1. FORCE VECTORS, VECTOR OPERATIONS & ADDITION COPLANAR FORCES
Objective:
a) Resolve a 2-D vector into components.
b) Add 2-D vectors using Cartesian vector notations.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

2. 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
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

3. 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”
In these PowerPoint presentations, a vector quantity is represented like this (in bold, italics, and yellow).
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

4. VECTOR OPERATIONS (Section 2.2)
Scalar Multiplication
and Division
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

5. VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE
Triangle method (always ‘tip to tail’):
How do you subtract a vector?
How can you add more than two concurrent vectors graphically ?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

6. RESOLUTION OF A VECTOR
“Resolution” of a vector is breaking up a vector into components It is kind of like using the parallelogram law in reverse.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

7. ADDITION OF A SYSTEM OF COPLANAR FORCES (Section 2.4)
We ‘resolve’ vectors into components using the x and y axis system.
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.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

8. F = Fx i + Fy j or F' = F'x i + ( F'y ) j
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.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

9. ADDITION OF SEVERAL VECTORS
Step 1 is to resolve each force into its components.
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 3 is to find the magnitude and angle of the resultant vector.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

10. An example of the process:
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
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

11. You can also represent a 2-D vector with a magnitude and angle.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

12. Given: Three concurrent forces acting on a tent post.
EXAMPLE
Given: Three concurrent forces acting on a tent post.
Find: The magnitude and angle of the resultant force.
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.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

13. F2 = {– 450 cos (45°) i + 450 sin (45°) j } N
EXAMPLE (continued)
F1 = {0 i j } N
F2 = {– 450 cos (45°) i sin (45°) j } N
= {– i j } N
F3 = { (3/5) 600 i + (4/5) 600 j } N
= { 360 i j } N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

14. Summing up all the i and j components respectively, we get,
EXAMPLE (continued)
Summing up all the i and j components respectively, we get,
FR = { (0 – ) i + ( ) j } N
= { i j } N
Using magnitude and direction:
FR = ((41.80)2 + (1098)2)1/2 = 1099 N
 = tan-1(1098/41.80) = 87.8° x y  FR
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

15. Given: Three concurrent forces acting on a bracket
GROUP PROBLEM SOLVING
Given: Three concurrent forces acting on a bracket
Find: The magnitude and angle of the resultant force.
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.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

16. GROUP PROBLEM SOLVING (continued)
F1 = { (5/13) 300 i + (12/13) 300 j } N
= { i j } N
F2 = {500 cos (30°) i sin (30°) j } N
= { i j } N
F3 = { 600 cos (45°) i  600 sin (45°) j } N
{ i  j } N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

17. GROUP PROBLEM SOLVING (continued)
Summing up all the i and j components respectively, we get,
FR = { ( ) i + ( – 424.3) j }N
= { i j } N x y  FR
Now find the magnitude and angle,
FR = ((972.7)2 + (102.7)2) ½ = N
= tan–1( / ) = 6.03°
From Positive x axis,  = 6.03°
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

18. CARTESIAN VECTORS AND THEIR ADDITION & SUBTRACTION
Objectives:
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
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

19. Many structures and machines involve 3-Dimensional Space.
APPLICATIONS
Many structures and machines involve 3-Dimensional Space.
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?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

20. 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?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

21. CARTESIAN UNIT VECTORS
For a vector A, with a magnitude of A, an unit vector is defined as uA = A / A .
Characteristics of a unit vector :
a) Its magnitude is 1.
b) It is dimensionless (has no units).
c) It points in the same direction as the original vector (A).
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.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

22. CARTESIAN VECTOR REPRESENTATION
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
The projection of vector A in the x-y plane is A´. The magnitude of A´ is found by using the same approach as a 2-D vector: A´ = (AX2 + AY2)1/2 .
The magnitude of the position vector A can now be obtained as
A = ((A´)2 + AZ2) ½ = (AX2 + AY2 + AZ2) ½
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

23. DIRECTION OF A CARTESIAN VECTOR
These angles are measured between the vector and the positive X, Y and Z axes, respectively. Their range of values are from 0° to 180°
The direction or orientation of vector A is defined by the angles ά, β, and γ.
Using trigonometry, “direction cosines” are found using
These angles are not independent. They must satisfy the following equation.
cos ²  + cos ²  + cos ²  = 1
This result can be derived from the definition of a coordinate direction angles and the unit vector. Recall, the formula for finding the unit vector of any position vector:
or written another way, u A = cos  i + cos  j + cos  k .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

24. ADDITION OF CARTESIAN VECTORS (Section 2.6)
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
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 .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

25. 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 .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

26. Given: Two forces F1 and F2 are applied to a hook.
EXAMPLE
Given: Two forces F1 and F2 are applied to a hook.
Find: The resultant force in Cartesian vector form.
Plan: G
Using geometry and trigonometry, write F1 and F2 in Cartesian vector form.
2) Then add the two forces (by adding x and y components).
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

27. Solution : First, resolve force F1.
Fx = 0 = 0 lb
Fy = 500 (4/5) = 400 lb
Fz = 500 (3/5) = 300 lb
Now, write F1 in Cartesian vector form (don’t forget the units!).
F1 = {0 i j k} lb
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

28. We are given only two direction angles,  and .
Now resolve force F2.
We are given only two direction angles,  and .
So we need to find the value of .
Recall that cos ² () + cos ² () + cos ² () = 1.
Now substitute what we know:
cos ² (30°) + cos ² () + cos ² (45) = 1.
Solving,  = 75.5° or 104.5°.
Since the vector is pointing in the
positive direction,  = 75.5°
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

29. So, using u A = cos  i + cos  j + cos  k .
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  i + cos  j + cos  k .
F2 = {800 cos (30°) i cos (75.5°) j  800 cos (45°) k )} lb
F2 = {712.8 i j  k } lb
Now, R = F1 + F2 or
R = {713 i j  308 k} lb
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

30. Given: The screw eye is subjected to two forces, F1 and F2 .
GROUP PROBLEM SOLVING
Given: The screw eye is subjected to two forces, F1 and F2 .
Find: The magnitude and the coordinate direction angles of the resultant force.
Plan:
1) Using the geometry and trigonometry, resolve and write F1 and F2 in the Cartesian vector form.
2) Add F1 and F2 to get FR .
3) Determine the magnitude and angles , ,  .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

31. GROUP PROBLEM SOLVING (continued)
First resolve the force F1 .
F1z = 450 sin 45° = N
F´ = 450 cos 45° = N
F1z F´
F’ can be further resolved as,
F1x =  sin 30° =  N
F1y = cos 30° = N
Now we can write:
F1 = { 159 i j k } N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

32. GROUP PROBLEM SOLVING (continued)
Now, resolve force F2.
First, we need to find the value of .
cos ² (45°) + cos ² (30) + cos ² () = 1
Solving, we get  = 120°
The force F2 can be represented in the Cartesian vector form as:
F2 = 600{ cos 45° i + cos 60° j + cos 120° k } N
= { i j – 300 k } N
F2 = { 424 i j  300 k } N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

33. GROUP PROBLEM SOLVING (continued)
So FR = F1 + F2 and
F1 = {– i j k } N
F2 = { i j – 300 k } N
FR = { i j k } N
Now find the magnitude and direction angles for the vector.
FR = ( ) ½ = = 634 N
 = cos-1 (FRx / FR) = cos-1 (265.2 / 634.0) = 65.3°
 = cos-1 (FRy / FR) = cos-1 (575.6 / 634.0) = 24.8°
 = cos-1 (FRz / FR) = cos-1 (18.20 / 634.0) = 88.4°
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

34. POSITION VECTORS & FORCE VECTORS
Objectives:
a) Represent a position vector in Cartesian coordinate form, from given geometry.
b) Represent a force vector directed along a line.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

35. APPLICATIONS
This awning is held up by three chains. What are the forces in the chains and how do we find their directions? Why would we want to know these things?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

36. 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 and B, in 3-D space. Let their coordinates be (XA, YA, ZA) and (XB, YB, ZB ), respectively.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

37. 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. ALWAYS subtract the “tail” coordinates from the “tip” coordinates!
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

38. FORCE VECTOR DIRECTED ALONG A LINE (Section 2.8)
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’s magnitude. So we need to:
a) Find the position vector, rAB , 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 .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

39. Given: The 420 N force along the cable AC.
EXAMPLE
Given: The 420 N force along the cable AC.
Find: The force FAC in the Cartesian vector form.
Plan:
1. Find the position vector rAC and the unit vector uAC.
2. Obtain the force vector as FAC = 420 N uAC .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

40. Now uAC = rAC/rAC and FAC = 420 uAC N = 420 (rAC/rAC )
EXAMPLE (continued)
As per the figure, when relating A to C, we will have to go 2 m in the x-direction, 3 m in the y-direction, and -6 m in the z-direction. Hence,
rAC = {2 i j  6 k} m.
(We can also find rAC by subtracting the coordinates of A from the coordinates of C.)
rAC = ( )1/2 = 7 m
Now uAC = rAC/rAC and FAC = 420 uAC N = 420 (rAC/rAC )
So FAC = 420{ (2 i + 3 j  6 k) / 7 } N
= {120 i j k } N
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

41. Given: Two forces are acting on a pipe as shown in the figure.
GROUP PROBLEM SOLVING
Given: Two forces are acting on a pipe as shown in the figure.
Find: The magnitude and the coordinate direction angles of the resultant force.
Plan:
1) Find the forces along CA and CB in the Cartesian vector form.
2) Add the two forces to get the resultant force, FR.
3) Determine the magnitude and the coordinate angles of FR.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

42. GROUP PROBLEM SOLVING (continued)
FCA = 100 lb (rCA/rCA)
FCA = 100 lb (–3 sin 40° i + 3 cos 40° j – 4 k)/5
FCA = (– i j – 80 k) lb
FCB = 81 lb (rCB/rCB)
FCB = 81 lb (4 i – 7 j – 4 k)/9
FCB = {36 i – 63 j – 36 k} lb
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

43. GROUP PROBLEM SOLVING (continued)
FR = FCA + FCB
= {– 2.57 i – j – 116 k} lb
FR = ( )
= lb = 117 lb
 = cos-1(–2.57/117.3) = 91.3°
 = cos-1(–17.04/117.3) = 98.4°
 = cos-1(–116/117.3) = 172°
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

44. DOT PRODUCT Objective: a) determine an angle between two vectors, and,
b) determine the projection of a vector along a specified line.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 2.9

45. APPLICATIONS
If the design for the cable placements required specific angles between the cables, how would you check this installation to make sure the angles were correct?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 2.9

46. APPLICATIONS
For the force F being applied to the wrench at Point A, what component of it actually helps turn the bolt (i.e., the force component acting perpendicular to the pipe)?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 2.9

47. The dot product of vectors A and B is defined as A•B = A B cos .
DEFINITION
The dot product of vectors A and B is defined as A•B = A B cos .
The angle  is the smallest angle between the two vectors and is always in a range of 0º to 180º.
Dot Product Characteristics:
1. The result of the dot product is a scalar (a positive or negative number).
2. The units of the dot product will be the product of the units of the A and B vectors.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 2.9

48. DOT PRODUCT DEFINITON (continued)
Examples: By definition, i • j = 0
i • i = 1
A • B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k)
= Ax Bx + AyBy + AzBz
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 2.9

49. USING THE DOT PRODUCT TO DETERMINE THE ANGLE BETWEEN TWO VECTORS
For the given two vectors in the Cartesian form, one can find the angle by
a) Finding the dot product, A • B = (AxBx + AyBy + AzBz ),
b) Finding the magnitudes (A & B) of the vectors A & B, and
c) Using the definition of dot product and solving for , i.e.,
 = cos-1 [(A • B)/(A B)], where 0º    180º .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 2.9

50. DETERMINING THE PROJECTION OF A VECTOR
You can determine the components of a vector parallel and perpendicular to a line using the dot product.
Steps:
1. Find the unit vector, uaa´ along line aa´
2. Find the scalar projection of A along line aa´ by
A|| = A • uaa = AxUx + AyUy + Az Uz
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 2.9

51. DETERMINING THE PROJECTION OF A VECTOR (continued)
3. If needed, the projection can be written as a vector, A|| , by using the unit vector uaa´ and the magnitude found in step 2.
A|| = A|| uaa´
4. The scalar and vector forms of the perpendicular component can easily be obtained by
A  = (A 2 - A|| 2) ½ and
A  = A – A||
(rearranging the vector sum of A = A + A|| )
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 2.9

52. EXAMPLE
Given: The force acting on the hook at point A.
Find: The angle between the force vector and the line AO, and the magnitude of the projection of the force along the line AO.
Plan:
1. Find rAO
2. Find the angle  = cos-1{(F • rAO)/(F rAO)}
3. Find the projection via FAO = F • uAO (or F cos  )

53. EXAMPLE (continued)
rAO = {1 i j  2 k} m
rAO = ( )1/2 = 3 m
F = { 6 i j k} kN
F = ( )1/2 = kN
F • rAO = ( 6)(1) + (9)(2) + (3)(2) = 18 kN m
 = cos-1{(F • rAO)/(F rAO)}
 = cos-1 {18 / (11.22 * 3)} = 57.67°

54. EXAMPLE (continued)
uAO = rAO/rAO = {( 1/3) i + (2/3) j + ( 2/3) k}
FAO = F • uAO = ( 6)( 1/3) + (9)(2/3) + (3)( 2/3) = 6.00 kN
Or: FAO = F cos  = cos (57.67°) = kN

55. GROUP PROBLEM SOLVING
Given: The force acting on the pole.
Find: The angle between the force vector and the pole, and the magnitude of the projection of the force along the pole AO.
Plan:
1. Find rAO
2. Find the angle  = cos-1{(F • rAO)/(F rAO)}
3. The find the projection via FAO = F • uAO or F cos 

56. GROUP PROBLEM SOLVING (continued)
rAO = {4 i  4 j  2 k} ft.
rAO = ( )1/2 = 6 ft.
Fz = 600 sin 60° = lb
F´ = 600 cos 60° = 300 lb
F = { 300 sin 30°i cos 30° j k} lb
F = { 150 i j k}lb
F = ( )1/2 = 600 lb
F • rAO = (150)(4)+(259.8)(4)+(519.6)( 2) =  2678 lb·ft

57. GROUP PROBLEM SOLVING (continued)
 = cos-1{(F • rAO)/(F rAO)}
 = cos-1{2678 /(600 × 6)}=138.1° 
uAO = rAO/rAO = {(4/6) i  (4/6) j (2/6) k}
FAO = F • uAO = ( 150)(4/6) + (259.8) (4/6) + (519.6) (2/6) =  446 lb
Or: FAO = F cos  = 600 cos (138.1°) =  446 lb