Solutions Manual for Meriam's Engineering Mechanics: Statics, 9th Edition (Meriam, Kraige & Bolton, 2018) | All Chapters 1–7 Covered

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, TABLE OF CONTENTS
Solutions Manual: Meriam's Engineering Mechanics: Statics, 9th
Edition
Authors: J. L. Meriam, L. G. Kraige, J. N. Bolton (2018)
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Chapter 1. Introduction to Statics
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Chapter 2. Force Systems
Chapter 3. Equilibrium
Chapter 4. Structures




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Chapter 5. Distributed Forces
Chapter 6. Friction
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Chapter 7. Virtual Work
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, Chapter 1
Solution 1.1


V Vx2 V y2 402 302 50 y
V 40i  30 j θy
n   0.8i  0.6 j x
V 50 θx
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V = 40i – 30j




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cosx  0.8, x  36.9
cosy  0.6, y  126.9
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Solution 1.2

V  V 1 V 2 But V  V1 V2 !




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96.9°
Graphically, V  16.4 units, x  83 3
V2 = β 4
12 units V1 = 10 units
V 2  102 122  2 1012 cos 96.9 θx
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Algebraically, 60° 36.9°



sin   sin 96.9 ,
V  16.51 units

  46.2
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12 16.51
x    36.9 46.2  36.9 83.0
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, Solution 1.3
Graphically, V  14.7 units, x  163
V2 (12 units)
36.9°
y
V 2  102 122  2 1012 cos 83.1 Vʹ =
53.1°
30° 30° V1
Algebraically,
V   14.67 units V2 + (–V1) θx
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sin   sin 83.1 ,   54.3 β
4 x

12 14.67
x  180  36.9    180  36.9  54.3
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83.1° –V1 (10 units)




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 162.6




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Solution 1.4
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F  1602  802 1202  215N
Fx  160  0.743,
cos x  x  42.0
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F 215 ED
Fy 80
cos y    0.371, y  68.2
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Fz 120
cos z    0.557, z  123.9
F 215
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