Engineering Mechanics:
Statics in SI Units, 12e
2
Force Vectors
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Chapter Objectives
• Parallelogram Law • Cartesian vector form • Dot product and an angle between two vectors
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Chapter Outline
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Scalars and Vectors Vector Operations Vector Addition of Forces Addition of a System of Coplanar Forces Cartesian Vectors Addition and Subtraction of Cartesian Vectors Position Vectors Force Vector Directed along a Line 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 (italics) e.g. 4
2.1 Scalars and Vectors
• Vector – A quantity that has magnitude and direction e.g. !
A
– Arrow above a letter !
A
– A magnitude is represented by – In this course, we sometimes use a bold font, A, for a vector and the magnitude is represented by 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
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2.2 Vector Operations
• Vector Addition -­‐ Two vectors addition, A and B, gives a vector R, can be done by parallelogram law -­‐ R can be obtained by triangle -­‐ R = A + B = B + A -­‐ For collinear vectors, i.e. A and B are on the same straight line.
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2.2 Vector Operations
• Vector Subtraction -­‐ Special case of addition, similar to vector addition e.g. R’ = A – B = A + ( -­‐ B )
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2.3 Vector Addition of Forces
Resultant Force • Parallelogram law • Resultant, FR = ( F1 + F2 )
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2.3 Vector Addition of Forces
Analysis Procedure • Parallelogram Law – Draw a parallelogram, parallel to the two forces – The resultant force is a diagonal of the parallelogram
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2.3 Vector Addition of Forces
-­‐ 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 Law Unknown: magnitude of FR and angle θ
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Solution
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Solution
Trigonometry Direction Φ of FR measured from the horizontal φ = 39.8! + 15!
= 54.8! ∠φ
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2.4 Addition of a System of Coplanar Forces
• Scalar Notation – Use scalar values and use axis x and y, consider positive and negative values accordingly
Fy
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2.4 Addition of a System of Coplanar Forces
• Cartesian Vector Notation – Caretesian vectors i and j for x and y, respectively – i and j is a unit vector –
Examples
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2.4 Addition of a System of Coplanar Forces
• Coplanar Force Resultants – Find components in x and y for all forces – Adding all the forces for each direction – The resultant force found by parallelogram or pythagorus –
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 Resultant – Find FR by 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, 27
2.5 Cartesian Vectors (3D)
• Right-­‐Handed Coordinate System – Right thumb z – Other fingers sweeping from x to y – z axis in 2D is pointing outwards from the paper
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2.5 Cartesian Vectors
• Rectangular Components of a Vector –
Vector A can be found in x, y and z –
Use parallelogram twice A = A’ + Az A’ = Ax + Ay A = Ax + Ay + Az
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2.5 Cartesian Vectors
• Unit Vector – Direction A find by unit vector – Magnitude 1 – uA = A / A ดังนี้ A = A uA 30
2.5 Cartesian Vectors
• Cartesian Vector Representations – 3 componenets A in i, j and k directions 31
2.5 Cartesian Vectors
• Magnitude of a Cartesian Vector 32
2.5 Cartesian Vectors
• Direction of a Cartesian Vector – Direction of A defined by α, β and γ – 0° ≤ α, β and γ ≤ 180 ° –
The direction cosines of A is
cos α =
Ax
A
cos γ =
Az
A
cos β =
Ay
A
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2.5 Cartesian Vectors
• Direction of a Cartesian Vector A = Axi + Ayj + AZk uA = A /A = (Ax/A)i + (Ay/A)j + (AZ/A)k A = Ax2 + Ay2 + Az2
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2.5 Cartesian Vectors
• Direction of a Cartesian Vector uA = cosαi + cosβj + cosγk – A = Ax2 + Ay2 + Az2
cos 2 α + cos 2 β + cos 2 γ = 1
= Acosαi + Acosβj + Acosγk = Axi + Ayj + AZk
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2.6 Addition and Subtraction of Cartesian Vectors
• Concurrent Force Systems – Resultant forces FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk
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Example 2.8
Express the force F as Cartesian vector. 37
Solution
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Solution
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2.7 Position Vectors
• x,y,z Coordinates –
Right-­‐handed coordinate system – All other points are relative to O
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2.7 Position Vectors
Position Vector – Position vector r used to indicate a location from a reference position – E.g. r = xi + yj + zk 41
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)k
or r = (xB – xA)i + (yB – yA)j + (zB –zA)k
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2.7 Position Vectors
• Direction and magnitude of a cable can be found by position vectors A and B • Position vector r • Angles α, β and γ •
Unit vector, u = r/r
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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) m
B (-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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2.8 Force Vector Directed along a Line
• For 3D, direction of F •
F = F u = F (r/r) •
F with unit (N) r with unit (m) (r/r) 47
2.8 Force Vector Directed along a Line
•
F is force along the cable -­‐ need x, y, z -­‐ find position vector r along a cable • Find unit vector u = r/r •
F = Fu 48
Example 2.13
The 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. 49
Solution
End points of the cord are A (0m, 0m, 7.5m) and B (3m, -­‐2m, 1.5m) Magnitude = length of cord AB Unit vector, 50
Solution
Force F has a magnitude of 350N, direction specified by u.
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2.9 Dot Product
•
Dot product of A and B written as A·∙B (A dot B) A·∙B = AB cosθ where 0°≤ θ ≤180° • The result is scalar
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2.9 Dot Product
• Laws of Operation 1. Commutative law A·∙B = B·∙A 2. Multiplication by a scalar a(A·∙B) = (aA)·∙B = A·∙(aB) = (A·∙B)a 3. Distribution law A·∙(B + D) = (A·∙B) + (A·∙D) 53
2.9 Dot Product
• Cartesian Vector Formulation -­‐ Dot product of Cartesian unit vectors i·∙i = (1)(1)cos0° = 1 i·∙j = (1)(1)cos90° = 0 -­‐ Similarly i·∙i = 1 j·∙j = 1 k·∙k = 1 i·∙j = 0 i·∙k = 0 j·∙k = 0 54
2.9 Dot Product
• Cartesian Vector Formulation – Dot product of 2 vectors A and B A·∙B = AxBx + AyBy + AzBz Application •
– Find angle between two vectors θ = cos-­‐1 [(A·∙B)/(AB)] –
Find vector parallel and perpendicular components 0°≤ θ ≤180° 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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