1. Engineering Mechanics: Statics
Chapter 2:
Force Vectors
Engineering Mechanics: Statics

2. Objectives
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.

3. Chapter Outline Scalars and Vectors Vector Operations
Vector Addition of Forces
Addition of a System of Coplanar Forces
Cartesian Vectors

4. Chapter Outline Addition and Subtraction of Cartesian Vectors
Position Vectors
Force Vector Directed along a Line
Dot Product

5. 2.1 Scalars and Vectors Scalar
– A quantity characterized by a positive or negative number
– Indicated by letters in italic such as A
Eg: Mass, volume and length

6. 2.1 Scalars and Vectors Vector
– A quantity that has both magnitude and direction
Eg: 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

7. 2.1 Scalars and Vectors 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

8. 2.1 Scalars and Vectors Example Magnitude of Vector = 4 units
Direction of Vector = 20° measured counterclockwise from the horizontal axis
Sense 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

9. 2.2 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

10. 2.2 Vector Operations
Multiplication and Division of a Vector by a Scalar
- Negative of a vector is found by multiplying the vector by ( -1 )
- Law of multiplication applies
Eg: A/a = ( 1/a ) A, a≠0

11. 2.2 Vector Operations Vector 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
- Communicative
Eg: R = A + B = B + A

12. 2.2 Vector Operations
Vector Addition

13. 2.2 Vector Operations Vector Addition
- Special case: Vectors A and B are collinear (both have the same line of action)

14. 2.2 Vector Operations Vector Subtraction - Special case of addition
Eg: R’ = A – B = A + ( - B )
- Rules of Vector Addition Applies

15. 2.2 Vector Operations 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

16. 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 resultant
Eg: Forces F1, F2 and F3 acts at a point O
- First, find resultant of
F1 + F2
- Resultant,
FR = ( F1 + F2 ) + F3

17. 2.3 Vector Addition of Forces
Example
Fa and Fb are forces exerting on the hook.
Resultant, Fc can be found using the parallelogram law
Lines parallel to a and b
from the heads of Fa and Fb are
drawn to form a parallelogram
Similarly, given Fc, Fa and Fb
can be found

18. 2.3 Vector Addition of Forces
Procedure for Analysis
Parallelogram 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

19. 2.3 Vector Addition of Forces
Procedure for Analysis
Parallelogram 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

20. 2.3 Vector Addition of Forces
Procedure for Analysis
Trigonometry
- 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

21. 2.3 Vector Addition of Forces
Procedure for Analysis
Trigonometry
- Magnitude of the two components can be determined by the law of sines

22. 2.3 Vector Addition of Forces
Example 2.1
The screw eye is subjected to two forces F1
and F2. Determine the
magnitude and direction
of the resultant force.

23. 2.3 Vector Addition of Forces
Solution
Parallelogram Law
Unknown: magnitude of
FR and angle θ

24. 2.3 Vector Addition of Forces
Solution
Trigonometry
Law of Cosines

25. 2.3 Vector Addition of Forces
Solution
Trigonometry
Law of Sines

26. 2.3 Vector Addition of Forces
Solution
Trigonometry
Direction Φ of FR measured from the horizontal

27. 2.3 Vector Addition of Forces
Example 2.2
Resolve the 1000 N ( ≈ 100kg) force
acting on the pipe into the components in the
x and y directions,
and (b) x’ and y
directions.

28. 2.3 Vector Addition of Forces
Solution
Parallelogram Law
From the vector diagram,

29. 2.3 Vector Addition of Forces
Solution
(b) Parallelogram Law

30. 2.3 Vector Addition of Forces
Solution
(b) Law of Sines
NOTE: A rough sketch drawn to scale will give some idea of the relative magnitude of the components, as calculated here.

31. 2.3 Vector Addition of Forces
Example 2.3
The 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.

32. 2.3 Vector Addition of Forces
Solution
Parallelogram Law

33. 2.3 Vector Addition of Forces
Solution
Law of Sines

34. 2.3 Vector Addition of Forces
Solution
By Law of Cosines or
Law of Sines
Hence, show that FAB
has a magnitude of 561N

35. 2.3 Vector Addition of Forces
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

36. 2.3 Vector Addition of Forces
Example 2.4
The 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.

37. 2.3 Vector Addition of Forces
Solution
(a) Parallelogram Law
Unknown: Forces F1 and F2
View Free Body Diagram

38. 2.3 Vector Addition of Forces
Solution
Law of Sines

39. 2.3 Vector Addition of Forces
Solution
(b) Minimum length of F2 occur when its line of action is perpendicular to F1. Hence when
F2 is a minimum

40. 2.3 Vector Addition of Forces
Solution
(b) From the vector diagram

41. 2.4 Addition of a System of Coplanar Forces
For 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.

42. 2.4 Addition of a System of Coplanar Forces
Scalar Notation
- x and y axes are designated positive and negative
- Components of forces expressed as algebraic scalars
Eg:
Sense of direction
along positive x and
y axes

43. 2.4 Addition of a System of Coplanar Forces
Scalar Notation
Eg:
Sense of direction
along positive x and
negative y axes

44. 2.4 Addition of a System of Coplanar Forces
Scalar Notation
- 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

45. 2.4 Addition of a System of Coplanar Forces
Cartesian 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

46. 2.4 Addition of a System of Coplanar Forces
Cartesian Vector Notation
F = Fxi + Fyj F’ = F’xi + F’y(-j)
F’ = F’xi – F’yj

47. 2.4 Addition of a System of Coplanar Forces
Coplanar 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

48. 2.4 Addition of a System of Coplanar Forces
Coplanar Force Resultants Example: Consider three coplanar forces
Cartesian vector notation
F1 = F1xi + F1yj
F2 = - F2xi + F2yj
F3 = F3xi – F3yj

49. 2.4 Addition of a System of Coplanar Forces
Coplanar Force Resultants Vector resultant is therefore
FR = F1 + F2 + F3
= F1xi + F1yj - F2xi + F2yj + F3xi – F3yj
= (F1x - F2x + F3x)i + (F1y + F2y – F3y)j
= (FRx)i + (FRy)j

50. 2.4 Addition of a System of Coplanar Forces
Coplanar Force Resultants If scalar notation are used
FRx = (F1x - F2x + F3x)
FRy = (F1y + F2y – F3y)
In all cases,
FRx = ∑Fx
FRy = ∑Fy
* Take note of sign conventions

51. 2.4 Addition of a System of Coplanar Forces
Coplanar 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

52. 2.4 Addition of a System of Coplanar Forces
Coplanar Force Resultants - Direction angle θ (orientation of the force) can be found by trigonometry

53. 2.4 Addition of a System of Coplanar Forces
Example 2.5
Determine x and y components of F1 and F2
acting on the boom. Express each force as a
Cartesian vector

54. 2.4 Addition of a System of Coplanar Forces
Solution
Scalar Notation
Hence, from the slope triangle

55. 2.4 Addition of a System of Coplanar Forces
Solution
Alt, by similar triangles
Similarly,

56. 2.4 Addition of a System of Coplanar Forces
Solution
Scalar Notation
Cartesian Vector Notation
F1 = {-100i +173j }N
F2 = {240i -100j }N

57. 2.4 Addition of a System of Coplanar Forces
Example 2.6
The link is subjected to two forces F1 and
F2. Determine the magnitude and orientation of the resultant force.

58. 2.4 Addition of a System of Coplanar Forces
Solution
Scalar Notation

59. 2.4 Addition of a System of Coplanar Forces
Solution
Resultant Force
From vector addition,
Direction angle θ is

60. 2.4 Addition of a System of Coplanar Forces
Solution
Cartesian Vector Notation
F1 = { 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 j}N

61. 2.4 Addition of a System of Coplanar Forces
Example 2.7
The end of the boom O is subjected to three
concurrent and coplanar forces. Determine
the magnitude and orientation of the
resultant force.

62. 2.4 Addition of a System of Coplanar Forces
View Free Body Diagram
Solution
Scalar Notation

63. 2.4 Addition of a System of Coplanar Forces
Solution
Resultant Force
From vector addition,
Direction angle θ is

64. 2.5 Cartesian Vectors Right-Handed Coordinate System
A 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

65. 2.5 Cartesian Vectors Right-Handed Coordinate System
- z-axis for the 2D problem would be perpendicular, directed out of the page.

66. 2.5 Cartesian Vectors Rectangular 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

67. 2.5 Cartesian Vectors 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 / A
So that
A = A uA

68. 2.5 Cartesian Vectors 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 direction
and sense of A

69. 2.5 Cartesian Vectors 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)

70. 2.5 Cartesian Vectors 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.

71. 2.5 Cartesian Vectors Magnitude of a Cartesian Vector
- From the colored triangle,
- From the shaded triangle,
- Combining the equations gives
magnitude of A

72. 2.5 Cartesian Vectors 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 °

73. 2.5 Cartesian Vectors Direction of a Cartesian Vector
- For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A

74. 2.5 Cartesian Vectors Direction of a Cartesian Vector
- For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A

75. 2.5 Cartesian Vectors Direction of a Cartesian Vector
- For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A

76. 2.5 Cartesian Vectors 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

77. 2.5 Cartesian Vectors Direction 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

78. 2.6 Addition and Subtraction of Cartesian Vectors
Example
Given: A = Axi + Ayj + AZk
and B = Bxi + Byj + BZk
Vector Addition
Resultant R = A + B
= (Ax + Bx)i + (Ay + By )j + (AZ + BZ) k
Vector Substraction
Resultant R = A - B
= (Ax - Bx)i + (Ay - By )j + (AZ - BZ) k

79. 2.6 Addition and Subtraction of Cartesian Vectors
Concurrent 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

80. 2.6 Addition and Subtraction of Cartesian 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

81. 2.6 Addition and Subtraction of Cartesian Vectors
Cosines of their values forms a unit vector u that acts in the direction of the rope
Force F has a magnitude of F
F = Fu = Fcosαi + Fcosβj + Fcosγk

82. 2.6 Addition and Subtraction of Cartesian Vectors
Example 2.8
Express the force F as Cartesian vector

83. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
Since two angles are specified, the third
angle is found by
Two possibilities exit, namely or

84. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
By inspection, α = 60° since Fx is in the +x direction
Given F = 200N
F = Fcosαi + Fcosβj + Fcosγk
= (200cos60°N)i + (200cos60°N)j
+ (200cos45°N)k
= {100.0i j k}N
Checking:

85. 2.6 Addition and Subtraction of Cartesian Vectors
Example 2.9
Determine the magnitude and coordinate
direction angles of resultant force acting on the ring

86. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
Resultant force
FR = ∑F
= F1 + F2
= {60j + 80k}kN
+ {50i - 100j + 100k}kN
= {50j -40k + 180k}kN
Magnitude of FR is found by

87. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
Unit vector acting in the direction of FR
uFR = FR /FR
= (50/191.0)i + (40/191.0)j + (180/191.0)k
= i j k
So that
cosα = α = 74.8°
cos β = β = 102°
cosγ = γ = 19.6°
*Note β > 90° since j component of uFR is negative

88. 2.6 Addition and Subtraction of Cartesian Vectors
Example 2.10
Express the force F1 as a Cartesian vector.

89. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
The angles of 60° and 45° are not coordinate
direction angles.
By two successive applications of
parallelogram law,

90. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
By trigonometry,
F1z = 100sin60 °kN = 86.6kN
F’ = 100cos60 °kN = 50kN
F1x = 50cos45 °kN = 35.4kN
F1y = 50sin45 °kN = 35.4kN
F1y has a direction defined by –j,
Therefore
F1 = {35.4i – 35.4j k}kN

91. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
Checking:
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 j k

92. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
α1 = 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

93. 2.6 Addition and Subtraction of Cartesian Vectors
Example 2.11
Two 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.

94. 2.6 Addition and Subtraction of Cartesian Vectors
View Free Body Diagram
Solution
Cartesian vector form
FR = F1 + F2
F1 = F1cosα1i + F1cosβ1j + F1cosγ1k
= (300cos45°N)i + (300cos60°N)j
+ (300cos120°N)k
= {212.1i + 150j - 150k}N
F2 = F2xi + F2yj + F2zk

95. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
Since FR has a magnitude of 800N and acts
in the +j direction
FR = F1 + F2
800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk
800j = ( F2x)i + (150 + F2y)j + ( F2z)k
To satisfy the equation, the corresponding
components on left and right sides must be equal

96. 2.6 Addition and Subtraction of Cartesian Vectors
Solution
Hence,
0 = F2x F2x = N
800 = F2y F2y = 650N
0 = 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°

97. 2.7 Position Vectors 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.

98. 2.7 Position Vectors x,y,z Coordinates
Eg: 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)

99. 2.7 Position Vectors 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

100. 2.7 Position Vectors Position Vector
Note 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)

101. 2.7 Position Vectors Position Vector
- Position vector maybe directed from point A to point B
- Designated by r or rAB
Vector addition gives
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

102. 2.7 Position Vectors 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

103. 2.7 Position Vectors
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

104. 2.7 Position Vectors
Angles, α, β and γ represent the direction of the cable
Unit vector, u = r/r

105. 2.7 Position Vectors 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.

106. 2.7 Position Vectors Solution Position vector
View Free Body Diagram
Solution
Position 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 r
u = r /r
= -3/7i + 2/7j + 6/7k

107. 2.7 Position Vectors Solution α = cos-1(-3/7) = 115°

108. 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)

109. 2.8 Force Vector Directed along a Line
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

110. 2.8 Force Vector Directed along a Line
Unit vector, u = r/r that defines the direction of both the chain and the force
We get F = Fu

111. 2.8 Force Vector Directed along a Line
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.

112. 2.8 Force Vector Directed along a Line
Solution
End 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

113. 2.8 Force Vector Directed along a Line
Solution
Force 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°

114. 2.9 Dot Product
Dot 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

115. 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)

116. 2.9 Dot Product Cartesian Vector Formulation
- Dot product of Cartesian unit vectors
Eg: i·i = (1)(1)cos0° = 1 and
i·j = (1)(1)cos90° = 0
- Similarly
i·i = 1 j·j = 1 k·k = 1
i·j = 0 i·k = 1 j·k = 1

117. 2.9 Dot Product Cartesian Vector Formulation
- Dot product of 2 vectors A and B
A·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

118. 2.9 Dot Product 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

119. 2.9 Dot Product Applications
- 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

120. 2.9 Dot Product 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

121. 2.9 Dot Product Applications
For 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

122. 2.9 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)

123. 2.9 Dot Product 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

124. 2.9 Dot Product
Example 2.16
The frame is subjected to a horizontal force
F = {300j} N. Determine the components of
this force parallel and perpendicular to the
member AB.

125. 2.9 Dot Product
Solution
Since
Then

126. 2.9 Dot Product Solution Since result is a positive scalar,
FAB has the same sense of
direction as uB. Express in
Cartesian form
Perpendicular component

127. 2.9 Dot Product Solution Magnitude can be determined
From F┴ or from Pythagorean
Theorem

128. Chapter Summary Parallelogram Law Addition of two vectors
Components 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

129. Chapter Summary Cartesian Vectors
Vector F resolved into Cartesian vector form
F = 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

130. Chapter Summary Cartesian Vectors Force and Position 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

131. Chapter Summary Force and Position Vectors Dot Product
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 vector
F = Fu = F(r/r)
Dot Product
Dot product between two vectors A and B
A·B = AB cosθ

132. Chapter Summary 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

133. Chapter Review

134. Chapter Review

135. Chapter Review

136. Chapter Review

137. Chapter Review

138. Chapter Review