mech chapter 02

8/3/2019 Mech Chapter 02

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Force Vectors2

Engineering Mechanics:

Statics in SI Units, 12e

Copyright 2010 Pearson Education South Asia Pte Ltd


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Chapter Outline

2.1 Scalars and Vectors

2.2 Vector Operations

2.3 Vector Addition of Forces

2.4 Addition of a System of Coplanar Forces2.5 Cartesian Vectors

2.6 Addition and Subtraction of Cartesian Vectors

2.7 Position Vectors

2.8 Force Vector Directed along a Line

2.9 Dot Product


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2.1 Scalars and Vectors (1/2)

Scalar A quantity characterized by a positive or negativenumber

Indicated by letters in italic such as Ae.g. Mass, volume and length


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2.1 Scalars and Vectors (2/2)

Vector A quantity that has magnitude and direction

e.g. Position, force and moment

Represent by a letter with an arrow over it,

Magnitude is designated as

In this subject, vector is presented as A and itsmagnitude (positive quantity) as A

A

A


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2.2 Vector Operations (1/3)

Multiplication and Division of a Vector by a Scalar- Product of vector A and scalar a = aA

- Magnitude =

- Law of multiplication applies e.g. A/a = ( 1/a ) A, a0

aA


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Vector Addition- Addition of two vectors A and B gives a resultant

vector R by the parallelogram law

- Result R can be found by triangle construction- Communicative e.g. R = A + B = B + A

- Special case: Vectors A and B are collinear(bothhave the same line of action)


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Vector Subtraction- Special case of addition

e.g. R = AB = A + ( - B )

- Rules of Vector Addition Applies



2.3 Vector Addition of Forces (1/3)

Finding a Resultant Force

Parallelogram lawis carried out to find the resultantforce

Resultant,

FR = ( F1 + F2 )




Procedure for Analysis

Parallelogram Law Make a sketch using the parallelogram law

2 components forces add to form the resultant force

Resultant force is shown by the diagonal of theparallelogram

The components is shown by the sides of theparallelogram




Procedure for Analysis Trigonometry

Redraw half portion of the parallelogram

Magnitude of the resultant force can be determinedby the law of cosines

Direction if the resultant force can be determined bythe law of sines

Magnitude of the two components can be determined bythe law of sines



Example 2.1

The screw eye is subjected to two forces, F1 and F2.Determine the magnitude and direction of the resultantforce.



Solution

Parallelogram Law

Unknown: magnitude of FR and angle


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Solution

Trigonometry

Law of Cosines

Law of Sines

NN

NNNNFR

2136.2124226.0300002250010000

115cos150100215010022

8.39

9063.06.212

150sin

115sin

6.212

sin

150

N

N

NN


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Solution

Trigonometry

Direction of FR measured from the horizontal

8.54

158.39


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TEST (2-1, 2-2, 2-3)

Resolve the force (F=450 N acting on the frame) intocomponents acting along members AB and AC, anddetermine the magnitude of each component.


N


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Force Vectors2





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Chapter Outline








2.9 Dot Product


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2.4 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

sinandcos FFFF

FFF

yx

yx


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2.4 Addition of a System of Coplanar Forces

Cartesian Vector Notation Cartesian unit vectors i and j are used to designate

the x and y directions

Unit vectors i and j have dimensionless magnitudeof unity ( = 1 )

Magnitude is always a positive quantity,represented by scalars Fxand Fy

jFiFF yx


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Example (2-4)

Determine x and y components of F1 and F2 acting on theboom. Express each force as a Cartesian vector.


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Solution (1) -- Scalar Notation

Scalar Notation

NNNF

NNNF

y

x

17317330cos200

10010030sin200

1

1


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Solution

By similar triangles we have

Scalar Notation:

N10013

5260

N24013

12260

2

2

y

x

F

F

NNF

NF

y

x

100100

240

2

2


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Solution (2) -- Cartesian Vector Notation

Cartesian Vector Notation:

NjiF

NjiF

100240

173100

2

1


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Coplanar Force Resultants To determine resultant of several coplanar forces:

Resolve force into x and y components

Addition of the respective components using scalaralgebra

Resultant force is found using the parallelogramlaw

Cartesian vector notation:

jFiFF

jFiFF

jFiFF

yx

yx

yx

333

222

111


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Coplanar Force Resultants Vector resultant () is therefore

If scalar notation are used

jFiFFFFF

RyRx

R

321

yyyRy

xxxRx

FFFF

FFFF

321

321


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Coplanar Force Resultants In all cases we have

Magnitude of FRcan be found by Pythagorean Theorem

yRy

xRx

FF

FF

Rx

Ry

RyRxRF

FFFF 1-22 tanand

* Take note of sign conventions


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Example (2-4)

The link is subjected to two forces F1 and F2. Determinethe magnitude and orientation of the resultant force.


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Solution I (Scalar Notation)

Scalar Notation:

N

NNF

FF

N

NNF

FF

Ry

yRy

Rx

xRx

8.582

45cos40030sin600

:

8.236

45sin40030cos600

:


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Solution I (Scalar Notation)

Resultant Force

From vector addition, direction angle is

N

NNFR

629

8.5828.23622

9.67

8.236

8.582

tan1

N

N


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Solution II (Cartesian Vector Notation)

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

The magnitude and direction of FRare determined in thesame manner as before.


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TEST (2-4)

If the magnitude of the resultant force acting on thebracket is to be 400 N directed along the u axis,determine the magnitude of F and its direction



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2.5 Cartesian Vectors Right-Handed Coordinate System

A rectangular or Cartesian coordinate system is saidto be right-handed provided:

Thumb of right hand points in the direction of thepositive z axis

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


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2.5 Cartesian Vectors

Rectangular Components of a Vector A vector A may have one, two or three rectangular

components along the x, yand zaxes, depending onorientation

By two successive application of the parallelogram law

A = A + AzA = Ax + Ay

Combing the equations,A can be expressed as

A = Ax + Ay + Az


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Unit Vector Direction of A can be specified using a unit vector

Unit vector has a magnitude of 1

If A is a vector having a magnitude of A 0, unitvector having the same direction as A is expressedby uA = A /A. So that

A = A uA


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Cartesian Vector Representations 3 components of A act in the positive i, j and k

directions

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 equationsgives magnitude of A

222zyx AAAA

22

' yx AAA

22' zAAA


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Direction of a Cartesian Vector Orientation of A is defined as the coordinate

direction angles , and measured between the

tail of A and the positive x, y and z axes 0 , and 180 The direction cosinesof A is

A

Axcos

A

Aycos

A

Azcos


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Direction of a Cartesian Vector Angles , and can be determined by the

inverse cosines

GivenA = 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 uA can also be expressed as

uA = cosi + cosj + cosk

Since and uA = 1, we have

A as expressed in Cartesian vector form isA = AuA= Acosi + Acosj + Acosk= Axi + Ayj + AZk

222

zyx AAAA

1coscoscos222


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Example (2-5)

Express the force F as Cartesian vector.


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Solution

Since two angles are specified, the third angle is found by

Two possibilities exit, namely

1205.0cos 1

605.0cos1

5.0707.05.01cos

145cos60coscos

1coscoscos

22

222

222


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Solution

By inspection, = 60 since Fx is in the +x direction

Given F= 200N

F = Fcosi + Fcosj + Fcosk

= (200cos60N)i + (200cos60N)j+ (200cos45N)k

= {100.0i + 100.0j +141.4k}N

Checking:

N

FFFF zyx

2004.1410.1000.100222

222


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Concurrent Force Systems Force resultant is the vector sum of all the forces in

the system

FR = F= Fxi+ Fyj+ Fzk


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TEST (2-5, 2-6)

Three force act on the ring. If the resultant force FR has amagnitude and direction as shown, determine themagnitude and the coordinate direction angles offorce F3.



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2.7 Position Vectors x,y,zCoordinates

Right-handed coordinate system

Positive zaxis points upwards, measuring the height ofan object or the altitude of a point

Points are measured relativeto the origin, O.


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Position Vector

Position vector r is defined as a fixed vector whichlocates a point in space relative to another point.

E.g. r = xi + yj + zk


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Position Vector

Vector addition gives rA + r = rB

Solving

r = rB

rAr = (xB xA)i + (yB yA)j + (zBzA)k


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Length and direction of cable AB can be found bymeasuring A and B using the x, y, zaxes

Position vector r can be established

Magnitude r represent the length of cable Angles, , and represent the direction of the cable

Unit vector, u = r/r


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Example (2-7)

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 = [-2m 1m]i + [2m 0]j + [3m (-3m)]k

= {-3i + 2j + 6k} m

Magnitude = 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


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TEST (2-7)

Determine the length of the rod and the positionvector directed from A to B. What is the angle ?



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2.8 Force Vector Directed along a Line In 3D problems, direction of F is specified by 2 points,

through which its line of action lies

F can be formulated as a Cartesian vector

F = Fu =F (r/r)

Note that F has units of forces (N)unlike r, with units of length (m)


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Force F acting along the chain can be presented as aCartesian vector by

- Establish x, y, zaxes

- Form a position vector r along length of chain Unit vector, u = r/rthat defines the direction of boththe chain and the force

We get F = Fu


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Example (2-8)

The man pulls on the cord with a force of 350N.Represent this force acting on the support A, as aCartesian vector and determine its direction.


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Solution

End points of the cord are A (0m, 0m, 7.5m) andB (3m, -2m, 1.5m)

r = (3m 0m)i + (-2m 0m)j + (1.5m 7.5m)k= {3i 2j 6k}m

Magnitude = length of cord AB

Unit vector,u = r /r

= 3/7i - 2/7j - 6/7k

mmmmr 7623 222


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Solution

Force F has a magnitude of 350N, direction specified byu.

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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TEST (2-8)

Determine the magnitude of the resultant force at A.



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Force Vectors2





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Chapter Outline








2.9 Dot Product


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2.9 Dot Product Dot product of vectors A and B is written as AB

(Read A dot B)

Define the magnitudes of A and B and the anglebetween their tails

AB = 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 Operation1. Commutative law

AB = BA

2. Multiplication by a scalar

a(AB) = (aA)B = A(aB) = (AB)a

3. Distribution law

A(B + D) = (AB) + (AD)


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2.9 Dot Product

Cartesian Vector Formulation - Dot product of Cartesian unit vectors

ii = (1)(1)cos0 = 1

ij = (1)(1)cos90 = 0

- Similarly

ii = 1 jj = 1 kk = 1

ij = 0 ik = 1 jk = 1


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2.9 Dot Product

Cartesian Vector Formulation Dot product of 2 vectors A and B

AB = (Ax i+ Ay j + Az k) (Bx i + Byj + Bz k)= Ax Bx (ii) + Ax By (ij) + Ax Bz (ik)

+ .

+ .

AB = AxBx + AyBy + AzBz


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TEST (2-9)

Determine the angle between the force (rAC) and theline AB(rAB)


= cos-1 [(AB)/(AB)]

AB = ABcos


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Example (2-9)

The frame is subjected to a horizontal force F = {300j} N.Determine the components of this force parallel andperpendicular to the member AB.


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Solution

Since

Thus

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

429.0857.0286.0

362

362

222


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Solution

Since result is a positive scalar, FAB has the same senseof direction as uB. Express in Cartesian form

Nkji

kjiN

uFF ABABAB

}1102205.73{

429.0857.0286.01.257


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Solution

Perpendicular component

Magnitude can be determinedfrom For from PythagoreanTheorem,

N

NN

FFF AB

155

1.25730022

22

Nkji

kjijFFF AB

}110805.73{

)1102205.73(300


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TEST (2-9)

Determine the components of the force acting paralleland perpendicular to the axis of the pole.



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END OF CHAPTER 2

mech chapter 02

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