ENGR-1100 Introduction to Engineering Analysis. Lecture 3

Transcription

1 ENGR-1100 Introduction to Engineering Analysis Lecture 3

2 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 along a line

3 DOT PRODUCT Today s Objective: Students will be able to use the vector dot product to: a) determine an angle between two vectors, and, b) determine the projection of a vector along a specified line. In-Class Activities: Applications / Relevance Dot product - Definition Angle Determination Determining the Projection

4 APPLICATIONS This ship s mooring line, connected to the bow, can be represented as a Cartesian vector. What are the forces in the mooring line and how do we find their directions? Why would we want to know these things?

5 APPLICATIONS (continued) This awning is held up by three chains. What are the forces in the chains and how do we find their directions? Why would we want to know these things?

6 A position vector is defined as a fixed vector that locates a point in space relative to another point. POSITION VECTOR 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.

7 POSITION VECTOR 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. ALWAYS subtract the tail coordinates from the tip coordinates!

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 s 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.

9 EXAMPLE Given: The 420 N force along the cable AC. Find: The force F AC in the Cartesian vector form. Plan: 1. Find the position vector r AC and its unit vector u AC. 2. Obtain the force vector as F AC = 420 N u AC.

10 EXAMPLE (continued) r AC = { (-6) 2 } 1/2 = 7 m (We can also find r AC by subtracting the coordinates of A from the coordinates of C.) Now u AC = r AC /r AC and F AC = 420 u AC = 420 (r AC /r AC ) So F AC = 420{ (2 i + 3 j 6 k) / 7 } N = {120 i j k } N As per the figure, when relating A to C, we will have to go 2 m in the x- direction, 3 m in the y-direction, and -6 m in the z-direction. Hence, r AC = {2 i + 3 j 6 k} m.

11 Plan: GROUP PROBLEM SOLVING Given: Two forces are acting on a flag pole as shown in the figure. F B = 700 N and F C = 560 N Find: The magnitude and the coordinate direction angles of the resultant force. 1) Find the forces along AB and AC 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.

12 GROUP PROBLEM SOLVING (continued) r AB = {2 i 3 j 6 k} m r AC = {3 i + 2 j 6 k} m r AB = {2 2 + (-3) 2 + (-6) 2 } 1/2 = 7 m r AC = { (-6) 2 } 1/2 = 7 m F AB = 700 (r AB / r AB ) N F AB = 700 ( 2 i 3 j 6 k) / 7 N F AB = (200 i 300 j 600 k) N F AC = 560 (r AC / r AC ) N F AC = 560 (3 i + 2 j 6 k) / 7 N F AC = {240 i j 480 k} N

13 GROUP PROBLEM SOLVING (continued) F R = F AB + F AC = {440 i 140 j 1080 k} N F R = { (-140) 2 + (-1080) 2 } 1/2 = N F R = 1175 N = cos -1 (440/1174.6) = 68.0 = cos -1 ( 140/1174.6) = 96.8 = cos -1 ( 1080/1174.6) = 156.8

14 APPLICATIONS If you know the physical locations of the four cable ends, how could you calculate the angle between the cables at the common anchor?

15 APPLICATIONS (continued) For the force F being applied to the wrench at Point A, what component of it actually helps turn the bolt (i.e., the force component acting perpendicular to arm AB of the pipe)?

16 DEFINITION The dot product of vectors A and B is defined as A B = A B cos. The angle is the smallest angle between the two vectors and is always in a range of 0 º to 180 º. Dot Product Characteristics: 1. The result of the dot product is a scalar (a positive or negative number). 2. The units of the dot product will be the product of the units of the A and B vectors.

17 DOT PRODUCT DEFINITON (continued) Examples: By definition, i j = 0 i i = 1 A B = (A x i + A y j + A z k) (B x i + B y j + B z k) = A x B x + A y B y + A z B z

18 USING THE DOT PRODUCT TO DETERMINE THE ANGLE BETWEEN TWO VECTORS For these two vectors in Cartesian form, one can find the angle by a) Find the dot product, A B = (A x B x + A y B y + A z B z ), b) Find the magnitudes (A & B) of the vectors A & B, and c) Use the definition of dot product and solve for, i.e., = cos -1 [(A B)/(A B)], where 0 º 180 º.

19 DETERMINING THE PROJECTION OF A VECTOR You can determine the components of a vector parallel and perpendicular to a line using the dot product. Steps: 1. Find the unit vector, u aa along line aa 2. Find the scalar projection of A along line aa by A = A u aa = A x U x + A y U y + A z U z

20 DETERMINING THE PROJECTION OF A VECTOR (continued) 3. If needed, the projection can be written as a vector, A, by using the unit vector u aa and the magnitude found in step 2. A = A u aa 4. The scalar and vector forms of the perpendicular component can easily be obtained by A = (A 2 - A 2 ) ½ and A = A A (rearranging the vector sum of A = A + A )

21 EXAMPLE Given: The force acting on the hook at point A. Find: The angle between the force vector and the line AO, and the magnitude of the projection of the force along the line AO. Plan: 1. Find r AO 2. Find the angle = cos -1 {(F r AO )/(F r AO )} 3. Find the projection via F AO = F u AO (or F cos )

22 EXAMPLE (continued) r AO = { 1 i + 2 j 2 k} m r AO = {(-1) (-2) 2 } 1/2 = 3 m F = { 6 i + 9 j + 3 k} kn F = {(-6) } 1/2 = kn F r AO = ( 6)( 1) + (9)(2) + (3)( 2) = 18 kn m = cos -1 {(F r AO )/(F r AO )} = cos -1 {18 / ( )} = 57.67

23 EXAMPLE (continued) u AO = r AO /r AO = ( 1/3) i + (2/3) j + ( 2/3) k F AO = F u AO = ( 6)( 1/3) + (9)(2/3) + (3)( 2/3) = 6.00 kn Or: F AO = F cos = cos (57.67 ) = 6.00 kn

24 ATTENTION QUIZ 1. The dot product can be used to find all of the following except. A) sum of two vectors B) angle between two vectors C) component of a vector parallel to another line D) component of a vector perpendicular to another line 2. Find the dot product of the two vectors P and Q. P = {5 i + 2 j + 3 k} m Q = {-2 i + 5 j + 4 k} m A) -12 m B) 12 m C) 12 m 2 D) -12 m 2 E) 10 m 2

25 GROUP PROBLEM SOLVING Given: The 300 N force acting on the bracket. Find: The magnitude of the projected component of this force acting along line OA Plan: 1. Find r OA and u OA 2. Find the angle = cos -1 {(F r OA )/(F r OA )} 3. Then find the projection via F OA = F u OA or F (1) cos

26 GROUP PROBLEM SOLVING (continued) F r OA = { i j k} m r OA = {( 0.450) } 1/2 = 0.60 m u OA = r OA / r OA = {-0.75 i j k }

27 GROUP PROBLEM SOLVING (continued) F F = 300 sin 30 = 150 N F = { 150 sin 30 i cos 30 j cos 30 k} N F = { 75 i j k} N F = {(-75) } 1/2 = 300 N

28 GROUP PROBLEM SOLVING (continued) F r OA = (-75) (-0.45) + (259.8) (0.30) + (129.9) (0.26) = N m = cos -1 {(F r OA )/(F r OA )} = cos -1 {145.5 /( )} = 36.1 The magnitude of the projected component of F along line OA will be F OA = F u OA = (-75)(-0.75) + (259.8) (0.50) + (129.9) (0.433) Or = 242 N F OA = F cos = 300 cos 36.1 = 242 N