1. Newton’s Laws Applications

2. Monday, October 9, 2006 Introduction to Friction

3. Announcements Exam corrections tomorrow, Thursday, Friday. Lunch Bunch Wednesday Homeroom: Federal Forms Due ASAP

4. Friction Friction is the force that opposes a sliding motion. Friction is due to microscopic irregularities in even the smoothest of surfaces. Friction is highly useful. It enables us to walk and drive a car, among other things. Friction is also dissipative. That means it causes mechanical energy to be converted to heat. We’ll learn more about that later.

5. Microscopic View W N Friction may or may not exist between two surfaces. The direction of friction, if it exists, is opposite to the direction object will slide when subjected to a horizontal force. It is always parallel to the surface. F push f (friction) Small view: Microscopic irregularities resist movement. Big view: Surfaces look perfectly smooth.

6. Friction depends on the normal force. The friction that exists between two surfaces is directly proportional to the normal force. Increasing the normal force increases friction; decreasing the normal force decreases friction. This has several implications, such as… Friction on a sloping surface is less than friction on a flat surface (since the normal force is less on a slope). Increasing weight of an object increases the friction between the object and the surface it is resting on. Weighting down a car over the drive wheels increases the friction between the drive wheels and the road (which increases the car’s ability to accelerate).

7. Static Friction This type of friction occurs between two surfaces that are not slipping relative to each other. f s   s N f s : static frictional force (N)  s : coefficient of static friction N: normal force (N)

8. f s <  s N is an inequality! The fact that the static friction equation is an inequality has important implications. Static friction between two surfaces is zero unless there is a force trying to make the surfaces slide on one another. Static friction can increase as the force trying to push an object increases until it reaches its maximum allowed value as defined by  s. Once the maximum value of static friction has been exceeded by an applied force, the surfaces begin to slide and the friction is no longer static friction.

9. Static friction and applied horizontal force Physics N W Force Diagram surface f s = 0 There is no static friction since there is no applied horizontal force trying to slide the book on the surface.

10. Static friction and applied horizontal force Physics N W Force Diagram surface Ffsfs 0 < f s <  s N and f s = F Static friction is equal to the applied horizontal force, and there is no movement of the book since  F = 0.

11. Static friction and applied horizontal force Physics N W Force Diagram surface Ffsfs f s =  s N and f s = F Static friction is at its maximum value! It is still equal to F, but if F increases any more, the book will slide.

12. Static friction and applied horizontal force Physics N W Force Diagram surface Ffkfk f s =  s N and f s < F Static friction cannot increase any more! The book accelerates to the right. Friction becomes kinetic friction, which is usually a smaller force.

13. Static friction on a ramp Physics N surface fsfs W x = mgsin  and N = mgcos  At maximum angle before the book slides, we can prove that  s = tan  W = mg  Without friction, the book will slide down the ramp. If it stays in place, there is sufficient static friction holding it there.

14. Static friction on a ramp Physics N surface fsfs f s = mgsin  and N = mgcos  At maximum angle before the book slides, we can prove that  s = tan  W = mg   F = 0 W x = f s mgsin  =  s mgcos   s = sin  cos  = tan  Assume  is maximum angle for which book stays in place. x  WxWx

15. Kinetic Friction This type of friction occurs between surfaces that are slipping past each other. f k =  k N f k : kinetic frictional force (N)  k : coefficient of kinetic friction N: normal force (N) Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces.

16. Sample Problem A 10-kg box rests on a ramp that is laying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. a) What is the maximum horizontal force that can be applied to the box before it begins to slide? b) What force is necessary to keep the box sliding at constant velocity?

17. Sample Problem A 10-kg wooden box rests on a ramp that is lying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp if a) no force horizontal force is applied to the box? b) a 20 N horizontal force is applied to the box? c) a 60 N horizontal force is applied to the box?

18. Tuesday, October 10, 2006 Coefficients of Friction Laboratory

19. Announcements Tomorrow HW Due Geometric Optics #6 Newton’s II #1 Free body diagrams and calculations from today’s laboratory Exam corrections Today, Thursday, Friday

20. Problem A 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp if a) the ramp is at a 25 o angle? b) the ramp is at a 45 o angle? c) what is the acceleration of the box when the ramp is at 45 o ?

21. Laboratory Determine the coefficients of static and kinetic friction between the wooden block (felt side) and the cart track. The only additional equipment you may use is a meter stick, a clamp, and a pole. For homework, calculate the coefficients of friction. Include diagrams (free-body), calculations, and results for each kind of friction.

22. Wednesday, October 11, 2006 More experiments to determine coefficients of friction.

23. Announcements HW due today: Newton’s II assignment #1 (folder) Geometric Optics #6 (folder) Free body diagrams and calculations from yesterday’s lab (pass it forward). Clicker Quiz Homework Questions from Newton’s II assignment #1

24. Laboratory Determine the coefficients of friction (static and kinetic) between the felt- covered wood block and the cart track using spring scales. Tomorrow you will turn in the following for each coefficient of friction: Free body diagram Calculation of coefficient of friction. Balances are available for measurement of mass.

25. Thursday, October 12, 2006 Tension and Strings and Springs

26. Announcements Turn in the lab from yesterday. HW due tomorrow: Newton’s #1 and #2. Lab from today. (Include free body diagram, data, and calculations). There will be no Lunch or Breakfast Bunch next week due to PSAT/NMSQT. However, your geometric optics homework must be turned in on or before next Wednesday.

27. Tension Tension is a pulling force that arises when a rope, string, or other long thin material resists being pulled apart without stretching significantly. Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string.

28. A physical picture of tension Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string.

29. Tension examples Note that the pulleys shown are magic! They affect the tension in any way, and serve only to bend the line of action of the force.

30. Sample problem A.A 1,500 kg crate hangs motionless from a crane cable. What is the tension in the cable? Ignore the mass of the cable. B.Suppose the crane accelerates the crate upward at 1.2 m/s 2. What is the tension in the cable now?

31. Springs (Hooke’s Law) The magnitude of the force exerted by a spring is proportional to the amount it is stretched. F = kx F: force exerted by the spring (N) k: force constant of the spring (N/m or N/cm) x: displacement from equilibrium (unstretched and uncompressed) position (m or cm) The direction of the force is back toward the equilibrium (or unstretched) position.

32. Sample problem A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?

33. Sample problem A 1.80 kg object is connected to a spring of force constant 120 N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60.

34. Laboratory Using the ramp at an angle, determine the coefficient of kinetic friction between the felt side of the block and the ramp by allowing the block to accelerated down the ramp. Tomorrow, you will turn in the appropriate free body diagrams and calculations.

35. Friday, October 13, 2006 Connected Objects

36. Announcements Turn in the lab from yesterday. HW due today: Newton’s #1 and #2. Lab from today is due Monday. (Include free body diagram, data, and calculations). There will be no Lunch or Breakfast Bunch next week due to PSAT/NMSQT. However, your geometric optics homework must be turned in on or before next Wednesday. Exam corrections should be completed today. My room will be unavailable at lunch all next week. (It’s my fall recommendation writing workshop!) I will be available every morning but Thursday.

37. Sample problem A 5.0 kg object (m 1 ) is connected to a 10.0 kg object (m 2 ) by a string. If a pulling force F of 20 N is applied to the 5.0 kg object as shown, A) what is the acceleration of the system? B) what is the tension in the string connecting the objects? (Assume a frictionless surface.)

38. Gravity A very common accelerating force is gravity. Here is gravity in action. The acceleration is g.

39. The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g. Slowing gravity down

40. Magic pulleys on a flat table Magic pulleys bend the line of action of the force without affecting tension. Frictionless table m1m1 m2m2 T m2gm2g N m1gm1g T -x x  F = ma m 2 g + T – T = (m 1 + m 2 )a a = m 2 g/(m 1 +m 2 )

41. Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find (a) the acceleration of each block. (b) the tension in the connecting string. Sample problem m1m1 m2m2

42. Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary. Sample problem m1m1 m2m2

43. Sample problem - solution m1m1 m2m2 T m2gm2g N m1gm1g T fsfs  F = 0 m 2 g - T + T – f s = 0 f s = m 2 g  s N = m 2 g  s m 1 g = m 2 g  s = m 2 /m 1 = 0.50

44. Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If  s = 0.30 and  k = 0.20, what is (a) the acceleration of each block? (b) the tension in the connecting string? Sample problem m1m1 m2m2 Note: we know from previous problem that the static friction is not enough to hold the blocks in place!

45. Sample problem – solution (a) m1m1 m2m2 T m2gm2g N m1gm1g T  F = ma m 2 g - T + T – f k = ma m 2 g -  k m 1 g = (m 1 + m 2 )a a = (m 2 -  k m 1 ) g/(m 1 + m 2 ) a = 2.0 m/s 2 fkfk

46. Sample problem – solution (b) m1m1 m2m2 T m2gm2g N m1gm1g T Using block 2  F = ma m 2 g - T = m 2 a T = m 2 (g – a) T = 40 N fkfk Using block 1  F = ma T - f k = m 1 a T = m 1 (a +  k g) T = 40 N Using that the acceleration is 2.0 m/s 2 from part a)

47. Laboratory Using the pulley, masses, friction block, cart (for added weight), and cart track, determine the coefficient of static and kinetic friction for the felt side of the wood block and the track. For static friction, assume  F = 0 and find the maximum value of static friction. For kinetic friction, you must base your results on an accelerating system. Photogate timers may be used to increase timing accuracy.

48. Monday, October 16, 2006 Ramps and Pulleys – together!

49. Announcements Turn in the lab from Friday. HW due FRIDAY: Newton’s #3-#6. Full lab report from today. You will have class workdays Wednesday and Thursday, but PLEASE don’t assume you can get it all done then! There will be no Lunch or Breakfast Bunch this week due to PSAT/NMSQT. Geometric Optics #6 must be turned in before Wednesday. My room will be unavailable at lunch this week, as I will be writing letters of recommendation. I will be available every morning but Thursday.

50. Magic pulleys on a ramp It’s a little more complicated when a magic pulley is installed on a ramp. m1m1 m2m2  F = ma m 2 g -T + T – m 1 gsin  = (m 1 +m 2 )a m 2 g – m 1 gsin  = (m 1 +m 2 )a a = (m 2 – m 1 sin  )g/(m 1 +m 2 )  m1gm1g NT T m2gm2g m 1 gsin  m 1 gcos 

51. Sample problem Two blocks are connected by a string as shown in the figure. What is the acceleration, assuming there is no friction? 10 kg 5 kg  

52. Sample problem - solution 10 kg 5 kg    F = ma m 2 g -T + T – m 1 gsin  = (m 1 +m 2 )a m 2 g – m 1 gsin  = (m 1 +m 2 )a a = (m 2 – m 1 sin  )g/(m 1 +m 2 ) a = [(5 – 10sin45 o )(9.8)]/15 a = - 1.35 m/s 2 m1gm1g NT T m2gm2g m 1 gsin  m 1 gcos 

53. Sample problem - solution 10 kg 5 kg    F = ma m 2 g -T + T – m 1 gsin  = (m 1 +m 2 )a m 2 g – m 1 gsin  = (m 1 +m 2 )a a = (m 2 – m 1 sin  )g/(m 1 +m 2 ) a = [(5 – 10sin45 o )(9.8)]/15 a = - 1.35 m/s 2 m1gm1g NT T m2gm2g m 1 gsin  m 1 gcos  How would this change if there is friction on the ramp?

54. Laboratory Determine the coefficients of static and of kinetic friction using a RAMP and PULLEY SYSTEM combined. For kinetic friction, timing may be done with photo-gate timers or with stopwatches. Do 4 trials: Static friction; tendency of the block to slide up the ramp. Static friction; tendency of the block to slide down the ramp. Kinetic friction; block accelerating up the ramp. Kinetic friction; block accelerating down the ramp. Full lab report will be collected Friday. All free body diagrams and calculations may be done by hand; all other information must be typed.

55. Tuesday, October 17, 2006 Uniform Circular Motion

56. Announcements HW due FRIDAY: Newton’s #3-#6. Full lab report from yesterday. No Lunch or Breakfast Bunch tomorrow. Geometric Optics #6 must be turned in by the end of the day. My room will be unavailable at lunch this week.

57. Uniform Circular Motion An object that moves at uniform speed in a circle of constant radius is said to be in uniform circular motion. Question: Why is uniform circular motion accelerated motion? Answer: Although the speed is constant, the velocity is not constant since an object in uniform circular motion is continually changing direction.

58. Centrifugal Force Question: What is centrifugal force? Answer: That’s easy. Centrifugal force is the force that flings an object in circular motion outward. Right? Wrong! Centrifugal force is a myth! There is no outward directed force in circular motion. To explain why this is the case, let’s review Newton’s 1 st Law.

59. Newton’s 1 st Law and cars When a car accelerates forward suddenly, you as a passenger feel as if you are flung backward. You are in fact NOT flung backward. Your body’s inertia resists acceleration and wants to remain at rest as the car accelerates forward. When a car brakes suddenly, you as a passenger feel as if you are flung forward. You are NOT flung forward. Your body’s inertia resists acceleration and wants to remain at constant velocity as the car decelerates.

60. You feel as if you are flung to the outside. You call this apparent, but nonexistent, force “centrifugal force”. You are NOT flung to the outside. Your inertia resists the inward acceleration and your body simply wants to keep moving in straight line motion! As with all other types of acceleration, your body feels as if it is being flung in the opposite direction of the actual acceleration. The force on your body, and the resulting acceleration, actually point inward. When a car turns

61. Centripetal Acceleration Centripetal (or center-seeking) acceleration points toward the center of the circle and keeps an object moving in circular motion. This type of acceleration is at right angles to the velocity. This type of acceleration doesn’t speed up an object, or slow it down, it just turns the object.

62. Centripetal Acceleration a c = v 2 /r a c : centripetal acceleration in m/s 2 v: tangential speed in m/s r: radius in meters v acac Centripetal acceleration always points toward center of circle!

63. Centripetal Force A force responsible for centripetal acceleration is referred to as a centripetal force. Centripetal force is simply mass times centripetal acceleration. F c = m a c F c = m v 2 / r F c : centripetal force in N v: tangential speed in m/s r: radius in meters FcFc Always toward center of circle!

64. Any force can be centripetal The name “centripetal” can be applied to any force in situations when that force is causing an object to move in a circle. You can identify the real force or combination of forces which are causing the centripetal acceleration. Any kind of force can act as a centripetal force.

65. Static friction As a car makes a turn on a flat road, what is the real identity of the centripetal force?

66. Tension As a weight is tied to a string and spun in a circle, what is the real identity of the centripetal force?

67. Gravity As the moon orbits the Earth, what is the real identity of the centripetal force?

68. Normal force with help from static friction As a racecar turns on a banked curve on a racing track, what is the real identity of the centripetal force?

69. Tension, with some help from gravity As you swing a mace in a vertical circle, what is the true identity of the centripetal force?

70. Gravity, with some help from the normal force When you are riding the Tennessee Tornado at Dollywood, what is the real identity of the centripetal force when you are on a vertical loop?

71. Sample problem A 1200-kg car rounds a corner of radius r = 45 m. If the coefficient of static friction between tires and the road is 0.93 and the coefficient of kinetic friction between tires and the road is 0.75, what is the maximum velocity the car can have without skidding?

72. Thursday, October 19, 2006 Class workday

73. Friday, October 20, 2006 Inquiry Based Lab

74. Announcements HW due TODAY: Newton’s #3-#6. Full lab report from Monday’s lab. My room will be unavailable at lunch today, as I will be writing letters of recommendation. I will be available every morning but Thursday.

75. Centripetal Force and Friction Lab Using only the hand strobe, a penny, stopwatch, chalk, and ruler, determine the coefficient of friction between a penny and the hand strobe. Rules, hints, and tips: –You may write on the hand strobe with the chalk. –You must use centripetal force in your analysis. –Can you spin the hand strobe at a gradually increasing rate? –Can you spin the hand strobe at a constant rate? –The lowest rotational speed necessary to make the penny fly off the hand strobe is an important number! How will you measure this speed with high accuracy? One report per group (handwritten) is due at the end of class. It should include a procedure, a good free body diagram, all data you collect, and a clear Newton’s 2 nd Law analysis. Names of group members must be on the report. You will be graded on how well you develop a procedure to do this analysis correctly, your application of Newton’s 2 nd Law, and your results.

76. Monday, October 30, 2006 Begin Exam Review for Newton’s Laws II

77. Announcements Federal Forms – does anyone have them. Free response exam Wednesday. Lunch bunch resumes Wednesday. Today, we will review Newton’s Laws by starting the packet. This MUST be completed by tomorrow when you come to class if you are going to be allowed to correct the free response exam given on Wednesday.

78. Sample problem You whirl a 2.0 kg stone in a horizontal circle about your head. The rope attached to the stone is 1.5 m long. a) What is the tension in the rope? (The rope makes a 10 o angle with the horizontal). b) How fast is the stone moving?

79. Monday, October 31, 2006 Exam Review

80. Announcements Federal Forms – does anyone have them? Free response exam tomorrow. Get out your free response packet for my inspection.