Engineering mechanics statics 4th edition Pytel Kiusalaas solution manual pdf

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@solutionmanual1
https://gioumeh.com/product/engineering-mechanics-statics-pytel-solution/
= 9:57 lb J
1.3
(a) 400 lbft = 400 lbft 4:448 N 1:0 lb
0:3048 m 1:0 ft
= 542 Nm J
1:0 m 1:0 mi 5280 ft
3600 s 1:0 h
= 1: 247 mi/h J
in.2 4:448 N
= 137:9 kPa J
14:593 kg 1:0 slug
39:37 in. 1:0 m
1.4
5280 ft 1:0 mi
0:3048 m 1:0 ft
1:0 gal 3:785 L
1.5
= 18 000 Nm = 18 kNm J
(b) E = 18 000 Nm = 18 000 Nm 0:2248 lb 1:0 N
3:281 ft 1:0 m
= 13 280 lbft J
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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1:0 m 1000 mm
= 8000 m/s J
3:281 ft 1:0 m
1:0 mi 5280 ft
3600 s 1:0 h
= 17 900 mi/h J
(a) The dimensions of x = At2 Bvt are
[L] = [A][T 2] [B][LT1][T ] ) [A] = [LT2] J [B] = [1] (dimensionless) J
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(b) The dimensions of x = AvteBt are
[L] = [A][LT1][T ]e[B][T ]
[B][T ] = [1] ) [B] = [T1] J [L] = [A][LT1][T ] ) [A] = [1] J
1.12 d4y
!2
!2
The argument of the sine function must be dimensionless: Bx
k
[F ] = [Akx2] = [A][FL1][L2] [A] = [L1] J
1.14
(a) 110 hp = 110 hp 550 lb ft/s 1:0 hp
= 60 500 lbft/s J
(b) 110 hp = 110 hp 0:7457 kW 1:0 hp
= 82:0 kW J
% of weight = F
117:7 100% = 5:10 108 % J
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= 8534 m = 8:534 km
On earth: We = GMem
GMem
1.18
GMem
(Re + h)2
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1.21
= 392:1 103 km = 392:1 106 m
F = G MeMm
= 1:904 1020 N J
3
31.0o
1.23
8 m/s 40o50o
v1 = 8 sin 40 = 5:14 m/s J v2 = 8 sin 50
= 6:13 m/s J
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P 90o
240 lb
Component parallel to AB : P = 240 cos 30 = 208 lb J Component pependicular to AB : Q = 240 sin 30 = 120 lb J
1.25
Pu = 20 sin 70
1.26
140o
Law of cosines: v = p 32 + 52 2(3)(5) cos 140 = 7:549 mi/h
Law of sines: 5
o25.2 7.55 mi/h
J
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1.28
= 11 671 lb J
= sin1(0:8438) = 57:54
11 670 lb 7.46o
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PAC = 360 sin 60
sin 80 = 317 lb J
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Law of cosines:
1.33
1.34
α
β
P 125o
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1.35
1.36
b = 200 sin 42:5
sin 21:3 = 372 m J
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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.
Law of cosines: R = p 2502 + 3002 2(250)(300) cos 60
= 278:4 N
Law of sines: 300
278:4 = 68:9
β γ
50 lb
30 lb
65 lb
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= cos1 652 + 502 + 302
2(50)(30) = 105:96
Law of sines: 30
65 = 26:3 J
! AB!AB = 5i+ 3j
288 lb
480 lb
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! AB!AB = 2:2i+ 7:5j+ 3k
8:372 = 0:2628i+ 0:8958j+ 0:3583k J
(b) v = 8AB = 8(0:2628i+ 0:8958j+ 0:3583k) = 2:10i+ 7:17j+ 2:87k m/s J
1.45
OA =
5:590 = 0:5367i+ 0:7156j+ 0:4472k
F = FOA = 320(0:5367i+ 0:7156j+ 0:4472k) = 172i+ 229j+ 143k N J
1.46
BA =
24:90 = 0:5623i 0:4016j 0:7229k
F = FAB = 160(0:5623i 0:4016j 0:7229k) = 90:0i 64:3j 115:7k lb J
1.47
=
1602 + 2202 + 702 = 0:5696i+ 0:7832j 0:2492k
v = v = 1400(0:5696i+ 0:7832j 0:2492k) = 797i+ 1096j 349k ft/s J
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!BA =p202 + 602 + 902 = 110:0 ft =
! BA!BA = 20i+ 60j 90k
110:0 = 0:1818i+ 0:5455j 0:8182k
Fx = Fx = 600 (0:1818) = 109 lb J Fy = Fy = 600(0:5455) = 327 lb J Fz = Fz = 600(0:8182) = 491 lb J
(b)
y = cos1 y = cos 1(0:5455) = 56:9 J
z = cos1 z = cos 1(0:8182) = 144:9 J
1.49
! AB!AB = 353j+ 5k5:831
= 18:01j+ 30:01k kN
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!AC =p52 + 32 + 52 = 7:681 m TAC = T
! AC!AC = 355i 3j+ 5k7:681
= 22:78i 13:67j+ 22:78k kN
= 22:78i+ (18:01 13:67)j+ (30:01 + 22:78)k = 22:8i 31:7j+ 52:8k kN J
1.50
!AB =p82 + 82 + 42 = 12:0 ft FAB = F
! AB!AB = F 8i 8j+ 4k12:0
! AC = 4i 8j+ 4k ft
!AC =p42 + 82 + 42 = 9:798 ft FAC = 200
! AC!AC = 2004i 8j+ 4k9:798
The resultant R lies in the yz-plane if
Rx = (FAB)x + (FAC)x = 0 F 8
12:0 200 4
+ 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
) R = 2:83(0:807i+ 0:496j+ 0:320k) kN J
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5i 12j 13
Because R = P+Q lies in x-direction, we have
Ry = 0 54 12 13 Q = 0 Q = 58:5 lb J
R = Rx = 72 + 58:5 5
13 = 94:5 lb J
1.54
Px +Qx = Rx P cos 30 Q sin 30 = 360 cos 25
Py +Qy = Ry P sin 30 Q cos 30 = 360 sin 25
Solution is: P = 717 lb J Q = 590 lb J
1.55
= cos1 2
3 sin 55
Q = 2 sin 55 + 3 sin 56:90 = 4:15 kN J
1.56
= 0:3841i+ 0:5121j 0:7682k
= 0:4082i+ 0:4082j 0:8165k
= 0:5547j 0:8321k
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Engineering mechanics statics 4th edition Pytel Kiusalaas solution manual pdf

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