1. Push and Pull 2-1-1 Newton’s Laws

2. Newton’s First Law An object at rest remains at rest, and an object in motion continues in motion with constant velocity (that is, constant speed in a straight line) unless it experiences a net external force. Also known as the “LAW OF INERTIA” INERTIA – tendency of an object to maintain its current state of motion

3. Do these guys have a lot of inertia? Inertia is a relative measurement… –More inertia = MORE MASS –More inertia = harder to CHANGE VELOCITY

4. Force A force is a push or pull that may change an object’s state of motion. –CONTACT FORCE – requires touching Normal, Tension, Friction –FIELD FORCE – no contact required (“Action over a distance” Gravity (Weight)

5. A block of wood is placed on a table and is motionless. What forces are acting on it? FgFg FNFN F g = WEIGHT a force pulling any object toward the CENTER OF THE EARTH F N = NORMAL FORCE a reaction force that any object exerts when pushed on

6. Net Force No NET FORCE on an object = EQUILIBRIUM… either –MOTIONLESS –MOVING WITH CONSTANT VELOCITY A “NET” or “UNBALANCED” force changes an object’s VELOCITY –This means that SPEED AND/OR DIRECTION change –Also called ACCELERATION

7. Net Force  Acceleration How much acceleration a net force causes an object to have depends on… –amount of FORCE greater force = GREATER ACCELERATION –amount of MASS greater mass = LESS ACCELERATION

8. Newton’s Second Law The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the mass of the object. Unit of force is the NEWTON (N)

9. Example A 2.0 kilogram box is pushed with a net force of 10. newtons. What is the acceleration experienced by the box? a = F net / m a = (10 N)/(2.0 kg) a = 5.0 m/s 2

10. Weight The force with which gravity pulls on an object.

11. Example What is the weight of an object with a mass of 30 kilograms? F g = mg F g = (30 kg)(9.81 m/s 2 ) F g = 294.3 kg·m/s 2 or 294.3 N

12. Newton’s Third Law “For every action, there is an equal and opposite reaction” FgFg FNFN FgFg FTFT

13. A firefighter directs a stream of water from a hose to the east. In what direction is the force on the hose? To the WEST Examples

14. A man getting out of a rowboat jumps north onto a dock. What happens to the boat? It moves SOUTH Examples

15. A 60 kilogram astronaut pushes against a 120 kilogram satellite with a force of 15 newtons. How much force does the satellite exert on the astronaut? 15 newtons Examples

16. End of 2.1.1 - PRACTICE

17. Drawing Pictures 2.1.2A Free Body Diagrams and Net Force

18. Free Body Diagrams VECTOR diagrams Show ALL FORCES acting on an object Must be properly labeled FgFg FNFN Equilibrium

19. Determining Net Force equilibrium  net force = 0  acceleration = 0 –MOTIONLESS or CONSTANT VELOCITY net force ≠ 0 –SPEEDING UP or SLOWING DOWN

20. Determining Net Force – Top View Consider each dimension separately then combine using Pythagorean Theorem. F 1 = +7.0 N XY +7.0 N 0 N Net force = +7.0 N a = 3.5 m/s 2 +7.0 N 2 kg object Not in notes

21. Determining Net Force – Top View Consider each dimension separately then combine using Pythagorean Theorem. XY +7.0 N -3.0 N 0 N+4.0 N F 2 = -3.0 N F 1 = +7.0 N Net force = +4.0 N a = 2.0 m/s 2 2 kg object Not in notes

22. Determining Net Force – Top View Consider each dimension separately then combine using Pythagorean Theorem. XY +7.0 N -3.0 N +3.0 N+4.0 N F 2 = -3.0 N F 1 = +7.0 N F 3 = +3.0 N +3.0 N Net force = 5.0 N a = 2.5 m/s 2 2 kg object Not in notes

23. Determining Net Force – Top View Consider each dimension separately then combine using Pythagorean Theorem. XY +7.0 N -3.0 N -2.0 N+4.0 N F 2 = -3.0 N F 1 = +7.0 N F 3 = +3.0 N +3.0 N -5.0 N F 4 = -5.0 N Net force = 4.5 N a = 2.25 m/s 2 2 kg object Notes

24. A 30 newton force and a 20 newton force act concurrently on an object. –What are the minimum/maximum net force can these two forces produce? At what angle between the forces does each happen? –What net force is produced when the angle between the two forces is 90°? Example #1 min = 10 N occurs at 180° max = 50 N occurs at 0° 36 N

25. Determining Net Force – Side View Consider each dimension separately. On a flat surface total vertical force = 0. XY +8.0 N 0 N+8.0 N F = +8.0 N -F g +F N frictionless Net force = +8.0 N a = 4.0 m/s 2 2 kg object Not in notes

26. Determining Net Force – Side View Consider each dimension separately. On a flat surface total vertical force = 0. Hor.Vert. +8.0 N -2.0 N 0 N+6.0 N F = +8.0 N -F g +F N Net force = +6.0 N a = 3.0 m/s 2 2 kg object F f = -2.0 N Notes

27. A 50 kilogram object is pushed along a flat, frictionless surface with a constant force of 100 newtons. –Sketch this object. Include all forces with labels and quantities. –What rate of acceleration will the crate experience? Example #2

28. Determining Net Force – Vertical Object moving straight up or down  total horizontal force = 0. Hor.Vert. -20 N 0 N Net force = -20 N a = -9.81 m/s 2 20 N object in freefall F g = -20 N Not in notes

29. What is the net force acting on a 3000 newton rocket if its engine produces an upward thrust of 4500 newtons? Example #3 FgFg F thrust 1500 N upward

31. End of 2.1.2A - PRACTICE

32. Drawing Pictures II 2.1.2B Forces on Angles

33. Redraw force vectors head-to-tail then draw resultant from start to finish. Resultant Forces on Angles – Top View

34. Redraw force vectors head-to-tail then draw resultant from start to finish. Resultant Forces on Angles – Top View

35. Redraw force vectors head-to-tail then draw resultant from start to finish. Resultant Forces on Angles – Top View

36. Redraw force vectors head-to-tail then draw resultant from start to finish. Resultant Forces on Angles – Top View

37. Equilibrants – Top View Equilibrant is the force that is added to a system to produce a net force of ZERO. –Opposite of the RESULTANT Resultant Equilibrant

38. Resultant Equilibrant Equilibrants – Top View

39. Resultant Equilibrant Equilibrants – Top View

40. Resultant Equilibrant Equilibrants – Top View

41. Determine net horizontal force. Vertical forces must total zero. F g = F Y + F N …. F N < F g HV FXFX 0 NFXFX Net force = F X a = F X /m F Forces on Angles – Side View FgFg FNFN FYFY FXFX +F Y -F g +F N

42. Determine net horizontal force. Vertical forces must total zero. F N = F Y + F g …. F N > F g F Forces on Angles – Side View FgFg FNFN FYFY FXFX HV FXFX 0 NFXFX Net force = F X a = F X /m -F Y -F g +F N

43. A 40 kilogram box is pulled across a smooth, frictionless surface with a 20 newton force that is 30° above horizontal. –What is the net force acting on the box? –What is the acceleration of the box? –How could this acceleration be increased? Example #1 F X = F cos θ = 20N cos 30° = 17.3 N a = F net /m a = 17.3 N / 40 kg a = 0.43 m/s 2 Decrease the angle of the pull or increase force.

44. A woman pushes a 30 kilogram lawnmower with a force of 15 newtons at an angle of 60° below horizontal. –What is the net force acting on the mower? –What is the acceleration of the mower? –How could this acceleration be decreased? Example #2 F X = F cos θ = 15N cos -60° = 7.5 N a = F net /m a = 7.5 N / 30 kg a = 0.25 m/s 2 Increase the angle of the pull or decrease force.

45. End of 2.1.2B - PRACTICE

46. Where the Rubber Meets the Road 2.1.3 Friction

47. A force that OPPOSES MOTION. What causes friction? Friction + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + 1.Attraction between protons and electrons in the object and surface

48. What causes friction? Friction 2. Small ridges and bumps at the microscopic level cause objects and surfaces to catch on one another.

49. Friction force strength –How “sticky” are the surfaces involved? Coefficient of friction (μ) – NO UNITS –How hard are the surfaces pushing together? NORMAL FORCE –Are the surfaces sliding against each other? KINETIC vs. STATIC Friction

50. Kinetic –Used ONLY for SLIDING objects Pulled along a surface Skidding Static –Used for any NON-SLIDING objects Not moving Rolling Which Coefficient?

51. A 20 newton wooden box is sliding across a wooden surface. What is the friction force acting on the box? F f = μF N F f = (0.3)(20 N) F f = 6.0 N Example #1

52. A 40 newton copper box is placed on a flat, steel table. What is the minimum force needed to make the box move? Example #2 F f = μF N F f = (0.53)(40 N) F f = 21 N

53. End of 2.1.3 - PRACTICE

54. 2.1.4 Inclined Planes Down the Slope

55. A tool used to move objects from one height to another. Allows for the movement of an object without lifting it directly against gravity. What is an Inclined Plane?

56. Principle Forces Gravitational Force (F g ) – STRAIGHT DOWN –Should be broken into components Perpendicular to incline – F g ┴ = F g cos θ Parallel to incline – F g ║ = F g sin θ FgFg Fg║Fg║ Fg┴Fg┴

57. Principle Forces Normal Force (F N ) – PERPENDICULAR TO PLANE –Always equal to F g ┴ FgFg Fg║Fg║ Fg┴Fg┴ FNFN

58. Principle Forces Friction Force (F f ) – OPPOSES MOTION –If F f < F g ║ then accelerates down plane –If F f < F g ║ then not moving or constant speed FgFg Fg║Fg║ Fg┴Fg┴ FNFN FfFf

59. 0° FgFg FNFN F g ║ = F g sin 0° = 0F g ┴ = F g cos 0° = F g

60. 15° FgFg Fg║Fg║ Fg┴Fg┴ FNFN

61. 30° FgFg Fg║Fg║ Fg┴Fg┴ FNFN

62. 45° FgFg Fg║Fg║ Fg┴Fg┴ FNFN

63. 60° FgFg Fg║Fg║ Fg┴Fg┴ FNFN

64. 90° FgFg F g ║ = F g sin 0° = F g F g ┴ = F g cos 0° = 0

65. As the angle increases NO CHANGE – F g –μ–μ INCREASES –Fg║–Fg║ DECREASES –F g ┴  F N  F f

66. Example A 50 newton box is sliding down a 30° incline at a constant speed. F N = F g ┴ = F g cos θ F N = 50 N cos 30° = 43 N What is the normal force acting on the box? 30°

67. Example A 50 newton box is sliding down a 30° incline at a constant speed. Since v = const… F f = F g ║ = F g sin θ F f = 50 N sin 30° = 25 N What is the friction force acting on the box? 30°

68. Example A 50 newton box is sliding down a 30° incline at a constant speed. F f = μF N μ = F f / F N = 25N / 43 N = 0.58 What is the coefficient of kinetic friction in this system? 30°

69. End of 2.1.4 - PRACTICE