1. Mechanics Kinematics Dynamics

2. Force

3. Fundamental Forces

5. Comparing Contact and Field Forces

6. Concept Check – Newton’s 1 st Law A book is lying at rest on a table. The book will remain there at rest because: 1.there is a net force but the book has too much inertia 2.there are no forces acting on it at all 3.it does move, but too slowly to be seen 4.there is no net force (  F= 0) on the book 5.there is a net force, but the book is too heavy to move

7. Concept Check – Newton’s 1 st Law A book is lying at rest on a table. The book will remain there at rest because: 1.there is a net force but the book has too much inertia 2.there are no forces acting on it at all 3.it does move, but too slowly to be seen 4.there is no net force (  F= 0) on the book 5.there is a net force, but the book is too heavy to move There are forces acting on the book, but the only forces acting are in the y-direction. Gravity acts downward, but the table exerts an upward force that is equally strong, so the two forces cancel, leaving no net force.

8. Concept Check – Newton’s 1 st Law (2) A hockey puck slides on ice at constant velocity. What is the net force acting on the puck? 1.more than its weight 2.equal to its weight 3.less than its weight but more than zero 4.depends on the speed of the puck 5.zero

9. Concept Check – Newton’s 1 st Law (2) A hockey puck slides on ice at constant velocity. What is the net force acting on the puck? 1.more than its weight 2.equal to its weight 3.less than its weight but more than zero 4.depends on the speed of the puck 5.zero The puck is moving at a constant velocity, and therefore it is not accelerating. Thus, there must be no net force acting on the puck. Follow-up: Are there any forces acting on the puck? What are they?

10. Concept Check – Newton’s 1 st Law (3) You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why? 1. a net force acted on it 2. no net force acted on it 3. it remained at rest 4. it did not move, but only seemed to 5. gravity briefly stopped acting on it

11. Concept Check – Newton’s 1 st Law (3) You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why? 1. a net force acted on it 2. no net force acted on it 3. it remained at rest 4. it did not move, but only seemed to 5. gravity briefly stopped acting on it The book was initially moving forward (since it was on a moving bus). When the bus stopped, the book continued moving forward, which was its initial state of motion, and therefore it slid forward off the seat. Follow-up: What is the force that usually keeps the book on the seat?

12. You kick a smooth flat stone out on a frozen pond. The stone slides, slows down and eventually stops. You conclude that: 1. the force pushing the stone forward finally stopped pushing on it 2. no net force acted on the stone 3. a net force acted on it all along 4. the stone simply “ran out of steam” 5. the stone has a natural tendency to be at rest Concept Check – Newton’s 1st Law (4)

13. You kick a smooth flat stone out on a frozen pond. The stone slides, slows down and eventually stops. You conclude that: 1. the force pushing the stone forward finally stopped pushing on it 2. no net force acted on the stone 3. a net force acted on it all along 4. the stone simply “ran out of steam” 5. the stone has a natural tendency to be at rest After the stone was kicked, no force was pushing it along! However, there must have been some force acting on the stone to slow it down and stop it. This would be friction. Concept Check – Newton’s 1st Law (4)

14. Newton’s First Law

16. Concept Check – Newton’s First Law Consider a cart on a horizontal frictionless table. Once the cart has been given a push and released, what will happen to the cart? 1. slowly come to a stop 2. continue with constant acceleration 3. continue with decreasing acceleration 4. continue with constant velocity 5. immediately come to a stop

17. Concept Check – Newton’s First Law Consider a cart on a horizontal frictionless table. Once the cart has been given a push and released, what will happen to the cart? 1. slowly come to a stop 2. continue with constant acceleration 3. continue with decreasing acceleration 4. continue with constant velocity 5. immediately come to a stop After the cart is released, there is no longer a force in the x-direction. This does not mean that the cart stops moving. It simply means that the cart will continue moving with the same velocity it had at the moment of release. The initial push got the cart moving, but that force is not needed to keep the cart in motion.

18. Concept Check – Newton’s Second Law We just decided that the cart continues with constant velocity. What would have to be done in order to have the cart continue with constant acceleration? 1. push the cart harder before release 2. push the cart longer before release 3. push the cart continuously 4. change the mass of the cart 5. it is impossible to do that

19. Concept Check – Newton’s Second Law We just decided that the cart continues with constant velocity. What would have to be done in order to have the cart continue with constant acceleration? 1. push the cart harder before release 2. push the cart longer before release 3. push the cart continuously 4. change the mass of the cart 5. it is impossible to do that In order to achieve a non-zero acceleration, it is necessary to maintain the applied force. The only way to do this would be to continue pushing the cart as it moves down the track. This will lead us to a discussion of Newton’s Second Law.

20. Concept Check – Newton’s 2 nd Law From rest, we step on the gas of our Ferrari, providing a force F for 4 seconds, speeding it up to a final speed v. If the applied force were only ½ F, how long would it have to be applied to reach the same final speed? 1.16 s 2.8 s 3.4 s 4.2 s 5.1 s v F

21. Concept Check – Newton’s 2 nd Law From rest, we step on the gas of our Ferrari, providing a force F for 4 seconds, speeding it up to a final speed v. If the applied force were only ½ F, how long would it have to be applied to reach the same final speed? 1.16 s 2.8 s 3.4 s 4.2 s 5.1 s therefore the acceleration is also half In the first case, the acceleration acts over time  t = 4 s to give velocity v = a  t. In the second case, the force is half, therefore the acceleration is also half, so to achieve the same final speed, the time must be doubled. v F

22. Concept Check – Newton’s 2 nd Law A force F acts on mass M for a time interval  t, giving it a final speed v. If the same force acts for the same time on twice the mass 2M, what would be the final speed of the bigger mass? 1. 4 v 2. 2 v 3. v 4. ½ v 5. ¼ v

23. Concept Check – Newton’s 2 nd Law A force F acts on mass M for a time interval  t, giving it a final speed v. If the same force acts for the same time on twice the mass 2M, what would be the final speed of the bigger mass? 1. 4 v 2. 2 v 3. v 4. ½ v 5. ¼ v so the acceleration is cut in half In the first case, the acceleration acts over time  t to give velocity v = a  t. In the second case, the mass is doubled, so the acceleration is cut in half; therefore, in the same time  t, the final speed will only be half as much.

24. Concept Check – Newton’s 2 nd Law A force F acts on mass m 1 giving acceleration a 1. The same force acts on a different mass m 2 giving acceleration a 2 = 2a 1. If m 1 and m 2 are glued together and the same force F acts on this combination, what is the resulting acceleration? 1. 3/4 a 1 2. 3/2 a 1 3. 1/2 a 1 4. 4/3 a 1 5. 2/3 a 1 F a1a1 m1m1 F m2m2 m1m1 a3a3 F a 2 = 2a 1 m2m2

25. Concept Check – Newton’s 2 nd Law A force F acts on mass m 1 giving acceleration a 1. The same force acts on a different mass m 2 giving acceleration a 2 = 2a 1. If m 1 and m 2 are glued together and the same force F acts on this combination, what is the resulting acceleration? 1. 3/4 a 1 2. 3/2 a 1 3. 1/2 a 1 4. 4/3 a 1 5. 2/3 a 1 Mass m 2 must be (1/2)m 1 because its acceleration was 2a 1 with the same force. Adding the two masses together gives (3/2)m 1, leading to an acceleration of (2/3)a 1 for the same applied force. F a1a1 m1m1 F m2m2 m1m1 a3a3 F a 2 = 2a 1 m2m2

26.  F represents the vector sum of all external forces acting on the object, or the net force. Newton’s 2 nd Law

27. Newton’s Third Law

28. Action and Reaction Forces If A acts on B, then B acts back on A

29. Concept Check – Newton’s 3 rd Law When you climb up a rope, the first thing you do is pull down on the rope. How do you manage to go up the rope by doing that?? 1. this slows your initial velocity, which is already upward 2. you don’t go up, you’re too heavy 3. you’re not really pulling down – it just seems that way 4. the rope actually pulls you up 5. you are pulling the ceiling down

30. Concept Check – Newton’s 3 rd Law When you climb up a rope, the first thing you do is pull down on the rope. How do you manage to go up the rope by doing that?? 1. this slows your initial velocity, which is already upward 2. you don’t go up, you’re too heavy 3. you’re not really pulling down – it just seems that way 4. the rope actually pulls you up 5. you are pulling the ceiling down When you pull down on the rope, the rope pulls up on you!! reactionrope on you you exerted on the rope When you pull down on the rope, the rope pulls up on you!! It is actually this upward force by the rope that makes you move up! This is the “reaction” force (by the rope on you) to the force that you exerted on the rope. And voilá, this is Newton’s 3 rd Law.

31. Concept Check – Newton’s 3 rd Law A small car collides with a large truck. Which experiences the greater impact force? 1. the car 2. the truck 3. both the same 4. it depends on the velocity of each 5. it depends on the mass of each

32. Concept Check – Newton’s 3 rd Law A small car collides with a large truck. Which experiences the greater impact force? 1. the car 2. the truck 3. both the same 4. it depends on the velocity of each 5. it depends on the mass of each

33. Concept Check – Newton’s 3 rd Law A small car collides with a large truck. Which experiences the greater acceleration? 1. the car 2. the truck 3. both the same 4. it depends on the velocity of each 5. it depends on the mass of each

34. Concept Check – Newton’s 3 rd Law A small car collides with a large truck. Which experiences the greater acceleration? 1. the car 2. the truck 3. both the same 4. it depends on the velocity of each 5. it depends on the mass of each F/m We have seen that both vehicles experience the same magnitude of force. But the acceleration is given by F/m so the car has the larger acceleration, since it has the smaller mass.

35. Concept Check – Newton’s 3 rd Law F F 12 F F 21 1. the bowling ball exerts a greater force on the ping-pong ball 2. the ping-pong ball exerts a greater force on the bowling ball 3. the forces are equal 4. the forces are zero because they cancel out 5. there are actually no forces at all In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare?

36. Concept Check – Newton’s 3 rd Law F F 12 F F 21 1. the bowling ball exerts a greater force on the ping-pong ball 2. the ping-pong ball exerts a greater force on the bowling ball 3. the forces are equal 4. the forces are zero because they cancel out 5. there are actually no forces at all In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare? forces The forces are equal and opposite by Newton’s 3 rd Law!

37. Concept Check – Newton’s 3 rd Law F F 12 F F 21 1. they do not accelerate because they are weightless 2. accelerations are equal, but not opposite 3. accelerations are opposite, but bigger for the bowling ball 4. accelerations are opposite, but bigger for the ping-pong ball 5. accelerations are equal and opposite In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the accelerations of the objects compare?

38. Concept Check – Newton’s 3 rd Law F F 12 F F 21 1. they do not accelerate because they are weightless 2. accelerations are equal, but not opposite 3. accelerations are opposite, but bigger for the bowling ball 4. accelerations are opposite, but bigger for the ping-pong ball 5. accelerations are equal and opposite In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the accelerations of the objects compare? smaller massbigger acceleration The forces are equal and opposite -- this is Newton’s 3 rd Law!! But acceleration is F/m and so the smaller mass has the bigger acceleration. Follow-up: Where will the balls meet if they are released from this position?

39. Newton’s 2 nd and 3 rd Laws FF m a = FF m a = FF m a = CarTruck FF m a = FF m a = FF m a = 2 nd LawYouEarth 2 nd Law

40. Although several forces are acting on this car, the vector sum of the forces is zero. Thus, the net force is zero, and the car moves at a constant velocity or remains at rest. Net Force

41. Force Diagrams In a force diagram, vector arrows represent all the forces acting in a situation. A free-body diagram shows only the forces acting on the object of interest—in this case, the car. Force Diagram Free-Body Diagram

42. Free Body Diagrams

43. Common Forces to Consider Gravity Normal Friction Push and Pull Tension

44. Free Body Diagrams

45. Free Body Diagram Flash Tutorial

46. Drawing a Free-Body Diagram FgFg FkFk FaFa FnFn y x Free Body Diagram Flash Tutorial x

47. Drawing a Free-Body Diagram FgFg FkFk FaFa FnFn y x Free Body Diagram Flash Tutorial x

48. Net Force equations from Free Body Diagrams Falling, no air resistance Falling through air, speeding up DescriptionFBDNet Force Equations

49. Weight F g = ma g = mga g = g = 9.81 m/s 2

50. Air Resistance

51. Net Force equations from Free Body Diagrams Falling at Terminal Velocity Falling, but slowing DescriptionFBDNet Force Equations

52. Net Force equations from Free Body Diagrams Elevator at rest Elevator moving up or down at constant velocity DescriptionFBDNet Force Equations

53. Net Force equations from Free Body Diagrams Elevator moving up, slowing down Elevator moving up, speeding up DescriptionFBDNet Force Equations

54. Net Force equations from Free Body Diagrams Elevator moving down, speeding up Elevator moving down, slowing DescriptionFBDNet Force Equations

55. Sample An elevator of mass _____ kg is slowing as it nears its final destination going up. If the tension in the rope is ______ N, determine the elevator’s acceleration. (2 force vectors) G: U: E: S:

56. Net Force equations from Free Body Diagrams At rest on a surface or sliding with no friction Pushed on a surface DescriptionFBDNet Force Equations

57. Friction

58. The quantity that expresses the dependence of frictional forces on the particular surfaces in contact is called the coefficient of friction, . The Coefficient of Friction

59. Coefficients of Friction Coefficient of static friction: Coefficient of kinetic friction:

60. Coefficients of Friction You will find this table in your text

61. In free-body diagrams, the force of friction is always parallel to the surface of contact. The force of kinetic friction is always opposite the direction of motion. To determine the direction of the force of static friction, use the principle of equilibrium. For an object in equilibrium, the frictional force must point in the direction that results in a net force of zero. Friction Forces in Free-Body Diagrams If moving, kinetic friction opposes motion If not moving, static friction is in a direction that provides equilibrium

62. Net Force Equations From Free Body Diagrams Sliding and slowing due to friction DescriptionFBDNet Force Equations

63. Net Force equations from Free Body Diagrams Pushed on a surface with friction DescriptionFBDNet Force Equations

64. Sample Problem m = 55 kg F a = 19 N to set in motion g = 9.81 m/s 2 a x = 0 a y = 0  s = ?

65. Solving Problems using Newton’s Laws Given Include Free Body Diagram Imagine x-y axis Choose + direction Unknown Equations ΣF = ma (or ΣF = 0 ) F k = μ k F n (or F s = μ s F n ) F g = mg Use components if necessary Solve system of equations if necessary Substitute only at the end Significant Figures

66. Net Force equations from Free Body Diagrams Pushed down at an angle on a surface (no friction) DescriptionFBDNet Force Equations

67. Net Force equations from Free Body Diagrams Pulled up at an angle no friction DescriptionFBDNet Force Equations

68. Net Force equations from Free Body Diagrams Pulled up at an angle with friction Pushed down at an angle with friction DescriptionFBDNet Force Equations

69. Chapter 4 Sample Problem m FaFa  kk g a y = 0 a x = ?

70. Chapter 4 Sample Problem (with numbers) m = 20.0 kg F a = 90 N  = 30.0°  k = 0.500 g = 9.81 m/s 2 a y = 0 a x = ?

71. Net Force equations from Free Body Diagrams Sliding down an incline (no friction) DescriptionFBDNet Force Equations

72. Net Force equations from Free Body Diagrams Sliding down an incline with friction DescriptionFBDNet Force Equations

73. Sample A _____ kg child-and-sled combination is sliding down a snowy slope at ____ degrees below the horizontal. If the coefficient of kinetic friction is ____, what is the acceleration? (3 force vectors) G: U: E: