ENGI 1313 Mechanics I

ENGI 1313 Mechanics I Lecture 06: Cartesian and Position Vectors

Chapter 2 Objectives • to review concepts from linear algebra • to sum forces, determine force resultants and resolve force components for 2D vectors using Parallelogram Law • to express force and position in Cartesian vector form • to introduce the concept of dot product

Lecture 06 Objectives • to further examine 3D Cartesian vectors • to define a position vector in Cartesian coordinate system • to determine force vector directed along a line

Example Problem 6-01 • Problem 2-77 (Hibbeler, 2007). The bolt is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 80 N, and  = 60° and  = 45°, determine the magnitudes of its components.

Example Problem 6-01 • Known • Find

Example Problem 6-01 (cont.) • Find Angle  • Find component magnitudes  = 60  = 45 Fz Fx Fy

Position Vectors – General • 3D Coordinates • Unique position in space • Right-hand coordinate system • A(4,2,-6) • B(0,2,0) • C(6,-1,4)

^ xi ^ zk ^ yj Position Vectors – Origin to a Point • Fixed vector locating a point P(x,y,z) in space relative to another point (origin) within a defined coordinate system. • Right-hand Cartesian coordinate system • Tip-to-tail vector component technique

Position Vector – General Case • Two Points in Space • Rectangular Cartesian coordinate system • Origin O • Point A and Point B z B(xB, yB, zB) A(xA, yA, zA) y O(0, 0, 0) x

rAB Recall “tip-to-tail” vector addition laws  rOB rOA Position Vector – General Case • Establish Position Vectors • From Point O to Point A (rOA = rA) • From Point O to Point B (rOB = rB) • From Point A to Point B(rAB = r ) z B(xB, yB, zB) A(xA, yA, zA) y O(0, 0, 0) x

^ ^ (xB – xA) i (yB – yA) j ^ (zB – zA) k rAB r = rAB rOB rOA Position Vector – General Case • Define Position Vector (rAB = r ) • “tip – tail” or B(xB, yB, zB) – A(xA, yA, zA) z B(xB, yB, zB) A(xA, yA, zA) y O(0, 0, 0) x

Comprehension Quiz 6-01 • Two points in 3D 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 • Answer: B  {-3 i - 3 j - 3 k} m

rPQ = -rQP Comprehension Quiz 6-02 • 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 = 2rQP • Answer: B z Q(xB, yQ, zQ) P(xP, yP, zP) y x

Comprehension Quiz 6-03 • If F is a force vector (N) and r is a position vector (m), what are the units of the expression • A) N • B) Dimensionless • C) m • D) Nm • E) The expression is algebraically illegal • Answer: A

Example 6-01 • Express the force vector FDA in Cartesian form • Known: • A(0,0,14) ft • D(2,6,0) ft • FDA = 400 lb

Example 6-01 (cont.) • Find Position Vector rDA • Through point coordinates rDA

Example 6-01 (cont.) • Find Position Vector |rDA|Magnitude rDA

Example 6-01 (cont.) • Find unit vector uDA uDA

Example 6-01 (cont.) • Find Unit Vector uDAMagnitude • Confirm unity uDA

Example 6-01 (cont.) • Find Force Vector FDA • or

Group Problem 6-01 • Find the resultant force magnitude and coordinate direction • Plan • Cartesian vector form of FCA and FCB • Sum concurrent forces • Obtain solution

Group Problem 6-01 (cont.) • Position Vectors and Magnitude • rCA • rCB

Group Problem 6-01 (cont.) • Force Vectors and Magnitude • FCA • FCB

Group Problem 6-01 (cont.) • Force Resultant Vector Magnitude & Orientation

Group Problem 6-01 (cont.) • Force Resultant Vector Magnitude & Orientation F1 F2 FR

Classification of Textbook Problems • Hibbeler (2007)

References • Hibbeler (2007) • http://wps.prenhall.com/esm_hibbeler_engmech_1

ENGI 1313 Mechanics I