1. S CALAR & V ECTOR By. Engr. Jamal uddin 1

2. S CALAR Aquantityissaidtobescalarifitiscompletely defined by its magnitude alone A quantity characterized by a positive or negative number Indicated by letters in italic such as A e.g.Mass, length,volume,temperature andtime etc are scalar 2 CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

3. V ECTORS A quantity that has magnitude and direction. e.g. Displacement, Velocity, Acceleration, Momentum & Force. Represent by a letter with an arrow over it, Magnitude is designated as In this subject, vector is presented as A and its magnitude (positive quantity) as A AA AA 3 CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

4. C OMPONENTS OF A V ECTOR A Head Tail Vector Designation 500 N Vector Magnitude 4 CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

5. V ECTOR O PERATIONS Multiplication and Division of a Vector by a Scalar - Product of vector A and scalar a = a A - Magnitude = - Law of multiplication applies e.g. A /a = ( 1/a ) A, a≠0 aA 5 CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

6. Vector Addition The addition of two vectors results in a resultant vector ( P + Q = R ) where R is a vector pointing from the starting point of P to the ending point of Q. Resultant Vector 6 V ECTOR O PERATIONS CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

7. V ECTOR O PERATIONS Vector Subtraction - Special case of addition e.g. R ’ = A – B = A + ( - B ) - Rules of Vector Addition Applies 7 CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

8. P ARALLELOGRAM L AW By drawing construction lines parallel to the vectors, the resultant vector goes from the point of origin to the intersection of the construction lines 8 CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

9. T RIANGLE M ETHOD Place the tail of B to the head of A. The Resultant ( R )canbefoundby of connectingtheTail resultantwithtailof A and Head of resultant to theHeadof B. formsthethird This vector knownasthe vector. resultant A B A + BA + B 9 Resultant CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

10. S UCCESSIVE T RIANGLE M ETHOD If there are multiple vectors to be added together, add the first two vectors to find the first resultant. Once the first Resultant ( R 1 ) is found, add the next vector to the resultant to find ( R 2 ). Can be repeated as many times as necessary to add all the vectors (it also does not matter what order they are added in, the end resultant will be the same). Addition of three or more vectors through repeated application of the triangle rule 10 CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

11. P OLYGON M ETHOD  Polygon method is similar to the Successive Triangle Method but no intermediate resultants are calculated 11 Graphically measure length and direction of R!! CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

12. 2.3 V ECTOR A DDITION OF F ORCES Finding a Resultant Force Parallelogram law is carried out to find the resultant force Resultant, F R = ( F 1 + F 2 ) 12 CHAPTER 2 Engineering MechanicsEngr. Mahesh Kumar

13. 2.3 V ECTOR A DDITION OF F ORCES Procedure for Analysis Parallelogram Law Make a sketch using the parallelogram law 2 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 13 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

14. 2.3 V ECTOR A DDITION OF F ORCES 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 Magnitude of the two components can be determined by the law of sines 14 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

15. E XAMPLE 2.1 The screw eye is subjected to two forces, F 1 and F 2. Determine the magnitude and direction of the resultant force. 15 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

16. S OLUTION Parallelogram Law Unknown: magnitude of F R and angle θ 16 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

17. S OLUTION Trigonometry Law of Cosines Law of Sines 150 N  212.6 N F   100N  2   150N  2  2  100N  150N  cos115  R  10000  22500  30000   0.4226   212.6N  213N  0.9063    39.8  39.8 sin115  212.6N 150N sin  sin   17 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

18. S OLUTION Trigonometry Direction Φ of F R measuredfrom the horizontal   39.8   15   54.8    18 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

19. E XAMPLE 2.2 Resolve the horizontal 600-lb force in Fig. 2–12 a into components acting along the u and v axes and determine the magnitudes of these components. 19 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

20. S OLUTION 20 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

21. S OLUTION 21 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

22. H OME W ORK Example 2.3 Fundamental Problems 2.1,2.2, 2.3 Exercise Problem 2.1 22 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

23. 2.4 A DDITION OF A S YSTEM OF C OPLANAR F ORCES Scalar Notation x and y axes are designated positive and negative Components of forces expressed as algebraic scalars F  F x  F y F x  F cos  andF y  F sin  23 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

24. 2.4 A DDITION OF A S YSTEM OF C OPLANAR F ORCES 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 ) Magnitude is always a positive quantity, represented by scalars F x and F y F  F x i  F y j CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

25. 2.4 A DDITION OF A S YSTEM OF C OPLANAR F ORCES 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 Cartesian vector notation: F 1  F 1x i  F 1y j F 2   F 2 x i  F 2 y j F 3  F 3x i  F 3 y j 25 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

26. 2.4 A DDITION OF A S YSTEM OF C OPLANAR F ORCES Coplanar Force Resultants Vector resultant is therefore F R  F 1  F 2  F 3 If scalar notation are used  Fi  Fj Fi  Fj RxRy  F1x  F2 x  F3x F1y  F2 y  F3 y F1x  F2 x  F3x F1y  F2 y  F3 y FRxFRyFRxFRy 26 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

27. 2.4 A DDITION OF A S YSTEM OF C OPLANAR F ORCES Coplanar Force Resultants In all cases we have Magnitude of F R can be found by Pythagorean Theorem   Fx  Fy  Fx  Fy FRxFRyFRxFRy Rx F  F 2 and   tan -1 Ry RyRxR F F 2F 2 FF * Take note of sign conventions CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

28. E XAMPLE 2.5 Determine x and y components of F 1 and F 2 acting on the boom. Express each force as a Cartesian vector. 28 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

29. S OLUTION Scalar Notation Hence, from the slope triangle, we have  200 cos 30N  173N  173N    200sin 30N   100N  100N  F1xF1yF1xF1y    12 12   5  5    tan1  tan1 29 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

30. S OLUTION By similar triangles we have Scalar Notation: F Cartesian Vector Notation: F    100i  173 j  N F  260  5   100N 13 13  F  260  12   240N 13 13  2 y2 y 2 x2 x      240N    100N  100N  F2 yF2 y 2 x2 x F   240i  100 j  N 2 1 30 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

31. E XAMPLE 2.6 The link is subjected to two forces F 1 and F 2. Determine the magnitude and orientation of the resultant force. 31 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

32. S OLUTION I Scalar Notation:  582.8N   600 sin 30  N  400 cos 45  N  Fy : Fy :  236.8N   600 cos 30  N  400 sin 45  N  Fx : Fx : FRyFRyFRyFRy FRxFRxFRxFRx 32 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

33. S OLUTION I Resultant Force R  629N From vector addition, direction angle θ is F   236.8N  2   582.8N  2  67.9   236.8N  582.8N   tan  tan 1      33 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

34. S OLUTION II Cartesian Vector Notation F 1 = { 600cos30 ° i + 600sin30 ° j } N F 2 = { -400sin45 ° i + 400cos45 ° j } N Thus, F R = F 1 + F 2 = (600cos30º - 400sin45º) i + (600sin30 º + 400cos45 º ) j = {236.8 i + 582.8 j }N The magnitude and direction of F R are determined i 3 n 4 the same manner as before. CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

35. E XERCISE 2-35 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. 35 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

36. ௫ = 00 sin ͵ 0 = 200 N ଶ௫ = ʹ 0 cos− = 176.78 N ௬ = 00 cos ͵ 0 = 200 N ଶ௬ = ʹ 0 sin− = -176.78 N 36 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

37. 37 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

38. H OME W ORK Example 2.7 Fundamental 2-7 to 2-9 Exercise 2.33 and 2.35 38

39. 2.5 C ARTESIAN V ECTORS Right-Handed Coordinate System A Cartesian coordinate system is often used to solve problems in 3 dimensions (3D). The coordinate system is right-handed : The thumb of the right hand points in the direction of the positive z axis when the right hand fingers curled about this axis and directed from the positive x towards the positive y axis. 39 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

40. 2.5 C ARTESIAN V ECTORS 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 40 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

41. 2.5 C ARTESIAN V ECTORS Unit Vector In 3D, the set of Cartesian unit vectors, i, j, k, is used to designate the directionsof the x, y, z axes, respectively. 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 A = A u A 41 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar that

42. 2.5 C ARTESIAN V ECTORS Cartesian Vector Representations Any vector A with scalar components A x, A y and A z can be written in the Cartesian vector form as: A = A x i + A y j + A Z k *Note themagnitude and direction of each components are separated, easing vector algebraic operations. 42 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

43. 2.5 C ARTESIAN V ECTORS Magnitude of a Cartesian Vector Combining the equations gives magnitude of A 2z2z 22 y AxAAxAA A A  From the colored triangle, A  From the shaded triangle, A'  A 2  A 2 xy A'2  A2A'2  A2 z 43 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

44. 2.5 C ARTESIAN V ECTORS Direction of a Cartesian Vector The direction of vector A is defined by the angles ǂ, ǃ, andmeasured between the tail of A and the positive x, y, z axes located at the tail of A. The angles ǂ, ǃ, and  are found from their direction cosines: A cos  Axcos   Azcos  Axcos   Az A A y cos   A cos 2   cos 2   cos 2   1 This means that only two of the angles ǂ, ǃ, and  have to be specified, the third can be found CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar 44

45. 2.5 C ARTESIAN V ECTORS Direction of a Cartesian Vector 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 u A is a unit vector in the direction of A 45 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

46. 2.5 C ARTESIAN V ECTORS Direction of a Cartesian Vector u A can also be expressed as u A = cos ǂ i + cos ǃ j + cos  k Sinceand u A = 1, we have cos 2   cos 2   cos 2   1 A as expressed in Cartesian vector form is A = A u A = A cos ǂ i + A cos ǃ j + A cos  k = A x i + A y j + A Z k 2z2z2 y A Ax  A AA Ax  A A 46 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

47. 2.6 A DDITION AND S UBTRACTION OF C ARTESIAN V ECTORS Theadditionorsubtractionoftwoor moremore vectors are simplified if the vectors are expressed in terms of their Cartesian components. To find the resultant of a concurrent force system, express each force as a Cartesian vector i, j, k components of all forces in the sy and add all the stem. CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

48. 48 F R = ∑ F = ∑ F x i + ∑ F y j + ∑ F z k 2.6 A DDITION AND S UBTRACTION OF C ARTESIAN V ECTORS

49. E XAMPLE 2.8 Express the force F as Cartesian vector. 49 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

50. S OLUTION Since two angles are specified, the third angle is found by cos   1   0.5  2   0.707  2   cos  1  0.5   60  Given F = 200N F = Fcos ǂ i + Fcos ǃ j + Fcos DŽ k = (200cos60ºN) i + (200cos60ºN) j + (200cos45ºN) k = {100.0 i + 100.0 j + 141.4 k }N cos 2   cos 2   cos 2   1 cos 2   cos 2 60   cos 2 45   0.5 0.5  1 1 50 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

51. S OLUTION Checking: F    100.0  2   100.0  2   141.4  2  200N Fx F FFx F F 2z2z y 2 51 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

52. E XAMPLE 2.11 Two forces act on the hook shown in Fig. 2–33 a. Specify themagnitude of F 2 anditscoordinatedirection angles so that the resultant force F R acts along the positive y axis and has a magnitude of 800 N. 52 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

53. S OLUTION 53 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

54. S OLUTION in Fig. 2–33 b, it is necessary that F R = F 1 + F 2 54 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

55. S OLUTION 55 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

56. S OLUTION 56 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

57. F UNDAMENTAL 2-14 Express the force as a Cartesian vector. 57 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

58. S OLUTION u A = cos ǂ i + cos ǃ j + cos  k u A = cos(135) i + cos(135) j + cos(60) k u A = 58 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

59. HOMEWORK Example 2.9 Fundamental 2-15 Exercise 2.63 to 2.65, 2.70 59 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

60. 2.7 P OSITION V ECTORS The position vector r is defined as a fixed vector which locates a point in space relative to another point. If r extends from the origin of coordinates O to a point P ( x, y, z ) then: r = x i + y j + z k 60 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

61. 2.7 P OSITION V ECTORS More generally, the position vector may be directed from point A to point B in space. The vector position is denoted by r (or r AB ). by the head-to-tail vector addition we have: r A + r = r B 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 61 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

62. 2.8 F ORCE V ECTOR D IRECTED ALONG A L INE In 3D problems, direction of F is specified by 2 points, through which its line of action lies F can be formulated as a Cartesian vector F = F u = F ( r / r ) Note that F has units of forces ( unlike r, with units of length (m) N) 62 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

63. 2.8 F ORCE V ECTOR D IRECTED ALONG A L INE Force F acting along the chain can be presented as a Cartesian vector by -Establish x, y, z axes -Form a position vector r along length of chain Unit vector, u = r / r that defines the direction of both the chain and the forc We get F = F u e 63 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

64. F UNDAMENTAL (2-20) Determine the length of the rod and the position vector directed from A to B. What is the angle θ ? 64 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

65. S OLUTION r A =4i+0j+0k r B =0i+2j+4k r AB = -4i+2j+4k  = 180-131.8 = 48.2 65  r    4  2   2  2   4m  2  6m AB = cos − −4 = 131.8 6 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

66. E XAMPLE 2.13 The man pulls on the cord with a force of 350N. Represent this force acting on the support A, as a Cartesian vector and determine its direction. 66 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

67. S OLUTION End points of the cord are A (0m, 0m, 7.5m) and B (3m, -2m, 1.5m) r = (3m – 0m) i + (-2m – 0m) j + (1.5m – 7.5m) k = {3 i – 2 j – 6 k }m Magnitude = length of cord AB r   3m  2    2m  2    6m  2  7m Unit vector, u = r / r = 3/7 i - 2/7 j - 6/7 k 67 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

68. S OLUTION Force F has a magnitude of 350N, direction specified by u. F = F u = 350N(3/7 i - 2/7 j - 6/7 k ) = {150 i - 100 j - 300 k } N ǂ = cos -1 (3/7) = 64.6 ° ǃ = cos -1 (-2/7) = 107 ° DŽ = cos -1 (- 6/7) = 149 ° 68 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

69. E XAMPLE 2.12 An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B. 69 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

70. S OLUTION Position vector r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m Magnitude = length of the rubber band r    3  2   2  2   6  2  7m Unit vector in the director of r u = r / r = -3/7 i + 2/7 j + 6/7 k 70 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

71. S OLUTION ǂ = cos -1 (-3/7) = 115 ° ǃ = cos -1 (2/7) = 73.4 ° DŽ = cos -1 (6/7) = 31.0 ° 71 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar

72. HOMEWORK Example 2.15 Fundamental 2-19,2.23 Exercise 2.90 72 CHAPTER 2Engineering Mechanics Engr. Mahesh Kumar