Newton’s
Newton’s Laws
Laws
MONDAY, September 14,
Re-introduction to
Newton’s 3 Laws


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!
What is Force?
Types of Forces

Contact forces: involve contact between
bodies.


Normal, Friction
Field forces: act without necessity of
contact.


Gravity, Electromagnetic, Strong, Weak
Dark side, light side, etc.
Question: Is there really any such
thing as a contact force?
A.
B.
C.
yea.
No.
I think this is a trick so I’m not
answering.
If the net force on a body is zero
then it must….
A.
B.
C.
D.
Be At rest
Have a Constant Velocity
Not be touching the ground
Not be a jedi
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
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 at rest stays at rest unless acted
upon by an external force
A body in motion stays in motion in a
straight line unless acted upon by an
external force.
Or.. Mass wants to do its thing.
This law is commonly applied to the horizontal
component of velocity, which is assumed not to
change during the flight of a projectile.
Note on Inertia




Inertia is a property of matter.
Forget any definition of mass that you
learned in Chemistry.
Mass is a measure of an objects inertia.
It is the resistance to changes in motion
An objects mass cannot change unless
you actually change the amount of
matter present
A.
B.
True- I believe in conservation of mass
False- I am a non-believer and think it’s
more complicated than that.
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.
Or Action Force = -Reaction Force
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
We normally use the term “weightless” to
describe the feeling that astronauts get when
they are in orbit or in simulators like the Vomit
Comet. What does “weightless” really mean
when applied to these situations?
A.
B.
C.
D.
You have no weight
You have weight but it is not opposed by
another force.
You have no mass
You have been eating dehydrated space food
so you don’t weigh as much.
Only Two forces, F1 and F2 act upon a body
of mass 3.0 kg. The body moves at
constant speed. What do you know must be
true?
A.
B.
C.
D.
The net force must point in the direction
of motion
The object must be weightless.
The vectors F1 and F2 must be 180°
opposite each other.
The sum of forces F1 and F2 must be
zero
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?
A.
B.
C.
D.
The body’s motion will not change.
The body’s motion will change.
Today is Saturday.
There must be another force.
A tug-of-war team ties a rope to a tree and pulls
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?
A.
B.
C.
D.
Less than 30,000 N
30,000 N
More than 30,000 N
The rope will break
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 (2nd law holds
for only inertial reference frames).
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.
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?
A.
B.
C.
D.
160 N
320 N
640 N
1280 N
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.
A.
B.
C.
D.
4.8 N 59° from horizontal
14.4 N 31° from horizontal
4.8 N 31° from horizontal
14.4 N 59° from horizontal
Wednesday, September 17, 2009
Types of Forces Commonly Found
in Newton’s Law Problems and
Practice
Normal force

The force that keeps one object from
“invading” another object.




Ex: 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.
It ALWAYS acts perpendicular to
contact!!! NOT always vertical!!!!
Normal Force
on Flat surface
Normal Force
on Flat surface
N
mg
Fn = -(mg) for objects
resting on horizontal
surfaces (no vertical
acceleration).
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

What will acceleration be in this
situation?
N
mgsin

mg
mgcos
A.
B.
C.
D.
E.
mgsin
mgcos
gsin
gcos
g
Normal Force on Ramp
N = -(mgcos)
N
How could you keep the
block from accelerating?
F
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
A Mass m hangs by a cable from the bottom of
an elevator. What is the tension in the cable
when the elevator is stopped at a floor?
A.
B.
C.
D.
E.
g
mg
Greater in
magnitude than
mg
Less magnitude
than mg
0
m
Tension (elevator)
T
M
Mg
The sum of the
forces must be
zero or the
mass would
accelerate!
A Mass m hangs by a cable from the bottom of
an elevator. What is the tension in the cable
when the elevator accelerates upward?
A.
B.
C.
D.
E.
g
mg
Greater in
magnitude than
mg
Less magnitude
than mg
0
m
Tension (elevator)
T
M
Mg
T must be
greater than
mg in order to
result in an
upward
acceleration!
A Mass m hangs by a cable from the bottom of
an elevator. What is the tension in the cable
when the elevator is moving at constant speed
between floors?
A.
B.
C.
D.
E.
g
mg
Greater in
magnitude than
mg
Less magnitude
than mg
0
m
A Mass m hangs by a cable from the bottom of
an elevator. What is the tension in the cable
when the elevator is slowing down at the top
floor?
A.
B.
C.
D.
E.
g
mg
Greater in
magnitude than
mg
Less magnitude
than mg
0
m
Tension (elevator)
T
M
Mg
Net force
must point
down in order
to accelerate
down
A Mass m hangs by a cable from the bottom of
an elevator. What will the tension in the cable
be if it comes attached from the elevator?
A.
B.
C.
D.
E.
g
mg
Greater in
magnitude than
mg
Less magnitude
than mg
0
m
Tension (elevator)
No opposing
force!
(weightless???)
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