unit 2 vector

7/30/2019 Unit 2 Vector

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JJ205ENGINEERING MECHANICS

COURSE LEARNING OUTCOMES :

Upon completion of this course, students should be able to:

CLO 1. apply the principles of statics and dynamics to solve

engineering problems (C3)CLO 2. sketch related diagram to be used in problem solving (C3)

CLO 3. study the theory of engineering mechanics to solve related

engineering problems in group (A3)


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JJ205 ENGINEERING MECHANICS

CHAPTER 2:

FORCE VECTORSCLO 1. apply the principles of statics and dynamics tosolve engineering problems (C3)

Prepared by:

THEEBENRAJ

2


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Objectives:

At the end of this chapter, student should be able to:

1. Understand scalars and vectorsa. Differentiate between scalars and vectors.

b. Distinguish free vectors, sliding vectors, fixed vectors.

2. Understand rectangular components

a. Explain two forces acting on a particle.3. Understand vectors and vector operations.

a. Calculate addition of vectors.

b. Calculate subtraction of vectors.

c. Determine resolution of vectors.

4. Understand the resultant force of coplanar forces byaddition.

a. Explain scalar notation.

b. Explain cartesian vector notation.

c. Determine coplanar forces and resultant force.


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Objectives:

5. Understand Cartesian vectors.

a. Explain right handed coordinate system.

b. Explain cartesian unit vector.

c. Apply cartesian vector representation.

6. Understand the magnitude of cartesian vector.

a. Determine the direction of cartesian vector.

7. Understand resultant cartesian vector by addition and

substraction.

a. Solve problems regarding concurrent force system.

8. Understand position vectors andx, y, z coordinates.a. Explain position vectors andx, y, z coordinate.


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Objectives:

9. Understand the force vector directed along the line.

a. Explain the force vector directed along the line.

b. Determine the force vector directed along the line.

10. Understand the dot product.

a. Apply laws of operation.i. Commutative law

ii. Multiplication by scalar

iii. Distributive law

b. Formulate cartesian vector formulation.


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Force On A Particle. Resultant of Two Forces

A force represents the action of one body on

another and generally characterized by its

point of application, its magnitude, and its

direction. The magnitude of the force is characterized by

a certain number of units.

SI units to measure the magnitude of a force arethe Newton (N).

Multiple: kilonewton (kN) = 1000N

6


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The direction of a force is defined by the line

of action and the sense of the force.

a)

b)

A 30

Magnitude of a force on particle A is 100N at 30.

A 30

The different sense of the force but have same magnitude and same

direction.

7


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Two forces, P and Qacting on a particle A:

Can be replaced by a single force R which has

the same effect on the particle.

R is the resultant of the forces P and Q.

The method is known as the parallelogram law

for the addition of two forces.

A

A

P

P

Q

Q

R=

A

R

8


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Example 2.1:

The two forces P and Q act on bolt A. Determine their resultant.

Solution:

a) Using graphical solution:

i. A parallelogram with sides equal to P and Q is drawn

to scale.

ii. The magnitude and direction of the resultant are

measured and found to be

iii. The triangle rule may also be used. Forces P and Q are drawn in tip-to-tail

fashion.

9


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Continue

b) Trigonometric Solution.

i. The triangle rule is again used; two sides and the included angle are known.

ii. We apply the law of cosines:

iii. Now, applying the law of sines, we write;

iv. Solving equation (1) for sin A, we have;

v. Using a calculator, we first compute the quotient,

then its are sines, and obtain;

(1)

10

i


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c) Alternative Trigonometric Solution.

i. We construct the right triangle BCD and compute;

ii. Then, using triangle ACD, we obtain;

iii. Again,

Continue

11


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Example 2.2 The screw eye in Figure (a) is subjected to

two forces, F1 and F2. Determine the

magnitude and direction of the resultantforce.

Solutions:

a) Parallelogram Law:

i. The parallelogram law of addition is shown

in Figure (b). The two unknowns are the

magnitude ofFR and the angle (theta).

12


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b) Trigonometry:

continue

13

-The vector triangle as shown in Fig. (c), is constructed from Fig. (b).

- FR is determined by using the law of cosines;

-The angle is determined by applying the law of sines;

-Thus, the direction (phi) ofFR, measured from the horizontal, is;


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Vectors

Vectors is defined as mathematicalexpressions possessing magnitude anddirection, which add according to theparallelogram law.

It is represented by arrows.

The magnitude of a vector defines the lengthof the arrow used to represent the vector.

14


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Types of vectors:

1) Fixed vectors:

cannot be moved without modifying the

conditions of the problem.

2) Free vectors:

couples, which represented by vectors whichmay be freely moved in space.

3) Sliding vectors:

forces acting on a rigid body, which represented

by vectors which can be moved, or slid, along

their lines of action.

15

i


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4) Equal vectors:

two vectors have the same magnitude and the same

direction, whether or not they also have the same point of

application. They are denoted by same letter.

5) Negative vectors:

vector having the same magnitude but opposite direction.

continue

PP

P

-P16


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Addition of Vectors

Vectors add according to the parallelogram law.

The sum of two vectors P and Qis obtained by attaching the

two vectors to the same point A and constructing a

parallelogram, using P and Qas two sides of the

parallelogram.

A

P

Q

The diagonal that passes through A representsthe sum of the vectors P and Qwhich denoted as

P + Q.

However, the magnitude of the vector P+Qis NOT

In general, equal to the sum (P+Q) of themagnitudes of the vectors P and Q.

Since that, we conclude that the addition of two

Vectors is commutative, write as:

P + Q = Q + P17

ti


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Triangle Rule is an alternative method for determining the

sum of two vectors from the parallelogram law.

From the only half of the parallelogram;

From the figures shown above, it confirms the fact thatvectors addition is commutative.

continue

A

P

Q

OR

A

P

Q

18


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Defined as the addition of the corresponding

negative vector.

P-Q representing the difference between the vectors

P and Q is obtained by adding to P the negative

vectorQ. we write;

P Q = P + (-Q)

Subtraction of Vectors

A

-Q

P

19


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The sum of three vectors P, Q, and S was obtained graphically.

The triangle rule was first applied to obtain the sum P+Qof thevectors P and Q.

It was applied again to obtain the sum of vectors P+Qand S.

For addition of vectors, Polygon Rule is applied by arranging the given vectors

in tip-to-tail fashion and connecting the tail of the first vector with the tip ofthe last one.

Coplanar Vectors

A

P

QS

A

P

QS

20

continue


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The result obtained would have been unchanged if the vectors

Q and S had been replaced by their sum Q + S. We may thus

write;

P + Q + S = (P + Q) + S = P + (Q + S)

which expresses the fact that vector addition is associative.

Recalling that vector addition has been shown, in the case of

two vectors, to be commutative, we write:P + Q + S = (P + Q) + S = S + (P + Q)

= S + (Q + P) = S + Q + P

continue

A

P

QS

21

continue


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This expression, as well as others which may be obtained in

the same way, shows that the order in which several vectorsare added together is immaterial.

A

P

QS

continue

SQ

P

22


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Components of the original force F, is a single force F acting on

a particle may be replaced by two or more forces which,together, have the same effect on the particle.

The process of substituting them for F is called resolving theforce F into components.

There are two cases of particular interest:

1) One of the two components.

a) P is known.

b) Second component, Q, is obtained by applying the triangle rule andjoining the tip of P to the tip ofF

c) Magnitude and direction ofQare determined graphically or bytrigonometry.

Resolution of A Force Into Components

A

P

Q

F

23

continue


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2) The line of action of each component is known.

a) The magnitude and sense of the components are obtained by

applying the parallelogram law and drawing lines, through the tip

ofF, parallel to the given lines of action.

b) This process leads to two well-defined components, P and Q,which can be determined graphically or computed

trigonometrycally by applying the law if sines.

continue

F

A

Q

P

24


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It will be found desirable to resolve a force into two

components which are perpendicular to each other. In figure below, the force F has been resolved into component

Fx along the x axis and a component Fy along the y axis.

The parallelogram drawn to obtain the two components is a

rectangle, and Fx and Fy are called rectangular components.

Addition of a System of Coplanar Forces

O

FFy

Fx

y

x

O

FFy

Fx

y

xOR

25

continue


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Cartesian Unit vectors:

Two vectors of unit magnitude, directed respectively along the

positive x and y axes, will be introduced at this point. They are

denoted by i andj.

Scalar components:

The scalars Fx and Fy of forces F.

Vector components:

The actual component forces Fx and Fy ofF.

j

y

x

continue

i

26

continue


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Cartesian Vector Notation

Note that the rectangular components Fx and Fy of a force F may be

obtained by multiplying respectively the unit vectors i andj by

appropriate scalars. We write;

Fx = Fxi Fy = Fyj

and express F as the Cartesian vector,

F = Fx

i+ Fy

j

O

Fj

y

x

continue

i

Fy = Fyj

Fx = Fxi

27

continue


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Scalar Notation. Indicates positive and negative Fx :

Positive Fx when the vector component Fx has the same sense as the unitvector i. (same sense as the positive x axis).

Negative Fx when Fx has the opposite sense.

The positive and negative Fy is same as Fx

Denoting by F the magnitude of the force F and by the anglebetween F and the x axis, measured counterclockwise from thepositive x axis.

This may express the scalar components of F as follows:

Fx = F cos Fy = F sin

continue

O

F

y

x

Fy

Fx

28

Example 2 3


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A force of 800 N is exerted on a bolt A as shown in Figure (a). Determine the

horizontal and vertical components of the force.

(a)

Example 2.3

The vector components ofF are:

Fx = -(655 N) i Fy = +(459 N)j

May write in Cartesian vector form:

F = -(655 N) i + (459 N)j

Example 2 4


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Resolve the 1000 N ( 100 kg) force acting on the pipe Fig. a, into

components in the (a)x andy directions, and (b)xandy directions.

Solution:

In each case the parallelogram law is used to resolve F into its two

components, and then the vector triangle is constructed to determine

the numerical results by trigonometry.

30

Example 2.4

continue


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Part (a)

The vector addition F = Fx + Fy is shown in Fig. b.

In particular, note that the length of the components is scaled along

the x and y axes by first constructing lines from the tip ofF parallel

to the axes in accordance with the parallelogram law.

From the vector triangle, Fig. c,

31

continue

continue


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Part (b)

The vector addition F = Fx + Fy is shown in Fig. d.

Note carefully how the parallelogram is constructed.

Applying the law of sines and using the data listed on the

vector triangle, Fig. e, yields:

32

continue

Example 2 5


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The force F acting on the frame shown in Fig.2-12a has a

magnitude of 500 N and is to be resolved into two

components acting along members AB and AC. Determine

the angle , measured below the horizontal, so that thecomponent FAC is directed from A toward C and has a

magnitude of 400 N.

Solution:

i. by using the parallelogram law, the vector addition of the

two components yielding the resultant is shown in Fig.b.

ii. Note carefully how the resultant force is resolved into two

components FAB and FAC, which have specified lines ofaction.

iii. The corresponding vector triangle is shown in Fig.c.

33

Example 2.5

continue


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iv. The angle can be determined by using the law of sines:

Hence;

34

continue


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Using this value for , apply the law of cosines or the law of sines and

show that FAB has a magnitude of 561 N.

Notice that F can also be directed at an angle above the horizontal, asshown in Fig. d, and still produce the required component FAC.

Show that in this case = 16.1 and FAB = 161 N.

35

Example 2 6


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The ring shown in Figure a is subjected to two forces, F1 and

F2. if it is required that the resultant force have a magnitude of

1 kN and be directed vertically downward, determine (a) themagnitudes ofF1 and F2 provided = 30, and (b) the

magnitudes ofF1 and F2 ifF2 is to be a minimum.

36

Example 2.6

continue


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Solution:

Part (a):

i. A sketch of the vector addition according to the

parallelogram law is shown in Fig. b.ii. From the vector triangle constructed in Fig. c, the

unknown magnitudes F1and F2 are determined by

using the law of sines:

37

continue


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Part (b):

i. If is not specified, then by the vector triangle, Fig. d,

F2 may be added to F1 in various ways to yield the

resultant 1000 N force.

ii. The minimum length or magnitude ofF2 will occur

when its line of action isperpendicularto F1.

iii. Any other direction, such as OA or OB, yields a larger

value for F2.

iv. Hence, when = 90 - 20 = 70, F2 is minimum.

v. From the triangle shown in Fig. e, it is seen that;

F1 = 1000 sin 70 N = 940 N

F2 = 1000 cos 70 N = 342 N

38

Coplanar Force Resultants.


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Coplanar Force Resultants.

To determine the resultant of several coplanar forces

1. Resolved each force into its x and y components.

2. The respective components are added using scalar algebra since

they are colinear.3. The resultant force is then formed by adding the resultants of the

x and y components using parallelogram law.

Example:

Given three concurrent forces below:

F1

F3

F2

x

y

To solve this problem using Cartesian

vector notation, each force is first

represented as a Cartesian vector, i.e;

F1 = F1x i + F1y j

F2 = - F2x i +F2y jF3 = F3x i -F3y j

The vector resultant is therefore;

FR = F1 + F2+ F3= F1x i + F1y j - F2x i +F2y j +F3x i -F3y j

= (F1x - F2x i +F3x )i +(F1y + F2y -F3y )j

= (FRx )i + (FRy )j 39

continue


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To solve this problem using Scalar notation,

from figure shown, sincex is positive to the

right andy is positive upward, we have;

The vector resultant is therefore;

FR = = (FRx )i + (FRy )j

In the general case, thex andy components of the resultant of any number

of coplanar forces can be represented symbolically by the algebraic sum ofthex andy components of all the forces;

F1y

F3xx

y

F2x

F2y

F1x

F3y

40

continue


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When applying these equations, it is important to use the sign convention

establish for the components.

Components having a directional sense along the positive coordinate

axes are considered positive scalars. Components having a directional sense along the negative coordinate

axes are considered negative scalars.

If this convention is followed, then the signs of the resultant components

will specify the sense of these components. For example; positive result indicates that the component has a

directional sense which is in positive coordinate direction.

FR

x

y

FRy

FRx

41

continue


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Once the resultant components are determined, they may be sketched along the x

and y axes in their proper direction, and the resultant force can be determined

from vector addition.

Then, the magnitude of FR can be found from the Pythagorean Theorem; which is:

Also, the direction angle , which specifies the orientation of the force isdetermined from trigonometry:

FR

x

y

FRy

FRx

42

Example 2 7


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43

Four forces act on bolt A as shown. Determine

the resultant of the forces on the bolt.

Example 2.7

Thus, the resultant R of the four forces is;

Force Magnitude, N x Component, N y Component, N

F1 150 + 129.9 + 75.0

F2 80 - 27.4 + 75.2F3 110 0 - 110.0

F4 100 + 96.6 - 25.9

Rx = +199.1 Ry= + 14.3

continue


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44

The magnitude and direction of the resultant may now be determined.

From the triangle shown, we have:

Example 2 8


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45

Example 2.8

Determine thex andy components ofF1 and F2 acting on the boom

shown in Fig. (a). Express each force as a Cartesian vector.

continue


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46

Solution:

Scalar Notation:

F1 is resolved into x and ycomponents using

parallelogram law as shown in

Fig. (b).

The magnitude of each

component is determined by

trigonometry.

Then, we have;

continue


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47

Solution:

F2 is resolved into x and y

components as shown in Fig.

(c).

The slope of the line of action

is indicated and could obtain

the angle :

Then, determine the

magnitudes of the components;

continue


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48

Solution:

The magnitude of the horizontal component, F2x, was

obtained by multiplying the force magnitude by the ratio of

the horizontal leg of the slope triangle divided by thehypotenuse.

The magnitude of the vertical component, F2y, was obtained

by multiplying the force magnitude by the ratio of thevertical leg divided by the hypotenuse.

Hence, using scalar notation;

continue


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49

Solution:

Cartesian Vector Notation:

The magnitudes and directions of the components of eachforce is determined.

Thus, express each force as a Cartesian vector;

Example 2.9


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50

Example 2.9

The link in Figure (a) below is subjected to two forces F1 and F2.

determine the magnitude and orientation of the resultant force.

continue


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51

Solution 1:

Scalar Notation: (*Can be solved usingparallelogram law.)

Each force is resolved into itsx

andy components, Figure (b).

These components are summed

algebraically.

The positive sense of thex and

y force components alongside

each equation is indicated;

continue


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52

Solution 1:

Scalar Notation: (*Can be solved usingparallelogram law.)

Magnitude of the resultant force,shown in Figure (c);

From the vector addition, Figure(c), the direction angle is;

continue


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53

Solution 2:

Cartesian Vector Notation:

From Figure (b), each force isexpressed as a Cartesian

vector;

Thus,

The magnitude and direction of FR are determined in the same

manner in solution 1.

Example 2.10


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54

p

The end of the boom O in Figure (a) below is subjected to three

concurrent and coplanar forces. Determine the magnitude and

orientation of the resultant force.

continue


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55

Solution :

Each force is resolved into itsx andy components, Figure (b).

Summing thex components;

The negative sign indicates that FRx acts to the left, as notedby the small arrow.

Summing the y components yields;

continue


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56

Solution :

Magnitude of the resultant force,

shown in Figure (c);

From the vector addition, Figure

(c), the direction angle is;


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Exercises

2.31. Determine thex andycomponents of the 800 N force

2.33. Determine the magnitude of force

F so that the resultant FR of the threeforces is as small as possible.

2.34. Determine the magnitude of the

resultant force and its direction,measured counterclockwise from the

positivex axis.

57

C t i V t


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Right handed coordinate system.

A right handed coordinate system willbe used for developing the theory ofvector algebra that follows.

A rectangular or Cartesian coordinatesystem is said to be right-handedprovided the thumb of the right handpoints in the direction of the positive zaxis when the right-hand fingers arecurled about this axis and directedfrom the positive x toward the positivey axis.

Cartesian Vectors

58

continue


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Rectangular Components of a Vector

A vector A may have one, two, or three rectangular

components along thex,y, andz coordinate axes,depending on how the vector is oriented relative to the

axes.

A

Ay

A

Ax

Az

z

y

x

When A is directed within an octant of

the

x, y, z frame, then, by two successiveapplications of the parallelogram law,

we may resolve the vector into

components as;

A = A + Az and A = Ax + Ay

Combining these equations, A

represented by the vector sum of its

three rectangular components;

A = Ax

+ Ay

+ Az

59

U i

continue


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Unit vector

Specified as the direction ofA since it has amagnitude of 1.

IfA is a vector having a magnitude A 0, then the unitvector having the same direction as A is representedby;

From equation 2.1, the unit vector will be dimensionlesssince the unit will cancel out.

Equation 2.2 therefore indicates that vector A may beexpressed in terms of both its magnitude and directionseparately

Eg: A positive scalar defines the magnitude ofA.

uA (a dimensionless vector) defines the direction and sence of A.

2.1

2.2

60

continue


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Cartesian unit vectors.

In 3D, the set of Cartesian unit vectors i,j,k, is used to

designate the directions of thex, y, z axesrespectively.

z

y

x

k

ji

The sense (or arrowhead)

of these

vectors will be describedanalytically by a plus or

minus sign, depending on

whether they are pointing

along the positive or

negativex, y, orz axes.

Figure shows the positive Cartesian unit vectors.

61

continue


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Cartesian Vector Representation.

Magnitude of Cartesian Vector

z

y

x

k

j

i

A

Az k

Ax i

Ay j

62

f

continue


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Direction of a Cartesian Vector

The orientation ofA is defined by the coordinate

direction angle (alpha), (beta), and (gamma),measured between the tail ofA and the positivex, y, z

axes located at the tail ofA.

Each of the angles will be between 0 and 180.z

y

x

A

Az

Ax

Ay 63

continue


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Direction cosine ofA

z

y

A

x

z

y

x

A

Ax

z

y

x

A

Ay

Az

64

continue


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A expressed in Cartesian vector form as:

Direction angles:

Since the magnitude of a vector is equal to the positive square

root of the sum of the squares of the magnitudes of its

components, and uA has a magnitude of 1.

65

Example 2.11


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Determine the magnitude and the coordinate direction angles

of the resultant force acting on the ring in Fig. a.

66

continue


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Solution: Since each force is represented in Cartesian vector form, the resultant force,

shown in Fig. b, is:

The magnitude ofFR is found from equation above;

67

continue


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The coordinate direction angles ,, are determined from

the components of unit vector acting in the direction ofFR.

So that,

These angles are shown in Figure b. in particular, note that

> 90 since the j component ofuFR is negative.

68


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x, y, z coordinates. Right-handed coordinate system is used to reference

the location of points in space.

In many technical books, to require the positive z axisto be directed upward (the zenith direction) so that ismeasures the height of an object or the altitude of thepoint.

Thex andy axes then lie in the horizontal plane.

Points in space are located relative to the origin ofcoordinates, O, by successive measurements along the

x, y, z axes.

Position Vectors

69

z


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y

x

4 m

4 m

6 m1 m

2 m

2 m

From the figure above, coordinate at point A:

xA = +4 m along thex-axis

yA = +2 m along they-axis

zA = -6 m along thez-axis

thus,

A (4,2,-6)

B (0,2,0)

C (6,-1,4)

A

BC

70

continue


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Position vector.

The position vector, r, is

defined as a fixed vectorwhich locates a point inspace relative to anotherpoint.

For example, ifr extendsfrom the origin ofcoordinates, O, to pointP(x,y,z), then, r can beexpressed in Cartesian

vector form as:

r =xi +yj+zk

A

O

x i

z

y

x

yj

r

zk

71

continue

z


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y

x

r

rA

rB

B (xB, yB, zB)

A (xA, yA, zA)

By the head-to-tail vector addition, we require:

rA +r = rBSolving for r and expressingrA andrB in cartesian vector

form as:

r = rB -rA= (xB i +yB j + zB k)(xA i + yA j + zA k)

72

z

continue


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y

x

r

rA

rB

B

A(xBxA)

(yByA)

(zBzA)

In other way in solving for r and expressingrA andrB in

Cartesian vector:

[+ i direction] [+j direction] [+ kdirection]

73

Example 2.12


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74

An elastic rubber band is attached to pointsA andB as shown in Fig.

(a). Determine its length and its direction measured fromA towardB.

continue


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75

Solution:

First, establish a position vector

A toB, Figure (b). The coordinates of the tail

A (1 m, 0, -3 m) are substracted

from the coordinates of the head

B (-2 m, 2 m, 3 m), which yields;

r = BA

= [-2 m1 m]i

+ [2 m0]j

+ [3 m(-3 m)]k

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

Solution:

continue


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76

Solution:

The components ofr can be determined directly by

realizing from fig. (a) that they represent the direction and

distance one must go along each axis in order to move fromA toB.

The magnitude ofr represents the length of the rubber

band.

Formulating a unit vector in the direction ofr, we have;

Solution:

continue


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77

Solution:

The components of this unit vector yield the coordinatedirection angles:

These angles are measured from thepositive axes of alocalized coordinate system placed at the tail ofr, point Aas shown in Fig. (c).

d d l l


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The direction of a force is specified by two pointsthrough which its line of action passes.

We can formulate F as a Cartesian vector by realizing

that it has the same direction and sense as the

position vector r directed from point A to point B on

the cord.

This common direction is specified by the unit vector

u=r/r. Hence, F = Fu = F (r/r)

Force vector directed along a line

z

y

x

A

B

F

Force F is directed along the cord AB.

78

Example 2.13


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79

The man shown in Figure (a) pulls on chord with a force of 350 N.

Represent this force, acting on the support A, as a Cartesian vector and

determine its direction.

Solution:

continue


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Solution:

Force F is shown in Figure (b). The direction of this vector,u, is determined from the position vector r, which extends

from A to B. The coordinates of the end points

of the cord are:

A (0, 0, 7.5 m)

B (3 m, -2 m, 1.5 m)

Forming the position vector, we

have;

r = BA

= (30) i + (-20)j + (1.57.5) k

= { 3i2j6k} m

80

Magnitude of r (represent the length of cord AB):

continue


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Magnitude ofr, (represent the length of cord AB):

Forming the unit vector that defines the direction and sense

of both r and F yields:

Since F has a magnitude of 350 N and a direction specified

by u, then,

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Coordinate direction angles:

continue


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82

Coordinate direction angles:

Measured between r (or F) and the positive axes of a

localized coordinate system with origin placed at A.

From the components of the unit vector:

Dot Product


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Dot product defines a particular method formultiplying two vectors and is used to solve the 3Dimensional problems.

Dot product of vectors A and B is A . B Defined as the product of the magnitudes of A and B and

the cosine of the angle between their tails.

Equation form:

A . B = AB cos

Where 0 180

Dot product is often referred to as the scalar product ofvectors; since the result is a scalar, NOT a vector.

Dot Product

A

B

(Eqn. 2.3)

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Laws of Operation

continue


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Laws of Operation

Commutative Law

Multiplication by a scalar

Distributive Law

It is easy to prove the first and second laws by using Eqn. 2.3

84

continue


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Cartesian Vector Formulation

Equation 2.1 may be used to find the dot product for

each of the Cartesian unit vectors. Example:

In similar manner:

Should not be memorized, but understood.

85

Consider now the dot product of two general vectors A and

continue


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Consider now the dot product of two general vectors A andB which are expressed in Cartesian vector form. We have:

Carrying out the dot-product operations, the final resultbecomes:

Thus, to determine the dot product of two Cartesian vectors,multiply their corresponding x,y,z components and sum theirproducts algebraically.

Since the result is a scalar, NOT to include any unit vector infinal result.

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R f


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References:

R.C. Hibbeler. (2004). Engineering MechanicsStatics Third Edition.

F.P. Beer, E.R. Johnston, Jr, E.R. Eisenberg.

(2004). Vector Mechanics for Engineers.Statics. Seventh Edition in SI Units.

unit 2 vector

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