2. Dynamics

3. If something is moving, then something MUST be pushing on it. Things don’t move all by themselves! Aristotle

4. Oh, baloney! Once an object is moving in a straight line, it’s going to keep right on moving, unless it runs into something. That is, of course, as long as there is NO friction! Friction is what slows things down. Galileo

5. Galileo… Galileo found that, ignoring the effect of the air, all freely falling objects had the SAME acceleration, regardless of their mass or the height from which they were dropped. The Law of Falling Bodies Born in Pisa 1564

6. Galileo described HOW things moved, but not WHY they moved that way. Galileo wrote, “the present does not seem to be the proper time to investigate the cause of the acceleration of natural motion….” Galileo died in 1642. When Isaac Newton, born Christmas day, 1642, began his studies of motion in the second half of the seventeenth century, that statement was no longer appropriate.

7. Because Galileo had been so effective in describing motion, Newton could turn his attention to dynamics. Dynamics is the study of why an object moves the way it does- why it starts to move instead of remaining at rest, why it speeds up or moves on a curved path, and why it comes to a stop. Newton’s most famous book was “Principia”, published in 1687. This book first listed what came to be known as Newton’s Three Laws of Motion

8. Forces Force: a push or a pull Forces are vectors. ( you push or pull in a direction!) There are Four Fundamental Forces in our universe 1.Gravitational Force 2.Electromagnetic Force 3.Strong Nuclear Force 4.Weak Force

9. The strongest of these forces is…. The Strong Nuclear Force (it holds the nucleus of all atoms together) The weakest of these forces is…. Gravity

10. The unit for Force is the Newton, N If someone was pushing on you with one Newton of Force, it wouldn’t hurt much. One Newton of Force is about the same as the weight of a quarter pound hamburger patty.

11. Inertia Inertia: an object’s resistance to a change in its motion Mass: a way to measure inertia unit: kg

12. Newton’s First Law of Motion: “The Law of Inertia” An object at rest remains at rest unless a net external force acts on it. An object in motion continues that motion unless a net external force acts on it.

13. Newton’s First Law of Motion: “The Law of Inertia” An object at rest remains at rest unless a net external force acts on it. An object in motion continues that motion unless a net external force acts on it. Newton’s law of inertia confirmed what Galileo concluded: Once an object is moving, it requires no additional force to keep it moving. It will continue to move in a straight line unless a NET force acts upon it. I knew it first!!

14. If there is a net external force acting on an object, it will accelerate... Newton’s Second Law is expressed as an equation: (I call this “the granddaddy of all physics equations”!) Second Law

16. Same Force on different mass….

17. Third Law For every force, there is an equal but opposite force. Action / Reaction forces

18. For every force, there is an equal but opposite force. Action / Reaction forces

19. For every force, there is an equal but opposite force. Action / Reaction forces

20. Action Force: Balloon pushes air downward Reaction Force: Air pushes balloon upward

21. The gun pushes on the bullet. The bullet pushes back on the gun!

23. The Action / Reaction forces do NOT act on the same object. I kick the wall (Action) The wall kicks me back (Reaction)

24. While driving down the road, an unfortunate bug strikes the windshield of a bus. The bug hit the bus and the bus hit the bug. Which of the two forces is greater: the force on the bug or the force on the bus? For every force, there is an EQUAL by opposite force!

25. A child pulls on a 5.0 kg wagon with a net force of 20.0 N. What is the wagon’s acceleration? m = 5.0 kg F net = 20.0 N a = F net / m a = 20.0 / 5.0 = a = 4 m/s 2

26. A dog pulling a man on a sled over ice can accelerate the sled at 2 m/s 2. If the sled and man have a combined mass of 150.0 kg, what was the net force on the sled? m = 150.0 kg a = 2 m/s 2 F net = ma F net = 150.0 kg x 2 m/s 2 = 300 N

27. When released, the lift provides a net force of 6 N on a 2 kg balloon. If it started at rest, how fast will it be moving in 4.0 s? F net = 6 Nm = 2 kg v o = 0t = 4 s v f = ? v f = v o + at, so we need “a”. a = F net / m a = 6 / 2= 3 m/s 2 v f = 0 + 3 x 4 v f = 12 m/s

28. A 1500 kg. car starts from rest and is moving at 10 m/s after 5.0 s. What was the net force on the car? m = 1500 kg V o = 0 V f = 10 m/s t = 5.0 s F net = ? F net = ma, so we need to find the acceleration. v f = v o + at a = ( v f – v o ) / t a = (10 – 0) / 5 = 2 m/s 2 F net = ma F net = 1500 kg x 2 m/s 2 F net = 3000 N

30. A boy pushing on a 15 kg lawnmower wants to produce an acceleration along the horizontal ground of 3 m/s 2. What force must he exert if his force is directed at an angle of 40 degrees measured from the horizontal? Force = mass x acceleration Force = 15 kg x 3 m/s 2 Horizontal Force = 45 N But…. Question: If the horizontal component of force is 45 N, what was the magnitude of the force the boy exerted at an angle of 40 degrees? Which trig function can be used to find the hypotenuse when the angle and adjacent side are known? Cos  = adj / hyp The boy’s force = 58.74 N 45 N 40 o Force = ?

31. Weight Weight, Wt. is the gravitational force on an object Weight = mass x gravity W = mg Since weight is a force, it is measured in Newtons, N Remember, “g” on Earth is 9.8 m/s 2

32. What is the weight of a 42 kg child on Earth? W = mgm = 42 kg W = 42 kg x 9.8 m/s 2 W = 411.6 N

33. What is the mass of a 15000 N car? W = mg m = W / g m = 15000 N / 9.8 m/s 2 m = 1530.6 kg

34. Remember, one Newton is not a very big force (about the same as a ¼ pound). So, your weight in Newtons is MUCH bigger than your weight in pounds! In fact, you would have to multiply your weight in pounds by 4.45 to get your weight in Newtons. How much do you weigh in Newton’s?

35. Even if you weigh 550 Newtons, You still wouldn’t be much of a Sumo Wrestler! (that’s only around 120 lbs)

36. More than one force can act on an object at the same time. For example, two people could push on a book at the same time. One person could push toward the left and the other could push toward the right. In this case the two forces would act against each other. What is the NET force? 2 N towards the left If the book had a mass of ½ kg, what would be its acceleration? a = F net / m a = 2 /.5 a = 4 m/s 2 (left)

37. What if the opposing forces were equal? What is the net Force? What is the acceleration?

38. What if the forces were in the same direction? What is the net force? If the book had a mass of 2 kg, what is its acceleration? a = F net / m a = 14 / 2 = 7 m/s 2

39. Free Body Diagrams Free-body diagrams are pictures used to show the relative magnitude and direction of all forces acting upon an object in a given situation. These diagrams are often used in physics. The length of the arrow in a free-body diagram sometimes is used to represent the magnitude (size) of the force. The direction of the arrow shows the direction that the force is acting. Each force arrow in the diagram is labeled to indicate the exact type of force. The object itself is either drawn as a box or squeezed down to a dot. The force arrows are always drawn pointing away from the center of the box. 3 kg 18N6N 15N

40. What are the unknown forces for the given net force?

41. Forces on an Airplane: if the forces are not “balanced”, there will be an acceleration! When Lift is larger than Weight, the plane will…. When Weight is larger than Lift, the plane will… When Thrust is larger than Drag, the plan will…. When Drag is larger than Thrust, the plane will…. When Thrust = Drag, the plane will… When Lift = Weight, the plane will….

42. We usually label forces as negative or positive. Forces upward are positive. Forces downward are negative. Forces to the right are positive. Forces to the left are negative + - - +

43. What is the net Force? - 6 N + 18 N = + 12 N What is the acceleration? a = F net / m 12 / 3 = 4 m/s 2 3 kg 18N6N

44. What is the net Force? - 6 N – 15 N + 18 N = - 3 N What is the acceleration? a = F net / m a = - 3 N / 3 kg = - 1 m/s 2 3 kg 18N6N 15N

45. A parachute provides a lift force of 400 N on a parachutist that weighs 500 N. What is the net force on him? -100 N What is his acceleration? a = F net / m What is his mass? a = -100 N / 50 kg a = -2 m/s 2

46. Tension Tension, T, is the force that cables, ropes, and strings pull with.

47. A child pulls up on a string that is holding 2 fish of total mass 5 kg. If he is providing a tension of 60 N, what is the net force on the fish? (use g = 10 m/s 2 ) F net = Tension – Weight F net = 60 N – 50 N F net = 10 N What is the acceleration of the fish? a = F net / m a = 10 / 5 a = 2 m/s 2

49. One child pulls up on a box with a force of 19 N. Another child pulls down on the box with a force of 5 N. What is the net Force? Hold on, there’s another force not drawn! The gravitational force of weight is pulling down! Wt = mg, (g = 10 m/s 2 ) Wt = 2 kg x 10 m/s 2 = 20 N Draw the weight vector also! Now, what is the net Force? Net force = +19 N – 5 N – 20 N = Net Force = - 6 N What is the acceleration? a = Fnet / m = a = - 6 / 2 a = -3 m/s 2 It will accelerate downward. 2 kg 19 N 5 N mg = 20 N

50. , “sigma” is a Greek letter that is used to signify “the sum of” Quite often, in Newton’s 2 nd Law, we write  F = ma instead of F net = ma

51. A child pulls a 5 kg bucket out of well with a rope. If the bucket accelerates upward at 1.2 m/s 2, what is the tension in the rope? m = 5 kg a = 1.2 m/s 2 T = ?  F = ma T – mg = ma T = ma + mg T = 5 x 1.2 + 5 x 9.8 T = 55 N T mg

53. Now for a challenging problem…

54. The “Normal” Force, N When an object is pressed against a surface, the surface pushes back. (That’s Newton’s 3 rd Law) This “push back” from the surface is called the Normal Force, N The word “normal” in math terminology means “perpendicular” The surface pushes back in a direction that is perpendicular to the surface.

55. If the box is not accelerating up or down, then the Normal force must by equal to the Weight Weight = mg Normal force, N

56. But… what if you were accelerating up or down? The Normal force would NOT be equal to your weight if you’re accelerating up or down. And… your weight seems to change!

57. A 50 kg woman steps on a scale in an elevator that accelerates upward at 1.5 m/s 2. What is her REAL weight? mg = 500 N What is her APPARENT weight?  F = ma N – mg = ma Her APPARENT weight is what she feels like she weighs, given by the Normal force, N. N = ma + mg N = 50 x 1.5 + 500 Apparent weight = 575 N mg N

58. Apparent Weight When you ride an elevator, you “feel” heavier or lighter than you actually weigh because of the acceleration of the elevator. Your “apparent weight” is found by taking your REAL weight, mg, and adding the term ma, where “a” is your acceleration Apparent weight = mg + ma

59. A man of mass 75 kg is standing inside an elevator. What is his weight (use g = 10 m/s 2 )? 750 N How heavy does he feel when the elevator has an acceleration of -2 m/s 2 ? Apparent weight = mg + ma Apparent weight = 75 x 10 + 75 x (-2) = 600N How heavy does he feel when the elevator is moving upward with a constant velocity? Since a = 0, he only feels his real weight- 750 N.

60. Do falling objects REALLY accelerate toward the Earth at 9.8 m/s 2 ? No, because of air resistance. Air resistance is a force that pushes up on an object as it falls. The faster you fall… The greater the air resistance. Eventually, the air resistance pushing up on you is just as large as your weight that is pulling down on you!!

62. The faster the man falls, the more air resistance pushes up on him. Eventually, there will be just as much air resistance pushing up on him as his weight pulling him down. What will be the NET force acting then? What will be his acceleration?

63. Once the air resistance pushing up is as large as the weight pushing down, the NET force acting on you is ZERO! If the net force is zero, what is your acceleration? ZERO! This doesn’t mean you stop in mid air. But it does mean that you stop accelerating! You still continue to fall towards the Earth, but you don’t pick up any more velocity- you continue to fall towards the Earth at the same velocity. This speed is called your “terminal velocity” You will reach your terminal velocity when Air resistance = your weight

64. Which one will have a faster terminal velocity? You don’t reach terminal velocity until the air resistance grows to as large a force as your weight. The more massive skydiver will have a faster terminal velocity and hit the ground at a faster speed

68. The “Normal” Force, N When an object is pressed against a surface, the surface pushes back. (That’s Newton’s 3 rd Law) This “push back” from the surface is called the Normal Force, N The word “normal” in math terminology means “perpendicular” The surface pushes back in a direction that is perpendicular to the surface.

69. If the box is not accelerating up or down, then the Normal force must by equal to the Weight Weight = mg Normal force, N

70. Friction, f A force that always opposes motion Depends on two things: the roughness of the surfaces and how hard they are pressed together. f =  N , mu- the “coefficient of friction” tells how rough the surfaces are. N, the Normal force tells how hard the surfaces are pressed together

71. Example: How large is the frictional force between 2 surfaces if the coefficient of friction is 0.2 and the Normal force is 80 N? f =  N f = 0.2 x 80 f = 16 N

72. There are two kinds of friction: “static friction” (not moving) must be overcome to initiate motion. “kinetic friction” must be overcome while an object is moving Static friction > Kinetic friction

73. Weight = mg Normal force, N F A = 30 Nm = 2 kg  = 0.4a = ? F net = maf =  N N = mg = 2 x 10 = 20N f =  N = 0.4(20) = 8N Horizon: F net = F A - f 30N – 8N = 22N a = F net / m a = 11 m/s 2 FAFA f =  N = 30 N You pull on a box with an applied force of 30 N. The coefficient of friction is 0.4. If the mass of the box is 2 kg, what is its acceleration? 1.Draw the box and all FOUR forces acting on it. 2.Write what you know and don’t know. 3.Write the equations, Fnet = ma and f =  N 4.Calculate the Normal force and the friction force. 5.Calculate the value of the net Force and then the acceleration.

74. Weight = mg Normal force, N F A = 25Nm = 2 kg  = 43  = 0.4 a = ?  F x = maf =  N  F x = -  N + F A cos  = ma (-0.4(36.65) + 25cos 43 ) / 2= a a = 1.81 m/s 2 FAFA f   F y = N – mg - Fsin  = ma = 0 N = mg + Fsin  

75. Weight = mg Normal force, N F A = 15 Nm = 2 kg  = 43  = 0.4a = ?  F x = maf =  N  F x = -  N + F A cos  = ma (-0.4(9.77) + 15cos 43) / 2 = a a = 3.53 m/s 2 FAFA f   F y = N – mg + Fsin  = ma = 0 N = mg – Fsin  

76. Weight = mg Normal force, N If the box is moving at constant velocity, there is no acceleration, Therefore the net force must be zero… so the horizontal forces must cancel each other.  N = Fcos  If you push hard enough to just get the box moving, the acceleration is zero in that case also, but the friction is static, not kinetic. FAFA f 

78. Spring Force If a mass is suspended from a spring, two forces act on the mass: its weight and the spring force. The spring force for many springs is given by F s = -kx, Where x is the distance the spring is stretched (or compressed) from its normal length and “k” is called the spring constant, which tells the stiffness of the spring. The stiffer the spring, the larger the spring constant. This relationship is known as “Hooke’s Law of Elasticity”. The spring force is negative because the spring always pulls in the opposite direction from which it is stretched or compressed.

79. Spring Force If the mass is at rest, then the two forces are balanced, so that kx = mg These balanced forces provide a quick way to determine the spring constant of a spring, or anything springy, like a rubber band: Hang a known mass from the spring and measure how much it stretches, then solve for “k” !

80. More examples… Xavier, who was stopped at a light, accelerated forward at 6 m/s 2 when the light changed. He had dice hanging from his rear view mirror. What angle did they make with the vertical during this acceleration? 1- draw free body diagram 2- write Newton’s 2 nd law for BOTH directions  F x = ma x Tsin  = ma x Tcos  = mg tan  = a/g  = 31.47 degrees   F y = ma y mg T Tcos  – mg = ma y = 0

81. An “Atwood’s Machine” Two masses of 5 kg and 2 kg are suspended from a massless, frictionless pulley, When released from rest, what is their acceleration? What is the Tension in the string? 1- Draw a free body diagram 2- Write Newton’s Second Law for EXTERNAL forces acting on the whole system.  F ext = m total a The Tension in the string is an “internal” force. The only “external” forces are the gravitational forces of weight, which oppose each other. So…  F ext = m 1 g – m 2 g = m total a Solving for the acceleration yields a = 4.2 m/s 2 Now, to find the Tension, we must “zoom in” on mass 2:  F= T – m 2 g = m 2 a T = m 2 a + m 2 g T = 28 N a +

83. Objects on Inclines- sliding down, no friction   mg The free-body diagram ALWAYS comes first: Draw the weight vector, mg Draw the Normal force vector. Are there any other forces??? Since the motion is parallel to the plane, ROTATE the axis from horizontal and vertical to “parallel” and “perpendicular”. Then draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. Does the Normal force balance with the force of weight, mg? NO! What force must balance the Normal force? N Does all of the weight, mg, pull the box down the incline? Write Newton’s Second Law for the forces on box parallel to the plane with “down” being negative.  F = ma -mgsin  = ma N = mgcos  No! Only mgsin 

84. Inclines: pushed upward, no friction   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma F A - mgsin  = ma FAFA

85. Inclines: pushed downward, no friction   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma - F A - mgsin  = ma FAFA

86. Inclines: With friction, case 1: at rest or sliding down   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma f - mgsin  = ma  mgcos  - mgsin  = ma  gcos  - gsin  = a f What is friction? f =  N What is N ? N = mgcos  f =  mgcos 

87. Inclines: With friction, case 2: pushed downward   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma f - mgsin  – F A = ma  mgcos  - mgsin  – F A = ma f What is friction? f =  N f =  mgcos  FAFA

88. Inclines: With friction, case 3: pushed upward   mg Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. N Write Newton’s Second Law for the box.  F = ma F A - mgsin  – f = ma F A  - mgsin  -  mgcos  = ma f What is friction? f =  N f =  mgcos  FAFA

89. Friction along an incline An object placed along an incline will eventually slide down if the incline is elevated high enough. The angle at which it slides depends on how rough the incline surface is. To find the angle where it slides:  Since it doesn’t move: mgsin  =  mgcos  Therefore: Tan  max  =  max, the coefficient of static friction Or The angle  max  = tan -1  max