E-Portfolio Mohd Fardzlee bin Abd Patah GB190077 MBV

DPP C2(b) -2
The vector is xi. The vector is yj. The vector is the sum of and , that is,
We now extend this to three dimensions to show how to construct the Cartesian form of a point P.
Define k to be a vector of length 1 in the direction of OZ. We now have the following picture.

Draw a perpendicular PT from P to the OZ axis.

In the rectangle OQPT,PQ and OT both have length z. The vector is zk. We know that = xi
+ yj. The vector , being the sum of the vectors and , is therefore

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This formula, which expresses in terms of i, j, k, x, y and z, is called the Cartesian

representation of the vector in three dimensions. We call x, y and z the components of
along the OX, OY and OZ axes respectively.
The formula

applies in all octants, as x, y and z run through all possible real values.

Length of a vector in terms of components .
The length of the vector in the XOY plane is

From the three dimensional picture below,

“cut out” the triangle OPQ to obtain:
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It can be seen that the length of the vector is = .

Addition and subtraction of vectors in terms of components
Given two vectors in Cartesian form

the sum + = is obtained by completing the parallelogram.

It can be proved that this is the same as the following calculation:

That is, the components of a sum are the sums of the components.
Notice that the parallelogram OQRP is part of a two dimensional plane sitting within three
dimensional space (in a tilted way like the slanting face of the roof of a house).

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Subtraction for the vector - , the rule

The rule for subtraction works in exactly the same way. Writing
above gives

The subtraction is illustrated below. Recall that = - = .

An example of the use of this rule is the calculation of the Cartesian form of the position vector
of a point P2 relative to a point P1. Suppose that in Cartesian form, = x1i + y1j + z1k and
= x2i + y2j + z2k.

Then

EXERCISE:
1. Write a force of 200 N at 20° to the horizontal in Cartesian form.
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2. A ship’s course is set to travel at 23 km/h, relative to the water, on a heading of 040°. A

current of 8 km/h is flowing from a bearing of 160°.
a) Write each vector as a Cartesian vector.
b) Determine the resultant velocity of the ship
REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor

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DPP C2(b) -2

Kolej Kolej Kemahiran Tinggi MARA
Masjid Tanah, Melaka.

INFORMATION SHEET

PROGRAMME : DIPLOMA IN AUTOMOTIVE ENGINEERING TECHNOLOGY
SESSION :
CODE/COURSE : OCTOBER – DECEMBER 2021 SEMESTER : 3
LECTURER : SHEET NO : 3
DKV 21273 STATICS & WEEK : 3
DYNAMICS
MOHD FARDZLEE BIN ABD
PATAH

TOPIC : FORCE VECTOR
SUB-TOPIC :
1.7Position Vectors
1.8Force Vector Directed Along a Line
1.9Dot Product

LEARNING After completing the course, students should be able to:
OUTCOME : 1. Apply Parallelogram Law to add forces and resolve into
components.
2. Express force and position in Cartesian vector form also
explain how to determine the vector's magnitude and
direction.
3. Understand the dot product in order to determine the
angle between two vectors or the projection of one
vector onto another.

CONTENT:

Position Vector
A vector that starts from the origin (O) is called a position vector.
In the following diagram, point A has the position vector a.

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Example:

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Force Vector Directed Along a Line
If a force is directed along a line, then we can represent the force vector in Cartesian coordinates
by using a unit vector and the force’s magnitude. So we need to:
Find the position vector, rBA, along two points on that line.
Find the unit vector describing the line’s direction, uBA= (rBA/rBA).
Multiply the unit vector by the magnitude of the force, F=FuBA.
Addition of a System of Coplanar Forces
The magnitude of a vector is always a (+ or -) quantity?
The magnitude of a vector is always a positive quantity.

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DPP C2(b) -2

Rectangular Components
Rectangular Components
A force resolved in to two (2D) or three (3D) components.
Rectangular components can be represented using either Scalar Notation or Cartesian Vector
Notation.
Scalar Notation
Scalar Notation
Rectangular Components of a force form a right triangle ∴ the magnitudes of the rectangular
components can be determined by SOH-CAH-TOA.
Scalar Notation Format: F = Fx + Fy
Cartesian Vector Notation
Cartesian Vector Notation
Rectangular Components of a force can be represented in terms of Cartesian unit vectors i and j
which are use to designate the x and y axis respectively.
Since the magnitude of each component of F is always a positive quantity represented by positive
scalars Fx and Fy, we can express F as a Cartesian vector.
Cartesian Vector Notation Format: F = Fx +Fy

Cartesian Vector Notation diagram
Coplanar Force Resultants
Coplanar Force Resultants

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To determine resultant of several coplanar forces:
- Resolve force into x and y components
- Addition of the respective components using scalar algebra
- Resultant force is found using the parallelogram law

Either Scalar Notation or Cartesian Vector Notation can be used to determine the resultant of
several coplanar forces by resolving each force into its x and y components.
F₁ = Fx₁ i + Fy₁ j
F₂ = Fx₂ i + Fy₂ j
F₃ = Fx₃ i + Fy₃ j
The vector resultant is therefore
FR = F₁ + F₂ + F₃
If Scalar Notation is used, we have
FRx = F₁x + F₂x + F₃x → FRx = ∑Fx
FRy = F₁y + F₂y + F₃y → FRy = ∑Fy

Coplanar Force Resultants diagram
Resolving the Magnitude and Direction of the Resultant Force Magnitude
FR = √(FR²x + FR²y)

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θ = tan⁻¹(FRy ÷ FRx)
Coordinate axes and Cartesian coordinates
By three dimensional space we mean the space we live in. To fix a point P in three dimensional
space requires a system of axes and three numbers. First select any point, call it the origin and
mark it as O. All measurements will from now on originate from this point O. Next place three
mutually perpendicular axes OX, OY , OZ through O. This axis system is drawn on a page like this:

Note that although it may not look it, the angles XOY , XOZ, Y OZ are all right-angles!
To fix any given point P in three dimensional space, we refer it to the axis system. Let us first show
the point P and the coordinate axes.

1. Drop a line from P perpendicular to the XOY plane (think of this plane as the floor), meeting
the XOY plane at a point Q (the foot of the perpendicular).
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2.

2. Now we are in the XOY plane with a point Q and an axis system in two dimensions.

3. Drop a perpendicular from Q to OX and OY .
4. Transfer the two-dimensional picture from 3 into the three dimensional diagram from 1.

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(Some of the right angles are marked: you should mark all the other right angles in the picture.)
5. Measure the lengths OR, OS, QP, and denote them by x, y, z respectively. We call the three
numbers (x,y,z), in the order given, the Cartesian coordinates of the point P.

Notice that the order in which the numbers are written is important: (1, 2, 1) and (2, 1, 1) are the
Cartesian coordinates of different points.

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To make the diagrams uncluttered, only half of each axis has been drawn. However, if the axes are
extended infinitely in both directions, it can be seen that this axis system creates eight octants in
space, just as the two dimensional axis system creates four quadrants in the plane. Points in the
eight different octants have Cartesian coordinates corresponding to all possible combinations of
positive and negative values of x, y and z. The figures above illustrate points in the octant
corresponding to positive values of x, y and z.
Cartesian form of a vector
We begin with two dimensions. We have the following picture illustrating how to construct the
Cartesian form of a point Q in the XOY plane.

Vectors i and j are vectors of length 1 in the directions OX and OY respectively.
The vector is xi. The vector is yj. The vector is the sum of and , that is,
We now extend this to three dimensions to show how to construct the Cartesian form of a point P.
Define k to be a vector of length 1 in the direction of OZ. We now have the following picture.

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Draw a perpendicular PT from P to the OZ axis.

In the rectangle OQPT,PQ and OT both have length z. The vector is zk. We know that = xi
+ yj. The vector , being the sum of the vectors and , is therefore

This formula, which expresses in terms of i, j, k, x, y and z, is called the Cartesian

representation of the vector in three dimensions. We call x, y and z the components of
along the OX, OY and OZ axes respectively.
The formula

applies in all octants, as x, y and z run through all possible real values.

Length of a vector in terms of components .
The length of the vector in the XOY plane is

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DPP C2(b) -2

From the three dimensional picture below,

“cut out” the triangle OPQ to obtain:

It can be seen that the length of the vector is = .

Addition and subtraction of vectors in terms of components
Given two vectors in Cartesian form

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DPP C2(b) -2
the sum + = is obtained by completing the parallelogram.

It can be proved that this is the same as the following calculation:

That is, the components of a sum are the sums of the components.
Notice that the parallelogram OQRP is part of a two dimensional plane sitting within three
dimensional space (in a tilted way like the slanting face of the roof of a house).

Subtraction for the vector - , the rule

The rule for subtraction works in exactly the same way. Writing
above gives

The subtraction is illustrated below. Recall that = - = .
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DPP C2(b) -2

An example of the use of this rule is the calculation of the Cartesian form of the position vector
of a point P2 relative to a point P1. Suppose that in Cartesian form, = x1i + y1j + z1k and
= x2i + y2j + z2k.

Then

EXERCISE:
1. Write a force of 200 N at 20° to the horizontal in Cartesian form.
2. A ship’s course is set to travel at 23 km/h, relative to the water, on a heading of 040°. A
current of 8 km/h is flowing from a bearing of 160°.
a) Write each vector as a Cartesian vector.
b) Determine the resultant velocity of the ship

REFERENCE :
1. Hibbeler, R.C & S.C Fan 1997. Engineering Mechanics, Statics. SI Ed. New Jersey:
Prentice Hall. ISBN 0135995981
2. Yusof Ahmad. 1999. Mekanik Statik. Cetakan Ketiga. Malaysia: Penerbit UTM Skudai Johor

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DPP B2

Kolej Kolej Kemahiran Tinggi MARA
Masjid Tanah, Melaka.

INFORMATION SHEET

PROGRAMME : DIPLOMA IN AUTOMOTIVE ENGINEERING TECHNOLOGY
SESSION :
CODE/COURSE : OCTOBER – DECEMBER 2021 SEMESTER : 3
LECTURER : SHEET NO : 1
DKV 21273 STATICS & WEEK : 1
DYNAMICS
MOHD FARDZLEE BIN ABD
PATAH

TOPIC : EQUILIBRIUM OF PARTICLE

SUB-TOPIC : 2.1 Condition for the Equilibrium of a Particle
2.2 The Free Body Diagram
2.3 Equilibrium Force
2.4 Coplanar Force
2.5 Three Dimensional Force Systems

LEARNING After completing the course, students should be able to:
OUTCOME : 1. Determine moment in force system resultant for 3D.
2. Determine forces and moments in equilibrium state.
3. Determine the centroids and moment of inertia of rigid body.
4. Analyze the motion of a body in curvilinear and angular.

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CONTENT:
KESEIMBANGAN ZARAH

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KESEIMBANGAN JASAD TEGAR

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