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BMCG 1113 STATICS
Dr. Isa Halim
Faculty of Manufacturing Engineering
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Learning Objectives
At the end of this chapter, students should be able to: Differentiate between scalars and vectors.
Apply Parallelogram (PG) Law, Scalar Notation, and
CartesianVector Notation toperform forces addition.
Calculate the Resultant Force (FR), its magnitudeanddirection.
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APPLICATIONS OF FORCE VECTORS
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Scalar
- A scalaris physical quantitythat has a magnitude
and does not depend on direction.- It does not changed by coordinate systemrotations.
-It can be indicated by a positive or negative number.
- Examples of scalar: mass, volume, time and length.
SCALARS AND VECTORS
http://en.wikipedia.org/wiki/Physical_quantityhttp://en.wikipedia.org/wiki/Directionhttp://en.wikipedia.org/wiki/Coordinate_systemhttp://en.wikipedia.org/wiki/Coordinate_systemhttp://en.wikipedia.org/wiki/Directionhttp://en.wikipedia.org/wiki/Physical_quantity8/11/2019 Chapter 02 Force Vectors
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Vector
- A vector is quantity that has both magnitude anddirection. For e.g., position, force and moment.
- Represented by a letter with an arrow over it
- Magnitude (positive quantity) is designated by
-In figure below, the vector is shown by an arrow.
- The length of arrow is magnitude.
- The angle () is direction.
- The arrow head is sense of direction.
SCALARS AND VECTORS cont)
A
A
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VECTOR OPERATIONS
Multiplication and Division of a Vector by a Scalar
- If vector (A) is multiplied by a positive scalar (a), its
magnitude increase. Magnitude =
- If vector (A) is multiplied by a negative scalar (-a), its
sense of direction changed.
aA
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Vector Addition (+)
- Addition of two vectors Aand Bgives a resultantvector Rby theparallelogram law.
- Resultant Rcan be found by triangle construction.
- Resultant e.g. R= A+ B= B+ A
- Special case when vectors Aand Bare collinear(bothhave the same line of action).
VECTOR OPERATIONS cont)
triangle construction
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VECTOR OPERATIONS cont)
Vector Subtraction (-)- Vector subtraction is special case of addition.
- So, the rules of Vector Addition also apply to vector
subtraction.
- Example: R= AB= A+ ( - B)
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VECTOR ADDITION OF FORCES
Finding a Resultant Force
Parallelogram lawis carried out to find the resultant
force
Resultant force, FR
FR= ( F1+ F2 )
Problem Parallelogram Triangle construction
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VECTOR ADDITION OF FORCES cont)
Procedures for Analysis: Step 1: Paral lelog ram Law.
Step A:Sketch a thin line from the
arrow head of F1, parallel to F2.
Step B:Sketch a thin line from thearrow head of F2, parallel to F1.
Step C: Sketch a diagonal line
from the origin of F1 and F2 to
intersection point of the two thin
lines.
- Resultantforce(FR) is shown by
the diagonal of the parallelogram.
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VECTOR ADDITION OF FORCES cont)
Procedures for Analysis:
Step 2: Tr igonometry.
Step A: Redraw half portion of the
parallelogram.
Step B:Apply the cosines law
to determine the magnitude of the
resultant force (FR) .
Step C:Apply the sines law to
determine the directionof the
resultant force (FR) .Step D:Apply the sines law to
determine the magnitude of the two
components.
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Copyright 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 1
The hook is subjected to two forces,F1and F2.
Determine:
1) The magni tudeof resultantforce (FR).
2) The direct ionof the resultantforce (FR) measured from thepositive x-axis.
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Copyright 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 1 cont)
Step 1: Parallelogram
Law
Unknown:
1) Magnitude of resultant
force, FR.
2) Direction of resultant
force, measured from
the positive x-axis.
x-axis
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EXAMPLE 1 cont)
Solution
Step 2: TrigonometryCosines law
Sines law
75.39
639.0sin
578.234
150sin
578.234sin
150
115sin
6.212
sin
150
N
N
NN
N
NNNNCFR
6.2124226.0300002250010000
115cos150100215010022
A
B
C
c
C
b
B
sinsin
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Copyright 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 1 cont)
Solution
Trigonometry
Direction of FR
measured from the
positive x-axis:
75.54
1575.39
)(
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ADDITION OF A SYSTEM OF COPLANAR FORCES
Scalar Notation
x and y axes are designated positive and negative
Components of forces expressed as algebraic
scalars
sin
and,cos
,
FF
FF
but
FFF
y
x
yx
Fy
cos
/cos
FF
FF
x
x
sin
/
FF
FFSin
y
y
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ADDITION OF A SYSTEM OF COPLANAR FORCES
cont)
Cartesian Vector Notation Cartesian unit vectors iandjare used to designate
the x and y directions.
Unit vectors iandjhave dimensionless magnitude
of unity ( = 1 ).
Magnitude is always a
positive quantity,
represented by
scalars Fxand Fy
jFiFF yx
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Coplanar Force ResultantsTo determine resultant of
several coplanar forces:
Resolve force into x and y
components. Addition of the respectivecomponents using scalaralgebra.
Resultant force is found
using the parallelogram law Cartesian vector notation on
top right:
jFiFF
jFiFF
jFiFF
yx
yx
yx
333
222
111
ADDITION OF A SYSTEM OF COPLANAR FORCES
cont)
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Definition of CoplanarCoplanar is a set of points, lines, or any othergeometrical shapes that lie on the same plane.
ADDITION OF A SYSTEM OF COPLANAR FORCES
cont)
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ADDITION OF SEVERAL VECTORS
Step 3: Find the magnitudeand
direction (angle)of the resultant
vector.
Step 1: Resolveeach force intoits components.
Step 2: Add all the x components
together. Then add all the y
componentstogether. These twototals become the resultant vector.
Coplanar Force Resultants
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Coplanar Force Resultants Vector resultant is therefore:
If scalar notation is used:
jFiF
FFFF
RyRx
R
321
yyyRy
xxxRx
FFFFFFFF
321
321
ADDITION OF A SYSTEM OF COPLANAR FORCES
cont)
F1F2
F3
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Coplanar Force Resultants In all cases we have
Magnitude of FRcan be found by Pythagorean Theorem:
yRy
xRx
FF
FF * Take note of sign conventions
ADDITION OF A SYSTEM OF COPLANAR FORCES
cont)
22
RyRxR FFF
Rx
Ry
F
FTherefore 1-tan,
Rx
Ry
F
Ftan
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EXAMPLE 1
Given: Three concurrent forces
acting on a bracket.Find: The magnitude and
direction (angle) of the
resultant force.
Solution Steps:
Step 1: Resolve the forces in their x-y components.
Step 2: Add the respective components to get the resultant vector.
Step 3: Find magnitude and angle from the resultant components.
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Solution 1 (using Cartesian vector notation)
Component x - axis y axis
F1 {15 sin 40i} kN= {9.642i} kN
{15 cos 40j} kN{11.49j} kN
F2 {-(12/13)26i} kN
= {-24i} kN
{(5/13)26j} kN
= {10j} kN
F3 { 36 cos 30i} kN
{ 31.18i} kN
{36 sin 30j} kN
{18j} kN
Summing up all the iandjcomponents respectively, we get,
FR= { (9.642 24 + 31.18) i+ (11.49 + 10 18)j} kN
= { 16.82i+ 3.49j} kN x
yF
R
3.49
16.82
Step 1:
Step 2:
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Solution 2 (using scalar notation)
Component x - axis y axis
F1 {15 sin 40} kN= {9.642} kN
{15 cos 40} kN= {11.49} kN
F2 {-(12/13)26} kN
= {-24} kN
{(5/13)26} kN
= {10} kN
F3 { 36 cos 30} kN
= { 31.18} kN
{36 sin 30} kN
= {18} kN
Summing up all the xaxis, we get,
FR(x)= { (9.642 24 + 31.18) } kN
= 16.82 kN
Summing up all the yaxis, we get,
FR(y)= { (11.49 + 10 18)} kN
= 3.49 kN
x
y
FR
3.49
16.82
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EXAMPLE(continued)
x
y
FR
FR= ((16.82)2+ (3.49)2)1/2= 17.2 kN
The angle:
= tan-1
(3.49/16.82) = 11.7
The magnitude:16.82
3.49Rx
Ry
RyRxR
F
F
FFFMagnitude
1-
22
tanDirection,
,
Formula:
Step 3:
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Copyright 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 2
Determine x and y
components of F1and
F2acting on the boom.
Express each force as
a Cartesian vector.
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Solution 3 (using scalar notation)
Component x - axis y axis
F1 {- 200 sin 30}N= -100 N
{200 cos 30}N= 173.205 N
F2 {260 (12/13)}N
= 240 N
{-260 (5/13)}N
= -100 N
Summing up all the xaxis, we get,
FR(x)= { (-100 + 240) } N
= 140 N
Summing up all the yaxis, we get,
FR(y)= { (173.205 100)} N
= 73.205 N
Step 2:
Step 1:
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Solution 3 (using Cartesian vector notation)
Component x - axis y axis
F1 {- 200 sin 30i}N= -100iN
{200 cos 30j}N= 173.205j N
F2 {260 (12/13)i}N
= 240i N
{-260 (5/13)j}N
= -100jN
Summing up all the iandjcomponents respectively, we get,
FR= { (-100 + 240) i+ (173.205 - 100)j} N
= { 140i+ 73.205j} N
Expression of each force in Cartesian vector:
F1= { (-100 ) i+ (173.205 )j} N
F2 = { 240i- 100j} N
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EXAMPLE 2 cont)
Scalar Notation:
Cartesian Vector Notation:
NNNF
NNNF
y
x
17317330cos200
10010030sin200
1
1
Solution (if you want to solve without table)
(for component F1)
NjiF 1731001
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EXAMPLE 2 cont)
Scalar Notation:
Cartesian Vector Notation:
N10013
5260
N240
13
12260
2
2
NF
NF
y
x
NNF
NF
y
x
100100
240
2
2
NjiF 1002402
13
12
N260
2
xF
Solution (if you want to solve without table)
(for component F2)Use triangle ratio.
13
5
N260
2
yF Use triangle ratio.
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EXAMPLE 3
The link is subjected to two forces F1
and F2. Determine the magnitude andorientation of the resultant force.
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Solution 3 (using scalar notation)
Component x - axis y axis
F1 {600 cos 30}N= 519.615 N
{600 sin 30}N= 300 N
F2 {-400 sin 45}N
= -282.843 N
{400 cos 45}N
= 282.843 N
Summing up all the xaxis, we get,
FR(x)= { (519.615 282.843) } N
= 236.772 NSumming up all the yaxis, we get,
FR(y)= { (300 + 282.843)} N
= 582.843 N
Step 1:
Step 2:
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Solution 3 (using Cartesian vector notation)
Component x - axis y axis
F1 {600 cos 30i}N= 519.615 iN
{600 sin 30j}N= 300jN
F2 {-400 sin 45i}N
= -282.843 iN
{400 cos 45j}N
= 282.843j N
Summing up all the iandjcomponents respectively, we get,
FR= { (519.615 282.843) i+ (300 + 282.843)j} N
= { 236.772i+ 582.843j} N
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EXAMPLE 3 cont)
Resultant Force
Direction angle is:
N
NNFR
1.629
843.582772.23622
89.67
772.236
843.582tan 1
N
N
Step 3:
22
RyRxR FFF
Rx
Ry
F
F1-tan
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CARTESIAN VECTORS 3D)
Right-HandedCoordinate System
A rectangular or Cartesian
coordinate system is said tobe right-handed provided:
Thumb of right hand
points in the direction of
the positive z axis
z-axis for the 2D problem
would be perpendicular,
directed out of the page.
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CARTESIAN VECTORS cont)
Rectangular Components of a Vector A vector Amay have one, two or three rectangular
components along thex, yand zaxes, depending on
orientation.
By two successive applicationsof the parallelogram law,
A= A+ Az
A= Ax+ Ay
Combing the equations,
Acan be expressed as
A= Ax+ Ay+ Az
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CARTESIAN VECTORS cont)
Unit Vector
Direction of Acan bespecified using a unit vector.
Unit vector has a magnitudeof 1.
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CARTESIAN VECTORS cont)
Cartesian Vector Representations
3 components of Aact in the positive i,jand k
directions
A=Axi+Ayj+AZk
*Note the magnitude and direction
of each components are separated,
easing vector algebraic operations.
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1. Magnitude of a Cartesian Vector From the colored triangle,
From the shaded triangle,
Combining the equations
gives magnitude of A
CARTESIAN VECTORS cont)
222zyx AAAA
22' yx AAA
22' zAAA
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CARTESIAN VECTORS cont)
2. Direction of a Cartesian Vector Orientation of Ais defined as the coordinate direction
angles , and measured between the tail of Aand
the positive x, y and z axes.
Angle of , and is between 0
to 180
The direction cosines of Aare:
A
Axcos
A
Aycos
A
Azcos
Note: cos = adjacent / hypotenuse
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CARTESIAN VECTORS cont)
Direction of a Cartesian Vector Angles , and can be determined by the
inverse cosines
Given
A=Axi+Ayj+AZk
then,
uA= A/A = (Ax/A)i+ (Ay/A)j+ (AZ/A)k
where 222 zyx AAAA
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Direction of a Cartesian Vector uAcan also be expressed as
uA= cosi+ cosj+ cosk
Since and uA= 1, we have
Aas expressed in Cartesian vector form is
A=A uA=A cosi+A cosj+A cosk
=Axi+Ayj+AZk
222
zyx AAAA
1coscoscos 222
CARTESIAN VECTORS cont)
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Addition and Subtraction of
Forces - Concurrent Force
Systems:
Resultant force (FR) is thevector summation of all
the forces in the system
FR= F
= Fxi+ Fyj+ Fzk
ADDITION AND SUBTRACTION OF
CARTESIAN VECTORS
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EXAMPLE 4
Question:
Express the force Fas Cartesian vector.
Given:
1.F = 200 N
2. 60
3. = 45
Solution Plan:
1.Find .2.Express all forces in
Cartesian notation.
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EXAMPLE 4 cont)
Since two angles are specified, the third angle can be found by:
Two possibilities of will exist:
1205.0cos 1
605.0cos 1
5.0707.05.01cos
145cos60coscos
1coscoscos
22
222
222
Solution
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EXAMPLE 4 cont)
Solution
By inspection, = 60 since Fx must be in the +x direction
Given F= 200N
F= F cos i+ F cos j + F cos k
= (200N cos 60)i+ (200N cos 60)j+ (200N cos 45)k
= {100.0i+ 100.0j+141.4k}N
Checking:
N
FFFF zyx
2004.1410.1000.100222
222
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Wing strut
APPLICATIONS
How can we
represent the force
along the wing
strut in a 3-DCartesian vector
form?
POSITION VECTORS
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POSITION VECTORS
x,y,zCoordinates Apply right-handed
coordinate system.
Positive zaxis points
upwards and downwardsis negative, measuring the
height of an object or
point.
Points are measured
relative to the origin, O.
Example:
Point A = 4xi+ 2yj6zk
Point B = 6xi1yj+ 4zk
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POSITION VECTORS cont)
Position Vector (r) Note: pls. dont get confuse with resultant force (FR) Position vector ris defined as a fixed vector which
locates a point in space relative to another point.
Example: r=xi+ yj+ zk
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POSITION VECTORS cont)
Position Vector (r)
Vector addition gives rA + r= rB
Solving using the triangle construction:
r= rBrA= (xBxA)i+ (yByA)j+ (zBzA)k
hence,
r= (xBxA)i+ (yByA)j+ (zBzA)k
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POSITION VECTORS cont)
Length and direction of cable AB can be found bymeasuring A and B using thex, y, zaxes.
Position vector (r)can be established.
Magnitude of rrepresents the length of cable AB.
Angles, , and represent the
direction of the cable.
Unit vector, u
u= position vector/ magnitudeu = r/r
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EXAMPLE 5
An elastic rubber band is
attached to points A and B.
Determine its length and its
direction measured from A
towards B.
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EXAMPLE 5 cont)
SolutionStep 1: Position vector (r)
r= (xBxA)i+ (yByA)j+ (zBzA)kr= [-2m1m]i+ [2m0]j+ [3m(-3m)]k
= {-3i+ 2j+ 6k}m
Step 2: Magnitude = length of the rubber band
Step 3: Unit vector in the direction of r
u= r/r= -3/7i+ 2/7j+ 6/7k
mr 7623 222
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cos =xi/r
cos = yj/rcos = zk /r
EXAMPLE 5 cont)
Solution
Step 4: Direction from A
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FORCE VECTOR F) DIRECTED ALONG A LINE
In 3D problems, directionof Fis specified by 2 points,
through which its line
of action lies.
Fcan be formulatedas a Cartesian vector:
F= F u =F (r/r)
Note that Fhas
the unit of force (N)unlike r, with the
unit of length (m)
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Force Facting along the chain can be presented as aCartesian vector by:
- Establishx, y, zaxes.
- Form a position vector ralong length of chain.
Unit vector, u= r/rthatdefines the directionof both the chainand the force.
We get F= Fu
*Note: F = in Cartesian vector (i,j, k)
F= magnitude of force (e.g.: 100 N, 30 kN)
FORCE VECTOR DIRECTED ALONG A LINE
cont)
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EXAMPLE 6
The man pulls on the cordwith a force of 350 N.
Questions:
1) Express the force acting on
the support A, as a Cartesianvector,
2) Determine its direction.
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EXAMPLE 6 cont)
Solution
End points of the cord are:
At support, A: (0m, 0m, 7.5m) and
At mans hold, B:(3m, -2m, 1.5m)
Step 1: Position vector (r)
r= (xBxA)i + (yByA)j + (zBzA)k
r= (3m0m)i + (-2m0m)j + (1.5m7.5m)kr= {3i2j6k}m
Step 2: Magnitude = length of cord AB
Step 3: Unit vector (u)
u= r/r
= 3/7i- 2/7j- 6/7k
mmmmr 7623222
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EXAMPLE 6 cont)
Solution
Force Fhas a magnitude of350 N, direction specified by u.
Step 4: Force vector (F) along the cord
F= Fu
= 350N(3/7i- 2/7j- 6/7k)= {150i- 100j- 300k}N
Step 5: Direction
= cos-1(3/7) = 64.6
= cos-1(-2/7) = 107
= cos-1(-6/7) = 149
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