scalar and - vector - - presentation SCALAR AND VECTOR

Transcription

1 and presentation SCLR ND VECTOR

2 Scalars Scalars are quantities which have magnitude without directioni Examples of scalars temperaturere mass kinetic energy time amount density charge

3 Vector 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 Example

4 Two ways to specify a vector It is either given by a magnitude, and a direction y x Or it is given in the x and y components as y x x y y x

5 y x y x x = cos y = sin The magnitude (length) of is found by using the Pythagorean Theorem = ( x 2 + y 2 ) The length of a vector The length of a vector clearly does not depend on its direction.

6 y x y x The direction of can be stated as tan = y / x =tan -1 (y / x)

7 Some Properties of Vectors Equality of Two Vectors Two vectors and B may be defined to be equal if they have the same magnitude and point in the same directions. i.e. = B B B B

8 Negative of a Vector The negative of vector is defined as giving the vector sum of zero value when added to. That is, + (- ) = 0. The vector and have the same magnitude but are in opposite directions. -

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

10 B = If = 0, therefore B = = 0, which is also known as a zero vector ( ) = = ( ) ( + ) = + Example

11 Vector ddition The addition of two vectors and B - will result in a third vector C called the resultant C = + B Geometrically (triangle method of addition) put the tail-end of B at the top-end of C connects the tail-end of to the top-end of B C B We can arrange the vectors as we like, as long as we maintain their length and direction Example

12 More than two vectors? x i x 5 x 4 x 3 x 1 x 2 x i = x 1 + x 2 + x 3 + x 4 + x 5 Example

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

14 Rules of Vector ddition commutative + B = B + B B

15 associative ( + B) + C = + (B + C) B B + B C B + C C ( + B) + C + (B + C)

16 distributive m( + B) = m + mb mb B + B m m( + B)

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

18 Unit Vectors a vector whose magnitude is 1 and dimensionlessi the magnitude of each unit vector equals a unity; that t is, i = j = k = 1 i a unit vector pointing in the x direction and defined as j a unit vector pointing in the y direction k a unit vector pointing in the z direction

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

20 Component of a Vector in 2-D vector can be resolved into two components x and y y- axis θ y x x- axis = x + y

21 The component of are x = x = cos θ x x y- axis y = y = sin θ The magnitude of y θ y = x2 + 2 y The direction of tan = y / x =tan -1 (y / x) x x- axis Example

22 The unit vector notation for the vector is written y- axis = x i + y j y j θ i x x- axis Example

23 Component of a Vector in 3-D vector can be resolved into three components x, y and z = x i + y j + z k z- axis z k i j y y- axis x x- axis

24 if = x i + y j + z k x yj z B = B x i + B y j + B z k x yj z + B = C sum of the vectors and B can then be obtained as vector C C = ( x i + y j + z k) + (B x i + B y j + B z k) C = ( x + B x )i+ ( y + B y )j + ( z + B z )k C = C x i + C y j + C z k Example

25 Dot product (scalar) of two vectors The definition: B = B cos θ θ B

26 Dot product (scalar product) properties: if θ = 90 0 (normal vectors) then the dot product is zero B = B cos 90 = 0 and i j = j k = i k = 0 if θ = 0 0 (parallel vectors) it gets its maximum value of 1 B = B cos 0 = 1 and i j = j k = i k = 1

27 the dot product is commutative. B = B. Use the distributive law to evaluate the dot product if the components are known B = ( x i + j + z k) y (B x i + B yj + B z k). B = ( x B x ) i.i + ( y B y ) j.j + ( z B z ) k.k. B = x B x + y B y + z B z Example

28 Cross product (vector) of two vectors The magnitude of the cross product given by C = x B = B sin θ C B θ 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

29 Cross product (vector product) properties: if θ = 0 0 (parallel vectors) then the cross product is zero x B = B sin 0 = 0 and i x i = j x j = k x k = 0 if θ = 90 0 (normal vectors) it gets its maximum value x B = B 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 kxi= - ixk=j Example

31 Measurement and Error

32 THE END

33 Vectors are represented by an arrow -B θ B

34 Conceptual Example * If B is added to, under what condition does the resultant vector + B have the magnitude equal to + B? Under what conditions is the resultant vector equal to zero?

35 Example (1Dimension) x 1 + x 2 x 1 = 5 x 1 x 2 x 1 + x 2 = 8 x 2 = 3 x = x 2 -x 1 x x 2 x 1 x = x 2 - x 1 = 2 MORE EXMPLE

36 Example 1 (2 Dimension) If the magnitude of vector and B are equal to 2 cm and 3 cm respectively, determine the magnitude and direction of the resultant vector, C for a) + B b) 2 + B B SOLUTION

37 Solution a) + B = 2 + B 2 = = b) 2 + B = (2)2 + B2 = 3.6 cm The vector direction tan θ = B / = 5.0 cm The vector direction tan θ = B / 2 θ = 56.3 θ = 36.9 MORE EXMPLE

38 Example 2 ( Vacation Trip) car travels 20.0 km due north and then 35.0 km in a direction 60 0 west of north. Find the magnitude and direction of the car s resultant displacement. SOLUTION

39 Solution The magnitude of R can be obtained using the law of cosines as in figure Since θ = = and C 2 = 2 + B 2 2B cos θ, we find that C = 2 + B 2 2B cos θ C = (20)(35) cos C = 48.2 km B 60 C θ β Continue

40 The direction of C measured from the northerly direction can be obtained from the sines law sin sin B C B 35.0 sin sin sin C 48.2 β = Therefore, the resultant displacement of the car is 48.2 km in direction west of north

41 Conceptual Example * If one component of a vector is not zero, can its magnitude be zero? Explain. MORE EXMPLE

42 Conceptual Example * If + B = 0, what can you say about the components of the two vectors?

43 Example 1 Find the sum of two vectors and B lying in the xy plane and given by = 2.0i + 2.0j and B = 2.0i 4.0j SOLUTION

44 Solution * Comparing the above expression for with the general relation = x i + y j, we see that x = 2.0 and y = 2.0. Likewise, i B x = 2.0, and B y = -4.0 Therefore, the resultant t vector C is obtained by using Equation C = + B + ( )i + ( )j = 4.0i -2.0j or C x = 4.0 C y = -2.0 The magnitude of C given by equation C = C 2 2 x + C y = 20 = 4.5 Exercise Find the angle θ that t C makes with the positive x axis

45 Example particle undergoes three consecutive displacements d 1 = (1.5i + 3.0j 1.2k) cm, d 2 = (2.3i 1.4j 3.6k) cm d 3 = (-1.3i + 1.5j) cm. Find the component and its magnitude.

46 Solution R = d 1 + d 2 + d 3 = ( )i + ( )j + ( )k = (2.5i + 3.1j 4.8k) cm That is, the resultant displacement has component R x = 2.5 cm R y = 3.1 cm and R z = -4.8 cm x y z Its magnitude is R = R 2 2 x2 + R y2 + R 2 z = 6. 2 cm

47 Example - 2D [headtotail] (2, 2) (1, 0) x 1 + x 2 = (1, 0) + (2, 2) = (3, 2) x 1 + x 2 x 2 x 1

48 Example - 2D [tailtotail] (2, 2) (1, 0) x 2 x 1 + x 2 (x 2 ) x 1 x 1 + x 2 = (1, 0) + (2, 2) = (3, 2) x 1 -x 2?

49 Example of 2D (subtraction) (2, 2) (1, 0) x 1 + x 2 x2 x 1

50 Example -2D for subtraction (2, 2) (1, 0) x 1 -x 2 = x 1 + (-x 2 ) x 1 -x 2 = (1, 0) - (2, 2) = (-1, -2) x 1 -x2 x 1 -x 2 2

51 Not given the components? 2 2 m 1 m 45 o X 2 = (x 2E, x 2N ) X 1 = (1, 0) = (2 2cos(45 o ), 2 2sin(45 o )) = (2, 2) Cosine rule: 1 m a 2 =b 2 + c 2-2bccos x 1 45 o = (1/ 2) = x 1 -x 2 2 m a 5 m 2 -x 2