vector operation & Force

1. Lecture #2 (ref Ch 2)
Vector Operation and Force Analysis
1
R. Michael PE 8/14/2012

2. Ch. 2 Key Concepts
 A little on vectors (and scalars)
 Finding Resultant Force (vector addition)
 Graphical Approach (sec 2.3)
 Triangle method (or successive triangle method)
 Parallelogram method
 Polygon method (good if more than one force)
 Finding resultants by resolving forces into components!!
(sec 2.4-Addition of a System of Coplanar Forces)
2

3. Scalar vs. Vector
 Scalar Quantity
 A mathematical expression possessing only
magnitude characterized by a positive or negative
number
 The following are classified as Scalar Quantities
 Mass
 Volume
 Length
3

4. Scalar vs. Vector
 Vector
 Physical quantity that requires both a magnitude
and a direction for its complete description.
 possessing magnitude and direction and must be
added using Vector Operations
 The following are classified as Vectors
 Displacements
 Velocities
 Accelerations
 Moments 4

5. SCALARS AND VECTORS
(Section 2.1)
Scalars Vectors
Examples: Mass, Volume Force, Velocity
Characteristics: It has a magnitude It has a magnitude
(positive or negative) and direction
Addition rule: Simple arithmetic Parallelogram law
Special Notation: None Bold font, a line, an
arrow or a “carrot”
5

6. Vector Notation
• In Slides and handouts Vectors will be
denoted as a BOLD letter.
• Example;
• (a+b) will denote a scalar addition
• (A+B) will denote a vector addition
• When hand writing a Vector use an arrow
over the letter to denote it is a Vector. [ ]
A
6

7. Vector Notation
 Vector Notation for Rectangular Components of a
Vector (Force) - Because the directional sense of the
axes of the rectangular coordinate system are known,
Rectangular Vector Components can be written in a
couple of different ways.
 Cartesian Vector Notation – Cartesian unit vectors (i and
j) are used to designate the x-axis and y-axis respectively
where F=Fxi+Fyj.
 Magnitude and Direction – Define the Vector by
magnitude, units, and angle it makes with respect to the x-
axis - F= 45N 38°
7
135 vs 45

8. Components of a Vector
A Head
Tail
Vector Designation
500 N
Vector Magnitude
8

9. Vector Operations
 Multiplication and Division of a Vector by a
Scalar
 Product of Vector (A) and Scalar (b) = bA = a
vector with the same direction as A but with the
magnitude multiplied by the scalar (b).
 Example – If a 500 lb force acting along the x-
axis is doubled, it becomes a 1000 lb force acting
along the x-axis.
9

10. Vector Addition
 Vectors can be compared to giving directions.
Go north 4 steps, Go east 3 steps.
 The vector would be defined as the arrow
pointing from where you started to where you
are now.
 The magnitude would be defined by how far
are you from where you started (not how far
you traveled to get there)
10

11. VECTOR OPERATIONS
(Section 2.2)
Scalar Multiplication
and Division
11

12. 2.3 APPLICATION OF VECTOR ADDITION
There are three concurrent forces
acting on the hook due to the chains.
We need to decide if the hook will fail
(bend or break)?
To do this, we need to know the
resultant force acting on the hook.
FR
12

13. Vector Addition
 The addition of two
vectors results in a
resultant vector
(P+Q=R) where R is a
vector pointing from
the starting point of P
to the ending point of
Q.
Resultant Vector
13

14. Parallelogram Law
 By drawing
construction lines
parallel to the vectors,
the resultant vector
goes from the point of
origin to the
intersection of the
construction lines
14

15. Triangle Method
 Place the tail of B to
the head of A. The
Resultant (R) can be
found by connecting
the Tail of A to the
Head of B. This forms
the third leg of the
triangle and the
resultant vector.
A
B
A + B
15
Resultant

16. Summary: VECTOR ADDITION USING EITHER THE
PARALLELOGRAM LAW OR TRIANGLE
Parallelogram Law:
Triangle method (always
‘tip to tail’):

17. Example #1
 Triangle Method – Find Resultant (mag and direction)
x
y
30°
45°
50 N
30 N
See solution in notes
17

18. 18

19. 19

24. Successive Triangle Method
 If there are multiple vectors to be added together,
add the first two vectors to find the first resultant.
Once the first Resultant (R1) is found, add the
next vector to the resultant to find (R2).
 Can be repeated as many times as necessary to
add all the vectors (it also does not matter what
order they are added in, the end resultant will be
the same).
24
See HO, normally just
resolve into
components

25. Polygon Method
 Polygon method is
similar to the
Successive Triangle
Method but no
intermediate resultants
are calculated
25
See HO, do example on
board
Graphically measure length and direction
of R!!

26. ADDITION OF A SYSTEM OF COPLANAR FORCES (Section 2.4) – Basically
finding resultant vectors by breaking forces up into components and adding!
• Each component of the vector is
shown as a magnitude and a
direction.
• The directions are based on the x and y axes. We use the
“unit vectors” i and j to designate the x and y axes.
• We ‘resolve’ vectors into
components using the x and y
axis system.

27. For example,
F = Fx i + Fy j or F' = F'x i + ( F'y ) j
The x and y axis are always perpendicular to each other.
Together, they can be directed at any inclination.

28. ADDITION OF SEVERAL VECTORS
Step 3 is to find the magnitude
and angle of the resultant vector.
• Step 2 is to add all the x-
components together, followed by
adding all the y components
together. These two totals are the
x and y components of the
resultant vector.
• Step 1 is to resolve each force
into its components.

29. Break the three vectors into components, then add them.
FR = F1 + F2 + F3
= F1x i + F1y j  F2x i + F2y j + F3x i  F3y j
= (F1x  F2x + F3x) i + (F1y + F2y  F3y) j
= (FRx) i + (FRy) j
An example of the process:

30. Remember! You can also represent a 2-D vector with a magnitude
and angle:
FR = (FRx) i + (FRy) j
Or, Cartesian Vector Notation:

31. 31
F2-8: Determine the magnitude and direction of the resultant force.
Side question: What additional force would you have to apply so the net force
acting on the hook was zero (think, same magnitude but opposite direction for
the resultant force found in step 1).

32. EXAMPLE
Plan:
a) Resolve the forces into their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Given: Three concurrent forces
acting on a tent post.
Find: The magnitude and
angle of the resultant
force.

33. EXAMPLE (continued)
F1 = {0 i + 300 j } N
F2 = {– 450 cos (45°) i + 450 sin (45°) j } N
= {– 318.2 i + 318.2 j } N
F3 = { (3/5) 600 i + (4/5) 600 j } N
= { 360 i + 480 j } N

34. EXAMPLE
(continued)
Summing up all the i and j components respectively, we get,
FR = { (0 – 318.2 + 360) i + (300 + 318.2 + 480) j } N
= { 41.80 i + 1098 j } N
x
y

FR
Using magnitude and direction:
FR = ((41.80)2 + (1098)2)1/2 = 1099 N
 = tan-1(1098/41.80) = 87.8°

35. GROUP PROBLEM SOLVING
Plan:
a) Resolve the forces into their x and y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Given: Three concurrent
forces acting on a
bracket
Find: The magnitude and
angle of the
resultant force.

36. F1 = { (5/13) 300 i + (12/13) 300 j } N
= { 115.4 i + 276.9 j } N
F2 = {500 cos (30°) i + 500 sin (30°) j } N
= { 433.0 i + 250 j } N
F3 = { 600 cos (45°) i  600 sin (45°) j } N
{ 424.3 i  424.3 j } N
GROUP PROBLEM SOLVING (continued)

37. GROUP PROBLEM SOLVING (continued)
Summing up all the i and j components respectively, we get,
FR = { (115.4 + 433.0 + 424.3) i + (276.9 + 250 – 424.3) j }N
= { 972.7 i + 102.7 j } N
Now find the magnitude and angle,
FR = ((972.7)2 + (102.7)2) ½ = 978.1 N
 = tan–1( 102.7 / 972.7 ) = 6.03°
From Positive x axis,  = 6.03°
x
y

FR
Do example on board

38. ATTENTION QUIZ
1. Resolve F along x and y axes and write it in
vector form. F = { ___________ } N
A) 80 cos (30°) i – 80 sin (30°) j
B) 80 sin (30°) i + 80 cos (30°) j
C) 80 sin (30°) i – 80 cos (30°) j
D) 80 cos (30°) i + 80 sin (30°) j
2. Determine the magnitude of the resultant (F1 + F2) force in N
when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N .
A) 30 N B) 40 N C) 50 N
D) 60 N E) 70 N
30°
x
y
F = 80 N

41. Rectangular Components of 3D Forces
 3D Force Vector – Vector defining a Force in
more than one Cartesian Plane defined by its
location and rectangular components
 Rectangular Components - Components that
fall along the Cartesian coordinate system
axes
 Coordinate Angles (θx, θy, θz)– The angle a
vector makes with the individual axes of the
Cartesian Coordinate System
41

42. APPLICATIONS
In this case, the power pole has guy
wires helping to keep it upright in
high winds. How would you
represent the forces in the cables
using Cartesian vector form?
Many structures and machines involve
3-Dimensional Space.
We’ll solve this later

43. APPLICATIONS (continued)
In the case of this radio tower, if you know the forces in the three cables,
how would you determine the resultant force acting at D, the top of the
tower?
We’ll solve this later

44. Coordinate Angles
44
Note, book uses a, b, g:
a
b g

45. Coordinate Angles
 The values of the three angles are not
independent, they are related by the identity:
 cos2(θx) + cos2(θy) + cos2(θz) = 1
45

46. Resolving a 3D Force Vector into its
Rectangular Components
 Given the magnitude of a force vector (F) and
its Coordinate angles (θx, θy, θz):
 Fx = Fcos(θx)
 Fy = Fcos(θy)
 Fz = Fcos(θz)
46
Note, book uses a, b, g:

47. Resultant of a 3D Force Vector from
its Rectangular Components
 If given the components of a 3D force (Fx, Fy, Fz),
the force can be determined by:
 Magnitude (F) = √(Fx
2+Fy
2+Fz
2)
 The Coordinate Angles of the Force Vector can
be found by
 cos(θx) = Fx/F
 cos(θy) = Fy/F
 cos(θz) = Fz/F
47
Do Example on board, then HO examples

48. Addition of 3D Force Vectors
 Forces are easy to add once they are broken
down into their rectangular components. The
components of the resultant force can be
found as follows:
 Rx=ΣFx
 Ry=ΣFy
 Rz=ΣFz
48

49. 3D Resultant Force
 The magnitude of the resultant force is equal
to the square root of the addition of the scalar
quantity of each leg squared:
 R = √(Rx
2+Ry
2+Rz
2)
 The Coordinate Angles of the resultant can be
found by:
 cos(θx) = Rx/R
 cos(θy) = Ry/R
 cos(θz) = Rz/R
49
See HO’s

50. ADDITION OF CARTESIAN VECTORS
(Section 2.6)
For example, if
A = AX i + AY j + AZ k and
B = BX i + BY j + BZ k , then
A + B = (AX + BX) i + (AY + BY) j + (AZ + BZ) k
or
A – B = (AX - BX) i + (AY - BY) j + (AZ - BZ) k .
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.

51. 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 + 30 k} N .

52. EXAMPLE
1) Using geometry and trigonometry, write F1 and F2 in
Cartesian vector form.
2) Then add the two forces (by adding x and y components).
G
Given: Two forces F1 and F2 are
applied to a hook.
Find: The resultant force in
Cartesian vector form.
Plan:

53. Solution :
First, resolve force F1.
Fz = 500 (3/5) = 300 lb
Fx = 0 = 0 lb
Fy = 500 (4/5) = 400 lb
Now, write F1 in Cartesian vector form
(don’t forget the units!).
F1 = {0 i + 400 j + 300 k} lb

54. Now resolve force F2.
We are given only two direction angles, a and g.
So we need to find the value of b.
Recall that cos ² (a) + cos ² (b) + cos ² (g) = 1.
Now substitute what we know:
cos ² (30°) + cos ² (b) + cos ² (45) = 1.
Solving, b = 75.5° or 104.5°.
Since the vector is pointing in the
positive direction, b = 75.5°

55. F2 = {800 cos (30°) i + 800 cos (75.5°) j  800 cos (45°) k )} lb
F2 = {712.8 i + 200.3 j  608.3 k } lb
Now, R = F1 + F2 or
R = {713 i + 600 j  308 k} lb
Now that we have the coordinate direction
angles, we can find uG and use it to
determine F2 = 800 uG lb.
So, using u A = cos a i + cos b j + cos g k .

61. 61
Good 3D problem.
Know resultant mag
and direction, F1
mag and direction,
F2 mag. Find F2
direction.