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### Scalars and vectors

• 1. CHAPTER 1.4: SCALAR AND VECTORCHAPTER 1.4: SCALAR AND VECTOR SCALAR AND VECTOR
• 2. Scalars are quantities which have magnitude without direction Examples of scalars • temperature • mass • kinetic energy • time • amount • density • charge Scalars
• 3. Vector A vector is a quantity that has both magnitude (size) and direction it is represented by an arrow whereby – the length of the arrow is the magnitude, and – the arrow itself indicates the direction The symbol for a vector is a letter with an arrow over it A ExampleExample
• 4. Two ways to specify a vector It is either given by • a magnitude A, and • a direction θ Or it is given in the x and y components as • Ax • Ay y x θ A A Ay x Ax Ay
• 5. y x AAx Ay A θ Ax = Acosθ Ay = Asinθ │A = (│ √ Ax 2 +Ay 2 ) The magnitude (length) of A is found by using the Pythagorean Theorem The length of a vector clearly does not depend on its direction.
• 6. y x AAx Ay A θ The direction of A can be stated as tan θ = Ay / Ax θ =tan-1 (Ay / Ax)
• 7. Some Properties of Vectors Equality of Two Vectors Two vectors A and B may be defined to be equal if they have the same magnitude and point in the same directions. i.e. A = B A B A A B B
• 8. Negative of a Vector The negative of vector A is defined as giving the vector sum of zero value when added to A . That is, A + (- A) = 0. The vector A and –A have the same magnitude but are in opposite directions. A -A
• 9. Scalar Multiplication The multiplication of a vector A by a scalar α - will result in a vector B B = α A - whereby the magnitude is changed but not the direction • Do flip the direction if α is negative
• 10. B = α A If α = 0, therefore B = α A = 0, which is also known as a zero vector β(αA) = αβA = α(βA) (β+α)A = αA + βA ExampleExample
• 11. The addition of two vectors A and B - will result in a third vector C called the resultant C = A + B A B C Geometrically (triangle method of addition) • put the tail-end of B at the top-end of A • C connects the tail-end of A to the top-end of B We can arrange the vectors as we like, as long as we maintain their length and direction Vector Addition ExampleExample
• 12. More than two vectors? x1 x5 x4 x3 x2 Σxi Σxi = x1 + x2 + x3 + x4 + x5 ExampleExample
• 13. Vector Subtraction Equivalent to adding the negative vector A -B A - B B A BC = A + (-B)C = ExampleExample
• 14. Rules of Vector Addition commutative A + B = B + A A B A + B B A A + B
• 15. associative (A + B) + C = A + (B + C) B C A B CA A + B (A + B) + C A + (B + C) B + C
• 16. distributive m(A + B) = mA + mB A B A + B mA mB m(A + B)
• 17. Parallelogram method of addition (tailtotail) A B A + B The magnitude of the resultant depends on the relative directions of the vectors
• 18. a vector whose magnitude is 1 and dimensionless the magnitude of each unit vector equals a unity; that is, = = = 1│ │ │ │ │ │ i a unit vector pointing in the x direction j a unit vector pointing in the y direction k a unit vector pointing in the z direction 〉〉〉 and defined as Unit Vectors k 〉 j 〉 i 〉
• 19. Useful examples for the Cartesian unit vectors [ i, j, ki, j, k ] - they point in the direction of the x, y and z axes respectively x y z ii jj kk
• 20. Component of a Vector in 2-D vector A can be resolved into two components Ax and Ay x- axis y- axis Ay Ax A θ A = Ax + Ay
• 21. The component of A are │Ax│ = Ax = A cos θ │Ay│ = Ay = A sin θ The magnitude of A A = √Ax 2 + Ay 2 tan θ = Ay / Ax θ =tan-1 (Ay / Ax) The direction of A x- axis y- axis Ay Ax A θ ExampleExample
• 22. The unit vector notation for the vector A is written A = Axi + Ayj x- axis y- axis Ax Ay θ A i j ExampleExample
• 23. Component of a Vector in 3-D vector A can be resolved into three components Ax , Ayand Az A Ax Ay Az z- axis y- axis x- axis i j k A = Axi + Ayj + Azk
• 24. if A = Axi + Ayj + Azk B = Bxi + Byj + Bzk A + B = C sum of the vectors A and B can then be obtained as vector C C = (Axi + Ayj + Azk) + (Bxi + Byj + Bzk) C = (Ax + Bx)i+ (Ay + By)j + (Az + Bz)k C = Cxi + Cyj + Czk ExampleExample
• 25. Dot product (scalar) of two vectors The definition: θ B A A · B = │A││B │cos θ
• 26. if θ = 900 (normal vectors) then the dot product is zero Dot product (scalar product) properties: if θ = 00 (parallel vectors) it gets its maximum value of 1 and i · j = j · k = i · k = 0|A · B| = AB cos 90 = 0 |A · B| = AB cos 0 = 1 and i · j = j · k = i · k = 1
• 27. A + B = B + A the dot product is commutative Use the distributive law to evaluate the dot product if the components are known A · B = (Axi + Ayj + Azk) · (Bxi + Byj + Bzk) A. B = (AxBx) i.i + (AyBy) j.j + (AzBz) k.k A . B = AxBx + AyBy + AzBz ExampleExample
• 28. Cross product (vector) of two vectors The magnitude of the cross product given by the vector product creates a new vector this vector is normal to the plane defined by the original vectors and its direction is found by using the right hand rule │C │= │A x B│ = │A││B │sin θ θ A BC
• 29. if θ = 00 (parallel vectors) then the cross product is zero Cross product (vector product) properties: if θ = 900 (normal vectors) it gets its maximum value and i x i = j x j = k x k = 0|A x B| = AB sin 0 = 0 |A x B| = AB sin 90 = 1 and i x i = j x j = k x k = 1
• 30. the relationship between vectors i , j and k can be described as i x j = - j x i = k j x k = - k x j = i k x i = - i x k = j ExampleExample
• 33. Vectors are represented by an arrow A - B B A A θ
• 34. Conceptual Example If B is added to A, under what condition does the resultant vector A + B have the magnitude equal to A + B ? Under what conditions is the resultant vector equal to zero? *
• 35. Example (1Dimension) x1 = 5 x1 x2 x2 = 3 ∆x = x2 - x1 = 2 x1 + x2 x1 x2 ∆x = x2 - x1 x1 + x2 = 8 MORE EXAMPLEMORE EXAMPLE
• 36. Example 1 (2 Dimension) If the magnitude of vector A and B are equal to 2 cm and 3 cm respectively , determine the magnitude and direction of the resultant vector, C for B A a) A + B b) 2A + B SOLUTIONSOLUTION
• 37. Solution a) |A + B| = √A2 + B2 = √22 + 32 = 3.6 cm The vector direction tan θ = B / A θ = 56.3 b) |2A + B| = √(2A)2 + B2 = √42 + 32 = 5.0 cm The vector direction tan θ = B / 2A θ = 36.9 MORE EXAMPLEMORE EXAMPLE
• 38. Example 2 (A Vacation Trip) A car travels 20.0 km due north and then 35.0 km in a direction 600 west of north. Find the magnitude and direction of the car’s resultant displacement. SOLUTIONSOLUTION
• 39. Solution The magnitude of R can be obtained using the law of cosines as in figure Since θ =1800 – 600 = 1200 and C2 = A2 + B2 – 2AB cos θ, we find that C = √A2 + B2 – 2AB cos θ C = √202 + 352 – 2(20)(35) cos 1200 C = 48.2 km C A B 60 θ β ContinueContinue
• 40. The direction of C measured from the northerly direction can be obtained from the sines law CB θβ sinsin = 629.0120sin 2.48 0.35 sinsin 0 === θβ C B β = 38.90 Therefore, the resultant displacement of the car is 48.2 km in direction 38.90 west of north
• 41. If one component of a vector is not zero, can its magnitude be zero? Explain. *Conceptual Example MORE EXAMPLEMORE EXAMPLE
• 42. Conceptual Example If A + B = 0, what can you say about the components of the two vectors? *
• 43. Example 1 Find the sum of two vectors A and B lying in the xy plane and given by A = 2.0i + 2.0j and B = 2.0i – 4.0j SOLUTIONSOLUTION
• 44. Solution Comparing the above expression for A with the general relation A = Axi + Ayj , we see that Ax= 2.0 and Ay= 2.0. Likewise, Bx= 2.0, and By= -4.0 Therefore, the resultant vector C is obtained by using Equation C = A + B + (2.0 + 2.0)i + (2.0 - 4.0)j = 4.0i -2.0j or Cx = 4.0 Cy = -2.0 The magnitude of C given by equation C = √Cx 2 + Cy 2 = √20 = 4.5 * Find the angle θ that C makes with the positive x axis Exercise
• 45. Example A particle undergoes three consecutive displacements d1 = (1.5i + 3.0j – 1.2k) cm, d2 = (2.3i – 1.4j – 3.6k) cm d3 = (-1.3i + 1.5j) cm. Find the component and its magnitude.
• 46. Solution R = d1 + d2 + d3 = (1.5 + 2.3 – 1.3)i + (3.0 – 1.4 + 1.5)j + (-1.2 – 3.6 + 0)k = (2.5i + 3.1j – 4.8k) cm That is, the resultant displacement has component Rx = 2.5 cm Ry = 3.1 cm and Rz = -4.8 cm Its magnitude is R = √ Rx 2 + Ry 2 + Rz 2 = 6. 2 cm
• 47. Example - 2D [headtotail] x1 x2 x1 + x2 (1, 0) (2, 2) x1 + x2 = (1, 0) + (2, 2) = (3, 2)
• 48. Example - 2D [tailtotail] x1 - x2? (x2) x1 x1 + x2x2 (1, 0) (2, 2) x1 + x2 = (1, 0) + (2, 2) = (3, 2)
• 49. Example of 2D (subtraction) (1, 0) (2, 2) x1 x2 x1 + x2
• 50. Example -2D for subtraction x1 -x2 x1 - x2 (1, 0) (2, 2) x1 - x2 = (1, 0) - (2, 2) = (-1, -2) x1 - x2 = x1 + (-x2)
• 51. Not given the components? 1 m 2√2 m 45o X1 = (1, 0) X2 = (x2E, x2N) = (2√2cos(45o ), 2√2sin(45o )) = (2, 2) x1 -x2 x1 - x2 2√2 m 1 m 45o Cosine rule: a2 =b2 + c2 - 2bccosA = 1 + 8 - 2√2(1/ √2) a = √5 m