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Chapter 5
More Applications of
Newton’s Laws
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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
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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
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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
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Static Friction



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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
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Active Figure
5.1
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Some Coefficients of Friction
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Fig 5.2
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Kinetic Friction


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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)
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Fig 5.3(b)&(c)
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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
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Fig 5.4
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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.
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A How is the coefficient of static friction related to
the critical angle  at which the block begins to move?

The block is sliding down
the plane, so friction acts
up the plane
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B How could fin the coefficient of kinetic friction?


This setup can be used
to experimentally
determine the
coefficient of friction
µ = tan 


For µs, use the angle
where the block just slips
For µk, use the angle
where the block slides
down at a constant
speed
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Friction Example 2


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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
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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
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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 0.30. Find the
acceleration of the two objects
and the tension in the string.
Fig 5.6
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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:
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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.
Fig 5.7
(a)
(b)
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5.2 Uniform Circular Motion


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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
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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
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Active Figure
5.9
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Centripetal Force

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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
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An object of mass 0.500 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?
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Solution The force that provides the centripetal acceleration
is the tension T exerted on the object, Newton’s 2nd law
gives us
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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
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A Find the speed of the object.
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
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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
r  L sin
v  Lg sin  tan 
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B Find the period of revolution, defined as the time
interval required to complete one revolution.
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
2 r
2 r
L cos 
t 

 2
v
g
rg tan 
Note that the period is independent of m!
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Horizontal (Flat) Curve

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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
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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
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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
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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?
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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
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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.
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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.
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A 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.
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
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B 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.
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
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Loop-the-Loop
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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
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Non-Uniform Circular Motion

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The acceleration and
force have tangential
components
produces the
centripetal acceleration
produces the
tangential acceleration

Fig 5.15
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Active Figure
5.15
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Vertical Circle with NonUniform Speed

The gravitational
force exerts a
tangential force on
the object

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Look at the
components of Fg
The tension at any
point can be found
Fig 5.17
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Top and Bottom of Circle
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The tension at the
bottom is a
maximum
The tension at the
top is a minimum
If Ttop = 0, then
Fig 5.17
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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.
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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.
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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
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5.4 Motion with Resistive Forces

Motion can be through a medium
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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
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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
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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

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R Proportional To v, Example

Analyzing the
motion results in
Fig 5.18(a)
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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
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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)
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R Proportional To v2
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For objects moving at high speeds through air,
the resistive force is approximately
proportional to the square of the speed
R = 1/2 DrAv2
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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
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R Proportional To

2
v,
example
Analysis of an object
falling through air
accounting for air
resistance
Fig 5.19
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2
v,
Terminal
Fig 5.19
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R Proportional To
Speed
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The terminal speed
will occur when the
acceleration goes to
zero
Solving the equation
gives
Some Terminal Speeds
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5.5 Fundamental Forces

Gravitational force

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Electromagnetic forces
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Between two charges
Nuclear force

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Between two objects
Between subatomic particles
Weak forces

Arise in certain radioactive decay processes
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Gravitational Force
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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
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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
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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
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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
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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
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5.6 Drag Coefficients of
Automobiles
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Reducing Drag of Automobiles
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Small frontal area
Smooth curves from the front

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The streamline shape contributes to a low
drag coefficient
Minimize as many irregularities in the
surfaces as possible

Including the undercarriage
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Fig 5.23
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