Ch02 Statics

Engineering Mechanics: StaticsEngineering Mechanics: Statics

Chapter 2: Force Vectors

Chapter 2: Force Vectors

ObjectivesObjectives

To show how to add forces and resolve them into components using the Parallelogram Law.

To express force and position in Cartesian vector form and explain how to determine the vector’s magnitude and direction.

To introduce the dot product in order to determine the angle between two vectors or the projection of one vector onto another.

Chapter OutlineChapter Outline

1. Scalars and Vectors2. Vector Operations3. Vector Addition of Forces4. Addition of a System of

Coplanar Forces5. Cartesian Vectors

Chapter OutlineChapter Outline

6. Addition and Subtraction of Cartesian Vectors7. Position Vectors8. Force Vector Directed along

a Line9. Dot Product

2.1 Scalars and Vectors2.1 Scalars and Vectors

Scalar – A quantity characterized by a positive or negative number – Indicated by letters in italic such as AEg: Mass, volume and length


Vector – A quantity that has both magnitude and directionEg: Position, force and moment – Represent by a letter with an arrow over it such as or A– Magnitude is designated as or simply A– In this subject, vector is presented as A and its magnitude (positive quantity) as A

A

A


Vector – Represented graphically as an arrow – Length of arrow = Magnitude of Vector – Angle between the reference axis and arrow’s line of action = Direction of Vector – Arrowhead = Sense of Vector


ExampleMagnitude of Vector = 4 unitsDirection of Vector = 20° measured counterclockwise from the horizontal axisSense of Vector = Upward and to the right

The point O is called tail of the vector and the point P is called the tip or head

2.2 Vector Operations2.2 Vector Operations

Multiplication and Division of a Vector by a Scalar- Product of vector A and scalar a = aA- Magnitude = - If a is positive, sense of aA is the same as sense of A- If a is negative sense of aA, it is opposite to the sense of A

aA


Multiplication and Division of a Vector by a Scalar- Negative of a vector is found by multiplying the vector by ( -1 )- Law of multiplication appliesEg: A/a = ( 1/a ) A, a≠0

2.2 Vector Operations2.2 Vector OperationsVector 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- CommunicativeEg: R = A + B = B + A


Vector Addition

2.2 Vector Operations2.2 Vector OperationsVector Addition

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


Vector Subtraction- Special case of additionEg: R’ = A – B = A + ( - B )- Rules of Vector Addition Applies


Resolution of Vector- Any vector can be resolved into two components by the parallelogram law- The two components A and B are drawn such that they extend from the tail or R to points of intersection

2.3 Vector Addition of Forces


When two or more forces are added, successive applications of the parallelogram law is carried out to find the resultantEg: Forces F1, F2 and F3 acts at a point O

- First, find resultant of F1 + F2

- Resultant,

FR = ( F1 + F2 ) + F3



Example Fa and Fb are forces exerting on the

hook.

Resultant, Fc can be found using the parallelogram law

Lines parallel to a and bfrom the heads of Fa and Fb are drawn to form a parallelogram Similarly, given Fc, Fa and Fb

can be found



Procedure for AnalysisParallelogram Law

- Make a sketch using the parallelogram law- Two components forces add to form the resultant force - Resultant force is shown by the diagonal of the parallelogram - The components is shown by the sides of the parallelogram



Procedure for AnalysisParallelogram Law

To resolve a force into components along two axes directed from the tail of the force- Start at the head, constructing lines parallel to the axes

- Label all the known and unknown force magnitudes and angles- Identify the two unknown components



Procedure for AnalysisTrigonometry

- Redraw half portion of the parallelogram- Magnitude of the resultant force can be determined by the law of cosines- Direction if the resultant force can be determined by the law of sines



Procedure for AnalysisTrigonometry

- Magnitude of the two components can be determined by the law of sines



Example 2.1The screw eye is subjected to two forces F1

and F2. Determine the magnitude and direction of the resultant force.



Solution Parallelogram LawUnknown: magnitude of FR and angle θ



Solution TrigonometryLaw of Cosines

N

N

NNNNFR

213

6.212

4226.0300002250010000

115cos1501002150100 22



Solution TrigonometryLaw of Sines

8.39sin

9063.06.212

150sin

115sin

6.212

sin

150

N

N

NN



Solution TrigonometryDirection Φ of FR measured from the horizontal

8.54

158.39



Example 2.2Resolve the 1000 N ( ≈ 100kg) force acting on the pipe into the components in the (a) x and y directions, (b) and (b) x’ and ydirections.



Solution(a) Parallelogram

Law

From the vector diagram,

yx FFF

NF

NF

y

x

64340sin1000

76640cos1000



Solution(b) Parallelogram

Law

'yx FFF



Solution(b) Law of Sines

NNF

NF

NNF

NF

y

y

x

x

108560sin

70sin1000

60sin

1000

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.



Example 2.3The force F acting on the frame has a magnitude of 500N and is to be resolved into two components acting along the members AB and AC. Determine the angle θ, measured below the horizontal, so that components FAC is directed from A towards C and has a magnitude of 400N.



SolutionParallelogram

LawACAB FFN 500



SolutionLaw of Sines

9.43

6928.0sin

60sin500

400sin

60sin

500

sin

400

N

N

NN



Solution

By Law of Cosines or Law of SinesHence, show that FAB

has a magnitude of 561N

1.769.4360180

,Hence



Solution F can be directed at an angle θ above the

horizontal to produce the component FAC. Hence, show that θ = 16.1° and FAB = 161N



Example 2.4The ring is subjected to two

forces F1 and F2. If it is required that the resultant force have a magnitude

of 1kN and be directed vertically downward, determine (a) magnitude of F1 and F2

provided θ = 30°, and (b) the magnitudes of F1 and F2 if

F2 is to be a minimum.



Solution(a) Parallelogram LawUnknown: Forces F1 and F2

View Free Body Diagram



SolutionLaw of Sines

NF

NF

NF

NF

446130sin

1000

20sin

643130sin

1000

30sin

2

2

1

1



Solution(b) Minimum length of F2

occur when its line of action is perpendicular to F1. Hence when

F2 is a minimum

702090



Solution(b) From the vector

diagram

NNF

NNF

34270cos1000

94070sin1000

2

1

2.4 Addition of a System of Coplanar

Forces


ForcesFor 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


Forces


ForcesScalar Notation

- x and y axes are designated positive and negative- Components of forces expressed as algebraic scalarsEg: Sense of direction along positive x and y axes

yx FFF


Forces



Eg: Sense of direction along positive x and negative y axes

yx FFF '''


Forces



- Head of a vector arrow = sense of the vector graphically (algebraic signs not used)- Vectors are designated using boldface notations- Magnitudes (always a positive quantity) are designated using italic symbols


Forces


ForcesCartesian Vector Notation

- Cartesian unit vectors i and j are used to designate the x and y directions- Unit vectors i and j have dimensionless magnitude of unity ( = 1 )

- Their sense are indicated by a positive or negative sign (pointing in the positive or negative x or y axis) - Magnitude is always a positive quantity, represented by scalars Fx and Fy


Forces


ForcesCartesian Vector Notation

F = Fxi + Fyj F’ = F’xi + F’y(-j)

F’ = F’xi – F’yj


Forces


ForcesCoplanar Force Resultants

To determine resultant of several coplanar forces:- Resolve force into x and y components- Addition of the respective components using scalar algebra - Resultant force is found using the parallelogram law


Forces


ForcesCoplanar Force Resultants

Example: Consider three coplanar forces

Cartesian vector notationF1 = F1xi + F1yj

F2 = - F2xi + F2yj

F3 = F3xi – F3yj

Coplanar Force ResultantsVector resultant is thereforeFR = F1 + F2 + F3

= F1xi + F1yj - F2xi + F2yj + F3xi – F3yj

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

= (FRx)i + (FRy)j


Forces


Forces


Forces


ForcesCoplanar Force ResultantsIf scalar notation are usedFRx = (F1x - F2x + F3x)

FRy = (F1y + F2y – F3y)

In all cases,FRx = ∑Fx

FRy = ∑Fy

* Take note of sign conventions


Forces


ForcesCoplanar Force Resultants- Positive scalars = sense of direction along the positive coordinate axes- Negative scalars = sense of direction along the negative coordinate axes- Magnitude of FR can be found by Pythagorean Theorem

RyRxR FFF 22


Forces


ForcesCoplanar Force Resultants- Direction angle θ (orientation of the force) can be found by trigonometry

Rx

Ry

F

F1tan


Forces


ForcesExample 2.5Determine x and y components of F1 and F2

acting on the boom. Express each force as a Cartesian vector


Forces


ForcesSolutionScalar Notation

Hence, from the slope triangle

NNNF

NNNF

y

x

17317330cos200

10010030sin200

1

1

12

5tan 1


Forces


ForcesSolutionAlt, by similar

triangles

Similarly,

NNF

N

F

x

x

24013

12260

13

12

260

2

2

NNF y 10013

52602


Forces


ForcesSolution

Scalar Notation

Cartesian Vector Notation

F1 = {-100i +173j }N

F2 = {240i -100j }N

NNF

NNF

y

x

100100

240240

2

2


Forces


ForcesExample 2.6The link is subjected to two forces F1

and F2. Determine the magnitude and

orientation of the resultant force.


Forces



N

NNF

FF

N

NNF

FF

Ry

yRy

Rx

xRx

8.582

45cos40030sin600

:

8.236

45sin40030cos600

:


Forces


ForcesSolutionResultant Force

From vector addition,Direction angle θ is

N

NNFR629

8.5828.236 22

9.67

8.236

8.582tan 1

N

N


Forces


ForcesSolutionCartesian Vector NotationF1 = { 600cos30°i + 600sin30°j } N

F2 = { -400sin45°i + 400cos45°j } N

Thus, FR = F1 + F2

= (600cos30°N - 400sin45°N)i + (600sin30°N + 400cos45°N)j

= {236.8i + 582.8j}N


Forces


ForcesExample 2.7The end of the boom O is subjected to three concurrent and coplanar forces. Determine the magnitude and orientation of the resultant force.


Forces



N

NNF

FF

NN

NNNF

FF

Ry

yRy

Rx

xRx

8.296

5

320045cos250

:

2.3832.383

5

420045sin250400

:


SolutionResultant Force

From vector addition,Direction angle θ is


Forces


Forces

N

NNFR485

8.2962.383 22

8.37

2.383

8.296tan 1

N

N

2.5 Cartesian Vectors2.5 Cartesian Vectors

Right-Handed Coordinate SystemA rectangular or Cartesian coordinate system is said to be right-handed provided:- Thumb of right hand points in the direction of the positive z axis when the right-hand fingers are curled about this axis and directed from the positive x towards the positive y axis


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

2.5 Cartesian Vectors2.5 Cartesian VectorsRectangular Components of a Vector

- A vector A may have one, two or three rectangular components along the x, y and z axes, depending on orientation- By two successive application of the parallelogram law

A = A’ + Az

A’ = Ax + Ay

- Combing the equations, A can be expressed as

A = Ax + Ay + Az


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, unit vector having the same direction as A is expressed by

uA = A / ASo that

A = A uA


Unit Vector- Since A is of a certain type, like force vector, a proper set of units are used for the description- Magnitude A has the same sets of units, hence unit vector is dimensionless- A ( a positive scalar) defines magnitude of A - uA defines the directionand sense of A


Cartesian Unit Vectors- Cartesian unit vectors, i, j and k are used to designate the directions of z, y and z axes- Sense (or arrowhead) of these vectors are described by a plus or minus sign (depending on pointing towards the positive or negative axes)


Cartesian Vector Representations- Three 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.


Magnitude of a Cartesian Vector - From the colored triangle,

- From the shaded triangle,

- Combining the equations givesmagnitude of A

222

22

22

'

'

zyx

yx

z

AAAA

AAA

AAA


Direction of a Cartesian Vector- Orientation of A is defined as the coordinate direction angles α, β and γ measured between the tail of A and the positive x, y and z axes- 0° ≤ α, β and γ ≤ 180 °


Direction of a Cartesian Vector- For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A

A

Axcos



A

Aycos



A

Azcos


Direction of a Cartesian Vector- Angles α, β and γ 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 222zyx AAAA

2.5 Cartesian Vectors2.5 Cartesian VectorsDirection of a Cartesian Vector

- uA can also be expressed as

uA = cosαi + cosβj + cosγk

- Since and magnitude of uA = 1,

- A as expressed in Cartesian vector form A = AuA = Acosαi + Acosβj + Acosγk

= Axi + Ayj + AZk

222zyx AAAA

1coscoscos 222

ExampleGiven: A = Axi + Ayj + AZk

and B = Bxi + Byj + BZk

Vector AdditionResultant R = A + B

= (Ax + Bx)i + (Ay + By )j + (AZ + BZ) kVector SubstractionResultant R = A - B

= (Ax - Bx)i + (Ay - By )j + (AZ - BZ) k

2.6 Addition and Subtraction of Cartesian

Vectors


Vectors


Vectors


VectorsConcurrent Force Systems

- Force resultant is the vector sum of all the forces in the system

FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk

where ∑Fx , ∑Fy and ∑Fz represent the algebraic sums of the x, y and z or i, j or k components of each force in the system


Vectors


Vectors

Force, F that the tie down rope exerts on the ground support at O is directed along the rope

Angles α, β and γ can be solved with axes x, y and z


Vectors


Vectors

Cosines of their values forms a unit vector u that acts in the direction of the rope

Force F has a magnitude of FF = Fu = Fcosαi + Fcosβj + Fcosγk


Vectors


VectorsExample 2.8Express the force F as Cartesian

vector


Vectors


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

Two possibilities exit, namelyor

605.0cos

5.0707.05.01cos

145cos60coscos

1coscoscos

1

22

222

222

1205.0cos 1


Vectors


VectorsSolutionBy inspection, α = 60° since Fx is in the +x

directionGiven F = 200N

F = Fcosαi + Fcosβj + Fcosγk = (200cos60°N)i + (200cos60°N)j

+ (200cos45°N)k = {100.0i + 100.0j + 141.4k}N

Checking:

N

FFFF zyx

2004.1410.1000.100 222

222


Vectors


Vectors Example 2.9 Determine the magnitude and

coordinate direction angles of resultant force

acting on the ring


Vectors


VectorsSolutionResultant force

FR = ∑F

= F1 + F2 = {60j + 80k}kN

+ {50i - 100j + 100k}kN = {50j -40k + 180k}kN

Magnitude of FR is found by

kN

FR

1910.191

1804050 222


Vectors


VectorsSolutionUnit 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 + 0.9422kSo that

cosα = 0.2617 α = 74.8° cos β = -0.2094 β = 102° cosγ = 0.9422 γ = 19.6°

*Note β > 90° since j component of uFR is negative


Vectors


VectorsExample 2.10Express the force F1 as a Cartesian

vector.


Vectors


VectorsSolutionThe angles of 60° and 45° are not

coordinate direction angles.

By two successive applications ofparallelogram law,


Vectors


VectorsSolutionBy trigonometry,

F1z = 100sin60 °kN = 86.6kNF’ = 100cos60 °kN = 50kNF1x = 50cos45 °kN = 35.4kNF1y = 50sin45 °kN = 35.4kN

F1y has a direction defined by –j, Therefore

F1 = {35.4i – 35.4j + 86.6k}kN

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


Vectors


Vectors

N

FFFF zyx

1006.864.354.35 222

21

21

211

Solutionα1 = cos-1(0.354) = 69.3°

β1 = cos-1(-0.354) = 111°

γ1 = cos-1(0.866) = 30.0°

Using the same method, F2 = {106i + 184j - 212k}kN


Vectors


Vectors


Vectors


VectorsExample 2.11Two forces act on the hook. Specify the coordinate direction angles of F2, so that the

resultant force FR acts along the positive y axis and has a magnitude of 800N.


Vectors


Vectors

SolutionCartesian vector formFR = F1 + F2

F1 = F1cosα1i + F1cosβ1j + F1cosγ1k

= (300cos45°N)i + (300cos60°N)j + (300cos120°N)k

= {212.1i + 150j - 150k}NF2 = F2xi + F2yj + F2zk



Vectors


VectorsSolutionSince FR has a magnitude of 800N and actsin the +j directionFR = F1 + F2

800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk

800j = (212.1 + F2x)i + (150 + F2y)j + (- 50 + F2z)k

To satisfy the equation, the corresponding components on left and right sides must be equal


Vectors


VectorsSolutionHence,

0 = 212.1 + F2x F2x = -212.1N 800 = 150 + F2y F2y = 650N 0 = -150 + F2z F2z = 150N

Since magnitude of F2 and its components are known, α1 = cos-1(-212.1/700) = 108° β1 = cos-1(650/700) = 21.8° γ1 = cos-1(150/700) = 77.6°

x,y,z Coordinates- Right-handed coordinate system- Positive z axis points upwards, measuring the height of an object or the altitude of a point- Points are measured relative to the origin, O.

2.7 Position Vectors2.7 Position Vectors

x,y,z CoordinatesEg: 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)


Position Vector- Position vector r is defined as a fixed vector which locates a point in space relative to another point. Eg: If r extends from the origin, O to point P (x, y, z) then, in Cartesian vector form

r = xi + yj + zk


Position VectorNote the head to tail vector addition of the three components

Start at origin O, one travels x in the +i direction,

y in the +j direction and z in the +k direction, arriving at point P (x, y, z)


2.7 Position Vectors2.7 Position Vectors Position Vector

- Position vector maybe directed from point A to point B - Designated by r or rAB

Vector addition givesrA + r = rB

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

or r = (xB – xA)i + (yB – yA)j + (zB –zA)k

Position Vector- The i, j, k components of the positive vector r may be formed by taking the coordinates of the tail, 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



Length and direction of cable AB can be found by measuring A and B using the x, y, z axes

Position vector r can be established

Magnitude r represent the length of cable


Angles, α, β and γ represent the direction of the cable

Unit vector, u = r/r


Example 2.12An elastic rubber band is attached to points A and

B. Determine its length and

its direction measured from

A towards B.


SolutionPosition 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



Solutionα = cos-1(-3/7) = 115°

β = cos-1(2/7) = 73.4°

γ = cos-1(6/7) = 31.0°

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 = F u = F (r/r)

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



Force F acting along the chain can be presented as a Cartesian vector by- Establish x, y, z axes- Form a position vector r along length of chain



Unit vector, u = r/r that defines the direction of both the chain and the force

We get F = Fu



Example 2.13The man pulls on the cord with a force of 350N. Represent this force

acting on the support A, as a Cartesian vector and determine its direction.



SolutionEnd points of the cord are A (0m, 0m,

7.5m) and B (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



SolutionForce F has a magnitude of 350N, direction specified by u

F = Fu = 350N(3/7i - 2/7j - 6/7k) = {150i - 100j - 300k} N

α = cos-1(3/7) = 64.6° β = cos-1(-2/7) = 107° γ = cos-1(-6/7) = 149°



Example 2.14The circular plate is partially supported by the cable AB. If the force of the cable on

the hook at A is F = 500N, express F as a Cartesian vector.



SolutionEnd points of the cable are (0m, 0m, 2m)

and B (1.707m, 0.707m, 0m)

r = (1.707m – 0m)i + (0.707m – 0m)j + (0m – 2m)k

= {1.707i + 0.707j - 2k}mMagnitude = length of cable AB

mmmmr 723.22707.0707.1 222

SolutionUnit vector,

u = r /r = (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k

= 0.6269i + 0.2597j – 0.7345kFor force F,

F = Fu = 500N(0.6269i + 0.2597j – 0.7345k) = {313i - 130j - 367k} N



SolutionChecking

Show that γ = 137° and indicate this angle on the diagram



N

F

500

367130313 222



Example 2.15The 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.



SolutionrAB = (4m – 0m)i + (0m – 0m)j + (0m –

4m)k = {4i – 4k}m

FAB = 100N (rAB/r AB)

= 100N {(4/5.66)i - (4/5.66)k} = {70.7i - 70.7k} N

mmmrAB 66.544 22




SolutionrAC = (4m – 0m)i + (2m – 0m)j + (0m – 4m)k

= {4i + 2j – 4k}m

FAC = 120N (rAB/r AB)

= 120N {(4/6)i + (2/6)j - (4/6)k} = {80i + 40j – 80k} N

mmmmrAC 6424 222



SolutionFR = FAB + FAC

= {70.7i - 70.7k} N + {80i + 40j – 80k} N = {150.7i + 40j – 150.7k} N

Magnitude of FR

N

FR

217

7.150407.150 222

2.9 Dot Product2.9 Dot ProductDot product of vectors A and B is

written as A·B (Read A dot B)Define the magnitudes of A and B and

the angle between their tails A·B = AB cosθ where 0°≤ θ

≤180°Referred to as scalar

product of vectors as result is a scalar

2.9 Dot Product2.9 Dot Product

Laws of Operation1. Commutative law

A·B = B·A2. Multiplication by a scalar

a(A·B) = (aA)·B = A·(aB) = (A·B)a

3. Distribution lawA·(B + D) = (A·B) + (A·D)


Cartesian Vector Formulation- Dot product of Cartesian unit vectorsEg: i·i = (1)(1)cos0° = 1 andi·j = (1)(1)cos90° = 0- Similarly

i·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 1 j·k = 1


Cartesian Vector Formulation- Dot product of 2 vectors A and BA·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)

= AxBx(i·i) + AxBy(i·j) + AxBz(i·k)

+ AyBx(j·i) + AyBy(j·j) + AyBz(j·k)

+ AzBx(k·i) + AzBy(k·j) + AzBz(k·k)

= AxBx + AyBy + AzBz

Note: since result is a scalar, be careful of including any unit vectors in the result


Applications- The angle formed between two vectors or intersecting lines

θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°

Note: if A·B = 0, cos-10= 90°, A is perpendicular to B

2.9 Dot Product2.9 Dot ProductApplications

- The components of a vector parallel and perpendicular to a line- Component of A parallel or collinear with line aa’ is defined by A║ (projection of A onto the line)

A║ = A cos θ

- If direction of line is specified by unit vector u (u = 1),

A║ = A cos θ = A·u


Applications- If A║ is positive, A║ has a directional sense same as u- If A║ is negative, A║ has a directional sense opposite to u- A║ expressed as a vector

A║ = A cos θ u = (A·u)u

ApplicationsFor component of A perpendicular to line aa’ 1. Since A = A║ + A┴,

then A┴ = A - A║

2. θ = cos-1 [(A·u)/(A)]then A┴ = Asinθ

3. If A║ is known, by Pythagorean Theorem


2||

2 AAA

2.9 Dot Product2.9 Dot Product For angle θ between the

rope and the beam A, - Unit vectors along the beams, uA = rA/rA

- Unit vectors along the ropes, ur=rr/rr

- Angle θ = cos-1 (rA.rr/rArr)

= cos-1 (uA· ur)


For projection of the force along the beam A - Define direction of the beam

uA = rA/rA

- Force as a Cartesian vector

F = F(rr/rr) = Fur

- Dot product F║ = F║·uA


Example 2.16The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB.


SolutionSince

Then

N

kjijuF

FF

kji

kjirr

u

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

362222

2.9 Dot Product2.9 Dot ProductSolutionSince result is a positive scalar,FAB has the same sense of

direction as uB. Express in Cartesian form

Perpendicular component

NkjikjijFFF

Nkji

kjiN

uFF

AB

ABABAB

}110805.73{)1102205.73(300

}1102205.73{

429.0857.0286.01.257

2.9 Dot Product2.9 Dot ProductSolutionMagnitude can be determined From F┴ or from Pythagorean

Theorem

N

NN

FFF AB

155

1.257300 22

22


Example 2.17The pipe is subjected to F = 800N. Determine the angle θ between F and pipe segment BA, and the magnitudes of the components of F, which are parallel and perpendicular to BA.


SolutionFor angle θrBA = {-2i - 2j + 1k}m

rBC = {- 3j + 1k}m

Thus,

5.42

7379.0

103

113202cos

BCBA

BCBA

rr

rr



SolutionComponents of F

N

kjikj

uFF

kji

kjirr

u

BAB

AB

ABAB

590

3.840.5060

31

32

32

0.2539.758

.

31

32

32

3)122(


SolutionChecking from trigonometry,

Magnitude can be determined From F┴

N

N

FFAB

590

5.42cos800

cos

NFF 5405.42sin800sin


SolutionMagnitude can be determined from F┴ or

from Pythagorean Theorem

N

FFF AB

540

590800 22

22

Chapter SummaryChapter Summary

Parallelogram LawAddition of two vectorsComponents form the side and

resultant form 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


Cartesian Vectors Vector F resolved into Cartesian vector

formF = Fxi + Fyj + Fzk

Magnitude of F

Coordinate direction angles α, β and γ are determined by the formulation of the unit vector in the direction of F

u = (Fx/F)i + (Fy/F)j + (Fz/F)k

222zyx FFFF


Cartesian Vectors Components of u represent cosα, cosβ and

cosγ These angles are related by

cos2α + cos2β + cos2γ = 1

Force and Position Vectors Position Vector is directed between 2 points Formulated by distance and direction moved

along the x, y and z axes from tail to tip


Force and Position Vectors For line of action through the two

points, it acts in the same direction of u as the position vector

Force expressed as a Cartesian vectorF = Fu = F(r/r)

Dot Product Dot product between two vectors A and

B A·B = AB cosθ


Dot Product Dot product between two vectors A and B

(vectors expressed as Cartesian form)A·B = AxBx + AyBy + AzBz

For angle between the tails of two vectors θ = cos-1 [(A·B)/(AB)]

For projected component of A onto an axis defined by its unit vector u

A = A cos θ = A·u

Chapter ReviewChapter Review

Ch02 Statics

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