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Dynamics
Dynamics
• The laws and causes of motion were first
properly formulated in the book _________
by ____________ ___________ in the year
__________. These laws were later modified
in 1905 and 1915 by _____________
______________.
Dynamics
• The laws and causes of motion were first
properly formulated in the book Principia by
Isaac Newton in the year 1687. These laws
were later modified in 1905 and 1915 by
Albert Einstein.
Word Definition for Dynamics: a study
of _______objects…
Accelerate or…
Don’t accelerate…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
• 3 possible forms ?…
Don’t accelerate…
• 2 possible states of motion
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
• 3 possible forms…
Speeding up
Slowing down
Changing direction
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
• 3 possible forms…
Speeding up
Slowing down
Changing direction
Newton says: Any of these
forms of acceleration are
caused by a non-zero
___________
or ________________ force.
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
• 3 possible forms…
Speeding up
Slowing down
Changing direction
Newton says: Any of these
forms of acceleration are
caused by a non-zero net
or unbalanced force.
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
• 3 possible forms…
Speeding up
Slowing down
Changing direction
Newton says: Any of these
forms of acceleration is
caused by a non-zero net
or unbalanced force. The
symbols for net force are …
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion…
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion… ?
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion… at rest or at
constant velocity
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion… at rest or at
constant velocity
• Newton’s first law : If an
object is not speeding up,
not slowing down or not
changing direction, then it
has an acceleration of
______ and a net force of
______ acting on it.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion… at rest or at
constant velocity
• Newton’s first law : If an
object is not speeding up,
not slowing down or not
changing direction, then it
has an acceleration of zero
and a net force of ______
acting on it.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion… at rest or at
constant velocity
• Newton’s first law : If an
object is not speeding up,
not slowing down or not
changing direction, then it
has an acceleration of zero
and a net force of zero
acting on it.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion… at rest or at
constant velocity
• Newton’s first law : If an
object is at ________ or
__________ _________,
then it has an acceleration
of zero and a net force of
zero acting on it.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of
motion… at rest or at
constant velocity
• Newton’s first law : If an
object is at rest or constant
velocity, then it has an
acceleration of zero and a
net force of zero acting on
it.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of motion…
at rest or at constant velocity
• Newton’s first law : If an
object is at rest or constant
velocity, then it has an
acceleration of zero and a net
force of zero acting on it. If
the net force is zero, there
could be ____ forces acting or
the forces are _____________.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of motion…
at rest or at constant velocity
• Newton’s first law : If an
object is at rest or constant
velocity, then it has an
acceleration of zero and a net
force of zero acting on it. If
the net force is zero, there
could be no forces acting or
the forces are _____________.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
• 2 possible states of motion…
at rest or at constant velocity
• Newton’s first law : If an
object is at rest or constant
velocity, then it has an
acceleration of zero and a net
force of zero acting on it. If
the net force is zero, there
could be no forces acting or
the forces are balanced.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
•
2 possible states of motion… at rest orat
constant velocity
•
Newton’s first law : If an object is at
rest or constant velocity, then it has an
acceleration of zero and a net force of
zero acting on it. If the net force is zero,
there could be no forces acting or the
forces are balanced.
•
The converse is also true If the net
force on an object is zero, then the
acceleration is ______ and the object
is either at ______ or moving at
__________ ____________.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
•
2 possible states of motion… at rest orat
constant velocity
•
Newton’s first law : If an object is at
rest or constant velocity, then it has an
acceleration of zero and a net force of
zero acting on it. If the net force is zero,
there could be no forces acting or the
forces are balanced.
•
The converse is also true If the net
force on an object is zero, then the
acceleration is zero and the object is
either at ______ or moving at
__________ ____________.
Word Definition for Dynamics: a study
of WHY objects…
Accelerate or…
Don’t accelerate…
•
2 possible states of motion… at rest orat
constant velocity
•
Newton’s first law : If an object is at
rest or constant velocity, then it has an
acceleration of zero and a net force of
zero acting on it. If the net force is zero,
there could be no forces acting or the
forces are balanced.
•
The converse is also true If the net
force on an object is zero, then the
acceleration is zero and the object is
either at rest or moving at constant
velocity
The three elements of Newton’s
second law in symbols
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ What does this mean?
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is
inversely proportional to the mass.
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is
inversely proportional to the mass.
ǀaǀ α 1/m
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is
inversely proportional to the mass.
ǀaǀ α 1/m What does this mean?
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is
inversely proportional to the mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is
inversely proportional to the mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
• Acceleration is in the same direction as the
________ ___________.
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly
proportional to the net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is
inversely proportional to the mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
• Acceleration is in the same direction as the
net force.
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly proportional to the
net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is inversely proportional
to the mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
1. Acceleration is in the same direction as the net force.
Note: #1 holds true only if the __________ is constant.
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly proportional to the
net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is inversely proportional
to the mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
1. Acceleration is in the same direction as the net force.
Note: #1 holds true only if the mass is constant.
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly proportional to the
net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is inversely proportional
to the mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
1. Acceleration is in the same direction as the net force.
Note: #1 holds true only if the mass is constant.
#2 holds true only if the _______ ______ is constant.
The three elements of Newton’s
second law in symbols
1. The size of the acceleration is directly proportional to the
net force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is inversely proportional
to the mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
1. Acceleration is in the same direction as the net force.
Note: #1 holds true only if the mass is constant.
#2 holds true only if the net force is constant.
The three elements of Newton’s
second law in symbols
1.
The size of the acceleration is directly proportional to the net
force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is inversely proportional to the
mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
• Acceleration is in the same direction as the net force.
Note: #1 holds true only if the mass is constant.
#2 holds true only if the net force is constant.
In SI mks, all three elements can be combined to form Newton’s
second law vector equation
a = ??????
The three elements of Newton’s
second law in symbols
1.
The size of the acceleration is directly proportional to the net
force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is inversely proportional to the
mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
• Acceleration is in the same direction as the net force.
Note: #1 holds true only if the mass is constant.
#2 holds true only if the net force is constant.
In SI mks, all three elements can be combined to form Newton’s
second law vector equation
a = Fnet / m
The three elements of Newton’s
second law in symbols
1.
The size of the acceleration is directly proportional to the net
force
ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y
1. The magnitude of the acceleration is inversely proportional to the
mass.
ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y
• Acceleration is in the same direction as the net force.
Note: #1 holds true only if the mass is constant.
#2 holds true only if the net force is constant.
In SI mks, all three elements can be combined to form Newton’s
second law vector equation
please!
a = Fnet / m Memorize
Other Forms of
Newton’s Second Law equation
Other Forms of
Newton’s Second Law equation
• a = Fnet / m
• What does Fnet = ?
Other Forms of
Newton’s Second Law equation
• a = Fnet / m
• Fnet = m a
Other Forms of
Newton’s Second Law equation
• a = Fnet / m
• Fnet = m a
Friday = Monday afternoon!
Other Forms of
Newton’s Second Law equation
• a = Fnet / m
• Fnet = m a
Friday = Monday afternoon!
• What does m = ?
Other Forms of
Newton’s Second Law equation
• a = Fnet / m
• Fnet = m a
Friday = Monday afternoon!
• m = ǀFnet ǀ / ǀaǀ
Other Forms of
Newton’s Second Law equation
• a = Fnet / m
• Fnet = m a
Friday = Monday afternoon!
• m = ǀFnet ǀ / ǀaǀ
• Note: Once we know “a” from the second law
dynamics equation, we can use _________
kinematics equations to find _____, _____,
____, and_____.
Other Forms of
Newton’s Second Law equation
• a = Fnet / m
• Fnet = m a
Friday = Monday afternoon!
• m = ǀFnet ǀ / ǀaǀ
• Note: Once we know “a” from the second law
dynamics equation, we can use four
kinematics equations to find v1 , v2, Δd, and Δt.
Other Forms of
Newton’s Second Law equation
•
•
•
•
a = Fnet / m
Fnet = m a Friday = Monday afternoon!
m = ǀFnet ǀ / ǀaǀ
Note: Once we know “a” from the second law dynamics
equation, we can use four kinematics equations to find v1 , v2,
Δd, and Δt.
• Also, if we know three kinematics quantities,
v1 , v2, Δd, or Δt, we can find “____” and then
find Fnet using the second law dynamics
equation.
Other Forms of
Newton’s Second Law equation
•
•
•
•
a = Fnet / m
Fnet = m a Friday = Monday afternoon!
m = ǀFnet ǀ / ǀaǀ
Note: Once we know “a” from the second law dynamics
equation, we can use four kinematics equations to find v1 , v2,
Δd, and Δt.
• Also, if we know three kinematics quantities,
v1 , v2, Δd, or Δt, we can find “ a ” and then
find Fnet using the second law dynamics
equation.
Other Forms of
Newton’s Second Law equation
•
•
•
•
a = Fnet / m
Fnet = m a Friday = Monday afternoon!
m = ǀFnet ǀ / ǀaǀ
Note: Once we know “a” from the second law dynamics
equation, we can use four kinematics equations to find v1 , v2,
Δd, and Δt.
• Also, if we know three kinematics quantities, v1 , v2, Δd, or Δt,
we can find “ a ” and then find Fnet using the second law
dynamics equation.
v1 , v2, Δd, and Δt ↔ a ↔ Fnet , m
4 kinematics equation
dynamics 2nd law
equation
Definition of the Newton:
Unit of Force
Definition of the Newton:
Unit of Force
• Start with
Definition of the Newton:
Unit of Force
• Start with
Fnet = m a
Definition of the Newton:
Unit of Force
• Start with
UNIT ?
Fnet = m a
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
=
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
=
UNIT ?
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
=
kilogram
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
=
kilogram
UNIT ?
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
=
kilogram meter/sec2
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
=
1 kilogram meter/sec2
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
=
kilogram meter/sec2
• In words, one Newton of unbalanced force is
applied when a one ___________ mass is
accelerated at __________
Definition of the Newton:
Unit of Force
• Start with
Newton
Fnet = m a
=
kilogram meter/sec2
• In words, one Newton of unbalanced force is
applied when a one kilogram mass is
accelerated at 1 meter/sec2
Mass vs Force
Mass vs Force
Mass
• ?
Force
• ?
Mass vs Force
Mass
• Amount of matter
Force
• ?
Mass vs Force
Mass
• Amount of matter
Force
• A push or pull
Mass vs Force
Mass
• Amount of matter
• SI unit = kilogram (kg)
Force
• A push or pull
Mass vs Force
Mass
• Amount of matter
• SI unit = kilogram (kg)
Force
• A push or pull
• SI unit = Newton = kg m/s2
Mass vs Force
Mass
• Amount of matter
• SI unit = kilogram (kg)
• scalar
Force
• A push or pull
• SI unit = Newton = kg m/s2
Mass vs Force
Mass
• Amount of matter
• SI unit = kilogram (kg)
• scalar
Force
• A push or pull
• SI unit = Newton = kg m/s2
• vector
Mass vs Force
•
•
•
•
Mass
Amount of matter
SI unit = kilogram (kg)
Scalar
Use a balance or collision
experiments to measure
Force
• A push or pull
• SI unit = Newton = kg m/s2
• Vector
Mass vs Force
•
•
•
•
Mass
Amount of matter
SI unit = kilogram (kg)
Scalar
Use a balance or collision
experiments to measure
•
•
•
•
Force
A push or pull
SI unit = Newton = kg m/s2
Vector
Use a spring scale to
measure
Newton’s Third Law of Motion
Newton’s Third Law of Motion
• For every action force on body A by body B,
there is an ___________ but ______________
reaction force on body B by body A.
Newton’s Third Law of Motion
• For every action force on body A by body B,
there is an equal but opposite reaction force
on body B by body A.
Newton’s Third Law of Motion
• For every action force on body A by body B, there is an
equal but opposite reaction force on body B by body A.
Mr. BIG-A
BIG’s Spring scale
= more, less or
equal to 122 N ?
rope
Little’s Spring
scale = 122 N
Mr. Little-B
Newton’s Third Law of Motion
• For every action force on body A by body B, there is an
equal but opposite reaction force on body B by body A.
Mr. BIG-A
BIG’s Spring scale =
122 N
rope
Little’s Spring
scale = 122 N
Mr. Little-B
Newton’s Third Law of Motion
• For every action force on body A by body B, there is an
equal but opposite reaction force on body B by body A.
Mr. BIG-A
BIG’s Spring scale =
122 N
Little’s Spring
scale = 122 N
rope
Action Force Statement: Mr BIG-A pulls on Mr Little-B
with a force of 122 N [W]
Mr. Little-B
Newton’s Third Law of Motion
• For every action force on body A by body B, there is an
equal but opposite reaction force on body B by body A.
Mr. BIG-A
BIG’s Spring scale =
122 N
Little’s Spring
scale = 122 N
rope
Action Force Statement: Mr BIG-A pulls on Mr Little-B
with a force of 122 N [W]
Reaction force statement: Mr Little-B pulls on
Mr BIG-A with a force of 122 N [E]
Mr. Little-B
Newton’s Third Law of Motion
• For every action force on body A by body B, there is an
equal but opposite reaction force on body B by body A.
Mr. BIG-A
Suppose the men are on ice. Which man will have the
greater acceleration and why?
BIG’s Spring scale =
122 N
Little’s Spring
scale = 122 N
rope
Action Force Statement: Mr BIG-A pulls on Mr Little-B
with a force of 122 N [W]
Reaction force statement: Mr Little-B pulls on
Mr BIG-A with a force of 122 N [E]
Mr. Little-B
Newton’s Third Law of Motion
• For every action force on body A by body B, there is an
equal but opposite reaction force on body B by body A.
Mr. BIG-A
If the men are on ice, Mr. Little will have the greater
acceleration because he has the smaller mass, a α 1/m,
and as m ↓ a ↑ for the same net force.
BIG’s Spring scale =
122 N
Little’s Spring
scale = 122 N
rope
Action Force Statement: Mr BIG-A pulls on Mr Little-B
with a force of 122 N [W]
Reaction force statement: Mr Little-B pulls on
Mr BIG-A with a force of 122 N [E]
Mr. Little-B
Important Notes
on Newton’s Third Law
Important Notes
on Newton’s Third Law
• This law only applies when there are two
bodies
Important Notes
on Newton’s Third Law
• This law only applies when there are two
bodies
• The action and reaction forces never act on
the same body. What would happen if action and
reaction forces did happen on the same body?
Important Notes
on Newton’s Third Law
• This law only applies when there are two
bodies
• The action and reaction forces never act on
the same body. Forces would always be balanced on
the same body, and there would be no acceleration ever!
Important Notes
on Newton’s Third Law
• This law only applies when there are two
bodies
• The action and reaction forces never act on
the same body. Forces would always be balanced on
the same body, and there would be no acceleration ever!
• The action force acts on one body and the
equal, reaction force acts on the other.
Important Notes
on Newton’s Third Law
• This law only applies when there are two bodies
• The action and reaction forces never act on the same
body.
• The action force acts on one body and the equal,
reaction force acts on the other. Forces would always be
balanced on the same body, and there would be no
acceleration ever!
• If the action and reaction forces are the only forces
acting, each body will accelerate according to
|a| α 1/m so that |a1| / |a2| = m2 / m1
Example: Here is an action force statement: A foot exerts
an applied force on a soccer ball at 83 N [forward]. What is
the reaction force statement?
Two-body diagram:
Reaction force statement: ?
Action:
83 N
[forward]
Example: Here is an action force statement: A foot exerts
an applied force on a soccer ball at 83 N [forward]. What is
the reaction force statement?
Two-body diagram:
Reaction: 83 N
[backward]
Action:
83 N
[forward]
Reaction force statement: The ball exerts an equal applied force
on the foot at 83 N [backward].
Example: Here is an action force statement: A foot exerts
an applied force on a soccer ball at 83 N [forward]. What is
the reaction force statement?
Two-body diagram:
Reaction: 83 N
[backward]
Action:
83 N
[forward]
Reaction force statement: The ball exerts an equal applied force
on the foot at 83 N [backward].
• What is the effect of the reaction force of the ball on the
foot?
Example: Here is an action force statement: A foot exerts
an applied force on a soccer ball at 83 N [forward]. What is
the reaction force statement?
Two-body diagram:
Reaction: 83 N
[backward]
Action:
83 N
[forward]
Reaction force statement: The ball exerts an equal applied force
on the foot at 83 N [backward].
• The applied net force and acceleration is backward on the foot, but the
velocity of the foot is forward, so the foot slows down.
You try this one!: Here is an action force statement: The planet
earth exerts a gravitational force of 1.4 N [down] on a falling
apple. What is the reaction force statement?
Reaction force statement: ?
apple
Planet
earth
Action:
1.4 N [down]
You try this one!: Here is an action force statement: The planet
earth exerts a gravitational force of 1.4 N [down] on a falling
apple. What is the reaction force statement?
Reaction force statement:
apple
Action:
1.4 N [down]
The apple exerts an equal
but opposite gravitational
force on the earth of 1.4 N [up]
Planet
earth
Reaction:
1.4 N [up]
You try this one!: Here is an action force statement: The planet
earth exerts a gravitational force of 1.4 N [down] on a falling
apple. What is the reaction force statement?
Reaction force statement:
apple
Action:
1.4 N [down]
The apple exerts an equal
but opposite gravitational
force on the earth of 1.4 N [up]
Planet
earth
Reaction: 1.4
N [up]
If the apple exerts the same force up on the earth, why
doesn't the earth accelerate up to meet the apple?
You try this one!: Here is an action force statement: The planet
earth exerts a gravitational force of 1.4 N [down] on a falling
apple. What is the reaction force statement?
Reaction force statement:
apple
Action:
1.4 N [down]
The apple exerts an equal
but opposite gravitational
force on the earth of 1.4 N [up]
Planet
earth
Reaction: 1.4
N [up]
The earth doesn't accelerate up to meet the apple because
the mass of the earth is so much more than the apple.
Note that |a| α 1/m, and as m ↑ a ↓ for the same net
force of 1.4 N.
You try this one!: Here is an action force statement: The planet
earth exerts a gravitational force of 1.4 N [down] on a falling
apple. What is the reaction force statement?
Reaction force statement:
apple
Action:
1.4 N [down]
The apple exerts an equal
but opposite gravitational
force on the earth of 1.4 N [up]
Planet
earth
Reaction: 1.4
N [up]
The earth doesn't accelerate up to meet the apple because the mass of the
earth is so much more than the apple. Note that |a| α 1/m, and as m ↑ a
↓ for the same net force of 1.4 N. If the earth has a mass of 6.0 X 1024 kg
and the apple has a mass of 0.14 kg, what is the ratio of the sizes of the
acceleration of the earth : the acceleration of the apple ?
You try this one!: Here is an action force statement: The planet
earth exerts a gravitational force of 1.4 N [down] on a falling
apple. What is the reaction force statement?
Reaction force statement:
apple
Action:
1.4 N [down]
The apple exerts an equal
but opposite gravitational
force on the earth of 1.4 N [up]
Planet
earth
Reaction: 1.4
N [up]
The earth doesn't accelerate up to meet the apple because the mass of the
earth is so much more than the apple. Note that |a| α 1/m, and as m ↑ a
↓ for the same net force of 1.4 N.
|aearth| / |aapple| = mapple/mearth = 0.14 kg /6.0 x 1024 kg=2.3 X 10-26
The Fundamental Forces
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = ?
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = ?
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
Caused by moving charges = ?
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
Caused by moving charges = magnetic force
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
Caused by moving charges = magnetic force
Charges could be moving or not moving depending on the frame of
reference, so the electrical and magnetic forces are actually different
forms of the same fundamental force = ?
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
Caused by moving charges = magnetic force
Charges could be moving or not moving depending on the frame of
reference, so the electrical and magnetic forces are actually different
forms of the same fundamental force = electromagnetic force
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
Caused by moving charges = magnetic force
Charges could be moving or not moving depending on the frame of
reference, so the electrical and magnetic forces are actually different
forms of the same fundamental force = electromagnetic force
3.
Caused by nucleons like the proton and neutron being very close
together ( < 10-15 m ) to prevent the protons in the nucleus from repelling
each other = ?
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
Caused by moving charges = magnetic force
Charges could be moving or not moving depending on the frame of
reference, so the electrical and magnetic forces are actually different
forms of the same fundamental force = electromagnetic force
3.
Caused by nucleons like the proton and neutron being very close
together ( < 10-15 m ) to prevent the protons in the nucleus from repelling
each other = strong nuclear force
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
Caused by moving charges = magnetic force
Charges could be moving or not moving depending on the frame of
reference, so the electrical and magnetic forces are actually different
forms of the same fundamental force = electromagnetic force
3.
Caused by nucleons like the proton and neutron being very close
together ( < 10-15 m ) to prevent the protons in the nucleus from repelling
each other = strong nuclear force
4. Turns one particle into another, like a neutron into a proton, by emitting
“beta” radiation (electrons, positrons, neutrinos) = ?
The Fundamental Forces
All the different kinds of pushes and pulls around us, or forces, can be
classified as belonging to four fundamental forces:
1. Caused by two masses near each other = force of gravity
2. Caused by two charges near each other = electrical force
Caused by moving charges = magnetic force
Charges could be moving or not moving depending on the frame of
reference, so the electrical and magnetic forces are actually different
forms of the same fundamental force = electromagnetic force
3.
Caused by nucleons like the proton and neutron being very close
together ( < 10-15 m ) to prevent the protons in the nucleus from repelling
each other = strong nuclear force
4.
Turns one particle into another, like a neutron into a proton, by emitting
“beta” radiation(electrons, positrons, neutrinos)=weak force
The Gravitational Force
• Caused by two masses near each other
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
• Fg α mM/d2
M
m
d
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
• Fg α mM/d2
M
m
d
Fg (M on m)
Fg (m on M)
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
• Fg α mM/d2
M
m
d
• As d ↑ 3 Fg ?
Fg (M on m)
Fg (m on M)
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
• Fg α mM/d2
M
m
d
• As d ↑ 3 Fg ↓ 9
Fg (M on m)
Fg (m on M)
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
• Fg α mM/d2
m
d
• As d ↑ 3 Fg ↓ 9
• As M ↑ 13 Fg ?
M
Fg (M on m)
Fg (m on M)
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
• Fg α mM/d2
m
d
• As d ↑ 3 Fg ↓ 9
• As M ↑ 13 Fg ↑ 13
M
Fg (M on m)
Fg (m on M)
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
• Fg α mM/d2
m
d
• As d ↑ 3 Fg ↓ 9
• As M ↑ 13 Fg ↑ 13
• Equation?
M
Fg (M on m)
Fg (m on M)
The Gravitational Force
• Caused by two masses near each other
M
• Always attractive (according to Newton)
• Fg α mM/d2
• As d ↑ 3 Fg ↓ 9
• As M ↑ 13 Fg ↑ 13
m
d
Fg (M on m)
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
Fg (m on M)
The Gravitational Force
• Caused by two masses near each other
• Always attractive (according to Newton)
• Fg α mM/d2
m
d
• As d ↑ 3 Fg ↓ 9
• As M ↑ 13 Fg ↑ 13
M
Fg (M on m)
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
Fg (m on M)
• If the two masses are small, the force of gravity is also small.
If the two masses are large, the force of gravity is also large.
The Gravitational Force
• Example: Find the magnitude of force of gravitational attraction
between a 67 kg girl and a 75 kg boy sitting 2.0 m apart.
The Gravitational Force
• Example: Find the magnitude of force of gravitational attraction
between a 67 kg girl and a 75 kg boy sitting 2.0 m apart.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
=?
The Gravitational Force
• Example: Find the magnitude of force of gravitational attraction
between a 67 kg girl and a 75 kg boy sitting 2.0 m apart.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11)(67)(75) / (2.0)2
The Gravitational Force
• Example: Find the magnitude of force of gravitational attraction
between a 67 kg girl and a 75 kg boy sitting 2.0 m apart.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11)(67)(75) / (2.0)2
= 8.4 X 10-8 N very, very small force
The Gravitational Force
• Try this: Find the magnitude of force of gravitational attraction
between the 2.0 x 1030 kg sun and the 6.0 X 1024 kg earth separated
by 1.5 X 1011 m.
The Gravitational Force
• Try this: Find the magnitude of force of gravitational attraction
between the 2.0 x 1030 kg sun and the 6.0 X 1024 kg earth separated
by 1.5 X 1011 m.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11)(2.0 x 1030)(6.0 X 1024) / (1.5 X 1011 )2
= 3.6 X 1022 N very, very large force
Proportionality and the
Force of Gravity
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
• Note the masses stay constant so... Fg α 1/d2
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
• Note the masses stay constant so... Fg α 1/d2
• As d ↑ 3, d2 changes how?
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
• Note the masses stay constant so... Fg α 1/d2
• As d ↑ 3, d2 ↑ 32
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
• Note the masses stay constant so... Fg α 1/d2
• As d ↑ 3, d2 ↑ 32 , Fg changes how?
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
• Note the masses stay constant so... Fg α 1/d2
• As d ↑ 3, d2 ↑ 32 , Fg ↓ 32
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
• Note the masses stay constant so... Fg α 1/d2
• As d ↑ 3, d2 ↑ 32 , Fg ↓ 32
• So the new force of gravity = ?
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
• Note the masses stay constant so... Fg α 1/d2
• As d ↑ 3, d2 ↑ 32 , Fg ↓ 32
• So the new force of gravity = 1000.0 N / 32 = ?
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is increased by a factor of three.
• Note the masses stay constant so... Fg α 1/d2
• As d ↑ 3, d2 ↑ 32 , Fg ↓ 32
• So the new force of gravity = 1000.0 N / 32 = 111.11 N
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is divided by a factor of five. Try this.
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is divided by a factor of five. Try this.
• Note the masses stay constant so... Fg α 1/d2
• As d ↓ 5, d2 ↓ 52 , Fg ↑ 52
• So the new force of gravity = 1000.0 N X 52 = 25000 N
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is divided by a factor of five, one mass is
increased by a factor of three, the other mass is divided by
four. Try this harder one.
Proportionality and the
Force of Gravity
• Example: Two objects in deep space exert a force of
gravitational attraction of 1000.0 N on each other. Using
Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the
new force of gravity if the following changes are made?
• The distance is divided by a factor of five, one mass is
increased by a factor of three, the other mass is divided by
four. Try this harder one.
• 1000.0 N X 52 X 3 / 4 = 18750 N
The Gravitational Force
• You try this: Find the magnitude of force of gravitational attraction
between the same 67.0 kg girl on the surface of the earth and the
planet earth itself. The distance between the centre of the earth
and the centre of the girl is very nearly equal to the radius of the
earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg.
The Gravitational Force
• You try this: Find the magnitude of force of gravitational attraction
between the same 67.0 kg girl on the surface of the earth and the
planet earth itself. The distance between the centre of the earth
and the centre of the girl is very nearly equal to the radius of the
earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11 ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2
= 657 N
The Gravitational Force
• You try this: Find the magnitude of force of gravitational attraction
between the same 67.0 kg girl on the surface of the earth and the
planet earth itself. The distance between the centre of the earth
and the centre of the girl is very nearly equal to the radius of the
earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11 ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2
= 657 N
• You know a short-cut formula to get the force of earth's gravity on
an object on the earth's surface. What is it?
The Gravitational Force
• You try this: Find the magnitude of force of gravitational attraction
between the same 67.0 kg girl on the surface of the earth and the planet
earth itself. The distance between the centre of the earth and the centre
of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m.
The mass of the earth is 6.00 X 1024 kg.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11 ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N
• *Fg = mg
• m = mass of surface object (kg)
• g = gravitational field strength or force of
earth's gravity on a one kilogram object (N/kg)
The Gravitational Force
• You try this: Find the magnitude of force of gravitational attraction
between the same 67.0 kg girl on the surface of the earth and the planet
earth itself. The distance between the centre of the earth and the centre
of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m.
The mass of the earth is 6.00 X 1024 kg.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11 ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N
• *Fg = mg
• m = mass of surface object (kg)
• g = gravitational field strength or force of a planet's gravity on a
one kilogram object (N/kg) On the earth's surface,
g = 9.81 N/kg [d] It is numerically equal to ag , but different.
The Gravitational Force
• You try this: Find the magnitude of force of gravitational attraction
between the same 67.0 kg girl on the surface of the earth and the planet
earth itself. The distance between the centre of the earth and the centre
of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m.
The mass of the earth is 6.00 X 1024 kg.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11 ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N
• *Fg = mg = ?
• m = mass of surface object (kg)
• g = gravitational field strength or force of a planet's gravity on a
one kilogram object (N/kg) On the earth's surface,
g = 9.81 N/kg [d] It is numerically equal to ag , but different.
The Gravitational Force
• You try this: Find the magnitude of force of gravitational attraction
between the same 67.0 kg girl on the surface of the earth and the planet
earth itself. The distance between the centre of the earth and the centre
of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m.
The mass of the earth is 6.00 X 1024 kg.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11 ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N
• *Fg = mg = (67 kg )(9.81 N/kg) [d] = ?
• m = mass of surface object (kg)
• g = gravitational field strength or force of a planet's gravity on a
one kilogram object (N/kg) On the earth's surface,
g = 9.81 N/kg [d] It is numerically equal to ag , but different.
The Gravitational Force
• You try this: Find the magnitude of force of gravitational attraction
between the same 67.0 kg girl on the surface of the earth and the planet
earth itself. The distance between the centre of the earth and the centre
of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m.
The mass of the earth is 6.00 X 1024 kg.
• Fg = GmM/d2 G = 6.67 X 10-11 Nm2/kg2
= (6.67 X 10-11 ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N
• *Fg = mg = (67 kg )(9.81 N/kg) [d] = 657 N [d]
• m = mass of surface object (kg)
• g = gravitational field strength or force of a planet's gravity on a
one kilogram object (N/kg) On the earth's surface,
g = 9.81 N/kg [d] It is numerically equal to ag , but different.
A Special Force of gravity
A Special Force of gravity
• The special force of a planet's gravity on a
surface object is called __________.
A Special Force of gravity: Weight
• The special force of a planet's gravity on a
surface object is called weight.
A Special Force of gravity: Weight
• The special force of a planet's gravity on a
surface object is called weight.
• From previous example, the short-cut formula
is... ?
A Special Force of gravity: Weight
• The special force of a planet's gravity on a
surface object is called weight.
• From previous example, the short-cut formula
is... Fg = mg
A Special Force of gravity: Weight
• The special force of a planet's gravity on a
surface object is called weight.
• From previous example, the short-cut formula
is... Fg = mg
• g depends on the planet
A Special Force of gravity: Weight
• The special force of a planet's gravity on a
surface object is called weight.
• From previous example, the short-cut formula
is... Fg = mg
• g depends on the planet
• Earth g ≈ 10.0 N/kg [d]
• Moon g ≈ 1.6 N/kg [d]
• Mars g ≈ 3.7 N/kg [d]
A Special Force of gravity: Weight
Try this.
Find the weight of the 67.0 kg girl on the surface of the moon (g ≈ 1.60
N/kg [d]) and on the surface of Mars (g ≈ 3.70 N/kg [d]).
A Special Force of gravity: Weight
Try this.
Find the weight of the 67.0 kg girl on the surface of the moon (g ≈ 1.60
N/kg [d]) and on the surface of Mars (g ≈ 3.70 N/kg [d]).
Moon
Fg = mg = (67.0 kg) ( 1.60 N/kg [d] ) = 107 N [d]
Mars
Fg = mg = (67.0 kg) ( 3.70 N/kg [d] ) = 248 N [d]
Remember we calculated the girl's weight on the earth = 657 N
A Special Force of gravity: Weight
Try this.
Find the weight of the 67.0 kg girl on the surface of the moon (g ≈ 1.60
N/kg [d]) and on the surface of Mars (g ≈ 3.70 N/kg [d]).
Moon
Fg = mg = (67.0 kg) ( 1.60 N/kg [d] ) = 107 N [d]
Mars
Fg = mg = (67.0 kg) ( 3.70 N/kg [d] ) = 248 N [d]
Remember we calculated the girl's weight on the earth = 657 N
Does weight depend on our location in the universe?
A Special Force of gravity: Weight
Try this.
Find the weight of the 67.0 kg girl on the surface of the moon (g ≈ 1.60
N/kg [d]) and on the surface of Mars (g ≈ 3.70 N/kg [d]).
Moon
Fg = mg = (67.0 kg) ( 1.60 N/kg [d] ) = 107 N [d]
Mars
Fg = mg = (67.0 kg) ( 3.70 N/kg [d] ) = 248 N [d]
Remember we calculated the girl's weight on the earth = 657 N
Yes, weight depends on our location in the universe. Look at our
calculations above.
Two formulas for Weight
Two formulas for Weight
• In our calculation of weight, it is easy to use
the short-cut formula: Fg = mg
Two formulas for Weight
• In our calculation of weight, it is easy to use
the short-cut formula: Fg = mg
• But we also calculated weight using the more
general formula: Fg = GmM/d2
Where d = rp (planet radius)
Two formulas for Weight
• In our calculation of weight, it is easy to use
the short-cut formula: Fg = mg
• But we also calculated weight using the more
general formula: Fg = GmM/d2
Where d = rp (planet radius)
• So m|g| = GmM/rp2
Two formulas for Weight
• In our calculation of weight, it is easy to use
the short-cut formula: Fg = mg
• But we also calculated weight using the more
general formula: Fg = GmM/d2
Where d = rp (planet radius)
• So m|g| = GmM/rp2
|g| = ?
Two formulas for Weight
• In our calculation of weight, it is easy to use
the short-cut formula: Fg = mg
• But we also calculated weight using the more
general formula: Fg = GmM/d2
Where d = rp (planet radius)
• So m|g| = GmM/rp2
|g| = GM/rp2
Two formulas for Weight
• In our calculation of weight, it is easy to use
the short-cut formula: Fg = mg
• But we also calculated weight using the more
general formula: Fg = GmM/d2
Where d = rp (planet radius)
• So m|g| = GmM/rp2
|g| = GM/rp2 Memorize and be able to derive.
Try this: Venus has a planetary radius of 6.05 X 106 m and a
mass of 4.83 X 1024 kg. Find the gravitational field strength on
Venus. |g| = GM/rp2
Try this: Venus has a planetary radius of 6.05 X 106 m and a
mass of 4.83 X 1024 kg. Find the gravitational field strength on
the surface of Venus.
|g| = GM/rp2 = (6.67 X 10-11)(4.83 X 1024 ) / (6.05 X 106)2
= 8.80 N/kg
Try this: Venus has a planetary radius of 6.05 X 106 m and a
mass of 4.83 X 1024 kg. Find the gravitational field strength on
the surface of Venus.
|g| = GM/rp2 = (6.67 X 10-11)(4.83 X 1024 ) / (6.05 X 106)2
= 8.80 N/kg
• What would our 67.0 kg girl weigh on the surface of Venus?
Try this: Venus has a planetary radius of 6.05 X 106 m and a
mass of 4.83 X 1024 kg. Find the gravitational field strength on
the surface of Venus.
|g| = GM/rp2 = (6.67 X 10-11)(4.83 X 1024 ) / (6.05 X 106)2
= 8.80 N/kg
• What would our 67.0 kg girl weigh on the surface of Venus?
Fg = mg = (67.0 kg) ( 8.80 N/kg [d]) = 590 N
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg[d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2
note r varies
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2
note r varies
• Which quantities remain constant?
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2
note r varies
• Which quantities remain constant?
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
• Note r is centre-to-centre
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
• What is the proportionality?
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
• As we go from the earth's surface
to two earth radii above its surface,
r
rearth
how does “ r “ change?
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
• r↑3
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
• r↑3
•
r2
↑
32
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
• r↑3
•
r2
↑
32
r
rearth
|g| ↓ 9
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
• r↑3
•
r2
↑
32
r
rearth
|g| ↓ 9
• Therefore |g| = ?
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
• r↑3
•
r2
↑
32
r
rearth
|g| ↓ 9
• Therefore |g| = 10.0 N/kg / 9
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
r
• r↑3
•
r2
↑
rearth
32
|g| ↓ 9
• Therefore |g| = 10.0 N/kg / 9
= 1.11 N/kg [d]
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Proportionality Method:
|g| = GM/rp2 = GM/r2 = k / r2
|g| α 1 / r2
• r↑3
•
r2
↑
|g| ↓ 9
32
Note that the
proportionality
method did not
require that we
know the earth
radius!
• Therefore |g| = 10.0 N/kg / 9
= 1.11 N/kg [d]
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Equation Method:
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Equation Method:
|g| = GM/r2 note r varies
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Equation Method:
|g| = GM/r2 note r varies
= (6.67 X 10-11)(6.00 X 1024 )
(
?
)2
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Equation Method:
|g| = GM/r2 note r varies
= (6.67 X 10-11)(6.00 X 1024 )
( 3 X 6.39 X 106 )2
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Equation Method:
|g| = GM/r2 note r varies
= (6.67 X 10-11)(6.00 X 1024 )
( 3 X 6.39 X 106 )2
= 1.09 N/kg [d] very close to
r
rearth
other method
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Need to be given Mearth = 6.00 X 1024 kg
rearth = 6.39 X 106 m
2rearth
• Equation Method:
|g| = GM/r2 note r varies
= (6.67 X 10-11)(6.00 X 1024 )
( 3 X 6.39 X 106 )2
= 1.09 N/kg [d] very close to
r
rearth
other method
• Note we did have to know
the earth radius with this method.
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Using both methods, to two sigs, we found
|g| = 1.1 N/kg [d]
2rearth
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Using both methods, to two sigs, we found
|g| = 1.1 N/kg [d]
• How is this helpful?
2rearth
r
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Using both methods, to two sigs, we found
|g| = 1.1 N/kg [d]
• How is this helpful?
2rearth
• It tells us that if we have a one kg
mass two earth radii above the earth's
r
surface, it will weigh 1.1 N [d].
rearth
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• Using both methods, to two sigs, we found
|g| = 1.1 N/kg [d]
• How is this helpful?
2rearth
• It tells us that if we have a one kg
mass two earth radii above the earth's
r
surface, it will weigh 1.1 N [d].
• It also tells us that if we have a
rearth
projectile two earth radii above
earth's surface, it will have an
acceleration of 1.1 m/s2 [d]
earth
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• |g| = 1.1 N/kg [d]
• It tells us that if we have a one kg mass two earth radii above
the earth's surface, it will weigh 1.1 N [d].
• It also tells us that if we have a projectile two earth radii above
earth's surface, it will have an acceleration of 1.1 m/s2 [d]
• Note the projectile may fall straight down, move in a parabola,
or move in a circle or ellipse. It depends on the horizontal
constant velocity of the projectile.
Harder Example: The gravitational field strength on the
earth's surface is 10.0 N/kg [d]. Find the gravitational field
strength two earth radii above the earth's surface.
• |g| = 1.1 N/kg [d]
• It tells us that if we have a one kg mass two earth radii above
the earth's surface, it will weigh 1.1 N [d].
• It also tells us that if we have a projectile two earth radii above
earth's surface, it will have an acceleration of 1.1 m/s2 [d]
• Note the projectile may fall straight down, move in a parabola,
or move in a circle in UCM or even an ellipse. It depends on
the horizontal constant velocity of the projectile or orbiting
body.
• Your teacher will discuss Newton's thought experiment with
you!
More on the
gravitational field strength = g
More on the
gravitational field strength = g
• Force of a planet's gravity on a unit mass
More on the
gravitational field strength = g
• Force of a planet's gravity on a unit mass
• Or, g = Fg / m
More on the
gravitational field strength = g
• Force of a planet's gravity on a unit mass
• Or, g = Fg / m
• Si unit is N/kg
More on the
gravitational field strength = g
•
•
•
•
Force of a planet's gravity on a unit mass
Or, g = Fg / m
Si unit is N/kg
Also gives the acceleration of a projectile or
object moving with UCM in m/s2
More on the
gravitational field strength = g
•
•
•
•
Force of a planet's gravity on a unit mass
Or, g = Fg / m
Si unit is N/kg
Also gives the acceleration of a projectile or
object moving with UCM in m/s2
• There is also an analogous idea called electric
field strength = Fe / q or
= electrical force per unit + charge
Fields in General
Fields in General
• Something that has some value at every point
in space
Fields in General
• Something that has some value at every point
in space
earth
Fields in General
• Something that has some value at every point
in space
Could be a scalar field like the
temperature of the earth's
atmosphere at any point in
the air
earth
Fields in General
• Something that has some value at every point
in space
Could be a scalar field like the
temperature of the earth's
atmosphere at any point in
the air
Could be a vector field, like
the gravitational field of the
earth or the electrical field
around a charged particle
earth
Fields in General
• Something that has some value at every point
in space
Could be a scalar field like the
temperature of the earth's
atmosphere at any point in
the air
Could be a vector field, like
the gravitational field of the
earth or the electrical field
around a charged particle
Some fields, like gravitational and electrical fields are
fundamental fields: they are not made of anything
else nor are they properties of other quantities like
“temperature fields”
earth
Try this: Centre-to centre, the moon is sixty earth radii away from the earth.
Given the mass of the earth is 6.00 X 1024 kg and its radius is 6.39 X 106 m, use
proportionality and the equation method to find the earth's gravitational
field strength at sixty earth radii from its centre.
Assuming the moon moves around the earth in UCM, what is the moon's
centripetal acceleration about the earth?
Try this: Centre-to centre, the moon is sixty earth radii away from the earth.
Given the mass of the earth is 6.00 X 1024 kg and its radius is 6.39 X 106 m, use
proportionality and the equation method to find the earth's gravitational
field strength at sixty earth radii from its centre.
Assuming the moon moves around the earth in UCM, what is the moon's
centripetal acceleration about the earth?
Proportionality:
As before, |g| α 1 / r2
As r ↑ 60
r2 ↑ 602
|g| ↓ 602
|g| ↓ 602
|g| = 10.0/ 602
= 0.00278 N/kg
Equation:
|g| = GM/r2
= (6.67 X 10-11)(6.00 X 1024 )
(60 X6.39 X 106)2
= 0.00272 N/kg
This means a one kg mass at the moon's
distance from the earth would feel a 0.00278 N
force of earth's gravity on it. It is also the value
of the centripetal acceleration of the moon in
UCM at 0.0027 m/s2.
Mass ≠ Weight
Mass
Weight
Mass ≠ Weight
Mass
• Amount of matter
Weight
Mass ≠ Weight
Mass
• Amount of matter
Weight
• Force of a planet's
gravity on a surface
object
Mass ≠ Weight
Mass
• Amount of matter
• SI unit = kg
Weight
• Force of a planet's
gravity on a surface
object
Mass ≠ Weight
Mass
• Amount of matter
• SI unit = kg
Weight
• Force of a planet's
gravity on a surface
object
• SI unit = Newton
Mass ≠ Weight
Mass
• Amount of matter
• SI unit = kg
• Does not depend on
location in the
universe
Weight
• Force of a planet's
gravity on a surface
object
• SI unit = Newton
Mass ≠ Weight
Mass
• Amount of matter
• SI unit = kg
• Does not depend on
location in the
universe
Weight
• Force of a planet's
gravity on a surface
object
• SI unit = Newton
• Does depend on
location in the
universe
Mass ≠ Weight
•
•
•
•
Mass
Amount of matter
SI unit = kg
Does not depend on
location in the
universe
scalar
Weight
• Force of a planet's
gravity on a surface
object
• SI unit = Newton
• Does depend on
location in the
universe
Mass ≠ Weight
•
•
•
•
Mass
Amount of matter
SI unit = kg
Does not depend on
location in the
universe
scalar
•
•
•
•
Weight
Force of a planet's
gravity on a surface
object
SI unit = Newton
Does depend on
location in the
universe
vector
Homework Practice:
New textbook
• New textbook: Read p70-p73 Define inertia. Re-do sample
problem #1 on p72
• do practice Q1 p72 check ans same page
• Re-do sample problem #1 p73 Q1, Q2 p74
check ans same page
• Try p 99 Q3,Q10,Q11 p100 Q1-Q5 Q10,Q11 check ans p715
• Review p288-295 Review and redo sample problems p290-292, Do
Q1-Q3 p293 check ans same page review and re-do sample
problems p294 Do Q1,2 p295 Do Q1-3, Q5-Q7 Q13 p296 Check ans
p715
Homework Practice:
Old textbook
• Read p77-p86 Define inertia, static equilibrium Review and redo
sample problems #1,#2 p79 Do Q2 p80 Q6 p81 check ans same
page
• Review and re-do sample problem #4 p82-83 Do Q10,11,13 p83
check ans same page Do q16 a to d q17,q18 p84 check ans same
page
• Review and re-do sample problem #5 p85 Do Q4,6,8 p87 check ans
p 781 Try p115 Q5,6,8,9,12,15 check p782
• Read p139-p143 Review and redo sample problem p143 do Q8-Q13
p143 check ans same page p144 Q4,5,6 check p782
• Read p274-p276 Review and redo sample problems p275
• Do Q2-Q6 p276 Check ans same page Do Q7 p277 chk p783