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EML 3004C
CHAPTER 9
Statics, Dynamics, and Mechanical
Engineering
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-1
EML 3004C
Objectives
Understand the concept of a Vector
Write a vector in component form
Understand the concepts of a couple and moment
Construct a free-body diagram of a physical system
Sections
9.1
9.2
9.3
9.4
9.5
9.6
Introduction
The Concept of a Vector
Forces, Couples, and Moments
Equilibrium and Free-body Diagrams
Frictional Forces
Motion of a Rigid Body
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-2
EML 3004C
9.1 Introduction
 Statics deals with bodies (Solid, Liquid, Gases) that are in
equilibrium with applied forces. At static equilibrium, a body is
at rest or in constant motion (no acceleration).
 Newton’s Law
 First: A body is at rest or in uniform motion (constant
velocity) along a straight-line unless acted upon by some
external (unbalanced) forces.
 Second: Every body accelerates in the direction of external
forces. The acceleration is proportional to the magnitude of
the force.
 Third: For every action, there is an equal and opposite
reaction.
 Statics is critical in the study of bodies under forces in the design.
Free body diagram isolates forces/bodies and is a critical tool.
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-3
EML 3004C
9.2 Concept of a Vector
 While scalar has only magnitude, e.g.: mass, temperature -vector
has magnitude and direction, e.g. velocity, force, moment
Head
Magnitude of the vector
Tail
 The above is a geometric representation of the magnitude of the
vector is unity, it is called a unit vector
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-4
EML 3004C
9.2.1Component of Vector
F Fx  Fy
=Fx ˆi+Fˆ y ˆj (2-D)
=Fx ˆi+Fy ˆj+Fz kˆ (3-D)
ˆi, ˆj, kˆ are unit vectors
Magnitude of F
F  Fx  Fy + Fz
2
Namas Chandra
Introduction to Mechanical engineering
2
2
Chapter 9-5
EML 3004C
9.2.2Component of Vector
If F=22 N is being pulled at 30, then
Fx = F sin  = 22 sin  = 11N 
Fy = F cos  = 22 cos  = 11 3N 
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-6
EML 3004C
9.2.3Direction Cosines of Vector
Let  ,  ,  be the angle of vector F has with coordinate axis
projection of F on x-axis
cos  =
magnitude of F
Fx
=
F
Fy
Fz
cos  =
; cos  =
F
F
Fx ˆ Fy ˆ
F
Unit vector c 
i+
j + z kˆ = cos ˆi+cos ˆj+cos kˆ
F
F
F
Note that cos 2  cos 2   cos 2  1
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-7
EML 3004C
9.2.4Addition/Subtraction of Vectors
 Since vectors have magnitude and directions only components
can be added
For example, adding 2iˆ to 3jˆ directly as (2+3) makes no sense
So always resolve components and then add
Example:
F  3iˆ + 5jˆ +8kˆ Fˆ  2iˆ + 4jˆ + 5kˆ
2
Fˆ3  Fˆ1 + Fˆ2 = ˆi + 9jˆ + 13k̂
Fˆ4  Fˆ1 - Fˆ2 = 5iˆ + ˆj + 3 kˆ
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-8
EML 3004C
9.3 Forces, Couples, and Moments
 Magnitude of a couple:
M = Fd
 Moment of a force
M = Force x Lever arm
 The direction of the moment is normal to the plane containing the
force and the arm.
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-9
EML 3004C
9.4 Equilibrium and Free-Body Diagram
 For static equilibrium, using Newton’s Second Law
F 
0 and  M = 0
 Free body diagrams show all the forces and boundary conditions
on a given body.
Use the following steps
 Isolate the body from surroundings and draw a simple
sketch
 Pick the right coordinate system (2D, 3D, Cartesian,
Cylindrical)
 Add all extreme forces (right direction) and internal forces
 Label all known quantities (magnitude/direction)
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-10
EML 3004C
9.5 Frictional forces
 A body on a surface (horizontal or sloped) is acted upon by
weight acting vertically down and a reaction normal to the surface
Using Coloumb’s Law for Static case
fs  s N
f s = frictional force
s  Static coefficient of friction
N = Normal force
 For bodies in motion,
fk  k N
 k  Kinetic frictional coefficient, usually lower than s
s and  k do not depend on the force, only on the pair of surfaces
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-11
EML 3004C
9.5.2 Frictional Forces
Given 15 Kg crate up 30
degree slope with  s  0.3
Find: Force required to push the 15 kg crate up the hill
Solution:
F  0
M  0
Draw the free body diagram
For sliding f = s N = .3 N
 F  m a  P-m g sin 30 + f =0
 F  m a  N -m g cos 30 =0
Now  M  0  f (0.1)+N (x) = 0; x= 0.03
x
x
y
y
0
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-12
EML 3004C
9.5.2 Frictional Forces..2
Given 15 Kg crate up 30
degree slope with  s  0.3
m
N=m g cos 30 = 15 kg  9.81 2 cos 30 = 127.4 N
s
P=m g sin 30 + f  P=111.8 N
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-13
EML 3004C
9.6 Motion of Rigid Body
If
 F  0,
then there is unbalanced force leading the acceleration and hence motion
Now
F  m a
 M  I
I - moment of inertia (kg  m 2 or slug  ft 2 )
 rad 
 - angular acceleration  2 
s 
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-14
EML 3004C
9.6.1 Crate Problem Revisited
If the friction =0.3 compute the dynamics of motion?
f = s N  0.3 N
Let s  0.3
F
y
 N - m g cos   0
N=m g cos 
F
x
 -m g sin  + f = m a x
f= N = 0.3N = 0.3 (m g cos 30)
Thus, a x = -g sin 30 + (0.3) g cos 30
= g (0.3 cos 30-sin 30)
m
(0.3 cos 30-sin 30)
sec 2
m
=2.36

2
sec
a x = 9.81
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-15
EML 3004C
CHAPTER 9…concludes
Statics, Dynamics, and Mechanical
Engineering
Namas Chandra
Introduction to Mechanical engineering
Chapter 9-16