Section 2.7 Position Vectors

Engineering Mechanics: Statics Fifteenth Edition Chapter 2 Force Vector Section 2.7 Position Vectors Section 2.8 Force Vector Directed Along a Line Copyright © 2022 Pearson Education, Inc. All Rights Reserved Position Vectors & Force Vectors Today’s Objectives: a. Represent a position vector in Cartesian coordinate form, from given geometry. b. Represent a force vector directed along a line. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Position Vector (1 of 2) A position vector is defined as a fixed vector that locates a point in space relative to another point. Consider two points, A and B, in 3-D space. Let their coordinates be (X A, Y A, Z A) and (X B, Y B, Z B), respectively. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Position Vector (2 of 2) The position vector directed from A to B, rAB , is defined as rAB = {( XB – X A ) i + ( YB – YA ) j + ( ZB – Z A ) k }m Please note that B is the ending point and A is the starting point. Always subtract the “tail” coordinates from the “tip” coordinates! Copyright © 2022 Pearson Education, Inc. All Rights Reserved Section 2.8: 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’s magnitude. So we need to: a. Find the position vector, rA B , along two points on that line. b. Find the unit vector describing the line’s direction, u AB  rAB =  rAB  .  c. Multiply the unit vector by the magnitude of the force, F = F u AB . Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example (1 of 3) Given: The 420 N force along the cable AC. Find: The force F A C in the Cartesian vector form. Plan: 1. Find the position vector rAC and its unit vector uAC 2. Obtain the force vector as FA C = 420 N uAC. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example (2 of 3) As per the figure, when relating A to C, we will have to go 2 m in the x-direction, 3 m in the ydirection, and −6 m in the zdirection. Hence, rAC = {2i + 3 j − 6k } m. (We can also find rAC by subtracting the coordinates of A from the coordinates of C.) Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example (3 of 3)  rAC = 22 + 32 + ( −6 ) 2  =7m 1 2 r r  Now uAC = rAC and FAC = 420 uAC = 420  AC   rAC  AC So FAC = 420  (2i + 3 j − 6k ) N  7  = 120i +180 j −360k N Copyright © 2022 Pearson Education, Inc. All Rights Reserved Concept Quiz (1 of 2) 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 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Concept Quiz (2 of 2) 2. If F and r are force and position vectors, respectively, in SI units, what are the units of the expression  r   F   ? F   A. B. C. D. E.  Newton Dimensionless Meter Newton − Meter The expression is algebraically illegal. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (1 of 3) Given: Two forces are acting on a flag pole as shown in the figure. FB = 560 N and FC = 700 N Find: The magnitude and the coordinate direction angles of the resultant force. Plan: 1. Find the forces along A B and A C 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 . Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (2 of 3) rAB = 2i − 3 j − 6k  m rAC = 3i + 2 j − 6k  m  = 3 rAB = 22 + ( −3 ) + ( −6 ) rAC 2 2 + 22 + ( −6 ) 2  1 2  1 2 2 =7m =7m   FAB = 560  rrAB  N FAC =700 rAC  N FAB = 560( 2i – 3 j – 6k ) FAC = 700(3i + 2 j – 6k ) FAB = (160i –240 j – 480k ) N FAC =300i + 200 j –600k  N  AB  7N  rAC  7N Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (3 of 3) FR = FAB + FAC = 460i –40 j – 1080k  N  FR = 4602 + ( −40 ) + ( −1080 ) 2 2  1 2 = 1174.6 N FR = 1175 Ν  = cos−1 460  o = 66.9   1175  −40  o = 92.0   1175   = cos−1 −1080  o = 157   1175   = cos−1 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (1 of 2) 1. The position vector rPQ is obtained by A. Coordinates of Q minus coordinates of the origin B. Coordinates of P minus coordinates of Q C. Coordinates of Q minus coordinates of P D. Coordinates of the origin minus coordinates of P Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (2 of 2) 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 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (1 of 2) 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. {3i + 3 j + 3k } m B. {–3i – 3 j – 3k } m C. {5i + 7 j + 9k } m D. {–3i + 3 j + 3k } m E. {4i + 5 j + 6k } m Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (2 of 2) 2. A force vector, F, directed along a line defined by PQ is given by A. F   F  rPQ   rPQ B. rPQ C.  rPQ  F  r  PQ  D. r  F  PQ   rPQ  Copyright © 2022 Pearson Education, Inc. All Rights Reserved Copyright This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. 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