Ch02 Statics- Pearson1 Hr

8/11/2019 Ch02 Statics- Pearson1 Hr

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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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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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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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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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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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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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Vector Addition


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Vector Addition

- Special case: VectorsAand Barecollinear(both have the same line ofaction)


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Vector Subtraction

- Special case of addition

Eg: R =AB=A+ ( - B)

- Rules of Vector Addition Applies


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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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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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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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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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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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Trigonometry

- Magnitude of the two components can bedetermined by the law of sines


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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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Solution

Parallelogram Law

Unknown: magnitude ofFRand angle


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Solution

Trigonometry

Law of Cosines

N

N

NNNNFR

213

6.212

4226.0300002250010000

115cos1501002150100 22


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Solution

Trigonometry

Law of Sines

8.39sin

9063.0

6.212

150sin

115sin

6.212

sin

150

N

N

NN


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Solution

Trigonometry

Direction of FRmeasured from the horizontal

8.54

158.39


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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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Solution

(a) Parallelogram Law

From the vector diagram,yx

FFF

NF

NF

y

x

64340sin1000

76640cos1000


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Solution

(b) Parallelogram Law

'yx FFF


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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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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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Solution

Parallelogram Law

ACAB FFN 500


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Solution

Law of Sines

9.43

6928.0sin

60sin

500

400sin

60sin500

sin400

N

N

NN


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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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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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Solution

(a) Parallelogram LawUnknown: Forces F1and F2

View Free Body Diagram


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Solution

Law of Sines

NF

NF

NF

NF

446

130sin1000

20sin

643

130sin

1000

30sin

2

2

1

1


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Solution

(b) Minimum length of F2occurwhen its line of action is

perpendicular to F1. Hencewhen

F2is a minimum

702090


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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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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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Scalar NotationEg:

Sense of directionalong positive x and

negative y axes

yx FFF '''


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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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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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Cartesian Vector NotationF= Fxi+ Fyj F = Fxi+ Fy(-j)

F = Fx

iFy

j


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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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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



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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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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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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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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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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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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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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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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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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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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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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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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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Solution

Scalar Notation

N

NNF

FF

NN

NNNF

FF

Ry

yRy

Rx

xRx

8.296

5

320045cos250

:

2.3832.383

5

420045sin250400

:


2 4 Additi f S t


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Solution

Resultant Force

From vector addition,

Direction angle is


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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Right-Handed Coordinate System

- z-axis for the 2D problem would beperpendicular, directed out of the page.


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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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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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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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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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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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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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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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Direction of a Cartesian Vector

- For angles , and (blue colored

triangles), we calculate the directioncosinesofA

A

Axcos


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- For angles , and (blue coloredtriangles), we calculate the directioncosinesofA

A

Aycos


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- For angles , and (blue coloredtriangles), we calculate the directioncosinesofA

A

Azcos


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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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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



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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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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



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Example 2.8

Express the force Fas Cartesian vector



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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



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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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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



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Example 2.10

Express the force F1as a Cartesian vector.



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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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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



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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


N

FFFF zyx

1006.864.354.35 222

2

1

2

1

2

11



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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




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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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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




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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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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

2 7 Position Vectors


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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)




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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




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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)




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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




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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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Angles, , and represent the direction

of the cable

Unit vector, u= r/r



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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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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




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Solution

= cos-1(-3/7) = 115

= cos-1(2/7) = 73.4

= cos-1(6/7) = 31.0

2.8 Force Vector Directed


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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



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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

2.8 Force Vector Directedl


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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.

2.8 Force Vector Directedl Li


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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


2.8 Force Vector Directedl i


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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

2.8 Force Vector Directedl Li


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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

2.9 Dot Product


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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

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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

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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


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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

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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

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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

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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

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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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Chapter Review


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Chapter Review


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Chapter Review


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Ch02 Statics- Pearson1 Hr