Chapter 02 Force Vectors

8/11/2019 Chapter 02 Force Vectors

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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_quantity


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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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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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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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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


cont)


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Definition of CoplanarCoplanar is a set of points, lines, or any othergeometrical shapes that lie on the same plane.


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


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


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)


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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EXAMPLE 2

Determine x and y

components of F1and

F2acting on the boom.

Express each force as

a Cartesian vector.


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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


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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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


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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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


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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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


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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Unit Vector

Direction of Acan bespecified using a unit vector.

Unit vector has a magnitudeof 1.


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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


222zyx AAAA

22' yx AAA

22' zAAA


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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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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



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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 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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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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Chapter 02 Force Vectors

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Transcript of Chapter 02 Force Vectors