Announcements. 2-D Vector Addition

Announcements 2-D Vector Addition Today s Objectives Understand the difference between scalars and vectors Resolve a 2-D vector into components Perform vector operations Class Activities Applications Scalar and vector definitions Triangle and Parallelogram laws Vector resolution Examples Engr221 Chapter 2 1

Applications There are four concurrent cable forces acting on the bracket. How do you determine the resultant force acting on the bracket? Scalars and Vectors Scalars o Magnitude - positive or negative o Addition rule - simple arithmetic o Special Notation - none o Examples speed, mass, volume Vectors o Magnitude and direction positive or negative o Addition rule parallelogram law o Special notation - Bold font, line, arrow, or carrot o Examples force, velocity Engr221 Chapter 2 2

Vector Operations Multiplication and division Vector Addition Engr221 Chapter 2 3

Vector Subtraction Resolution of a Vector Resolution of a vector is breaking it up into components. It is kind of like using the parallelogram law in reverse. Engr221 Chapter 2 4

Triangle Laws Textbook Problem 2-4 Determine the magnitude of the resultant force F R = F 1 + F 2 and its direction, measured counterclockwise from the positive u axis. Answer: F r = 605N θ = 85.4 Engr221 Chapter 2 5

Textbook Problem 2-16 Resolve the force F 2 into components acting along the u and v axes and determine the magnitudes of the components. Answer: F 2v = 77.6N F 2u = 150N Summary Understand the difference between scalars and vectors Resolve a 2-D vector into components Perform vector operations Engr221 Chapter 2 6

Announcements Cartesian Vector Notation Today s Objectives Resolve 2-dimensional vectors into x and y components Find the resultant of a 2-dimensional force system Express vectors in Cartesian form Class Activities Cartesian Vector Notation Addition of vectors Examples Engr221 Chapter 2 7

Cartesian Vector Notation 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, respectively. Cartesian Vector Notation - continued For example, F = F x i + F y j 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. Engr221 Chapter 2 8

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. Addition of Several Vectors - continued Step 3 is to find the magnitude and angle of the resultant vector. Engr221 Chapter 2 9

Magnitude and Angle You can also represent a 2-D vector with a magnitude and an angle. Example A 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 Engr221 Chapter 2 10

Example A - continued F 1 F 2 F 3 = { 15 sin 40 i + 15 cos 40 j } kn = { 9.642 i + 11.49 j } kn = { -(12/13)26 i + (5/13)26 j } kn = { -24 i + 10 j } kn = { 36 cos 30 i 36 sin 30 j } kn = { 31.18 i 18 j } kn Example A - continued Summing up all the i and j components respectively, we get, F R = { (9.642 24 + 31.18) i + (11.49 + 10 18) j } kn = { 16.82 i + 3.49 j } kn y F R F R = ((16.82) 2 + (3.49) 2 ) 1/2 = 17.2 kn θ = tan -1 (3.49/16.82) = 11.7 θ x Engr221 Chapter 2 11

Example B 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 Example B - continued F 1 = { (4/5) 850 i - (3/5) 850 j } N = { 680 i - 510 j } N F 2 = { -625 sin(30 ) i - 625 cos(30 ) j } N = { -312.5 i - 541.3 j } N F 3 = { -750 sin(45 ) i + 750 cos(45 ) j } N { -530.3 i + 530.3 j } N Engr221 Chapter 2 12

Example B - continued Summing up all the i and j components respectively, we get, F R = { (680-312.5-530.3) i + (-510-541.3 + 530.3) j }N = { - 162.8 i - 521 j } N F R = ((162.8) 2 + (521) 2 ) ½ = 546 N θ = tan 1 (521/162.8) = 72.64 or θ y x From the positive x axis: θ = 180 + 72.64 = 253 F R Textbook Problem 2-55 If F 2 = 150 lb and θ = 55º, determine the magnitude and orientation, measured clockwise from the positive x axis, of the resultant force of the three forces acting on the bracket. Answer: F R = 161 lb θ = 38.3 Engr221 Chapter 2 13

Summary Resolve 2-dimensional vectors into x and y components Find the resultant of a 2-dimensional force system Express vectors in Cartesian form Announcements Homework feedback Graph paper Problem statements Week of prayer schedule Engr221 Chapter 2 14

3 Dimensional Vectors Today s Objectives Represent a 3-D vector in a Cartesian coordinate system Find the magnitude and coordinate angles of a 3-D vector Add vectors is 3-D space Class Activities Applications Unit vector definition 3-D vector terms Adding vectors Examples Applications Many problems in real-life involve 3-Dimensional space. How do you represent each of the cable forces in Cartesian vector form? Engr221 Chapter 2 15

Applications - continued Given the forces in the cables, how do you determine the resultant force acting at D, the top of the tower? Applications - continued Engr221 Chapter 2 16

Right Handed Coordinate System A Cartesian coordinate system is said to be right-handed provided the thumb of the right hand points in the direction of the positive z axis when the right-hand fingers are curled about this axis and directed from the positive x toward the positive y axis. A Unit Vector For a vector A with a magnitude of A, a unit vector U A is defined as U A = 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. Engr221 Chapter 2 17

3-D Cartesian Vector Terminology Consider a box with sides A X, A Y, and A Z meters long. The vector A can be defined as A = (A X i + A Y j + A Z 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 = (A X2 + A Y2 ) 1/2. The magnitude of the vector A can now be obtained as: A = ((A ) 2 + A Z2 ) ½ = (A X2 + A Y2 + A Z2 ) ½ Terminology - continued The direction or orientation of vector A is defined by the angles α, β, and γ. These angles are measured between the vector and the positive x, y and z axes, respectively. Their values range from 0 to 180. Using trigonometry, direction cosines, and coordinate direction angles are found using the formulas: Engr221 Chapter 2 18

Terminology - continued These angles are not independent. They must satisfy the following equation: cos ² α + cos ² β + cos ² γ = 1 This result can be derived from the definition of coordinate direction angles and the unit vector. Recall, the formula for finding the unit vector is: or written another way: u A = cos α i + cos β j + cos γ k Cartesian Vector Math 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 = A X i + A Y j + A Z k and B = B X i + B Y j + B Z k then A + B = (A X + B X ) i + (A Y + B Y ) j + (A Z + B Z ) k and A B = (A X - B X ) i + (A Y - B Y ) j + (A Z - B Z ) k Engr221 Chapter 2 19

Important Notes Sometimes 3-D vector information is given as: a) Magnitude and coordinate direction angles, or b) Magnitude and projection angles You should be able to use both types of information to change the representation of the vector into the Cartesian form, i.e., F = {10 i 20 j + 30 k} N Example A G Given: Two forces F and G are applied to a hook. Force F is shown in the figure and it makes a 60 angle with the x-y plane. Force G is pointing up and has a magnitude of 80 lb with α = 111 and β = 69.3. Find: The resultant force in Cartesian vector form. Plan: 1) Using geometry and trigonometry, write F and G in Cartesian vector form 2) Add the two forces together Engr221 Chapter 2 20

Example A - continued Solution : First, resolve force F. F z = 100 sin 60 = 86.60 lb F' = 100 cos 60 = 50.00 lb F x = 50 cos 45 = 35.36 lb F y = 50 sin 45 = 35.36 lb Now, you can write: F = {35.36 i 35.36 j + 86.60 k} lb Example A - continued Next, resolve force G. We are given only α and β. We need to find the value of γ. Recall the formula cos ² (α) + cos ² (β) + cos ² (γ) = 1. Now substitute what we know. We have cos ² (111 ) + cos ² (69.3 ) + cos ² (γ) = 1. Solving, we get γ = 30.22 or 120.2. Since the vector is pointing up, γ = 30.22 G Now using the coordinate direction angles, we can get U G, and determine G from the formula: G = 80U G lb. G = {80 ( cos (111 ) i + cos (69.3 ) j + cos (30.22 ) k )} lb G = {- 28.67 i + 28.28 j + 69.13 k } lb Finally, find the resultant vector R = F + G or R = {6.69 i 7.08 j + 156 k} lb Engr221 Chapter 2 21

Example B Given: The screw eye is subjected to two forces. Find: The magnitude and the coordinate direction angles of the resultant force. Plan: 1) Using geometry and trigonometry, write F 1 and F 2 in Cartesian vector form 2) Add F 1 and F 2 to get F R 3) Determine the magnitude and α, β, and γ of F R Example B - continued F 1z F First resolve force F 1. F 1z = 300 sin 60 = 259.8 N F = 300 cos 60 = 150.0 N F can be further resolved as, F 1x = -150 sin 45 = -106.1 N F 1y = 150 cos 45 = 106.1 N Now we can write : F 1 = {-106.1 i + 106.1 j + 259.8 k } N Engr221 Chapter 2 22

Example B - continued Force F 2 can be represented in Cartesian vector form as: F 2 = 500{ cos 60 i + cos 45 j + cos 120 k } N = { 250 i + 353.6 j 250 k } N F R = F 1 + F 2 = { 143.9 i + 459.6 j + 9.81 k } N F R = (143.9 2 + 459.6 2 + 9.81 2 ) ½ = 481.7 = 482 N α = cos -1 (F Rx / F R ) = cos -1 (143.9/481.7) = 72.6 β = cos -1 (F Ry / F R ) = cos -1 (459.6/481.7) = 17.4 γ = cos -1 (F Rz / F R ) = cos -1 (9.81/481.7) = 88.8 Questions 1. What is not true about the unit vector, U A? A) It is dimensionless B) Its magnitude is one C) It always points in the direction of positive x axis D) It always points in the direction of vector A 2. If F = {10 i + 10 j + 10 k} N and G = {20 i + 20 j + 20 k } N, then F + G = { } N A) 10 i + 10 j + 10 k B) 30 i + 20 j + 30 k C) -10 i - 10 j - 10 k D) 30 i + 30 j + 30 k Engr221 Chapter 2 23

Textbook Problem 2-68 Determine the magnitude and coordinate direction angles of the resultant force. Answers: F R = 369 N α = 19.5 β = 20.6 γ = 69.5 Textbook Problem 2-78 Two forces F 1 and F 2 act on the bolt. If the resultant force F r has a magnitude of 50 lb and coordinate direction angles α = 110º and β = 80º as shown, determine the magnitude of F 2 and its coordinate direction angles. Answers: F 2 = 32.4 lb α 2 = 122º β 2 = 74.5º γ 2 = 144º Engr221 Chapter 2 24

Summary Represent a 3-D vector in a Cartesian coordinate system Find the magnitude and coordinate angles of a 3-D vector Add vectors (forces) in 3-D space Announcements Engr221 Chapter 2 25

Review Vector: A = (A X i + A Y j + A Z k) Direction cosines: Trig. Law: cos ² α + cos ² β + cos ² γ = 1 Unit vector: or written another way: u A = cos α i + cos β j + cos γ k Position and Force Vectors Today s Objectives Represent a position vector in Cartesian vector form using given geometry Represent a force vector directed along a line Class Activities Applications Position vectors Force vectors Examples Engr221 Chapter 2 26

Applications How can we represent the force along the wing strut in a 3-D Cartesian vector form? Wing strut A position vector is defined as a fixed vector that locates a point in space relative to another point. Position Vector Consider two points, A & B, in 3D space. Let their coordinates be (X A, Y A, Z A ) and (X B, Y B, Z B ), respectively. The position vector directed from A to B, r AB, is defined as r AB = {( X B X A ) i + ( Y B Y A ) j + ( Z B Z A ) k } m Note that B is the ending point and A is the starting point. Always subtract the tail coordinates from the tip coordinates. Engr221 Chapter 2 27

Force Vector Along a Line If a force is directed along a line, then the force can be represented 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, u AB = (r AB /r AB ) c) Multiply the unit vector by the magnitude of the force, F = F u AB Example A Given: 400 lb force along the cable DA Find: Plan: The force F DA in Cartesian vector form 1. Find the position vector r DA and the unit vector u DA 2. Obtain the force vector as F DA = 400 lb u DA Engr221 Chapter 2 28

r DA Example A - continued We can find r DA by subtracting the coordinates of D from the coordinates of A. D = (2, 6, 0) ft and A = (0, 0, 14) ft r DA = {(0 2) i + (0 6) j + (14 0) k} ft = {-2 i 6 j + 14 k} ft We can also calculate r DA by realizing that when relating D to A, we will have to go -2 ft in the x-direction, -6 ft in the y-direction, and +14 ft in the z-direction. = (2 2 + 6 2 + 14 2 ) 0.5 = 15.36 ft u DA = r DA /r DA and F DA = 400 u DA lb F DA = 400{(-2 i 6 j + 14 k)/15.36} lb = {-52.1 i 156 j + 365 k} lb Example B Given: Two forces are acting on a pipe as shown in the figure Find: Plan: The magnitude and the coordinate direction angles of the resultant force 1) Find the forces along CA and CB in Cartesian vector form 2) Add the two forces to get the resultant force, F R 3) Determine the magnitude and the coordinate angles of F R Engr221 Chapter 2 29

Example B - continued F CA = 100 lb{r CA /r CA } F CA = 100 lb(-3 sin 40 i + 3 cos 40 j 4 k)/5 F CA = {-38.57 i + 45.96 j 80 k} lb F CB = 81 lb{r CB /r CB } F CB = 81 lb(4 i 7 j 4 k)/9 F CB = {36 i 63 j 36 k} lb F R = F CA + F CB = {-2.57 i 17.04 j 116 k} lb F R = (2.57 2 + 17.04 2 + 116 2 ) 1/2 = 117.3 lb = 117 lb α = cos -1 (-2.57/117.3) = 91.3, β = cos -1 (-17.04/117.3) = 98.4 γ = cos -1 (-116/117.3) = 172 Questions 1. Two points in 3-D space have coordinates of P(1, 2, 3) and Q (4, 5, 6) meters. The position vector r QP is given by A) {3 i + 3 j + 3 k} m B) {- 3 i 3 j 3 k} m C) {5 i + 7 j + 9 k} m D) {- 3 i + 3 j + 3 k} m E) {4 i + 5 j + 6 k} m 2. P and Q are two points in 3-D space. How are the position vectors r PQ and r QP related? A) r PQ = r QP B) r PQ = - r QP C) r PQ = 1/r QP D) r PQ = 2 r QP Engr221 Chapter 2 30

Textbook Problem 2.90 Determine the magnitude and coordinate direction angles of the resultant force. Answers: F R = 52.2 lb α = 87.8º β = 63.7º γ = 154º Summary Represent a position vector in Cartesian vector form using given geometry Represent a force vector directed along a line Engr221 Chapter 2 31

Announcements Homework Excuses. Not! Due date and time Graph paper Do solutions on one side only Problem statements Dot Product Today s Objectives Determine the angle between two vectors Determine the projection of a vector along a specified line Class Activities Applications Dot product Angle determination Projection determination Examples Engr221 Chapter 2 32

Applications For this geometry, can you determine angles between the pole and the cables? For force F at Point A, what component of it (F 1 ) acts along the pipe OA? What component (F 2 ) acts perpendicular to the pipe? Dot Product Definition The dot product of vectors A and B is defined as A B = AB cos θ Angle θ is the smallest angle between the two vectors and is always in a range of 0 º to 180 º 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 Engr221 Chapter 2 33

Dot Product Definition - continued Examples: i j = 0 i.e. (1)(1)cos(90 ) i i = 1 i.e. (1)(1)cos(0 ) A B = (A x i + A y j + A z k) (B x I + B y j + B z k) = (A x B x (i i) + A x B y (i j) + A x B z (i k) +.) = A x B x + A y B y + A z B z Angle Between Two Vectors For two vectors in Cartesian form, one can find the angle by: a) Finding the dot product, A B = (A x B x + A y B y + A z B z ) b) Finding the magnitudes (A & B) of the vectors A & B c) Using the definition of dot product and solving for θ, i.e. θ = cos -1 [(A B)/(A B)], where 0 º θ 180 º Engr221 Chapter 2 34

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, U aa along the line aa 2. Find the scalar projection of A along line aa by A = A U = A x U x + A y U y + A z U z Projection of a Vector - continued 3. If needed, the projection can be written as a vector, A, by using the unit vector U aa and the magnitude found in step 2. A = A U aa 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 ) Engr221 Chapter 2 35

Example A Given: The force acting on the pole A Find: Plan: The angle between the force vector and the pole, and the magnitude of the projection of the force along the pole OA. 1. Find r OA 2. θ = cos -1 {(F r OA )/(F r OA )} 3. F OA = F u OA or F cos θ Example A - continued r OA = {2 i + 2 j 1 k} m r OA = (2 2 + 2 2 + 1 2 ) 1/2 = 3 m F = {2 i + 4 j + 10 k} kn F = (2 2 + 4 2 + 10 2 ) 1/2 = 10.95 kn A F r OA = (2)(2) + (4)(2) + (10)(-1) = 2 kn m θ = cos -1 {(F r OA )/(F r OA )} θ = cos -1 {2/(10.95 3)} = 86.5 u OA = r OA /r OA = {(2/3) i + (2/3) j (1/3) k} F OA = F u OA = (2)(2/3) + (4)(2/3) + (10)(-1/3) = 0.667 kn or F OA = F cos θ = 10.95 cos(86.51 ) = 0.667 kn Engr221 Chapter 2 36

Example B 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 r AO 2. θ = cos -1 {(F r AO )/(F r AO )} 3. F OA = F u AO or F cos θ Example B - continued r AO = {-3 i + 2 j 6 k} ft r AO = (3 2 + 2 2 + 6 2 ) 1/2 = 7 ft F = {-20 i + 50 j 10 k} lb F = (20 2 + 50 2 + 10 2 ) 1/2 = 54.77 lb F r AO = (-20)(-3) + (50)(2) + (-10)(-6) = 220 lb ft θ = cos -1 {(F r AO )/(F r AO )} θ = cos -1 {220/(54.77 7)} = 55.0 u AO = r AO /r AO = {(-3/7) i + (2/7) j (6/7) k} F AO = F u AO = (-20)(-3/7) + (50)(2/7) + (-10)(-6/7) = 31.4 lb or F AO = F cos θ = 54.77 cos(55.0 ) = 31.4 lb Engr221 Chapter 2 37

Questions 1. The dot product of two vectors P and Q is defined as A) P Q cos θ B) P Q sin θ C) P Q tan θ D) P Q sec θ θ P Q 2. The dot product of two vectors results in a quantity. A) scalar B) vector C) complex D) zero Questions 1. If a dot product of two non-zero vectors is 0, then the two vectors must be to each other. A) parallel (pointing in the same direction) B) parallel (pointing in the opposite direction) C) perpendicular D) cannot be determined 2. Find the dot product of the two vectors P and Q. P = {5 i + 2 j + 3 k} m Q = {-2 i + 5 j + 4 k} m A) -12 m B) 12 m C) 12 m 2 D) -12 m 2 E) 10 m 2 Engr221 Chapter 2 38

Textbook Problem 2-127 Determine the angle θ between the edges of the sheet-metal bracket. Answer: 82.0 Summary Determine an angle between two vectors Determine the projection of a vector along a specified line Engr221 Chapter 2 39