1. 1 Chapter Objectives Parallelogram Law Cartesian vector form Dot product

2. 2 Chapter Outline 1.Scalars and Vectors 2.Vector Operations 3.Vector Addition of Forces 4.Addition of a System of Coplanar Forces 5.Cartesian Vectors 6.Addition and Subtraction of Cartesian Vectors 7.Position Vectors 8.Dot Product

3. 3 2.1 Scalars and Vectors Scalar – A quantity characterized by a positive or negative number – Indicated by letters in italic such as A e.g.

4. 4 2.1 Scalars and Vectors Vector – A quantity that has magnitude and direction e.g. – Vector – Magnitude

5. 5 2.2 Vector Operations Multiplication and Division of a Vector by a Scalar - Product of vector A and scalar a - Magnitude = - Law of multiplication applies e.g. A/a = ( 1/a ) A, a≠0

6. Vector Addition - R = A + B = B + A - collinear 6 2.2 Vector Operations

7. 7 Vector Subtraction - Special case of addition e.g. R’ = A – B = A + ( - B )

8. 8 2.3 Vector Addition of Forces Parallelogram law Resultant, F R = ( F 1 + F 2 )

9. 9 2.3 Vector Addition of Forces Trigonometry –law of cosines –law of sines

10. 10 Example 2.1 The screw eye is subjected to two forces, F 1 and F 2. Determine the magnitude and direction of the resultant force.

11. 11 Solution Parallelogram Law Unknown: magnitude of F R and angle θ

12. 12 Solution

13. 13 Solution Trigonometry Direction Φ of F R measured from the horizontal

14. 14 2.4 Addition of a System of Coplanar Forces Scalar Notation

15. 15 2.4 Addition of a System of Coplanar Forces Cartesian Vector Notation –use i and j for x and y direction –The magnitude of i and j is one

16. 16 2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants – Find componenets in x and y –Add in each direction –Resultant is from parallelogram –Cartesian vector notation:

17. 17 2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants

18. 18 2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants –We can show that –Magnitude of F R from Pythagorus

19. 19 Example 2.5 Determine x and y components of F 1 and F 2 acting on the boom. Express each force as a Cartesian vector.

20. 20 Solution Scalar Notation Cartesian Vector Notation

21. 21 Solution By similar triangles we have Scalar Notation: Cartesian Vector Notation:

22. 22 Example 2.6 The link is subjected to two forces F 1 and F 2. Determine the magnitude and orientation of the resultant force.

23. 23 Solution I Scalar Notation:

24. 24 Solution I Resultant Force From vector addition, direction angle θ is

25. 25 Solution II Cartesian Vector Notation Thus,

26. 26 2.5 Cartesian Vectors (3D) Right-Handed Coordinate System - thumb represents z – the rest, sweeping from x to y

27. 27 2.5 Cartesian Vectors Unit Vector –Vector A can be described by a unit vector –u A = A / A A = A u A Unit vector for x, y, z

28. 28 2.5 Cartesian Vectors Cartesian Vector Representations –A can be written by i, j and k directions

29. 29 2.5 Cartesian Vectors Direction of a Cartesian Vector –The direction of A is defined by α, β and γ angle between A and x, y and z –0° ≤ α, β and γ ≤ 180 ° –The direction cosines of A is

30. 30 2.5 Cartesian Vectors Direction of a Cartesian Vector A = Au A = Acosαi + Acosβj + Acosγk = A x i + A y j + A Z k

31. 31 Example 2.8 Express the force F as Cartesian vector.

32. 32 Solution

33. 33 Solution Notice, α = 60 º since F x is in +x From F = 200N

34. 34 2.7 Position Vectors x,y,z Coordinates –Right-handed coordinate system –O is a reference

35. 35 2.7 Position Vectors Position Vector –Position vector r is a vector to identify a location of a point relative to other points –E.g. r = xi + yj + zk

36. 36 2.7 Position Vectors Position Vector (between 2 points) –Vector addition r A + r = r B –Solving r = r B – r A = (x B – x A )i + (y B – y A )j + (z B –z A )k or r = (x B – x A )i + (y B – y A )j + (z B –z A )k

37. 37 Example 2.12 An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B. A (1, 0, -3) m B (-2, 2, 3) m

38. 38 Solution Position vector Magnitude = length of the rubber band Unit vector in the director of r

39. 39 Solution

40. 40 2.9 Dot Product Dot product of A and B can be written as A·B A·B = AB cosθ where 0°≤ θ ≤180° The result is scalar

41. 41 2.9 Dot Product Laws of Operation 1. Commutative law A·B = B·A 2. Multiplication by a scalar a(A·B) = (aA)·B = A·(aB) = (A·B)a 3. Distribution law A·(B + D) = (A·B) + (A·D)

42. 42 2.9 Dot Product Cartesian Vector Formulation - Dot product of Cartesian unit vectors i·i = (1)(1)cos0° = 1 i·j = (1)(1)cos90° = 0 - Similarly i·i = 1j·j = 1k·k = 1 i·j = 0i·k = 0j·k = 0

43. 43 2.9 Dot Product Cartesian Vector Formulation –Dot product of 2 vectors A and B A·B = A x B x + A y B y + A z B z Dot product can be used for –Finding angles between two vectors θ = cos -1 [(A·B)/(AB)] 0°≤ θ ≤180° –Finding a vector on the direction of a unit vecotr A a = A cos θ = A·u

44. 44 Example 2.17 The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB. A (0, 0, 0) B (2, 6, 3)

45. 45 Solution Since Thus

46. 46 Solution Since result is a positive scalar, F AB has the same sense of direction as u B. Express in Cartesian form Perpendicular component

47. 47 Solution Magnitude can be determined from F ┴ or from Pythagorean Theorem, or

48. 48 4.2 Cross Product Cross product of A and B C = A X B C = AB sinθ

49. 49 4.2 Cross Product C is perpendicular to the plane containing A and B C = A X B = (AB sin θ)u C u C is a unit vector

50. 50 4.2 Cross Product Laws of Operations 1. Commutative law A X B ≠ B X A But ่ A X B = - (B X A) Cross product B X A B X A = -C

51. 51 4.2 Cross Product Laws of Operations 2. Multiplication by a Scalar a( A X B ) = (aA) X B = A X (aB) = ( A X B )a 3. Distributive Law A X (B + D) = (A X B) + (A X D) And (B + D) X A = (B X A) + (D X A)

52. 52 4.2 Cross Product Cartesian Vector Formulation i jk

53. 53 4.2 Cross Product Cartesian Vector Formulation A more compact determinant in the form as

54. 54 Example 4.4 Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector. A (0, 5, 0) B (4, 5, -2)

55. 55 Solution Position vectors are directed from point O to each force as shown. These vectors are The resultant moment about O is