# Physics 131: Lecture 1 Agenda for Today

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```					         Chapter 6: Force and Motion II
In this chapter, we will take a rigorous look at three particular
forces. The first is friction, which is surprisingly difficult to
analyze. The second is the drag force, which affects all
moving objects other than those in a vacuum. Finally, we will
discuss the forces that arise during uniform circular motion.

PC131/PC151: Chapter 6, Pg 1
Contents

• Friction
• Drag
• Uniform Circular Motion

PC131/PC151: Chapter 6, Pg 2
Friction
•   Consider the following “thought experiments”…
• A book slides across a horizontal surface. The book slows and
then stops. This means that the book must have an acceleration
parallel to the surface, in the direction opposite to the book’s
velocity.
• By Newton’s 2nd law, this means that a force exists in the
direction opposite to the book’s velocity. This is the frictional
force.
• Now, push horizontally on the book to make it move at a constant
velocity. The force that you are applying can not be the only
force on the book, otherwise it would be accelerating in the
direction of your push. By Newton’s 2nd law, there must be a
second force, directed opposite to your push, that exactly
balances it. This is the frictional force.

PC131/PC151: Chapter 6, Pg 3
Friction
•   Here is a second, slightly more complex situation.
• In fig (a), a block rests on a tabletop. As we have
seen, there are two forces acting on it; the 

gravitational force Fg and the normal force FN . They
must be perfectly balanced, since the block is not
accelerating.

• In fig. (b), we exert a force F on the block,
attempting to pull it to the left. In response, a
                                      
frictional force f s is directed to the right, balancing F

so that the block still does not move. f s is called the
static frictional force.

• In figs (c-d), the magnitude of the applied force F

increases. In response to this, f s also increases so
that the block remains at rest.

PC131/PC151: Chapter 6, Pg 4
Friction
• In fig (e), the applied force reaches a certain
magnitude that the block begins to slide; for a brief
moment it accelerates to the left. The frictional
force that opposes this motion is called the kinetic
                    
frictional force, f k . Note that F and f k are not
balanced at this moment, which is evident since the
block is accelerating.

• In response to the onset of motion, you reduce the applied force, as in
                                                         
fig. (f). When F  f k , the block moves at constant velocity. Note that f k

is constant, regardless of the applied force F (actually, it’s
approximately constant, since the microscopic properties of
the surface have a slight variation).
•   The figure to the right is an example of the
measured magnitude of the frictional force
during steps (a)-(f).

PC131/PC151: Chapter 6, Pg 5
Friction – Physical Description
•   The frictional force is the vector sum of a vast number of forces
acting between the surface atoms of the two objects in contact.
•   If we take two highly-polished (i.e. atomically smooth) and perfectly
clean metal surfaces and bring them together in a vacuum (so that no
particles are trapped between them), they can cold-weld together
instantly, so that no force can cause them to slide apart. This is
because the number of atom-to-atom interactions is enormous.

•   When there are surface irregularities, the
proportion of the surface that is actually in contact
is much smaller, and thus the frictional force is
weaker, as shown in the figure.
•   As the surfaces slide, there isa continuous breaking
and re-forming of welds, so f k varies slightly.

PC131/PC151: Chapter 6, Pg 6
Friction - Properties

•   When an object presses against a surface and a force Fattempts to
slide the object in one direction, the resulting frictional force has
three properties:

• If the object does not move, then the static frictional force f s and

the component of F that is parallel to the surface balance each
other; they are equal in magnitude, with opposite directions.

• The magnitude of f shas a maximum value, f s ,max , that is given by
f s ,m ax   s FN

where μs is the coefficient of static friction and FN is the
magnitude of the normal force on the object from the surface. If

the magnitude of the component of F that is parallel to the surface
exceeds f s ,max , then the object begins to slide along the surface.
This property will become more clear when we look at some
sample problems.

PC131/PC151: Chapter 6, Pg 7
Friction - Properties
• If the object begins to slide along the surface, the magnitude of
the frictional force rapidly decreases to a value fk, given by
f k   k FN

where μk is the coefficient of kinetic friction. During the sliding,

a kinetic frictional force f k with a magnitude given by the above
equation opposes the motion.

•   It should come as no surprise that the frictional force is proportional
to the magnitude of the normal force, since the latter describes how
firmly the object presses against the surface.
•   Note however that the proportionality between frictional and normal
forces only refers to their magnitudes; the frictional force is always
directed along the surface while the normal force is always directed
perpendicular to the surface.

PC131/PC151: Chapter 6, Pg 8
Friction - Properties
•   The coefficients μs and μk are dimensionless, and are determined
experimentally. Their values depend on the materials that make up
both the object and the surface, as well as the particular physical
preparation of the materials (are they atomically smooth? weathered?
heat-treated?)
•   The field of study that deals with friction (as well as lubrication and
wear) is called tribology. It is generally considered a sub-discipline
of mechanical engineering.

PC131/PC151: Chapter 6, Pg 9
Problem #1: Sliding a Block
PROBLEM               (Sample Problem, pp. 120-121)
A block of mass m = 3.0 kg is pulled along a floor
by a force F of magnitude 12.0 N. The force is
directed at an adjustable angle θ above the vertical,
where θ can vary from 0 to 90 degrees.
The coefficient of kinetic friction between the
block and the floor is μk = 0.40.
What angle θ results in the maximum value of the
block’s acceleration magnitude a?

SOLUTION
If there was no friction, then the answer would be obvious; the angle

should be zero. This way, the component of F that accelerates the block
along the floor would be maximized. When friction is accounted for, we

must note that having a slight upward component to F acts to reduce the
normal force acting on the block by the floor. This in turn reduces the
frictional force that opposes the acceleration.
PC131/PC151: Chapter 6, Pg 10
Problem #1: Sliding a Block
(Sample Problem, pp. 120-121)

Because the block is moving, a kinetic frictional force
acts on it: fk = μkFN. We need to find FN. The top
figure is a free-body diagram showing the forces on the
block along the vertical axis. These forces are the
gravitational force, the normal force, and the
vertical component of the applied force, which is
illustrated in the bottom figure and has a magnitude of
Fy= F sin θ.

We can now write Newton’s 2nd law for the vertical component of motion,
keeping in mind that the block does not accelerate vertically:
FN  F sin   mg  ma  0
and therefore, FN  mg  F sin 

PC131/PC151: Chapter 6, Pg 11
Problem #1: Sliding a Block
(Sample Problem, pp. 120-121)
Now let’s look at the horizontal motion. The free-body
diagram on the right shows the two forces along this
axis; the x-component of the applied force, Fx= F cos θ,
and the frictional force fk. which opposes the direction
of motion and is thus directed to the left.

Newton’s 2nd law for the horizontal motion gives us
F cos   k FN  ma
The value a – the acceleration of the block – is what we are seeking.
Substituting our calculated value of FN and solving for a, we get
F               F      
a  cos   k  g  sin  
m               m      
In this expression, F, m, and μk are constant. We now know acceleration a
as a function of θ.
PC131/PC151: Chapter 6, Pg 12
Problem #1: Sliding a Block
(Sample Problem, pp. 120-121)
To find the value of θ that maximizes a, we take the derivative of a with
respect to θ and set the result equal to zero:
da    F           F
  sin    k cos  0
d    m           m

A bit of rearranging - the F / m term cancels out - results in
tan    k
and, since μk = 0.40, we find that the acceleration will be maximum if
  tan 1 (0.40 )  21 .8
It is worth noting that the optimum angle is a function only of μk and does
not depend on the block’s mass or the applied force F.

PC131/PC151: Chapter 6, Pg 13
Problem #2: Two Blocks and a Pulley
(Problem 6.28)
PROBLEM
Consider the figure on the right, in which two
blocks are connected by a frictionless and
massless pulley. The mass of block A is mA = 10
kg, and the coefficient of kinetic friction between
block A and the incline is μk = 0.20. The angle θ
of the incline is 30 degrees.

You observe that block A is sliding down the incline at a constant speed.
What is the mass of block B?
SOLUTION
There are a few different forces at work here. Gravity is acting on both
blocks, tension in the rope is pulling on both blocks, and block A
experiences kinetic friction (opposing its downward slide). There’s also a
normal force upon block A, oriented perpendicular to the incline.

PC131/PC151: Chapter 6, Pg 14
Problem #2: Two Blocks and a Pulley
(Problem 6.28)

We can draw free-body diagrams for each of the
blocks separately: 
For a downhill slide, f k
must be directed uphill;
its value is negative for
the orientation shown.

For block A we take the +x direction to be uphill and the +y direction to
be in the direction of the normal force. For block B, the +x direction is
vertically downward. As with the problem in the last chapter, this is done
so that the motion of both blocks can be properly coupled (when A moves
uphill, B moves downward by the same amount). The +y direction for
block B is irrelevant, as no forces upon or motion of B take place in any
direction other than vertically.

PC131/PC151: Chapter 6, Pg 15
Problem #2: Two Blocks and a Pulley
(Problem 6.28)

The problem stipulates that block A slides
downhill at a
constant speed. This
implies that block B
moves upward at a
constant speed. Both
blocks have zero
acceleration.
We can now write Newton’s 2nd law for three different acceleration
components of motion, the x- and y-motion of block A and the x-motion
of block B. With all acceleration equal to zero, these equations
T  f k  m A g sin   0
FN  m A g cos  0
mB g  T  0
PC131/PC151: Chapter 6, Pg 16
Problem #2: Two Blocks and a Pulley
(Problem 6.28)
What we have then is three equations in three unknowns (T, FN, and mB).
Note that since fk = μkFN, it is not considered as an independent unknown
quantity.
Those of you who have taken linear algebra have learned a slick way to
solve such a problem (it involves a 3 x 3 matrix). However, because at
least one of the equations (all three, in fact) involve only two of the three
unknowns, this particular problem can be solved by simple back-
substitution. I will go through the details in class. The result is

mB  mA sin    k cos   3.3 kg

PC131/PC151: Chapter 6, Pg 17
The Drag Force

•   When an object moves through a fluid, it experiences a drag force D
that points in the direction in which the fluid flows relative to the
object.
•   A fluid is anything that can flow; that is, a gas or a liquid.

•   The magnitude of D depends on several factors, most importantly:
• the density of the fluid (air produces less drag than water)
• the velocity of the object (a faster-moving object will be “hit” by
more of the fluid particles in a given time interval)
• the shape of the object (a blunt object like a sphere is affected
more than a slender object like a javelin…this is the basis of
aerodynamics)

PC131/PC151: Chapter 6, Pg 18
The Drag Force
•   In PC131/PC151, we will only examine blunt objects (approximated
as spheres), traveling through still air.

•   In this case, the magnitude of D can be expressed as
1
D  CAv2
2
Here, C is the drag coefficient, which is determined experimentally.
ρ is the density of the air (mass per unit volume), v is the object’s
velocity, and A is the object’s effective cross-sectional area; that is,
the area of a cross-section of the object, taken perpendicular to the
direction of its velocity.
•   We consider C to be a constant for a given object. This is not strictly
true; if the velocity varies over a wide range, then C will vary as
well. We will ignore this fact.
•   Typically, C is in the range from 0.4 to 1.0. It is dimensionless (you
should be able to determine this by the above equation).

PC131/PC151: Chapter 6, Pg 19
The Drag Force
•   Athletes are well aware of the drag force. To minimize D, they try
to minimize their effective cross-sectional area.
•   Sky divers, on the other hand, want to enjoy the ride as long as
possible, so they attempt to increase A.

PC131/PC151: Chapter 6, Pg 20
The Drag Force – Terminal Speed
•   Our previous analysis of free-fall indicates that an object falling in a
vacuum will accelerate indefinitely; acceleration is constant, velocity
increases linearly with time, and position increases quadratically
with time.
•   When we account for drag, this is no longer the case. We can show
this using Newton’s 2nd law.
•   Imagine that you drop an object from the top of a tall building.
                        
Because v is directed downward, D is directed upward. Its
magnitude D is initially zero, since v is zero. As v increases, so does
D. In this case, Newton’s 2nd law (in 1D, the vertical axis) tells us
that
D  Fg  ma

that is, the net force on the object equals its mass multiplied by its
acceleration.

PC131/PC151: Chapter 6, Pg 21
The Drag Force – Terminal Speed
•   As the object accelerates, its velocity will eventually reach a point
where the drag force and the gravitational force balance each
other perfectly. As a result, the acceleration will be zero, and the
object will fall with a constant speed. We call this the terminal
speed, vt.
•   To find vt, we need only to set a = 0 in Newton’s second law, and
insert the formula for D:

1                       2 Fg
CAvt  Fg  0  vt 
2

2                       CA
•   In order to use this formula, we must be able to find an expression
for Fg. If we are given the mass m of the object, then we can simply
use Fg = mg. Many problems only state the physical properties of
the object, in which case a little more work is necessary.
•   If we throw an object downward with an initial speed that is greater
than vt, it will slow down.
PC131/PC151: Chapter 6, Pg 22
The Drag Force – Terminal Speed
•   Another measure of the drag force is the 95% distance. This is the
distance through which the object must fall from rest to reach 95%
of its terminal speed. We won’t be concerned with this measure, but
those of you going on to PC235 will eventually understand its
meaning.

The table below lists some typical terminal speeds in air.
Object                     Term. Speed (m/s)        95% Distance (m)
Shot (from shot put)              145                        2500
Skydiver                           60                         430
Baseball                           42                         210
Tennis ball                        31                         115
Ping-pong ball                      9                          10
Raindrop (1.5 mm rad.)              7                          6
Parachutist                         5                          3

PC131/PC151: Chapter 6, Pg 23
Problem #3: Falling Raindrop
(Sample Problem, p. 123)
PROBLEM
A raindrop with radius R = 1.5 mm falls from a cloud that is at
height h = 1200 m above the ground. The drag coefficient C for the
raindrop is 0.60, and we assume that the drop remains spherical.
The density of water is ρw = 1000 kg/m3 and the density of air is ρa =
1.2 kg/m3.
What is the terminal speed of the raindrop?

SOLUTION
First of all, note that there are two different densities specified in
this problem, for the raindrop and for the air. The latter is what we
simply called ρ in the previous slides; this describes the medium that
produces the drag. The density of the water is necessary to calculate
the mass of the raindrop (and hence the gravitational force).

PC131/PC151: Chapter 6, Pg 24
Problem #3: Falling Raindrop
(Sample Problem, p. 123)
We previously showed that (with ρ = ρa )

2 Fg
vt 
C aA
Because the raindrop is assumed spherical, its effective cross-sectional area
A is πR2.
As for the gravitational force, we know that Fg = mg, where m is the mass
of the raindrop. This mass is the density ρw times the volume, which is
(4/3)πR3 for a sphere. Thus, we have
4 3
Fg  R  w g
All together,
3

2 Fg 8R 3  wg   8R wg
vt                             7.4 m/s
C aA   3C aR  2
3C a
PC131/PC151: Chapter 6, Pg 25
Problem #3: Falling Raindrop
(Sample Problem, p. 123)
Note that the height of the cloud did not enter into the equation; the table
shown a few slides ago indicates that the raindrop reaches terminal speed
long before it hits the ground.

What would be the drop’s speed just before impact if there were no drag
force?

In the absence of drag, we are back to the case of 1D constant acceleration
(with magnitude g). Recall from chapter 2 that in this case, the speed
reached by an object that starts from rest and falls through a distance h is

v  2gh
For h = 1200 m, this speed is 153 m/s, or 550 km/h (!)

PC131/PC151: Chapter 6, Pg 26
Uniform Circular Motion
•   In chapter 4, it was mentioned that when an object moves in a circle
(or a circular arc) of radius R at a constant speed v, it is said to be in
uniform circular motion. We also showed that such an object has a
centripetal acceleration, directed toward the center of the circle,
with a magnitude of
v2
a
R
•   Newton’s 2nd law dictates that a centripetal force must be
responsible for this acceleration. We can investigate the nature of
this force by way of some examples.

PC131/PC151: Chapter 6, Pg 27
Uniform Circular Motion
•   As a first example, consider the figure on the
right. A hockey puck moves with constant speed
v along a circle of radius R. It is tied to a string
which is looped around a central peg.
•   Here, the centripetal force is the tension in the
string, which pulls radially inward on the puck.

•   For a second example, imagine that you are on board a space shuttle
that orbits the earth. In this case, the centripetal force is gravity,
(toward the center of the earth).

PC131/PC151: Chapter 6, Pg 28
Uniform Circular Motion
•   As a third example, imagine sitting in the back seat of a car. The car
turns left, and you slide across the seat toward the right, jamming up
against the inside of the passenger-side door.
•   In this example, the turn of the car is only possible because the
tendency of the car to continue in a straight line is opposed by the
frictional force between the tires and the road. This frictional force is
directed radially inward; it is centripetal. The driver hopes that the
maximum static friction is sufficient to keep the car from sliding.
•   As for your motion in the back seat, your tendency to move in a
straight line is initially opposed by friction between the car seat and
your @\$\$. But apparently the maximum static friction could not
provide enough centripetal force; this is why you slid to the right.
Now, the centripetal force is provided by the car door, which is forcing
you to the left (radially inward).

PC131/PC151: Chapter 6, Pg 29
Uniform Circular Motion
•   These examples indicate that a centripetal force is not a specific type
of force such as friction, tension, etc. For any situation,
A centripetal force accelerates an object by changing the direction
of its velocity without changing its speed.
•   From Newton’s 2nd law, we can write the magnitude F of a centripetal
force as
mv 2
F
R
•   An object undergoing uniform circular motion has constant values of v
and R, and therefore the magnitude of F is constant. However, its
direction is continuously changing so that it is directed inward.
•   Later on in the course, we will re-visit uniform circular motion in the
context of angular momentum.

PC131/PC151: Chapter 6, Pg 30
Problem #4: The Gravitron Ride
•   You may be familiar with the Gravitron carnival ride (the text
refers to it as the Rotor). This is essentially a large, hollow cylinder
that spins around a central axis (we will ignore the fact that the walls
are actually sloped). The riders stand against the inner wall of the
cylinder, and are “pinned” to the wall even after the floor drops out
from beneath them.

•   This ride relies on two
particular effects –
centripetal force and
friction. Let’s take a close
look at what exactly goes
on inside the cylinder…

PC131/PC151: Chapter 6, Pg 31
Problem #4: The Gravitron Ride
•   Let’s assume that the cylinder’s radius R is 5.1 m, and that the
coefficient of static friction between the rider and the wall is
μs = 0.40. Generally the walls have either a thick carpeting or a
vinyl padding. This isn’t just for comfort, it’s to provide a suitably
large μs.
PROBLEM
What minimum speed v must the cylinder (and
the rider) have if the rider is not to fall when
the floor drops?

SOLUTION

There are three forces to consider here. Fg 
downward, f s
attempts to accelerate the rider 
opposes this acceleration, and FN is the
centripetal (normal) force.
PC131/PC151: Chapter 6, Pg 32
Problem #4: The Gravitron Ride
•   We start by analyzing the forces along a vertical axis at the position
of the rider. Because we are interested in the case where the rider
remains pinned to the wall without slipping, the vertical acceleration
is zero, and Newton’s second law reads

f s  mg  0
where m is the mass of the rider.
•   Then, because fs = μsFN, we can write
 s FN  mg  0  FN  mg /  s
•   Next, we will analyze the forces along a radial
axis that extends from the center of the
cylinder through the rider. Newton’s 2nd law
 v2 
FN  ma  m  
R
                        PC131/PC151: Chapter 6, Pg 33
Problem #4: The Gravitron Ride
•   Combining the previous two equations, we find that
gR
v     7.2 m/s
s
That is, for speeds of 7.2 m/s or greater, the static frictional force is
sufficiently large to prevent slippage due to the gravitational force.
For speeds slower than this, the rider is only weakly pressed into the
wall, and static friction is overwhelmed by gravity.
•   Notice that the result is independent of the rider’s mass. A big fat
guy has no more difficulty remaining pinned to the wall than a small
child. The increase in the gravitational force for a large mass is
exactly counteracted by the increase in the centripetal (normal) force
(and hence in the static frictional force, which is proportional to FN).

PC131/PC151: Chapter 6, Pg 34
Problem #5: Angle of a Banked Road
•   In the example of the car turning around a corner, it was mentioned
that a suitable magnitude of static friction must exist between the
wheels and the road so that the car does not slide.

• In many cases, roads are banked at an
angle in order to increase the maximum
possible slip-free speed of the cars. It is
possible to bank the road at an angle such
that no friction is required to keep the car
from slipping.
PROBLEM
For a car traveling with speed v around a
curve of radius R, determine a formula for
the angle θ at which the road should be
banked so that no friction is required.

PC131/PC151: Chapter 6, Pg 35
Problem #5: Angle of a Banked Road
SOLUTION
Even though the road is banked, the car still moves along a
horizontal circle, so the centripetal acceleration remains horizontal.

A free-body diagram for the car is shown to
the right. There are two forces present, the
gravitational force (directed downward, as
always) and the normal force (directed
perpendicular to the surface). As we have
assumed in the problem statement, there are
no frictional forces present.

Since there is no vertical motion, Newton’s
2nd law tells us that
FN cos  mg  0
PC131/PC151: Chapter 6, Pg 36
Problem #5: Angle of a Banked Road
mg
Therefore,     FN 
cos
As for the horizontal motion, Newton’s 2nd
law reveals that
v2
FN sin   ma  m
R

Combining these two equations, we get

mg            v2                v2
sin   m         or tan 
cos           R                 gR
Thus, for an angle θ = tan-1 (v2/gR), a car
can make the turn at a speed of v.

PC131/PC151: Chapter 6, Pg 37
Video: Physics of Ski Jumping
The YouTube video below (which contains audio) describes how ski
jumpers handle concepts of drag and lift. We don’t really talk about lift in
PC131/PC151; when air flows underneath an angled surface, it produces a
net upward force on that surface. This is what allows aircraft to fly.

Videos are not
embedded in this
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PC131/PC151: Chapter 6, Pg 38
Video: Physics of Bobsleigh
The YouTube video below (which contains audio) describes how drag,
resistance, and centripetal acceleration all play a role in driving a
bobsleigh.

Videos are not
embedded in this
document - internet
access is required to
play them

PC131/PC151: Chapter 6, Pg 39

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