Engineering Mechanics: Statics, 4th Edition by Andrew Pytel and Jaan Kiusalaas – Complete Solutions Manual | Verified Step-by-Step Solutions for All Chapters | Vector Mechanics, Equilibrium, Force Systems, Moments, Centroids, and Friction Problems | Essen

All 10 Chapters Covered




SOLUTIONS

, 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




@@
SeSiesim ii s ol a tio n
ici
sm ci s o l a ti on

, Chapter 1
1.1




1.2

W = ggV = (7850)(9.81) π(0.042)(0.110) = 42.58 N
0.2248 lb
W = 42.58 N × = 9.57 lb (
1.0 N

1.3
4.448 N 0.3048 m
(a) 400 lb·ft = 400 lb·ft × × = 542 N·m (
1.0 lb 1.0 ft
6 m 0.3048 ft 1.0 mi 3600 s
(b) 6 m/s = × × × = 1. 247 mi/h (
s 1.0 m 5280 ft 1.0 h

20 lb × 4.448 N × 1.0 in.2
= 1. 379 × 105 N/m2
(c) 20 lb/in.2 = 1.0 lb
in.2 645.2 × 10—6 m2
= 137.9 kPa (

500 slug 14.593 kg 39.37 in.
(d) 500 slug/in. = × × = 2. 87 × 105 kg/m (
in. 1.0 slug 1.0 m

1.4
30 mi 5280 ft 0.3048 m 1.0 gal
30 mi/gal = × × ×
gal 1.0 mi 1.0 ft 3.785 L
= 12 760 m/L = 12.76 km/L (

1.5
1 m 2 kg · m2 kg · m (m)
(a) E = (1000 kg) 6 s = 18 000 s 2
= 18 000 s2
2
= 18 000 N·m = 18 kN·m (
0.2248 lb 3.281 ft
×
(b) E = 18 000 N·m = 18 000 N·m × 1.0 N 1.0 m
= 13 280 lb·ft (




© 2017 Cengage Learning®. May not be scanned, copied

, 1.6




1.7




1.8
8 mm 1.0 m 1.0 µs
(a) 8 mm/µs = × × = 8000 m/s (
µs 1000 mm 10—6 s

8000 m 3.281 ft 1.0 mi 3600 s
(b) 8000 m/s = × × × = 17 900 mi/h (
s 1.0 m 5280 ft 1.0 h

1.9




1.10


G
= [A] G2 + [B] [G] [T ]
T2
1 1
) [A] = ( [B] = (
GT 2 T3


1.11
(a) The dimensions of x = At2 — Bvt are
[G] = [A][T 2] — [B][GT —1][T ]
) [A] = [GT —2 ] ( [B] = [1] (dimensionless) (




@@
SeSies im
s m 2
iciii cs iosloaltaiotinon


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

, (b) The dimensions of x = Avte—Bt are
[G] = [A][GT —1][T ]e[B][T ]
[B][T ] = [1] ) [B] = [T —1] (
[G] = [A][GT —1][T ] ) [A] = [1] (

1.12

d4y G
= = [G—3]
dx4 G4
c2g [T —2][MG —1] M
y = [G] =
D [FG2] T 2FG2
Substituting [F ] = MGT —2 —see Eq. (1.2b)— we get
c2g y M T2
= = [G 3] Q.E.D.
D T 2G2 MG
Substituting [F ] = MGT —2 —see Eq. (1.2b)— we get
c2g y M T2
= = [G 3] Q.E.D.
D T 2G2 MG

1.13
The argument of the sine function must be dimensionless:
Bx G
= [1] [B][G] = [1] [B] = [FG—2] (
k F
[F ] = [Akx2] = [A][FG—1][G2] [A] = [G—1] (

1.14
550 lb · ft/s
(a) 110 hp = 110 hp × = 60 500 lb ft/s
· (
1.0 hp

(b) 110 hp = 110 hp × 0.7457 kW = 82.0 kW (
1.0 hp

1.15
mAmB (12)(12)
F = G= (6.67 × 10 —11) = 6.003 × 10 8
N
R 2 0.42
W = mg = (12)(9.81) = 117.7 N
F 6.003 × 10—8 8
% of weight = × 100% = × 100% = 5.10 × 10 % (
W 117.7




@@
SeSies im
s m 3
iciii cs iosloaltaiotinon


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

, 1.16




1.1
7 0.3048 m
h = (28 000 ft) = 8534 m = 8.534 km
1.0 ft

GMem GMem
On earth: We = At elevation h: W =
R 2e (Re + h)2
R2 63782
W = We e = 170 = 169.5 lb ►
(Re + h)2 (6378 + 8.534)2

1.18

GMm GMe
gm = ge =
2
Rm R e2
gm ________ MmR2 e 1
= 0.07348(6378)
2
= = 0.1658 ≈ Q.E.D.
ge 2
MeR m 6
5.974(1737)2

1.19




1.20
GMem GMem
On earth: We = At elevation h: W =
R 2e (Re + h)2




@@
SeSies im
s m 4
iciii cs iosloaltaiotinon


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

, We GMem
W = = GMem (R + h)2 = 10R2
e e
10 (Re + h)2 10R2e
(6378 + h)2 = 10(6378)2 h = 13 790 km (

1.2
1
R = Re + Rm + d = 6378 + 1737 + 384 × 103
= 392.1 × 103 km = 392.1 × 106 m

MeM m —11 5.974 × 1024 (0.07348 × 1024)
F = G = 6.67 × 10
R2 (392.1 × 106)2
20
= 1.904 × 10 N (

1.22

90o
40o
 v
50o


√ 2 3
v= 5 + 32 = 5.83 m/s = tan—1 = 31.0○
5

|v1 + v2| = 5.83 m/s
31.0o

1.23

90o
v1 v2

8 m/s
v1 = 8 sin 40○ = 5.14 m/s ( v2 = 8 sin 50○ = 6.13 m/s (




@@
SeSies im
s m 5
iciii cs iosloaltaiotinon


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

, 1.24

240 lb
30o
P
90o
Q

Component parallel to AB : P = 240 cos 30○ = 208 lb (
Component pependicular to AB : Q = 240 sin 30○ = 120 lb (

1.25
u
50o 60o
Pv 70o 20 kN
50o 60o
v Pu

Pv Pu 20
= =
sin 70 ○ sin 50○
sin 60○
sin 60○
Pv = 20 = 22.6 kN (
sin 50○
sin 70○
Pu = 20 = 24.5 kN (
sin 50○
1.26

140o v
5 mi/h
 40o
3 mi/h

√
Law of cosines: v = 32 + 52 — 2(3)(5) cos 140○ = 7.549 mi/h
5 7.549
Law of sines: = sin = 0.4257 = 25.2○
sin sin 140○

7.55
o
25.2mi/h (




@@
SeSies im
s m 6
iciii cs iosloaltaiotinon


© 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
65o 35o
8000 lb 80o
P

P 8000 sin 65○
= P = 8000 = 12 640 lb (
sin 65○ sin 35○ sin 35○

1.28
 R
25o 
80o 10 000 lb
8000 lb

Law of cosines:
√
R = 80002 + 10 0002 — 2(8000)(10 000) cos 80○
= 11 671 lb (

Law of sines:
11 10 000
10 000 sin = sin 80○ = 0.8438
671 11 671
sin =
sin 80○
= sin—1(0.8438) = 57.54○
8 = 90○ — 25○ — 57.54○ = 7.46○ (

7.46o
11 670 lb




@@
SeSies im
s m 7
iciii cs iosloaltaiotinon


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

, 1.29




1.30




1.31

PAB
60o
360 lb 80o
40o P
AC


Law of sines:
360 PAB PAC
= =
sin 80○ sin 40○ sin 60○
360 sin 40○
PAB = = 235 lb (
sin 80○
360 sin 60○
PAC = = 317 lb (
sin 80○




@@
SeSies im
s m 8
iciii cs iosloaltaiotinon


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