1. Engineering Mechanics: Statics Chapter 2: Force Vectors Chapter 2: Force Vectors

2. Objectives To show how to add forces and resolve them into components using the Parallelogram Law. To express force and position in Cartesian vector form and explain how to determine the vector’s magnitude and direction. To introduce the dot product in order to determine the angle between two vectors or the projection of one vector onto another.

3. Chapter Outline Scalars and Vectors Vector Operations Vector Addition of Forces Addition of a System of Coplanar Forces Cartesian Vectors

4. Chapter Outline Addition and Subtraction of Cartesian Vectors Position Vectors Force Vector Directed along a Line Dot Product

5. 2.1 Scalars and Vectors Scalar – A quantity characterized by a positive or negative number – Indicated by letters in italic such as A Eg: Mass, volume and length

6. 2.1 Scalars and Vectors Vector – A quantity that has both magnitude and direction Eg: Position, force and moment – Represent by a letter with an arrow over it such as or A – Magnitude is designated as or simply A – In this subject, vector is presented as A and its magnitude (positive quantity) as A

7. 2.1 Scalars and Vectors Vector – Represented graphically as an arrow – Length of arrow = Magnitude of Vector – Angle between the reference axis and arrow’s line of action = Direction of Vector – Arrowhead = Sense of Vector

8. 2.1 Scalars and Vectors Example Magnitude of Vector = 4 units Direction of Vector = 20 ° measured counterclockwise from the horizontal axis Sense of Vector = Upward and to the right The point O is called tail of the vector and the point P is called the tip or head

9. 2.2 Vector Operations Multiplication and Division of a Vector by a Scalar - Product of vector A and scalar a = aA - Magnitude = - If a is positive, sense of aA is the same as sense of A - If a is negative sense of aA, it is opposite to the sense of A

10. 2.2 Vector Operations Multiplication and Division of a Vector by a Scalar - Negative of a vector is found by multiplying the vector by ( -1 ) - Law of multiplication applies Eg: A/a = ( 1/a ) A, a≠0

11. 2.2 Vector Operations Vector Addition - Addition of two vectors A and B gives a resultant vector R by the parallelogram law - Result R can be found by triangle construction - Communicative Eg: R = A + B = B + A

12. 2.2 Vector Operations Vector Addition

13. 2.2 Vector Operations Vector Addition - Special case: Vectors A and B are collinear (both have the same line of action)

14. 2.2 Vector Operations Vector Subtraction - Special case of addition Eg: R’ = A – B = A + ( - B ) - Rules of Vector Addition Applies

15. 2.2 Vector Operations Resolution of Vector - Any vector can be resolved into two components by the parallelogram law - The two components A and B are drawn such that they extend from the tail or R to points of intersection

16. 2.3 Vector Addition of Forces When two or more forces are added, successive applications of the parallelogram law is carried out to find the resultant Eg: Forces F 1, F 2 and F 3 acts at a point O - First, find resultant of F 1 + F 2 - Resultant, F R = ( F 1 + F 2 ) + F 3

17. 2.3 Vector Addition of Forces Example F a and F b are forces exerting on the hook. Resultant, F c can be found using the parallelogram law Lines parallel to a and b from the heads of F a and F b are drawn to form a parallelogram Similarly, given F c, F a and F b can be found

18. 2.3 Vector Addition of Forces Procedure for Analysis Parallelogram Law - Make a sketch using the parallelogram law - Two components forces add to form the resultant force - Resultant force is shown by the diagonal of the parallelogram - The components is shown by the sides of the parallelogram

19. 2.3 Vector Addition of Forces Procedure for Analysis Parallelogram Law To resolve a force into components along two axes directed from the tail of the force - Start at the head, constructing lines parallel to the axes - Label all the known and unknown force magnitudes and angles - Identify the two unknown components

20. 2.3 Vector Addition of Forces Procedure for Analysis Trigonometry - Redraw half portion of the parallelogram - Magnitude of the resultant force can be determined by the law of cosines - Direction if the resultant force can be determined by the law of sines

21. 2.3 Vector Addition of Forces Procedure for Analysis Trigonometry - Magnitude of the two components can be determined by the law of sines

22. 2.3 Vector Addition of Forces 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.

23. 2.3 Vector Addition of Forces Solution Parallelogram Law Unknown: magnitude of F R and angle θ

24. 2.3 Vector Addition of Forces Solution Trigonometry Law of Cosines

25. 2.3 Vector Addition of Forces Solution Trigonometry Law of Sines

26. 2.3 Vector Addition of Forces Solution Trigonometry Direction Φ of F R measured from the horizontal

27. 2.3 Vector Addition of Forces Example 2.2 Resolve the 1000 N ( ≈ 100kg) force acting on the pipe into the components in the (a) x and y directions, (b) and (b) x’ and y directions.

28. 2.3 Vector Addition of Forces Solution (a) Parallelogram Law From the vector diagram,

29. 2.3 Vector Addition of Forces Solution (b) Parallelogram Law

30. 2.3 Vector Addition of Forces Solution (b) Law of Sines NOTE: A rough sketch drawn to scale will give some idea of the relative magnitude of the components, as calculated here.

31. 2.3 Vector Addition of Forces Example 2.3 The force F acting on the frame has a magnitude of 500N and is to be resolved into two components acting along the members AB and AC. Determine the angle θ, measured below the horizontal, so that components F AC is directed from A towards C and has a magnitude of 400N.

32. 2.3 Vector Addition of Forces Solution Parallelogram Law

33. 2.3 Vector Addition of Forces Solution Law of Sines

34. 2.3 Vector Addition of Forces Solution By Law of Cosines or Law of Sines Hence, show that F AB has a magnitude of 561N

35. 2.3 Vector Addition of Forces Solution F can be directed at an angle θ above the horizontal to produce the component F AC. Hence, show that θ = 16.1 ° and F AB = 161N

36. 2.3 Vector Addition of Forces Example 2.4 The ring is subjected to two forces F 1 and F 2. If it is required that the resultant force have a magnitude of 1kN and be directed vertically downward, determine (a) magnitude of F 1 and F 2 provided θ = 30 °, and (b) the magnitudes of F 1 and F 2 if F2 is to be a minimum.

37. 2.3 Vector Addition of Forces Solution (a) Parallelogram Law Unknown: Forces F 1 and F 2 View Free Body Diagram

38. 2.3 Vector Addition of Forces Solution Law of Sines

39. 2.3 Vector Addition of Forces Solution (b) Minimum length of F 2 occur when its line of action is perpendicular to F 1. Hence when F 2 is a minimum

40. 2.3 Vector Addition of Forces Solution (b) From the vector diagram

41. 2.4 Addition of a System of Coplanar Forces For resultant of two or more forces: Find the components of the forces in the specified axes Add them algebraically Form the resultant In this subject, we resolve each force into rectangular forces along the x and y axes.

42. 2.4 Addition of a System of Coplanar Forces Scalar Notation - x and y axes are designated positive and negative - Components of forces expressed as algebraic scalars Eg: Sense of direction along positive x and y axes

43. 2.4 Addition of a System of Coplanar Forces Scalar Notation Eg: Sense of direction along positive x and negative y axes

44. 2.4 Addition of a System of Coplanar Forces Scalar Notation - Head of a vector arrow = sense of the vector graphically (algebraic signs not used) - Vectors are designated using boldface notations - Magnitudes (always a positive quantity) are designated using italic symbols

45. 2.4 Addition of a System of Coplanar Forces Cartesian Vector Notation - Cartesian unit vectors i and j are used to designate the x and y directions - Unit vectors i and j have dimensionless magnitude of unity ( = 1 ) - Their sense are indicated by a positive or negative sign (pointing in the positive or negative x or y axis) - Magnitude is always a positive quantity, represented by scalars F x and F y

46. 2.4 Addition of a System of Coplanar Forces Cartesian Vector Notation F = F x i + F y jF’ = F’ x i + F’y(-j) F’ = F’ x i – F’ y j

47. 2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants To determine resultant of several coplanar forces: - Resolve force into x and y components - Addition of the respective components using scalar algebra - Resultant force is found using the parallelogram law

48. 2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants Example: Consider three coplanar forces Cartesian vector notation F 1 = F 1x i + F 1y j F 2 = - F 2x i + F 2y j F 3 = F 3x i – F 3y j

49. Coplanar Force Resultants Vector resultant is therefore F R = F 1 + F 2 + F 3 = F 1x i + F 1y j - F 2x i + F 2y j + F 3x i – F 3y j = (F 1x - F 2x + F 3x )i + (F 1y + F 2y – F 3y )j = (F Rx )i + (F Ry )j 2.4 Addition of a System of Coplanar Forces

50. Coplanar Force Resultants If scalar notation are used F Rx = (F 1x - F 2x + F 3x ) F Ry = (F 1y + F 2y – F 3y ) In all cases, F Rx = ∑F x F Ry = ∑F y *Take note of sign conventions

51. 2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants - Positive scalars = sense of direction along the positive coordinate axes - Negative scalars = sense of direction along the negative coordinate axes - Magnitude of F R can be found by Pythagorean Theorem

52. 2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants - Direction angle θ (orientation of the force) can be found by trigonometry

53. 2.4 Addition of a System of Coplanar Forces 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

54. 2.4 Addition of a System of Coplanar Forces Solution Scalar Notation Hence, from the slope triangle

55. 2.4 Addition of a System of Coplanar Forces Solution Alt, by similar triangles Similarly,

56. 2.4 Addition of a System of Coplanar Forces Solution Scalar Notation Cartesian Vector Notation F 1 = {-100i +173j }N F 2 = {240i -100j }N

57. 2.4 Addition of a System of Coplanar Forces Example 2.6 The link is subjected to two forces F 1 and F 2. Determine the magnitude and orientation of the resultant force.

58. 2.4 Addition of a System of Coplanar Forces Solution Scalar Notation

59. 2.4 Addition of a System of Coplanar Forces Solution Resultant Force From vector addition, Direction angle θ is

60. 2.4 Addition of a System of Coplanar Forces Solution Cartesian Vector Notation F 1 = { 600cos30°i + 600sin30 ° j } N F 2 = { -400sin45°i + 400cos45 ° j } N Thus, F R = F 1 + F 2 = (600cos30°N - 400sin45°N)i + (600sin30 ° N + 400cos45 ° N)j = {236.8i + 582.8j}N

61. 2.4 Addition of a System of Coplanar Forces Example 2.7 The end of the boom O is subjected to three concurrent and coplanar forces. Determine the magnitude and orientation of the resultant force.

62. 2.4 Addition of a System of Coplanar Forces Solution Scalar Notation View Free Body Diagram

63. Solution Resultant Force From vector addition, Direction angle θ is 2.4 Addition of a System of Coplanar Forces

64. 2.5 Cartesian Vectors Right-Handed Coordinate System A rectangular or Cartesian coordinate system is said to be right-handed provided: - Thumb of right hand points in the direction of the positive z axis when the right-hand fingers are curled about this axis and directed from the positive x towards the positive y axis

65. 2.5 Cartesian Vectors Right-Handed Coordinate System - z-axis for the 2D problem would be perpendicular, directed out of the page.

66. 2.5 Cartesian Vectors Rectangular Components of a Vector - A vector A may have one, two or three rectangular components along the x, y and z axes, depending on orientation - By two successive application of the parallelogram law A = A’ + A z A’ = A x + A y - Combing the equations, A can be expressed as A = A x + A y + A z

67. 2.5 Cartesian Vectors Unit Vector - Direction of A can be specified using a unit vector - Unit vector has a magnitude of 1 - If A is a vector having a magnitude of A ≠ 0, unit vector having the same direction as A is expressed by u A = A / A So that A = A u A

68. 2.5 Cartesian Vectors Unit Vector - Since A is of a certain type, like force vector, a proper set of units are used for the description - Magnitude A has the same sets of units, hence unit vector is dimensionless - A ( a positive scalar) defines magnitude of A - u A defines the direction and sense of A

69. 2.5 Cartesian Vectors Cartesian Unit Vectors - Cartesian unit vectors, i, j and k are used to designate the directions of z, y and z axes - Sense (or arrowhead) of these vectors are described by a plus or minus sign (depending on pointing towards the positive or negative axes)

70. 2.5 Cartesian Vectors Cartesian Vector Representations - Three components of A act in the positive i, j and k directions A = A x i + A y j + A Z k *Note the magnitude and direction of each components are separated, easing vector algebraic operations.

71. 2.5 Cartesian Vectors Magnitude of a Cartesian Vector - From the colored triangle, - From the shaded triangle, - Combining the equations gives magnitude of A

72. 2.5 Cartesian Vectors Direction of a Cartesian Vector - Orientation of A is defined as the coordinate direction angles α, β and γ measured between the tail of A and the positive x, y and z axes - 0 ° ≤ α, β and γ ≤ 180 °

73. 2.5 Cartesian Vectors Direction of a Cartesian Vector - For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A

74. 2.5 Cartesian Vectors Direction of a Cartesian Vector - For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A

75. 2.5 Cartesian Vectors Direction of a Cartesian Vector - For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A

76. 2.5 Cartesian Vectors Direction of a Cartesian Vector - Angles α, β and γ can be determined by the inverse cosines - Given A = A x i + A y j + A Z k - then, u A = A /A = (A x /A)i + (A y /A)j + (A Z /A)k where

77. 2.5 Cartesian Vectors Direction of a Cartesian Vector - u A can also be expressed as u A = cosαi + cosβj + cosγk - Since and magnitude of u A = 1, - A as expressed in Cartesian vector form A = Au A = Acosαi + Acosβj + Acosγk = A x i + A y j + A Z k