DYNAMICS
16th Edition
Russell C. Hibbeler
SOLUTIONS MANUAL – PART 1
Solutions and Answers to Selected Problems
CHAPTERS 1-5
General Principles • Force Vectors • Particle Equilibrium • Force System Resultants •
Rigid Body Equilibrium
ISBN: 9780135426388
, PART-1
SOLUTIONS
CH 1 to 5
CONTENTS
Chapter 1 – General Principles
Chapter 2 – Force Vectors
Chapter 3 – Equilibrium of a Particle
Chapter 4 – Force System Resultants
Chapter 5 – Equilibrium of a Rigid Body
,CHAPTER NO. 01: GENERAL PRINCIPLES
1–1.
Round off the following numbers to three significant
figures: (a) 3.455 55 m, (b) 45.556 s, (c) 5555 N, (d) 4525 kg.
SOLUTION
(a) 3.46 m (b) 45.6 s (c) 5.56 kN (d) 4.52 Mg Ans:
a) 3.46 m
b) 45.6 s
c) 5.56 kN
d) 4.52 Mg
1–2.
Represent each of the following combinations of units in
the correct SI form using an appropriate prefix: (a) kN>ms,
(b) Mg>mN, (c) MN>(kg # ms).
SOLUTION
a) kN>ms = 103 N>10 - 6 s = GN>s Ans:
b) Mg>mN = 106 g>10 - 3 N = Gg>N a) GN>s
b) Gg>N
c) MN>(kg # ms) = 106 N>kg(10 - 3 s) = GN>(kg # s) c) GN>(kg # s)
1–3.
Represent each of the following combinations of units in the
correct SI form using an appropriate prefix: (a) Mg>mm,
(b) mN>ms, (c) mm # Mg.
SOLUTION
103 kg 106 kg
a) Mg>mm = -3
= = Gg>m
10 m m
Ans:
10-3 N
103 N a) Gg>m
b) mN>ms = -6
= = kN>s
10 s s b) kN>s
c) mm # kg
c) mm # Mg = 310- 6 m4 # 3103 kg4 = 10-3 m # kg
= mm # kg
1
, *1–4.
What is the weight in newtons of an object that has a mass
of (a) 8 kg, (b) 0.04 g, (c) 760 Mg?
SOLUTION
a) W = 9.81(8) = 78.5 N Ans:
a) W = 78.5 N
b) W = 9.81(0.04) ( 10 -3
) = 0.392 ( 10 ) N = 0.392 mN
-3
b) W = 0.392 mN
c) W = 9.81(760) ( 103 ) = 7.46 ( 106 ) N = 7.46 MN c) W = 7.46 MN
1–5.
Represent each of the following as a number between 0.1
and 1000 using an appropriate prefix: (a) 45 320 kN,
(b) 568(105) mm, (c) 0.005 63 mg.
SOLUTION
a) 45 320 kN = 45.3 MN Ans:
a) 45.3 MN
b) 568 ( 105 ) mm = 56.8 km
b) 56.8 km
c) 0.005 63 mg = 5.63 mg c) 5.63 mg
1–6.
Wood has a density of 4.70 slug>ft 3. What is it density
expressed in SI units?
SOLUTION
(1 ft)3(14.59 kg)
(4.70 slug>ft 3) e f = 2.42 Mg>m3
(0.3048 m)3(1 slug)
2