1. Chapter 6 Lecture

2. Chapter 6 Dynamics I: Motion Along a Line
Chapter Goal: To learn how to solve linear force-and-motion problems.
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3. Chapter 6 Preview
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4. Chapter 6 Preview
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5. Chapter 6 Preview
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6. Chapter 6 Preview
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7. Chapter 6 Preview
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8. Chapter 6 Preview
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9. Chapter 6 Reading Quiz
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10. Reading Question 6.1 Newton’s first law can be applied to
Static equilibrium.
Inertial equilibrium.
Dynamic equilibrium.
Both A and B.
Both A and C.
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11. Reading Question 6.1 Newton’s first law can be applied to
Static equilibrium.
Inertial equilibrium.
Dynamic equilibrium.
Both A and B.
Both A and C.
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12. Reading Question 6.2 Mass is An intrinsic property. A force.
A measurement.
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13. Reading Question 6.2 Mass is An intrinsic property. A force.
A measurement.
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14. Reading Question 6.3 Gravity is An intrinsic property. A force.
A measurement.
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15. Reading Question 6.3 Gravity is An intrinsic property. A force.
A measurement.
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16. Reading Question 6.4 Weight is An intrinsic property. A force.
A measurement.
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17. Reading Question 6.4 Weight is An intrinsic property. A force.
A measurement.
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18. Reading Question 6.5 The coefficient of static friction is
Smaller than the coefficient of kinetic friction.
Equal to the coefficient of kinetic friction.
Larger than the coefficient of kinetic friction.
Not discussed in this chapter.
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19. Reading Question 6.5 The coefficient of static friction is
Smaller than the coefficient of kinetic friction.
Equal to the coefficient of kinetic friction.
Larger than the coefficient of kinetic friction.
Not discussed in this chapter.
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20. Reading Question 6.6 The force of friction is described by
The law of friction.
The theory of friction.
A model of friction.
The friction hypothesis.
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21. Reading Question 6.6 The force of friction is described by
The law of friction.
The theory of friction.
A model of friction.
The friction hypothesis.
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22. Reading Question 6.7
When an object moves through the air, the magnitude of the drag force on it
Increases as the object’s speed increases.
Decreases as the object’s speed increases.
Does not depend on the object’s speed.
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23. Reading Question 6.7
When an object moves through the air, the magnitude of the drag force on it
Increases as the object’s speed increases.
Decreases as the object’s speed increases.
Does not depend on the object’s speed.
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24. Reading Question 6.8 Terminal speed is Equal to the speed of sound.
The minimum speed an object needs to escape the earth’s gravity.
The speed at which the drag force cancels the gravitational force.
The speed at which the drag force reaches a minimum.
Any speed which can result in a person’s death.
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25. Reading Question 6.8 Terminal speed is Equal to the speed of sound.
The minimum speed an object needs to escape the earth’s gravity.
The speed at which the drag force cancels the gravitational force.
The speed at which the drag force reaches a minimum.
Any speed which can result in a person’s death.
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26. Chapter 6 Content, Examples, and QuickCheck Questions
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27. Equilibrium
An object on which the net force is zero is in equilibrium.
If the object is at rest, it is in static equilibrium.
If the object is moving along a straight line with a constant velocity it is in dynamic equilibrium.
The requirement for either type of equilibrium is:
The concept of equilibrium is essential for the engineering analysis of stationary objects such as bridges.
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28. QuickCheck 6.1
The figure shows the view looking down onto a sheet of frictionless ice. A puck, tied with a string to point P, slides on the ice in the circular path shown and has made many revolutions. If the string suddenly breaks with the puck in the position shown, which path best represents the puck’s subsequent motion?
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29. QuickCheck 6.1 Newton’s first law!
The figure shows the view looking down onto a sheet of frictionless ice. A puck, tied with a string to point P, slides on the ice in the circular path shown and has made many revolutions. If the string suddenly breaks with the puck in the position shown, which path best represents the puck’s subsequent motion?
Newton’s first law!
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30. Problem-Solving Strategy: Equilibrium Problems
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31. Problem-Solving Strategy: Equilibrium Problems
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32. QuickCheck 6.2
A ring, seen from above, is pulled on by three forces. The ring is not moving. How big is the force F?
20 N
10cos N
10sin N
20cos N
20sin N
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33. QuickCheck 6.2
A ring, seen from above, is pulled on by three forces. The ring is not moving. How big is the force F?
20 N
10cos N
10sin N
20cos N
20sin N
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34. Example 6.2 Towing a Car up a Hill
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35. Example 6.2 Towing a Car up a Hill
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36. Example 6.2 Towing a Car up a Hill
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37. Example 6.2 Towing a Car up a Hill
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38. Example 6.2 Towing a Car up a Hill
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39. QuickCheck 6.3
A car is parked on a hill. Which is the correct free-body diagram?
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40. QuickCheck 6.3
A car is parked on a hill. Which is the correct free-body diagram?
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41. QuickCheck 6.4
A car is towed to the right at constant speed. Which is the correct free-body diagram?
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42. QuickCheck 6.4
A car is towed to the right at constant speed. Which is the correct free-body diagram?
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43. Using Newton’s Second Law
The essence of Newtonian mechanics can be expressed in two steps:
The forces on an object determine its   acceleration , and
The object’s trajectory can be determined by   using in the equations of kinematics.
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44. Problem-Solving Strategy: Dynamics Problems
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45. Problem-Solving Strategy: Dynamics Problems
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46. QuickCheck 6.5
The cart is initially at rest. Force is applied to the cart for time t, after which the car has speed v. Suppose the same force is applied for the same time to a second cart with twice the mass. Friction is negligible. Afterward, the second cart’s speed will be v 2v 4v 1 4 1 2
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47. QuickCheck 6.5
The cart is initially at rest. Force is applied to the cart for time t, after which the car has speed v. Suppose the same force is applied for the same time to a second cart with twice the mass. Friction is negligible. Afterward, the second cart’s speed will be v 2v 4v 1 4 1 2
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48. Example 6.3 Speed of a Towed Car
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49. Example 6.3 Speed of a Towed Car
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50. Example 6.3 Speed of a Towed Car
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51. Example 6.3 Speed of a Towed Car
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52. QuickCheck 6.6
The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
n > mg.
n = mg.
n < mg.
n = 0.
Not enough information to tell.
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53. QuickCheck 6.6 Upward acceleration requires a net upward force.
The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
n > mg.
n = mg.
n < mg.
n = 0.
Not enough information to tell.
Upward acceleration requires a net upward force.
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54. Mass: An Intrinsic Property
A pan balance, shown in the figure, is a device for measuring mass.
The measurement does not depend on the strength of gravity.
Mass is a scalar quantity that describes an object’s inertia.
Mass describes the amount of matter in an object.
Mass is an intrinsic property of an object.
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55. Gravity: A Force
Gravity is an attractive, long-range force between any two objects.
The figure shows two objects with masses m1 and m2 whose centers are separated by distance r.
Each object pulls on the other with a force:
where G = 6.67 × 10−11 N m2/kg2 is the gravitational constant.
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56. Gravity: A Force
The gravitational force between two human-sized objects is very small.
Only when one of the objects is planet-sized or larger does gravity become an important force.
For objects near the surface of the planet earth:
where M and R are the mass and radius of the earth, and g = 9.80 m/s2.
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57. Gravity: A Force
The magnitude of the gravitational force is FG = mg, where:
The figure shows the free-body diagram of an object in free fall near the surface of a planet.
With , Newton’s second law predicts the acceleration to be:
All objects on the same planet, regardless of mass, have the same free-fall acceleration!
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58. Weight: A Measurement
You weigh apples in the grocery store by placing them in a spring scale and stretching a spring.
The reading of the spring scale is the magnitude of Fsp.
We define the weight of an object as the reading Fsp of a calibrated spring scale on which the object is stationary.
Because Fsp is a force, weight is measured in newtons.
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59. Weight: A Measurement
A bathroom scale uses compressed springs which push up.
When any spring scale measures an object at rest,
The upward spring force exactly balances the downward gravitational force of magnitude mg:
Weight is defined as the magnitude of Fsp when the object is at rest relative to the stationary scale:
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60. QuickCheck 6.7
An astronaut takes her bathroom scales to the moon, where g = 1.6 m/s2. On the moon, compared to at home on earth:
Her weight is the same and her mass is less.
Her weight is less and her mass is less.
Her weight is less and her mass is the same.
Her weight is the same and her mass is the same.
Her weight is zero and her mass is the same.
Answer: B
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61. QuickCheck 6.7
An astronaut takes her bathroom scales to the moon, where g = 1.6 m/s2. On the moon, compared to at home on earth:
Her weight is the same and her mass is less.
Her weight is less and her mass is less.
Her weight is less and her mass is the same.
Her weight is the same and her mass is the same.
Her weight is zero and her mass is the same.
Answer: B
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62. Weight: A Measurement
The figure shows a man weighing himself in an accelerating elevator.
Looking at the free-body diagram, the y-component of Newton’s second law is:
The man’s weight as he accelerates vertically is:
You weigh more as an elevator accelerates upward!
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63. QuickCheck 6.8
A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the scale reads
> 490 N.
490 N.
< 490 N but not 0 N.
0 N.
Answer: B
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64. QuickCheck 6.8
A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the scale reads
> 490 N.
490 N.
< 490 N but not 0 N.
0 N.
Answer: B
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65. QuickCheck 6.9
A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the student’s weight is
> 490 N.
490 N.
< 490 N but not 0 N.
0 N.
Answer: B
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66. QuickCheck 6.9
A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the student’s weight is
> 490 N.
490 N.
< 490 N but not 0 N.
0 N.
Weight is reading of a scale on which the object is stationary relative to the scale.
Answer: B
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67. Weightlessness The weight of an object which accelerates vertically is
If an object is accelerating downward with ay = –g, then w = 0.
An object in free fall has no weight!
Astronauts while orbiting the earth are also weightless.
Does this mean that they are in free fall?
This question will be answered in Chapter 8. (The answer is yes!)
Astronauts are weightless as they orbit the earth.
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68. QuickCheck 6.10
A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. Sadly, the elevator cable breaks. What is the student’s weight during the few second it takes the student to plunge to his doom?
> 490 N.
490 N.
< 490 N but not 0 N.
0 N.
Answer: B
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69. QuickCheck 6.10
A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. Sadly, the elevator cable breaks. What is the student’s weight during the few second it takes the student to plunge to his doom?
> 490 N.
490 N.
< 490 N but not 0 N.
0 N.
Answer: B
The bathroom scale would read 0 N. Weight is reading of a scale on which the object is stationary relative to the scale.
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70. QuickCheck 6.11
A 50-kg astronaut (mg = 490 N) is orbiting the earth in the space shuttle. Compared to on earth:
His weight is the same and his mass is less.
His weight is less and his mass is less.
His weight is less and his mass is the same.
His weight is the same and his mass is the same.
His weight is zero and his mass is the same.
Answer: B
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71. QuickCheck 6.11
A 50-kg astronaut (mg = 490 N) is orbiting the earth in the space shuttle. Compared to on earth:
His weight is the same and his mass is less.
His weight is less and his mass is less.
His weight is less and his mass is the same.
His weight is the same and his mass is the same.
His weight is zero and his mass is the same.
Answer: B
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72. Static Friction A shoe pushes on a wooden floor but does not slip.
On a microscopic scale, both surfaces are “rough” and high features on the two surfaces form molecular bonds.
These bonds can produce a force tangent to the surface, called the static friction force.
Static friction is a result of many molecular springs being compressed or stretched ever so slightly.
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73. Static Friction
The figure shows a person pushing on a box that, due to static friction, isn’t moving.
Looking at the free-body diagram, the x-component of Newton’s first law requires that the static friction force must exactly balance the pushing force:
points in the direction opposite to the way the object would move if there were no static friction.
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74. Static Friction Static friction acts in response to an applied force.
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75. QuickCheck 6.12
A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation:
fs > T.
fs = T.
fs < T.
fs = smg.
fs = 0.
Answer: B
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76. QuickCheck 6.12
A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation:
fs > T.
fs = T.
fs < T.
fs = smg.
fs = 0.
Newton’s first law.
Answer: B
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77. Static Friction
Static friction force has a maximum possible size fs max.
An object remains at rest as long as fs < fs max.
The object just begins to slip when fs = fs max.
A static friction force fs > fs max is not physically possible.
  where the proportionality constant μs is called the coefficient of static friction.
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78. QuickCheck 6.13
A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force.
Does the box move?
Yes No
Not enough information to say.
Answer: B
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79. QuickCheck 6.13
A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force.
Does the box move?
Yes No
Not enough information to say.
30 N < fs max = 40 N
Answer: B
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80. Kinetic Friction
The kinetic friction force is proportional to the magnitude of the normal force:
where the proportionality constant μk is called the coefficient of kinetic friction.
The kinetic friction direction is opposite to the velocity of the object relative to the surface.
For any particular pair of surfaces, μk < μs.
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81. QuickCheck 6.14
A box is being pulled to the right over a rough surface. T > fk, so the box is speeding up. Suddenly the rope breaks.
What happens? The box
Stops immediately.
Continues with the speed it had when the rope broke.
Continues speeding up for a short while, then slows and stops.
Keeps its speed for a short while, then slows and stops.
Slows steadily until it stops .
Answer: B
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82. QuickCheck 6.14
A box is being pulled to the right over a rough surface. T > fk, so the box is speeding up. Suddenly the rope breaks.
What happens? The box
Stops immediately.
Continues with the speed it had when the rope broke.
Continues speeding up for a short while, then slows and stops.
Keeps its speed for a short while, then slows and stops.
Slows steadily until it stops.
Answer: B
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83. QuickCheck 6.15
A box is being pulled to the right at steady speed by a rope that angles upward. In this situation:
n > mg.
n = mg.
n < mg.
n = 0.
Not enough information to judge the size of the normal force.
Answer: B
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84. QuickCheck 6.15
A box is being pulled to the right at steady speed by a rope that angles upward. In this situation:
n > mg.
n = mg.
n < mg.
n = 0.
Not enough information to judge the size of the normal force.
Answer: B
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85. QuickCheck 6.16 You’ll have to work this one out. Don’t just guess!
A box is being pulled to the right by a rope that angles upward. It is accelerating. Its acceleration is
(cosksin) – kg. T m
(cosksin) – kg. T m
You’ll have to work this one out.
Don’t just guess!
(sinkcos) – kg. T m
Answer: B
 – kg. T m
cos – kg. T m
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86. QuickCheck 6.16
A box is being pulled to the right by a rope that angles upward. It is accelerating. Its acceleration is
(cosksin) – kg. T m
(cosksin) – kg. T m
(sinkcos) – kg. T m
Answer: B
 – kg. T m
cos – kg. T m
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87. Rolling Motion
If you slam on the brakes so hard that the car tires slide against the road surface, this is kinetic friction.
Under normal driving conditions, the portion of the rolling wheel that contacts the surface is stationary, not sliding.
If your car is accelerating or decelerating or turning, it is the static friction of the road on the wheels that provides the net force which accelerates the car.
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88. Rolling Friction
A car with no engine or brakes applied does not roll forever; it gradually slows down.
This is due to rolling friction.
The force of rolling friction can be calculated as:
where μr is called the coefficient of rolling friction.
The rolling friction direction is opposite to the velocity of the rolling object relative to the surface.
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89. Coefficients of Friction
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90. A Model of Friction
The actual causes of friction involve microscopic surface properties and molecular bonds.
Experiments show that reasonable predictions are produced by a model of friction — a simplification of reality:
Here “motion” means “motion relative to the surface.”
Forces of kinetic and rolling friction are proportional to the normal force of the surface on the object.
The maximum static friction force is proportional to the normal force of the surface on the object.
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91. A Model of Friction
The friction force response to an increasing applied force.
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92. Causes of Friction All surfaces are very rough on a microscopic scale.
When two surfaces are pressed together, the high points on each side come into contact and form molecular bonds.
The amount of contact depends on the normal force n.
When the two surfaces are sliding against each other, the bonds don’t form fully, but they do tend to slow the motion.
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93. Drag
The air exerts a drag force on objects as they move through the air.
Faster objects experience a greater drag force than slower objects.
The drag force on a high-speed motorcyclist is significant.
The drag force direction is opposite the object’s velocity.
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94. Drag
For normal-sized objects on earth traveling at a speed v which is less than a few hundred meters per second, air resistance can be modeled as:
A is the cross-section area of the object.
ρ is the density of the air, which is about 1.2 kg/m3.
C is the drag coefficient, which is a dimensionless number that depends on the shape of the object.
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95. Drag
Cross-section areas for objects of different shape.
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96. Example 6.7 Air Resistance Compared to Rolling Friction
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97. Example 6.7 Air Resistance Compared to Rolling Friction
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98. Example 6.7 Air Resistance Compared to Rolling Friction
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99. Example 6.7 Air Resistance Compared to Rolling Friction
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100. Terminal Speed
The drag force from the air increases as an object falls and gains speed.
If the object falls far enough, it will eventually reach a speed at which D = FG.
At this speed, the net force is zero, so the object falls at a constant speed, called the terminal speed vterm.
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101. Terminal Speed
The figure shows the velocity-versus-time graph of a falling object with and without drag.
Without drag, the velocity graph is a straight line with ay = –g.
When drag is included, the vertical component of the velocity asymptotically approaches –vterm.
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102. Example 6.10 Make Sure the Cargo Doesn’t Slide

103. Example 6.10 Make Sure the Cargo Doesn’t Slide
MODEL
Let the box, which we’ll model as a particle, be the object of interest.
Only the truck exerts contact forces on the box.
The box does not slip relative to the truck.
If the truck bed were frictionless, the box would slide backward as seen in the truck’s reference frame as the truck accelerates.
The force that prevents sliding is static friction.
The box must accelerate forward with the truck: abox = atruck.
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104. Example 6.10 Make Sure the Cargo Doesn’t Slide

105. Example 6.10 Make Sure the Cargo Doesn’t Slide

106. Example 6.10 Make Sure the Cargo Doesn’t Slide

107. Example 6.10 Make Sure the Cargo Doesn’t Slide

108. Chapter 6 Summary Slides

109. General Strategy
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110. General Strategy
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111. Important Concepts
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112. Important Concepts