52512172511General-Physics-1-VECTOR.pptx

Physics

1. Physics 1: VECTORS Portia a. egken . E d.D

2. Vectors A quantity that has both Magnitude and Direction Examples : Wind Boat or aircraft travel Forces in physics

3. Geometrically A directed line segment Initial point Terminal point Vector Notation An arrow over a letter

4. Vector Notation An arrow over two letters The magnitude (length) of a vector is notated with double vertical lines

5. Equivalent Vectors Have both same direction and same magnitude Given points The components of a vector - Ordered pair of terminal point with initial point at (0,0) -

6. Find the Vector Given P 1 (0, -3) and P 2 (1, 5) Show vector representation in <x, y> format for ● ( 1 – 0 , 5 – (-3) ) = ( 1 , 8 )

7. Try these P 1 (4,2) and P 2 (-3, -3 ) (-7 , -5) (0 , -2) ▪ (- 3 - 4 , -3 - 2 ) = ▪ P 4 (3 , -2 and P 2 (3 , 0 ) ▪ (3 - 3 , - 2 - 0 ) = ▪ (x 2 – x 1 , y 2 – y 1 )

8. Fundamental Vector Operations Given vectors V = <a, b>, W = <c, d > Magnitude Scalar multiplication – changes the magnitude, not the direction 3V = <3a, 3b>

9. Parallelogram Method : (Vector Addition) Draw the vectors so that their initial points coincide. Then draw lines to form a complete parallelogram. The diagonal from the initial point to the opposite vertex of the parallelogram is the resultant .

10. Parallelogram Method : (Vector subtraction)

11. Vector SUBTRACTION The difference of two vectors is the result of adding a negative vector A – B = - A + (-B)

12. Vector Addition / Subtraction To add or subtract two vectors, add or subtract the corresponding components . Let u⃗ =⟨u 1 ,u 2 ⟩ and v⃗ =⟨v 1 ,v 2 ⟩ be two vectors . Then, the sum of u⃗ and v⃗ is the vector u ⃗+v ⃗ = (u 1 + v 1 , u 2 + v 2 ) - The difference of u⃗ and v⃗ is u⃗ + v ⃗ = u⃗ + (-v⃗) = u 1 - v 1 , u 2 - v 2

13. Vector Addition / Subtraction Find (a) u⃗ +v⃗ and (b) u⃗ −v⃗ u = ( u 1 , u 2 ) v = (v 1 , v 2 ) if u⃗ =⟨ 3 , 4 ⟩ and v⃗ =⟨ 5 , − 1⟩ . Substitute the given values of u 1 , u 2 , v 1 and v 2 into the definition of vector addition. u ⃗ + v ⃗ = (u 1 + v 1 , u 2 + v 2 ) = ( 3 + 5 , 4 + (-1) = ( 8 , 3 )

14. Find u ⃗ − v ⃗ - Rewrite the difference u⃗ − v ⃗ as a sum u⃗ +(−v⃗ ) . We will need to determine the components of − v ⃗ . - From the definition of scalar multiplication, we have -v = -1(v 1 , v 2 ) = -1( 5, -1) = (- 5, 1)

15. Now add the components of u⃗ and −v⃗ . u⃗ + (- v ⃗) = (3 + (-5) , (4 +1) -v = -1(v 1 , v 2 ) = (- 5, 1) = (- 2, 5) u = ( u 1 , u 2 ) = (3 , 4 ) if u⃗ =⟨3 , 4⟩ and v⃗ =⟨5 , −1⟩

16. Vector Addition / Subtraction Find (1) a⃗ + b⃗ and (2) a⃗ − b⃗ if a ⃗ =⟨ 6 , 4 ⟩ and b ⃗ =⟨ -2 , −5⟩ Substitute the given values of a 1 , a 2 , b 1 and b 2 into the definition of vector addition. a ⃗+ b ⃗ = ( a 1 + b 1 , a 2 + b 2 ) = ( 6 +(- 2)) , 4 + (-5)) = ( 4 , -1)

17. Now add the components of a⃗ and −b⃗ a ⃗ + (- b ⃗ ) = ( a 1 + b 1 , a 2 + b 2 ) a⃗ + (- b⃗) = (6 + 2 , (4 +5) -b = -1(b 1 , b 2 ) = ( 2 , 5 ) = ( 8, 9 ) a = (a 1 , a 2 ) = (6 , 4 ) if a⃗ =⟨6 , 4⟩ and b⃗ =⟨ -2 , −5⟩

18. The magnitudes of two vectors U and V are equal to 5 and 10 respectively. Vector U makes an angle of 20° with the positive direction of the x-axis and vector V makes an angle of 80° with the positive direction of the x-axis. Both angles are measured counterclockwise . Find the magnitudes and directions of vectors U + V and U - V.

19. The magnitudes of two vectors U and V are equal to 5 and 10 respectively . Vector U makes an angle of 20° with the positive direction of the x-axis and vector V makes an angle of 80° with the positive direction of the x-axis. Both angles are measured counterclockwise . +x - x 20 o U V 8 o

20. Let us first use the magnitudes and directions to find the components of vectors U and V . ( Note: x =cos , y = sin) U   → = (5 cos (20°) , 5 sin(20 °)) = ( 5 , 2 ) V   → = (10 cos (80°) , 10 sin(80 °)) = ( 2 , 10 ) Magnitude and direction of vector U + V U  → +  V → = (5 cos (20°) , 5 sin(20°)) + (10 cos (80°) , 10 sin(80°)) = (5 cos (20°) + 10 cos (80°) , 5 sin(20°)+10 sin(80°)) = (5 + 2 , 2 + 10 ) = ( 7 , 12) = ( x , y )

21. = (5 + 2 , 2 + 10 ) = ( 7 , 12) = ( x , y )   U  → +  V → =  √  ( 7 + 12)   2   ≈ 13.22 If θ is the angle in standard position (angle between vector U+V and x-axis positive direction) of vector U + V, then tan(θ) = y- component x- component = tan -1 (12 / 7) = 59.7

22. RESULTANT VECTOR The sum of two vectors is represented by as single vector A. GRAPHICAL METHOD B. ANALYTICAL METHOD / COMPONENT METHOD

23. GRAPHICAL METHOD - Using ruler and protractor - Resultant Velocity , V R - Resultant displacement , d R - Resultant Force , F R

24. Example : The ship sails 25km north Vector diagram N E W S scale: 10km = 1cm 25km = 2.5cm

25. Example : The ship sails 20 km south then 15 km east Vector diagram N E W S scale: 10km = 1cm d R = 2.5cm d R = 25km

26. Example 2: A person walks 9 blocks east and 5 blocks north. What is his resultant displacement? d R = ?

27. 29 Example 2: A person walks 9 blocks east and 5 blocks north. What is his resultant displacement?

28. d R = ? 29

29. d R = 10.3cm, 29 north of east 29 Example 2: A person walks 9 blocks east and 5 blocks north. What is his resultant displacement?

30. Example : Carlito was observing an ant that crawled along a tabletop . With a piece of chalk he followed the path. He determined the ant’s displacement using a ruler and protractor The displacement were as follows: d 1 = 2 cm east d 2 = 3.5 cm, 32 north of east d 3 = 2.3 cm , 22 west of north Find the Resultant Displacement , d R

31. Vector diagram N E W S d 1 = 2 cm east d 2 = 3.5 cm, 32 north of east d 2 = 2.3 cm , 22 west of north

32. i n cm inches

33. Each line represent 10

34. d 1 = 2 cm east

35. d 1 = 2 cm east d 2 = 3.5 cm, 32 north of east 32

36. d 1 = 2 cm east d 2 = 3.5 cm, 32 north of east

37. d 1 = 2 cm east d 2 = 3.5 cm, 32 north of east N E d 2 = 2.3 cm , 22 west of north 22

38. d 1 = 2 cm east d 2 = 3.5 cm, 32 north of east d 2 = 2.3 cm , 22 west of north N W

39. d 1 = 2 cm east d 2 = 3.5 cm, 32 north of east d 2 = 2.3 cm , 22 west of north N W

40. d 1 = 2 cm east d 2 = 3.5 cm, 32 north of east d 2 = 2.3 cm , 22 west of north N W d R = 5.5cm

41. Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction 49.0º north of east. Then, she walks 23.0 m heading 15.0º north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east. Example 3

42. d 1 = 25.0 m in a direction 49.0º north of east. d 2 = 23.0 m heading 15.0º north of east. d 3 = 32.0 m in a direction 68.0° south of east.

43. d 1 = 25.0 m in a direction 49.0º north of east. d 2 = 23.0 m heading 15.0º north of east. d 3 = 32.0 m in a direction 68.0° south of east.

44. COMPONENT METHOD /ANALYTICAL METHOD A x and A y are defined to be the components of A along the x– and y-axes. The three vectors A, A x , and A y form a right triangle:

45. COMPONENT METHOD /ANALYTICAL METHOD To find Ax and Ay, its x– and y-components, we use the following relationships for a right triangle.

46. Ax = A cos ϴ Ay = A sin ϴ Example 2: A person walks 9 blocks east and 5 blocks north. What is his resultant displacement? A   = 10.3 blocks and  θ  = 29.1º , so that A x   = A cos ϴ = (10.3 blocks) (cos 29.1º) A y   = A sin ϴ = (10.3 blocks) (sin 29.1º) = 9 blocks = 5 blocks

47. A = √ A x 2 + A y 2 Finally, the direction θ = tan – 1( ) A = √ 9 2 + 5 2 = 10.3   θ = tan – 1( )   = 29.1

48. Vector A represents the first leg of a walk in which a person walks 53.0 m in a direction 20.0º north of east. Vector B represents the second leg, a displacement of 34.0 m in a direction 63.0º north of east.

49. Vector A = 53.0 m , 20.0º north of east. Vector B = 34.0 m , 63.0º north of east. A ​ x ​​ = A cos θ ​ ​​= ( 53.0 m )( cos 20.0 ​ ∘ ​​) ​ = ( 53.0 m )(0.940 ) = 49.8  m​​ A ​ ​y = A sin θ ​ = ( 53.0 m )(sin 20.0 ​ ∘ ​​) ​ = ( 53.0 m )( 0.342) = 18.1m ​​

50. Vector A = 53.0 m , 20.0º north of east. Vector B = 34.0 m , 63.0º north of east. B ​ x ​​ = B cos θ ​ ​​ = ( 34.0 m )( cos 63.0 ​ ∘ ​​) ​ = ( 34.0 m )(0.454 ) = 15.4  m​​ B ​ y ​​ = B sin θ ​ ​= ( 34.0 m )(sin 63.0 ​ ∘ ​​) ​ = ( 34.0 m )(0.891 ) = 30.3  m​​

51. The x– and y-components of the resultant are thus Rx = Ax + Bx = 49.8 m + 15.4 m = 65.2 m Ry = Ay + By = 18.1 m + 30.3 m = 48.4 m. Now we can find the magnitude of the resultant by using the Pythagorean theorem :

52. R = √ (65.2) 2 + ( 48.4) 2 R = 81.2 m R = √ R x 2 + R y 2

53. V 1 = 25km/hr , 30 o NW V 2 = 60 km/hr , 40 NE V 3 = 55 km/hr , 35 WS V R = ? Velocity x - component y - component V 1 V 2 V 3 V R 25 ( cos 30) = -21.65 25 ( sin 30) = 12.50 60 ( cos 40) = 45.96 60 ( sin 40) = 38.57 55 ( cos 35) = - 45.05 55 ( sin 35) = -31.55 = - 20.74 = 19.52

54. Velocity x - component y - component V 1 V 2 V 3 V R 25 ( cos 30) = -21.65 25 ( sin 30) = 12.50 60 ( cos 40) = 45.96 60 ( sin 40) = 38.57 55 ( cos 35) = - 45.05 55 ( sin 35) = -31.55 = - 20.74 = 19.52 V R = √ (-20.74) 2 + (19.52) 2 28.48 km/h

55. A boy walks around a block and ends up at his starting point. If each of the four sides of the block is 100m long, What is his resultant displacement? d 1 d 2 d 3 d 4 d R = 0

56. Displacement X -component Y-component d 1 d 2 d 3 d 4 d R 100m 100m -100m -100m

57. A motorcycle is driven 60km west, then 35km south and 35 km, 30 west of south. Find the total displacement using component method.