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Chapter Objectives

• Parallelogram Law• Cartesian vector form• Dot product

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Chapter Outline

1. Scalars and Vectors2. Vector Operations3. Vector Addition of Forces4. Addition of a System of Coplanar Forces5. Cartesian Vectors6. Addition and Subtraction of Cartesian Vectors7. Position Vectors 8. Dot Product

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2.1 Scalars and Vectors

• Scalar – A quantity characterized by a positive or negative number

– Indicated by letters in italic such as A e.g.

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2.1 Scalars and Vectors

• Vector – A quantity that has magnitude and direction e.g. – Vector

– MagnitudeA

A

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2.2 Vector Operations

• Multiplication and Division of a Vector by a Scalar- Product of vector A and scalar a - Magnitude = - Law of multiplication applies e.g. A/a = ( 1/a ) A, a≠0

• Vector Addition - R = A + B = B + A

- collinear

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2.2 Vector Operations

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2.2 Vector Operations

• Vector Subtraction - Special case of additione.g. R’ = A – B = A + ( - B )

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2.3 Vector Addition of Forces

• Parallelogram law

• Resultant, FR = ( F1 + F2 )

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2.3 Vector Addition of Forces

• Trigonometry– law of cosines– law of sines

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Example 2.1

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

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Solution

Parallelogram LawUnknown: magnitude of FR and angle θ

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Solution

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Solution

TrigonometryDirection Φ of FR measured from the horizontal

8.54

158.39

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2.4 Addition of a System of Coplanar Forces

• Scalar Notation

yF

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2.4 Addition of a System of Coplanar Forces

• Cartesian Vector Notation – use i and j for x and y direction– The magnitude of i and j is one

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2.4 Addition of a System of Coplanar Forces

• Coplanar Force Resultants– Find componenets in x and y – Add in each direction– Resultant is from parallelogram– Cartesian vector notation:

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2.4 Addition of a System of Coplanar Forces

• Coplanar Force Resultants

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2.4 Addition of a System of Coplanar Forces

• Coplanar Force Resultants– We can show that

– Magnitude of FR from Pythagorus

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Example 2.5

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

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Solution

Scalar Notation

Cartesian Vector Notation

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Solution

By similar triangles we have

Scalar Notation:

Cartesian Vector Notation:

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Example 2.6

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

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Solution I

Scalar Notation:

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Solution I

Resultant Force

From vector addition, direction angle θ is

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Solution II

Cartesian Vector Notation

Thus,

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2.5 Cartesian Vectors (3D)

• Right-Handed Coordinate System- thumb represents z– the rest, sweeping from x to y

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2.5 Cartesian Vectors

• Unit Vector– Vector A can be described by a unit vector– uA = A / A

A = A uA

Unit vector for x, y, z

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2.5 Cartesian Vectors

• Cartesian Vector Representations– A can be written by i, j and k directions

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2.5 Cartesian Vectors

• Direction of a Cartesian Vector – The direction of A is defined by α, β and γ angle

between A and x, y and z– 0° ≤ α, β and γ ≤ 180 °– The direction cosines of A is

A

AxcosA

Aycos

A

Azcos

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2.5 Cartesian Vectors

• Direction of a Cartesian Vector

A = AuA = Acosαi + Acosβj + Acosγk = Axi + Ayj + AZk

222zyx AAAA

1coscoscos 222

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Example 2.8

Express the force F as Cartesian vector.

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Solution

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Solution

Notice, α = 60º since Fx is in +x

From F = 200N

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2.7 Position Vectors

• x,y,z Coordinates– Right-handed coordinate system– O is a reference

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2.7 Position Vectors

Position Vector– Position vector r is a vector to identify a location

of a point relative to other points– E.g. r = xi + yj + zk

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2.7 Position Vectors

Position Vector (between 2 points)– Vector addition rA + r = rB

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

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Example 2.12

An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B.

A (1, 0, -3) mB (-2, 2, 3) m

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Solution

Position vector

Magnitude = length of the rubber band

Unit vector in the director of r

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Solution

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2.9 Dot Product

• Dot product of A and B can be written as A·B A·B = AB cosθ where 0°≤ θ ≤180°

• The result is scalar

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2.9 Dot Product

• Laws of Operation1. Commutative law

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

a(A·B) = (aA)·B = A·(aB) = (A·B)a3. Distribution law

A·(B + D) = (A·B) + (A·D)

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2.9 Dot Product

• Cartesian Vector Formulation- Dot product of Cartesian unit vectors

i·i = (1)(1)cos0° = 1i·j = (1)(1)cos90° = 0

- Similarlyi·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 0 j·k = 0

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2.9 Dot Product

• Cartesian Vector Formulation– Dot product of 2 vectors A and B

A·B = AxBx + AyBy + AzBz

Dot product can be used for – Finding angles between two vectors

θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°

– Finding a vector on the direction of a unit vecotr

Aa = A cos θ = A·u

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Example 2.17

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

A (0, 0, 0)B (2, 6, 3)

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Solution

Since

Thus

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Solution

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

Perpendicular component

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Solution

Magnitude can be determined from F ┴ or from Pythagorean Theorem,

or

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4.2 Cross Product

• Cross product of A and B C = A X B

C = AB sinθ

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4.2 Cross Product

• C is perpendicular to the plane containing A and B

C = A X B = (AB sin θ)uC

uC is a unit vector

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4.2 Cross Product

Laws of Operations1. Commutative law

A X B ≠ B X A

But ่ A X B = - (B X A)

• Cross product B X A

B X A = -C

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4.2 Cross Product

Laws of Operations2. Multiplication by a Scalar

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

3. Distributive Law A X (B + D) = (A X B) + (A X D)

And (B + D) X A = (B X A) + (D X A)

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4.2 Cross Product

Cartesian Vector Formulation

0kkijkjik

0jjkijikj

0iijkikji

kji

kkjkik

kjjjij

kijiii

kjikjiBA

)()()(

)()()(

)()()(

)()()(

)()(

xyyxzxxzyzzy

zzyzxz

zyyyxy

zxyxxx

zyxzyx

BABABABABABA

BABABA

BABABA

BABABA

BBBAAA

i

jk

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4.2 Cross Product

Cartesian Vector Formulation• A more compact determinant in the form as

zyx

zyx

BBB

AAA

kji

BA

kji )()()( xyyxzxxzyzzy BABABABABABA

yx

yx

BB

AA

ji

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Example 4.4

Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector.

A (0, 5, 0)B (4, 5, -2)

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Solution

Position vectors are directed from point O to each force as shown.These vectors are

The resultant moment about O is