MULUNGUSHI UNIVERSITY
DEPARTMENT OF ENGINEERING
EGM 251: STATICS
CHAP 1: INTRODUCTION TO STATICS
BY Eng. LILANDA R.
CHAPTER OBJECTIVES
To provide an introduction to the basic quantities and
idealizations of mechanics.
To give a statement of Newton’s Laws of Motion and
Gravitation.
To review the principles for applying the SI system of units.
To examine the standard procedure for performing numerical
calculations.
To present a general guide for solving problems.
1.1 Mechanics
Mechanics is a branch of physics that is concerned with the behahavior
of physical bodies subject to Force, F of Displacement,s
1. Statics
Rigid Body 2. Dynamics
1. Thermodynamics
Mechanics Deformable Body 2. Solid Mechanics
3. Mechanics of Materials
1. Aerodynamics
Fluid Mechanics
2. Fluid Dynamics
Statics – The study of Rigid bodies at rest or at a constant velocity. It
concerns the determination of internal and external forces forces acting
on a body.
a=0 v = constant or v = 0
1.2 Fundamental Concepts
a. Basic Quantities
Length – used to locate a body in space or describe the size of a
body.
Time – conceived as the succesion of events.
Mass – a measure of a quantity of matter used to compare the
action of one body with another.
Force – considered as a ‘’push’’ or ‘’pull’’ exerted by one body on
another.
b. Idealizations
Models or idealizations are used in mechanics in order to simplify
application of the theory.
Particle – A particle has a mass, but a size that can be negleced.
Rigid Body – 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. (Has mass and
size)
Concentrated Force – A concentrated force represents a load acting on
a point on a boby.
c. Newton’s Three Laws of Motion
First Law – A particle at rest or moving with constant velocity tends to
remain in this state provided the particle is NOT subjected to an
unbalaced force.
𝛴𝐹 = 0
Second Law – A particle acted upon by an unbalanced Force
experiences an acceleration that has the SAME direction as the Force
and a magnitude that is directly propotional to the Force.
Third Law – The mutual Forces of action and Reaction between two
particles are equal, opposite, and colinear.
𝐹𝐴 = 𝐹𝐵
FA = -FB
𝛴𝐹 = FA – FB = 0
d. Newton’s Law of Gravitational Attraction
This Law was pospulated by Newton and governs the gravitational
attraction between any two particles.
m1m2
𝐹=𝐺 2
r
e. Weight
Weight is the Force exerted on an object with mass due to Earth’s gravity
W = mg g = 9.81 m/s2
1.3 Units of Measurement
Unit conversion
The table below provides a set of direct conversion factors between FPS
and SI units for the basic quantities.
1.4 The International System of Units
Prefixes
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.
Rules to use
• Quantities defined by several units which are multiple of one another
are separated by a dot to avoid cofusion with prefix notoation.
e.g N = Kg.m/s2
• The exponetial power on a unit having a prefix refers to both the unit
and its prefix.
e.g μN2 = (μN)2 = μN x μN ; mm2 = mm . mm
• With the exception of the base unit the kilogram, in general avoid the
use of a prefix in the denominator of composite units.
e.g do not write N/mm, but rather KN/m
• When performing calculations, represent the numbers in terms of base or
derived units by converting all prefixes to powers of 10. The final result
should then be expressed using a single prefix. Also, after calculation, it is
best to keep numerical values between 0.1 and 1000; otherwise, a suitable
prefix should be chosen.
e.g (50 kN)(60 nm) = [50(103)N][60(10-9)m]
= 3000 (10-6)N.m
= 3(10-3) N.m
= 3mN.m
1.5 Numerical calculations
In numerical calculations, it is important that the answers to any problem
be reported with both justifiable accuracy and appropriate significant
figures.
a. Dimensional Homogeneity. The terms of any equation used to
describe a physical process must be dimensionally homogeneous: that
is, each term must be expressed in the same units.
b. Significant figures. The number of significant figures contained in any
number determinnes the accuracy of the number.
Number Significant Digits
425.0 3
4250.0 4
42500.0 5
Note: when zeroes occur at the end of a whole number, it may be unclear
as to how many significant figures the number represent. Therefoe, to
avoid these ambiguities, Engineering Notation is used.
c. Engineering Notation. Round off to appriopraite significant digits and
then express in power of ten (multiples of 103)
Examples a. 42500 42.5 x 103
b. 0.00821 8.21 x 10-3
c. 000582 0.582 x 10-3 or 582 x 10-6
Round off numbers.
Generally we round off to three significant digits
Rules and examples
>5 3.5587 Round up 3.56
<5 3.5547 Round down 3.55
0dd 10.75 Round up 10.8
Even 10.25 Round down 10.2
Zero 10.05 Round up 10.1
Note. DO NOT round off intermediate calculations. Only the Final answer
should be rounded off.
1.6 General procedure of Analysis
• Read the problem carefully and try to correlate the actual physical
situation with the theory studied.
• Tabulate the problem data and draw any necessary diagrams.
• Apply the relevant principles, generally in the mathematical form. When
writing any equation, be sure they are dimensionally homogeneous.
• Solve the necessary equations, and report the answer with no more than
three significant figures.
Chapter 1 Examples
1. Convert 2 km/h to m/s. How many ft/s is this?
Solution
We know that 1km = 1000m and 1h = 3600, therore
2 𝑘𝑚 1000 𝑚 1ℎ
2 km/h = ( )( )
h km 3600 s
= 0.555555
= 0.556 m/s
Part b
We know that 1 ft = 0.3048 m. Thus
0.556 𝑚 1 𝑓𝑡
0.556 m/s = ( )( )
s 0.3048 m
= 1.82414...
= 1.82 ft/s
2. Evaluate each of the following and express with SI units having an
appropriate prefix.
a. (50 mN)(6 GN)
b. (400 mm)(0.6 MN)2
c. 45 MN3 /900Gg
Solutions
a. (50 mN)(6 GN)
(50 mN)(6 GN) = (50 x 10-3 N)(6 x 109)
= 300 x 106 N2
= 300 kN2
Note: kN2 = (kN)2 = 106N2
b. (400 mm)(0.6 MN)2
= [400(10-3)m][0.6(106)N]2
= [400(10-3)m][0.36(1012)N2]
= 144(109)m.N2
=144Gm.N2
c. 45 MN3/900Gg
45 𝑀𝑁3 45(106𝑁)3
= =
900 G𝑔 900(106) 𝑘𝑔
= 50(109) N3/kg
= 50 kN3/kg
