1. Engineering Mechanics : STATICS
BDA10203
Lecture #05
Group of Lecturers
Faculty of Mechanical and Manufacturing Engineering
Universiti Tun Hussein Onn Malaysia

3. 3D VECTORS (Section 2.5) Today’s Objectives:
Students will be able to :
a) Represent a 3-D vector in a Cartesian coordinate system.
b) Find the magnitude and coordinate angles of a 3-D vector
c) Add vectors (forces) in 3-D space
Learning Topics:
Applications / Relevance
A unit vector
3-D vector terms
Adding vectors

4. APPLICATIONS
Many problems in real-life involve 3-Dimensional Space.
How will you represent each of the cable forces in Cartesian vector form?

5. 3D CARTESIAN VECTOR
The unit vectors in the Cartesian axis system are i, j, and k. They are unit vectors along the positive x, y, and z axes respectively.
Consider a box with sides AX, AY, and AZ meters long.
The vector A can be defined as
A = (AX i + AY j + AZ k) m
Note:Separating the magnitude and
direction of each components simplifies
the operation

6. 3D Cartesian Vectors
- By two successive application of the parallelogram law
A = A’ + Az
A’ = Ax + Ay
- Combining the equations, A can be expressed as
A = Ax + Ay + Az
The magnitude of this projection, A´, is found by using the same approach as a 2-D vector: |A´| = (AX2 + AY2)1/2 .
For 3D vector;

7. MAGNITUDE OF CARTESIAN VECTOR
Once individual vectors are written in Cartesian form, it is easy to add or subtract them. The process is essentially the same as when 2-D vectors are added.
For example, if
F1 = F1X i + F1Y j + F1Z k and
F2 = F2X i + F2Y j + F2Z k ,
then
F1 + F2 = (F1X + F2X) i + (F1Y + F2Y) j + (F1Z + F2Z) k
|A | = [ (F1X + F2X)2 i + (F1Y + F2Y)2 j + (F1Z + F2Z)2 k ]1/2 AX Ay Az

8. DIRECTION OF CARTESIAN VECTOR
The direction or orientation of vector A is defined by the angles , , and .
These angles are not independent. They must satisfy the
following equation.
cos ²  + cos ²  + cos ²  = 1

9. MAGNITUDE & DIRECTION OF CARTESIAN VECTOR
or written another way,
u A = cos  i + cos  j + cos  k .
A as expressed in Cartesian vector form
Vector, A = |A |uA
= |A | cosα i + |A| cosβ j + |A| cosγ k
Direction
magnitude
F = |F| cosα i + |F |cosβ j + | F| cosγ k

10. IMPORTANT NOTES Sometimes 3-D vector information is given as:
a) Magnitude and the coordinate direction angles, or
b) Magnitude and projection angles.
You should be able to use both these types of information to change the representation of the vector into the Cartesian form, i.e.,
F = {10 i – 20 j k} N .

11. Example 2.9
Determine the magnitude and coordinate
direction angles of resultant force acting on the ring

12. Solution
Resultant force
FR = ∑F
= F1 + F2
= {60j + 80k}kN
+ {50i - 100j + 100k}kN
= {50i -40j + 180k}kN
Magnitude of FR is found by

13. 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

14. Express the force F as Cartesian vector
How to change from Magnitude to Vector
Example 2.8
Express the force F as Cartesian vector
Hint:
The answer should be in this form
A = (AX i + AY j + AZ k)

15. Solution
Since two angles are specified, the third
angle is found by
Two possibilities exit, namely or

16. F = Fcosαi + Fcosβj + Fcosγk
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
Vector, F = {100.0i j k}N
Note: You can check the answer :

17. (a)Express the force F1 as a Cartesian vector.
WHAT IF COORDINATE DIRECTION ANGLES ARE NOT GIVEN?
Example 2.10
(a)Express the force F1 as a Cartesian vector.
(b) Prove that F2 = {106i + 184j - 212k} kN
( c) Direction of F1
(d) Find the resultant Force

18. The angles of 60° and 45° are not coordinate direction angles.
Solution (a)
The angles of 60° and 45° are not coordinate
direction angles.
Hint:Resolve the force into Fx, Fy & Fz
The answer should be in this form
A = (AX i + AY j + AZ k)
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

19. Unit vector acting in the direction of F1 u1 = F1 / |F1|
Solution (c )
F1 = {35.4i – 35.4j k}kN
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
α1 = cos-1(0.354) = 69.3°
β1 = cos-1(-0.354) = 111°
γ1 = cos-1(0.866) = 30.0°

20. Magnitude of FR is found by
Solution (d)
Resultant force
FR = ∑F
= F1 + F2
= {35.4i -35.4j k}kN
+ {106i +184j – 212 k}kN
= {141.4i j k}kN
Magnitude of FR is found by
F1 = {35.4i – 35.4j k}kN
F2 = {106i + 184j - 212k} kN

21. uFR = FR / |FR | |FR |= {(141.4)2 + (148.6)2 + (-123.4)2}1/2 =239 kN
= (141.4/239)i +(148.6/239)j -(123.4/239)k
= 0.59i j k
α1 = cos-1(0.59) = 53.84°
β1 = cos-1(0.62) = 51.68°
γ1 = cos-1(-0.52) = 148.6°

22. IN CLASS TUTORIAL (GROUP PROBLEM SOLVING)
Given: The screw eye is subjected to two forces.
Find: The magnitude and the coordinate direction angles of the resultant force.
Plan:
1) Using the geometry and trigonometry, write F1 and F2 in the Cartesian vector form.
2) Add F1 and F2 to get FR .
3) Determine the magnitude and , ,  .

23. GROUP PROBLEM SOLVING (continued)
First resolve the force F1 .
F1z = 300 sin 60° = N
F´ = 300 cos 60° = N
F1z F´
F’ can be further resolved as,
F1x = -150 sin 45° = N
F1y = 150 cos 45° = N
Now we can write :
F1 = { i j k } N

24. GROUP PROBLEM SOLVING (continued)
The force F2 can be represented in the Cartesian vector form as:
F2 = 500{ cos 60° i + cos 45° j cos 120° k } N
= { 250 i j – 250 k } N
FR = F1 + F2
= { i j k } N
FR = ( ) ½ = = 482 N
 = cos-1 (FRx / FR) = cos-1 (143.9/481.7) = 72.6°
 = cos-1 (FRy / FR) = cos-1 (459.6/481.7) = 17.4°
 = cos-1 (FRz / FR) = cos-1 (9.81/481.7) = 88.8°

25. 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.

26. Solution
Hint: Since FR has a magnitude of 800N and acts in the +j direction
FR = F1 + F2 = 800 j
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

27. Solution
FR = F1 + F2 = 800 j
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

28. 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°

29. HOMEWORK TUTORIAL Q1 (2.81):
Determine the magnitude and direction of the force
F=(3.6 kN) i+(1.2 kN )j −(1.5 kN )k.
Q2 (2.82):
Determine the magnitude and direction of the force
F =(400 N )i − (1200 N ) j +(300 N )k.

30. HOMEWORK TUTORIAL (continued)
Q3 (2.93):
Find the magnitude and direction of the resultant of the two forces shown knowing that P = 2.25 kN and Q = 2.7 kN.

31. HOMEWORK TUTORIAL (continued)
Q4 (2-61):
The stock S mounted on the lathe is subjected to a force F, which is caused by the die D. Determine the coordinate direction angle β and express the force as a Cartesian vector.
Given:
F = 60N
α = 60deg
γ = 30deg

32. HOMEWORK TUTORIAL (continued)
Q5 (2-66):
The shaft S exerts three force components on the die D. Find the magnitude and direction of the resultant force. Force F2 acts within the octant shown.
Given :
F1 = 400N
F2 = 300N
F3 = 200N
α2 = 60°
γ2 = 60°
c = 3
d = 4

33. HOMEWORK TUTORIAL (continued)
Q6 (2-74):
The eye bolt is subjected to the cable force F which has a component Fx along the x axis, a component Fz along the z axis, and a coordinate direction angle β. Determine the magnitude of F.
Given:
Fx = 60N
Fz = −80N
β = 80°

34. HOMEWORK TUTORIAL (continued)Q7

35. HOMEWORK TUTORIAL (continued)Q8