APPLICATION OF VECTOR ADDITION

There are four concurrent cable forces acting on the bracket. How do you determine the resultant force acting on the bracket ?

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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

Simple arithmetic

Parallelogram law

Addition rule: Special Notation:

None

Bold font, a line, an arrow or a “carrot”

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VECTOR OPERATIONS (Section 2.2)

Scalar Multiplication and Division

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VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE Parallelogram Law:

Triangle method (always ‘tip to tail’):

How do you subtract a vector? How can you add more than two concurrent vectors graphically ? Chungnam National University

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.

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CARTESIAN VECTOR NOTATION (Section 2.4) • We ‘ resolve’ vectors into components using the x and y axes 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.

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For example, F = Fx i + Fy j

or F' = F'x i + F'y j

The x and y axes are always perpendicular to each other. Together,they can be directed at any inclination. Chungnam National University

ADDITION OF SEVERAL VECTORS • Step 1 is to resolve each force into its components • Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector. § Step 3 is to find the magnitude and angle of the resultant vector.

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Example of this process,

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You can also represent a 2-D vector with a magnitude and angle.

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EXAMPLE Given: Three concurrent forces acting on a bracket. Find: The magnitude and angle of the resultant force. Plan: a) Resolve the forces in their x-y components. b) Add the respective components to get the resultant vector. c) Find magnitude and angle from the resultant components. Chungnam National University

EXAMPLE (continued) F1 = { 15 sin 40° i + 15 cos 40° j } kN = { 9.642 i + 11.49 j } kN F2 = { -(12/13)26 i + (5/13)26 j } kN = { -24 i + 10 j } kN F3 = { 36 cos 30° i – 36 sin 30° j } kN = { 31.18 i – 18 j } kN

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EXAMPLE (continued) Summing up all the i and j components respectively, we get, FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN = { 16.82 i + 3.49 j } kN y FR =

((16.82)2

+

(3.49)2)1/2

FR

= 17.2 kN

f = tan-1(3.49/16.82) = 11.7°

f x

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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 in their x-y components. b) Add the respective components to get the resultant vector. c) Find magnitude and angle from the resultant components. Chungnam National University

GROUP PROBLEM SOLVING (continued)

F1 = { (4/5) 850 i - (3/5) 850 j } N = { 680 i - 510 j } N F2 = { -625 sin(30°) i - 625 cos(30°) j } N = { -312.5 i - 541.3 j } N F3 = { -750 sin(45°) i + 750 cos(45°) j } N { -530.3 i + 530.3 j } N

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GROUP PROBLEM SOLVING (continued) Summing up all the i and j components respectively, we get, FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N = { - 162.8 i - 521 j } N y

FR = ((162.8)2 + (521)2) ½ = 546 N f=

tan–1(521/162.8)

= 72.64°

or

From Positive x axis q = 180 + 72.64 = 253 °

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f

FR

x

3 – D 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

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APPLICATIONS

Many problems in real-life involve 3-Dimensional Space.

How will you represent each of the cable forces in Cartesian vector form?

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APPLICATIONS (continued) Given the forces in the cables, how will you determine the resultant force acting at D, the top of the tower?

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A UNIT VECTOR 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. 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. Chungnam National University

3-D CARTESIAN VECTOR TERMINOLOGY

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 the vector A in the x-y plane is A´. The magnitude of this projection, 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 Chungnam National University

TERMS (continued) The direction or orientation of vector A is defined by the angles a, b, and g. 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° Using trigonometry, “direction cosines” are found using the formulas

These angles are not independent. They must satisfy the following equation. cos ² a + cos ² b + cos ² g = 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 a i + cos b j + cos g k . Chungnam National University

ADDITION/SUBTRACTION OF 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 .

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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 .

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EXAMPLE Given:Two forces F and G are applied to a hook. Force F is shown in G the figure and it makes 60° angle with the X-Y plane. Force G is pointing up and has a magnitude of 80 N with a = 111° and b = 69.3°. Find: The resultant force in the Cartesian vector form. Plan: 1) Using geometry and trigonometry, write F and G in the Cartesian vector form. 2) Then add the two forces. Chungnam National University

Solution : First, resolve force F. Fz = 100 sin 60° = 86.60 N F' = 100 cos 60° = 50.00 N Fx = 50 cos 45° = 35.36 N Fy = 50 sin 45° = 35.36 N Now, you can write: F = {35.36 i – 35.36 j + 86.60 k} N

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Now resolve force G. We are given only a and b. Hence, first we need to find the value of g. Recall the formula cos ² (a) + cos ² (b) + cos ² (g) = 1. Now substitute what we know. We have cos ² (111°) + cos ² (69.3°) + cos ² (g) = 1. Solving, we get g = 30.22° or 120.2°. Since the vector is pointing up, g = 30.22° Now using the coordinate direction angles, we can get UG, and determine G = 80 UG N. G = {80 ( cos (111°) i + cos (69.3°) j + cos (30.22°) k )} N G = {- 28.67 i + 28.28 j + 69.13 k } N Now, R = F + G or R = {6.69 i – 7.08 j + 156 k} N Chungnam National University

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 a, b, g . Chungnam National University

GROUP PROBLEM SOLVING (continued) First resolve the force F1 .

F1z

F1z = 300 sin 60° = 259.8 N F ´

F´ = 300 cos 60° = 150.0 N F’ can be further resolved as, F1x = -150 sin 45° = -106.1 N F1y = 150 cos 45° = 106.1 N

Now we can write : F1 = {-106.1 i + 106.1 j + 259.8 k } N

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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 + 353.6 j – 250 k } N FR = F1 + F2 = { 143.9 i + 459.6 j + 9.81 k } N FR = (143.9 2 + 459.6 2 + 9.81 2) ½ = 481.7 = 482 N a = cos-1 (FRx / FR) = cos-1 (143.9/481.7) = 72.6° b = cos-1 (FRy / FR) = cos-1 (459.6/481.7) = 17.4° g = cos-1 (FRz / FR) = cos-1 (9.81/481.7) = 88.8° Chungnam National University

DOT PRODUCT (Section 2.9)

Today’s Objective: Students will be able to use the dot product to a) determine an angle between two vectors, and, b) determine the projection of a vector along a specified line.

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READING QUIZ 1. The dot product of two vectors P and Q is defined as

P q

A) P Q cos q

B) P Q sin q Q

C) P Q tan q

D) P Q sec q

2. The dot product of two vectors results in a _________ quantity. A) scalar

B) vector

C) complex

D) zero

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APPLICATIONS For this geometry, can you determine angles between the pole and the cables?

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DEFINITION

The dot product of vectors A and B is defined as A•B = A B cos q. Angle q 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. Chungnam National University

DOT PRODUCT DEFINITON (continued) Examples:

i•j = 0 i•i = 1

A•B = =

(Ax i + Ay j + Az k) • (Bx i + By j + Bz k) Ax Bx +

AyBy

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+ AzBz

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 q, i.e., q = cos-1 [(A • B)/(A B)], where 0º £ q £ 180º . Chungnam National University

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 • U = AxUx + AyUy + Az Uz Chungnam National University

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|| )

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EXAMPLE Given: The force acting on the pole Find:

A

Plan: 1. Get rOA 2. q = cos-1{(F • rOA)/(F rOA)} 3. FOA = F • uOA or F cos q Chungnam National University

The angle between the force vector and the pole, and the magnitude of the projection of the force along the pole OA.

EXAMPLE (continued) rOA = {2 i + 2 j – 1 k} m rOA = (22 + 22 + 12)1/2 = 3 m F = {2 i + 4 j + 10 k}kN F = (22 + 42 + 102)1/2 = 10.95 kN

A

F • rOA = (2)(2) + (4)(2) + (10)(-1) = 2 kN·m q = cos-1{(F • rOA)/(F rOA)} uOA = rOA/rOA

q = cos-1 {2/(10.95 * 3)} = 86.5° = {(2/3) i + (2/3) j – (1/3) k}

FOA = F • uOA = (2)(2/3) + (4)(2/3) + (10)(-1/3) = 0.667 kN Or FOA = F cos q = 10.95 cos(86.51°) = 0.667 kN

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