Engineering Mechanics : STATICS BDA10203 Lecture #06 Group of Lecturers Faculty of Mechanical and Manufacturing Engineering Universiti Tun Hussein Onn Malaysia

POSITION & FORCE VECTORS (Sections 2.7 - 2.8) Today’s Objectives: Students will be able to : a) Represent a position vector in Cartesian coordinate form, from given geometry. b) Represent a force vector directed along a line. Learning Topics: Applications / Relevance Write position vectors Write a force vector

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 & B, in 3-D space. Let their coordinates be (XA, YA, ZA) and ( XB, YB, ZB ), respectively. 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. So ALWAYS subtract the “tail” coordinates from the “tip” coordinates!

NOTE that r has unit of length POSITION VECTOR The position vector rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m Magnitude Unit vector, u = r /r = rAB / NOTE that r has unit of length

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)

EXAMPLE Given: A position vector r acting from point A(3m,5m,6m) to point B(5m,-2m,1m) Find: Represent the position vector in Cartesian vector form and determine its direction angles and find the distance between point A and B. Plan: 1) Represent the position vector in Cartesian vector form 2) Find the magnitude of the position vector 3) Use trigonometry to find the direction angle

EXAMPLE Position vector, rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m = {( 5 – 3 ) i + ( -2 – 5) j + ( 1 – 6 ) k }m = {2i - 7j - 5k }m y x z A(3,5,6) B(5,-2,1) 5 3 -2 1 6 Note that Start at point A (3,5,6), one travels 2m in the +i direction, 7m in the -j direction and 5m in the -k direction, arriving at point B (5,-2,1) Distance between point A and B, Total 8.83m from A to B

EXAMPLE Unit vector, u = r /r Solving the equations and we will get the answers as below α = 76.9°, β = 142.4°, γ = 124.5°

IN CLASS TUTORIAL (GROUP PROBLEM SOLVING) Given: Position vector as shown in the figure Find: Express the position vector r in the Cartesian vector form, its magnitude and coordinate direction angle. Plan: 1) Represent the position vector in Cartesian vector form 2) Find the magnitude of the position vector 3) Use trigonometry to find the direction angle

GROUP PROBLEM SOLVING (continued) A rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m = {( 0 – 3 ) i + ( -8 – 4) j + ( 4 – 0 ) k }m = {-3i - 12j + 4k }m Distance between point A and B,

GROUP PROBLEM SOLVING (continued) Solving the equations and we will the answers as below α = 103.3°, β = 157.4°, γ = 72.1°

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.

r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m Solution Position vector r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m Note that it took 3m in –x direction, 2m in +y direction and 6m in +j dirention to arrive at B from A point Magnitude = length of the rubber band Unit vector in the direction of r u = r /r = -3/7i + 2/7j + 6/7k

Solution α = cos-1(-3/7) = 115° β = cos-1(2/7) = 73.4° γ = cos-1(6/7) = 31.0° Note: These angles are measured from the positive axes of a localized coordinate system placed at the tail of r

HOMEWORK TUTORIAL Q1 (2-84): Determine the length of the connecting rod AB by first formulating a Cartesian position vector from A to B and then determining its magnitude. Given: b = 400mm a = 125mm α = 25°

HOMEWORK TUTORIAL (continued) Q2 (2-85): Determine the length of member AB of the truss by first establishing a Cartesian position vector from A to B and then determining its magnitude. Given: a = 1.2m b = 0.8m c = 0.3m d = 1.5m, Length CB θ = 40deg

Problem 2-83 Express the position vector r in Cartesian vector form; then determine its magnitude and coordinate direction angles.

Force Vector Directed along a Line 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 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, uAB = (rAB/rAB). c) Multiply the unit vector by the magnitude of the force, F = F uAB .

APPLICATIONS How can we represent the force along the wing strut in a 3-D Cartesian vector form? Wing strut

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 =F (r /r ) Note that F has unit of force, unlike r which has units of length

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.

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

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.

Solution End points of the cable are A (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

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

Solution Checking Show that γ = 137° and indicate this angle on the diagram

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.

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

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

Solution FR = FAB + FAC = {70.7i - 70.7k} N + {80i + 40j – 80k} N = {150.7i + 40j – 150.7k} N Magnitude of FR

EXAMPLE Given: The window is held open by cable AB as shown in the figure Find: Length of the cable and express the F force acting at A along the cable as a Cartesian vector Plan: 1) Represent the position vector in Cartesian vector form 2) Find the magnitude of the position vector 3) Using a unit vector, express the F force as a Cartesian vector

EXAMPLE rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }mm = {( 0 – 300cos30°) i + ( 150 – 500) j + (250 + 300sin30°) k }mm = {-259.8i - 350j + 400k }mm Distance between point A and B,

EXAMPLE Use a unit vector to express F in the form of Cartesian vector F = FuAB = F(rAB/rAB). F = 30 {(-259.8 /591.6)i + (-350/591.6)j + (400/591.6)k} = 30{-0.44i – 0.59j + 0.68k} = {-13.2i - 17.7j + 20.4k} N

IN CLASS TUTORIAL (GROUP PROBLEM SOLVING) Given: The cable attached to the tractor at B exerts a force of 3500 N on the framework. Find: Express this force as a Cartesian vector Plan: 1) Represent the position vector in Cartesian vector form 2) Find the magnitude of the position vector 3) Using a unit vector, express the F force as a Cartesian vector

GROUP PROBLEM SOLVING (continued) rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m = {( 15sin20° – 0 ) i + ( 15cos20° – 0) j + ( 0 – 10.5 ) k }m = {5.13i + 14.1j - 10.5k }m Distance between point A and B,

GROUP PROBLEM SOLVING (continued) Use a unit vector to express F in the form of Cartesian vector F = FuAB = F(rAB/rAB). F = 3500 {(5.13 /18.31)i + (14.1/18.31)j + (-10.5/18.31)k} = 3500{0.28i + 0.77j – 0.57k} = {980i + 2695j - 1995k} N

CONCEPT QUIZ 1. P and Q are two points in a 3-D space. How are the position vectors rPQ and rQP related? A) rPQ = rQP B) rPQ = - rQP C) rPQ = 1/rQP D) rPQ = 2 rQP 2. If F and r are force vector and position vectors, respectively, in SI units, what are the units of the expression (r * (F / F)) ? A) Newton B) Dimensionless C) Meter D) Newton - Meter E) The expression is algebraically illegal. Answers: 1. B 2. C

ATTENTION QUIZ 1. Two points in 3 – D space have coordinates of P (1, 2, 3) and Q (4, 5, 6) meters. The position vector rQP 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 Answers 1. B 2. C 2. Force vector, F, directed along a line PQ is given by A) (F/ F) rPQ B) rPQ/rPQ C) F(rPQ/rPQ) D) F(rPQ/rPQ)

READING QUIZ 1. A position vector, rPQ, is obtained by A) Coordinates of Q minus coordinates of P B) Coordinates of P minus coordinates of Q C) Coordinates of Q minus coordinates of the origin D) Coordinates of the origin minus coordinates of P 2. A force of magnitude F, directed along a unit vector U, is given by F = ______ . A) F (U) B) U / F C) F / U D) F + U E) F – U Answers: 1. A 2. A

HOMEWORK TUTORIAL Q1 (2-93): The plate is suspended using the three cables which exert the forces shown. Express each force as a Cartesian vector. Given: FBA = 3.5kN FCA = 5kN FDA = 4kN a = 0.3m b = 0.3m c = 0.6m d = 1.4m e = 0.3m f = 0.3m g = 0.2m

HOMEWORK TUTORIAL (continued) Q2 (2-94): The engine of the lightweight plane is supported by struts that are connected to the space truss that makes up the structure of the plane. The anticipated loading in two of the struts is shown. Express each of these forces as a Cartesian vector. Given: F1 = 4kN F2 = 6kN a = 0.15m b = 0.15m c = 0.9m d = 0.75m e = 0.15m f = 0.9m

HOMEWORK TUTORIAL (continued) Q3 (2-104) : The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles α,β,γ of the resultant force. Units Used: kN := 1000N Given: x = 20m, a = 16m y = 15m, b = 18m F1 = 600N, c = 6m F2 = 400N, d = 4m F3 = 800N, e = 24m