2012. Fall Semester
Chungbuk National University
School of Civil Engineering
1
Chapter 12: Introduction
How do Engineers design
and construct the devices
we uses?
Via the studying of
mechanics, with
mathematical models, to
analyze and predict the
behaviors of these physical
systems
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Chapter 12: Introduction
Dynamics is one of the sciences underlying the
design of all machines. For instance, to design and
program an industrial robot, engineers must
analyze its motion using the principles of dynamics.
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Chapter Outline:
 Introduction
 Engineering & Mechanics
 Learning Mechanics
 Fundamental Concepts
 Units
 Newtonian Gravitation
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12.1 Engineering & Mechanics
Mechanics
(Study of forces and their effects)
Statics
Dynamics
(Objects in Equilibrium)
(Objects in Motion)
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12.1 Engineering & Mechanics
 Statics
Mechanical and civil engineers use the
equilibrium equations derived in statics to design
structures.
 Dynamics
Civil engineers use the equations derived in
dynamics to analyze the responses of buildings
to earthquakes.
Aerospace engineers use it to determine the
trajectories of satellites.
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12.2 Learning Mechanics
1. Problem Solving
• Identify the information given. Understand the
physical system involved.
• Develop a strategy. Identify all principles &
equations to be applied. Draw diagrams to
visualize the problem.
• Predict an answer. Try solving the equations.
Do a reality check and compare against the
prediction to see if it is reasonable.
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12.2 Learning Mechanics
2. Calculators & computers
• Use of scientific calculators and software
programmes on the computer, e.g. Matlab can
help solve the complicated algebraic equations
for the numerical solutions to the problems.
• Sections headed “Computational Mechanics”
contains examples and problems that are
suitable for solution with a programmable
calculator or a computer.
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12.2 Learning Mechanics
3. Engineering Applications
• Design aspect problems – require to choose
the values of parameters to satisfy stated
design criteria.
• Safety aspect problems – require to evaluate
the safety of devices and choose the values of
parameters to satisfy stated safety
requirements.
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12.3 Fundamental Concepts
1. Numbers
• Significant Digits – refers to the number of
meaningful digits in a number, counting to the
right starting with the first non-zero digit. E.g.
7.360 and 0.007360 are each stated to four
significant digits.
• If a number is the result of a measurement, the
significant digits it contains are limited by the
accuracy of the measurement.
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12.3 Fundamental Concepts
1. Numbers
• When using calculators, avoid rounding-off
errors for intermediate results to maximize
accuracy.
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12.3 Fundamental Concepts
2. Space and Time
• Space refers to the 3-D universe we lived in.
Measured by defining the length or distance
between two points in space. S.I unit: meter
(m) or U.S customary unit: foot (ft).
• Time is measured by the intervals between
repeatable events. S.I and U.S customary unit:
seconds (s).
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12.3 Fundamental Concepts
2. Space and Time
• Others are velocity (m/s or ft/s) and
acceleration (m/s2 or ft/s2).
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12.3 Fundamental Concepts
3. Newton’s 1st Law of Motion
• When the sum of the forces acting on a
particle is zero, its velocity is constant.
• If the particle is initially stationary, it will
remain stationary.
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12.3 Fundamental Concepts
3. Newton’s 2nd Law of Motion
• When the sum of the forces acting on a
particle is not zero, the sum of the forces is
equal to the rate of change of linear
momentum of the particle.
• If the mass is constant, the sum of the forces
(F) is equal to the product of the mass of the
particle (m) and its acceleration (a):
F  ma
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12.3 Fundamental Concepts
3. Newton’s 3rd Law of Motion
• The forces exerted by two particles on each
other are equal in magnitude and opposite in
direction.
• Notice from Newton’s 2nd Law,
F ( N)  m (kg)  a (m/s2 )
• Thus we can determined either of the
unknown parameter value if given the other
parameters are known.
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12.4 Units
1. International System of Units (S.I. Units)
1 N  (1 kg)(1 m/s2 )  1 kg  m/s2
Base units
Derived unit
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12.4 Units
1. International System of Units (S.I. Units)
Prefix
nanomicromillikilomegagiga-
Abbreviation
n
µ
m
k
M
G
Multiple
10-9
10-6
10-3
103
106
109
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12.4 Units
2. U.S. Customary Units
• Note that 1 lb = (1 slug )(1 ft/s2)
Base SI units
Derived unit
• Rewriting in terms of mass, we have
1 slug  1 lb  s /ft
2
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12.4 Units
3. Angular Units
s
θ
R
• An angle is defined to be
the ratio of the part of the
circumference of the
subtended by θ to the
radius of the circle.
S

R
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12.4 Units
3. Angular Units
s
θ
R
• Angles (θ) are expressed in
radians (rad) or degree (°):
2 rad  360
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12.4 Units
4. Conversion of Units
For example,
 5280 ft  1 h 
1 mi/h  1 mi/h 

  1.47 ft/s
 1 mi  3600 s 
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Example 12.1 – Converting Units of
Pressure
 Question
The pressure exerted at a point of the hull of
the deep sea submersible vehicle in Fig. 12.3
is 3.00x106 Pa (pascals) where it is defined
that 1 Pa = 1 N/m2.
Determine the pressure in
pounds per square foot.
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Example 12.1 – Converting Units of
Pressure
 Strategy
From Table 12.2, 1 pound = 4.448 newtons
and 1 foot = 0.3048 meters.
 Solution
The pressure (to 3 sig. digits) is
1 lb  0.3048 m 
3  10 N/m  3  10 N/m 


 4.448 N  1 ft 
6
2
6
2
2
 62,700 lb/ft 2
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Example 12.2 – Determining Units
from an Equation
 Question
Suppose that in Einstein’s equation,
E = mc2
the mass m is in kg, velocity of light is in m/s.
a) What are the SI units of E?
b) If the value of E in SI units is 20, what is its
value in U.S Customary base units?
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Example 12.2 – Determining Units
from an Equation
 Strategy
a) Since we know the units in terms of m and c,
we can deduce the units of E from the given
equation.
b) We can use the unit conversions for mass
and length in Table 12.2 to convert E from SI
units to U.S Customary base units.
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Example 12.2 – Determining Units
from an Equation
 Solution
a) From the equation for E, E = (m kg)(c m/s2)2.
hence SI units of E is kgm2/s2.
b) From Table 12.2, 1 slug = 14.59 kg and
1 ft = 0.3408 m. Therefore,
1 ft 
2 2
2 2  1 slug 
1 kg  m /s  1 kg  m /s 


 14.59 kg  0.3048 m 


 0.738 slug  ft 2 /s 2
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Example 12.2 – Determining Units
from an Equation
 Solution
c) The value of E in U.S Customary base
units is
E = (20)(0.738) = 14.8 slug·ft2/s2
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12.5 Newtonian Gravitation
Newton postulated that the
gravitational force between
2 particles of mass m1
and m2 that are separated
at a distance r is
F
Gm1m2
r
2
12.1
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12.5 Newtonian Gravitation
G is the universal gravitational constant and r
is the distance from the particle to the center of
sphere.
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12.5 Newtonian Gravitation
Although the earth is not a homogeneous
sphere, we can use Eqn (12.1) to approximate
the weight of an object of mass m due to the
gravitational attraction of the earth,
F
GmmE
r
2
12.2
where mE is the mass of the earth and r is the
distance from the center of the earth to the
object.
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12.5 Newtonian Gravitation
When an object’s weight is the only force
acting on it, the resulting acceleration is called
the acceleration due to gravity. From
Newton’s 2nd Law, W = ma and Eqn (12.2),
this acceleration due to gravity is
a
GmE
r
2
12.3
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12.5 Newtonian Gravitation
Denoting the acceleration due to gravity at sea
level by g = 9.81 m/s2 and the radius of the
earth by RE and from Eqn (12.3), we have
GmE  gRE2
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12.5 Newtonian Gravitation
Hence the acceleration due to gravity at a
distance r from the center of the earth at sea
level is
ag
RE2
r
2
12.4
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12.5 Newtonian Gravitation
Since the weight of the object W = ma, the
weight of an object at a distance r from the
center of the earth is
W  mg
RE2
r2
12.5
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12.5 Newtonian Gravitation
At sea level (r = RE), the weight of the object in
terms of its mass is given by the simple
relation
W  mg
12.6
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Example 12.3 – Determining an
Object’s Weight
 Question
When the Mars Exploration Rover was fully
assembled, its mass was 180 kg. The
acceleration due to gravity at the surface of
Mars is 3.68 m/s2 and the radius of Mars is
3390 km.
a) What was the rover’s weight when it was at
sea level on Earth?
b) What was the rover’s weight on the surface
of Mars?
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Example 12.3 – Determining an
Object’s Weight
 Question
c) The entry phase began when the
spacecraft reached the Mars atmospheric
entry interface point at 3522 km from the
center of Mars.
What was the
rover’s weight
at that point?
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Example 12.3 – Determining an
Object’s Weight
 Strategy
1. The rover’s weight at sea level is given by
Eqn (12.6) with g = 9.81 m/s2.
2. Similarly, the rover’s weight on Mars can
be determined using Eqn (12.6) with g =
3.68 m/s2.
3. To determine the rover’s weight as it
began the entry phase, we can write an
equation for Mars equivalent to Eqn (12.5).
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Example 12.3 – Determining an
Object’s Weight
 Solution
a) The weight at sea level is
W  mg
 (180 kg )(9.81 m/s 2 )
 1770 N 397 lb .
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Example 12.3 – Determining an
Object’s Weight
 Solution
b) Let gM = 3.68 m/s2 be the acceleration due
to gravity at the surface of Mars. Then the
weight of the rover on the surface of Mars is
W  mg M
 (180 kg )(3.68 m/s 2 )
 662 N 149 lb .
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Example 12.3 – Determining an
Object’s Weight
 Solution
c) Let RM = 3390 km be the radius of Mars.
From Eqn (12.5), the rover’s weight when it
is 3522 km above on the center of Mars is
W
2
RM
mg M 2
r
2


3
,
390
,
000
m
 (180 kg )(3.68 m/s 2 )
2
3,522,000 m 
 614 N 138 lb .
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