1. 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

2. 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

3. 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

4. 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

5. 2.6 Addition and Subtraction of Cartesian Vectors
Example 2.8
Express the force F as Cartesian vector

6. 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

7. 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:

8. 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

9. 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

10. 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

11. 2.6 Addition and Subtraction of Cartesian Vectors
Example 2.10
Express the force F1 as a Cartesian vector.

12. 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,

13. 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

14. 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

15. 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

16. 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.

17. 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

18. 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

19. 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°

20. 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.

21. 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)

22. 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

23. 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)

24. 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

25. 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

26. 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

27. 2.7 Position Vectors
Angles, α, β and γ represent the direction of the cable
Unit vector, u = r/r

28. 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.

29. 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

30. 2.7 Position Vectors Solution α = cos-1(-3/7) = 115°

31. 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)

32. 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

33. 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

34. 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.

35. 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

36. 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°

37. 2.8 Force Vector Directed along a Line
Example 2.14
The circular plate is
partially supported by
the cable AB. If the
force of the cable on the
hook at A is F = 500N,
express F as a
Cartesian vector.

38. 2.8 Force Vector Directed along a Line
Solution
End points of the cable are (0m, 0m, 2m) and B
(1.707m, 0.707m, 0m)
r = (1.707m – 0m)i + (0.707m – 0m)j
+ (0m – 2m)k
= {1.707i j - 2k}m
Magnitude = length of cable AB

39. 2.8 Force Vector Directed along a Line
Solution
Unit vector,
u = r /r
= (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k
= i j – k
For force F,
F = Fu
= 500N(0.6269i j – k)
= {313i - 130j - 367k} N

40. 2.8 Force Vector Directed along a Line
Solution
Checking
Show that γ = 137° and
indicate this angle on the
diagram

41. 2.8 Force Vector Directed along a Line
Example 2.15
The roof is supported by
cables. If the cables exert
FAB = 100N and FAC = 120N
on the wall hook at A,
determine the magnitude of
the resultant force acting at A.

42. 2.8 Force Vector Directed along a Line
View Free Body Diagram
Solution
rAB = (4m – 0m)i + (0m – 0m)j + (0m – 4m)k
= {4i – 4k}m
FAB = 100N (rAB/r AB)
= 100N {(4/5.66)i - (4/5.66)k}
= {70.7i k} N

43. 2.8 Force Vector Directed along a Line
Solution
rAC = (4m – 0m)i + (2m – 0m)j + (0m – 4m)k
= {4i + 2j – 4k}m
FAC = 120N (rAB/r AB)
= 120N {(4/6)i + (2/6)j - (4/6)k}
= {80i + 40j – 80k} N

44. 2.8 Force Vector Directed along a Line
Solution
FR = FAB + FAC
= {70.7i k} N + {80i + 40j – 80k} N
= {150.7i + 40j – 150.7k} N
Magnitude of FR

45. 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

46. 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)

47. 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

48. 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

49. 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

50. 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

51. 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

52. 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

53. 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)

54. 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

55. 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.

56. 2.9 Dot Product
Solution
Since
Then

57. 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

58. 2.9 Dot Product Solution Magnitude can be determined
From F┴ or from Pythagorean
Theorem

59. 2.9 Dot Product
Example 2.17
The pipe is subjected to F = 800N. Determine the
angle θ between F and pipe segment BA, and the
magnitudes of the components of F, which are
parallel and perpendicular to BA.

60. 2.9 Dot Product Solution For angle θ rBA = {-2i - 2j + 1k}m
View Free Body Diagram
Solution
For angle θ
rBA = {-2i - 2j + 1k}m
rBC = {- 3j + 1k}m
Thus,

61. 2.9 Dot Product
Solution
Components of F

62. 2.9 Dot Product Solution Checking from trigonometry,
Magnitude can be determined
From F┴

63. 2.9 Dot Product Solution Magnitude can be determined from F┴ or from
Pythagorean Theorem

64. 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

65. 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

66. 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

67. 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

68. 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

69. Chapter Review

70. Chapter Review

71. Chapter Review

72. Chapter Review

73. Chapter Review

74. Chapter Review