ENGR-1100 Introduction to
Engineering Analysis
Section 4
Instructor: Professor Suvranu De
Office: JEC 5002
Office Ph: x6096
E-mail: des@rpi.edu
Office hours: Tuesday and Friday 2:00-3:00 pm
Course Coordinator: Mohamed Aboul-Seoud
aboulm@rpi.edu x2317
Teaching assistants: TA
• Ademola Akinlalu (Graduate)
– Office hours: M 3P-5PM;T 10AM -12Noon;
W: 12Noon-2PM; R 3PM-5PM; F 10AM-12
Noon
– Office: JEC 1022
– E-mail: akinla@rpi.edu
• Ji Ming Hong (Undergraduate)
hong.j.10@gmail.com
Some Important Points
• Studio course (combined lesson & problem session)
• ALL CLASSES ARE
CLOSED-LAPTOP unless otherwise
stated.
• Bring your relevant textbook, calculators, pencil,
engineering computation paper to class
EVERYDAY
• Importance of laptop/MATLAB to solve
complicated problems
• Important tools: syllabus, 2 textbooks (listed in
syllabus), laptop, pencil and, engineering
computation paper
Course websites
• Course official web site:
http://www.rpi.edu/dept/core-eng/WWW/IEA
• My website for the course:
http://www.rpi.edu/~des/IEA2012Spring.html
• McGraw-Hill Connect website (for Home works) for
this section:
http://connect.mcgraw-hill.com/class/2012-iea-4
• McGraw-Hill Connect help:
http://create.mcgraw-hill.com/wordpressmu/success-academy/
Course handouts
• Course syllabus: download from
http://www.rpi.edu/~des/IEA2012Spring.html
• Supplementary information: download from
http://www.rpi.edu/~des/IEA2012Spring.html
• Connect quick steps: download from
http://www.rpi.edu/~des/IEA2012Spring.html
• Matlab tutorial: download from
http://www.rpi.edu/~des/IEA2012Spring.html
Course format
• Mini lectures
• Daily Class Activities (CA) 5% (drop 4 lowest
grades). NO makeup for CA.
• Daily Homeworks (HW). 15% (drop 2 lowest
grades). HWs due next day of class 12 noon. NO
LATE SUBMISSIONS!
• Three mid term exams (2/15, 3/21, 4/18) in SAGE
3510: 2@20% + 1@15%, total 55%
– Exam times: Wednesday 8 – 9:50 am
– Make-up exams (2/22 , 3/28, 4/25) in TBD 5:00-6:50pm
– Grade challenges must be within a week (6-8pm Exam 1:
2/20, 2/21; Exam 2: 3/26, 3/27; Exam 3: 4/23, 4/24)
– No make-ups for missed make-ups!
• 1 final exam (time: TBA) : 25%
Course objectives
 Formulation and solution of static equilibrium problems
for particles and rigid bodies.
 A bit of linear algebra: solution of sets of linear equations
as they arise in mechanics and matrix operations.
 Use your laptop (running Matlab) for the manipulation of
vector quantities and solution of systems of equations (only
as an aid to completely solve “realistic” problems)
Lecture outline
• Newton’s laws
• Units of measurement
• Vectors
Mechanics
Mechanics is the branch of science that deals with the
state of rest or motion of bodies under the action of forces
Mechanics
Mechanics of
rigid bodies
Mechanics of
deformable bodies
Mechanics of
fluids
In this class we will exclusively deal with the mechanics of
rigid bodies.
Few basic principles but exceedingly wide applications
Very large
Mechanics
Statics
Net force=0
Dynamics
Net force 0
In this class we will deal with the statics of
rigid bodies.
Very small
Modeling
Physical Problem
Physical idealizations: particles,
rigid body, concentrated forces, etc.
Physical Model
Physical laws: Newton’s laws
Applied to each interacting body
(free body diagram)
Mathematical model
(set of equations)
No!
Does answer
make sense?
YES!
Happy 
Solution of equations: Using
pen+paper/own code/ canned
software like Matlab
Physical Idealizations
Continuum:
For most engineering applications assume
matter to be a continuous distribution rather than a
conglomeration of particles.
Rigid body: A continuum that does not undergo any
deformation.
Particle: No dimensions, only has mass. Important
simplifying assumption for situation where mass is
more important than exactly how it is distributed.
Point force: A body transmits force to another
through a finite area of contact. But it is sometimes
easier to assume that a finite force is transmitted
through an infinitesimal area.
Newton’s Laws of Motion
Law I: (Principle of equilibrium of forces) A
particle remains at rest or continues to move in
a straight line with uniform velocity (this is what
we mean by being “in equilibrium”) if there is no
unbalanced force acting on it.






F  F1  F2  F3  ...... Fn  0
•Inertial reference frame
•Necessary condition for equilibrium
•Foundation of Statics
Vector equation
Newton’s Laws of Motion
Law II: (Nonequilibrium of forces) The
acceleration of a particle is proportional to the
resultant force acting on it and is in the direction of
this force.


Vector equation
F  ma





• F  F1  F2  F3  ...... Fn is
the
force acting on a particle of mass ‘m’.

F
resultant

a
• Foundation of Dynamics
• Necessary
condition for equilibrium corresponding
 
to a  0
Newton’s Laws of Motion
Law III: (Principle of action and reaction) If
one body exerts a force on a second body, then
the second body exerts a force on the first body
that is (1) equal in magnitude, (2) opposite in
direction and (3) collinear (same line of action).
Faction   Freaction
• EXTREMELY IMPORTANT to keep this in
mind when working out problems!!
Newton’s Laws of Motion
Force (F)
Pencil
Force (F)
W=mg weight of pencil
Table
R (force acting
on the pencil)
R (force acting
on the table)
• Need to isolate the bodies and consider the
forces acting on them (Free Body Diagram).
• Be careful about which force in the pair we
are talking about!
Law of gravitation
Two bodies of mass M and m are mutually attracted
to each other with equal and opposite forces F and –F
of magnitude F given by the formula:
.
where r is the distance between the center
of mass of the two bodies; and G is the
Universal Gravitational Constant.
G=3.439(10-8)ft3/(slug*s2) in the U.S
customary system of units.
G=6.673(10-11)m3/(kg*s2)
units
in SI system of
m
F=G
Mm
r2
M
F
r
Mass and weight
The mass m of a body is an absolute quantity.
The weight W of a body is the gravitational attraction
exerted on the body by the earth or by another
massive body such as another planet.
At the surface of the earth:
Where: Me is the mass of the earth.
re is the mean radius of the earth
Me
At sea level and latitude 45
g=G
re2
F=G
Me.m
re 2
=mg
0
g =32.17 ft/s2 = 9.807 m/s2
Units of measurement
• The U.S customary system of units (the
British gravitational system)
• Base units are foot (ft) for length, the pound (lb) for
force, and the second (s) for time
• Pound is defined as the weight at sea level and altitude
of 450 of a platinum standard
• The international system of units (SI)
• Three class of units
• (1) base units
• (2) derived units
• (3) derived units with special name
Base units
Quantity
Unit
Symbol
Length
meter
m
Mass
kilogram
kg
time
second
s
Derived units
Quantity
Unit
Symbol
Area
Square meter
m2
Volume
Cubic meter
m3
Linear velocity
Meter per second
m/s
Derived units with special name
Quantity
Unit
Symbol
Plane angle
radian
rad
Solid angle
steradian
sr
SI / U.S. customary units
conversion
Quantity
U.S. customary to SI SI to U.S. customary
Length
1 ft = 0.3048 m
1 m = 3.281 ft
Velocity
1 ft/s = 0.304 m/s
1 m/s = 3.281 ft/s
Mass
1 slug = 14.59 kg
1 kg = 0.06854 slug
Scalar and vectors
• A scalar quantity is completely described
by a magnitude (a number).
-Examples: mass, density, length, speed, time,
temperature.
• A vector quantity has
1. Magnitude
2. Direction (expressed by the line of action + sense)
3. Obey parallelogram law of addition
-Examples: force, moment, velocity, acceleration.
We will represent vectors by bold face symbols (e.g., F)
in the lecture. But, when you write, you can use the
symbol with an arrow on top (e.g., F )
Vectors: geometric representation
A vector is geometrically represented as a line segment
with an arrow indicating direction
F
Line of action
Head
Tail
Direction of arrow
direction of vector
Length of arrow
magnitude of vector
Question: What is a vector having the same magnitude and line of action, but opposite
sense?
Operations on Vectors:
Multiplication by scalars
magnitude (2F)
magnitude (F)
2F
F
magnitude (2F)
magnitude (F)
-F
-2F
nF, n is a scalar (negative or positive, integer or fraction)
n can be a fraction less than 1, can n be 0?
Operations on Vectors: Adding
vectors using Parallelogram Rule
Task: Add two vectors ( P and Q ) to obtain a “resultant” vector (R)
that has the same effect as the original vectors
Vectors are added using the Parallelogram law
P
P

R
Q
Q
R=P+Q=Q+P
•To obtain the resultant, add two vectors using parallelogram law
•Addition of vectors is commutative (order does not matter)
Vectors in rectangular coordinate
systems- two dimensional
y
v2
(v1,v2)
v
v1
O
x
If the tail of the vector (v) is at the origin, then the
coordinates of the terminal point (head) (v1,v2) are called
the Cartesian components of the vector.
V = v1 i + v2 j
Or,
v=(v1,v2)
Vectors in rectangular coordinate
systems- multiplication by a
scalar
y
2v2
(2v1,2v2)
2v
2v1
x
O
The components of the vector 2v are (2v1, 2v2)
The sum of two vectors – by adding
components (two dimensional )
y
v2
(v1+w1,v2+w2)
(w1,w2)
w
w2
v
v1
(v1,v2)
w1
Just add the x- and y-components
Or,
v+w=(v1+w1,v2+w2)
v + w = (v1 + w1 )i + (v2 + w2 ) j
x
Vectors in rectangular coordinate
systems- Three dimensional
z
v
v2
v3
O
x
(v1,v2,v3)
v2
y
v1
(v1,v2,v3) are the coordinates of the terminal point (head) of
vector v
The sum of two vectors – rectangular
components (Three dimensional )
z
(a1,a2,a3)
a
O
y
b
(b1,b2,b3)
x
a+b=(a1 +b1,a2+b2, a3 +b3)
Vectors with initial point not at the
origin (VERY IMPORTANT!!)
z
P1(x1 ,y1 ,z1) u
P2(x2 ,y2 ,z2)
w
O
v
y
w  OP1  ( x1 , y1 , z1 )
x
Hence
v  OP 2  ( x2 , y2 , z2 )
w+u v
 u= v-w
u = PP
1 2  ( x2  x1 , y2  y1 , z2  z1 )
Coordinates of
head minus
coordinates of
tail
Vectors with initial point not at the
origin (VERY IMPORTANT!!)
z
z2-z1
P1(x1 ,y1 ,z1)
x2-x1
O
x
u
P2(x2 ,y2 ,z2)
y2-y1
y
The components are the projections of the vector
along the x-, y- and z-axes
Example
Find the components of the vector having initial
point P1 and terminal point P2
P1(-1,0,2), P2(0,-1,0)
Solution:
Head (P2) minus tail (P1)
v= (0-(-1),-1-0,0-2)=(1,-1,-2)
Vector arithmetic
If u,v,w are vectors in 2- or 3-space and k and l are scalar,
then the following relationship holds:
(a) u+v=v+u
(b) u+0=0+u=u
(c) k(lu)=(kl)u
(d) (k+l)u=ku+lu
(e) (u+v)+w=u+(v+w)
(f) u+(-u)=0
(g) k(u+v)= ku+ kv
(h) 1u=u
Class assignment: (on a separate piece
of paper with your name and RIN on top) please
submit to TA at the end of the lecture
1. Find the component of the vector having initial point P1 and
terminal point P2
(a) P1 = (-5,0), P2 = (-3,1)
2. Let u = (-3,1,2), v = (4,0,-8) and w = (6,-1,-4). Find the x, y and z
components of:
(a) 6u + 2v
(b) -3(v – 8w)
IEA wisdom “Success in IEA is proportional to the
number of problems solved.”