C2 Lecture 2.ppt

Microsoft PowerPoint - C2 Lecture 2.ppt [Compatibility Mode]LEARNING OUTCOMES
1. To show how to add forces and resolve them into components using
the Parallelogram Law.
2. To express force and position in Cartesian vector form and explain
how to determine the vector’s magnitude and direction.
PROF. DR. SHAHRUDDIN BIN MAHZAN@MOHD ZIN
SCALARS AND VECTORS
– Indicated by letters in italic such as A
E.g.: Mass, volume and length
Scalar Multiplication and Division
SCALARS AND VECTORS
E.g.: Position, force and moment
– Represented by a letter with an arrow over it such as or A
– Magnitude is designated as or simply A
In this subject, vector is presented as A and its magnitude (positive quantity) as A
– Represented graphically as an arrow
– Length of arrow = Magnitude of Vector
– Angle between the reference axis and arrow’s line of action = Direction of Vector
– Arrowhead = Sense of Vector
VECTOR OPERATIONS
- Product of vector A and scalar a = aA
- Magnitude =
- If a is positive, sense of aA is the same as sense of A
- If a is negative sense of aA, it is opposite to the sense of A
- Negative of a vector is found by multiplying the vector by ( -1 )
- Law of multiplication applies
VECTOR ADDITION USING THE PARALLELOGRAM LAW OR TRIANGLE CONSTRUCTION
Parallelogram Law:
Triangle method (always ‘tip to tail’):
How do you subtract a vector? How can you add more than two concurrent vectors graphically ?
RESOLUTION OF VECTOR
“Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse.
VECTOR ADDITION OF FORCES
When two or more forces are added, successive applications of the parallelogram law is carried out to find the resultant
E.g.: Forces F1, F2 and F3 acts at a point O
- First, find resultant of F1 + F2
- Resultant, FR = ( F1 + F2 ) + F3
Procedure for Analysis - Parallelogram Law
To resolve a force into components along two axes directed from the tail of the force
- Start at the head, constructing lines parallel to the axes - Label all the known and unknown force magnitudes and angles - Identify the two unknown components
Procedure for Analysis – Trigonometry
- Redraw half portion of the parallelogram
- Magnitude of the resultant force can be determined by the law of cosines
- Direction of the resultant force can be determined by the law of sines
Example 1
The screw eye is subjected to two forces F1 and F2. Determine the magnitude and
direction of the resultant force.
Parallelogram Law Unknown: magnitude of FR and angle θ
Trigonometry - Law of Cosines
VECTOR ADDITION OF FORCESVECTOR ADDITION OF FORCES
Example 2
The force F acting on the frame has a magnitude of 500N and is to be resolved into two components acting along the members AB and AC. Determine the angle θ, measured below the horizontal, so that components FAC is directed from A towards C and has a magnitude of 400N.
Parallelogram Law
VECTOR ADDITION OF FORCESVECTOR ADDITION OF FORCES
By Law of Cosines or Law of Sines
Hence, show that FAB has a magnitude of 561N
θθ ∠=−−= 1.769.4360180
ADDITION OF A SYSTEM OF COPLANAR FORCESADDITION OF A SYSTEM OF COPLANAR FORCES
For resultant of two or more forces:
Find the components of the forces in the specified axes
Add them algebraically
Form the resultant
In this subject, we resolve each force into rectangular forces along the x and y axes.
yx FFF +=
-x and y axes are designated positive and negative
-Components of forces expressed as algebraic scalars
yx FFF += Eg:
Sense of direction along positive x and negative y axes
E.g.:
)'('' yx FFF −+=
ADDITION OF A SYSTEM OF COPLANAR FORCESADDITION OF A SYSTEM OF COPLANAR FORCES
If scalar notation are used
FRx = (F1x - F2x + F3x) FRy = (F1y + F2y – F3y)
In all cases, FRx = ∑Fx FRy = ∑Fy
*Take note of sign conventions
SCALAR NOTATIONSCALAR NOTATION
CARTESIAN VECTOR NOTATIONCARTESIAN VECTOR NOTATION
Each component of the vector is shown as a magnitude and a direction.
We ‘ resolve’ vectors into components using the x and y axes system
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.
Cartesian unit vectors i and j are used to designate the x and y directions
F = Fxi + Fyj F’ = F’xi + F’y(-j)
F’ = F’xi – F’yj
The x and y axes are always perpendicular to each other. Together,they can be directed at any inclination.
• Step 1 is to resolve each force into its components
• Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector.
• Step 3 is to find the magnitude and angle of the resultant vector
F1 = F1xi + F1yj F2 = - F2xi + F2yj F3 = F3xi – F3yj
Example 3
Determine x and y components of F1 and F2 acting on the boom. Express each force as a Cartesian vector
Scalar Notation
By similar triangles

= −
Example 4
The link is subjected to two forces F1 and F2. Determine the magnitude and
orientation of the resultant force.
Scalar Notation
Resultant Force
Cartesian Vector Notation
Thus, FR = F1 + F2
= {236.8i + 582.8j}N
Given: Three concurrent forces acting on a bracket.
Find: The magnitude and angle of the resultant force.
a) Resolve the forces in their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Plan:
F1 = { 15 sin 40° i + 15 cos 40° j } kN
= { 9.642 i + 11.49 j } kN
F2 = { -(12/13)26 i + (5/13)26 j } kN
= { -24 i + 10 j } kN
F3 = { 36 cos 30° i – 36 sin 30° j } kN
= { 31.18 i – 18 j } kN
Summing up all the i and j components respectively, we get,
FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN
= { 16.82 i + 3.49 j } kN
x
y
φ
FR
φ = tan-1(3.49/16.82) = 11.7°
ATTENTION QUIZ
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 (F1 + F2) force in N when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N .
A) 30 N B) 40 N C) 50 N
D) 60 N E) 70 N
30°
HOMEWORK TUTORIAL
Q1(2-32) : Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis.
Given: F1 = 70N F2 = 50N F3 = 65N θ = 30° φ = 45°
HOMEWORK TUTORIAL
Q2 (2-33): Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.
Given: F1 = 50N F2 = 35N α = 120° β = 25°
HOMEWORK TUTORIAL
Q3 (2-35) :
Three forces act on the bracket. Determine the magnitude and direction θ of F1
so that the resultant force is directed along the positive x' axis and has a
magnitude of FR.
HOMEWORK TUTORIAL
Q4 (2.26) :
Member BD exerts on member ABC a force P directed along line BD. Knowing that P
must have a 960-N vertical component, determine
(a)The magnitude of the force P,
(b) its horizontal component.
Q5 (2.35):
Knowing that α = 35°, determine the resultant and the direction of the three
forces shown.
HOMEWORK TUTORIAL

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C2 Lecture 2.ppt