
x + 4x4
27
Z 6
x
6 in.
6 in.
x2
3
dx y = 12
x
Z 6
12
0
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 =
Z
Z
dIx =
A
Z b
0
Z b
1 3
1
y dx =
3
3
Z b
x
b
h3
0
Z b
x
Iy =
x dA =
x (y dx) =
x h
b
A
0
0
2
2
2
b
6
dx =
2
h3 x7
bh3
=
J
3b6 7 0
21
b
h x5
b3 h
dx = 2
=
J
b
5 0
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 =
(802 )
(602 )
+ 160(20)
= 7598 mm2
2
2
(802 ) 4(80)
(602 ) 4(60)
Ai y i =
+ (160 20)( 10)
2
3
2
3
= 0:16533 106 mm3
0:16533 106
Ai y i
=
= 21:76 mm
y =
A
7598
Ix = Ix Ay 2 = 11:422 106 7598(21:762 ) = 7:82 106 mm4 J
A =
Ai =
3 in. 4.5 in.
9.24
y
1
2
3
6 in.
6 in.
x
y
4
3 in.
x
y (in.)
2.5
4.5
1.5
1.0
Ay (in3 )
56.25
60.75
27.00
18:00
126.00
12 in.
Part
1
2
3
4
Sum
2
A (in )
22.5
13.5
18.0
18:0
36.0
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
2
bh3
3
3
1
1
R
+ Ay 2 =
R
R +
R2
36
36
2
2
8
3
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
Axy =
R4
8
= R4
1
8
4
9
=
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 =
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
y
=
=
=
=
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.
b
R
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
=
106 mm4
=
5:927
I1
I2
= b + R = (10:667 + 5:927)
= b R = (10:667 5:927)
sin 2 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 30
o
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
=
Ix 1 + A1 d21
Iy
=
Ixy
=
104
12
0 due to symmetry
(Iy )1
(Iy )2 =
(Ix )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
I1
Ix
= Iy = 2
Ixy
=
=
=
I2
=
=
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
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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
106 = 5:078
3:999
106 mm4
2
2
1:944
sin 1
= 60:99
2:223
+ 1:94402
106 = 2:223
= 90
= 29:01
106 mm4
= 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 mm4
106 = 16:59 106 mm4 J
106 = 4:74 106 mm4 J
5:120
= 59:75
) 1 = 29:9
5:927
J
2 = 29:9 + 90 = 119:9
sin 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
=
20 2
x
3
Z 90
0
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
=
2L sin
=
L
cos
2
2
xC = L cos
2
yG =
L
sin
4
2
U
=
2W
P
=
W
tan
J
2
2
2
L
sin
4
2
+ P L cos
2
=0
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=
P yD
yD
xB
= (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'
o θ
A
55
W
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
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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
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10.25
2.4 in.
WB = 0.4 lb
x
δyA
A
B
WA = 0.8 lb
δyB
A
B
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
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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 BC m
0:075 BC m
Equating the expressions for rB : 0:6
xC = 0:2(4
U
Cx
) = 0:8
m
= 0:15
BC m
yE = 0:075(4
) = 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 = 4
m
m
0:8Cx )
=0
499
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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 AC
rD = rC = 36 AC
36
rD
=
AC = 2:25
AC
DF =
16
16
32 DF = 32(2:25 AC ) = 72:0 AC
= P rB 4000 rF = [P (12)
= 24 000 lb J
4000(72)]
AC = 0
10.29
500
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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
:
:
10
rC = 15
rD = 10
DF = 15
= P rA
NF rF = P (15
=
18NF )
(15P
=0
in.
DF
in.
rA = 15
in.
rF = 12 DF in.
DF = 1:5
)
NF (12)(1:5 )
15
15
NF =
P =
(60) = 50:0 lb J
18
18
10.31
501
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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
:
rC
Bar CD rotates about D
= C1
=
AB
30 AB
rB
=
= 0:4511 AB
:
BC =
66:50
BE
= CE BC = 61:48(0:4511 AB ) = 27:73 AB
27:73 AB
rC
=
= 0:6162 AB
:
CD =
45
45
Bar BC rotates about E
U
C1
C2
rB = 30
C2
AB
CD = [C1
C2 (0:6162)]
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
:
:
rA = 21
0
rB
= 15
in.
rB = 9
in.
BD in.
3 0
r = rB
5 B
3
(15
5
BD ) = 9
BD =
= 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
1
= W L sin + k(L sec
2
1:5L)2
dV
d
= W L cos + kL2 (sec
1:5) sec tan = 0
W
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
cos3 = 0:5
= W ( 3 sin
3 sec2 tan )
3
d2 V
d 2
1:5 tan )
1:5 sec2 )
= W (3 cos
=
1.5 tanθ
t
3f C
3 sinθ
h
= W ( 3 sin 37:5
=37:5
=
5:48W < 0
= 37:5
J
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
509
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
<p J
R
2
10.45
510
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10.46
Choose the
= 0 position of rod AB as the datum.
Vg = W
dV
d
d2 V
d 2
L
sin
2
Ve =
1
2
k (L sin )
2
L
cos + kL2 sin cos
2
L
W sin + kL2 (cos2
2
= W
=
sin2 )
Equilibrium position:
dV
=0
d
sin =
W
2kL
Stability of the equilibrium position:
"
L W
d2 V
2
= W
+ kL 1
2 2kL
d 2
W
2kL
d2 V
Stable if
>0
d 2
= sin 1
2
W
2kL
W
J
2kL
2
#
=
4L2 k 2 3W 2
4k
p
Lk
3
>
J
W
2
10.47
511
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L/2
10.48
a
L/2
α
α
θ
a
Geometry:
+2
=
L
2
L0
Choosing the
V =
180
= 90
2
= a sin = a sin 90
= a cos
2
2
p
=
2a (free length of spring)
= 90 position as the datum:
1
W a cos + k(L
2
dV
d
L0 )2 =
1
W a cos + k 2a cos
2
2
= W a sin + k 2a cos
= W a sin + ka2
d2 V
d 2
d2 V
d 2
=0
p
2
2a
p
2a ( a sin )
2
p
sin + 2 sin
2
2
!
2
= W a cos + ka2
cos +
cos
2
2
p !
2
= W a + ka2
1+
= W a 0:2929ka2
2
p
For stability we need
W a > 0:2929ka2
W
> 0:2929 J
ka
512
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10.49
10.50
L
W
y
L = 2R cos
θ
k
R
θ
y = L sin = 2R cos sin = R sin 2
513
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1
= W y + k(L
2
V
1
L0 )2 = W R sin 2 + k(2R cos
2
1
= W R sin 2 + kR2 (2 cos
2
dV
d
=
R)2
1)2
2RW cos 2 + R2 k(2 cos
1)( 2 sin )
2RW cos 2 + 2R2 k( sin 2 + sin )
W
= 2R2 k
cos 2
sin 2 + sin
kR
=
W
d2 V
= 2R2 k
2
sin 2
kR
d 2
With W=(kR) = 0:5 and = 23:91 we get
dV
= 2R2 k(0:5 cos 47:82
d
d2 V
d 2
2 cos 2 + cos
sin 47:82 + sin 23:91 )
=
2R2 k ( sin 47:82
=
2R2 k( 1:170) < 0 Unstable J
0 In equilibrium J
2 cos 47:82 + cos 23:91 )
10.51
V
dV
d
d2 V
d 2
a
1
2
cos + k (b sin )
2
2
Wa
Wa
kb2
sin + kb2 sin cos =
sin +
sin 2
2
2
2
Wa
d2 V
Wa
cos + kb2 cos 2
+ kb2
=
2
2
2
d
=0
= W
=
=
Stability exists if
Wa
< kb2
2
k>
Wa
10(24)
=
= 3:33 lb/in. J
2b2
2(62 )
10.52
514
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*10.53
s
h
4 ft
V =
15 lb θ/2
15h + 30(2 cos
θ/2
2 ft
2 ft
30 lb
4)
dV
=
d
15
h + s = const (length of rope)
dh
h + 2(4 sin ) = const
+ 4 cos = 0
2
d
2
dV
=
d
15
4 cos
60 sin
2
dh
d
60 sin
dh
=
d
d2 V
=
d 2
30 sin
cos
2 sin
2
4 cos
2
60 cos
Equilibrium:
dV
d
=
0
cos
=
0 or 1
=
180
2
Stability at
d2 V
=
d 2
J
sin = 0
2
2 sin
or
2
2
cos
2
=0
=0
2
= 60 J
= 180 :
d2 V
=
d 2
Stability at
cos
30 sin 90
60 cos 180 = 30 > 0 Stable J
= 60 :
30 sin 30
60 cos 60 =
30
60 =
45 < 0 Unstable J
515
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10.54
10.55
516
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10.56
10.57
517
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10.58
518
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10.59
10.60
519
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10.61
520
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10.62
521
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.