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06 newton's law of motion
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### Dynamicssv.pptx

1. 1. Mechanics Mechanics Kinematics Dynamics
2. 2. Force
3. 3. Fundamental Forces
4. 4. Comparing Contact and Field Forces
5. 5. Concept Check – Newton’s 1st Law A book is lying at rest on a table. The book will remain there at rest because: 1. there is a net force but the book has too much inertia 2. there are no forces acting on it at all 3. it does move, but too slowly to be seen 4. there is no net force (SF= 0) on the book 5. there is a net force, but the book is too heavy to move
6. 6. Concept Check – Newton’s 1st Law A book is lying at rest on a table. The book will remain there at rest because: 1. there is a net force but the book has too much inertia 2. there are no forces acting on it at all 3. it does move, but too slowly to be seen 4. there is no net force (SF= 0) on the book 5. there is a net force, but the book is too heavy to move There are forces acting on the book, but the only forces acting are in the y-direction. Gravity acts downward, but the table exerts an upward force that is equally strong, so the two forces cancel, leaving no net force.
7. 7. Concept Check – Newton’s 1st Law (2) A hockey puck slides on ice at constant velocity. What is the net force acting on the puck? 1. more than its weight 2. equal to its weight 3. less than its weight but more than zero 4. depends on the speed of the puck 5. zero
8. 8. Concept Check – Newton’s 1st Law (2) A hockey puck slides on ice at constant velocity. What is the net force acting on the puck? 1. more than its weight 2. equal to its weight 3. less than its weight but more than zero 4. depends on the speed of the puck 5. zero The puck is moving at a constant velocity, and therefore it is not accelerating. Thus, there must be no net force acting on the puck. Follow-up: Are there any forces acting on the puck? What are they?
9. 9. Concept Check – Newton’s 1st Law (3) You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why? 1. a net force acted on it 2. no net force acted on it 3. it remained at rest 4. it did not move, but only seemed to 5. gravity briefly stopped acting on it
10. 10. Concept Check – Newton’s 1st Law (3) You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why? 1. a net force acted on it 2. no net force acted on it 3. it remained at rest 4. it did not move, but only seemed to 5. gravity briefly stopped acting on it The book was initially moving forward (since it was on a moving bus). When the bus stopped, the book continued moving forward, which was its initial state of motion, and therefore it slid forward off the seat. Follow-up: What is the force that usually keeps the book on the seat?
11. 11. You kick a smooth flat stone out on a frozen pond. The stone slides, slows down and eventually stops. You conclude that: 1. the force pushing the stone forward finally stopped pushing on it 2. no net force acted on the stone 3. a net force acted on it all along 4. the stone simply “ran out of steam” 5. the stone has a natural tendency to be at rest Concept Check – Newton’s 1st Law (4)
12. 12. You kick a smooth flat stone out on a frozen pond. The stone slides, slows down and eventually stops. You conclude that: 1. the force pushing the stone forward finally stopped pushing on it 2. no net force acted on the stone 3. a net force acted on it all along 4. the stone simply “ran out of steam” 5. the stone has a natural tendency to be at rest After the stone was kicked, no force was pushing it along! However, there must have been some force acting on the stone to slow it down and stop it. This would be friction. Concept Check – Newton’s 1st Law (4)
13. 13. Newton’s First Law
14. 14. Newton’s First Law
15. 15. Concept Check – Newton’s First Law Consider a cart on a horizontal frictionless table. Once the cart has been given a push and released, what will happen to the cart? 1. slowly come to a stop 2. continue with constant acceleration 3. continue with decreasing acceleration 4. continue with constant velocity 5. immediately come to a stop
16. 16. Concept Check – Newton’s First Law Consider a cart on a horizontal frictionless table. Once the cart has been given a push and released, what will happen to the cart? 1. slowly come to a stop 2. continue with constant acceleration 3. continue with decreasing acceleration 4. continue with constant velocity 5. immediately come to a stop After the cart is released, there is no longer a force in the x-direction. This does not mean that the cart stops moving. It simply means that the cart will continue moving with the same velocity it had at the moment of release. The initial push got the cart moving, but that force is not needed to keep the cart in motion.
17. 17. Concept Check – Newton’s Second Law We just decided that the cart continues with constant velocity. What would have to be done in order to have the cart continue with constant acceleration? 1. push the cart harder before release 2. push the cart longer before release 3. push the cart continuously 4. change the mass of the cart 5. it is impossible to do that
18. 18. Concept Check – Newton’s Second Law We just decided that the cart continues with constant velocity. What would have to be done in order to have the cart continue with constant acceleration? 1. push the cart harder before release 2. push the cart longer before release 3. push the cart continuously 4. change the mass of the cart 5. it is impossible to do that In order to achieve a non-zero acceleration, it is necessary to maintain the applied force. The only way to do this would be to continue pushing the cart as it moves down the track. This will lead us to a discussion of Newton’s Second Law.
19. 19. SF represents the vector sum of all external forces acting on the object, or the net force. Newton’s 2nd Law a F  S 1 a m  F a m S  F ma S  x x F ma S  y y F ma S 
20. 20. Concept Check – Newton’s 3rd Law When you climb up a rope, the first thing you do is pull down on the rope. How do you manage to go up the rope by doing that?? 1. this slows your initial velocity, which is already upward 2. you don’t go up, you’re too heavy 3. you’re not really pulling down – it just seems that way 4. the rope actually pulls you up 5. you are pulling the ceiling down
21. 21. Concept Check – Newton’s 3rd Law When you climb up a rope, the first thing you do is pull down on the rope. How do you manage to go up the rope by doing that?? 1. this slows your initial velocity, which is already upward 2. you don’t go up, you’re too heavy 3. you’re not really pulling down – it just seems that way 4. the rope actually pulls you up 5. you are pulling the ceiling down When you pull down on the rope, the rope pulls up on you!! It is actually this upward force by the rope that makes you move up! This is the “reaction” force (by the rope on you) to the force that you exerted on the rope. And voilá, this is Newton’s 3rd Law.
22. 22. Newton’s Third Law
23. 23. Action and Reaction Forces If A acts on B, then B acts back on A
24. 24. Concept Check – Newton’s 3rd Law A small car collides with a large truck. Which experiences the greater impact force? 1. the car 2. the truck 3. both the same 4. it depends on the velocity of each 5. it depends on the mass of each
25. 25. Concept Check – Newton’s 3rd Law A small car collides with a large truck. Which experiences the greater impact force? 1. the car 2. the truck 3. both the same 4. it depends on the velocity of each 5. it depends on the mass of each
26. 26. Concept Check – Newton’s 3rd Law A small car collides with a large truck. Which experiences the greater acceleration? 1. the car 2. the truck 3. both the same 4. it depends on the velocity of each 5. it depends on the mass of each
27. 27. Concept Check – Newton’s 3rd Law A small car collides with a large truck. Which experiences the greater acceleration? 1. the car 2. the truck 3. both the same 4. it depends on the velocity of each 5. it depends on the mass of each We have seen that both vehicles experience the same magnitude of force. But the acceleration is given by F/m so the car has the larger acceleration, since it has the smaller mass.
28. 28. Concept Check – Newton’s 3rd Law F12 F21 1. the bowling ball exerts a greater force on the ping-pong ball 2. the ping-pong ball exerts a greater force on the bowling ball 3. the forces are equal 4. the forces are zero because they cancel out 5. there are actually no forces at all In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare?
29. 29. Concept Check – Newton’s 3rd Law F12 F21 1. the bowling ball exerts a greater force on the ping-pong ball 2. the ping-pong ball exerts a greater force on the bowling ball 3. the forces are equal 4. the forces are zero because they cancel out 5. there are actually no forces at all In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare? The forces are equal and opposite by Newton’s 3rd Law!
30. 30. Concept Check – Newton’s 3rd Law F12 F21 1. they do not accelerate because they are weightless 2. accelerations are equal, but not opposite 3. accelerations are opposite, but bigger for the bowling ball 4. accelerations are opposite, but bigger for the ping-pong ball 5. accelerations are equal and opposite In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the accelerations of the objects compare?
31. 31. Concept Check – Newton’s 3rd Law F12 F21 1. they do not accelerate because they are weightless 2. accelerations are equal, but not opposite 3. accelerations are opposite, but bigger for the bowling ball 4. accelerations are opposite, but bigger for the ping-pong ball 5. accelerations are equal and opposite In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the accelerations of the objects compare? The forces are equal and opposite -- this is Newton’s 3rd Law!! But acceleration is F/m and so the smaller mass has the bigger acceleration. Follow-up: Where will the balls meet if they are released from this position?
32. 32. Newton’s 2nd and 3rd Laws SF m a = SF m a = SF m a = Car Truck SF m a = SF m a = SF m a = 2nd Law You Earth 2nd Law
33. 33. Free Body Diagrams (force of gravity) g F (normal force) n F (force of kinetic friction) k F (applied force = pull) a F • Represent object with a dot • Draw arrows to represent forces • Label forces based on nature • Draw vectors proportionately • Show ONLY forces on ONE object
34. 34. Common Forces to Consider Gravity Normal Friction Push and Pull Tension (force of surface on object) n F (force of gravity) g F or (force of friction - kinetic or static) k s F (applied force = push or pull) a F (force due to rope or chain) T F Fg = mg g = 9.81 m/s2
35. 35. Free Body Diagrams FT,1 FT,2 Fg FT Fg Fg Fg FT,1 FT,2
36. 36. SF represents the vector sum of all external forces acting on the object, or the net force. Newton’s 2nd Law a F  S 1 a m  F a m S  F ma S  x x F ma S  y y F ma S 
37. 37. Solving Problems involving Force and Motion 1. Given • List Variables • Draw a Free Body Diagram (FBD) 2. Unknown • Identify the unknown variable 3. Equation • Newton’s Laws (2nd or 1st) • Write out equation from the FBD • Solve for the Unknown
38. 38. Net Force equations from Free Body Diagrams (force of gravity) g F 0 x F S    y g y F F ma   S   Falling, no air resistance (force of gravity) g F (force of air resistance) r F 0 x F S    y g r y F F F ma   S    Falling through air, speeding up Description FBD Net Force Equations
39. 39. Air Resistance 2 air resistance F v 
40. 40. Net Force equations from Free Body Diagrams (force of gravity) g F (force of air resistance) r F 0 x F S    0 y g r y F F F ma   S     Falling at Terminal Velocity (force of gravity) g F (force of air resistance) r F 0 x F S    y g r y F F F ma   S    Falling, but slowing Description FBD Net Force Equations
41. 41. Net Force equations from Free Body Diagrams (force of gravity) g F (tension in rope) T F 0 x F S    0 y T g F F F   S    Elevator at rest (force of gravity) g F (tension in rope) T F 0 x F S    0 y T g F F F   S    Elevator moving up or down at constant velocity Description FBD Net Force Equations
42. 42. Net Force equations from Free Body Diagrams (force of gravity) g F (tension in rope) T F 0 x F S    y T g y F F F ma   S    Elevator moving up, slowing down (force of gravity) g F (tension in rope) T F 0 x F S    y T g y F F F ma   S    Elevator moving up, speeding up Description FBD Net Force Equations
43. 43. Net Force equations from Free Body Diagrams g F T F 0 x F S  Elevator moving down, speeding up g F T F 0 x F S    y g T y F F F ma   S    Elevator moving down, slowing Description FBD Net Force Equations   y g T y F F F ma   S   
44. 44. Sample An elevator of mass _____ kg is slowing as it nears its final destination going up. If the tension in the rope is ______ N, determine the elevator’s acceleration. (2 force vectors) G: U: E: S: S: All of the problems in this section are without numbers because the focus is on the skill of constructing a well-reasoned solution rather than simply obtaining a correct numerical answer. Once you solve some of the problems, we will include some numbers in class. g F T F ? a  T m g F   y T g y F F F ma   S    T g y F F a m   T F mg m   (since ) g F mg 
45. 45. Problems – 2 The ball from Q1 is falling straight down, but now consider that there is a ____N force or air resistance. Draw an FBD and determine the mass’s acceleration. (2 force vectors) G: U: E: S: S: ? a  r m g F   y g r y F F F ma   S    g r y F F a m   r mg F m   g F r F
46. 46. Problems – 3 An elevator (mass = ____kg) is held at rest from a vertical cable. Determine the tension in the cable. (2 force vectors) G: U: E: S: S: ? T F  2 0m s y m g a    0 y T g y F F F ma   S     0 T g F F   g F T F T g F F mg   
47. 47. Net Force equations from Free Body Diagrams g F n F 0 x F S    0 y n g F F F   S    At rest on a surface or sliding with no friction g F n F   x a x F F ma   S     0 y n g F F F   S    Pushed on a surface a F Description FBD Net Force Equations
48. 48. Friction Sliding friction is electrostatic. The force of friction depends on only 2 factors: • The texture of the surfaces in contact (coefficient of friction - 𝜇) • The force pushing the surfaces together (normal force - 𝐹𝑛) Friction can be: • Static (prevents an object from sliding) • Kinetic (acts while an object is sliding) In general, Static > Kinetic
49. 49. Friction f n F F   The force of friction is simply the product of the coefficient of friction and the normal force. Static friction ranges in value up to a maximum: Kinetic friction has a constant value of: k k n F F   ,max S s n F F  
50. 50. Coefficients of Friction You will find this table in your text on p. 138.
51. 51. In free-body diagrams, the force of friction is always parallel to the surface of contact. Kinetic friction is always opposite the direction of motion. Static friction must point in the direction that results in a net force of zero. Friction Forces in Free-Body Diagrams g F n F (kinetic friction) k F If moving, kinetic friction opposes motion If not moving, static friction is in a direction that provides equilibrium g F n F (static friction) S F a F a F Not moving
52. 52. Net Force Equations From Free Body Diagrams g F n F   x k x F F ma   S      0 y n g F F F   S    Sliding and slowing due to friction (kinetic friction) k F k x F ma   k n x F ma    n g F F  Description FBD Net Force Equations
53. 53. Net Force equations from Free Body Diagrams g F n F a F   or x a k s x F F F ma   S      0 y n g F F F   S    or k s F F Pushed on a surface with friction n g F F  Description FBD Net Force Equations
54. 54. Sample Problem n F m = 55 kg Fa = 19 N to set in motion g = 9.81 m/s2 ax = 0 ay = 0 s = ?   ,max 0 x a s F F F   S      0 y n g F F F   S    a s F F  max , 0.035 s   a F g F ,max s F n s s F F max ,   n g F F mg   ,max s a s n F F F mg       2 19 N 55 kg 9.81m s  (since ) g F mg 
55. 55. Problem – 10 A box of mass ______ kg is at rest on a horizontal floor. If the coefficient of static friction is ____, find the force needed to make the box begin moving. (4 force vectors) G: U: E: ? a F  0 0 s x y m g a a      0 x a s F F F   S    n F g F s F (since ) g F mg  (since ) s s n F F     0 y n g F F F   S    n g F F mg   a F a s F F  a s n F F   a s F mg  
56. 56. Net Force equations from Free Body Diagrams g F n F cos x a x F F ma  S   sin 0 n a g F F F     Pushed down at an angle on a surface (no friction) a F  a F  cos a F  sin a F  g F n F   , x a x x F F ma   S     , 0 y n a y g F F F F   S     Description FBD Net Force Equations
57. 57. Net Force equations from Free Body Diagrams g F n F cos x a x F F ma  S   sin 0 n a g F F F     Pulled up at an angle no friction a F  a F  cos a F  sin a F  g F n F   , x a x x F F ma   S     , 0 y n a y g F F F F   S     Description FBD Net Force Equations
58. 58. Net Force equations from Free Body Diagrams n F a F g F k F  n F a F g F k F    , x a x k x F F F ma   S      , 0 y n a y g F F F F   S       cos x a k x F F F ma    S      sin 0 y n a g F F F F    S     Pulled up at an angle with friction Pushed down at an angle with friction cos a F  sin a F  cos a F  sin a F  cos a k x F F ma    sin 0 n a g F F F     Description FBD Net Force Equations
59. 59. Solving Problems using Newton’s Laws Given Include Free Body Diagram Choose + direction Unknown Equations ΣF = ma (or ΣF = 0 ) Fk = μkFn (or Fs = μsFn) Fg = mg Use components if necessary Solve system of equations if necessary Substitute only at the end Significant Figures n F a F g F k F +→ Σ𝐹𝑥 = 𝐹𝑎 − 𝐹𝑘 = 𝑚𝑎𝑥 +↑ Σ𝐹𝑦 = 𝐹𝑛 − 𝐹 𝑔 = 𝑚𝑎𝑦 𝐹 𝑔 = 𝑚𝑔 𝐹𝑘 = 𝜇𝑘𝐹𝑛
60. 60. Chapter 4 Sample Problem n F m Fa  k g ay = 0 ax = ?   cos x a k x F F F ma    S      sin 0 y n a g F F F F    S     a F g F k F   sin a g n F F F   sin n a F mg F    (since ) k k n F F   m F F a k a x    cos m F F a n k a x     cos   cos sin a k a x F mg F a m       n F g F k F cos a F  sin a F  (since ) g F mg  , , sin (upward) cos (to the right) a y a a x a F F F F    
61. 61. Problem – 12 An already-sliding hockey puck of mass _______ grams is pushed horizontally by a hockey stick with a force of _______ N in the direction of its motion. The coefficient of kinetic friction is ______. Find the acceleration of the puck. (4 force vectors) G: U: E: ? x a  0 a k y m g F a     x a k x F F F ma   S    a k x F F a m   n F a F g F k F (since ) g F mg  (since ) k k n F F   a k x F mg a m      0 y n g F F F   S    n g F F mg   a k n F F m   
62. 62. Problem – 18 A box of mass ______ kg is moving along a horizontal floor at a constant velocity of ______ m/s to the right. A girl is pushing down-and-to-the-right on the box with a force of ______ N @ ______ degrees below the horizontal. Find the coefficient of kinetic friction. (4 force vectors) G: U: E: ? k   0 0 a x y m g F a a      cos 0 x a k F F F    S    (since ) g F mg  (since ) k k n F F     sin 0 y n a g F F F F    S     cos k a F F   n F a F g F k F  cos a F  sin a F  cos k n a F F    cos a k n F F    sin n a g F F F    cos sin a a F F mg     sin n a F F mg   
63. 63. Objects on an Incline n F g F k F x y 𝜃 𝜃 n F g F k F x y 𝜃 n F g F k F 𝜃 x y X-axis is parallel to surface; y-axis is  to surface. Force of gravity is straight down Normal force is  to the surface (and will no longer be opposite to gravity) The angle of the incline is equal to the angle between gravity and y-axis • Find and use the components of gravity Friction (when present) will always be along x-axis
64. 64. Net Force equations from Free Body Diagrams n F g F  x y   , 0 y n g y F F F   S      , x g x x F F ma   S   Sliding down an incline (no friction) g F  x y sin g F  cos g F  , g x F , g y F sin x g x F F ma  S   cos 0 y n g F F F  S    Description FBD Net Force Equations cos n g F F  
65. 65. Net Force equations from Free Body Diagrams n F g F k F  x y   , 0 y n g y F F F   S      , x g x k x F F F ma   S    Sliding down an incline with friction g F  x y sin g F  cos g F  , g x F , g y F sin x g k x F F F ma  S    cos 0 y n g F F F  S    Description FBD Net Force Equations cos n g F F  
66. 66. Sample A _____ kg child-and-sled combination is sliding down a snowy slope at ____ degrees below the horizontal. If the coefficient of kinetic friction is ____, what is the acceleration? (3 force vectors) G: U: E: ? x a  0 k y m g a      sin x g k x F F F ma    S    (since ) g F mg  (since ) k k n F F     cos 0 y n g F F F    S    cos cos n g F F mg     n F k F x y sin g F  cos g F  sin g k x F F a m    sin g k n x F F a m     sin cos k mg mg m      sin cos k g g      n F g F k F x y g F

### Notes de l'éditeur

• Kinematics The study of how things move
How far? How Fast? How long?
Dynamics The study of why things move
Force causes motion and changes in motion
• A force is an action exerted on an object which may change the object’s state of rest or motion.

Forces can cause accelerations.

The SI unit of force is the newton, N.

Forces can act through contact or at a distance.
• The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object’s mass.

F = ma
net force = mass  acceleration

• If two objects interact, the magnitude of the force exerted on object 1 by object 2 is equal to the magnitude of the force simultaneously exerted on object 2 by object 1, and these two forces are opposite in direction.

In other words, for every action, there is an equal and opposite reaction.

Because the forces coexist, either force can be called the action or the reaction.
• Action-reaction pairs do not imply that the net force on either object is zero.
The action-reaction forces are equal and opposite, but either object may still have a net force on it.

Consider driving a nail into wood with a hammer. The force that the nail exerts on the hammer is equal and opposite to the force that the hammer exerts on the nail. But there is a net force acting on the nail, which drives the nail into the wood.
• Physicists represent forces on an object by drawing a Free Body Diagram. This is simply a representation of an object and a graphical representation of all the forces acting on that object. Simply put, in a free body diagram, all the forces acting on the given object are represented as arrows. The size of the arrow is reflective of the magnitude of the force. The direction of the arrow reveals the direction in which the force acts. Each force arrow in the diagram is labeled to indicate the type of force.

It is customary in a free-body diagram to not draw the object, but instead to represent it simply by a box  or a small circle  and to draw the force arrows from the center of the box or circle outward in the direction in which the force is acting. A free body diagram showing the forces on a box being pushed across the floor is shown here, where Fg is the force due to gravity, Fn is the normal force of the floor pushing up on the object, Fa is the external applied force exerted by the pusher, and Fk is the force of sliding kinetic friction
• The first force we will investigate is that due to gravity, and we'll call it the gravitational force. We know that the acceleration due to gravity (if on Earth) is approximately g = 9.8 m/s2 . The force, by Newton's Second Law is F = m g

The normal force one which prevents objects from 'falling' into whatever it is they are sitting upon. It is always perpendicular to the surface with which an object is in contact. For example, if there is a crate on the floor, then we say that the crate experiences a normal force by the floor; and because of this force, the crate does not fall into the floor. The normal force on the crate points upward, perpendicular to the floor.

Related to the normal force is the frictional force. The two are related because they are both due to the surface in contact with the body. Whereas the normal force was perpendicular to the surface, the frictional force is parallel. Furthermore, friction opposes motion, and so its vector always points away from the direction of movement. Sliding or contact friction is either kinetic or static.

Another force which may act on an object could be any physical push or pull. This could be caused by a person pushing a crate on the floor, a child pulling on a wagon, or in the case of our example, the wind pushing on the ship. This is often referred to as an applied force.

Tension in an object results if pulling force act on its ends, such as in a rope used to pull a boulder. If no forces are acting on the rope, say, except at its ends, and the rope itself is in equilibrium, then the tension is the same throughout the rope.
• The gravitational force (Fg) exerted on an object by Earth is a vector quantity, directed toward the center of Earth.

The magnitude of this force (Fg) is a scalar quantity called weight.

Weight changes with the location of an object in the universe.

Calculating weight at any location:
Fg = mag
ag = free-fall acceleration at that location

Calculating weight on Earth's surface:
ag = g = 9.81 m/s2
Fg = mg = m(9.81 m/s2)

• The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object’s mass.

F = ma
net force = mass  acceleration

• Static friction is a force that resists the initiation of sliding motion between two surfaces that are in contact and at rest.

Kinetic friction is the force that opposes the movement of two surfaces that are in contact and are sliding over each other.

Kinetic friction is always less than the maximum static friction.
• Static friction is a force that resists the initiation of sliding motion between two surfaces that are in contact and at rest.

Kinetic friction is the force that opposes the movement of two surfaces that are in contact and are sliding over each other.

Kinetic friction is always less than the maximum static friction.