Transcript Review - prettygoodphysics

Newton’s Laws
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Tuesday, September 18, 2007
Re-introduction to
Newton’s 3 Laws
What is Force?


A push or pull on an object.
Unbalanced forces cause an object to
accelerate…




To speed up
To slow down
To change direction
Force is a vector!
Types of Forces

Contact forces: involve contact between
bodies.


Field forces: act without necessity of
contact.


Normal, Friction
Gravity, Electromagnetic, Strong, Weak
Question: Is there really any such thing as
a contact force?
Forces and Equilibrium



If the net force on a body is zero, it is in
equilibrium.
An object in equilibrium may be moving
relative to us (dynamic equilibrium).
An object in equilibrium may appear to be
at rest (static equilibrium).
Galileo’s Thought Experiment
© The Physics Classroom, Tom Henderson 1996-2007
Galileo’s Thought Experiment
This thought experiment lead to
Newton’s First Law.
© The Physics Classroom, Tom Henderson 1996-2007
Newton’s First Law



The Law of Inertia.
A body in motion stays in motion in a
straight line unless acted upon by an
external force.
This law is commonly applied to the
horizontal component of velocity, which is
assumed not to change during the flight of
a projectile.
Newton’s Second Law




A body accelerates when acted upon by
a net external force
The acceleration is proportional to the
net (or resultant) force and is in the
direction which the net force acts.
This law is commonly applied to the
vertical component of velocity.
SF = ma
Newton’s Third Law


For every action there exists an equal and
opposite reaction.
If A exerts a force F on B, then B exerts a
force of -F on A.
Commonly Confused Terms



Inertia: or the resistance of an object to
being accelerated
Mass: the same thing as inertia (to a
physicist).
Weight: gravitational attraction
inertia = mass  weight
Sample Problem – 1st Law
• Two forces, F1 = (4i – 6j + k) N and F2 = (i – 2j - 8k) N, act
upon a body of mass 3.0 kg as it is moving at constant
speed. What do you know must be true?
Sample Problem – 2nd Law
• Two forces, F1 = (4i – 6j + k) N and F2 = (i – 2j - 8k) N, act
upon a body of mass 3.0 kg. No other forces act upon the
body at this time. What do you know must be true?
Sample Problem – 3rd Law
• A tug-of-war team ties a rope to a tree and pulls hard
horizontally to create a tension of 30,000 N in the rope.
Suppose the team pulls equally hard when, instead of a
tree, the other end of the rope is being pulled by another
tug-of-war team such that no movement occurs. What is
the tension in the rope in the second case?
Working a 2nd Law Problem


Working Newton’s 2nd Law Problems is
best accomplished in a systematic fashion.
The more complicated the problem, the
more important it is to have a general
procedure to follow in working it.
2nd Law Procedure
1.
2.
3.
4.
5.
6.
7.
Identify the body to be analyzed.
Select a reference frame, stationary or
moving, but not accelerating
Draw a force or free body diagram.
Set up ΣF = ma equations for each
dimension.
Use kinematics or calculus where necessary
to obtain acceleration.
Substitute known quantities.
Calculate the unknown quantities.
Wednesday, September 19, 2007
Types of Forces Commonly Found
in Newton’s Law Problems
Sample Problem

A 5.00-g bullet leaves the muzzle of a rifle with a speed of
320 m/s. The bullet is accelerated by expanding gases
while it travels down the 0.820 m long barrel. Assuming
constant acceleration and negligible friction, what is the
force on the bullet?
Sample Problem

A 3.00 kg mass undergoes an acceleration given
by a = (2.50i + 4.10j) m/s2. Find the resultant
force F and its magnitude.
Normal force




The force that keeps one object from
invading another object.
Our weight is the force of attraction of our
body for the center of the planet.
We don’t fall to the center of the planet.
The normal force keeps us up.
Normal Force
on Flat surface
Normal Force
on Flat surface
N
mg
N = mg for objects
resting on horizontal
surfaces.
Normal Force on Ramp

Normal Force on Ramp
N = mgcos
N
mgsin

mg
The normal force is
perpendicular to angled
ramps as well. It’s usually
equal to the component
of weight perpendicular
to the surface.
mgcos

Normal Force on Ramp
What will acceleration
be in this situation?
N
SF= ma
mgsin = ma
mgsin gsin = a
N = mgcos

mg
mgcos

Normal Force on Ramp
N = mgcos
N
How could you keep the
block from accelerating?
T
mgsin

mg
mgcos

Tension
A pulling force.
 Generally exists in a rope, string, or
cable.
 Arises at the molecular level, when a
rope, string, or cable resists being
pulled apart.

Tension (static 2D)
The horizontal and vertical components of
the tension are equal to zero if the system
is not accelerating.
30o
45o
1
2
3
15 kg
Tension (static 2D)
The horizontal and vertical components of
the tension are equal to zero if the system
is not accelerating.
30o
T3
mg
45o
1
2
3
15 kg
SFx =
T1 0
T2
SFy =
T3 0
Tension (elevator)
What about when
an elevator is
accelerating
upward?
M
Tension (elevator)
T
M
Mg
What about when
an elevator is
accelerating
upward?
Tension (elevator)
What about when
the elevator is
moving at
constant velocity
between floors?
M
Tension (elevator)
T
M
Mg
What about when
the elevator is
moving at
constant velocity
between floors?
Tension (elevator)
What about when
the elevator is
slowing at the
top floor?
M
Tension (elevator)
T
M
Mg
What about when
the elevator is
slowing at the
top floor?
Tension (elevator)
What about if the
elevator cable
breaks?
M
Tension (elevator)
What about if the
elevator cable
breaks?
M
Mg
Pulley problems
Magic pulleys simply bend the coordinate system.
m1
m2
Pulley problems
Magic pulleys simply bend the coordinate system.
N
T
T
m1g
-x
m1
SF = ma
m2g
= (m1+m2)a
m2g
m2
x
Pulley problems
All problems should be started from a
force diagram.

m2
Pulley problems
Tension is determined by examining one
SF = (m1+m2)a
block or the other
T
T
N
m1g

m2
m2g – T+T – m1gsin
= (m1+m2)a
m2g
Pulley problems
Tension is determined by examining one
block or the other
T
T
N
m1g

m2
SF2 = m2a
m2g - T = m2a
SF1 = m1a
T-m1gsin = m1a
m2g
Atwood machine


m2
m1

A device for measuring g.
If m1 and m2 are nearly the
same, slows down freefall such
that acceleration can be
measured.
Then, g can be measured.
Atwood machine


T
m1
m1g
m2
T

A device for measuring g.
If m1 and m2 are nearly the
same, slows down freefall such
that acceleration can be
measured.
Then, g can be measured.
m2g
SF = ma
m2g-m1g = (m2+m1)a
Thursday, September 20,
2007
Post-Test Review
Friday, September 21, 2007
Workday
Exam Corrections
Homework Review
Atwood Machine Mini-lab



Draw diagram of apparatus in lab book.
Record all data.
Calculate g.
Easy Problem
How fast will the block be sliding at the bottom of the
frictionless ramp?
5.0 kg
20o
L = 12 m
Easy Problem
How high up the frictionless ramp will the block slide?
v = 12.0 m/s
5.0 kg
20o
Tuesday, September 25, 2007
Friction
Moderate Problem
Describe acceleration of the 5 kg block. Table and pulley are
magic and frictionless.
1.0 kg
20o
Friction
Friction opposes a sliding motion.
 Static friction exists before sliding
occurs
 (fs  sN).
 Kinetic friction exists after sliding
occurs
 f k = k N

Friction on flat surfaces
y
y
x
Draw a free body diagram for
a braking car.
x
Draw a free body diagram for
a car accelerating from rest.
But we don’t want cars to skid!


Why don’t we?
Let’s use DataStudio to see if we can
detect the difference in magnitude
between static and kinetic friction.
Wednesday, September 26, 2007
Friction Lab
Thursday, September 27, 2007
Friction on a ramp

Sliding down

Sliding up
Friction is always parallel to surfaces….
A 1.00 kg book is held against a wall by pressing it
against the wall with a force of 50.00 N. What must
be the minimum coefficient of friction between the
book and the wall, such that the book does not
slide down the wall?
f
F
N
W
(0.20)
Problem #1
Assume a coefficient of static friction of 1.0 between tires and
road. What is the minimum length of time it would take to
accelerate a car from 0 to 60 mph?
Problem #2
Assume a coefficient of static friction of 1.0 between tires and road and a
coefficient of kinetic friction of 0.80 between tires and road. How far
would a car travel after the driver applies the brakes if it skids to a stop?
Centripetal Force


Inwardly directed force which causes a
body to turn; perpendicular to velocity.
Centripetal force always arises from
other forces, and is not a unique kind of
force.




Sources include gravity, friction, tension,
electromagnetic, normal.
ΣF = ma
a = v2/r
ΣF = m v2/r
Highway Curves
z
R
Friction turns the vehicle
r
Normal force turns the vehicle
Sample problem
• Find the minimum safe turning radius for a car traveling
at 60 mph on a flat roadway, assuming a coefficient of
static friction of 0.70.
Sample problem
• Derive the expression for the period best banking angle
of a roadway given the radius of curvature and the likely
speed of the vehicles.
Friday, September 28, 2007
More on Circular Motion
Conical Pendulum
z
T = 2p L cos 
g
L
r
T
For conical
pendulums,
centripetal
force is
mg provided by a
component o
the tension.
Sample problem
• Derive the expression for the period of a conical
pendulum with respect to the string length and radius of
rotation.
Non-uniform Circular Motion



Consider circular motion in which either speed
of the rotating object is changing, or the forces
on the rotating object are changing.
If the speed changes, there is a tangential as
well as a centripetal component to the force.
In some cases, the magnitude of the
centripetal force changes as the circular
motion occurs.
Sample problem
You swing a 0.25-kg rock in a vertical circle on a 0.80 m long rope
at 2.0 Hz. What is the tension in the rope a) at the top and b) at
the bottom of your swing?
Monday, October 1
Time-dependent forces
Sample problem
A 40.0 kg child sits in a swing supported by 3.00 m long chains. If
the tension in each chain at the lowest point is 350 N, find a) the
child’s speed at the lowest point and b) the force exerted by the
seat on the child at the lowest point.
Sample problem
A 900-kg automobile is traveling along a hilly road. If it is to
remain with its wheels on the road, what is the maximum speed it
can have as it tops a hill with a radius of curvature of 20.0 m?
Non-constant Forces
•
•
•
Forces can vary with time.
Forces can vary with velocity.
Forces can vary with position.
Calculus Concepts for
Forces that Vary with Time
•
Differentiation
•
•
•
the tangent (or slope) of a function
position -> velocity -> acceleration
Integration
•
•
the area under a curve
acceleration -> velocity -> position
Evaluating Integrals
If a(t) = tn
then
tn dt
= tn+1 / (n+1) + C
t
t
tn dt
0
= tn+1 / (n+1)
0
Sample problem

Consider a force that is a function of time:
F(t) = (3.0 t – 0.5 t2)N

If this force acts upon a 0.2 kg particle at rest for 3.0
seconds, what is the resulting velocity and position of the
particle?
Sample problem

Consider a force that is a function of time:
F(t) = (16 t2 – 8 t + 4)N

If this force acts upon a 4 kg particle at rest for 1.0 seconds,
what is the resulting change in velocity of the particle?
Drag Forces

Drag forces
slow an object down as it passes
through a fluid.
 act in opposite direction to velocity.
 are functions of velocity.
 impose terminal velocity.

Drag as a Function of Velocity
 fD
= bv + cv2
 b and c depend upon
 shape
and size of object
 properties of fluid
b
is important at low velocity
 c is important at high velocity
Drag Force in Free Fall
fD
fD
fD
mg
mg
mg
mg
when fD
equals
mg,
terminal
velocity
has been
reached
Drag Force in Free Fall
FD = bv + c v2
FD
for fast moving objects
for slow moving objects
FD = c v2
FD = b v
c = 1/2 D r A
Where
D = drag coefficient
r = density of fluid
A = cross-sectional area
mg
Sample Problem: Slow moving objects

FD = bv
mg
Show that vT = mg/b
Sample Problem: Slow Moving Objects

Show that v(t) = (mg/b)(1 – e
FD = bv
mg
–bt/m)
Sample Problem: Fast moving objects

FD =
1/2DrAv2
mg
Show that vT = (2mg / DrA)1/2
Sample Problem: Fast moving objects

Derive an expression for the velocity of the
object as a function of time
FD =
1/2DrAv2
mg