Engineering Mechanics: Statics

Engineering Mechanics :
STATICS
DDA1013
Lecture #02
By,
Shaiful Rizal Bin Masrol
University Tun Hussein Onn Malaysia (UTHM),
Faculty of Mechanical and Manufacturing,
Department of Mechanical Engineering
Expected Lecture Date : 31/12 / 07

2
2D VECTOR ADDITION
Today’s Objective:
Students will be able to :
a) Resolve a 2-D vector into components
b) Add 2-D vectors using Cartesian vector
notations.
Learning topics:
• Application of adding forces

• Parallelogram law
• Resolution of a vector using
Cartesian vector notation (CVN)
• Addition using CVN
3
READING QUIZ
1. Which one of the following is a scalar quantity?

A) Force B) Position C) Mass D) Velocity
2. For vector addition you have to use ______ law.

A) Newton’s Second
B) the arithmetic
C) Pascal’s
D) the parallelogram
4
APPLICATION OF VECTOR ADDITION
There are four

concurrent cable forces
acting on the bracket.
How do you determine
the resultant force
acting on the bracket ?
5
SCALARS AND VECTORS (Section 2.1)
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 or
an arrow
6
VECTOR OPERATIONS (Section 2.2)
Scalar Multiplication
and Division
7
VECTOR ADDITION USING EITHER THE
PARALLELOGRAM LAW OR TRIANGLE
Parallelogram Law:
Triangle method
(always ‘tip to tail’):
How do you subtract a vector?

How can you add more than two
concurrent vectors graphically ?
8
RESOLUTION OF A VECTOR
“Resolution” of a vector is breaking up a vector into components. It
is kind of like using the parallelogram law in reverse.
9
CARTESIAN VECTOR NOTATION (Section 2.4)
• We ‘ resolve’ vectors into

components using the x and y axes
system
• Each component of the vector is

shown as a magnitude and a
direction.
• 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.
10
For example,
F = Fx i + Fy j or F' = F'x i + F'y j
The x and y axes are always perpendicular to each

other. Together,they can be directed at any inclination.
11
ADDITION OF SEVERAL VECTORS
• 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
12
Example of this process,
13
You can also represent a 2-D vector with a magnitude and
angle.
14
EXAMPLE
Given: Three concurrent forces
acting on a bracket.
Find: The magnitude and
angle of the resultant
force.
Plan:
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.
15
EXAMPLE (continued)
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
16
EXAMPLE (continued)
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
y
FR
FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN
 = tan-1(3.49/16.82) = 11.7° 
x
17
CONCEPT QUIZ
1. Can you resolve a 2-D vector along two directions, which are not
at 90° to each other?
A) Yes, but not uniquely.
B) No.
C) Yes, uniquely.
2. Can you resolve a 2-D vector along three directions (say at 0, 60,
and 120°)?
A) Yes, but not uniquely.
B) No.
C) Yes, uniquely.
18
IN CLASS TUTORIAL
Given: Three concurrent

forces acting on a
bracket
Find: The magnitude and
angle of the
resultant force.
Plan:
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.
19
IN CLASS TUTORIAL (continued)
F1 = { (4/5) 850 i - (3/5) 850 j } N

= { 680 i - 510 j } N
F2 = { -625 sin(30°) i - 625 cos(30°) j } N
= { -312.5 i - 541.3 j } N
F3 = { -750 sin(45°) i + 750 cos(45°) j } N
{ -530.3 i + 530.3 j } N
20
IN CLASS TUTORIAL (continued)
Summing up all the i and j components respectively, we get,

FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N
= { - 162.8 i - 521 j } N
y
FR = ((162.8)2 + (521)2) ½ = 546 N

x
= tan–1(521/162.8) = 72.64° or
From Positive x axis  = 180 + 72.64 = 253 ° FR
21
ATTENTION QUIZ
1. Resolve F along x and y axes and write it in
vector form. F = { ___________ } N
y
A) 80 cos (30°) i - 80 sin (30°) j x
B) 80 sin (30°) i + 80 cos (30°) j
C) 80 sin (30°) i - 80 cos (30°) j 30°
F = 80 N
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 22
23

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Engineering Mechanics :

Expected Lecture Date : 31/12 / 07

• Application of adding forces

1. Which one of the following is a scalar quantity?

2. For vector addition you have to use ______ law.

There are four

Addition rule: Simple arithmetic Parallelogram law

How do you subtract a vector?

• We ‘ resolve’ vectors into

• Each component of the vector is

• The directions are based on the x and y axes. We use the

The x and y axes are always perpendicular to each

• Step 1 is to resolve each force

F1 = { 15 sin 40° i + 15 cos 40° j } kN

Summing up all the i and j components respectively, we get,

Given: Three concurrent

a) Resolve the forces in their x-y components.

F1 = { (4/5) 850 i - (3/5) 850 j } N

Summing up all the i and j components respectively, we get,

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