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UNIT I Basic Concepts of Engineering Mechanics
Lecture 1 : Fundamental Principles, Resolution and Composition of forces and equilibrium of
particles and Pr inciple of transmissibility
Instructional Objectives:
After completing the course, you should be able to analyze forces and find out the resultant
forces in two and three dimensions.
Mechanics: Oldest of the Physical Sciences
Archi medes (287-212 BC): Principles of Lever and Buoyancy!
1.1 Definitions:
Sci ence i s the knowledge that comes fr om observing f acts about th e uni verse careful l y,
carrying out experiments and making statements that are always true in particular
conditions .
The word science comes from the Latin "scientia," meaning knowledge. Science refers to a
system of acquiring knowledge. This system uses observation and experimentation to
describe and explain natural phenomena.
Engineering is the appli cation of mathematics and sci ence to th e desi gn and manuf actur e
of i tems that benefi t h umanity.
En gineeri ng desi gn may be defi ned as a process of devi si ng a system, component or
process to meet desi red needs.
Mechanics i s a br anch of the physi cal sciences that i s concern ed wi th the state of rest or
moti on of bodi es subjected to the acti on of f orce.
En gineeri ng mechani cs i s the branch of engin eeri ng that appli es pri nci ples of mechani cs
to desi gn.
Engineering is the process of putting things together to get a desired outcome, technology is
engineering that has been proven to work over and over again. Technology is always
improving due to the advances in Engineering.
Engineering is creation of a new facility, machinery or establishment to produce goods,
materials and life needs. Hence it is detailing and creating hence should include design,
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procurement, construction and start-up. Technology is proprietary knowledge (know-how)
once given the establishment/facility can be created with the help of engineering
1.2 Rigid-body Mechanics:
A basic requirement for the study of the mechanics of deformable bodies and the
mechanics of fluids
Essential for the des ign and analysis of many types of structural members, mechanica l
components, electrical devices, etc, encountered in engineering.
A rigid body does not deform under load.
Statics: deals with equilibrium of bodies under action of forces (bodies may be either
at rest or move with a constant velocity).
Dynamics: deals with motion of bodies (accelerated motion)
1.3 Fundamental Concepts :
Length (Space): needed to locate position of a point in space, & describe size of the
physical system distances, geometric properties
Time : measure of succession of events basic quantity in Dynamics
Mass : quantity of matter in a body measure of inertia of a body (its resistance to
change in velocity)
Force: represents the action of one body on another characterized by its magnitude,
direction of its action, and its point of application Force is a vector quantity
Mechanics
Mechanics ofFluids
Mechanics ofSolids
Mechanics of DeformableBodies
Mechanics of RigidBodies
Statics Dynamics
Kinematics Kinetics
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1.4. Newtonian Mechanics:
Length, Time and Mass are absolute concepts independent of each other.
Force is a derived concept not independent of the other fundamental concepts. Force
acting on a body is related to the mass of the body and the variation of its velocity
with time.
Force can also occur between bodies that are physically separated (Ex: gravitational,
electrical, and magnetic forces).
Important Points:
Mass is a property of matter that does not change from one location to another.
Weight refers to the gravitational attraction of the earth on a body or quantity of mass.
Its magnitude depends upon the elevation at which the mass is located.
Weight of a body is the gravitational force acting on it.
1.5 Idealizations :
Idealizations are needed to simplify application of the theory
Continuum: Continuous distribution of mass or matter with no void or empty spaces.
Particle : A body with mass but with dimensions that can be neglected.
Size of earth is insignificant compared to the size
of its orbit. Earth can be modeled as a particle
when studying its orbital motion
Rigid Body: A combination of large number of particles in which all particles
remain at a fixed distance (practically) from one another before and after applying a
load.
Material properties of a rigid body are not required to be considered when
analyzing the forces acting on the body. In most cases, actual deformations
occurring in structures, machines, mechanisms, etc. are relatively small, and
rigid body assumption is suitable for analysis.
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Concentrated Force : Effect of a loading which is assumed to act at a point (CG) on a
body.
Provided the area over which the load is applied is very small compared to
the overall size of the body.
Example: Contact Force between a wheel and ground.
1.6 Newtons Three Laws of Motion :
Basi s of formul ati on of ri gid body mechani cs.
First Law: A particle originally at rest, or moving in a straight line with constant
velocity, tends to remain in this state provided the particle is not subjected to an
unbalanced force.
First law contains the principle of the
equilibrium of forces main topic of concern in
Statics.
Second Law: A particle of mass acted upon by an unbalanced force experiences
acceleration that has the same direction as the force and a magnitude that is directly
proportional to the force.
Accelerated Motion.
Second Law forms the basis for most of the analysis in Dynamics.
Third Law: The mutual forces of action and reaction between two particles are equal,
opposite, and collinear.
Action-Reaction
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Third law is basic to our unde rstanding of Force Forces always occur
in pairs of equal and opposite forces.
1.7 Newtons Law of Gravitational Attraction:
Weight of a body (gravitational force acting on a body) is required to be computed in Statics
as well as Dynamics. This law governs the gravitational attraction between any two particles.
F = mutual force of attract ion between two particles
G = universal constant of gravitation
Experiments G = 6.673x10 -11 m3/ (kg.s 2)
Rotation of Earth is not taken into account
1, 2 = masses of two particles,
=
Weight of a body (gravitational force acting on a body) is required to be computed in Statics
as well as Dynamics.
Weight of a Body: If a particle is located at or near the
surface of the earth, the only significant gravitational force
is that between the earth and the particle.
=
=
= =
whe re r = distance between the earths center and the particle
and = = acceleration due to gravity (9.81m/s 2)
= . , = , .
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1.8 Units:
Unit is defined as the numerical standard used to measure the qualitative dimension of a
physical quantity. When mass or force, length and time are accepted as basic quantities, then
all other quantities are secondary or derived quantities in terms of these basic quantities.
Four Fundamental Quantities
Quantity Dimensional
Symbol
SI Unit
Unit Symbol
Mass M Kilogram kg
Length L Meter M
Time T Second s
Force F Newton N
= N = kg m/ s 2
= N = kg m/s 2
1 Newton is the force required to give a mass of 1 kg an acceleration of 1 m/s 2
Metric System (SI)
SI System offers major advantages re lat ive to the FPS system
Widely used throughout the world
Use one basic unit for length meter; while FPS uses many basic
units inch, foot, yard, mile
SI based on multiples of 10, which makes it easier to use & learn
whereas FPS is complicated, for example.
SI system 1 meter = 100 centimeters, 1 kilometer = 1000
meters, etc
FPS system 1 foot = 12 inches, 1 yard = 3 feet, 1 mile =
5280 feet, etc.
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U.S. Customary System (FPS)
Force (lb) = mass (s lugs) acceleration (ft/sec 2 )
Thus (s lugs) = lb.sec 2/ft
Therefore 1 slug is the mass which is given an acceleration of 1 ft/sec 2 when
acted upon by a force of 1 lb
Conversion of Units
Convert ing from one system of unit to another;
The standard value of g (gravitational acceleration)
SI units, = . /
FPS units, = . /
Units Prefixes
Multiple/Submultiple Exponential Prefix SI
1 000 000 000 10 9 giga G
1 000 000 10 6 mega M
1 000 10 3 kilo k
0.001 10 -3 milli m
0.000 001 10 -6 micro
0.000 000 001 10 -9 nano n
1.9 Scalars and Vectors:
Scalars: only magnitude is associated.
Ex: time, volume, density, speed, energy, mass
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Vectors : possess direction as well as magnitude, and must obey the parallelogram law of
addition (and the triangle law).
Equivalent Vector: = + (Vector Sum)
Ex: displacement, velocity, acce leration, force, moment, momentum
A Vector V can be written as: = V n
V = magnitude of V
n = unit vector whose magnitude is one and whose direction coincides with that of V
Unit vector can be formed by dividing any vector, such as the geometric position vector, by
its length or magnitude
Vectors represented by Bold and Non-Italic letters, V (or rightwards arrows above, V)
Magnitude of vectors represented by Non-Bold, Italic letters (V)
Free Vector : whose action is not confined to or associated with a unique
line in space e.g. Movement of a body without rotation.
Sliding Vector : has a unique line of action in space but
not a unique point of application e.g. External force on a
rigid body
Principle of Transmissibility states that theconditions of equilibrium or conditions of motionof a rigid body will remain unchanged if a force
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acting at a give point of the rigid body is rep laced by a force of the same magnitudeand same direction, but acting at a different point, provided that the two forces have
the same line of action .Important in Rigid Body Mechanics
Fixed Vector : for which a unique point of application isspecified. e.g.: Action of a force on deformable body.
Vector Addition: Procedure for Analysis
Parallelogram Law (Graphical):
Resultant Force (diagona l)
Components (sides of parallelogram)
Algebraic Solution:
Using the coordinate system
Trigonometry (Geometry):
Resultant Force and Components from Law of Cosines and Law of Sines
Law of Sines: Consider the triangle ABC inscribed in a circle with centre O and radius .
From basic geometry, we know that
A = CAB = DOB.
From the right triangle DOB, we deduce
sin A = =
By repeating the same argument for angles B and C, we
obtain the Law of Sines.
= =
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Law of Cosines : Consider the following tr iangle ABC.
From the right triangle ADC, we deduce
2 + 2 = 2 (i)
and = = (ii)
From the right triangle BDC, we deduce
( ) 2 + 2 = 2 2 = 2 2 + 2 + 2 (iii)
Substituting (i) and (ii) in (iii), we get = +
1.10 Force Systems:
Force: Magnitude (P), direction (arrow) and point of application (point A) is important
Change in any of the three specifications will alter the effect on the bracket.
Force is a Fixed Vector
In case of rigid bodies, line of action of force is important (not its point of application if we
are interested in only the resultant external effects of the force),
we will treat most forces as-
External effect: Forces applied (applied force); Forces exerted
by bracket, bolts, Foundation (reactive force).
Inte rnal effect: Deformation, strain pattern permanent strain;
depends on material properties of bracket, bolts, etc.
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Concurrent force
Forces are said to be concurrent at a point if their lines of action intersect at that point.
, are concurrent forces; R will be on same plane; = +
(Apply Principle of Transmissibility)
Components and Projections of Force
Components of a Force are not necessarily equal to the Projections of the Force unless
the axes on which the forces are projected are orthogonal (perpendicular to each
other).
1 and 2 are the components of R and = +
and are perpendicular projections on axes
and , respectively.
+ unless and are perpendicular to
each other.
Problem 1.1
The two forces act on a bolt at A . Determine
their resultant.
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SOLUTION:
Graphical solution - construct a parallelogram with sides in the same direction as P
and Q and lengths in proport ion. Graphically evaluate the resultant which is
equivalent in direction and proportional in magnitude to the diagonal.
Graphical solution -1: A parallelogram with sides equal
to P and Q is drawn to scale. The magnitude and
direction of the resultant or of the diagonal to the
parallelogram are measured.
Graphical solution-2: A triangle is drawn with P and Q
head-to-tail and to scale. The magnitude and direction of
the resultant or of the third side of the triangle are measured
35 N98 R
Trigonometric solution - use the triangle rule for vector addition in conjunction with
the law of cosines and law of sines to find the resultant.
Trigonometric solution - Apply the triangle rule.From the Law of Cosines,
155cos N60 N402 N60 N40cos2
22
222 B PQQ P R
N73.97 R
From the Law of Sines,
A
A RQ
B A
R B
Q A
20
04.15 N73.97
N60155sinsinsin
sinsin
= .
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The angle for minimum tension in rope 2 isdetermined by applying the Triangle Rule andobserving the effect of variations in .
The minimum tension in rope 2 occurs when T1 andT2 are perpendicular.
2 = 5000 sin 30 2 = 2500
1 = 5000 cos30 2 = 4330
= 90 30 = 60
1.11 Rectangular Components of a Force: Unit Vectors:
May resolve a force vector into perpendicular components so that
the resulting parallelogram is a rectangle. and are referred to as
rectangular vector components and
= +
Define perpendicular unit vectors and which are parallel to the x and y axes
Vector components may be expressed as products of the unit
vectors with the scalar magnitudes of the vector components.
= +
and are referred to as the scalar components of
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1.12 Addition of Forces by Summing Components
Want to find the resultant of 3 or more concurrent forces.
S Q P R
Resolve each force into rectangular components
+ = + + + + +
= ( + + ) + ( + + )
The scalar components of the resultant are equal tothe sum of the corresponding scalar components of the
given forces.
x
x x x x
F
S Q P R
y
y y y y
F
S Q P R
To find the resultant magnitude and direction
x
y y x R
R R R R 122 tan
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Problem 1.3
Four forces act on bolt A as shown.
Determine the resultant of the force onthe bolt.
Solution:
Steps: Resolve each force into rectangular components. Determine the components of the resultant by adding the corresponding force components. Calculate the magnitude and direction of the resultant.
1. Resolve each force into rectangular components
9.256.96100
0.11001102.754.2780
0.759.129150
4
3
2
1
F
F F
F
comp ycomp xmag force
1.199 x R 3.14 y R
2. Determine the components of the resultant by adding the corresponding force components.
3. Calculate the magnitude and direction
22 3.141.199 R = 199.6 = 199.6
N1.199 N3.14
tan 1.4
1.13 Equilibrium of a Particle
When the resultant of all forces acting on a particle is zero, the particle is inequilibrium .
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Newtons First Law : If the resultant force on a particle is zero, the particle willremain at rest or will continue at constant speed in a straight line.
Particle acted upon by two forces:
- equal magnitude - same line of action - opposite s ense
Particle acted upon by three or more forces:
-
graphical solution yields a closed polygon
- algebraic solution
00
0
y x F F
F R
1.14 Rectangular Components in Space:
With the angles between and the axes,
k ji
F
k ji F
k F j F i F F
F F F F F F
z y x
z y x
z y x
z z y y x x
coscoscos
coscoscos
coscoscos
is a unit vector along the line of action of F
and z y x cosand,cos,cos are the direction cosines for F .
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Di rection of the force i s defi ned bythe l ocati on of two poi nts,
222111 ,,and,, z y x N z y x M
d Fd
F d
Fd F
d Fd
F
k d jd id d
F F
z z d y yd x xd
k d jd id
N M d
z z
y y
x x
z y x
z y x
z y x
1
and joiningvector
121212
Problem 1.4
The tension in the guy wire is 2500 N.Determine:
a) Components F x , F y , F z of the force actingon the bolt at A,
b) the angles q x , q y , q z defining the direct ionof the force
SOLUTION:
Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B.
Apply the unit vector to determine the components of the force acting on A. Noting that the components of the unit vector are the direction cos ines for the vector,
calculate the corresponding angles.
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Step 1: Determine the unit vector pointing from Atowards B.
m3.94
m30m80m40
m30m80m40222
AB
k ji AB
k ji
k ji
318.0848.0424.0
3.9430
3.9480
3.9440
Step 2: Determine the components of the force.
k ji
k ji
F F
N795 N2120 N1060
318.0848.0424.0 N2500
Step 3: Noting that the components of the unit vector are the direction cosines for thevector, calculate the corresponding angles.
k ji
k ji z y x
318.0848.0424.0
coscoscos
5.71
0.32
1.115
z
y
x