1. Ch6.1 - Forces Newton's 1st Law - object in motion will remain in that motion, an object at rest stays at rest. - Will not change unless a net force acts on it. - Rest is just a special case of motion.

2. Newton's 1st Law - object in motion will remain in that motion, an object at rest stays at rest. - Will not change unless a net force acts on it. - Rest is just a special case of motion. Inertia - tendency of an object to resist change. Ch6.1 - Forces

3. Newton's 1st Law - object in motion will remain in that motion, an object at rest stays at rest. - Will not change unless a net force acts on it. - Rest is just a special case of motion. Inertia - tendency of an object to resist change. Force - push or pull Contact force - touches object Field force - long range, acts at a distance Equilibrium - net force on an object is zero (object can be at rest or moving at constant speed) Free body diagram - drawing of object with all forces labeled on it. Ch6.1 - Forces

4. Net Force - what’s left over after all forces we added up. Force is a vector, so the vectors have to be added vectorally! Ex1) What is the net force in each: a. F 1 F 2 b. F 1 F 2 c. F 1 F 2 Ch6.1 - Forces

5. Net Force - what’s left over after all forces we added up. Force is a vector, so the vectors have to be added vectorally! Ex1) What is the net force in each: a. F 1 F 2 b. F 1 F 2 c. F 1 F 2 Ch6.1 - Forces

10. Lab6.1 – Inertia - due tomorrow - Ch6 HW#1 due at beginning of period

11. Ch6 HW#1 1- 8 1. Draw: c. A book on the desk while you hand is pushing down on it. d. A ball, just as the string holding it breaks. 2. 2 horiz forces 225N and 165N pushing on crate in same direction. Find F net. 3. 2 horiz forces 225N and 165N pushing on crate in opp directions. Find F net.

12. Ch6 HW#1 1- 8 F N 1. Draw: c. A book on the desk while you hand is pushing down on it. d. A ball, just as the string holding it breaks. F g F Hand c. F net = F g + F Hand – F N d. F net = F g 0 = F g 2. 2 horiz forces 225N and 165N pushing on crate in same direction. Find F net. 3. 2 horiz forces 225N and 165N pushing on crate in opp directions. Find F net.

13. Ch6 HW#1 1- 8 F N 1. Draw: c. A book on the desk while you hand is pushing down on it. d. A ball, just as the string holding it breaks. F g F Hand c. F net = F g + F Hand – F N d. F net = F g 0 = F g 2. 2 horiz forces 225N and 165N pushing on crate in same direction. Find F net. F N F = 165N F = 225N F net = 165N + 225N = 390NF g 3. 2 horiz forces 225N and 165N pushing on crate in opp directions. Find F net.

14. Ch6 HW#1 1- 8 F N 1. Draw: c. A book on the desk while you hand is pushing down on it. d. A ball, just as the string holding it breaks. F g F Hand c. F net = F g + F Hand – F N d. F net = F g 0 = F g 2. 2 horiz forces 225N and 165N pushing on crate in same direction. Find F net. F N F = 165N F = 225N F net = 165N + 225N = 390NF g 3. 2 horiz forces 225N and 165N pushing on crate in opp directions. Find F net. F N F = 165N F = 225N F net = 225N – 165N = 60N to the rightF g

15. 5. Calc the force you exert on the floor while you stand. (1 lb =.454kg) Does the force change if you lie down? 6. A cable pulls a crate at constant speed across the floor. 8. A rope lowers a bucket at constant speed.

16. 5. Calc the force you exert on the floor while you stand. (1 lb =.454kg) Does the force change if you lie down? F N 150 lb.454 kg= 68 kg x 9.8 m/s 2 1 lb = 680 N F g 6. A cable pulls a crate at constant speed across the floor. 8. A rope lowers a bucket at constant speed.

17. 5. Calc the force you exert on the floor while you stand. (1 lb =.454kg) Does the force change if you lie down? F N 150 lb.454 kg= 68 kg x 9.8 m/s 2 1 lb = 680 N F g 6. A cable pulls a crate at constant speed across the floor. F N F net = F g – F N 0 = F g – F N F f F T F g = F N F g 8. A rope lowers a bucket at constant speed.

18. 5. Calc the force you exert on the floor while you stand. (1 lb =.454kg) Does the force change if you lie down? F N 150 lb.454 kg= 68 kg x 9.8 m/s 2 1 lb = 680 N F g 6. A cable pulls a crate at constant speed across the floor. F N F net = F g – F N 0 = F g – F N F f F T F g = F N F g 8. A rope lowers a bucket at constant speed. F T F net = F g – F T 0 = F g – F T F g = F T F g

19. Ch6.2 – Mass vs. Weight -All objects in Free Fall accelerate at g= 9.8 m/s F = m. a

20. Ch6.2 – Mass vs. Weight -All objects in Free Fall accelerate at g= 9.8 m/s F = m. a Force of gravity: Fg= m. g Weight – measure of the force of gravity (Newtons, N) Mass – amount of matter present (kilograms, kg.) Ex1) My mass is 80 kg, a. What’s my weight? b. What’s my mass on the moon? c. If gravity on the moon is 1/6g, what’s my weight on the moon?

21. If you are accelerating up or down (like in an elevator), your weight changes. - If it accelerates upward, you squish the scale more & you appear to weigh more. F = m(g + a)

22. If you are accelerating up or down (like in an elevator), your weight changes. - If it accelerates upward, you squish the scale more & you appear to weigh more. F = m(g + a) - If it accelerates downward, the scale expands & you apparently weigh less. F = m(g – a)

23. If you are accelerating up or down (like in an elevator), your weight changes. - If it accelerates upward, you squish the scale more & you appear to weigh more. F = m(g + a) - If it accelerates downward, the scale expands & you apparently weigh less. F = m(g – a) Ex2) Your mass is 75kg, you stand on a scale in an elevator not moving. a. What is your weight? b. If the elevator accelerates UP at 2 m / s 2, what's your weight? c. If it is moving at a constant speed what's your weight? d. Accelerates down at 2 m / s 2 ? e. Accelerates down at 5 m / s 2 ? f. Accelerates down at 9.8 m / s 2 ?

24. If you are accelerating up or down (like in an elevator), your weight changes. - If it accelerates upward, you squish the scale more & you appear to weigh more. F = m(g + a) - If it accelerates downward, the scale expands & you apparently weigh less. F = m(g – a) Ex2) Your mass is 75kg, you stand on a scale in an elevator not moving. a. What is your weight? F g = m(g+a)= (75kg)(9.8+0)= 735N b. If the elevator accelerates UP at 2 m / s 2, what's your weight? F g = m(g+a)= (75kg)(9.8+2)= 885N c. If it is moving at a constant speed what's your weight? F g = m(g+a)= (75kg)(9.8+0)= 735N d. Accelerates down at 2 m / s 2 ? F g = m(g-a)= (75kg)(9.8-2)= 600N e. Accelerates down at 5 m / s 2 ? F g = m(g-a)= (75kg)(9.8-5)= 375N f. Accelerates down at 9.8 m / s 2 ? F g = m(g-a)= (75kg)(9.8-9.8)= 0N Apparent weightlessness!

25. Ex3) A 50kg bucket is lifted by a rope that has a max tension of 550N. It started at rest, and after 3m is moving at 3 m / s, will the rope break? Ch6 HW#2 9 – 13

26. Lab6.2 – Newton’s 2 nd Law of Skateboards - due tomorrow - Ch6 HW#2 due at beginning of period

27. Ch6 HW#2 9 – 13 9. Draw. A skydiver falls downward through the air at a constant velocity. 10. Draw. A rope lifts a bucket accelerating it upward. 11. Draw. A rocket blasts off and its vertical velocity increases with time. (No air drag.) 12. You weigh 585N on Earth. a. Mass? b. Weight on moon? (g=1.60 m / s 2 ) 13. Mass of 60kg. What does scale read: a. Elevator not moving? b. Up @ const speed? c. If the elevator were moving downward and then began slowing to a stop, scale reads ________. d. If the elevator were moving upward and then began slowing to a stop, scale reads ________. e. Slows at 2.0 m / s 2 while moving upward? f. Speeds up at 2.0 m / s 2 while moving downward?

28. Ch6 HW#2 9 – 13 F air 9. Draw. A skydiver falls downward through the air at a constant velocity. 10. Draw. A rope lifts a bucket accelerating it upward. 11. Draw. A rocket blasts off and its vertical velocity F g increases with time. (No air drag.) 12. You weigh 585N on Earth. a. Mass? b. Weight on moon? (g=1.60 m / s 2 ) 13. Mass of 60kg. What does scale read: a. Elevator not moving? b. Up @ const speed? c. If the elevator were moving downward and then began slowing to a stop, scale reads ________. d. If the elevator were moving upward and then began slowing to a stop, scale reads ________. e. Slows at 2.0 m / s 2 while moving upward? f. Speeds up at 2.0 m / s 2 while moving downward?

29. Ch6 HW#2 9 – 13 F air F T 9. Draw. A skydiver falls downward through the air at a constant velocity. 10. Draw. A rope lifts a bucket accelerating it upward. 11. Draw. A rocket blasts off and its vertical velocity F g increases with time. (No air drag.) 12. You weigh 585N on Earth. a. Mass? F g b. Weight on moon? (g=1.60 m / s 2 ) 13. Mass of 60kg. What does scale read: a. Elevator not moving? b. Up @ const speed? c. If the elevator were moving downward and then began slowing to a stop, scale reads ________. d. If the elevator were moving upward and then began slowing to a stop, scale reads ________. e. Slows at 2.0 m / s 2 while moving upward? f. Speeds up at 2.0 m / s 2 while moving downward?

30. Ch6 HW#2 9 – 13 F air F T F rocket 9. Draw. A skydiver falls downward through the air at a constant velocity. 10. Draw. A rope lifts a bucket accelerating it upward. 11. Draw. A rocket blasts off and its vertical velocity F g increases with time. (No air drag.) 12. You weigh 585N on Earth. a. Mass? F g F g b. Weight on moon? (g=1.60 m / s 2 ) 13. Mass of 60kg. What does scale read: a. Elevator not moving? b. Up @ const speed? c. If the elevator were moving downward and then began slowing to a stop, scale reads ________. d. If the elevator were moving upward and then began slowing to a stop, scale reads ________. e. Slows at 2.0 m / s 2 while moving upward? f. Speeds up at 2.0 m / s 2 while moving downward?

31. Ch6 HW#2 9 – 13 F air F T F rocket 9. Draw. A skydiver falls downward through the air at a constant velocity. 10. Draw. A rope lifts a bucket accelerating it upward. 11. Draw. A rocket blasts off and its vertical velocity F g increases with time. (No air drag.) 12. You weigh 585N on Earth. a. Mass?58.5kg F g F g b. Weight on moon? (g=1.60 m / s 2 ) F gm = (58.5kg)(1.60 m / s 2 ) = 93.6N 13. Mass of 60kg. What does scale read: a. Elevator not moving? b. Up @ const speed? c. If the elevator were moving downward and then began slowing to a stop, scale reads ________. d. If the elevator were moving upward and then began slowing to a stop, scale reads ________. e. Slows at 2.0 m / s 2 while moving upward? f. Speeds up at 2.0 m / s 2 while moving downward?

32. Ch6 HW#2 9 – 13 F air F T F rocket 9. Draw. A skydiver falls downward through the air at a constant velocity. 10. Draw. A rope lifts a bucket accelerating it upward. 11. Draw. A rocket blasts off and its vertical velocity F g increases with time. (No air drag.) 12. You weigh 585N on Earth. a. Mass?58.5kg F g F g b. Weight on moon? (g=1.60 m / s 2 ) F gm = (58.5kg)(1.60 m / s 2 ) = 93.6N 13. Mass of 60kg. What does scale read: a. Elevator not moving? F g = m(g+a)= (60kg)(10+0)= 600N b. Up @ const speed? F g = m(g+a)= (60kg)(10+0)= 600N c. If the elevator were moving downward and then began slowing to a stop, scale reads _more_. d. If the elevator were moving upward and then began slowing to a stop, scale reads _less_. e. Slows at 2.0 m / s 2 while moving upward? f. Speeds up at 2.0 m / s 2 while moving downward?

33. Ch6 HW#2 9 – 13 F air F T F rocket 9. Draw. A skydiver falls downward through the air at a constant velocity. 10. Draw. A rope lifts a bucket accelerating it upward. 11. Draw. A rocket blasts off and its vertical velocity F g increases with time. (No air drag.) 12. You weigh 585N on Earth. a. Mass?58.5kg F g F g b. Weight on moon? (g=1.60 m / s 2 ) F gm = (58.5kg)(1.60 m / s 2 ) = 93.6N 13. Mass of 60kg. What does scale read: a. Elevator not moving? F g = m(g+a)= (60kg)(10+0)= 600N b. Up @ const speed? F g = m(g+a)= (60kg)(10+0)= 600N c. If the elevator were moving downward and then began slowing to a stop, scale reads _more_. d. If the elevator were moving upward and then began slowing to a stop, scale reads _less_. e. Slows at 2.0 m / s 2 while moving upward? F g = m(g-a)= (60kg)(10-2) = 480N f. Speeds up at 2.0 m / s 2 while moving downward? F g = m(g-a)= (60kg)(10-2) = 480N

34. Ch6.3 – Friction Kinetic Friction- (F f,k ) Moving Friction F f,k = µ k × F N. Coefficient Normal force of kinetic friction (usually equal & opposite to F g ) Ex1) Calculate the kinetic friction for the steel blades of ice skates on ice given the total mass of the skater is 76 kg and µk =.06

35. Ch6.3 – Friction Kinetic Friction- (F f,k ) Moving Friction F f,k = µ k × F N. Coefficient Normal force of kinetic friction (usually equal & opposite to F g ) Ex1) Calculate the kinetic friction for the steel blades of ice skates on ice given the total mass of the skater is 76 kg. and µk =.06 (F g = (75kg.)(10 m / s 2 )= 750N) F F,k FgFg FNFN F F,k = µ k × F N =(.06)F g =(.06)(750N)= 45N

36. Ex2) What is the coefficient of friction for steel on steel without lubrication, if the mass on the block on top is 10kg, and the friction force is measured to be 30N.

37. Ex2) What is the coefficient of friction for steel on steel without lubrication, if the mass on the block on top is 10kg, and the friction force is measured to be 30N. F f,k = µ k × F N 30N = µ k (100N) µ k =.3

38. Static Friction- (F f,s ) Not moving friction (prevents motion from happening) F f,s = µ s × F N. Coefficient Normal force of static friction

39. Static Friction- (F f,s ) Not moving friction (prevents motion from happening) F f,s = µ s × F N. Coefficient Normal force of static friction Static Friction can vary between 0 and this maximum value Ex3) A person pushes on a wooden crate on a wood floor, with a force of 50N. If the crate has a mass of 25kg, and µ s = 0.5, will the crate move?

40. Static Friction- (F f,s ) Not moving friction (prevents motion from happening) F f,s = µ s × F N. Coefficient Normal force of static friction Static Friction can vary between 0 and this maximum value Ex3) A person pushes on a wooden crate on a wood floor, with a force of 50N. If the crate has a mass of 25kg, and µ s = 0.5, will the crate move? F N F f,s = µ s × F N. = (.5)(250N) F f,s F = 50N = 125N F g Friction depends on: 1. the types of surfaces (µ) 2. the force pressing the surfaces together (F N ) (not speed or surface area)

41. Lab ex) Your group weighs a block with a spring scale and finds it to be 15N. The force required to pull it at constant speed is 12N. What is the coefficient of friction between the block and table? Ch6 HW #3 14 – 17

42. Lab ex) Your group weighs a block with a spring scale and finds it to be 15N. The force required to pull it at constant speed is 12N. What is the coefficient of friction between the block and table? F N F f,k = µ k × F N 12N = (µ k )(15N) F f,s F = 12N µ k = 0.8 F g Ch6 HW #3 14 – 17

43. Lab6.3A – Calculating Coefficients of Friction - due tomorrow - Ch6 HW#3 due at beginning of period

44. Ch6 HW#3 14 – 17 14. Calculate the kinetic friction for the rubber soled shoe on wet concrete. mass of 55kg, µ k = 0.40. 15.What is the coefficient of static friction for you (55kg) in your rubber soled shoes if the friction force required to get you sliding is 350N.

45. Ch6 HW#3 14 – 17 14. Calculate the kinetic friction for the rubber soled shoe on wet concrete. mass of 55kg, µ k = 0.40. F N F f,k = µ k × F N F f,k = (.4)(550N) F f,k F F f,k = 220N F g 15.What is the coefficient of static friction for you (55kg) in your rubber soled shoes if the friction force required to get you sliding is 350N.

46. Ch6 HW#3 14 – 17 14. Calculate the kinetic friction for the rubber soled shoe on wet concrete. mass of 55kg, µ k = 0.40. F N F f,k = µ k × F N F f,k = (.4)(550N) F f,k F F f,k = 220N F g 15.What is the coefficient of static friction for you (55kg) in your rubber soled shoes if the friction force required to get you sliding is 350N. F N F f,s = µ s × F N 350N = (µ s )(550N) F f,s F = 350N µ s = 0.64 F g

47. 16. A person pushes on a wooden crate on a wood floor. Crate has a mass of 45kg, µ s = 0.50, what force necessary to get it started moving? 17. Your lab group weighs a block with a spring scale and finds it weighs 9.0N. The force required to move it at constant speed across light sand paper is 15.5N. What is µ k ?

48. 16. A person pushes on a wooden crate on a wood floor. Crate has a mass of 45kg, µ s = 0.50, what force necessary to get it started moving? F N F f,s = µ s × F N F f,s = (.5)(450N) F f,s F F f,s = 225N F g 17. Your lab group weighs a block with a spring scale and finds it weighs 9.0N. The force required to move it at constant speed across light sand paper is 15.5N. What is µ k ?

49. 16. A person pushes on a wooden crate on a wood floor. Crate has a mass of 45kg, µ s = 0.50, what force necessary to get it started moving? F N F f,s = µ s × F N F f,s = (.5)(450N) F f,s F F f,s = 225N F g 17. Your lab group weighs a block with a spring scale and finds it weighs 9.0N. The force required to move it at constant speed across light sand paper is 15.5N. What is µ k ? F N F f,k = µ k × F N 15.5N = (µ k )(9N) F f,k F µ k = 1.72 F g

50. Friction- Day 2 Ex1) A lab group pulls a 6kg ball at constant speed across a table with a force of 15N. What’s the coefficient of friction? F= 15N F f,k FNFN FgFg

51. Friction- Day 2 Ex1) A lab group pulls a 6kg ball at constant speed across a table with a force of 15N. What’s the coefficient of friction? F= 15N F F,K FNFN FgFg F net = F – F f,k m. a = 15N – μ k ×F N 0 = 15 – μ k ×60N μ k =.25

52. Ex2) Force of 2.5N is required to get a.4kg. block sliding. A force of 2N is required to keep it moving at a constant speed. Find µ s and µ k.

53. Ex2) Force of 2.5N is required to get a.4kg. block sliding. A force of 2N is required to keep it moving at a constant speed. Find µ s and µ k. FNFN F = 2.5N F F,S F F,S = μ s ×F N 2.5N= μ s × 4N μ s =.63 F g = 4N

54. Ex2) Force of 2.5N is required to get a.4kg. block sliding. A force of 2N is required to keep it moving at a constant speed. Find µ s and µ k. FNFN F = 2.5N F g = 4N F F,S F F,S = μ s ×F N 2.5N= μ s × 4N μ s =.63 FNFN F g = 4N F = 2N F F,K F F,K = µ k × F N 2N = µ k × 4N μ =.5

55. Ex3) A person pushes a crate of 20kg with a force of 101N. μ s = 0.50. Will the Crate move?

56. Ex3) A person pushes a crate of 20kg with a force of 101N. μ s = 0.50. Will the Crate move? F N F f,s = μ s x F N F f,s F=101N F g =200N = (.5)(200N) = 100N Yes! What force will be needed to keep it moving if μ k = 0.20?

57. Ex3) A person pushes a crate of 20kg with a force of 101N. μ s = 0.50. Will the Crate move? F N F f,s = μ s x F N F f,s F=101N F g =200N = (.5)(200N) = 100N Yes! What force will be needed to keep it moving if μ k = 0.20? F f,k = μ k x F N = (.2)(200N) = 40N If the person still pushes with 101N, what is the acceleration of the crate?

58. Ex3) A person pushes a crate of 20kg with a force of 101N. μ s = 0.50. Will the Crate move? F N F f,s = μ s x F N F f,s F=101N F g =200N = (.5)(200N) = 100N Yes! What force will be needed to keep it moving if μ k = 0.20? F f,k = μ k x F N = (.2)(200N) = 40N If the person still pushes with 101N, what is the acceleration of the crate? F net = F – F f,k m. a = 101N – 40N (20kg). a = 61N a = 3.1 m / s 2

59. Lab ex) A 0.25kg ball rolls off a ramp and down a piece of carpet, and rolls 5.5m during a time of 3.0sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t 2. v f = v i + at 3. F = m. a 4. F f,k = μ k x F N a = -___ d= 5.5mv f = 0 t= 3 sec. FNFN FgFg F F,k

60. Lab ex) A 0.25kg ball rolls off a ramp and down a piece of carpet, and rolls 5.5m during a time of 3.0sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t5.5 = ½(v i + 0)3 v i = 3.7 m / s 2. v f = v i + at 3. F = m. a 4. F f,k = μ k x F N a = -___ d= 5.5mv f = 0 t= 3 sec. FNFN FgFg F F,k

61. Lab ex) A 0.25kg ball rolls off a ramp and down a piece of carpet, and rolls 5.5m during a time of 3.0sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t5.5 = ½(v i + 0)3 v i = 3.7 m / s 2. v f = v i + at 0 m / s = 3.7 m / s + a(3s)a = - 1.23 m / s 2 3. F = m. a 4. F f,k = μ k x F N a = - 1.23 m / s 2 d= 5.5mv f = 0 t= 3 sec. FNFN FgFg F F,k

62. Lab ex) A 0.25kg ball rolls off a ramp and down a piece of carpet, and rolls 5.5m during a time of 3.0sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t5.5 = ½(v i + 0)3 v i = 3.7 m / s 2. v f = v i + at 0 m / s = 3.7 m / s + a(3s)a = - 1.23 m / s 2 3. F = m. aF = (0.25kg)(-1.23 m / s 2 )F = - 0.3N 4. F f,k = μ k x F N a = - 1.23 m / s 2 d= 5.5mv f = 0 t= 3 sec. FNFN FgFg F F,k

63. Lab ex) A 0.25kg ball rolls off a ramp and down a piece of carpet, and rolls 5.5m during a time of 3.0sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t5.5 = ½(v i + 0)3 v i = 3.7 m / s 2. v f = v i + at 0 m / s = 3.7 m / s + a(3s)a = - 1.23 m / s 2 3. F = m. aF = (0.25kg)(-1.23 m / s 2 )F = - 0.3N 4. F f,k = μ k x F N 0.3N = μ k x (2.5N) μ k = 0.12 a = - 1.23 m / s 2 d= 5.5mv f = 0 t= 3 sec. FNFN FgFg F F,k Ch6 HW#4 18,19

64. Lab6.3B – Sliding Friction - due tomorrow - Ch6 HW#4 due at beginning of period

65. Lab6.3C – Friction on Carpet - due tomorrow

66. Ch6 HW#4 18,19 18. Ex3) A person pushes a crate of 30kg with a force of 25N. μ s = 0.78. Will the Crate move? F N F f,s = μ s x F N F f,s F=25N F g =300N Will the crate move if μ s = 0.04? If the person still pushes with 25N, what is the acceleration of the crate? (μ k = 0.04 also)

67. Ch6 HW#4 18,19 18. Ex3) A person pushes a crate of 30kg with a force of 25N. μ s = 0.78. Will the Crate move? F N F f,s = μ s x F N F f,s F=25N F g =300N = (.78)(300N) = 234N No! Will the crate move if μ s = 0.04? If the person still pushes with 25N, what is the acceleration of the crate? (μ k = 0.04 also)

68. Ch6 HW#4 18,19 18. Ex3) A person pushes a crate of 30kg with a force of 25N. μ s = 0.78. Will the Crate move? F N F f,s = μ s x F N F f,s F=25N F g =300N = (.78)(300N) = 234N No! Will the crate move if μ s = 0.04? F f,s = μ s x F N = (.04)(300N) = 12NYes! If the person still pushes with 25N, what is the acceleration of the crate? (μ k = 0.04 also)

69. Ch6 HW#4 18,19 18. Ex3) A person pushes a crate of 30kg with a force of 25N. μ s = 0.78. Will the Crate move? F N F f,s = μ s x F N F f,s F=25N F g =300N = (.78)(300N) = 234N No! Will the crate move if μ s = 0.04? F f,s = μ s x F N = (.04)(300N) = 12NYes! If the person still pushes with 25N, what is the acceleration of the crate? (μ k = 0.04 also) F net = F – F f,k m. a = 25N – 12N (30kg). a = 13N a = 0.43 m / s 2

70. 19. A 14kg ball rolls off a ramp and down a piece of carpet, and rolls 25.2m during a time of 4.5sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t 2. v f = v i + at 3. F = m. a 4. F f,k = μ k x F N a = - ____ m / s 2 d= 25.2m v f = 0 t= 4.5 sec. FNFN FgFg F F,k

71. 19. A 14kg ball rolls off a ramp and down a piece of carpet, and rolls 25.2m during a time of 4.5sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t25.2 = ½(v i + 0)4.5 v i = 11.2 m / s 2. v f = v i + at 3. F = m. a 4. F f,k = μ k x F N a = - ____ m / s 2 d= 25.2m v f = 0 t= 4.5 sec. FNFN FgFg F F,k

72. 19. A 14kg ball rolls off a ramp and down a piece of carpet, and rolls 25.2m during a time of 4.5sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t25.2 = ½(v i + 0)4.5 v i = 11.2 m / s 2. v f = v i + at 0 m / s = 11.2 m / s + a(4.5s)a = - 2.4 m / s 2 3. F = m. a 4. F f,k = μ k x F N a = - ____ m / s 2 d= 25.2m v f = 0 t= 4.5 sec. FNFN FgFg F F,k

73. 19. A 14kg ball rolls off a ramp and down a piece of carpet, and rolls 25.2m during a time of 4.5sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t25.2 = ½(v i + 0)4.5 v i = 11.2 m / s 2. v f = v i + at 0 m / s = 11.2 m / s + a(4.5s)a = - 2.4 m / s 2 3. F = m. aF = (0.25kg)(-2.4 m / s 2 )F = - 33.6N 4. F f,k = μ k x F N a = - 2.4 m / s 2 d= 25.2m v f = 0 t= 4.5 sec. FNFN FgFg F F,k

74. 19. A 14kg ball rolls off a ramp and down a piece of carpet, and rolls 25.2m during a time of 4.5sec. Find the friction between the carpet and ball. (Takes 4 steps:) 1. d = ½(v i +v f ). t25.2 = ½(v i + 0)4.5 v i = 11.2 m / s 2. v f = v i + at 0 m / s = 11.2 m / s + a(4.5s)a = - 2.4 m / s 2 3. F = m. aF = (0.25kg)(-2.4 m / s 2 )F = - 33.6N 4. F f,k = μ k x F N 33.6N = μ k x (140N)μ k = 0.24 d= 25.2m v f = 0 t= 4.5 sec. FNFN FgFg F F,k a = - 2.4 m / s 2

75. Ch6.4 – Periodic Motion Simple harmonic motion – An object oscillates back and forth. Period - how long it takes to complete one cycle Amplitude - Its maximum distance from rest position. Equilibrium - rest position, all forces cancelling out. 1. A Pendulum

76. Ch6.4 – Periodic Motion Simple harmonic motion – An object oscillates back and forth. Period - how long it takes to complete one cycle Amplitude - Its maximum distance from rest position. Equilibrium - rest position, all forces cancelling out. 1. A Pendulum Period of a pendulum: FTFT FgFg FTFT FgFg F Net FTFT

77. Ch6.4 – Periodic Motion Simple harmonic motion – An object oscillates back and forth. Period - how long it takes to complete one cycle Amplitude - Its maximum distance from rest position. Equilibrium - rest position, all forces cancelling out. 1. A Pendulum Depends on string length Period of a pendulum: and accl of gravity. Pendulums don’t care about the mass attaached! FTFT FgFg FTFT FgFg F Net FTFT

78. Ex1) What is the length of a pendulum with a period of 2 seconds?

80. 2. Springs – amount they stretch is proportional to the mass attached. F = k. d elasticspring distance force constant displaced Ex2) A 100g mass is attached to a spring and it stretches 10 cm. What is the value of the spring constant?

81. Period of a Spring springs care about the mass attached and the stretchiness of the spring to determine the time to oscillate. Ex3) What is the spring constant for a spring that has a period of 0.63s when a mass of 100g is attached?

82. Period of a Spring springs care about the mass attached and the stretchiness of the spring to determine the time to oscillate. Ex3) What is the spring constant for a spring that has a period of 0.63s when a mass of 100g is attached? Resonance – with a small force applied, an object will oscillate at its natural frequency. Keep applying force, amplitude increases tremendously. Ch6 HW#5 20 – 24

83. Lab6.4 – Periodic Motion - due tomorrow - Ch6 HW#5 due at beginning of period

84. Ch6 HW#5 20 – 24 20. Period of a pendulum with length of 0.10m? 21. Length of pendulum with period of 5s? 22. Accl of gravity on planet where pendulum of 1m length has a period of 1 sec? 23. Mass of 0.250kg suspended from a spring that stretches 0.05m. Find spring constant. 24. A 0.01kg mass suspended from a spring with a spring constant of 10 N/m. How much stretch?

85. Ch6 HW#5 20 – 24 20. Period of a pendulum with length of 0.10m? 21. Length of pendulum with period of 5s? 22. Accl of gravity on planet where pendulum of 1m length has a period of 1 sec? 23. Mass of 0.250kg suspended from a spring F = k. d that stretches 0.05m. Find spring constant. 2.5N = k. (.05m) k = 24. A 0.01kg mass suspended from a spring with F = k. d a spring constant of 10 N/m. How much stretch? 0.1N = (10N/m). (d) d =

86. Ch6.5 – Newton’s 3 rd Law - For every action (force) there is an equal and opposite reaction (force) - Forces always come in pairs (Interaction pairs) F A on B = F B on A Ex1) A 100kg football player pushes off his 70kg girlfriend while ice skating. If her acceleration is 6 m / s 2, what is his?

87. Ch6.5 – Newton’s 3 rd Law - For every action (force) there is an equal and opposite reaction (force) - Forces always come in pairs (Interaction pairs) F A on B = F B on A Ex1) A 100kg football player pushes off his 70kg girlfriend while ice skating. If her acceleration is 6 m / s 2, what is his? F G on B = F B on G m B × a= m G × a 100a = (40)(6) a= 2.4

88. Ex2) A 50kg person jumps upwards off Earth’s surface w/ an acceleration of 3 m / s 2. What is the acceleration of the Earth downward? (M E = 6×10 24 kg)

89. Ex2) A 50kg person jumps upwards off Earth’s surface w/ an acceleration of 3 m / s 2. What is the acceleration of the Earth downward? (M E = 6×10 24 kg) F E on P = F P on E m p ×a = m E ×a (50)(3) = (6×10 24 )(a) a = 2.5×10 -23 m / s 2

90. Ex3) A 300kg astronaut is skydiving on the moon. If he free falls at 1.6 m/s 2, how fast does the moon accelerate up? (M m = 7×10 22 kg)

91. Ex3) A 300kg astronaut is skydiving on the moon. If he free falls at 1.6 m/s 2, how fast does the moon accelerate up? (M m = 7×10 22 kg) F M on P = F P on M m p ×a = m M ×a (300)(1.6) = (7×10 22 )(a) a = 7×10 -21 m / s 2

92. Ex4) A 1000kg truck starts from rest and after 3 sec is travelling 12 m/s. By how much does the earth accelerate its spin? F T on E F E on T

93. Ex4) A 1000kg truck starts from rest and after 3 sec is travelling 12 m/s. By how much does the earth accelerate its spin? v f = v i + at 12m/s = 0 + a(3) F T on E F E on T a = 4 m / s 2

94. Ex4) A 1000kg truck starts from rest and after 3 sec is travelling 12 m/s. By how much does the earth accelerate its spin? v f = v i + at 12m/s = 0 + a(3) F T on E F E on T a = 4 m / s 2 F E on T = F T on E m T ×a = m E ×a (1000)(4) = (6×10 24 )(a) a = 7×10 -22 m / s 2 Ch6 HW#6 25 – 28

95. 25. 7kg rock free falls, what is accl of earth? 26. 8000kg truck brakes on 50,000kg asteroid with a deceleration of 20 m / s 2, what is accl of asteroid?

96. Ch6 HW#6 25 – 28 25. 7kg rock free falls, what is accl of earth? F E on R = F R on E m R ×a = m E ×a 7)(9.8) = (6×10 24 )(a) a = 1×10 -23 m / s 2 26. 8000kg truck brakes on 50,000kg asteroid with a deceleration of -20 m / s 2, what is accl of asteroid?

97. Ch6 HW#6 25 – 28 25. 7kg rock free falls, what is accl of earth? F E on R = F R on E m R ×a = m E ×a 7)(9.8) = (6×10 24 )(a) a = 1×10 -23 m / s 2 26. 8000kg truck brakes on 50,000kg asteroid with a deceleration of -20 m / s 2, what is accl of asteroid? F A on T = F T on A m T ×a = m A ×a (8000)(-20) = (50000)(a) a = -3.2 m / s 2

98. 27. 65kg boy in tug-o-war with 45kg girl, she accl at 3 m / s 2, what is his accl? 28. What is the force on the boy?

99. 27. 65kg boy in tug-o-war with 45kg girl, she accl at 3 m / s 2, what is his accl? F G on B = F B on G m B ×a = m G ×a (65)(a) = (45)(3) a = 2 m / s 2 28. What is the force on the boy? v i = 30m/s v f = 0m/s t = 0.005s a = ?

100. 27. 65kg boy in tug-o-war with 45kg girl, she accl at 3 m / s 2, what is his accl? F G on B = F B on G m B ×a = m G ×a (65)(a) = (45)(3) a = 2 m / s 2 28. What is the force on the boy? v i = 30m/s v f = v i + at v f = 0m/s 0 = 30 + a(0.005) t = 0.005s a = -6000 m / s 2 a = ?

101. 27. 65kg boy in tug-o-war with 45kg girl, she accl at 3 m / s 2, what is his accl? F G on B = F B on G m B ×a = m G ×a (65)(a) = (45)(3) a = 2 m / s 2 28. What is the force on the boy? v i = 30m/s v f = v i + at v f = 0m/s 0 = 30 + a(0.005) t = 0.005s a = -6000 m / s 2 a = ? F Glove on Ball = F Ball on Glove m B ×a = m G ×a (0.145)(-6000) = F = -870N

102. Ch6.6 – Coupled Motion Ex1) Find the acceleration of this system:. 5kg 1kg

103. Ex2) Find the acceleration of this system:. 4kg 2kg

104. Ex3) Find the acceleration of this system:. 10kg 3kg

105. Air Drag - Friction with the air DOES depend on speed (Faster you fall, the more air has to get out of your way.) - DOES depend on surface area. (Same reason) F net Accl Velocity Ch6 HW#7 29 – 32

106. Lab6.5 – Coupled Motion - due tomorrow - Ch6 HW#7 due at beginning of period

107. Ch6 HW#7 29 – 32 29) Find the acceleration of this system: F net = F g3 – F T + F T – F g2. 3kg 2kg

108. Ch6 HW#7 29 – 32 29) Find the acceleration of this system: F net = F g3 – F T + F T – F g2 m T. a = 30N – 20N F T (5kg)a = 10N F T F g2 a = 2 m / s 2 F g3. 3kg 2kg

109. Ch6 HW#7 29 – 32 30) Find the acceleration of this system: F net = F g5 – F T + F T – F g3. 5kg 3kg

110. Ch6 HW#7 29 – 32 30) Find the acceleration of this system: F net = F g5 – F T + F T – F g3 m T. a = 50N – 30N F T (8kg)a = 20N F T a = 2.5 m / s 2 F g3 F g5. 5kg 3kg

111. 31) Find the acceleration of this system: F net = F g3 – F T + F T 3kg 2kg.

112. 31) Find the acceleration of this system: F T F g2 F net = F g3 – F T + F T F g3 m T. a = 30N (5kg)a = 30N a = 6 m / s 2 3kg 2kg.

113. 32) Find the acceleration of this system: F net = F g5 – F T + F T 5kg 3kg.

114. 32) Find the acceleration of this system: F T F g3 F net = F g5 – F T + F T F g5 m T. a = 50N (8kg)a = 50N a = 6.25 m / s 2 5kg 3kg.

115. Ch6 Rev 1. A rock is dropped from a bridge into a valley. The Earth pulls on the rock and accl it downward. According to Newton’s 3 rd Law, the rock must also be pulling on the Earth, yet the Earth doesn’t seam to accl. Explain. 2. State Newton’s 3 Laws. 3. A 873kg dragster starting from rest attains a speed of 26.3 m/s in 0.59s. a. Find accl v f = v i + at b. Find net force F = m. a 4. A bowling ball has a mass of 8.0kg. What is its weight?

116. Ch6 Rev 1. A rock is dropped from a bridge into a valley. The Earth pulls on the rock and accl it downward. According to Newton’s 3 rd Law, the rock must also be pulling on the Earth, yet the Earth doesn’t seam to accl. Explain. 2. State Newton’s 3 Laws. 3. A 873kg dragster starting from rest attains a speed of 26.3 m/s in 0.59s. a. Find accl v f = v i + at 26.3 m / s = 0 m / s + a(0.59s)a = 44.6 m / s 2 b. Find net force F = m. a = (873kg)(44.6 m / s 2 ) = 38,910N 4. A bowling ball has a mass of 8.0kg. What is its weight?

117. Ch6 Rev 1. A rock is dropped from a bridge into a valley. The Earth pulls on the rock and accl it downward. According to Newton’s 3 rd Law, the rock must also be pulling on the Earth, yet the Earth doesn’t seam to accl. Explain. 2. State Newton’s 3 Laws. 3. A 873kg dragster starting from rest attains a speed of 26.3 m/s in 0.59s. a. Find accl v f = v i + at 26.3 m / s = 0 m / s + a(0.59s)a = 44.6 m / s 2 b. Find net force F = m. a = (873kg)(44.6 m / s 2 ) = 38,910N 4. A bowling ball has a mass of 8.0kg. What is its weight? F = m. g = (8.0kg)(9.8 m / s 2 )= 78.4N

118. Ch6 Rev 5. Your new motorcycle weighs 2450N. What is its mass? 6. A pendulum has a length of 0.67m. Find its period. 7. What is the length of a pendulum with a period of 2.5s?

119. Ch6 Rev 5. Your new motorcycle weighs 2450N. What is its mass? F = m. g 2450N= (m)(9.8 m / s 2 ) m = 250kg 6. A pendulum has a length of 0.67m. Find its period. 7. What is the length of a pendulum with a period of 2.5s?

120. Ch6 Rev 5. Your new motorcycle weighs 2450N. What is its mass? F = m. g 2450N= (m)(9.8 m / s 2 ) m = 250kg 6. A pendulum has a length of 0.67m. Find its period. 7. What is the length of a pendulum with a period of 2.5s?

121. Ch6 Rev 5. Your new motorcycle weighs 2450N. What is its mass? F = m. g 2450N= (m)(9.8 m / s 2 ) m = 250kg 6. A pendulum has a length of 0.67m. Find its period. 7. What is the length of a pendulum with a period of 2.5s?

122. Ch6 Rev 8. What is the mass of an object hanging from a spring if the period of oscillation is 3s, and the spring constant has a value of 50 N/m. 9. If a mass weighing 20N is suspended from a spring causing it to stretch 15cm, what is the value of the spring constant?

123. Ch6 Rev 8. What is the mass of an object hanging from a spring if the period of oscillation is 3s, and the spring constant has a value of 50 N/m. 9. If a mass weighing 20N is suspended from a spring causing it to stretch 15cm, what is the value of the spring constant?

124. Ch6 Rev 8. What is the mass of an object hanging from a spring if the period of oscillation is 3s, and the spring constant has a value of 50 N/m. 9. If a mass weighing 20N is suspended from a spring causing it to stretch 15cm, what is the value of the spring constant? F = k. d 20N = k. (.15m) k = 133 N/m

125. Ch6 Rev 10. If you use a horizontal force of 30N to pull a 12.0kg wooden crate across a floor at constant velocity, what is the coefficient of kinetic friction between the crate and floor? F f,k = μ k x F N F N F f,k F=12N F g =120N 11. Calculate the kinetic friction force when a 200kg sled slides across a frozen lake, coefficient of kinetic friction = 0.06.

126. Ch6 Rev 10. If you use a horizontal force of 30N to pull a 12.0kg wooden crate across a floor at constant velocity, what is the coefficient of kinetic friction between the crate and floor? F f,k = μ k x F N F N 30N = μ k x 120N F f,k F=30N F g =120N μ k = 0.25 11. Calculate the kinetic friction force when a 200kg sled slides across a frozen lake, coefficient of kinetic friction = 0.06. F f,k = μ k x F N F N F f,k F=? F g =2000N

127. Ch6 Rev 10. If you use a horizontal force of 30N to pull a 12.0kg wooden crate across a floor at constant velocity, what is the coefficient of kinetic friction between the crate and floor? F f,k = μ k x F N F N 30N = μ k x 120N F f,k F=30N F g =120N μ k = 0.25 11. Calculate the kinetic friction force when a 200kg sled slides across a frozen lake, coefficient of kinetic friction = 0.06. F f,k = μ k x F N F N F f,k = (0.06)(2000)N F f,k F=? F g =2000N F f,k = 120N

128. Ch6 Rev 12. A person pushes a 25kg box with a force of 26N, coefficient of static friction is 0.43. a. Will the crate move? F f,s = μ s x F N F N F f,s F=26N F g =250N b. If coefficient decreases to 0.10, will it move? c. If the person still pulls with 26N, and the coefficient of kinetic friction is 0.05, what is the accl?

129. Ch6 Rev 12. A person pushes a 25kg box with a force of 26N, coefficient of static friction is 0.43. a. Will the crate move? F f,s = μ s x F N F N = (.43)(250N) F f,s F=26N F g =250N = 108N no! b. If coefficient decreases to 0.10, will it move? c. If the person still pulls with 26N, and the coefficient of kinetic friction is 0.05, what is the accl?

130. Ch6 Rev 12. A person pushes a 25kg box with a force of 26N, coefficient of static friction is 0.43. a. Will the crate move? F f,s = μ s x F N F N = (.43)(250N) F f,s F=26N F g =250N = 108N No! b. If coefficient decreases to 0.10, will it move? F f,s = μ s x F N = (.10)(250N) = 25NYes! c. If the person still pulls with 26N, and the coefficient of kinetic friction is 0.05, what is the accl?

131. Ch6 Rev 12. A person pushes a 25kg box with a force of 26N, coefficient of static friction is 0.43. a. Will the crate move? F f,s = μ s x F N F N = (.43)(250N) F f,s F=26N F g =250N = 108N No! b. If coefficient decreases to 0.10, will it move? F f,s = μ s x F N = (.10)(250N) = 25NYes! c. If the person still pulls with 26N, and the coefficient of kinetic friction is 0.05, what is the accl? F net = F – F f,k m. a = 26N – μ k x F N (25kg). a = 13N – (.05)(250N) a = 0.02 m / s 2

132. Ch6 Rev 13. A 60kg skydiver jumps from an airplane. a. What is the force of the earth on skydiver? b. What is the force of the skydiver on earth? c. If the mass of the earth is 6x10 24 kg, what is its accl?

133. Ch6 Rev 13. A 60kg skydiver jumps from an airplane. a. What is the force of the earth on skydiver? F E on P  F g = m. g = (60kg)(9.8 m / s 2 ) = 588N F E on P b. What is the force of the skydiver on earth? F P on E c. If the mass of the earth is 6x10 24 kg, what is its accl?

134. Ch6 Rev 13. A 60kg skydiver jumps from an airplane. a. What is the force of the earth on skydiver? F E on P  F g = m. g = (60kg)(9.8 m / s 2 ) = 588N F E on P b. What is the force of the skydiver on earth? F P on E F P on E = F E on P F P on E = 588N(Same!) c. If the mass of the earth is 6x10 24 kg, what is its accl?

135. Ch6 Rev 13. A 60kg skydiver jumps from an airplane. a. What is the force of the earth on skydiver? F E on P  F g = m. g = (60kg)(9.8 m / s 2 ) = 588N F E on P b. What is the force of the skydiver on earth? F P on E F P on E = F E on P F P on E = 588N(Same!) c. If the mass of the earth is 6x10 24 kg, what is its accl? F P on E = m E. a 588N = (6x10 24 kg). a a = 9.8x10 -23 m / s 2

136. Ch6 Rev 14. What is the accl of this system?15. What is the accl of this system? F net = F g3 – F T + F T – F g1. 3kg 1kg. 7kg 2kg

137. Ch6 Rev 14. What is the accl of this system?15. What is the accl of this system? F net = F g3 – F T + F T – F g1 F net = F g7 – F T + F T – F g2 m T. a = 30N – 10N F T (4kg)a = 20N F T F g1 a = 5 m / s 2 F g3. 3kg 1kg. 7kg 2kg

138. Ch6 Rev 14. What is the accl of this system?15. What is the accl of this system? F net = F g3 – F T + F T – F g1 F net = F g7 – F T + F T – F g2 m T. a = 30N – 10N m T. a = 70N – 20N F T F T (4kg)a = 20N (9kg)a = 50N F T F T F g1 a = 5 m / s 2 F g2 a = 5.6 m / s 2 F g3 F g7. 3kg 1kg. 7kg 2kg

139. Lab6.5A – Newton’s Third Law – Balloon Rockets Materials: 1 balloon, one straw cut to any length, one piece of tape, and a long string. Procedure: 1. Your lab group decides on the design. Use the low friction string first. 2. Once you are successful in getting the balloon across the room on the string, repeat the procedure on the thick string. Analysis: 1. Use Newton’s third law to explain how the balloon rocket is able to move across the room. Include a free-body diagram that shows the interaction-pair of forces exhibiting Newton’s third law.

140. 100 90 80 70 vel 60 (m/s) 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 time (sec) F N F fk F hangar F g