POSITION VECTORS & FORCE VECTORS 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. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

APPLICATIONS How can we represent the force along the wing strut in a 3-D Cartesian vector form? Wing strut Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

The position vector directed from A to B, r AB , is defined as 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! Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

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 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

Given: 400 lb force along the cable DA. EXAMPLE Given: 400 lb force along the cable DA. Find: The force FDA in the Cartesian vector form. Plan: Find the position vector rDA Find the unit vector uDA. 3. Obtain the force vector as FDA = 400 lb uDA . Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

"Walk" from point D to point A. EXAMPLE (cont.) Step 1: find rDA … Two approaches: "Walk" from point D to point A. Subtract the coordinates of point D from those of point A. rDA = rA – rD = {14 k} – {2 i + 6 j} = {-2 i – 6 j + 14 k} ft. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

Step 2: find uDA = (rDA/rDA): EXAMPLE (cont.) rDA = {-2 i – 6 j + 14 k} ft. Step 2: find uDA = (rDA/rDA): rDA = (22 + 62 + 142)0.5 = 15.36 ft uDA = (-2 i – 6 j + 14 k)/15.36 Step 3: find FDA: uDA = rDA/rDA and FDA = 400 uDA lb FDA = 400{(-2 i – 6 j + 14 k)/15.36} lb = {-52.1 i – 156 j + 365 k} lb Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

EXAMPLE 2 : Find the distance between points A and B. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

XA = -5 cos(60°) cos(35°) = -2.048 km EXAMPLE (cont.) Step 1: find A and B: XA = -5 cos(60°) cos(35°) = -2.048 km YA = -5 cos(60°) sin(35°) = -1.434 km ZA = 5 sin (60°) = 4.330 km XB = 2 cos(25°) sin(40°) = 1.165 km YB = 2 cos(25°) cos(40°) = 1.389 km ZB = -2 sin (25°) = -0.8452 km giving A(-2.048, -1.434, 4.330) km, B(1.165, 1.389, -0.8452) km Step 2: find rAB … “tip minus tail” rAB = {(1.165-(-2.048)) i + (1.389-(-1.434)) j + (-0.8452-4.330) k} km = { 3.213 i + 2.823 j – 5.175 k } km Step 3: find rAB – length of the position vector and distance A-B rAB = (3.2132 + 2.8232 + 5.1752)0.5 = 6.714 km Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

1. A position vector, rPQ, is obtained by CONCEPT 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. 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 Answers: 1. A 2. B Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8

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

GROUP PROBLEM SOLVING (continued) FCA = 100 lb{rCA/rCA} FCA = 100 lb(-3 sin 40° i + 3 cos 40° j – 4 k)/5 FCA = {-38.57 i + 45.96 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 FR = FCA + FCB = {-2.57 i – 17.04 j – 116 k} lb FR = (2.572 + 17.042 + 1162)0.5 = 117.3 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

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 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) Answers 1. B 2. C Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.7,2.8