Learning Objective

2. Learning Objective If an object is subjected to several forces that have different magnitudes and act in different directions, how can the magnitude and direction of the resulting total force on the object be determined? This chapter reviews vector operations, express vectors in terms of components, and present examples of engineering applications of vectors.

3. Chapter Outline Scalars & Vectors Components in Two Dimensions Components in Three Dimensions Dot Products Cross Products

4. Scalars & Vectors • Scalar – a physical quantity that is completely described by a real number • E.g. Time, mass • Vector – both magnitude (nonnegative real number) & direction • E.g. Position of a point in space relative to another point, forces • Represented by boldfaced letters: U, V, W, … • Magnitude of vector U = |U|

5. Scalars & Vectors • Graphical representation of vectors: arrows • Direction of arrow = direction of vector • Length of arrow magnitude of vector • Example: • rAB = position of point B relative to point A • Direction of rAB =direction from point A to point B

6. Scalars & Vectors • |rAB| = distance between 2 points

7. Scalars & Vectors • Vector Addition: • When an object undergoes a displacement (moves from 1 location in space to another) • Displacement vector: U • Direction of U =direction of displacement • |U| = distance the book moves

8. Scalars & Vectors • 2nd displacement V • Final position of book is the same whether we give it displacement U then V, or vice versa • U and V equivalent to a single displacement W:U + V = W

9. Scalars & Vectors • Definition of Vector Addition: • Vector from tail of U to head of V • Triangle rule • Sum is independent of the order in which the vectors are placed head to tail • Parallelogram rule

10. Scalars & Vectors • Vector addition is commutative: U + V = V + U (2.1) • Vector addition is associative: (U + V) + W = U + (V + W) (2.2) • If the sum of 2 or more vectors = 0, they form a closed polygon

11. Scalars & Vectors • Example: • Vector rAC from A to C is the sum of rAB & rBC

12. Scalars & Vectors • Product of a Scalar & a Vector: • Product of scalar (real number) a & vector U = vector aU • Magnitude = |a||U| , where |a| is the absolute value of the scalar a • Direction of aU is the same as direction of U when a is positive • Direction of aU is opposite to direction of U when a is negative

13. Scalars & Vectors • Division of a vector U by a scalar a:

14. Scalars & Vectors • The product is associative with respect to scalar multiplication: a(bU) = (ab)U (2.3) • The product is distributive with respect to scalar addition: (a + b)U = aU +bU (2.4) • The product is distributive with respect to vector addition: a(U + V) = aU +aV (2.5)

15. Scalars & Vectors • Vector Subtraction: U – V = U + (1)V (2.6) • Unit Vectors: • Magnitude = 1 • Specifies a direction • If a unit vector e & a vector U have the same direction: U = |U|e

16. Example 2.1Vector Operations The magnitudes of the vectors shown are |U| = 8 and |V| = 3. The vector V is vertical. Graphically determine the magnitude of the vector U + 2V.

17. Example 2.1 (continued) Strategy By drawing the vectors to scale and applying the triangle rule for addition, we can measure the magnitude of the vector U + 2V.

18. Example 2.1 (continued) Solution Drawing the vectors U and 2V to scale, place them head to tail.

19. Example 2.1 (continued) Solution The measured value of |U + 2V| is 13.0.

20. Example 2.1 (continued) Practice Problem The magnitudes of the vectors shown are |U| = 8 and |V| = 3. The vector Visvertical. Graphically determine the magnitude of the vector U 2V. Answer: |U  2V| = 5.7

21. Example 2.2Adding Vectors(refer to textbook)

22. Components in Two Dimensions • Vectors are much easier to work with when expressed in terms of mutually perpendicular vector components: • Consider vector U: • Place a cartesian coordinate system so that the vector U is parallel to the x-y plane • U = sum of perpendicular vector components Ux & Uy that are parallel to the x & y axes: U = Ux + Uy

23. Components in Two Dimensions • Introduce a unit vector i defined to point in the direction of the positive x axis & a unit vector j defined to point in the direction of the positive y axis: U = Uxi + Uyj (2.7) where Ux&Uyare scalar components of U • Magnitude of U is given in terms of its components by the Pythagorean theorem: (2.8)

24. Components in Two Dimensions • Manipulating Vectors in Terms of Components: • Sum of 2 vectors U & V: U + V = (Uxi + Uyj) + (Vxi + Vyj) = (Ux + Vx)i + (Uy + Vy)j (2.9) • Graphically:

25. Components in Two Dimensions • Manipulating Vectors in Terms of Components: • Product of number a & vector U: aU = a(Uxi + Uyj) = aUxi + aUyj

26. Components in Two Dimensions • Position Vectors in Terms of Components: • Consider point A with coordinates (xA, yA) & point B with coordinates (xB, yB)

27. Components in Two Dimensions • Position Vectors in Terms of Components: • Let rAB be the vector that specifies the position ofB relative to A: rAB = (xBxA)i + (yByA)j (2.10)

28. Example 2.3Determining Components The cable from point A to point B exerts a 900-N force on the top of the television transmission tower that is represented by the vector F. Express F in terms of components using the coordinate system shown.

29. Example 2.3 (continued) Strategy We will determine the components of the vector F in two ways. In the first method, we will determine the angle between F and the y axis and use trigonometry to determine the components. In the second method, we will use the given slope of the cable AB and apply similar triangles to determine the components of F.

30. Example 2.3 (continued) Solution First Method Determine the angle between Fand the y axis:

31. Example 2.3 (continued) Solution Use trigonometry to determine F in terms of its components:

32. Example 2.3 (continued) Solution Second Method Using the given dimensions, calculate the distance from A to B:

33. Example 2.3 (continued) Solution Use similar triangles to determine the components of F: so

34. Example 2.3 (continued) Practice Problem The cable from point A to point B exerts a 900-N force on the top of the television transmission tower that is represented by the vector F. Suppose that you change the placement of point B so that the magnitude of the y component of F is three times the magnitude of the x component of F. Express F in terms of its components. How far along the x axis from the origin of the coordinate system should B be placed? Answer: F = 285i - 854j (N). Place point B at 26.7 m from the origin.

35. Example 2.4Determining Components in Terms of an Angle (refer to textbook)Example 2.5Determining an Unknown Vector Magnitude (refer to textbook)

36. Components in Three Dimensions • Review of drawing objects in 3 dimensions: • A cube viewed with the line of sight • perpendicular to a face • An oblique view of the cube • A cartesian coordinate system aligned with the edges of the cube • 3-D representation of the coordinate system

37. Components in Three Dimensions • Right-handed coordinate system: • Express vector U in terms of vector components Ux, Uy & Uz parallel to the x, y & z axes respectively: U = Ux +Uy + Uz (2.11)

38. Components in Three Dimensions • Introducing unit vectors i, j & kthat point in thepositive x,y&zdirections, U can be expressed in terms of scalar components: U = Uxi +Uyj + Uz k (2.12)

39. Components in Three Dimensions • Magnitude of a Vector in Terms of Components: • Consider a vector U & its vector components:

40. Components in Three Dimensions • Magnitude of a Vector in Terms of Components: • From the right triangles formed by vectors Uy, Uz & their sum Uy + Uz: |Uy + Uz|2 = |Uy|2 + |Uz|2 (2.13) • The vector U is the sum of the vectors Ux & Uy+Uz. The 3 vectors form a right triangle: |U|2 = |Ux|2 + |Uy + Uz|2

41. Components in Three Dimensions • Magnitude of a Vector in Terms of Components: • Substituting Eqn (2.13): |U|2 = |Ux|2 + |Uy| + |Uz|2 = Ux2+Uy2+ Uz2 • Thus, magnitude of vector U: (2.14)

42. Components in Three Dimensions • Direction Cosines: • One way to describe the direction of a vector is by specifying the angles x, y& zbetween the vector & the positive coordinate axes: Ux = |U| cos x,Uy = |U| cos y,Uz = |U| cos z (2.15)

43. Components in Three Dimensions • Direction Cosines:

44. Components in Three Dimensions • Direction Cosines: • Direction cosines: cosx, cosy & cosz • Direction cosines satisfy the relation: cos2x + cos2y + cos2z= 1 (2.16) • Suppose that e is a unit vector with the same direction as U: U = |U| e • In terms of components: Uxi + Uyj + Uzk= |U| (exi + eyj + ezk)

45. Components in Three Dimensions • Direction Cosines: • Thus: Ux = |U| ex,Uy = |U| ey,Uz = |U| ez • By comparing these equations to Eqn (2.15): cos x = ex,cos y = ey,cos z = ez • The direction cosines of a vector U are the components of a unit vector with the same direction as U

46. Components in Three Dimensions • Position Vectors in Terms of Components: • Consider point A with coordinates (xA, yA, zA) & point B with coordinates (xB, yB, zB)

47. Components in Three Dimensions • Position Vectors in Terms of Components: • The position vector rAB from A to B: rAB= (xB xA)i + (yB yA)j + (zB zA)k (2.17)

48. Components in Three Dimensions • Components of Vector Parallel to a Given Line: • In 3-D applications, the direction of a vector U is often defined by specifying the coordinates of 2 points A & B on a line that is parallel to U • Determine position vector rAB using Eqn (2.17)

49. Components in Three Dimensions • Components of Vector Parallel to a Given Line: • Divide rAB by its magnitude  unit vector eAB that points from A to B • eAB has the same direction as U • Determine U as the product of its magnitude &eAB: U = |U| eAB

50. Example 2.6Direction Cosines The coordinates of point C of the truss are xC = 4 m, yC = 0, zC = 0,and the coordinates of point D are xD = 2 m, yD = 3 m, zD = 1 m. What are the direction cosines of the position vector rCD from point C to point D?