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. 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. 2.2 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

9. 2.2 Vector Operations

10. 2.2 Vector Operations Vector Subtraction - Special case of addition
Eg: R’ = A – B = A + ( - B )
- Rules of Vector Addition Applies

11. 2.3 Vector Addition Forces
When two or more forces are added, successive applications of the parallelogram law is carried out to find the resultant
Example
Forces F1, F2 and F3 acts at a point O
a) First, find resultant of F1 + F2
Solution;
Resultant,
FR = ( F1 + F2 ) + F3

12. Procedure of Analysis
Parallelogram Law
Trigonometry

13. Example 1 The screw eye is subjected to two forces F1 and F2.
Determine the magnitude and the direction of a
Resultant force.

14. Solution 1: Parallelogram Law
Unknown: magnitude of FR and angle θ

15. Solution 2: Trigonometry
Law of Cosines

16. Law of Sines
Direction Φ of FR measured from the horizontal

17. 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.

18. 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

19. 2.4 Addition of a System of Coplanar Forces
Scalar Notation
Eg:
Sense of direction
along positive x and
negative y axes

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 2.4 Addition of a System of Coplanar Forces
Coplanar Force Resultants - Direction angle θ (orientation of the force)
can be found by trigonometry

27. Example 2
The link is subjected to two forces F1 and F2.
Determine the magnitude and orientation of the
resultant force.

28. Solution
Scalar Notation

29. Solution
Resultant Force
From vector addition,
Direction angle θ is

30. 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

31. 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

32. 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

33. 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

34. 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

35. 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)

36. 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.

37. 2.5 Cartesian Vectors Magnitude of a Cartesian Vector
- From the colored triangle,
- From the shaded triangle,
- Combining the equations gives
magnitude of A

38. 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 °

39. 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

40. 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

41. 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

42. 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

43. 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

44. 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

45. 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

46. 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

47. 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

48. 2.6 Addition and Subtraction of Cartesian Vectors
Example 3
Express the force F as Cartesian vector

49. 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

50. 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:

51. 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.

52. 2.7 Position Vectors x,y,z Coordinates Example 4:
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)

53. 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

54. 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)

55. 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

56. 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

57. 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

58. 2.7 Position Vectors
Angles, α, β and γ represent the direction of the cable
Unit vector, u = r/r

59. 2.7 Position Vectors Example 5 An elastic rubber band is
attached to points A and B.
Determine its length and its
direction measured from A
towards B.

60. 2.7 Position Vectors 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

61. 2.7 Position Vectors Solution α = cos-1(-3/7) = 115°

62. 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)

63. 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

64. 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

65. 2.8 Force Vector Directed along a Line
Example 6
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.

66. 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

67. 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°

68. 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

69. 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)

70. 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

71. 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

72. 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

73. 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

74. 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

75. 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)

76. 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

77. 2.9 Dot Product Example 7 The frame is subjected to a horizontal force
F = {300j} N. Determine the components of
this force parallel and perpendicular to the
member AB.

78. 2.9 Dot Product
Solution
Since
Then

79. 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

80. 2.9 Dot Product Solution Magnitude can be determined
From F┴ or from Pythagorean
Theorem