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Chapter 1 : Introduction
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Introduction
 What
is mechanics?
 Fundamental quantities of mechanics
 Units of Measurements
 Dimensional consideration
 Newton’s law
 Mass and Weight
 Significance of numerical results
 Free body diagram
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What is mechanics?

The study of response of bodies to action of forces

Mechanics:
– Rigid body mechanics
– Mechanics of deformable bodies
– Fluid mechanics
Key subjects:
– Forces, reactions, and stresses
– Strength of materials
– Structure mechanics
– Free Body Diagrams (FBD)

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Some Quantities of Mechanics






Space and Length
– Space = a geometric region where the physical event takes place  coordinate systems
– Length = a measure of the size of the space
Time:
– The interval between two events
– Examples: before and after the application of a load
Matter
– Substance that occupies the space or a body
– Contributes to the “resistance” of the body under loading
– Mass = the quantification of matter
A force
– An action of one body upon another body
– The two bodies may be in contact or separated
– Gravitational force  separated bodies
– Action and reaction  bodies in contacts
A particle:
– It is often used as a representation of a body
– It may have mass but not size or shape
Rigid-body concepts
– Idealization of bodies/particles/collection of particles
– The shape and size of a rigid body remain constant under loadings
– In reality, bodies will be deformed and/or displaced under loadings
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Units of Measurements



Measurement of physical quantities require “standards”
– “Standard” is arbitrarily defined (but must be agreed)
– Two systems of “standards” are commonly used
The US Customary System of Units
– USCS
– British units
– Foot – pound (for force) - second
The International System of Units
– SI units
– MKS units
– Meter-kilogram-second
– Metric system
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International System of Units

Base Units (Table 1-2)
– Meter for length, kg for mass, etc.

Supplementary Units (Table 1-3)
– Only rad and sterad for angles
– Often regarded as non dimensional

Derived Units (Table 1-4)
– Based on the physical law of the quantity
– Combinations of base units and/or supplementary units
– m2 for area, m/s for speed, etc
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Dimensional Homogeneity

An equation is dimensionally homogeneous if the form of
the equation does not depend on the units of measurement.
– Example:
These equation are valid regardless of the unit system that is being
used (provided that the unit system is NOT mixed)
» Area = width x length
» s = a x t2
s = distance, a = acceleration, t = time
» Force = mass x acceleration
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Dimensional consideration



Physical quantities can be expressed dimensionally in terms
of three fundamental quantities: mass (M), length (L), and
time (T)
The dimensions of quantities other than the fundamental
quantities follow from the physical laws
Examples:
– Area  L2
– Velocity  L/T
– Force  ML/T2
– Work = Force x Distance  ML2/T2
– Power = Rate of Work  ML2/T3
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Commonly used physical quantities and their dimensions
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Dimensional Homogeneity


Also implies that the LHS and RHS of the equation must have the same
dimension
Example:
– The formulation of the non-dimensional angle of twist (θ ) of a circular
beam is given by
» T = moment of force = ML2/T2
» L = length of the beam = L
TL
» J = Torsional inertia = L4
θ=
2
» G = Shearing modulus = M/L/T
GJ
≈ θ = angle of twist = dimensionless
– We can show that the LHS and RHS have the same dimension
TL
ML2 / T 2 L
θ=
= 4
=1
2
GJ L M /( LT )
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Newton’s Laws of Motion

Law 1: In the absence of external forces, a particle originally at rest or
moving with constant velocity will remain at rest or continue to move with a
constant velocity along a straight line
∑

v = constant
Law 2: If an external force acts on a particle, the particle will be accelerated
in the direction of the force and the magnitude of the acceleration will be
directly proportional to the force and inversely proportional to the mass of
particle
1
∑

F = 0 ⇒ a = 0,
F ∝a a∝
m
Law 3: For every action there is an equal and opposite reaction. The forces
of action and reaction between bodies are equal in the magnitude, opposite
in direction, and collinear
Action = −Reaction
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Law of Gravitation
 Law
that governs the mutual attraction between
two bodies (in contact or separated)
F = Force magnitude
G = universal gravitational constant
r = distance between the center of the mass
m = mass of the body
m1m2
F =G 2
r
G = 3.439e-8 ft3/(slug.s2)
G = 6.673e-11 m3/(kg.s2)
e+n = 10+n
r
= center of the mass
m1
m2
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Example:

Determine the magnitude of gravitational force exerted between
two bodies of 20 and 50 kgs when they are 1 m, 1 cm and 1 mm
apart.
– G = 6.673e-11 m3/(kg.s2)

Answer:
– F = G 20*50/12 = 6.673e-8 N
– F = G 20*50/0.012 = G 20*50/1e-22 = 6.673e-4 N
– F = G 20*50/0.0012 = G 20*50/1e-32 = 6.673e-2 N
» N = Newton = kg.m/s2

Gravitational force in general can be ignored
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Mass and Weight

Mass (m)
– The absolute quantification of matter occupied in a body
– It is independent of the position of the body in space and the
surrounding forces

Weight (W)
– Gravitational attraction exerted by the earth on a body
– Generally speaking, it depends on the location on the earth surface
m1m2
F =G 2
r
Set:
m1 = mass of the earth (me)
m2 = any mass on the earth surface (m)
r = mean radius of the earth (re)
g = G me/re2
W = mg
g = 9.81 m/s2
g = 32.2 ft/s2
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Example:


The weight of an object on the earth surface was found to be
98100 N. Find its weight on the moon and on 200,000 km above
the earth surface (half way to the moon).
– Mass of the earth = me = 5.976e24 kg
– Mean radius of the earth = re = 6370 km
– Mass of the moon = mm = 7.350e22 kg
– Mean radius of the moon = rm = 1740 km
– G = 6.673e-11 m3/(kg.s2)
m1m2
F =G 2
Answer:
r
– Utilize the gravitational law
– Remember: the mass of the object (m) is constant everywhere
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Answer
mass of the object is given by m = Wg = 98100
= 10000kg
9.81
 The weight on 200000 km above the earth surface is
given by
m×m
 The
F =W = G
e
2
(re + s )
1.00e4 × 5.976e24
(6.370e6 + 2.00e5) 2
= 92384.89 → 9.24e4 or 92400 N
= 6.673e − 11×
 The
weight of the object on the moon surface is
F =W = G
m × mm
2
rm
1.00e4 × 7.350e22
1.740e6 2
= 16199.81 → 1.620e4 or 16200 N
= 6.673e − 11×
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Significance of Numerical results

The accuracy of the known physical data
– The accuracy of analysis results cannot be more than the input of the
analysis

The accuracy of the model
– The accuracy of analysis results is limited by the choice of the
model

The accuracy of the computations
– The number of significant figures given by the computers does not
represent the true accuracy because of the two limitation above

Bottom lines:
– Limit the results to four significant figures
– Try to avoid round-off error when performing calculation
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Significant figures

Please read last paragraph on p.18

Retain four significant figures if the leading figure is small: 12.34 and 2.345

Retain three significant figures if the leading figure is large:
0.0876

Rounded down if the digits being dropped < half of the last significant figure
retained:
7654 7650 because 4 < ½*10

Rounded up if the digits being dropped > half of the last significant figure
retained:
123.456 123.5 because 0.056 > ½*0.1

If the digits being dropped = ½*last figure retained, then
– Keep it unchanged when the last figure is an odd number
– Rounded up if the last figure is an even number
– Example: 12,345 and 12,355 both become 12,350
6780 and
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Reading assignments
 Section
1-5: Problem Solving
 Section 1-6: Significance of Numerical Results
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Free body diagram (FBD)

A collection of bodies of
interest separated from all other
interacting bodies and with all
external forces applied
Pushing force
Frictions
Frictions
Weight
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