2. Homework Review (Ch. 1 & 2) 1-6 (pg. 15) 1-8 (pg. 15) 2-10 (pg. 28) 2-32 (pg. 39) 2-57 (pg. 42)  Any questions???

3. Quiz (10 minutes, only, calc. OK) 12 5 30° 26 kN 30 kN Determine the magnitude of the resultant force and its direction, measured counterclockwise from the negative x axis.

4. Solution

5. ATTENTION Review 1. What is not true about a unit vector, U A ? A) It is dimensionless. B) Its magnitude is one. C) It always points in the direction of positive X- axis. D) It always points in the direction of vector A. 2. If F = {10 i + 10 j + 10 k} N and G = {20 i + 20 j + 20 k } N, then F + G = { ____ } N A) 10 i + 10 j + 10 k B) 30 i + 20 j + 30 k C) -10 i - 10 j - 10 k D) 30 i + 30 j + 30 k

6. 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. In-Class Activities: Applications / Relevance Write Position Vectors Write a Force Vector Group Problem

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

8. 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 (X A, Y A, Z A ) and ( X B, Y B, Z B ), respectively. The position vector directed from A to B, r AB, is defined as r AB = {( X B – X A ) i + ( Y B – Y A ) j + ( Z B – Z A ) 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!

9. 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, r AB, along two points on that line. b) Find the unit vector describing the line’s direction, u AB = (r AB /r AB ). c) Multiply the unit vector by the magnitude of the force, F = F u AB.

10. EXAMPLE Given: 400 lb force along the cable DA. Find: The force F DA in the Cartesian vector form. Plan: 1. Find the position vector r DA and the unit vector u DA 2. Obtain the force vector as F DA = 400 lb u DA

11. EXAMPLE (continued) The figure shows that when relating D to A, we will have to go -2 ft in the x-direction, -6 ft in the y-direction, and +14 ft in the z-direction. Hence, r DA = {-2 i – 6 j + 14 k} ft. We can also find r DA by subtracting the coordinates of D from the coordinates of A. r DA = (2 2 + 6 2 + 14 2 ) 0.5 = 15.36 ft u DA = r DA /r DA and F DA = 400 u DA lb F DA = 400{(-2 i – 6 j + 14 k)/15.36} lb = {-52.1 i – 156 j + 365 k} lb

12. CONCEPT Review 1. If P and Q are two points in a 3-D space. How are the position vectors r PQ and r QP related? A) r PQ = r QP B) r PQ = - r QP C) r PQ = 1/r QP D) r PQ = 2 r QP

13. 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, F R. 3) Determine the magnitude and the coordinate angles of F R.

14. GROUP PROBLEM SOLVING (continued) F CA = 100 lb{r CA /r CA } F CA = 100 lb(-3 sin 40° i + 3 cos 40° j – 4 k)/5 F CA = {-38.57 i + 45.96 j – 80 k} lb F CB = 81 lb{r CB /r CB } F CB = 81 lb(4 i – 7 j – 4 k)/9 F CB = {36 i – 63 j – 36 k} lb F R = F CA + F CB = {-2.57 i – 17.04 j – 116 k} lb F R = (2.57 2 + 17.04 2 + 116 2 ) = 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°

15. Homework (Due Thurs, 9/9) 2-75 (p. 54) 2-82 (p. 55) 2-89 (p. 64) 2-96 (p. 66) 2-134 (p. 78)