1. Chapter 7 Forces and Motion In Two Dimensions

2. Equilibrium An object is in equilibrium when the Net Force on the object is zero.  F Net = 0  Acceleration = 0  Constant Velocity  Velocity = 0

3. Equilibrium in 2 Dimensions  F x = 0  F y = 0

4. Static Equilibrium Objects are not moving.  F Net = 0  Acceleration = 0  Velocity = 0

5. Static Equilibrium 50N 20N F Net = 0 F x = 0 F y = 0

6. 2kg 60°

7. 2kg F g = mg F g = 2kgx9.8m/s 2 F g = 20N F1F1 F2F2

8. F1F1 F2F2 60° x direction F x = -F 1x + F 2x 0 = -F 1x + F 2x F 1x = F 2x F 1 cos60°= F 2 cos 60° F 1 = F 2 = F

9. F g = 20N F1F1 F2F2 60° y direction F y = F 1y + F 2y - F g F g = F 1 sin60°+ F 2 sin 60° 0 = F 1y + F 2y - F g F g = F 1y + F 2y F g = Fsin60°+ Fsin 60° F g = 2Fsin60°

10. __F g __ 2sin60° = F _20N__ 2sin60° = F 11.5N = F

11. Homework Finish Work Sheet

12. 50kg FtFt FPFP FgFg

13. FtFt FPFP FgFg

14. FtFt FPFP FgFg 30° F g = 50kg·9.8m/s 2 F g = mg F g = 500N F tx F ty

15. F x = F px - F tx X - ComponentsY- Components 0 = F p - F tx F px = F tx F p = F t cos30° F y = F py - F g 0 = F py - F g F ty = F g F t sin30° = F g F g sin30° F t =

16. X - ComponentsY- Components F p = F t cos30° F g sin30° F t = 500N sin30° F t = F t = 1000N F p = 1000cos30° F p = 866N

17. Homework Finish Work Sheet

19. Incline Plane FgFg FNFN F N =F g

20. Incline Plane FNFN FgFg θ FfFf

21. FNFN FgFg θ FfFf

22. FNFN FgFg θ FfFf F y =F N – F g cosθ F x =F f – F g sinθ F N = F g cosθ

23. FNFN θ FfFf FgFg

24. θ FgFg FfFf θ FNFN

25. θ FgFg FfFf θ FNFN F gy F gx F y =F N – F g cosθ F x =F f – F g sinθ F N = F g cosθ

26. Problem In a block/inclined plane system, the inclined plane makes an angle of 60° with the ground. The coefficient of friction is 0.5. If the block has a mass of 1.02kg, what is the net force on the block? What is the blocks acceleration?

27. Incline Plane FNFN FgFg θ=60° FfFf 1.02kg F g =mg F g =1.02kg·9.8m/s 2 =10N Ff=μFNFf=μFN

28. θ FgFg FfFf θ FNFN

29. FgFg FfFf 60° FNFN F gy F gx F x =F f – F gx F y =F N – F g cosθ F y =F N – F gy F x =F f – F g sinθ

30. F y = F N – F g cosθ F N = F g cosθ 0 = F N – F g cosθ F N = 10cos60° F N = 5N

31. F x =F f – F g sinθ F f = μF N F x =μF N – F g sinθ F x =0.5·5N – 10N·sin60° F x =2.5 – 8.7N F x = –6.2N Net Force of 6.2N down the incline!!

32. F Net = ma 6.1m/s 2 = a F Net m = a 6.2N 1.02kg = a 6.1m/s 2 down the incline!!

33. Projectile Motion

35. Vertical Component Horizontal Component

36. Projectile Motion Projectile motion is the combination of two independent motions, the motion in the x direction and the motion in the y direction. These two motions are usually independent of each other.

40. Projectile Motion y component x component

41. Problem Solving Strategy 1. Break up the problem into two interconnected one-dimensional problems. y component x component

42. Problem Solving Strategy 2. Vertical motion (y component) is exactly that of an object being dropped or thrown straight up or down. (g - gravity!!!!!)

43. Problem Solving Strategy 3. Horizontal motion (x component) is the same as solving constant velocity problems.

44. Problem Solving Strategy 4. Vertical (y) and horizontal (y) components are connected by the variable time (t). Solving for time in one dimension, x or y, automatically gives you the time for the other dimension.

45. d = d 0 +1/2(v+v 0 )t v 2 = v 0 2 +2a(d-d 0 ) v = v 0 + at d = d 0 +v 0 t + ½at 2 *Basic Equations*

46. Problem: A ball is kicked horizontally, with a velocity of 25m/s, off a 122.5m high cliff. How far from the cliff did the ball land?

47. Sketch the Problem y component x component v = 25 m/s y = 122.5 m

48. y component (up is +) d = d 0 +v 0 t + ½at 2 d = ½at 2 -122.5m = ½(-9.8m/s 2 )t 2 = t 2 2(-122.5m) (-9.8m/s 2 )

49. y component (up is +) = t 2(-122.5m) (-9.8m/s 2 ) √ 5.0s = t Use this to solve for distance in the x direction!!!

50. x component d = d 0 + v 0 t + ½at 2 d = v 0 t d = (25m/s)(5s) d = 125m

51. Projectiles Launched at an Angle θ Range

52. Launch Problem A football is thrown with a speed 15m/s at an angle of 60° with the horizontal. How far is the football thrown?

53. v = 15 m/s 60° Sketch the Problem y component x component

54. 1. Known x 0 = 0 y 0 = 0 v 0 = 15m/s θ 0 = 60° a = -g =-9.8m/s 2 2. Unknown v y v x d x d y t

55. 3. Find the x/y components of the velocity. v x0 = (15m/s)cos60° v x0 = 7.5m/s v y0 = (15m/s)sin60° v y0 = 13m/s

56. 4. Break up the x/y components and find time, then distance.

57. y - component v y = v 0 + at v y will be zero at the top (9.8m/s 2 )t = 13m/s 0 = 13m/s + (-9.8m/s 2 )t 13m/s 9.8m/s 2 t =

58. 13m/s 9.8m/s 2 t = t = 1.3s This is the time for half the trip!! t total = 2.6s

59. x component d x = d 0 +v x0 t + ½at 2 d x = v x0 t d x = (7.5m/s)(2.6s) d x = 19.5m

60. y - component Let’s find the height!! d y = d y0 +v y0 t + ½at 2 d y = v y0 t + ½at 2 d y = 13m/s(1.3s) + ½(-9.8/s 2 )(1.3s) 2 d y = 16.9m – 8.3m d y = 8.6m

61. Circular Motion

62. v r v = dtdt d = 2πr t = T T is the Period: the time it takes to make one revolution.

63. v = dtdt 2πrT2πrT

64. a c = v2rv2r 4π2rT24π2rT2 Centripetal Acceleration

65. Centripetal Force Centripetal Force is the force toward the center of the circle that keeps an object moving in a circle.

66. Centripetal Force FcFc

67. F c = ma c Centripetal Force 4π2rT24π2rT2 F c = m ( )

68. Example A 15g whistle is being swung on a lanyard 0.30m long. If one revolution takes 0.5s, what is the centripetal force?

69. Example

70. m = 15g =.015kg Given: F c = ? Find: T = 0.5s

71. 4π2rT24π2rT2 F c = m ( ) 4π 2 (.3m) (0.5s) 2 F c =.015m ( ) F c = 0.7N

72. Homework Worksheet Due: 12/14/06

74. Universal Gravity

75. What does gravity depend on? Mass Distance G

76. m1m1 m2m2 m 1 m 2 F ∞

77. r

80. r 1 r 2

81. m1m1 m2m2 F ∞ r m 1 m 2 r 2

82. F = m 1 m 2 r 2 G G – Gravitational Constant

83. G – 6.67 X 10 -11 Nm 2 /kg 2

84. F = Mm r2 r2 G M m

85. What is the magnitude of the gravitational force that acts on each particle, m 1 is 12kg and m 2 is 25kg and the two are 1.2m away. m1m1 m2m2 r

86. F = Gm 1 m 2 r2 r2 F= (6.67x10 -11 Nm 2 /kg 2 )(12kg)(25kg) (1.2m) 2 F= 1.4 x 10 -8 N

87. Acceleration due the Gravity(g) F = GM Earth m r2 r2 mg = GM Earth m r2 r2

88. Acceleration due the Gravity(g) g = GM Earth r2 r2 r

89. g= (6.67x10 -11 Nm 2 /kg 2 )(5.97x10 24 kg) (6.38x10 6 m) 2 g= 9.8m/s 2 g = GM Earth r2 r2

90. Homework Page: 194 Prob: 20,27,28,35,37 Due: 12/18/06 Test: 12/21/06

91. Satellites 1km/s.6mi/s

92. Satellites

93. Satellites – 17500mi/hr

94. Lets find a satellite speed!!! Fc = Fg r

95. ma c = Gm E m r 2 Gm E r 2 = v2rv2r

96. Gm E r = v2v2 = v √

97. Homework Worksheets Page: 194 Prob: 23,24,25,50,51 Due: 12/20/06 Test: 12/21/06

98. Homework Review Worksheet Due: 12/21/06 Test: 12/21/06