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SCALAR AND VECTOR

SCALAR AND VECTOR. By: Engr. Hinesh Kumar Lecturer I.B.T, LUMHS, Jamshoro. Scalars. Scalars are quantities which have magnitude without direction. Examples of scalars. time amount density charge. temperature mass kinetic energy. Vector. A vector is a quantity that has both

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SCALAR AND VECTOR

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  1. SCALAR AND VECTOR By: Engr. Hinesh Kumar Lecturer I.B.T, LUMHS, Jamshoro

  2. Scalars Scalars are quantities which have magnitude without direction. Examples of scalars • time • amount • density • charge • temperature • mass • kinetic energy

  3. Vector A vector is a quantity that has both magnitude(size) anddirection. • it is represented by an arrow whereby • the length of the arrow is the magnitude, and • the arrow itself indicates the direction

  4. A Contd…. • The symbol for a vector is a letter with an arrow over it. • All vectors have head and tail.

  5. y A A  x Ax y A Ay x Two ways to specify a vector • It is either given by • a magnitude A, and • a direction  • Or it is given in the • x and y components as • Ax • Ay

  6. Ax y A A Ay  x Ax = Acos Ay = Asin The magnitude (length) of A is found by using the Pythagorean Theorem │A │ =√ (Ax2+Ay2) The length of a vector clearly does not depend on its direction.

  7. Ax y A A Ay  x The direction of A can be stated as tan  = Ay / Ax  =tan-1(Ay / Ax)

  8. Vector Representation of Force • Force has both magnitude and direction and therefore can be represented as a vector.

  9. Vector Representation of Force • The figure on the left shows 2 forces in the same direction therefore the forces add. The figure on the right shows the man pulling in the opposite direction as the cart and forces are subtracted.

  10. A B A A B B 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

  11. A -A 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.

  12. Applications of Vectors VECTOR ADDITION– If 2 similar vectors point in the SAME direction, add them. • Example: A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started? + 54.5 m, E 30 m, E Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION. 84.5 m, E

  13. Vector Addition C = A + B C A B The addition of two vectors A and B - will result in a third vector C called the resultant • 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 Example

  14. More than two vectors? x4 x5 xi x3 x2 xi = x1 + x2 + x3 + x4 + x5 x1 Example

  15. Applications of Vectors VECTOR SUBTRACTION - If 2 vectors are going in opposite directions, you SUBTRACT. • Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement relative to where he started? 54.5 m, E - 30 m, W 24.5 m, E

  16. A - B A -B B C = C = A B A + (-B) Vector Subtraction Equivalent to adding the negative vector Example

  17. Scalar Multiplication The multiplication of a vector A by a scalar  - will result in a vector B B =  A - whereby the magnitude is changed but not the direction • Do flip the direction if  is negative

  18. B =  A If  = 0, therefore B =  A = 0, which is also known as a zero vector (A) = A = (A) (+)A = A + A Example

  19. A + B A + B A B B A Rules of Vector Addition • commutative A + B = B + A

  20. B C A B C A (A + B) + C = A + (B + C) A + B B + C A + (B + C) (A + B) + C • associative

  21. m(A + B) = mA + mB A B A + B mA mB m(A + B) • distributive

  22. A + B A B Parallelogram method of addition (tailtotail) The magnitude of the resultant depends on the relative directions of the vectors

  23. Unit Vectors    k j i a vector whose magnitude is 1 and dimensionless the magnitude of each unit vector equals a unity; that is, │ │= │ │= │ │= 1 and defined as  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  

  24. y j x i k z Useful examples for the Cartesian unit vectors [i, j, k] - they point in the direction of the x, yand z axes respectively

  25. A = Ax + Ay y- axis A Ay θ x- axis Ax Component of a Vector in 2-D • vector A can be resolved into two components Axand Ay

  26. y- axis │Ax│ = Ax = A cos θ │Ay│ = Ay = A sin θ θ x- axis A Ay A = √Ax2 + Ay2 Ax The component of A are The magnitude of A The direction of A tan  = Ay / Ax  =tan-1(Ay / Ax)

  27. Ay j A θ i Ax A = Axi + Ayj The unit vector notation for the vector Ais written y- axis x- axis

  28. Ay Ax A y- axis Az j i x- axis k z- axis Component of a Unit Vector in 3-D • vector A can be resolved into three components Ax, Ay and Az A = Axi + Ayj + Azk

  29. A = Axi + Ayj + Azk if B = Bxi + Byj + Bzk A + B = C sum ofthe 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

  30. A · B = │A││B │cos θ A θ B Dot product (scalar) of two vectors The definition:

  31. |A · B| = AB cos 90 = 0 |A · B| = AB cos 0 = 1 Dot product (scalar product) properties: if θ = 900 (normal vectors) then the dot product is zero and i · j = j · k = i · k = 0 • if θ = 00 (parallel vectors) it gets its maximum • value of 1 and i · j = j · k = i · k = 1

  32. A + B = B + A A · B = (Axi + Ayj + Azk) · (Bxi + Byj + Bzk) A. B = (AxBx) i.i + (AyBy) j.j + (AzBz) k.k A . • the dot product is commutative • Use the distributive law to evaluate the dot product if the components are known

  33. │C │= │A x B│ = │A││B │sin θ A θ B C 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

  34. |A x B| = AB sin 0 = 0 and i x i = j x j = k x k = 0 |A x B| = AB sin 90 = 1 and i x i = j x j = k x k = 1 Cross product (vector product) properties: if θ = 00 (parallelvectors) then the cross product is zero • if θ = 900 (normal vectors) it gets its maximum • value

  35. 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

  36. A B 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 • A + B • 2A + B

  37. |A + B| = √A2 + B2 • = √22 + 32 • = 3.6 cm • The vector direction • tan θ = B / A • θ = 56.3 • |2A + B| = √(2A)2 + B2 • = √42 + 32 • = 5.0 cm • The vector direction • tan θ = B / 2A • θ = 36.9 Solution

  38. Example 2 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

  39. C = √Cx2 + Cy2 = √20 = 4.5 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 Exercise Find the angle θ that C makes with the positive x axis

  40. Example - 2D [headtotail] (2, 2) (1, 0)

  41. x2 x1 Solution x1 + x2 = (1, 0) + (2, 2) = (3, 2) x1 + x2

  42. Example - 2D [tailtotail] (2, 2) (1, 0)

  43. (x2) x2 x1 + x2 x1 Solution x1 + x2 = (1, 0) + (2, 2) = (3, 2) x1 - x2?

  44. Example of 2D (Subtraction) (2, 2) (1, 0)

  45. Solution x1 - x2 = x1 + (-x2) x1 - x2 = (1, 0) - (2, 2) = (-1, -2) x1 x1 - x2 -x2

  46. Example -2D for subtraction (2, 2) (1, 0)

  47. Assignment If one component of a vector is not zero, can its magnitude be zero? Explain and Prove it. 1 If A + B = 0, what can you say about the components of the two vectors? 2 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. 3

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