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1
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
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