SOLUTIONS MANUAL For Engineering Mechanics Statics and Dynamics SI Units (Global Edition) 15 Edition

Engineering Mechanics Statics and Dynamics SI Units (Global Edition) 15e Russell C. Hibbeler (Solu�ons Manual All Chapters, 100% Original Verified, A+ Grade) All Chapters Solu�ons Manual Complete files download link at the end of this file.

Part 1: Statics: Ch 1-11: Page 2-103 Part 2: Dymamics: Ch 12-22: Page 104-204

Part 1: Statics Chapter 1

1

2

2

3

kN/μs,

Mg/mN,

MN/(kg∙ms).

3

*1-4.

4

5

5

6

6

7

7

*1-8.

8

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Chapter 2

2

q = 78.6°

Ans: FR = 8.67 kN, f = 3.05º

Ans: FR = 10.5 kN, f = 17.5°

Ans: T = 6.57 kN, q = 30.6°

50 N

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F3

30i N

2

Express F1, F2, and F3 as Cartesian vectors.

y

F1 F2

15 kN

40

26 kN 13 5 12

SOLUTION

x 30

F1 = {15 sin 40°i + 15 cos 40°j} kN = { 9.64i + 11.5j} kN

Ans.

F3

36 kN

F2 = -26(12>13)i + 26(5>13)j = {−24.0i + 10.0j } kN

Ans.

F3 = 36 cos 30°i - 36 sin 30°j = {31.2i − 18.0j } kN

Ans.

Ans: F1 = {9.64i + 11.5j} kN F2 = {−24.0i + 10.0j } kN F3 = {31.2i − 18.0j} kN

Fz

as they currently exist. exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

as they currently exist. exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

as they currently exist. exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

as they currently exist. exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

g2 > 90°, thus, g2 = cos−1(−0.5) = 120°(Ans.). By resolving F1 and F2 into their x,y, and

g = 120°

as they currently exist. exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

as they currently exist. exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

The pipe is subjected to the force F, which has components acting along the x, y, z axes. If the magnitude of F is 12 kN, and a = 120° and g = 45°, determine the magnitudes of its three components.

Fz F Fx

x

SOLUTION

Fy y

cos2 a + cos2 b + cos2 g = 1 cos2 120° + cos2 b + cos2 45° = 1 cos b = + 0.5

cos b = {0.5

b = 60°

Fx = F cos a = |12 cos 120°| = 6 kN

Ans.

Fy = F cos b = |12 cos 60°| = 6 kN

Ans.

Fz = F cos g = |12 cos 45°| = 8.49 kN

Ans.

Ans: Fx = 6 kN Fy = 6 kN Fz = 8.49 kN

Part 2: Dymamics Chapter 12

1

2

sp

3.00 m/s 2.00 m/s2

3

4

1

,

32.0 m/s 67.0 m 66.0 m

5

avg

∆t

8.00 m avg

−12.0 m/s2

6

7

8

aA

s

9

10

11

12

13

322 m/s 19.3 s

14

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0

0

0

2.4503 s = 2.45 s

6.6667 s = 6.67 s

t|s = 60 m = 2.45 s tstop = 6.67 s

15

16

17

2(1.8)

t1

18

19

20

21

22

23

24

25

26

27

28

12–28

12–28

12–28

29

30

31

32

33

Height h

h

34

35

36

s

s

4.00

4.00

37

For 0 ≤ t < 12 s, aA = 20 m/s2, A = (20t) m/s, sA = (10t2) m For 12 s < t ≤ 20 s, aA = 0, A = 240 m/s, sA = 240(t − 6) m For 0 ≤ t < 10 s, aB = 15 m/s2, B = (15t) m/s, sB = (7.5t2) m For 10 s < t ≤ 20 s, aB = 0, B = 150 m/s, sB = 150(t − 5) m ∆s = 1.11 km

38

−4.00 4.00 3.00 1.00 3.00

39

40

For 0 ≤ t < 10 s, = (0.4t2) m=s For 10 s < t ≤ 11.2 s, = (8t − 40) m=s 11.2 s

41

42

0.89443

0.89443

(2t2) m

0.2236

s 0.2236

(−2t2 + 1.788t − 0.2) m

max

= 0.894 m/s, t' = 0.447 s

For 0 ≤ t < 0.224 s, s = (2t2) m; For 0.224 s < t ≤ 0.447 s, s = (−2t2 + 1.79t − 0.2) m

43

44

45

46

m

For 0 … t 6 9 s, 3>2

v = (4t ) m>s 8 5>2 s = a t bm 5 For 9 s 6 t … 14 s, v = (2t 2 - 18t + 108) m>s 2 s = a t 3 - 9t 2 + 108t - 340b m 3

47

,

,

,

For 0 ≤ t < 4 s, a = 3.50 m/s2; For 4 s < t < 5 s, a = 0; For 5 s < t ≤ 8 s, a = 4.00 m/s2

48

49

t

Ans.

(0.800t) m/s

t Ans.

t Ans.

Ans: For 0 ≤ t < 30 s, v = (0.800t) m/s, a = 0.800 m/s2; For 30 s < t ≤ 40 s, = 24 m/s, a = 0

50

51

0.100

52

, 8.00 ,

53

54

55

For 0 ≤ 4 s,

(0.625t2) m

,

(5t −10) m

, −1.00 m/s2

(15t − 0.5t2 − 60) m

,

,

, −1.00

56

30.0 −1.00 48.0

57

Ans.

Ans.

Ans.

Ans.

187.5} m s

58

0.200

Ans.

Ans.

Ans. For 0 ≤ t < 30 s, a = 0.200 m/s2; For 30 s < t ≤ 48 s, a = −0.333 m/s2

59

( )

(

m

)

m

s = 144 m For 0 ≤ t < 30 s, s = (t2/10) m; For 30 s < t ≤ 48 s, s = (−t2/6 + 16t − 240) m

60

= 4t

Ans.

Ans.

Ans.

= 5t

Ans.

Ans.

61

For 0 … t 6 10 s, aA = 4 m/s2, A = (4t) m/s, sA = (2t2) m For 10 s 6 t … 15 s, aA = 0, A = 40.0 m/s, sA = 40(t − 5) m For 0 … t 6 5 s, aB = 5 m/s2, B = (5t) m/s, sB = (2.5t2) m For 5 s 6 t … 15 s, aB = 0, B = 25.0 m/s, sB = 25(t − 2.5) m

62

63

64

(t3 − 3t2 + 2t) m (3t2 − 6t + 2) m/s (6t − 6) m/s2

Ans.

Ans. 1.577 s

−0.385

0.4226 s

Ans: = (3t2 − 6t + 2) m/s, a = (6t − 6) m/s2

65

66

s s 2.5 s, −6.25 m

2.5

.

When

Figs.

−6.00

67

68

m,

Ans.

Ans.

m,

69

. Using the

For 0 ≤ s < 300 m,

For 300 m < s ≤ 600 m,

70

Ans.

400

100

Ans.

Ans.

Ans.

For 0 ≤ t < 10 s, s = (t2) m For 10 s < t ≤ 16.9 s, s = (13.5et/5) m For 0 ≤ s < 100 m, a = 2 m/s2 For 100 m < s ≤ 400 m, a = (0.04s) m/s2 t = 16.9 s

71

25.0

72

73

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74

4

75

76

77

78

a = 80.2 m/s2 (x, y, z) =

79

80

81

82

1.00 km=s

83

84

85

(−40i)

{−1.12i}

{1.20i + 1.60j} {−1.12i}

86

87

88

θA = 30.5°

89

90

.

91

92

93

94

45.0

95

Since sin 2(45° − f) = sin 2(45° + f) = cos 2f, R|θ = 45° − f = R|θ = 45° + f. (QED)

96

yB = −1.2 m. Thus,

97

98

Substituting

4.5446 s

99

100

101

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