Module
1
Fundamental Concepts
OVERVIEW
First, what is Mechanics? Mechanics is a branch of physical
science which deals with the state of rest or motion of bodies that
subjected to the action of forces. In general, this subject can be
subdivided
into
three
branches:
rigid-body
mechanics,
deformable -body mechanics, and fluid mechanics. Rigid-body
mechanics, which is the focus of this course, can be divided into two
branches: statics and dynamics.
This course, Statics, is the branch of mechanics that deals
with the study of forces acting on a body in equilibrium - either the
body at rest or in uniform motion.
In this module, we will have a review on the pre -requisite
concepts – basic quantities, models, Newton’s Laws of Motion and
gravitational attraction, weight, SI units and vector operations , and
an introduction on force vectors and its operations .
LEARNING OUTCOMES
1.
2.
3.
4.
At the end of this module, you should be able to:
state newton’s laws of motion
calculate the weight of an object ;
perform calculations using SI and U.S. units , and;
perform operations on force vectors .
INTRODUCTION
Did you know that the statics developed very early in
history? This is because even from simple measurements of
geometry and force, its principles can be applied.
Engr. KISSA P. BANAWIS
(+63)-9162417934
kissapujante@gmail.com
One of the ancient heroes of engineering is Archimedes
(287-212 B.C.) who studied the principles of the pulley,
inclined plane, and wrench which greatly helped in the
construction during this period .
LESSON 1: REVIEW ON BASIC CONCEPTS
Let us first discuss the basic concepts which are prerequisites
in the succeeding topics.
1.1 BASIC QUANTITIES
Space. It is the geometric region occupied by bodies whose
positions are described by linear and angular measurements relative
to a coordinate system. For two -dimensional problems, only two
coordinates are required (as shown in Figure 1.a). For three dimensional problems, three independent coordinates are needed
(as shown in Figure 1.b) .
y
y
P(x, y)
P(x, y, z)
x
x
z
Figure 1.a
Figure 1.b
Length. It is used to locate the position of a
point in space and thereby describe the size of a
physical system. Once a standard unit of length is
defined, one can then use it to define distances
and geometric properties of a body as multiples of
this unit. Referring to figure 1.a, the lengths of the
point P along the x and y -axes are x and y units,
respectively. Hence, distance of point P from the
y-axis is x units and its distance from the x-axis is
y units.
Mass. It is a measure of a quantity of matter
that is used to compare the action of one body with
that of another. This property manifests itself as a
gravitational attraction between two bodies . It also
provides a measure of the inertia of a body or the
resistance of matter to a change in velocity.
Force. It is the action of one body on another. A force tends
to move a body in the direction of its action. It is often described as
a "push" or "pull" exerted by one body on another. A force is
completely characteriz ed by its magnitude, direction, and point of
application.
Page 2
1.2 MODELS
The following models are very helpful in simplifying the
analysis of a system of forces. If a certain property is negligible, a
body may be represented by these models in order to solve a
particular problem.
Particle. A particle has a mass, but a size t hat can be
neglected. For example, the size of the earth is insignificant
compared to the size of its orbit, and therefore the earth can be
modeled as a particle when studying its orbital motion. When a body
is idealized as a particle, the principles of me chanics reduce to a
rather simplified form since the geometry of the body will not be
involved in the analysis of the problem.
Rigid Body. A body is considered rigid when small internal
deformations do not essentially affect a body under a certain
external load. A rigid body can be considered as a combination of a
large number of particles in which all the particles remain at a fixed
distance from one another, both before and after applying a load. In
this course, structures, machines, mechanisms an d the like are
assumed to be rigid -bodies which means actual deformations are
relatively small. Thus, in the analysis of forces affecting a rigid
body, material properties will not be considered . But this can be
studied in a more advanced course, mechani cs of deformable
bodies.
Concentrated Force. A concentrated force represents the
effect of a loading which is assumed to act at a point on a body. We
can represent a load by a concentrated force, provided the area over
which the load is applied is very small compared to the overall size
of the body. An example would be the contact force between a wheel
and the ground.
Fw
A
Fw
A
1.3 NEWTON’S LAWS OF MOTION
Sir Isaac Newton was the first to state correctly the basic
laws governing the motion of a particle and to demonstrate
their validity.
First Law (Law of Inertia). 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.
Page 3
F2
F1
v
Equilibrium
F3
Second Law (Law of Acceleration). 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.
F = ma
F
a
Accelerated Motion
Third Law (Law of Interaction). The mutual forces of action
and reaction between two particles are equal, opposite, and
collinear.
Force of B on A
F
Force of A on B
F
A
B
Action - Reaction
1.4 NEWTON'S LAW OF GRAVITATIONAL ATTRACTION
Shortly after formulating his three laws of motion. Newton
postulated a law governing the gravitational attraction between any
two particles. Stated mathematically
F = G
𝒎𝟏 𝒎𝟐
𝒓𝟐
Where:
F
= force of gravitation between the two particles
G = universal constant of gravitation; according
experimental evidence,
= 66.73(10 - 1 2 ) m 3 /(kg·s 2 )
m 1 , m 2 = mass of each of the two particles
r
= distance between the two particles
to
1.5 WEIGHT
Page 4
Weight will be the only gravitational force considered in our
study of mechanics. From the previous equation, we can develop
an approximate expression for finding the weight W of a particle
having a mass m 1 = m. If we assume the earth to be a non -rotating
sphere of constant density and having a mass m 2 = M c ,
then if r is the distance between the earth's center and the particle,
we have
W = G
Letting g = GM c
/r 2
𝑚 𝑀𝑐
𝑟2
W = mg
where g = acceleration due to gravity. Since it depends on r. then
the weight of a body is not an absolute quantity. Instead, its
magnitude is determined from where the measurement was made.
Note:
The quantity that you get when you
weigh yourself on a balance is just your
MASS. To get your WEIGHT, you need to
multiply your mass with the gravitational
acceleration value where your mass was
taken.
So when you are filling up the personal
data sheet form with the item Weight (kg),
you need not correct it, but you should
know that it is just your MASS that the
form requires.
1.6 SYSTEM OF UNITS
This is a review on the system of units, focusing mainly on SI
Units and U.S. Customary Units which ar e the systems recognized
worldwide.
SI Units. The International System of units, abbreviated SI
after the French "Système International d'Unites" is a modern
version of the metric system which has received worldwide
recognition. The unit of force, called a newton (N), is equal to a force
required to give 1 kilogram of mass an acceleration of 1 m/s 2 (N =
kg-m/s 2 ).
If the weight of a body located at the "standard location" is
to be determined in newtons, where
g = 9.806 65 m/s 2 ;
however, for calculations, let us use the value g = 9.81 m/s 2 .
U.S. Customary. If the measurements arc made at the
“standard location," gravitational acceleration,
g =
32.2 ft/s 2 .
Page 5
Table 1 shows the SI and U.S. Customary Units for quantities
of mass, length, time and force.
Table 1.
SI UNITS
QUANTITY
Mass
Length
Time
Force
UNIT
SYMBOL
kilogram
meter
second
newton
Kg
m
s
N or (𝑘𝑔𝑠2∙ 𝑚)
U.S.
CUSTOMARY
UNIT SYMBOL
UNITS 𝑙𝑏 ∙ 𝑠 2
(
)
slug
𝑓𝑡
foot
ft
second
sec
pound
lb
1.7 CONVERSION OF UNITS
It is very important to know how to convert units. Conversion
can be done within a system of unit or from one system to another.
Here are some of the conversion values:
1 ft = 12 in. (inches)
5280 ft = 1 mi (mile)
1000 lb = 1 kip (kilo-pound)
2000 lb = 1 ton
Table 2 shows the conversion from Foot Pound Second (FPS) or
the US Customary Units to SI Units.
Table 2.
Quantity
Force
Mass
Length
Un i t of
Me as ur em e nt
(FP S) EQ U A L
(U S Culb
eq u a ls
Un i t of
Me as ur em e nt ( SI)
4.448 N
14.59 kg
slug
ft
0.304 8 m
1.8 THE INTERNATIONAL SYSTEM OF UNITS
This is to give emphasis on the International System of Units
because we will use this f or the most of our discussion in this course.
The next part is the presentation on the rules of its use and some of
the terminologies related to engineering mechanics.
When a numerical quantity is either very large or very small,
the units used to define its size may be modified by using a prefix.
Table 3 shows some of the prefixes .
Page 6
Expanded Form
1 000 000 000
1 000 000
1 000
0.001
Table 3.
Exponential
Form
109
106
103
10–3
Prefix SI Symbol
gigaG
megaM
kilok
millim
0.000 001
0.000 000 001
10–6
10–9
micronano-
μ
n
LEARNING EXERCISE No. 1
I – What is the First Law of Motion? Give one real life situation
showing this law.
II - Determine the weight in Newtons of a car whose mass is 1800
kg. Convert the mass of the car to slugs and then determine its
weight in pounds.
III - What is the weight in both Newtons and pounds of a 95 -kg beam?
IV – Express the following measurements to the identified S.I. units.
Unit
Measurement
Prefix
SI
2 000 000 grams
gigakilomilli1.95 gigameter
megananomicro-
Page 7
LESSON 2: FORCE VECTORS
2.1 SCALARS AND VECTORS
We use two kinds of quantities in mechanics —scalars and
vectors.
Scalar. Scalar quantities are those with which only a
magnitude is associated. Examples of scalar quantities are time,
volume, density, speed, energy, and mass.
Vector. Vector quantities, on the other hand, possess
direction as well as magnit ude, and must obey the parallelogram law
of addition as described later in this article. Examples of vector
quantities are displacement, velocity, acceleration, force, moment,
and momentum.
A vector is shown graphically by an arrow. The length of the
arrow represents the magnitude of the vector , and the angle θ
between the vector and a fixed axis defines the direction of its
line of action. The head or tip of the arrow indicates the sense
of direction of the v ector.
magnitude
A
sense
A
θ
To denote vector quantities , these are typed as bold face
letters such as A, and its magnitude of the vector is italicized , A. For
handwritten work, it is often convenient to denote a vector quantity
by simply drawing an arrow on top of it, 𝐴⃗.
2.2 VECTOR OPERATIONS
Multiplication and Division of a Vector by a Scalar. If a
vector is multiplied by a positive scalar, its magnitude is increased
by that amount. When multiplied b y a negative scalar it will also
change the directional sense of the vector.
A
3A
-A
- 2A
Vector Addition.
1. Parallelogram Law of Addition. One of the methods
in adding vector quantities is through Parallelogram
B
Method. Let us adding two component vectors A and
B. This will result in getting the resultant R. The
following procedure should be followed:
A
A
Page 8
2A
B
• First join the tails of the components at a point so
that it makes them concurrent.
A
P
B
• From the head of
another line from
These two lines
adjacent sides of
B, draw a line parallel to A. Draw
the head of A that is parallel to B.
intersect at point P to form the
a parallelogram.
A
R=A+B
P
B
2. Triangle Rule. This is a method of addition of vectors
where vector B is added to vector A by connecting the
head of A to the tail of B. The resultant R extends from
the tail of A to the head of B. Also, R can also be
obtained by adding A to B. Therefore, the vectors can
be added in either order, i.e., R=A+B=B+A.
B
A
•The diagonal of this parallelogram that extends to P
forms R. which then represents the resultant vector
R = A + B.
R=A+B
OR
R=B+A
B
A
B
If the two vectors A and B are collinear or both have
the same line of action, you can just have an algebraic
or scalar addition
R = A + B.
A
R=A+B
Vector Subtraction. The resultant of the difference between
two vectors A and B of the same type may be expressed as R’ = A B = A + (-B). In subtracting vectors, rules of vector addition apply.
-B
P
A
B
R=A-B
R=A-B
A
A
-B
2.3 VECTOR ADDITION OF FORCES
In analyzing problems involving addition
following procedures should be followed:
F1
of
forces,
the
Parallelogram Law
FR
• Adding two “component” forces F 1 and F 2 according to the
parallelogram law , yields a resultant force F R that forms the diagonal
of the parallelogram.
F2
v
u
F
Fv
Fu
v
u
F
θ
Fv
Page 9
Fu
• If a force F is to be resolved into components along two axes u
and v, then start at the head of force F and construct lines parallel
to the axes, thereby forming the parallelogram. The sides of the
parallelogram represent the components F u and F v .
• Label all the known and unknown force magnitudes and the
angles on the sketch and identify the two unknowns as the
magnitude and direction of F R or the magnitudes of its components .
Trigonometry
Fv
F
θ
Fu
• Redraw a half portion of the parallelogram to illustrate the
triangular head -to-tail addition of the components.
• From this triangle, the magnitude of the resultant force can be
determined using the law of cosines, and its direction is determined
from the law of sines. The magnitudes of two force components are
determined from the law of sines.
Cosine Law
A
c
b
C
B a
𝑐 = √𝑎2 + 𝑏 2 − 2𝑎𝑏 𝑐𝑜𝑠 𝐶
Sine Law
𝑎
sin 𝐴
=
𝑏
sin 𝐵
=
𝑐
sin 𝐶
ILLUSTRATIVE EXAMPLE S
2.1. If the magnitude of the resultant force (F R ) is to be 9 kN
directed along the positive x-axis, determine the magnitude of
force T acting on the eyebolt and its angle θ.
Solution:
Let us use the parallelogram method
reflecting the resultant force as described
in the problem. The tensions of the ropes
T and 8 kN, as components, should be the
sides of the parallelogram.
To determine the unknown quantities, use
the triangle rule.
To solve for T,
use the Cosine law
T = √(9)2 + (8)2 – 2(9)(8) cos 45°
T = 6.57 kN
To get the value of the angle θ, let us use the Sine law
8
6.57
=
sin (90° − 𝜃)
sin 45°
6.57 sin (90° − 𝜃) = 8 sin 45°
Page 10
From your lesson in Plane Trigonometry, you have learned that
𝐬𝐢𝐧(𝟗𝟎° − 𝜽) = 𝐜𝐨𝐬 𝜽. Hence,
6.57 cos 𝜃
= 8 sin 45°
8 sin 45°
𝜃 = arc cos ( 6.57 )
𝜽 = 30.57°
2.2. Determine the magnitude of the resultant force acting on
the bracket and its direction measured counterclockwise from
the positive u-axis.
[Solution:
Let us again use parallelogram method in locating the
resultant force graphically. It is assumed that it is located
below the positive u-axis. But it can be otherwise, the result
will, later, show if your assumption is correct. Next, let us use the
triangle rule as shown.
Note: As you have observed in the previous figures, wherever the
name of the axis is located, that is the positive side of that axis.
This will be observed in all our other examples.
Let us solve for the magnitude of F R using the Cosine law
F R = √(200)2 + (150)2 – 2(200)(150) cos 75°
F R = 216.73 lb
To get the value of the angle α, let us use the Sine law
200
216.73
=
sin 𝛼
sin 75°
216.73 sin 𝛼
= 200 sin 75°
200 𝑠𝑖𝑛75°
𝛼 = 𝑎𝑟𝑐 𝑠𝑖𝑛 ( 216.73 )
𝛼 = 63.05°
Let us now solve the direction of the resultant from the positive
u-axis (ϕ),
𝛼 = 60° + ϕ
ϕ = 𝛼 − 60° = 63.05° − 60°
ϕ = 3.05°
This only means that our assumption that the resultant lies below
the positive u-axis is correct.
2.3. Resolve F 1 and F 2 into components along the u and v
axes and determine the magnitudes of these components.
Solution:
Let us start in finding the components of F 1 with the aid
of parallelogram and triangle methods.
Page 11
Using sine law:
250
𝐹1𝑢
=
sin 105°
sin 45°
F 1 u = 183.01 N
250
𝐹1𝑣
=
sin 105°
sin 30°
F 1 v = 129.41 N
Next, let us find the components of F 2 .
Using sine law:
150
𝐹2𝑢
=
sin 75°
sin 75°
F 2 u = 150 N
150
𝐹2𝑣
=
sin 75°
sin 30°
F 2 v = 77.65 N
LEARNING EXERCISE No. 2
I – If ϕ = 20°, F 1 = 15 kN, and
resultant force is to be directed
positive y-axis, determine the
magnitude of the resultant force
be a minimum. Also, what is F 2
angle θ?
the
along the
II – Combine the two forces P
which act on the fixed
at B, into a single equivalent
and
T,
structure
force R.
III - The log is being
two tractors A and B.
the magnitudes of the
forces F A and F B if it
that
the
resultant
a magnitude F R = 15
directed along the x
= 20°.
Page 12
if F 2 is to
and
the
towed
by
Determine
two towing
is required
force have
kN and be
axis. Set θ
IV - Three cables pull on the
that they create a resultant
a magnitude of 900 lb. If two
are
subjected
to
known
shown
in
the
figure,
angle θ of the third cable so
magnitude of force F in this
minimum. All forces lie in the
What is the magnitude of F?
find the resultant of the two
pipe
such
force having
of the cables
forces,
as
determine the
that
the
cable
is a
x-y
plane.
Hint:
First
known forces.
FEEDBACK
How was the discussion so far? Can you follow the solution
procedure in the illustrative problems?
You may have noticed that Lesson 1 is just a brief review of
the important concepts in your Physics 1 that are very useful in this
course, Statics of Rigid Bodies. For learning exercise 1, first law of
Newton was asked because I would like you to reflect and visualize
this law, since in Statics, equilibrium is the main focus. You should
also be able to memorize and familiarize all the conversion factors
in order to answer the other questions.
For learning exercise 2, you should keep in mind the basic
steps in solving these problems: parallelogram method, triangular
rule, and the laws of cosine and sine. In the first problem, if you
have set the value of θ = 90°, then you are correct that at this angle
the value of F 2 is at minimum. You may countercheck this by
assigning other values of θ.
In the fourth problem, you should have set the resultant of the
two known forces to be collinear with force F. This is where F is at
its minimum value.
Make sure to analyze carefully your
parallelogram in order for you to get the correct internal angle
needed in using sine law. From your Plane Geometry, sum of
internal angles of a parallelogram is 360° and opposite internal
angles are equal.
Page 13
Another means to solve these problems, although less
practical and less accurate , is by graphical method. You
just have to make sure to use the proper scaling to arrive
at a precise answer. Personally, I suggest that you resort to this
method to just countercheck your computed values and if you have
spare time.
SUMMARY
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Here is a short recap of what you have learned in Module 1.
Statics is the study of bodies that are at rest or move with constant
velocity.
A particle has a mass but a size that can be neglected.
A rigid body does not deform under load.
Concentrated forces are assumed to act at a point on a body.
Newton's three laws of motion should be memorized. First Law is
the main focus of Statics.
Mass is measure of a quantity 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.
In the SI system the unit of force, the newton, is a derived unit.
The meter, second, and k ilogram are base units.
Prefixes G, M, k, m, μ, and n are used to represent large and small
numerical quantities. Their exponential size should be known,
along with the rules for using the SI units.
A scalar is a positive or negative number.
A vector is a quantity that has a magnitude, direction, and sense.
Multiplication or division of a vector by a scalar will change the
magnitude of the vector. The sense of the vector will change if the
scalar is negative.
As a special case, if the vectors are collinear, the resultant is
formed by an algebraic or scalar addition.
In vector addition of forces, use parallelogram method and
triangular rule. Then use cosine and sine law to solve for the
unknown quantities .
SUGGESTED READINGS
https://mathalino.com/reviewer/engineering mechanics/principles -statics
https://www.physicsclassroo.com/class/vectors/Lesson 1/Vector-Resolution
REFERENCES
Hibbeler, R.C., Engineering Mechanics, Statics, Amazon, 14th
Edition
Kraige, L. G. and Meriam, I. L., Engineering Mechanics:
STATICS, 5 t h Edition, John Wiley & Sons, Inc., 2002
Page 14