em102 topic 2 - force vectors
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EM102 Engineering Statics
Topic 2: Force Vectors
Scalars and vectors,
vector operations,
cartesian vectors,
position vectors,
equilibrium of a particle.
Dr. Elango Natarajan
Assistant Professor
Aug 2015
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In-Class activities:• Reading Quiz
• Application of Adding Forces
•Resolution of a Vector Using
Cartesian Vector Notation (CVN)• Addition Using CVN
• Law of mechanics
•Resultant of two forces system
•Resultant of three forces system
•Resultant of multi forces system
Objective:
Students will be able to : a) Resolve a 2-D vector into
components.
b) Add 2-D vectors using
Cartesian vector notations.
FORCE VECTORS, VECTOR OPERATIONS &ADDITION COPLANAR FORCES
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Scalars and vectors
Scalar is a quantity characterized by a positive or
negative number.
Example: Mass, Volume and Length
Vector is a quantity that has both a magnitude and a
direction.
Example: Position, force and moment.
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Scalars Vectors
Examples: Mass, Volume Force, Velocity
Characteristics: It has a magnitude It has a magnitude
(positive or negative) and direction
Addition rule: Simple arithmetic Parallelogram law
Special Notation: None Bold font, a line, an
arrow or a “carrot”
In these PowerPoint presentations, a vector quantity is represented li ke this (in bold,
italics, and red).
SCALARS AND VECTORS(Section 2.1)
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2. For vector addition, you have to use ______ law.A) Newton’s Second
B) the arithmetic
C) Pascal’s
D) the parallelogram
READING QUIZ
1. Which one of the following is a scalar quantity?
A) Force B) Position
C) Mass D) Velocity
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There are three concurrentforces acting on the hook
due to the chains.
We need to decide if the
hook will fail (bend or break).
To do this, we need to know
the resultant or total forceacting on the hook as a
result of the three chains.
FR
APPLICATION OF VECTOR ADDITION
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VECTOR OPERATIONS (Section 2.2)
Scalar Multiplicationand Division
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FORCE SYSTEM
Before we proceed further with vectors, I would like
to introduce you the force systems.
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Force
• Force is an action that can change motion.
• Force is what we call a push or a pull.
•
Force is any action that has the ability to change anobject’s motion.
• Forces can be used to increase the speed of an object,
decrease the speed of an object, or change the direction in
which an object is moving.
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Force - Examples
• an object’s weight (self weight)
• tension in a rope
• friction
• attraction between an electron and proton
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Force
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Gravitational Force
Bodies don’t have to be in contact to exert forces on each other,
e.g., gravity.
Gravity is attraction between any two bodies.
It is weakest but most dominant.
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Net force
When more than one force acts on a body, the net force (resultant
force) is the vector combination of all the forces, i.e., the “net effect.”
When forces act in the same line, we can just add or subtract their
magnitudes to find the net force.
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Net force Examples
F1
F2
F3
Fnet
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Net force - Example
2 kg
15 N 32 N
10 N
This is an example of collinear force system
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Addition of Collinear force vectors
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Inertia
• Inertia is a term used to measure the ability of an object to resist a
change in its state of motion.
• An object with a lot of inertia takes a lot of force to start or stop;
• An object with a small amount of inertia requires a small amount
of force to start or stop.
• The word “inertia” comes from the Latin word inertus, which can
be translated to mean “lazy”.
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Inertia
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Moment of a force
The Moment of a force is a measure of its tendency to cause a body to
rotate about a specific point or axis.
This is different from the tendency for a body to move, or translate, in
the direction of the force.
In order for a moment to develop, the force must act upon the body in
such a manner that the body would begin to twist.
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Moment
This occurs every time a force is applied so that it does not pass
through the centroid of the body.
A moment is due to a force not having an equal and opposite force
directly along it's line of action.
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Moment - More Examples
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Couple
A couple is defined as two forces (coplaner) having the same
magnitude, line of action but in opposite sense.
It has pure rotation.
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Force system and Static Equilibrium
A force system is a collection of forces acting at specified
locations
When all the forces that act upon an object are balanced, then the
object is said to be in a state of equilibrium.
The forces are considered to be balanced if the rightward forces are
balanced by the leftward forces and the upward forces are balanced
by the downward forces.
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Static Equilibrium
This however does not necessarily mean that all the forces
are equal to each other.
If an object is at equilibrium, then the forces are balanced.
Balanced is the key word that is used to describe equilibrium
situations.
Thus, the net force is zero and the acceleration is 0 m/s2.
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Types of force system
• Coplanar force systems have all the forces acting in in one plane.
• They may be concurrent, parallel, non-concurrent or non-parallel.
Concurrent coplanar
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Resolution of a Vector Using
Cartesian Vector Notation (CVN)
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VECTOR ADDITION USING EITHER THEPARALLELOGRAM LAW OR TRIANGLE
Parallelogram Law:
A and B are joined at their tails, Parallel lines are drawn from the head of each
vector intersect at a common point, thereby forming the adjacent sides of a
parallelogram.
The resultant R is a diagonal of the parallelogram.
R = A + B = B + A.
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VECTOR ADDITION USING EITHER THEPARALLELOGRAM LAW OR TRIANGLE
Triangle method(always ‘tip to tail’):
A and B are joined in a “head – to- tail” fashion (by connectionthe head of A to the tail of B).
The resultant R extends from the tail of A to the head of B.
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VECTOR SUBTRACTION USING EITHER THEPARALLELOGRAM LAW OR TRIANGLE
R' = A - B = A + (-B)
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“Resolution” of a vector is breaking up a vector into components.
It is kind of like using the parallelogram law in reverse.
RESOLUTION OF A VECTOR
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RESOLUTION OF A VECTOR - procedure
• If F in Fig. (a) is to be resolved into components acting along
the lines u and v , one starts at the head of F and extends a
line parallel to u until it intersects v .
• Likewise, a line parallel to v is drawn from the head of F to
the point of intersection with u.
• Two components F u and F
v are then drawn such that they
extend from the tail of F to the points of intersection as
shown in Fig. (b).
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ADDITION OF A SYSTEM OF COPLANAR FORCES(Section 2.4)
• Each component of the vector is
shown as a magnitude and adirection.
• The directions are based on the x and y axes. We use the
“unit vectors” i and j to designate the x and y-axes.
• We ‘resolve’ vectors intocomponents using the x and y-axis
coordinate system.
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For example,
F = Fx i + Fy j or F ' = F'x i + ( F'y ) j
The x and y-axis are always perpendicular to each other.
Together, they can be “set” at any inclination.
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• Step 2 is to add all the x-
components together, followed by
adding all the y-componentstogether. These two totals are the
x and y-components of the
resultant vector.
• Step 1 is to resolve each forceinto its components.
ADDITION OF SEVERAL VECTORS
• Step 3 is to find the magnitude
and angle of the resultant
vector.
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Break the three vectors into components, then add them.
F R = F 1 + F 2 + F 3
= F1x i + F1y j F2x i + F2y j + F3x i F3y j
= (F1x F2x + F3x) i + (F1y + F2y F3y) j
= (FRx) i + (FRy) j
An example of the process:
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You can also represent a 2-D vector witha magnitude and angle.
1
tan
Ry
Rx
F
F
2 2
R Rx Ry F F F
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1. Resolve F along x and y axes and write it in
vector form. F = { ___________ } N
A) 80 cos (30°) i – 80 sin (30°) j
B) 80 sin (30°) i + 80 cos (30°) j
C) 80 sin (30°) i – 80 cos (30°) j
D) 80 cos (30°) i + 80 sin (30°) j
2. Determine the magnitude of the resultant (F 1 + F 2) force in N
when F 1 = { 10 i + 20 j } N and F 2 = { 20 i + 20 j } N .
A) 30 N B) 40 N C) 50 N
D) 60 N E) 70 N
30°
xy
F = 80 N
ATTENTION QUIZ
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Example 1
The link shown in Fig. is subjected to two forces F1 and
F2. Determine the magnitude and orientation of the
resultant force using Cartesian vector notation.
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Example 1 - solution
• Resolving forces as shown:
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Example 1 - solution
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Example 2 – solve yourself
Determine the
magnitude and
direction of theresultant force by
resolution of forces.
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Example 2 - solution
Answer:
F 1 = 1.41i-1.41j
F 2 = -3i-5.196jF R=-1.59i-6.606j
Magnitude of the resultant force is 6.79 kN
Direction of the resultant force is 76.47
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Plan:
a) Resolve the forces into their x-y components. b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
EXAMPLE 3
Given: Three concurrent forces
acting on a tent post.
Find: The magnitude and
angle of the resultant
force by resolution of
forces.
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F 1 = {0 i + 300 j } N
F 2 = { – 450 cos (45°) i + 450 sin (45°) j } N
= { – 318.2 i + 318.2 j } N
F 3 = { (3/5) 600 i + (4/5) 600 j } N
= { 360 i + 480 j } N
EXAMPLE 3 - solution
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Summing up all the i and j components respectively, we get,
F R = { (0 – 318.2 + 360) i + (300 + 318.2 + 480) j } N
= { 41.80 i + 1098 j } N
x
y
FR Using magnitude and direction:
FR = ((41.80)2 + (1098)2)1/2 = 1099 N
= tan-1(1098/41.80) = 87.8°
EXAMPLE 3 - solution
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Plan:
a) Resolve the forces into their x and y-components. b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Exercise 4 – solve yourself
Given: Three concurrent
forces acting on a bracket.
Find: The magnitude and
angle of the resultant
force. Show the
resultant in a sketch.
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F 1 = {850 (4/5) i 850 (3/5) j } N
= { 680 i 510 j } N
F 2 = {- 625 sin (30°) i 625 cos (30°) j } N
= {- 312.5 i 541.3 j } N
F 3 = {-750 sin (45°) i + 750 cos (45°) j } N
{- 530.3 i + 530.3 j } N
Exercise 4 – solution
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Summing all the i and j components, respectively, we get,
F R = { (680 312.5 530.3) i + (510 541.3 + 530.3) j }N
= { 162.8 i 520.9 j } N
Now find the magnitude and angle,
FR = (( 162.8)2 + ( 520.9)2) ½ = 546 N
= tan – 1
( 520.9 / 162.8 ) = 72.6°
From the positive x-axis, = 253°
Exercise 4 – solution
FR
x
y
-520.9
-162.8
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Exercise 5 – solve yourself
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Exercise 5 - solution
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Exercise 5 - solution
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Exercise 5 - solution
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Exercise 5 - solution
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Exercise 6 – solve yourself
Given: Three concurrent
forces acting on a
bracket
Find: The magnitude and
angle of the
resultant force.
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Exercise 6 - solution
• F 1 = [ 300 (5/13) i + 300 (12/13) j ] N
= [ 115.4 i + 276.9 j ] N
F 2 = [500 cos (30°) i + 500 sin (30°) j ] N= [ 433.0 i + 250 j ] N
F 3 = [ 600 cos (45°) i 600 sin (45°) j ] N
= [ 424.3 i 424.3 j ] N
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Exercise 6 - solution
Summing up all the i and j components respectively, weget
F R = [ (115.4 + 433.0 + 424.3)i + (276.9 + 250 - 424.3)] j
= [972.7 i + 102.7 j ] N
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You’ll have more similar exercises in Tutorial class.
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Further more…
Law of Mechanics
Solving problems on two forces system
Solving problems on three forces system
Solving problems on multi forces system
L f M h i
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Law of Mechanics
Newton’s First Law of Motion
Second Law of Motion
Third Law of Motion
Gravitational Law of Attraction
Parallelogram Law of Forces
Sine Law
Cosine Law
Lami’s Theorem
Principle of Transmissibility of Force
Fi l f N
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First law of Newton
“An object in motion tends to stay in motion; an object at rest tends to stay
at rest.”
Fi t L f N t
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First Law of Newton
•A moving body will continue moving in the samedirection with the same speed until some net forceacts on it.
•A body at rest will remain at rest unless a net forceacts on it.
•Summing it up: It takes a net force to change abody’s velocity .
S d l f N t
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Second law of Newton
• F net = m a
•The acceleration (a) an object is directly proportion
to the net force acting on it.
•Mass (m) is the constant of proportionality.
M f M l ti
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More force – More acceleration
Thi d l f N t
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Third law of Newton
“For every action there is an equal and oppositereaction.”
G it ti l l f tt ti
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Gravitational law of attraction
This law states that two particles of mass m1 and m2
are attracted towards each other along the line
connecting them with a force whose magnitude F is
proportional to the product of their masses and
inversely proportional to the square of the distance(r ) between them.
G it ti l l f tt ti
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Gravitational law of attraction
= (
)
where G is the universal constant of constant ofgravitation and its value is
(66.73±0.03) × 10-12 m3/kgs2
Parallelogram law of forces (two forces system)
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Parallelogram law of forces (two forces system)
If two forces F1 and F2 acting at a point to be
represented in magnitude and direction by the two
adjacent sides of a parallelogram, then their
resultant is represented in magnitude and direction
by the diagonal of the parallelogram at that point.
Parallelogram law of forces
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Parallelogram law of forces
Application of Parallelogram law
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Application of Parallelogram law
Application of Parallelogram law
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Application of Parallelogram law
Sine law (three forces system)
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Sine law (three forces system)
=
=
The law of sines relates two sides and the angles opposite
them. So any time you have two angles (and then can
easily figure out the third), it's easy to use the law of sines:
Angle-Side-Angle or Angle-Angle-Side.
Cosine law (three forces system)
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Cosine law (three forces system)
The law of cosines relates 3 sides and an
angle. So if you know 3 sides, you can use it,
or if you know two sides and an angle, you
can find the third side. However, because of
the form of the equation, if you have anangle that's not between the two sides, you
get a quadratic equation in that case--a bit
messy.
So the law of cosines is best for Side-Side-
Side and Side-Angle-Side.
=
+
− 2
=
+
− 2
=
+
− 2
Lami’s theorem (three forces system)
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Lami s theorem (three forces system)
If three forces acting at a point are in equilibrium,
each force will be proportional to the sine of the
angle between the other two forces.
Lami’s theorem
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Lami s theorem
1
=
2
=
3
Principle of transmissibility of forces
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Principle of transmissibility of forces
The state of rest or motion of a rigid body is
unaltered if a force acting on the body is replaced by
another force of the same magnitude and direction
but acting anywhere on the body along the line of
action of the replaced force.
Principle of transmissibility of forces
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Principle of transmissibility of forces
Two forces system
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Two forces system
Two forces system
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Two forces system
Two forces system
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Two forces system
Exercise 1
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Exercise 1
Determine the magnitude
and direction of the
resultant force.
Exercise 1 - solution
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Exercise 1 solution
Exercise 1- solution
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Exercise 1 solution
Exercise 1- solution
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Exercise 1 solution
Exercise 2 - do it yourself
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Exercise 2 do it yourself
Exercise 2 - solution
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Exercise 2 solution
Exercise 2 - solution
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e c se so u o
Exercise 3 – do it yourself
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y
Exercise 3 - solution
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• Resultant force FR = 629 N
• Angle of resultant force = 37.89
• Angle of resultant force from x axis = 30+37.89 = 67.89
Exercise 4 – do it yourself
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y
Exercise 4 - solution
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Exercise 4 -solution
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Exercise 5 – do it yourself
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y
Exercise 5 - solution
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Exercise 5 - solution
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Exercise 6 – do it yourself
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Exercise 6 -solution
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Exercise 6 - solution
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Exercise 7 – do it yourself
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Exercise 8 – do it yourself
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Exercise 9 - do it yourself
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Exercise 10 - do it yourself
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Exercises to be done in tutorial classes
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• Exercises on representation of forces in Cartesian vector.
• Exercises on two forces system – using Parallelogram law
•
Exercises on three forces system – using Parallelogram law orresolution forces
• Exercises on multi forces system – using resolution of forces.
TOPIC 2 FORCE VECTORS COMPLETED
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