1. Engineering Mechanics : STATICS
BDA10203
Lecture #06
Group of Lecturers
Faculty of Mechanical and Manufacturing Engineering
Universiti Tun Hussein Onn Malaysia

3. POSITION & FORCE VECTORS (Sections 2.7 - 2.8)
Today’s Objectives:
Students will be able to :
a) Represent a position vector in Cartesian coordinate form, from given geometry.
b) Represent a force vector directed along a line.
Learning Topics:
Applications / Relevance
Write position vectors
Write a force vector

4. POSITION VECTOR
A position vector is defined as a fixed vector that locates a point in space relative to another point.
Consider two points, A & B, in 3-D space. Let their coordinates be (XA, YA, ZA) and ( XB, YB, ZB ), respectively.
The position vector directed from A to B, r AB , is defined as
r AB = {( XB – XA ) i + ( YB – YA ) j ( ZB – ZA ) k }m
Please note that B is the ending point and A is the starting point. So ALWAYS subtract the “tail” coordinates from the “tip” coordinates!

5. NOTE that r has unit of length
POSITION VECTOR
The position vector rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m Magnitude
Unit vector, u = r /r
= rAB /
NOTE that r has unit of length

6. 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)

7. EXAMPLE
Given: A position vector r acting from point A(3m,5m,6m) to point B(5m,-2m,1m)
Find: Represent the position vector in Cartesian vector form and determine its direction angles and find the distance between point A and B.
Plan:
1) Represent the position vector in Cartesian vector form
2) Find the magnitude of the position vector
3) Use trigonometry to find the direction angle

8. EXAMPLE Position vector,
rAB = {( XB – XA ) i + ( YB – YA ) j ( ZB – ZA ) k }m
= {( 5 – 3 ) i + ( -2 – 5) j ( 1 – 6 ) k }m
= {2i - 7j - 5k }m y x z
A(3,5,6)
B(5,-2,1) 5 3 -2 1 6
Note that
Start at point A (3,5,6), one travels
2m in the +i direction,
7m in the -j direction and
5m in the -k direction, arriving at point B (5,-2,1)
Distance between point A and B,
Total 8.83m from A to B

9. EXAMPLE Unit vector, u = r /r
Solving the equations and we will get the answers as below
α = 76.9°, β = 142.4°, γ = 124.5°

10. IN CLASS TUTORIAL (GROUP PROBLEM SOLVING)
Given: Position vector as shown in the figure
Find: Express the position vector r in the Cartesian vector form, its magnitude and coordinate direction angle.
Plan:
1) Represent the position vector in Cartesian vector form
2) Find the magnitude of the position vector
3) Use trigonometry to find the direction angle

11. GROUP PROBLEM SOLVING (continued)A
rAB = {( XB – XA ) i + ( YB – YA ) j ( ZB – ZA ) k }m
= {( 0 – 3 ) i + ( -8 – 4) j ( 4 – 0 ) k }m
= {-3i j + 4k }m
Distance between point A and B,

12. GROUP PROBLEM SOLVING (continued)
Solving the equations and we will the answers as below
α = 103.3°, β = 157.4°, γ = 72.1°

13. 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.

14. r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m
Solution
Position vector
r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k
= {-3i + 2j + 6k}m
Note that it took 3m in –x direction, 2m in +y direction and 6m in +j dirention to arrive at B from A point
Magnitude = length of the rubber band
Unit vector in the direction of r
u = r /r
= -3/7i + 2/7j + 6/7k

15. Solution
α = cos-1(-3/7) = 115°
β = cos-1(2/7) = 73.4°
γ = cos-1(6/7) = 31.0°
Note: These angles are measured from the positive axes of a localized coordinate system placed at the tail of r

16. HOMEWORK TUTORIAL Q1 (2-84):
Determine the length of the connecting rod AB by first formulating a Cartesian position vector from A to B and then determining its magnitude.
Given:
b = 400mm
a = 125mm
α = 25°

17. HOMEWORK TUTORIAL (continued)
Q2 (2-85):
Determine the length of member AB of the truss by first establishing a Cartesian position vector from A to B and then determining its magnitude.
Given:
a = 1.2m
b = 0.8m
c = 0.3m
d = 1.5m, Length CB
θ = 40deg

18. Problem 2-83
Express the position vector r in Cartesian vector form;
then determine its magnitude and coordinate direction angles.

19. Force Vector Directed along a Line
If a force is directed along a line, then we can represent the force vector in Cartesian Coordinates by using a unit vector and the force magnitude. So we need to:
a) Find the position vector, r AB , along two points on that line.
b) Find the unit vector describing the line’s direction, uAB = (rAB/rAB).
c) Multiply the unit vector by the magnitude of the force, F = F uAB .

20. APPLICATIONS
How can we represent the force along the wing strut in a 3-D Cartesian vector form?
Wing strut

21. 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
=F (r /r )
Note that F has unit of force, unlike r which has units of length

22. 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.

23. 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

24. 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°

25. 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.

26. Solution
End points of the cable are A (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

27. 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

28. Solution
Checking
Show that γ = 137° and
indicate this angle on the
diagram

29. 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.

30. 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

31. 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

32. Solution
FR = FAB + FAC
= {70.7i k} N + {80i + 40j – 80k} N
= {150.7i + 40j – 150.7k} N
Magnitude of FR

33. EXAMPLE
Given: The window is held open by cable AB as shown in the figure
Find: Length of the cable and express the F force acting at A along the cable as a Cartesian vector
Plan:
1) Represent the position vector in Cartesian vector form
2) Find the magnitude of the position vector
3) Using a unit vector, express the F force as a Cartesian vector

34. EXAMPLE rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }mm
= {( 0 – 300cos30°) i + ( 150 – 500) j + ( sin30°) k }mm
= {-259.8i j k }mm
Distance between point A and B,

35. EXAMPLE Use a unit vector to express F in the form of Cartesian vector
F = FuAB = F(rAB/rAB).
F = 30 {( /591.6)i + (-350/591.6)j + (400/591.6)k}
= 30{-0.44i – 0.59j k}
= {-13.2i j k} N

36. IN CLASS TUTORIAL (GROUP PROBLEM SOLVING)
Given: The cable attached to the tractor at B exerts a force of 3500 N on the framework.
Find: Express this force as a Cartesian vector
Plan:
1) Represent the position vector in Cartesian vector form
2) Find the magnitude of the position vector
3) Using a unit vector, express the F force as a Cartesian vector

37. GROUP PROBLEM SOLVING (continued)
rAB = {( XB – XA ) i + ( YB – YA ) j ( ZB – ZA ) k }m
= {( 15sin20° – 0 ) i + ( 15cos20° – 0) j ( 0 – ) k }m
= {5.13i j k }m
Distance between point A and B,

38. GROUP PROBLEM SOLVING (continued)
Use a unit vector to express F in the form of Cartesian vector
F = FuAB = F(rAB/rAB).
F = 3500 {(5.13 /18.31)i + (14.1/18.31)j + (-10.5/18.31)k}
= 3500{0.28i j – 0.57k}
= {980i j k} N

39. CONCEPT QUIZ
1. P and Q are two points in a 3-D space. How are the position vectors rPQ and rQP related?
A) rPQ = rQP B) rPQ = - rQP
C) rPQ = 1/rQP D) rPQ = 2 rQP
2. If F and r are force vector and position vectors, respectively, in SI units, what are the units of the expression (r * (F / F)) ?
A) Newton B) Dimensionless
C) Meter D) Newton - Meter
E) The expression is algebraically illegal.
Answers:
1. B
2. C

40. ATTENTION QUIZ
1. Two points in 3 – D space have coordinates of P (1, 2, 3) and Q (4, 5, 6) meters. The position vector rQP is given by
A) {3 i j k} m
B) {- 3 i – 3 j – 3 k} m
C) {5 i j k} m
D) {- 3 i j k} m
E) {4 i j k} m
Answers
1. B
2. C
2. Force vector, F, directed along a line PQ is given by
A) (F/ F) rPQ B) rPQ/rPQ
C) F(rPQ/rPQ) D) F(rPQ/rPQ)

41. READING QUIZ 1. A position vector, rPQ, is obtained by
A) Coordinates of Q minus coordinates of P
B) Coordinates of P minus coordinates of Q
C) Coordinates of Q minus coordinates of the origin
D) Coordinates of the origin minus coordinates of P
2. A force of magnitude F, directed along a unit vector U, is given by F = ______ .
A) F (U)
B) U / F
C) F / U
D) F + U
E) F – U
Answers:
1. A
2. A

42. HOMEWORK TUTORIAL Q1 (2-93):
The plate is suspended using the three cables which exert the forces shown. Express each force as a Cartesian vector.
Given:
FBA = 3.5kN
FCA = 5kN
FDA = 4kN
a = 0.3m
b = 0.3m
c = 0.6m
d = 1.4m
e = 0.3m
f = 0.3m
g = 0.2m

43. HOMEWORK TUTORIAL (continued)
Q2 (2-94):
The engine of the lightweight plane is supported by struts that are connected to the space truss that makes up the structure of the plane. The anticipated loading in two of the struts is shown. Express each of these forces as a Cartesian vector.
Given:
F1 = 4kN
F2 = 6kN
a = 0.15m
b = 0.15m
c = 0.9m
d = 0.75m
e = 0.15m
f = 0.9m

44. HOMEWORK TUTORIAL (continued)
Q3 (2-104) :
The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles α,β,γ of the resultant force.
Units Used:
kN := 1000N
Given:
x = 20m, a = 16m
y = 15m, b = 18m
F1 = 600N, c = 6m
F2 = 400N, d = 4m
F3 = 800N, e = 24m