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General Principles 1
Engineering Mechanics: Statics in SI Units, 12e
Copyright © 2010 Pearson Education South Asia Pte Ltd 1
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Objectives
• Basic quantities and idealizations of mechanics • Newton’s Laws of Motion and Gravitation • Principles for applying the SI system of units • Standard procedures for performing numerical
calculations • General guide for solving problems
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Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Mechanics 2. Fundamental Concepts 3. Units of Measurement 4. The International System of Units 5. Numerical Calculations 6. General Procedure for Analysis
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Copyright © 2010 Pearson Education South Asia Pte Ltd
1.1 Mechanics
• Mechanics can be divided into 3 branches: - Rigid-body Mechanics - Deformable-body Mechanics - Fluid Mechanics
• Rigid-body Mechanics deals with
- Statics - Dynamics
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1.1 Mechanics
• Statics – Equilibrium of bodies Ø At rest Ø Move with constant velocity • Dynamics – Accelerated motion of bodies
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1.2 Fundamentals Concepts
Basic Quantities 1. Length
- locate the position of a point in space 3. Mass
- measure of a quantity of matter 4. Time
- succession of events 5. Force
- a “push” or “pull” exerted by one body on another
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1.2 Fundamentals Concepts
Idealizations 1. Particles
- has a mass and size can be neglected
2. Rigid Body - a combination of a large number of particles
3. Concentrated Force - the effect of a loading
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1.2 Fundamentals Concepts
Newton’s Three Laws of Motion • First Law
“A particle originally at rest, or moving in a straight line with constant velocity, will remain in this state provided that the particle is not subjected to an unbalanced force”
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1.2 Fundamentals Concepts
Newton’s Three Laws of Motion • Second Law
“A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force”
maF =
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1.2 Fundamentals Concepts
Newton’s Three Laws of Motion • Third Law
“The mutual forces of action and reaction between two particles are equal and, opposite and collinear”
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1.2 Fundamentals Concepts
Newton’s Law of Gravitational Attraction Me: mass of Earth
Weight: Letting yields
221
rmm
GF =F = force of gravitation between two particles G = universal constant of gravitation m1,m2 = mass of each of the two particles r = distance between the two particles
2rmMGW e=
2/ rGMg e= mgW =
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1.2 Fundamentals Concepts
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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–13. Two particles have a mass of 8 kg and 12 kg,respectively. If they are 800 mm apart, determine the forceof gravity acting between them. Compare this result withthe weight of each particle.
01 Ch01 1-6.indd 401 Ch01 1-6.indd 4 6/12/09 8:02:56 AM6/12/09 8:02:56 AM
F = G m1m2
r2
Where G = 66.73(10–12) m3/(kg · s2)
W1 = 8(9.81) = 78.5 N Ans
W2 = 12(9.81) = 118 N Ans
F = 66.73(10–12) = 10.0(10–9) N = 10.0 nN Ans8(12)(0.8)2! "
Copyright © 2010 Pearson Education South Asia Pte Ltd
1.3 Units of Measurement
SI Units • Stands for Système International d’Unités • F = ma is maintained only if
– 3 of the units, called base units, are defined – 4th unit is derived from the equation
• SI system specifies length in meters (m), time in seconds (s) and mass in kilograms (kg)
• Force unit, Newton (N), is derived from F = ma
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1.3 Units of Measurement
Name Length Time Mass Force
International Systems of Units (SI)
Meter (m) Second (s) Kilogram (kg) Newton (N)
⎟⎠⎞
⎜⎝⎛
2.smkg
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Copyright © 2010 Pearson Education South Asia Pte Ltd
1.3 Units of Measurement
• At the standard location, g = 9.806 65 m/s2
• For calculations, we use g = 9.81 m/s2
• Thus,
W = mg (g = 9.81m/s2) • Hence, Ø a body of mass 1 kg has a weight of 9.81 N, Ø a body of mass 2 kg has a weight of 19.62 N.
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Copyright © 2010 Pearson Education South Asia Pte Ltd
1.4 The International System of Units
Prefixes • For a very large or small numerical quantity, units can
be modified by using a prefix
• Each represent a multiple or sub-multiple of a unit Eg: 4,000,000 N = 4000 kN (kilo-newton) = 4 MN (mega- newton) 0.005m = 5 mm (milli-meter)
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1.4 The International System of Units
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1.5 Numerical Calculations
Dimensional Homogeneity • Each term must be expressed in the same units • Regardless of how the equation is evaluated, it
maintains its dimensional homogeneity • All terms can be replaced by a consistent set of units
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1.5 Numerical Calculations
Significant Figures • Accuracy of a number is specified by the number of
significant figures it contains • A significant figure is any digit including zero
e.g. 5604 and 34.52 have four significant numbers • When numbers begin or end with zero, we make use
of prefixes to clarify the number of significant figures e.g. 400 as one significant figure would be 0.4(103)
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1.5 Numerical Calculations
Rounding Off Numbers • Accuracy obtained would never be better than the
accuracy of the problem data • Calculators or computers involve more figures in the
answer than the number of significant figures in the data
• Calculated results should always be “rounded off” to an appropriate number of significant figures
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1.5 Numerical Calculations
Calculations • Retain a greater number of digits for accuracy • Work out computations so that numbers that are
approximately equal • Round off final answers to three significant figures
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Copyright © 2010 Pearson Education South Asia Pte Ltd
1.6 General Procedure for Analysis
• To solve problems, it is important to present work in a logical and orderly way as suggested:
1. Correlate actual physical situation with theory 2. Draw any diagrams and tabulate the problem data 3. Apply principles in mathematics forms 4. Solve equations which are
dimensionally homogenous 5. Report the answer with
significance figures 6. Technical judgment
and common sense 22
Copyright © 2010 Pearson Education South Asia Pte Ltd 23
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example
Convert to 2 km/h to m/s. Solution
m/s 556.0s 3600
h 1km
m 1000hkm 2 km/h 2 =⎟
⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=
Remember to round off the final answer to three significant figures.
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