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
respectively read
                        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
Basketball                         20                          47
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,
    which from your point of view is always pointing radially inward
    (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
    reads
                                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
                                                        document - internet
                                                        access is required to
                                                        play them




                                                        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.




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                                                     PC131/PC151: Chapter 6, Pg 39

				
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posted:11/3/2012
language:English
pages:39