2.6 Addition and Subtraction of Cartesian Vectors Example Given: A = Axi + Ayj + AZk and B = Bxi + Byj + BZk Vector Addition Resultant R = A + B = (Ax + Bx)i + (Ay + By )j + (AZ + BZ) k Vector Substraction Resultant R = A - B = (Ax - Bx)i + (Ay - By )j + (AZ - BZ) k

2.6 Addition and Subtraction of Cartesian Vectors Concurrent Force Systems - Force resultant is the vector sum of all the forces in the system FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk where ∑Fx , ∑Fy and ∑Fz represent the algebraic sums of the x, y and z or i, j or k components of each force in the system

2.6 Addition and Subtraction of Cartesian Vectors Force, F that the tie down rope exerts on the ground support at O is directed along the rope Angles α, β and γ can be solved with axes x, y and z

2.6 Addition and Subtraction of Cartesian Vectors Cosines of their values forms a unit vector u that acts in the direction of the rope Force F has a magnitude of F F = Fu = Fcosαi + Fcosβj + Fcosγk

2.6 Addition and Subtraction of Cartesian Vectors Example 2.8 Express the force F as Cartesian vector

2.6 Addition and Subtraction of Cartesian Vectors Solution Since two angles are specified, the third angle is found by Two possibilities exit, namely or

2.6 Addition and Subtraction of Cartesian Vectors Solution 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 = {100.0i + 100.0j + 141.4k}N Checking:

2.6 Addition and Subtraction of Cartesian Vectors Example 2.9 Determine the magnitude and coordinate direction angles of resultant force acting on the ring

2.6 Addition and Subtraction of Cartesian Vectors Solution Resultant force FR = ∑F = F1 + F2 = {60j + 80k}kN + {50i - 100j + 100k}kN = {50j -40k + 180k}kN Magnitude of FR is found by

2.6 Addition and Subtraction of Cartesian Vectors Solution Unit vector acting in the direction of FR uFR = FR /FR = (50/191.0)i + (40/191.0)j + (180/191.0)k = 0.1617i - 0.2094j + 0.9422k So that cosα = 0.2617 α = 74.8° cos β = -0.2094 β = 102° cosγ = 0.9422 γ = 19.6° *Note β > 90° since j component of uFR is negative

2.6 Addition and Subtraction of Cartesian Vectors Example 2.10 Express the force F1 as a Cartesian vector.

2.6 Addition and Subtraction of Cartesian Vectors Solution The angles of 60° and 45° are not coordinate direction angles. By two successive applications of parallelogram law,

2.6 Addition and Subtraction of Cartesian Vectors Solution 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 + 86.6k}kN

2.6 Addition and Subtraction of Cartesian Vectors Solution Checking: 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 - 0.354j + 0.866k

2.6 Addition and Subtraction of Cartesian Vectors Solution α1 = cos-1(0.354) = 69.3° β1 = cos-1(-0.354) = 111° γ1 = cos-1(0.866) = 30.0° Using the same method, F2 = {106i + 184j - 212k}kN

2.6 Addition and Subtraction of Cartesian Vectors 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.

2.6 Addition and Subtraction of Cartesian Vectors View Free Body Diagram Solution Cartesian vector form 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

2.6 Addition and Subtraction of Cartesian Vectors Solution Since FR has a magnitude of 800N and acts in the +j direction FR = F1 + F2 800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk 800j = (212.1 + F2x)i + (150 + F2y)j + (- 50 + F2z)k To satisfy the equation, the corresponding components on left and right sides must be equal

2.6 Addition and Subtraction of Cartesian Vectors Solution Hence, 0 = 212.1 + F2x F2x = -212.1N 800 = 150 + F2y F2y = 650N 0 = -150 + 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°

2.7 Position Vectors x,y,z Coordinates - Right-handed coordinate system - Positive z axis points upwards, measuring the height of an object or the altitude of a point - Points are measured relative to the origin, O.

2.7 Position Vectors x,y,z Coordinates Eg: For Point A, xA = +4m along the x axis, yA = -6m along the y axis and zA = -6m along the z axis. Thus, A (4, 2, -6) Similarly, B (0, 2, 0) and C (6, -1, 4)

2.7 Position Vectors Position Vector - Position vector r is defined as a fixed vector which locates a point in space relative to another point. Eg: If r extends from the origin, O to point P (x, y, z) then, in Cartesian vector form r = xi + yj + zk

2.7 Position Vectors Position Vector Note the head to tail vector addition of the three components Start at origin O, one travels x in the +i direction, y in the +j direction and z in the +k direction, arriving at point P (x, y, z)

2.7 Position Vectors Position Vector - Position vector maybe directed from point A to point B - Designated by r or rAB Vector addition gives rA + r = rB Solving r = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)k or r = (xB – xA)i + (yB – yA)j + (zB –zA)k

2.7 Position Vectors Position Vector - The i, j, k components of the positive vector r may be formed by taking the coordinates of the tail, A (xA, yA, zA) and subtract them from the head B (xB, yB, zB) Note the head to tail vector addition of the three components

2.7 Position Vectors Length and direction of cable AB can be found by measuring A and B using the x, y, z axes Position vector r can be established Magnitude r represent the length of cable

2.7 Position Vectors Angles, α, β and γ represent the direction of the cable Unit vector, u = r/r

2.7 Position Vectors Example 2.12 An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B.

2.7 Position Vectors Solution Position vector View Free Body Diagram Solution Position vector r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m Magnitude = length of the rubber band Unit vector in the director of r u = r /r = -3/7i + 2/7j + 6/7k

2.7 Position Vectors Solution α = cos-1(-3/7) = 115°

2.8 Force Vector Directed along a Line In 3D problems, direction of F is specified by 2 points, through which its line of action lies F can be formulated as a Cartesian vector F = F u = F (r/r) Note that F has units of forces (N) unlike r, with units of length (m)

2.8 Force Vector Directed along a Line Force F acting along the chain can be presented as a Cartesian vector by - Establish x, y, z axes - Form a position vector r along length of chain

2.8 Force Vector Directed along a Line Unit vector, u = r/r that defines the direction of both the chain and the force We get F = Fu

2.8 Force Vector Directed along a Line Example 2.13 The man pulls on the cord with a force of 350N. Represent this force acting on the support A, as a Cartesian vector and determine its direction.

2.8 Force Vector Directed along a Line Solution End points of the cord are A (0m, 0m, 7.5m) and B (3m, -2m, 1.5m) r = (3m – 0m)i + (-2m – 0m)j + (1.5m – 7.5m)k = {3i – 2j – 6k}m Magnitude = length of cord AB Unit vector, u = r /r = 3/7i - 2/7j - 6/7k

2.8 Force Vector Directed along a Line Solution Force F has a magnitude of 350N, direction specified by u F = Fu = 350N(3/7i - 2/7j - 6/7k) = {150i - 100j - 300k} N α = cos-1(3/7) = 64.6° β = cos-1(-2/7) = 107° γ = cos-1(-6/7) = 149°

2.8 Force Vector Directed along a Line Example 2.14 The circular plate is partially supported by the cable AB. If the force of the cable on the hook at A is F = 500N, express F as a Cartesian vector.

2.8 Force Vector Directed along a Line Solution End points of the cable are (0m, 0m, 2m) and B (1.707m, 0.707m, 0m) r = (1.707m – 0m)i + (0.707m – 0m)j + (0m – 2m)k = {1.707i + 0.707j - 2k}m Magnitude = length of cable AB

2.8 Force Vector Directed along a Line Solution Unit vector, u = r /r = (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k = 0.6269i + 0.2597j – 0.7345k For force F, F = Fu = 500N(0.6269i + 0.2597j – 0.7345k) = {313i - 130j - 367k} N

2.8 Force Vector Directed along a Line Solution Checking Show that γ = 137° and indicate this angle on the diagram

2.8 Force Vector Directed along a Line Example 2.15 The roof is supported by cables. If the cables exert FAB = 100N and FAC = 120N on the wall hook at A, determine the magnitude of the resultant force acting at A.

2.8 Force Vector Directed along a Line View Free Body Diagram Solution rAB = (4m – 0m)i + (0m – 0m)j + (0m – 4m)k = {4i – 4k}m FAB = 100N (rAB/r AB) = 100N {(4/5.66)i - (4/5.66)k} = {70.7i - 70.7k} N

2.8 Force Vector Directed along a Line Solution rAC = (4m – 0m)i + (2m – 0m)j + (0m – 4m)k = {4i + 2j – 4k}m FAC = 120N (rAB/r AB) = 120N {(4/6)i + (2/6)j - (4/6)k} = {80i + 40j – 80k} N

2.8 Force Vector Directed along a Line Solution FR = FAB + FAC = {70.7i - 70.7k} N + {80i + 40j – 80k} N = {150.7i + 40j – 150.7k} N Magnitude of FR

2.9 Dot Product Dot product of vectors A and B is written as A·B (Read A dot B) Define the magnitudes of A and B and the angle between their tails A·B = AB cosθ where 0°≤ θ ≤180° Referred to as scalar product of vectors as result is a scalar

2.9 Dot Product Laws of Operation 1. Commutative law A·B = B·A 2. Multiplication by a scalar a(A·B) = (aA)·B = A·(aB) = (A·B)a 3. Distribution law A·(B + D) = (A·B) + (A·D)

2.9 Dot Product Cartesian Vector Formulation - Dot product of Cartesian unit vectors Eg: i·i = (1)(1)cos0° = 1 and i·j = (1)(1)cos90° = 0 - Similarly i·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 1 j·k = 1

2.9 Dot Product Cartesian Vector Formulation - Dot product of 2 vectors A and B A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk) = AxBx(i·i) + AxBy(i·j) + AxBz(i·k) + AyBx(j·i) + AyBy(j·j) + AyBz(j·k) + AzBx(k·i) + AzBy(k·j) + AzBz(k·k) = AxBx + AyBy + AzBz Note: since result is a scalar, be careful of including any unit vectors in the result

2.9 Dot Product Applications - The angle formed between two vectors or intersecting lines θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180° Note: if A·B = 0, cos-10= 90°, A is perpendicular to B

2.9 Dot Product Applications - The components of a vector parallel and perpendicular to a line - Component of A parallel or collinear with line aa’ is defined by A║ (projection of A onto the line) A║ = A cos θ - If direction of line is specified by unit vector u (u = 1), A║ = A cos θ = A·u

2.9 Dot Product Applications - If A║ is positive, A║ has a directional sense same as u - If A║ is negative, A║ has a directional sense opposite to u - A║ expressed as a vector A║ = A cos θ u = (A·u)u

2.9 Dot Product Applications For component of A perpendicular to line aa’ 1. Since A = A║ + A┴, then A┴ = A - A║ 2. θ = cos-1 [(A·u)/(A)] then A┴ = Asinθ 3. If A║ is known, by Pythagorean Theorem

2.9 Dot Product For angle θ between the rope and the beam A, - Unit vectors along the beams, uA = rA/rA - Unit vectors along the ropes, ur=rr/rr - Angle θ = cos-1 (rA.rr/rArr) = cos-1 (uA· ur)

2.9 Dot Product For projection of the force along the beam A - Define direction of the beam uA = rA/rA - Force as a Cartesian vector F = F(rr/rr) = Fur - Dot product F║ = F║·uA

2.9 Dot Product Example 2.16 The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB.

2.9 Dot Product Solution Since Then

2.9 Dot Product Solution Since result is a positive scalar, FAB has the same sense of direction as uB. Express in Cartesian form Perpendicular component

2.9 Dot Product Solution Magnitude can be determined From F┴ or from Pythagorean Theorem

2.9 Dot Product Example 2.17 The pipe is subjected to F = 800N. Determine the angle θ between F and pipe segment BA, and the magnitudes of the components of F, which are parallel and perpendicular to BA.

2.9 Dot Product Solution For angle θ rBA = {-2i - 2j + 1k}m View Free Body Diagram Solution For angle θ rBA = {-2i - 2j + 1k}m rBC = {- 3j + 1k}m Thus,

2.9 Dot Product Solution Components of F

2.9 Dot Product Solution Checking from trigonometry, Magnitude can be determined From F┴

2.9 Dot Product Solution Magnitude can be determined from F┴ or from Pythagorean Theorem

Chapter Summary Parallelogram Law Addition of two vectors Components form the side and resultant form the diagonal of the parallelogram To obtain resultant, use tip to tail addition by triangle rule To obtain magnitudes and directions, use Law of Cosines and Law of Sines

Chapter Summary Cartesian Vectors Vector F resolved into Cartesian vector form F = Fxi + Fyj + Fzk Magnitude of F Coordinate direction angles α, β and γ are determined by the formulation of the unit vector in the direction of F u = (Fx/F)i + (Fy/F)j + (Fz/F)k

Chapter Summary Cartesian Vectors Force and Position Vectors Components of u represent cosα, cosβ and cosγ These angles are related by cos2α + cos2β + cos2γ = 1 Force and Position Vectors Position Vector is directed between 2 points Formulated by distance and direction moved along the x, y and z axes from tail to tip

Chapter Summary Force and Position Vectors Dot Product For line of action through the two points, it acts in the same direction of u as the position vector Force expressed as a Cartesian vector F = Fu = F(r/r) Dot Product Dot product between two vectors A and B A·B = AB cosθ

Chapter Summary Dot Product Dot product between two vectors A and B (vectors expressed as Cartesian form) A·B = AxBx + AyBy + AzBz For angle between the tails of two vectors θ = cos-1 [(A·B)/(AB)] For projected component of A onto an axis defined by its unit vector u A = A cos θ = A·u

Chapter Review

Chapter Review

Chapter Review

Chapter Review

Chapter Review

Chapter Review