1. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 10/16 do now A racing car is moving around the circular track of radius 300 meters shown. At the instant when the car’s velocity is directed due east, its acceleration is directed due north and has a magnitude of 3 m/s 2. When viewed from above, the car is moving 1.Clockwise at 30 m/s 2.Clockwise at 10 m/s 3.Counterclockwise at 30 m/s 4.Counterclockwise at 10 m/s 5.With constant velocity

2. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley PowerPoint ® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman Lectures by James Pazun Chapter 4 Newton’s Laws of Motion

3. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Goals for Chapter 4 To visualize force as a vector To find the net force acting on a body and apply Newton’s First Law To study mass, acceleration, and their application to Newton’s Second Law To calculate weight and compare/contrast it with mass To see action–reaction pairs and study Newton’s Third Law

4. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 4-1 force and interactions What are the properties of force(s)?

5. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley There are four common types of forces The normal force— When an object rests or pushes on a surface, the surface pushes back. Frictional forces—In addition to the normal force, surfaces can resist motion along the surface.

6. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley There are four common types of forces II Tension forces—When a force is exerted through a rope or cable, the force is transmitted through that rope or cable as a tension. Weight—Gravity’s pull on an object. This force can act from large distances.

7. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley What are typical sizes for common forces?—Table 4.1

8. The unit of force Force is a quantity which is measured using the standard metric unit known as the Newton. A Newton is abbreviated by a "N." To say "10.0 N" means 10.0 Newton of force. One Newton is the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s. Thus, the following unit equivalency can be stated:

9. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Force is a vector quantity Use a vector arrow to indicate magnitude and direction of the force.

10. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Use the net (overall) force—Figure 4.4 Several forces acting on a point have the same effect as their vector sum acting on the same point.

11. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Decomposing a force into components F x and F y are the parallel and perpendicular components of a force to a sloping surface. Use F*Cos θ and F*Sin θ operations to find force components.

12. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Notation and method for the vector sum—Figure 4.7 We refer to the vector sum or resultant as the “sum of forces” R = F 1 + F 2 + F 3 … F n = Σ F. Use Tan θ = R y /R x and R = (R x 2 + R y 2 ) 1/2.

13. It’s more convenient to describe a force F in terms of its x and y components F x and F y Our coordinate axes does not have to be vertical and horizontal. Note: we draw a wiggly line through the force vector F to show that we have replaced it by its x and y components.

14. Example 4.1 superposition of forces Three professional wrestlers are fighting over the same champion's belt. As viewed from above, they apply the three horizontal forces the belt that are shown in the figure. The magnitudes of the three forces are F 1 = 250 N, F 2 = 50 N, F 3 = 120 N. Find the x and y components of the net force on the belt, and find the magnitude and direction of the net force.

15. Test your understanding 4.1 The figure shows a force F acting on a crate. With the x- and y- axes shown in the figure, which statement about the components of the gravitational force that the earth exerts on the crate (the crate’s weight) is correct? 1.The x- and y-components are both positive 2.The x-components is zero and the y-component is positive 3.The x-component is negative and the y-component is positive 4.The x- and y-components are both negative 5.The x-component is zero and the y-component is negative 6.The x-component is positive and the y-component is negative.

16. 10/17 Do now A 50 Newton weight is suspended by two cords as shown in the figure above. Then tension in the slanted cord is 1.50 N 2.100 N 3.150 N 4.200 N 5.250 N

17. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 4.2 Newton’s First Law – a body in equilibrium Simply stated—“objects at rest tend to stay at rest, objects in motion stay in motion.” More properly, “A body acted on by no net force moves with constant velocity and zero acceleration.”

18. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Newton’s First Law II—Figure 4.10 Figure 4.10 shows an unbalanced force causing an acceleration and balanced forces resulting in no motion.

19. Example 4.2 – zero net force means constant velocity In the classic 1950 science fiction film X-M, a spaceship is moving in the vacuum of outer space, far from any planet, when its engine dies. As a result, the spaceship slows down and stops, what does Newton's first law say about this event? According the Newton’s 1 st Law, an object in motion will remain in motion. The spaceship will not slow down, it will travel at constant velocity forever.

20. Example 4.3: constant velocity means zero net force You are driving a Porsche Carrera GT on a straight testing track at a constant speed of 150 km/h. you pass a 1971 Volkswagen beetle doing a constant 75 km/h. For which car is the net force greater? Since both cars are in equilibrium because their velocities are constant, therefore the net force on each car is zero.

21. Suppose you are in a bus that is traveling on a straight road and speeding up. If you could stand in the aisle on roller skates, you would start moving backward relative to the bus as the bus gains speed. If instead the bus was slowing to a stop, you would start moving forward down the aisle. In either case, it looks as though Newton’s first law is not obeyed; there is no net force acting on you, yet your velocity change. What wrong?

22. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Inertial frames of reference—Figure 4.11 When a car turns and a rider continues to move, the rider perceives a force.

23. Inertial frame of reference The bus is accelerating with respect to the earth and in not a suitable frame of reference for Newton’s 1 st law. A frame of reference in which Newton's 1 st law is valid is called an inertial frame of reference. The earth is at least approximately an inertial frame of reference, but the bus is not. Because Newton’s first law is used to determine what we mean by an inertial frame of reference, it is sometimes called the law of inertia.

24. Test your understanding 4.2 In which of the following situations is there zero net force on the body? 1.An airplane flying due north at a steady 120 m/s and at a constant altitude. 2.A car driving straight up a hill with a 3 o slope at a constant 90 km/h 3.A hawk circling at a constant 20 km/h at a constant height of 15 m above an open field 4.A box with slick, frictionless surfaces in the back of a truck as the truck accelerates forward on a level road at 5 m/s 2.

25. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Newton’s Second Law—Figure 4.13 An unbalanced force (or sum of forces) will cause a mass to accelerate.

26. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley An object undergoing uniform circular motion Centripetal force is the net force. We have already seen the centripetal acceleration. But, if we measure the mass in motion, Newton’s Second Law allows us to calculate the centripetal force.

27. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley The relationship of F, m, and a Acceleration is directly proportional to the applied force and always in the same directions as the net force.

28. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley The relationship of F, m, and a redux acceleration is inversely proportional to the object’s mass.

29. Suppose we apply a constant net force ∑F to a body having a known mass m 1 and we find an acceleration of magnitude a 1. We then apply the same force to another body having an unknown mass m 2, and we find an acceleration of magnitude a 2. For the same net force, the ratio of the masses of two bodies is the inverse of the ratio of their acceleration.

30. Applying Newton's second law Since the equation is a vector equation, we will use it in component form, with a separate equation for each component of force and the corresponding acceleration: Note: 1. ∑F means the sum of all external forces. 2. the equation is only valid if mass is constant. 3. the equation is only valid in the inertial frame of reference

31. caution ma is not a force. The vector ma is equal to the vector sum of all the forces acting on the body (∑F)

32. Example 4.4 determining acceleration from force A worker applies a constant horizontal force with magnitude 20 N to a box with mass 40 kg resting on a level floor with negligible friction. What is the accelerations of the box?

33. Example 4.5 determining force from acceleration A waitress shoves a ketchup bottle with mass 0.45 kg to the right along a smooth, level lunch counter. The bottle leaves her hand moving at 2.8 m/s, then slows down as it slides because of the constant horizontal friction force exerted on it by the counter top. It slides a distance of 1.0 m before coming to rest. What are the magnitude and direction of the friction force acting on the bottle?

34. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Newtons, kilograms, pounds, and slugs—Table 4.2 Table 4.2 rightly points out that the pound is a force. The popular culture refers to it as a weight (which is actually a slug). The Dyne is actually a cgs version of the Newton (sometimes used with fine work on tiny objects).

35. Test your understanding Rank the following situations in order of the magnitude of the object’s acceleration, from lowest to highest. Are there any cases that have the same magnitude of acceleration? a.A 2.0 kg object acted on by a 2.0 N net force; b.A 2.0 kg object acted on by an 8.0 N net force. c.An 8.0 kg object acted on by a 2.0 N net force. d.An 8.0 kg object acted on by a 8.0 N net force. c, d = a, b

36. 10/25 do now A small box is on ramp tilted at an angle θ above the horizontal. The box may be subject to the following forces: friction (f), gravity (mg) and normal force (N). Draw a free body diagram best represents the box if it is at rest on the ramp. N mg f 54%

37. 4.4 mass and weight Mass characterizes the inertial properties of a body. The greater the mass, the greater the force needed to cause a given acceleration; ∑F = ma Weight is a force exerted on a body the pull of the earth. Bodies having large mass also have large weight.

38. The force that makes the body accelerate downward at 9.8 m/s 2 is its weight. A body with mass m has weight of magnitude of w w = mg –The magnitude w of a body’s weight is directly proportional to its mass m. –The weight of a body is a force, a vector quantity, w = mg

39. Caution: A body’s weight acts at all times

40. Example 4.6 Net force and acceleration in free fall A one-euro coin was dropped from rest from the Leaning Tower of Pisa. If the coin falls freely, so that the effects of the air are negligible, how does the net force on the coin vary as it falls?

41. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley g, and hence weight, is only constant on earth, at sea level On Earth, g depends on your altitude. On other planets, gravity will likely have an entirely new value. Mass doesn’t change, g and weight changes with location.

42. Measuring mass and weight The easiest way to measure the mass of a body is to measure its weight, often by comparing with a standard. Two bodies that have the same weight at a particular location also have the same mass. The equal-arm balance can determine with great precision when the weights of two bodies are equal and hence their masses are equal. In outer space, we can compute the mass as the ratio of force to acceleration.

43. Example 4.7 mass and weight

44. Test your understanding of section 4.4 Suppose an astronaut landed on a planet where g = 19.6 m/s 2. Compare to Earth, would it be easier, harder, or just as earth for her to walk around? Would it be easier, harder, or just as easy for her to catch a ball that is moving horizontally at 12 m/s (assume that the astronaut’s spacesuit is a light-weight model that doesn’t impede her movements in any way.)

45. 10/18 Do now (ap 2009 exam) The velocity v of an elevator moving upward between adjacent floors is shown as a function of time t in the graph. At which of the following times is the force exerted by the elevator floor on a passenger the least? v (m/s) t (s) 1 2 3 4 5 6 7 8 F norm F grav A. 1 s B. 3 s C. 4 s D. 5 s E. 6 s 29% correct

46. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 4.5 Newton’s Third Law A force acting on a body is always the result of its interaction with another body, so forces always comes in pairs. Exerting a force on a body results in a force back upon you.

47. CAUTION: the two forces in an action-reaction pair act on different bodies. Unlike in Newton's 1 st or 2 nd Law, which involve the forces act on one body. The action and reaction forces can be contact forces or long-range forces. When you drop a ball, both the ball and the earth accelerate toward each other. The net force on each body ahs the same magnitude, but the earth’s acceleration is microscopically small because its mass is so great. Nevertheless it does move!

48. Example 4.8 which force is greater? After your sports car breaks down, you start to push it to the nearest repair shop. While the car is starting to move, how does the force you exert on the car compare to the force the car exerts on you? How do these forces compare when your are pushing the car along at a constant speed? In both cases, the force you exert on the car is equal in magnitude and opposite in direction to the force the car exerts on you.

49. Example 4.9 applying Newton's 3 rd law – object at rest An apple sits on a table in equilibrium. What forces act on it? What is the reaction force to each of the forces acting on the apple? What are the action-reaction pairs?

50. Example 4.10 applying Newton's 3 rd law – object in motion A stonemason drags a marble block across a floor by pulling on a rope attached to the block. The block may or may not be in equilibrium. How are the various force related? What are the action-reaction pairs?

51. Action: Man on Rope Reaction: Rope on Man Action: Rope on Box Reaction: Box on Rope Man on Rope Box on Rope Man on Rope Rope on Box Action-Reaction Pair Not Action-Reaction Pair

52. Example 4.11 We saw in example 4.10 that the stonemason pulls as hard on the rope-block combination as that combination pulls back on him. Why, then, does the block move while the stonemason remains stationary?

53. Check your understanding 4.5 You are driving your car on a country road when a mosquito splatters itself on the windshield. Which has the greater magnitude: 1.The force that the car exerted on the mosquito? 2.The force that the mosquito exerted on the car? Why is the mosquito splattered while the car is undamaged? They are the same in magnitude and opposite in direction. The car’s mass is much bigger than the mosquito, therefore, its acceleration a = F net / mass is much smaller than that of the mosquito.

54. 4.6 free-body diagrams 1.Newton’s first and seconds laws apply to a specific body. Newton’s 1 st law, ∑F = 0, (equilibrium) Newton’s 2 nd law, ∑F = ma, (non equilibrium) 2.Only forces acting on the body matter. 3.Free-body diagram are essential to help identify the relevant forces. “free” of its surroundings, with vectors drawn to show the magnitudes and directions of all the forces applied to the body by the various other bodies that interact with it. 4.Label all the forces acting on the body.

55. Free-body diagrams—Figure 4.30 A sketch then an accounting of forces

56. Test your understanding 4.6 The buoyancy force shown in the figure is one half of an action-reaction pair. What force is the other half of this pair? 1.The weight of the swimmer; 2.The forward thrust force 3.The backward drag force 4.The downward force that the swimmer exerts on the water 5.The backward force that the swimmer exerts on the water by kicking.

57. Exercise 4.33 A truck is pulling a car on a horizontal highway using a horizontal rope. The car is in neutral gear, so we can assume that there is no appreciable friction between its tires and the highway. As the truck is accelerating to highway speeds, draw a free-body diagram of a.The car b.The truck c.What force accelerates this system forward? Explain how this force originates. As the tires of the truck push backwards on the highway surface as they rotate, the roadway pushes forward on the tires.

58. 10/26 do now A small box is on ramp tilted at an angle θ above the horizontal. The box may be subject to the following forces: friction (f), gravity (mg) and normal force (N). Draw a free body diagram best represents the box if it is sliding down the ramp at constant speed. N mg f 54%

59. objective Chapter 4 review Chapter 5 Homework: 4.27, 4.29, 4.31

64. Do now (ap 2009 exam) The velocity v of an elevator moving upward between adjacent floors is shown as a function of time t in the graph. At which of the following times is the force exerted by the elevator floor on a passenger the least? v (m/s) t (s) 1 2 3 4 5 6 7 8 F norm F grav A. 1 s B. 3 s C. 4 s D. 5 s E. 6 s 29% correct

65. 10/20 do now A small box is on ramp tilted at an angle θ above the horizontal. The box may be subject to the following forces: friction (f), gravity (mg), applied parallel to the incline (F p ) and normal force (N). Draw a free body diagram best represents the box if it is accelerating up the ramp. N mg FpFp f 96%

66. A. B. C. D. E.