1. More Applications of Newton’s Laws
Chapter 5
More Applications of
Newton’s Laws

3. 5.1 Forces of Friction
When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion
This is due to the interactions between the object and its environment
This resistance is called the force of friction

4. Forces of Friction, cont.
The force of static friction靜磨擦, ƒs, is generally greater than the force of kinetic friction動摩擦, ƒk
The coefficient of friction (µ) depends on the surfaces in contact
Friction is proportional to the normal force
ƒs £ µs n and ƒk= µk n
These equations relate the magnitudes of the forces, they are not vector equations

5. Forces of Friction, final
The direction of the frictional force is opposite the direction of motion and parallel to the surfaces in contact
The coefficients of friction are nearly independent of the area of contact

6. Static Friction Static friction acts to keep the object from moving
If increases, so does
If decreases, so does
ƒs  µs n where the equality holds when the surfaces are on the verge of slipping
Called impending motion

7. Active Figure
5.1
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8. Some Coefficients of Friction

9. Fig 5.2

10. Kinetic Friction
The force of kinetic friction acts when the object is in motion
Although µk can vary with speed, we shall neglect any such variations
ƒk = µk n
Fig 5.3(a)

11. Fig 5.3(b)&(c)

12. Friction in Newton’s Laws Problems
Friction is a force, so it simply is included in the SF in Newton’s Laws
The rules of friction allow you to determine the direction and magnitude of the force of friction

13. Fig 5.4

20. The following is a simple method of measuring coefficients of friction
The following is a simple method of measuring coefficients of friction. Suppose a block is placed on a rough surface inclined relative to the horizontal, as shown in Figure 5.5. The incline angle is increased until the block starts to move.

21. How is the coefficient of static friction related to
the critical angle  at which the block begins to move? A
The block is sliding down the plane, so friction acts up the plane

23. How could fin the coefficient of kinetic friction? B
This setup can be used to experimentally determine the coefficient of friction
µ = tan q
For µs, use the angle where the block just slips
For µk, use the angle where the block slides down at a constant speed

24. Friction Example 2
Image the ball moving downward and the cube sliding to the right
Both are accelerating from rest
There is a friction force between the cube and the surface
Fig 5.6

25. Friction Example 2, cont
Two objects, so two free body diagrams are needed
Apply Newton’s Laws to both objects
The tension is the same for both objects
Fig 5.6

26. A ball and a cube are connected by a light string that passes over a frictionless light pulley, as in Figure 5.6a. The coefficient of kinetic friction between the cube and the surface is Find the acceleration of the two objects and the tension in the string.
Fig 5.6

28. Now we apply Newton’s second law to the ball m2 moving in the vertical direction. We choose the positive direction downward for the ball:

30. A warehouse worker places a crate on a sloped surface that is inclined at 30.0° with respect to the horizontal (Fig. 5.7a). If the crate slides down the incline with an acceleration of magnitude g/3, determine the coefficient of kinetic friction between the crate and the surface.
(a)
(b)
Fig 5.7

33. 5.2 Uniform Circular Motion
A force, , is directed toward the center of the circle
This force is associated with an acceleration, ac
Applying Newton’s Second Law along the radial direction gives
Fig 5.8

34. Uniform Circular Motion, cont
A force causing a centripetal acceleration acts toward the center of the circle
It causes a change in the direction of the velocity vector
If the force vanishes, the object would move in a straight-line path tangent to the circle
Fig 5.9

35. Active Figure
5.9
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38. Centripetal Force
The force causing the centripetal acceleration is sometimes called the centripetal force
This is not a new force, it is a new role for a force
It is a force acting in the role of a force that causes a circular motion

39. An object of mass kg is attached to the end of a cord whose length is 1.50 m. The object is whirled in a horizontal circle as in Figure 5.8. If the cord can withstand a maximum tension of 50.0 N, what is the maximum speed the object can have before the cord breaks?

40. Solution The force that provides the centripetal acceleration is the tension T exerted on the object, Newton’s 2nd law gives us

41. A small object of mass m is suspended from a string of length L
A small object of mass m is suspended from a string of length L. The object revolves in a horizontal circle of radius r with constant speed v, as in Figure 5.11a. (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.)
Fig 5.11a

42. Find the speed of the object.A
Solution The free-body diagram for the object of mass m is shown in Figure 5.11b.
Because the object does not accelerate in the vertical direction, we model it as a particle in equilibrium in the vertical direction:
Fig 5.11b

43. In the horizontal direction, we have a centripetal acceleration so we model the object as a particle under a net force. From Newton’s 2nd law we have

44. Find the period of revolution, defined as the time interval required to complete one revolution.B
Solution The object is traveling at constant speed around its circular path. Because the object travels a distance of 2r (the circumference of the circular path) in a time interval t equal to the period of revolution, we find
Note that the period is independent of m!

45. Horizontal (Flat) Curve
The force of static friction supplies the centripetal force
The maximum speed at which the car can negotiate the curve is
Note, this does not depend on the mass of the car

46. A 1500-kg car moving on a flat, horizontal road negotiates a curve whose radius is 35.0 m (Fig. 5.12a). If the coefficient of static friction between the tires and the dry pavement is 0.523, find the maximum speed the car can have to make the turn successfully.
Fig. 5.12a

47. Solution The car is an extended object with four friction forces acting on it, one on each wheel, but we shall model it as a particle with only one net friction force.
Figure 5.12b shows a free-body diagram for the car. From Newton’s second law in the horizontal direction, we have
Fig. 5.12b

50. A civil engineer wishes to redesign the curved roadway in Interactive Example 5.7 in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car moving at the designated speed can negotiate the curve even when the road is
covered with ice. Such a curve is usually
banked, meaning that the roadway is tilted
toward the inside of the curve. Suppose
the designated speed for the curve is to be
13.4 m/s (30.0 mi/h) and the radius of the
curve is 35.0 m. At what angle should the
curve be banked?

51. Solution On a level (unbanked) road, the force that causes the centripetal acceleration is the force of static friction between car and road.
If the road is banked at an angle , as in Figure 5.13, the normal force has a horizontal component nx pointing toward the center of the curve. Because the curve is to be designed so that the force of static friction is zero, only the component causes the centripetal acceleration. Hence, Newton’s
2nd law for the radial direction gives

53. If a car rounds the curve at a speed less than 13
If a car rounds the curve at a speed less than 13.4 m/s, friction is needed to keep it from sliding down the bank (to the left in Fig. 5.13). A driver who attempts to negotiate the curve at a speed greater than 13.4 m/s has to depend on friction to keep from sliding up the bank (to the right in Fig. 5.13). The banking angle is independent of the mass of the vehicle negotiating the curve.

54. A pilot of mass m in a jet aircraft executes a “loop-the-loop” maneuver as illustrated in Figure 5.14a. The aircraft moves in a vertical circle of radius 2.70 km at a constant speed of 225 m/s.

55. Determine the force exerted by the seat on the pilot at the bottom of the loop. Express the answer in terms of the weight mg of the pilot. A
Solution The free-body diagram for the pilot at the bottom of the loop is shown in Figure 5.14b. The forces acting on the pilot are the downward gravitational force and the upward normal force exerted by the seat on the pilot. Newton’s 2nd law for the radial (upward) direction gives
Figure 5.14b

57. Determine the force exerted by the seat on the pilot at the top of the loop. Express the answer in terms of the weight mg of the pilot. B
Solution The free-body diagram for the pilot at the top of the loop is shown in Figure 5.14c. At this point, both the gravitational force and the force exerted by the seat on the pilot act downward, so the net force downward that provides the centripetal acceleration
Figure 5.14c

58. Loop-the-Loop This is an example of a vertical circle
At the bottom of the loop, the upward force experienced by the object is greater than its weight
At the top of the circle, the force exerted on the object is less than its weight

60. Non-Uniform Circular Motion
The acceleration and force have tangential components
produces the centripetal acceleration
produces the tangential acceleration
Fig 5.15

61. Active Figure
5.15
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62. Vertical Circle with Non-Uniform Speed
The gravitational force exerts a tangential force on the object
Look at the components of Fg
The tension at any point can be found
Fig 5.17

63. Top and Bottom of Circle
The tension at the bottom is a maximum
The tension at the top is a minimum
If Ttop = 0, then
Fig 5.17

64. A small sphere of mass m is attached to the end of a cord of length R, which rotates under the influence of the gravitational force and the force exerted by the cord in a vertical circle about a fixed point O, as in Figure 5.17a. Let us determine the tension in the cord at any instant when the speed of the sphere is v and the cord makes an angle  with the vertical.

65. Solution First, note that the speed is not uniform because a tangential component of acceleration arises from the gravitational force on the sphere. Although this example is similar to Example 5.9, it is not identical. From the free-body diagram in Figure 5.17a, we see that the only forces acting on the sphere are the gravitational force and the force exerted by the cord.

66. Applying Newton’s 2nd law for the tangential direction gives
= dv/dt
Applying Newton’s second law to the forces in the radial direction (for which the outward direction is positive), we find

68. 5.4 Motion with Resistive Forces
Motion can be through a medium
Either a liquid or a gas
The medium exerts a resistive force, , on an object moving through the medium
The magnitude of depends on the medium
The direction of is opposite the direction of motion of the object relative to the medium
nearly always increases with increasing speed

69. Motion with Resistive Forces, cont
The magnitude of can depend on the speed in complex ways
We will discuss only two
is proportional to v
Good approximation for slow motions or small objects
is proportional to v2
Good approximation for large objects

70. R Proportional To v The resistive force can be expressed as
b depends on the property of the medium, and on the shape and dimensions of the object
The negative sign indicates is in the opposite direction to

71. R Proportional To v, Example
Analyzing the motion results in
Fig 5.18(a)

72. R Proportional To v, Example, cont
Initially, v = 0 and dv/dt = g
As t increases, R increases and a decreases
The acceleration approaches 0 when R ® mg
At this point, v approaches the terminal speed of the object

73. Terminal Speed To find the terminal speed, let a = 0
Solving the differential equation gives
t is the time constant and t = m/b
Fig 5.18(b)

77. R Proportional To v2
For objects moving at high speeds through air, the resistive force is approximately proportional to the square of the speed
R = 1/2 DrAv2
D is a dimensionless empirical quantity that is called the drag coefficient
r is the density of air
A is the cross-sectional area of the object
v is the speed of the object

78. R Proportional To v2, example
Analysis of an object falling through air accounting for air resistance
Fig 5.19

79. R Proportional To v2, Terminal Speed
The terminal speed will occur when the acceleration goes to zero
Solving the equation gives
Fig 5.19

80. Some Terminal Speeds

82. 5.5 Fundamental Forces Gravitational force Electromagnetic forces
Between two objects
Electromagnetic forces
Between two charges
Nuclear force
Between subatomic particles
Weak forces
Arise in certain radioactive decay processes

83. Gravitational Force
Mutual force of attraction between any two objects in the Universe
Inherently the weakest of the fundamental forces
Described by Newton’s Law of Universal Gravitation

84. Electromagnetic Force
Binds atoms and electrons in ordinary matter
Most of the forces we have discussed are ultimately electromagnetic in nature
Magnitude is given by Coulomb’s Law

85. Strong Force
The force that binds the nucleons to form the nucleus of an atom
Attractive force
Extremely short range force
Negligible for r > ~10-14 m
For a typical nuclear separation, the nuclear force is about two orders of magnitude stronger than the electrostatic force

86. Weak Force Tends to produce instability in certain nuclei
Short-range force
About 1034 times stronger than gravitational force
About 103 times stronger than the electromagnetic force

87. Unifying the Fundamental Forces
Physicists have been searching for a simplification scheme that reduces the number of forces
1987 – Electromagnetic and weak forces were shown to be manifestations of one force, the electroweak force
The nuclear force is now interpreted as a secondary effect of the strong force acting between quarks

88. 5.6 Drag Coefficients of Automobiles

90. Reducing Drag of Automobiles
Small frontal area
Smooth curves from the front
The streamline shape contributes to a low drag coefficient
Minimize as many irregularities in the surfaces as possible
Including the undercarriage

91. Fig 5.23