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Engineering Mechanics: Statics
Chapter 2:
Force Vectors
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Objectives
To show how to add forces and resolve theminto components using the Parallelogram Law.
To express force and position in Cartesianvector form and explain how to determine thevectors magnitude and direction.
To introduce the dot product in order todetermine the angle between two vectors orthe projection of one vector onto another.
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Chapter Outline
Scalars and Vectors
Vector Operations
Vector Addition of ForcesAddition of a System of Coplanar
Forces
Cartesian Vectors
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Chapter Outline
Addition and Subtraction ofCartesian Vectors
Position Vectors Force Vector Directed along a Line
Dot Product
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2.1 Scalars and Vectors
Scalar
A quantity characterized by a positive or
negative numberIndicated by letters in italic such asA
Eg: Mass, volume and length
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2.1 Scalars and Vectors
VectorA quantity that has both magnitude and
directionEg: Position, force and moment
Represent by a letter with an arrow over itsuch as orA
Magnitude is designated as or simply A
In this subject, vector is presented asAand itsmagnitude (positive quantity) as A
A
A
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2.1 Scalars and Vectors
Vector
Represented graphically as an arrow
Length of arrow = Magnitude ofVector
Angle between the reference axis
and arrows line of action = Direction ofVector
Arrowhead = Sense of Vector
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2.1 Scalars and Vectors
Example
Magnitude of Vector = 4 units
Direction of Vector = 20measuredcounterclockwise from the horizontal axis
Sense of Vector = Upward and to the right
The point O is called tail
of the vector and the point
P is called the tipor head
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2.2Vector Operations
Multiplication and Division of a Vectorby a Scalar- Product of vectorA and scalar a = aA
- Magnitude =
- If a is positive, sense of aAis the same assense ofA
- If a is negative sense of
aA, itis opposite to the
sense ofA
aA
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2.2Vector Operations
Multiplication and Division of aVector by a Scalar
- Negative of a vector is found by multiplyingthe vector by ( -1 )
- Law of multiplication applies
Eg:A/a = ( 1/a )A, a0
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2.2Vector Operations
Vector Addition- Addition of two vectorsAand Bgives aresultant vector Rby the parallelogram
law- Result Rcan be found by triangleconstruction
- CommunicativeEg: R=A+ B= B+A
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2.2Vector Operations
Vector Addition
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2.2Vector Operations
Vector Addition
- Special case: VectorsAand Barecollinear(both have the same line ofaction)
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2.2Vector Operations
Vector Subtraction
- Special case of addition
Eg: R =AB=A+ ( - B)
- Rules of Vector Addition Applies
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2.2Vector Operations
Resolution of Vector- Any vector can be resolved into twocomponents by the parallelogram law
- The two componentsAand Bare drawn suchthat they extend from the tail or Rto points ofintersection
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2.3Vector Addition of Forces
When two or more forces are added,successive applications of theparallelogram lawis carried out to find the
resultantEg: Forces F1, F2and F3acts at a point O
- First, find resultant of
F1 + F2- Resultant,
FR= ( F1+ F2 ) + F3
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2.3Vector Addition of Forces
Example
Faand Fbare forces exerting on the hook.
Resultant, Fccan be found using theparallelogram law
Lines parallel to a and b
from the heads of Faand Fbaredrawn to form a parallelogram
Similarly, given Fc, Faand Fbcan be found
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2.3Vector Addition of Forces
Procedure for Analysis
Parallelogram Law
- Make a sketch using the parallelogram law- Two components forces add to form theresultant force
- Resultant force is shown by the diagonal of the
parallelogram
- The components is shown by the sides of theparallelogram
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2.3Vector Addition of Forces
Procedure for AnalysisParallelogram Law
To resolve a force into components alongtwo axes directed from the tail of the force
- Start at the head, constructing linesparallel to the axes
- Label all the known and unknown forcemagnitudes and angles
- Identify the two unknown components
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2.3Vector Addition of Forces
Procedure for Analysis
Trigonometry
- Redraw half portion of the parallelogram- Magnitude of the resultant force can bedetermined by the law of cosines
- Direction if the resultant force can bedetermined by the law of sines
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2.3Vector Addition of Forces
Procedure for Analysis
Trigonometry
- Magnitude of the two components can bedetermined by the law of sines
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2.3Vector Addition of Forces
Example 2.1
The screw eye is subjected to two forces F1
and F2. Determine the
magnitude and direction
of the resultant force.
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2.3Vector Addition of Forces
Solution
Parallelogram Law
Unknown: magnitude ofFRand angle
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2.3Vector Addition of Forces
Solution
Trigonometry
Law of Cosines
N
N
NNNNFR
213
6.212
4226.0300002250010000
115cos1501002150100 22
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2.3Vector Addition of Forces
Solution
Trigonometry
Law of Sines
8.39sin
9063.0
6.212
150sin
115sin
6.212
sin
150
N
N
NN
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2.3Vector Addition of Forces
Solution
Trigonometry
Direction of FRmeasured from the horizontal
8.54
158.39
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2.3Vector Addition of Forces
Example 2.2Resolve the 1000 N ( 100kg) forceacting on the pipe into the components in the(a) x and y directions,(b) and (b) x and ydirections.
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2.3Vector Addition of Forces
Solution
(a) Parallelogram Law
From the vector diagram,yx
FFF
NF
NF
y
x
64340sin1000
76640cos1000
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2.3Vector Addition of Forces
Solution
(b) Parallelogram Law
'yx FFF
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2.3Vector Addition of Forces
Solution
(b) Law of Sines
NNF
NF
NNF
NF
y
y
x
x
108560sin
70sin1000
60sin1000
70sin
6.88460sin
50sin1000
60sin
1000
50sin
'
'
NOTE: A rough sketch drawn to scale will give some idea of the
relative magnitude of the components, as calculated here.
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2.3Vector Addition of Forces
Example 2.3The force Facting on the framehas a magnitude of 500N and is
to be resolved into two componentsacting along the members AB and
AC. Determine the angle ,measured below the horizontal,
so that components FACis directedfrom A towards C and has amagnitude of 400N.
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2.3Vector Addition of Forces
Solution
Parallelogram Law
ACAB FFN 500
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2.3Vector Addition of Forces
Solution
Law of Sines
9.43
6928.0sin
60sin
500
400sin
60sin500
sin400
N
N
NN
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2.3Vector Addition of Forces
Solution
By Law of Cosines or
Law of SinesHence, show that FAB
has a magnitude of 561N
1.769.4360180
,Hence
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2.3Vector Addition of Forces
Solution
Fcan be directed at an angle above the horizontalto produce the component FAC. Hence, show that
= 16.1and FAB= 161N
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2.3Vector Addition of Forces
Solution
(a) Parallelogram LawUnknown: Forces F1and F2
View Free Body Diagram
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2.3Vector Addition of Forces
Solution
Law of Sines
NF
NF
NF
NF
446
130sin1000
20sin
643
130sin
1000
30sin
2
2
1
1
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2.3Vector Addition of Forces
Solution
(b) Minimum length of F2occurwhen its line of action is
perpendicular to F1. Hencewhen
F2is a minimum
702090
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2.3Vector Addition of Forces
Solution
(b) From the vector
diagram
NNF
NNF
34270cos1000
94070sin1000
2
1
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2.4 Addition of a Systemof 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
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2.4 Addition of a Systemof Coplanar Forces
Scalar Notation- x and y axes are designated positive andnegative
- Components of forces expressed as algebraicscalars
Eg:
Sense of direction
along positive x and
y axes
yx FFF
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2.4 Addition of a Systemof Coplanar Forces
Scalar NotationEg:
Sense of directionalong positive x and
negative y axes
yx FFF '''
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2.4 Addition of a Systemof Coplanar Forces
Scalar Notation- Head of a vector arrow = sense of the
vector graphically (algebraic signs notused)
- Vectors are designated using boldface
notations- Magnitudes (always a positive quantity)are designated using italic symbols
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2.4 Addition of a Systemof Coplanar Forces
Cartesian Vector Notation- Cartesian unit vectors iandjare used todesignate the x and y directions
- Unit vectors iandjhave dimensionlessmagnitude of unity ( = 1 )
- Their sense are indicated by a positive ornegative sign (pointing in the positive ornegative x or y axis)
- Magnitude is always a positive quantity,represented by scalars Fxand Fy
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2.4 Addition of a Systemof Coplanar Forces
Cartesian Vector NotationF= Fxi+ Fyj F = Fxi+ Fy(-j)
F = Fx
iFy
j
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2.4 Addition of a Systemof Coplanar Forces
Coplanar Force ResultantsTo determine resultant of severalcoplanar forces:
- Resolve force into x and ycomponents
- Addition of the respective
components using scalar algebra- Resultant force is found using theparallelogram law
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2.4 Addition of a Systemof Coplanar Forces
Coplanar Force ResultantsExample: Consider three coplanar
forces
Cartesian vector notation
F1= F1xi+ F1yjF2= - F2xi+ F2yj
F3= F3xiF3yj
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Coplanar Force ResultantsVector resultant is therefore
FR= F1+ F2+ F3
= F1xi+ F1yj - F2xi+ F2yj + F3xiF3yj
= (F1x- F2x + F3x)i+ (F1y + F2yF3y)j
= (FRx)i+ (FRy)j
2.4 Addition of a Systemof Coplanar Forces
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2.4 Addition of a Systemof Coplanar Forces
Coplanar Force ResultantsIf scalar notation are usedFRx= (F1x- F2x + F3x)
FRy= (F1y + F2yF3y)
In all cases,
FRx
= Fx
FRy= Fy* Take note of sign conventions
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2.4 Addition of a Systemof Coplanar Forces
Coplanar Force Resultants- Positive scalars = sense of directionalong the positive coordinate axes
-Negative scalars = sense of directionalong the negative coordinate axes
- Magnitude of FRcan be found by
Pythagorean Theorem
RyRxR FFF 22
dd f
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2.4 Addition of a Systemof Coplanar Forces
Coplanar Force Resultants- Direction angle (orientation of theforce) can be found by trigonometry
Rx
Ry
F
F1tan
2 dd f S
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2.4 Addition of a Systemof Coplanar Forces
Example 2.5
Determine x and y components of F1and F2
acting on the boom. Express each force as a
Cartesian vector
2 4 Addi i f S
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2.4 Addition of a Systemof Coplanar Forces
Solution
Scalar Notation
Hence, from the slope
triangle
NNNFNNNF
y
x
17317330cos20010010030sin200
1
1
12
5tan
1
2 4 Addi i f S
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2.4 Addition of a Systemof Coplanar Forces
Solution
Alt, by similar triangles
Similarly,
NNF
N
F
x
x
24013
12260
13
12
260
2
2
NNF y 10013
52602
2 4 Additi f S t
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2.4 Addition of a Systemof Coplanar Forces
Solution
Scalar Notation
Cartesian Vector Notation
F1= {-100i +173j }N
F2= {240i -100j }N
NNF
NNF
y
x
100100
240240
2
2
2 4 Additi f S t
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2.4 Addition of a Systemof Coplanar Forces
Example 2.6
The link is subjected to two forces F1and
F2. Determine the magnitude andorientation of the resultant force.
2 4 Additi f S t
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2.4 Addition of a Systemof Coplanar Forces
Solution
Scalar Notation
N
NNF
FF
N
NNFFF
Ry
yRy
Rx
xRx
8.582
45cos40030sin600
:
8.236
45sin40030cos600:
2 4 Additi f S t
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2.4 Addition of a Systemof Coplanar Forces
Solution
Resultant Force
From vector addition,
Direction angle is
N
NNFR
629
8.5828.236 22
9.67
8.236
8.582tan 1
N
N
2 4 Additi f S t
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2.4 Addition of a Systemof Coplanar Forces
Solution
Cartesian Vector Notation
F1= { 600cos30i+ 600sin30j} N
F2= { -400sin45i+ 400cos45j} N
Thus,
FR= F1+ F2= (600cos30N - 400sin45N)i+(600sin30N + 400cos45N)j
= {236.8i+ 582.8j}N
2 4 Additi f S t
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2.4 Addition of a Systemof Coplanar Forces
Example 2.7
The end of the boom O is subjected to three
concurrent and coplanar forces. Determine
the magnitude and orientation of the
resultant force.
2 4 Additi f S t
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2.4 Addition of a Systemof Coplanar Forces
Solution
Scalar Notation
N
NNF
FF
NN
NNNF
FF
Ry
yRy
Rx
xRx
8.296
5
320045cos250
:
2.3832.383
5
420045sin250400
:
View Free Body Diagram
2 4 Additi f S t
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Solution
Resultant Force
From vector addition,
Direction angle is
2.4 Addition of a Systemof Coplanar Forces
N
NNFR
485
8.2962.383 22
8.37
2.383
8.296tan 1
N
N
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2.5 Cartesian Vectors
Right-Handed Coordinate SystemA rectangular or Cartesian coordinatesystem is said to be right-handed
provided:- Thumb of right hand points
in the direction of the positive
z axis when the right-handfingers are curled about this
axis and directed from the
positive x towards the positive y axis
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2.5 Cartesian Vectors
Right-Handed Coordinate System
- z-axis for the 2D problem would beperpendicular, directed out of the page.
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2.5 Cartesian Vectors
Rectangular Components of a Vector- A vectorAmay have one, two or threerectangular components along the x, yand zaxes, depending on orientation
- By two successive application of theparallelogram law
A=A +AzA =Ax+Ay
- Combing the equations, A can beexpressed as
A=Ax+Ay+Az
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2.5 Cartesian Vectors
Unit Vector- Direction ofAcan be specified using a unitvector
- Unit vector has a magnitude of 1- IfAis a vector having a magnitude ofA 0,unit vector having the same direction asAisexpressed by
uA=A/ASo that
A=A uA
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2.5 Cartesian Vectors
Unit Vector- SinceA is of a certain type, like forcevector, a proper set of units are used for the
description- MagnitudeA has the same sets of units,hence unit vector is dimensionless
-A( a positive scalar)
defines magnitude ofA
- uAdefines the direction
and sense ofA
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2.5 Cartesian Vectors
Cartesian Unit Vectors
- Cartesian unit vectors, i,jand kare usedto designate the directions of z, yand zaxes
- Sense (or arrowhead) of these
vectors are described by a plus
or minus sign (depending on
pointing towards the positive
or negative axes)
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2.5 Cartesian Vectors
Cartesian Vector Representations- Three components ofAact in the positive i,jand kdirections
A=Axi+Ayj+AZk
*Note the magnitude and
direction of each components
are separated, easing vector
algebraic operations.
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2.5 Cartesian Vectors
Magnitude of a Cartesian Vector- From the colored triangle,
- From the shaded triangle,
- Combining the equations gives
magnitude ofA
222
22
22
'
'
zyx
yx
z
AAAA
AAA
AAA
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2.5 Cartesian Vectors
Direction of a Cartesian Vector- Orientation of A is defined as thecoordinate direction angles , and
measured between the tail of A and thepositive x, y and z axes
- 0 , and 180
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2.5 Cartesian Vectors
Direction of a Cartesian Vector
- For angles , and (blue colored
triangles), we calculate the directioncosinesofA
A
Axcos
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2.5 Cartesian Vectors
Direction of a Cartesian Vector
- For angles , and (blue coloredtriangles), we calculate the directioncosinesofA
A
Aycos
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2.5 Cartesian Vectors
Direction of a Cartesian Vector
- For angles , and (blue coloredtriangles), we calculate the directioncosinesofA
A
Azcos
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2.5 Cartesian Vectors
Direction of a Cartesian Vector- Angles , and can be determined by theinverse cosines
- GivenA=Axi+Ayj+AZk
- then,
uA=A/A
= (Ax/A)i+ (Ay/A)j+ (AZ/A)k
where222
zyx AAAA
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2.5 Cartesian Vectors
Direction of a Cartesian Vector- uAcan also be expressed as
uA= cosi+ cosj+ cosk
- Since and magnitude of uA= 1,
-Aas expressed in Cartesian vector formA=AuA
=Acosi+Acosj+Acosk
=Axi+Ayj+AZk
222
zyx AAAA
1coscoscos 222
2 6 Addition and Subtraction
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ExampleGiven:A=Axi+Ayj+AZk
and B= Bxi+ Byj+ BZk
Vector AdditionResultant R=A+ B
= (Ax + Bx)i+ (Ay+ By )j+ (AZ + BZ)k
Vector SubstractionResultant R=A- B
= (Ax - Bx)i+ (Ay- By )j+ (AZ - BZ)k
2.6 Addition and Subtractionof Cartesian Vectors
2 6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
Concurrent Force Systems- Force resultant is the vector sum of allthe forces in the system
FR= F= Fxi+ Fyj+ Fzk
where Fx, Fy and Fzrepresent thealgebraic sums of the x, yand zor i,jor kcomponents of each force in the system
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2 6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
Cosines of their values forms a unit vector uthat
acts in the direction of the rope Force Fhas a magnitude of F
F= Fu= Fcosi+ Fcosj + Fcosk
2 6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
Example 2.8
Express the force Fas Cartesian vector
2 6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
Solution
Since two angles are specified, the third
angle is found by
Two possibilities exit, namely
or
605.0cos
5.0707.05.01cos
145cos60coscos
1coscoscos
1
22
222
222
1205.0cos 1
2 6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
SolutionBy inspection, = 60since Fxis in the +x
direction
Given F= 200NF= Fcosi+ Fcosj + Fcosk
= (200cos60N)i+ (200cos60N)j
+ (200cos45N)k
= {100.0i+ 100.0j+141.4k}N
Checking:
N
FFFF zyx
2004.1410.1000.100 222
222
2 6 Addition and Subtraction of
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2.6 Addition and Subtraction ofCartesian Vectors
Example 2.9
Determine the magnitude and coordinate
direction angles of resultant force acting onthe ring
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2 6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
SolutionUnit vector acting in the direction of FR
uFR= FR/FR
= (50/191.0)i+ (40/191.0)j+(180/191.0)k
= 0.1617i- 0.2094j+ 0.9422k
So that
cos= 0.2617 = 74.8
cos = -0.2094 = 102
cos= 0.9422 = 19.6
*Note > 90sincejcomponent of uFR is negative
2 6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
Example 2.10
Express the force F1as a Cartesian vector.
2 6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
Solution
The angles of 60and 45are not coordinate
direction angles.
By two successive applications of
parallelogram law,
2.6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
SolutionBy trigonometry,
F1z= 100sin60 kN = 86.6kN
F= 100cos60
kN = 50kNF1x= 50cos45 kN = 35.4kNF1y= 50sin45 kN = 35.4kN
F1yhas a direction defined byj,Therefore
F1= {35.4i35.4j+ 86.6k}kN
2.6 Addition and Subtraction
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SolutionChecking:
Unit vector acting in the direction of F1
u1= F1/F1
= (35.4/100)i- (35.4/100)j+ (86.6/100)k
= 0.354i- 0.354j+ 0.866k
2.6 Addition and Subtractionof Cartesian Vectors
N
FFFF zyx
1006.864.354.35 222
2
1
2
1
2
11
2.6 Addition and Subtraction
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Solution
1= cos-1(0.354) = 69.3
1= cos-1(-0.354) = 111
1= cos-1(0.866) = 30.0
Using the same method,
F2= {106i+ 184j- 212k}kN
2.6 Addition and Subtractionof Cartesian Vectors
2.6 Addition and Subtraction
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2.6 Addition and Subtractionof Cartesian Vectors
Example 2.11
Two forces act on the hook. Specify the
coordinate direction angles of F2
, so that the
resultant force FRacts along the positive yaxis
and has a magnitude of 800N.
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2.6 Addition and Subtractionof Cartesian Vectors
Solution
Cartesian vector form
FR= F1+ F2F1= F1cos1i+ F1cos1j + F1cos1k
= (300cos45N)i+ (300cos60N)j
+(300cos120N)k
= {212.1i+ 150j-150k}N
F2= F2xi+ F2yj + F2zk
View Free Body Diagram
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2.6 Addition and Subtractionof Cartesian Vectors
Solution
Since FRhas a magnitude of 800N and acts
in the +jdirection
FR= F1+ F2800j = 212.1i+ 150j-150k+ F2xi+ F2yj + F2zk
800j = (212.1 + F2x)i+ (150 + F2y)j + (-50 + F2z)k
To satisfy the equation, the correspondingcomponents on left and right sides must be equal
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2.6 Addition and Subtractionof Cartesian Vectors
SolutionHence,0 = 212.1 + F2x F2x= -212.1N800 = 150 + F2y F2y= 650N
0 =-150 + F2z F2z= 150N
Since magnitude of F2and its componentsare known,
1= cos-1
(-212.1/700) = 1081= cos
-1(650/700) = 21.81= cos
-1(150/700) = 77.6
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x,y,zCoordinates- Right-handed coordinate system
- Positive zaxis points upwards, measuring
the height of an object or the altitude of apoint
- Points are measured relative to theorigin, O.
2.7 Position Vectors
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x,y,zCoordinatesEg: For Point A, xA= +4m along the x axis,yA= -6m along the y axis and zA= -6m
along the z axis. Thus, A (4, 2, -6)Similarly, B (0, 2, 0) and C (6, -1, 4)
2.7 Position Vectors
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Position Vector
- Position vector ris defined as a fixed vectorwhich locates a point in space relative to another
point.Eg: If rextends from the
origin, O to point P (x, y, z)
then, in Cartesian vector
form
r= xi+ yj+ zk
2.7 Position Vectors
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Position VectorNote the head to tail vector addition of the
three components
Start at origin O, one travels x in the +i direction,y in the +j direction and z in the +k direction,
arriving at point P (x, y, z)
2.7 Position Vectors
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2.7 Position Vectors
Position Vector- Position vector maybe directed from point A topoint B
- Designated by ror rAB
Vector addition gives
rA + r= rB
Solvingr= rBrA= (xBxA)i + (yByA)j + (zBzA)k
or r= (xBxA)i + (yByA)j + (zBzA)k
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Position Vector- The i,j, kcomponents of the positive vector rmay be formed by taking the coordinates of thetail, A (xA, yA, zA) and subtract them from the
head B (xB, yB, zB)
Note the head to tail vector addition of the
three components
2.7 Position Vectors
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2.7 Position Vectors
Length and direction ofcable AB can be found bymeasuring A and B using
the x, y, zaxes
Position vector rcan beestablished
Magnitude r representthe length of cable
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2.7 Position Vectors
Angles, , and represent the direction
of the cable
Unit vector, u= r/r
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2.7 Position Vectors
Example 2.12
An elastic rubber band is
attached to points A and B.Determine its length and
its
direction measured from Atowards B.
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2.7 Position Vectors
Solution
Position vector
r = [-2m1m]i + [2m0]j + [3m(-3m)]k
= {-3i + 2j + 6k}mMagnitude = length of the rubber band
Unit vector in the director of ru= r/r
= -3/7i+ 2/7j+ 6/7k
mr 7623 222
View Free Body Diagram
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2.7 Position Vectors
Solution
= cos-1(-3/7) = 115
= cos-1(2/7) = 73.4
= cos-1(6/7) = 31.0
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along a Line
In 3D problems, direction of Fis specified by2 points, through which its line of action lies
Fcan be formulated as a Cartesian vector
F= F u =F (r/r)
Note that Fhas units of
forces (N) unlike r, withunits of length (m)
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along a Line
Force Facting along the chain can be
presented as a Cartesian vector by- Establish x, y, zaxes
- Form a position vector ralong length ofchain
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along a Line
Unit vector, u= r/rthat defines the direction
of both the chain and the force
We get F= Fu
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along a Line
Example 2.13
The man pulls on the cord
with a force of 350N.Represent this force acting
on the support A, as a
Cartesian vector anddetermine its direction.
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along a Line
Solution
End points of the cord are A (0m, 0m, 7.5m)
and B (3m, -2m, 1.5m)
r= (3m0m)i+ (-2m0m)j+ (1.5m7.5m)k
= {3i2j6k}m
Magnitude = length of cord AB
Unit vector, u= r/r
= 3/7i- 2/7j- 6/7k
mmmmr 7623 222
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along a LineSolutionForce Fhas a magnitude of 350N, direction
specified by u
F= Fu
= 350N(3/7i- 2/7j- 6/7k)
= {150i- 100j- 300k} N
= cos-1(3/7) = 64.6= cos-1(-2/7) = 107
= cos-1(-6/7) = 149
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along a Line
Example 2.14
The circular plate is
partially supported by
the cable AB. If the
force of the cable onthe
hook at A is F= 500N,express Fas a
Cartesian vector.
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along a Line
SolutionEnd points of the cable are (0m, 0m, 2m) and B
(1.707m, 0.707m, 0m)
r= (1.707m0m)i+ (0.707m0m)j+ (0m2m)k
= {1.707i+ 0.707j- 2k}m
Magnitude = length of cable AB mmmmr 723.22707.0707.1 222
2.8 Force Vector Directed
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SolutionUnit vector,
u= r/r
= (1.707/2.723)i+ (0.707/2.723)j(2/2.723)k= 0.6269i+ 0.2597j0.7345k
For force F,
F= Fu= 500N(0.6269i+ 0.2597j0.7345k)
= {313i- 130j- 367k} N
along a Line
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SolutionChecking
Show that = 137and
indicate this angle on the
diagram
along a Line
N
F
500
367130313 222
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along a Line
Example 2.15
The roof is supported by
cables. If the cables exert
FAB= 100N and FAC= 120N
on the wall hook at A,
determine the magnitude of
the resultant force acting at
A.
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along a Line
Solution
rAB= (4m0m)i+ (0m0m)j+ (0m4m)k
= {4i4k}m
FAB= 100N (rAB/r AB)
= 100N {(4/5.66)i - (4/5.66)k}
= {70.7i- 70.7k} N
mmmrAB 66.544 22
View Free Body Diagram
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along a Line
Solution
rAC= (4m0m)i+ (2m0m)j+ (0m4m)k
= {4i+ 2j4k}m
FAC= 120N (rAB/r AB)
= 120N {(4/6)i + (2/6)j - (4/6)k}= {80i+ 40j80k} N
mmmmrAC 6424 222
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along a Line
Solution
FR= FAB+ FAC
= {70.7i- 70.7k} N + {80i+ 40j80k} N
= {150.7i+ 40j150.7k} N
Magnitude of FR
N
FR
2177.150407.150
222
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2.9 Dot Product
Dot product of vectorsAand Bis writtenasAB(ReadAdot B)
Define the magnitudes ofA and Band the
angle between their tailsAB=ABcos where 0180
Referred to as scalar
product of vectors as
result is a scalar
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2.9 Dot Product
Laws of Operation
1. Commutative law
AB= BA2. Multiplication by a scalar
a(AB)= (aA)B=A(aB) = (AB)a
3. Distribution lawA(B+ D) = (AB) + (AD)
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2.9 Dot Product
Cartesian Vector Formulation
- Dot product of Cartesian unit vectors
Eg: ii= (1)(1)cos0= 1 and
ij= (1)(1)cos90= 0
- Similarly
ii= 1 jj= 1 kk= 1ij= 0 ik= 1 jk= 1
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2.9 Dot Product
Cartesian Vector Formulation- Dot product of 2 vectorsAand B
AB= (Axi+ Ayj+ Azk) (Bxi+ Byj+ Bzk)
= AxBx(ii) + AxBy(ij) + AxBz(ik)+ AyBx(ji) + AyBy(jj) + AyBz(jk)
+ AzBx(ki) + AzBy(kj) + AzBz(kk)
= AxBx + AyBy+ AzBzNote: since result is a scalar, be careful of includingany unit vectors in the result
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2.9 Dot Product
Applications
- The angle formed between two vectors orintersecting lines
= cos-1[(AB)/(AB)] 0180
Note: ifAB= 0, cos-10= 90,Ais
perpendicular to B
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2.9 Dot Product
Applications- The components of a vector parallel andperpendicular to a line
- Component ofAparallel or collinear with line aa isdefined byA(projection ofAonto the line)
A= A cos
- If direction of line is specified by unit vector u (u=1),
A= A cos =Au
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2.9 Dot Product
Applications- IfAis positive,Ahas a directionalsense same as u
- IfAis negative,Ahas a directionalsense opposite to u
-Aexpressed as a vector
A= A cos u= (Au)u
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ApplicationsFor component of A perpendicular to line aa
1. SinceA=A+A,
thenA=A-A2. = cos-1[(Au)/(A)]
thenA=Asin
3. IfAis known, by Pythagorean Theorem
2.9 Dot Product
2
||
2 AAA
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2.9 Dot Product
For angle between therope and the beam A,
- Unit vectors along thebeams, u
A= r
A/r
A
- Unit vectors along theropes, ur=rr/rr
- Angle = cos-1
(rA.rr/rArr)
= cos-1(uA ur)
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9 ot oduct
For projection of the forcealong the beam A
- Define direction of the beam
uA= rA/rA- Force as a Cartesian vector
F= F(rr/rr) = Fur
- Dot productF= FuA
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Example 2.16
The frame is subjected to a horizontal force
F= {300j} N. Determine the components of
this force parallel and perpendicular to the
member AB.
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Solution
Since
Then
N
kjijuF
FF
kji
kji
r
ru
B
AB
B
BB
1.257
)429.0)(0()857.0)(300()286.0)(0(
429.0857.0286.0300.
cos
429.0857.0286.0362
362
222
2.9 Dot Product
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SolutionSince result is a positive scalar,
FABhas the same sense of
direction as uB. Express in
Cartesian form
Perpendicular component
NkjikjijFFF
Nkji
kjiN
uFF
AB
ABABAB
}110805.73{)1102205.73(300
}1102205.73{
429.0857.0286.01.257
2.9 Dot Product
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SolutionMagnitude can be determined
From F or from Pythagorean
Theorem
N
NN
FFF AB
1551.257300
22
22
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Example 2.17
The pipe is subjected to F= 800N. Determine the
angle between Fand pipe segment BA, and the
magnitudes of the components of F, which areparallel and perpendicular to BA.
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SolutionFor angle
rBA= {-2i- 2j+ 1k}m
rBC= {- 3j+ 1k}mThus,
5.42
7379.0103
113202cos
BCBA
BCBA
rr
rr
View Free Body Diagram
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Solution
Components of F
N
kjikj
uFF
kji
kji
r
ru
BAB
AB
ABAB
590
3.840.5060
31
32
320.2539.758
.
3
1
3
2
3
2
3
)122(
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Solution
Checking from trigonometry,
Magnitude can be determined
From F
N
N
FFAB
540
5.42cos800
cos
NFF 5405.42sin800sin
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Solution
Magnitude can be determined from F or from
Pythagorean Theorem
N
FFF AB
540
590800 22
22
Chapter Summary
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p y
Parallelogram LawAddition of two vectors
Components form the side and resultantform the diagonal of the parallelogram
To obtain resultant, use tip to tail addition
by triangle rule To obtain magnitudes and directions, use
Law of Cosines and Law of Sines
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Cartesian VectorsVector Fresolved into Cartesian vector form
F= Fxi+ Fyj+ Fzk
Magnitude of F
Coordinate direction angles , and aredetermined by the formulation of the unitvector in the direction of F
u= (Fx/F)i+ (Fy/F)j + (Fz/F)k
222
zyx FFFF
Chapter Summary
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Cartesian Vectors Components of urepresent cos, cosand cos
These angles are related by
cos2+ cos2+ cos2= 1
Force and Position Vectors Position Vector is directed between 2 points
Formulated by distance and direction movedalong the x, y and z axes from tail to tip
Chapter Summary
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Force and Position Vectors For line of action through the two points, it
acts in the same direction of uas the
position vector Force expressed as a Cartesian vector
F= Fu= F(r/r)
Dot Product Dot product between two vectorsAand B
AB=ABcos
Chapter Summary
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Dot Product Dot product between two vectorsAand B
(vectors expressed as Cartesian form)
AB= AxBx + AyBy+ AzBz For angle between the tails of two vectors
= cos-1[(AB)/(AB)]
For projected component ofAonto an axisdefined by its unit vector u
A= A cos =Au
Chapter Review
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Chapter Review
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Chapter Review
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