Download Chapter 06 Lecture Slides

Chapter 6: More on
Newtonian Mechanics
Cases to study:
Frictional forces
The drag force and terminal speed
Uniform circular motion
Vertical circular loop
6.2 Friction
Frictional forces are very common in our
everyday lives
Examples:
1.  If you send a book sliding down a horizontal surface, the
book will finally slow down and stop.
2.  If you push a heavy crate and the crate does not move,
then the applied force must be counteracted by frictional
forces.
6.2 Frictional Force: Let’s look at an example, the motion of a
crate with applied forces
There is no attempt at sliding.
No motion.
NO FRICTION
Force F attempts sliding but is balanced
by the frictional force.
No motion.
STATIC FRICTION (fs)
Force F is now stronger but is still
balanced by the frictional force.
No motion.
LARGER STATIC FRICTION
Force F is now even stronger but is still
balanced by the frictional force.
No motion.
EVEN LARGER STATIC FRICTION
6.2 Frictional Force: Let’s look at an example, the motion of a
crate with applied forces
Finally, the applied force has
overwhelmed the static frictional
force.
Block slides and accelerates.
WEAK KINETIC FRICTION
To maintain the speed, weaken force
F to match the weak frictional force.
SAME WEAK KINETIC FRICTION
6.2 Frictional Force: Summary
fs is the static
frictional force
fk is the kinetic
frictional force
1. Static frictional force can only match growing applied force
– Multiple values!
2. Kinetic frictional force has only one value (no
matching).
February 27
6.3 Properties of friction
Property 1. If the body does not move, then the static frictional force and
the component of F that is parallel to the surface balance each other. They
are equal in magnitude, and is fs directed opposite that component of F.
W=Mg
6.3 Properties of friction: Static
Property 2. The magnitude of has a maximum value fs,max that is given by
where µs is the coefficient of static friction. This is usually larger than its kinetic
counterpart µk.
It is an empirical property of the contacting materials.
One way to measure the
coefficient of static friction µk
6.3 Properties of friction: kinetic
Property 3. If the body begins to slide along the surface, the magnitude of the
frictional force rapidly decreases to a value fk given by
where µk is the coefficient of kinetic friction. Thereafter, during the sliding, a
kinetic frictional force fk opposes the motion.
It is also an empirical property of the contacting materials.
Friction:
(1)  Depend on Surface roughness and contact area
(2)  Chemical bonding between the surfaces – EM
interaction
Before the example: One dimensional motion with constant
acceleration
x-x0=v0t+at2/2 … (2-15)
v-v0=at
… (2-11)
(1) Eliminate t:
v2-v02=2a(x-x0) … (2-16)
(2) Eliminate a:
(x-x0)=(v0+v)t/2 … (2-17)
(3) Eliminate v0:
(x-x0)=vt-at2/2 … (2-18)
Sample Problem
Sample Problem
A kinetic frictional force on the car
from the road  The constant
acceleration a, directed opposite the
direction of the car s motion. This
results in:
The minus sign indicates the direction
of the kinetic frictional force.
Calculations: The frictional force has the magnitude
fk = µkFN,
where FN is the magnitude of the normal force on the car from the
road. Because the car is not accelerating vertically, FN= mg.
Thus, fk =µkFN = µkmg
Sample Problem
a = - fk/m = -µkmg/m = -µkg,
where the minus sign indicates that
the acceleration is in the negative
direction. Use (2-16)
v 2 = v 2 o + 2a( x − xo )
where (x-xo) = 290 m, and the final
speed is 0. Solving for vo,
vo = 2 µ k g( x − xo ) = 58 m / s
We assumed that v = 0 at the far end
of the skid marks.
How many miles per hour?
58 m/s = (58)(3600)/(1000) (0.62)~130 mi/h !
Another example: How to find the maximum value for a given
function?
Sample Problem, friction applied at an angle
Two factors:
1. Kinetic friction: proportional
to the normal force
2. Net force along the surface
Good if we can find out the
function of acceleration of θ
Sample Problem, Free body diagram. … X-, Y-direction
Y-direction
X-direction
Sample Problem: Fn and Acceleration
To find out the function of acceleration
of θ.
1.  We need the net force in the xdirection
2.  We need the kinetic friction
3.  We need the normal force
fk
?? But it’s what we
want
Sample Problem: X-direction, the acceleration
6.4: The drag force and terminal speed
When there is a relative velocity between a fluid/gas and a body (either
because the body moves through the fluid/gas or because the fluid/gas
moves past the body), the body experiences a drag force that opposes
the relative motion and points in the direction in which the fluid/gas
flows relative to the body.
Sometime,
we want it small
Sometime,
we want it big
Sorry! – You have to
move down in order to
enjoy its lifting you up.
6.4: Drag force and terminal speed
A body (e.g. blunt, like a baseball, rather than e.g. slender, like a
javelin) moves in fluid, and the relative motion is fast enough so that
the air becomes turbulent behind the body, the Drag Force is
approximated by
ρ: the air density (mass per volume),
A: the effective cross-sectional area of the body (the area of a cross
section taken perpendicular to the velocity v),
C: is the drag coefficient .
Because of the
term v2 …
Terminal speed
6.4: Drag force: A few notes
6.4: Terminal speed in air (or other fluid)
From Newton s second law along in
the vertical direction
where m is the mass of the body.
Eventually, a = 0, and the body then
falls at a constant speed, called the
terminal speed vt .
So, it is NOT an constant acceleration!
March 2.
6.4: The terminal speed
(6-16)
6.4: Drag force and terminal speed
Some typical values of terminal speed
Sample problem, terminal speed
Sample problem, terminal speed
Sample problem, terminal speed
x-x0=v0t+at2/2 … (2-15)
v-v0=at
… (2-11)
(2) Eliminate a:
(x-x0)=(v0+v)t/2 … (2-17)
(1) Eliminate t:
v2-v02=2a(x-x0) … (2-16)
(3) Eliminate v0:
(x-x0)=vt-at2/2 … (2-18)
6.5: Uniform circular motion
A body moving with speed v in uniform circular motion feels a
centripetal acceleration directed towards the center of the circle of
radius R.
Examples:
1.  When a car moves in the circular arc, it has an acceleration
that is directed toward the center of the circle.
•  The frictional force on the tires from the road provide
the centripetal force responsible for that.
2.  Space shuttle moving around the earth, both the rider and the
shuttle are in uniform circular motion and have accelerations
directed toward the center of the circle.
•  Centripetal forces, causing these accelerations, are
gravitational pulls exerted by Earth and directed
radially inward, toward the center of Earth.
6.5: Uniform circular motion
From Newton s 2nd Law:
Sample problem: Vertical circular loop
What does
this mean?
Sample problem: Vertical circular loop
Is the game safe?
-  Why the water does
not fall when the
bucket is upside down?
-  When the water starts
to spill?
-  Does he need to swirl
the bucket faster when
the bucket has more
water?
Vertical circular loop: Force analysis
Sample problem: Vertical circular loop
Sample problem: Vertical circular loop
Is the game safe?
vmin = gR
-  Why the water does not fall when
the bucket is upside down?
 vy = 0, in touch, TN,y≠0, downward.
-  When the water starts to spill?
 TN,y =0, no touch, vy ≠ 0, downward.
-  Does he need to swirl the bucket
faster than vmin when the bucket
has more water?
 No
Do we need a higher
speed at the bottom
for bigger mass?
March 4.
Example: Vertical circular loop
Is the game safe?
vmin = gR
-  Why the water does not fall when
the bucket is upside down?
 vy = 0, in touch, TN,y≠0, downward.
-  When the water starts to spill?
 TN,y =0, no touch, vy ≠ 0, downward.
-  Does he need to swirl the bucket
faster than vmin when the bucket
has more water?
 No
Sample problem, car in flat circular turn
a) If the car is on the verge of
sliding out of the turn when its
speed is 28.6m/s, what is the
magnitude of negative lifting FL
acting downward on the car?
Let’s think like a physicist …. Free body diagram + force analysis
Combining to cancel FN 
Sample problem, car in flat circular turn, cont.
(b) The magnitude FL of the negative lift on a car depends on
the square of the car s speed v2, just as the drag force does.
Thus, the negative lift on the car here is greater when the car
travels faster. What is the magnitude of the negative lift for a
speed of 90 m/s?
Sample problem, car in flat circular turn, cont.
Calculations: We can write a ratio of the negative lift FL,90 at v =90 m/s
to our result for the negative lift FL at v =28.6 m/s as
Using FL =663.7 N ,
Another example on p. 128
“Car in banked circular turn”
Problem for extra credits
Due:
3:00 PM Friday, March 6.
The 2nd in-class exam: this Friday
- Get ready
PHYS-211
- Good luck
Spring
Break
Review
&
Some exercises
Chapter 5: Review
Learning Objectives
5.01 Identify that a force is a
vector quantity and thus has
both magnitude and direction
and also components.
5.02 Given two or more forces
acting on the same particle, add
the forces as vectors to get the
net force.
5.03 Identify Newton's first and
second laws of motion.
5.04 Identify inertial reference
frames.
5.05 Sketch a free-body diagram
for an object, showing the
object as a particle and drawing
the forces acting on it as vectors
anchored to the particle.
5.06 Apply the relationship
between net force on an object,
its mass, and the produced
acceleration.
5.07 Identify that only external
forces on an object can cause
the object to accelerate.
5.2.2. Which one of the following statements concerning Newtonian
mechanics is false?
a) Newtonian mechanics was developed by Isaac Newton.
b) Newtonian mechanics can be considered a special case of either Special
Relativity or Quantum Mechanics, which are more comprehensive
theories.
c) Newtonian mechanics can be used to describe the motion of galaxies.
d) Newtonian mechanics provides an understanding of the relationship
between forces and accelerations.
e) Newtonian mechanics applies to all objects, regardless of their speed or
acceleration.
5.2.2. Which one of the following statements concerning Newtonian
mechanics is false?
a) Newtonian mechanics was developed by Isaac Newton.
b) Newtonian mechanics can be considered a special case of either Special
Relativity or Quantum Mechanics, which are more comprehensive
theories.
c) Newtonian mechanics can be used to describe the motion of galaxies.
d) Newtonian mechanics provides an understanding of the relationship
between forces and accelerations.
e) Newtonian mechanics applies to all objects, regardless of their speed or
acceleration.
5.4.4. Complete the following statement: An inertial reference frame is
one in which
a) the frame is accelerating.
b) Newton s laws of motion are valid.
c) the acceleration due to gravity is greater than zero m/s2.
d) Newton s third law of motion is not valid.
e) the coordinate axes are rotating.
5.4.4. Complete the following statement: An inertial reference frame is
one in which
a) the frame is accelerating.
b) Newton s laws of motion are valid.
c) the acceleration due to gravity is greater than zero m/s2.
d) Newton s third law of motion is not valid.
e) the coordinate axes are rotating.
ur
5.6.1. A net forcer F is required to give
ur an object with mass m an
acceleration a . If a net force 6 F is applied to an object with mass
2m, what is the acceleration on this object?
r
a) a
r
b) 2 a
r
c) 3 a
r
d) 4 a
r
e) 6 a
ur
5.6.1. A net forcer F is required to give
ur an object with mass m an
acceleration a . If a net force 6 F is applied to an object with mass
2m, what is the acceleration on this object?
r
a) a
r
b) 2 a
r
c) 3 a
r
d) 4 a
r
e) 6 a
5.7.2. During gym class, a boy climbs a vertical rope and temporarily
stops half way between the floor below and the ceiling above.
Consider the following forces: (1) gravity, (2) normal force, (3)
friction, and (4) tension. At that moment, ignoring any effects due
to the surrounding air, which of these forces are acting on the boy?
a) 1, 2, 3, and 4
b) 1, 2, and 3 only
c) 1 and 2 only
d) 1 and 3 only
e) 1 and 4 only
5.7.2. During gym class, a boy climbs a vertical rope and temporarily
stops half way between the floor below and the ceiling above.
Consider the following forces: (1) gravity, (2) normal force, (3)
friction, and (4) tension. At that moment, ignoring any effects due
to the surrounding air, which of these forces are acting on the boy?
a) 1, 2, 3, and 4
b) 1, 2, and 3 only
c) 1 and 2 only
d) 1 and 3 only
e) 1 and 4 only
5.9.1. Note the following situations:
In which case will the magnitude of the normal force on the block be equal to (Mg +
F sin θ)?
a) case 1 only
b) case 2 only
c) both cases 1 and 2
d) both cases 2 and 3
e) cases 1, 2, and 3
5.9.1. Note the following situations:
In which case will the magnitude of the normal force on the block be equal to (Mg +
F sin θ)?
a) case 1 only
b) case 2 only
c) both cases 1 and 2
d) both cases 2 and 3
e) cases 1, 2, and 3
Chapter 6: Review
Learning Objectives
6.01 Distinguish between friction
in a static situation and a kinetic
situation.
6.02 Determine direction and
magnitude of a frictional force.
6.03 For objects on horizontal,
vertical, or inclined planes in
situations involving friction,
draw free-body diagrams and
apply Newton's second law.
6.2.1. Which one of the following statements concerning the static and
kinetic frictional forces is correct?
a) When an object is stationary, both static and kinetic frictional forces are
acting on it.
b) When an object is stationary, only the kinetic frictional force acts on it.
c) When an object is sliding, only the static frictional force acts on it.
d) The static frictional force acts on an object when it is stationary and the
kinetic frictional force acts on it when it is sliding.
e) Static and kinetic frictional forces act in the same direction as the normal
force.
6.2.1. Which one of the following statements concerning the static and
kinetic frictional forces is correct?
a) When an object is stationary, both static and kinetic frictional forces are
acting on it.
b) When an object is stationary, only the kinetic frictional force acts on it.
c) When an object is sliding, only the static frictional force acts on it.
d) The static frictional force acts on an object when it is stationary and the
kinetic frictional force acts on it when it is sliding.
e) Static and kinetic frictional forces act in the same direction as the normal
force.
6.3.1. Two identical blocks are pulled along a rough surface as suggested in the
figure. Which one of the following statements is false?
a) The coefficient of kinetic friction is the same in each case.
b) A force of the same magnitude is needed to keep each block moving.
c) The normal force exerted on the blocks by the surface is the same for both blocks.
d) The magnitude of the force of kinetic friction is greater for the block on the right.
e) A force of the same magnitude was required to start each block moving.
6.3.1. Two identical blocks are pulled along a rough surface as suggested in the
figure. Which one of the following statements is false?
a) The coefficient of kinetic friction is the same in each case.
b) A force of the same magnitude is needed to keep each block moving.
c) The normal force exerted on the blocks by the surface is the same for both blocks.
d) The magnitude of the force of kinetic friction is greater for the block on the right.
e) A force of the same magnitude was required to start each block moving.
6.5.3. A bicycle racer is traveling at constant speed v around a circular
track. The centripetal acceleration of the bicycle is ac. What
happens to the centripetal acceleration of the bicycle if the speed is
doubled to 2v?
a) The centripetal acceleration increases to 4ac.
b) The centripetal acceleration decreases to 0.25ac.
c) The centripetal acceleration increases to 2ac.
d) The centripetal acceleration decreases to 0.5ac.
e) The centripetal acceleration does not change.
6.5.3. A bicycle racer is traveling at constant speed v around a circular
track. The centripetal acceleration of the bicycle is ac. What
happens to the centripetal acceleration of the bicycle if the speed is
doubled to 2v?
a) The centripetal acceleration increases to 4ac.
b) The centripetal acceleration decreases to 0.25ac.
c) The centripetal acceleration increases to 2ac.
d) The centripetal acceleration decreases to 0.5ac.
e) The centripetal acceleration does not change.
6.5.7. Which one of the following forces holds a car on a frictionless
banked curve?
a) the horizontal component of the normal force
b) the vertical component of the car's weight
c) the vertical component of the normal force
d) the horizontal component of the car's weight
e) the reaction force to the car's weight
6.5.7. Which one of the following forces holds a car on a frictionless
banked curve?
a) the horizontal component of the normal force
b) the vertical component of the car's weight
c) the vertical component of the normal force
d) the horizontal component of the car's weight
e) the reaction force to the car's weight
6.3.3. A crate of mass m is at rest on a horizontal frictionless surface.
Another identical crate is placed on top of it. Assuming
that there
r
is no slipping of the top crate as a horizontal force F is applied to
the bottom crate, determine an expression for the magnitude of the
static frictional force acting on the top crate.
a)
f =F
b)
f =
c)
f =F−
d)
f =
mg
2
e)
f =
F
2
F
− mg
2
mg
2
6.3.3. A crate of mass m is at rest on a horizontal frictionless surface.
Another identical crate is placed on top of it. Assuming
that there
r
is no slipping of the top crate as a horizontal force F is applied to
the bottom crate, determine an expression for the magnitude of the
static frictional force acting on the top crate.
a)
f =F
b)
f =
c)
f =F−
d)
f =
mg
2
e)
f =
F
2
F
− mg
2
mg
2
And of course …
x-x0=v0t+at2/2 … (2-15)
v-v0=at
… (2-11)
(2) Eliminate a:
(x-x0)=(v0+v)t/2 … (2-17)
(1) Eliminate t:
v2-v02=2a(x-x0) … (2-16)
(3) Eliminate v0:
(x-x0)=vt-at2/2 … (2-18)
6.3.5. On a rainy evening, a truck is driving along a straight, level road
at 25 m/s. The driver panics when a deer runs onto the road and
locks the wheels while braking. If the coefficient of friction for
the wheel/road interface is 0.68, how far does the truck slide
before it stops?
a) 55 m
b) 47 m
c) 41 m
d) 36 m
e) 32 m
6.3.5. On a rainy evening, a truck is driving along a straight, level road
at 25 m/s. The driver panics when a deer runs onto the road and
locks the wheels while braking. If the coefficient of friction for
the wheel/road interface is 0.68, how far does the truck slide
before it stops?
a) 55 m
b) 47 m
c) 41 m
d) 36 m
e) 32 m
Fk=0.68mg
Fk=-ma
a=-0.68g
V0=25 m/s
V1 = 0 m/s
v2-v02=2a(x-x0)
(x-x0) = v2-v02/2a
(x-x0) =252/[2(0.68)g]
(x-x0) =46.89