Transcript LAWS OF Chapter Six: Laws of Motion 6.1 Newton’s First Law 6.2 Newton’s Second Law 6.3 Newton’s Third Law and Momentum.

LAWS OF
Chapter Six: Laws of Motion
6.1 Newton’s First Law
6.2 Newton’s Second Law
6.3 Newton’s Third Law and
Momentum
Chapter 6.1 Learning Goals
Describe how forces cause changes
in motion.
Demonstrate and describe Newton’s
first law.
Explain the meaning of net force.
6.1 Force changes motion
 A force is a push or pull, or any
action that is able to change motion.
6.1 Law of inertia
 Newton’s first law says that objects
continue the motion they already
have unless they are acted on by a
net force.
 If the net force is zero, an object at
rest will stay at rest.
 If an object is acted upon by
unbalanced forces, its motion will
change.
6.1 Net force
Newton’s first law is
often written in
terms of the net
force:
“An object at rest
will stay at rest and
an object in motion
will continue in
motion at constant
According to these vectors, in
velocity UNLESS
there is a net force.” what direction is the net force?
6.1 Force changes motion
 Forces can be used to increase or
decrease the speed of an object, or
to change the direction an object is
moving.
6.1 Law of inertia
 Inertia is the
property of an
object that resists
changes in motion.
 Objects with more
mass have more
inertia and are
more resistant to
changes in their
motion.
Which ball has more
inertia?
Solving Problems
A car drives along the highway at constant
velocity. Find the car’s weight and the
friction force if the engine produces a
force of 2,000 newtons between the tires
and the road and the normal force on the
car is 12,000 N.
Solving Problems
1. Looking for:
 …weight of car in newtons, force due to
friction
2. Given:
 …ForceN = 12,000N (up);
 …ForceE = 2,000N (forward)
3. Relationships:
 Newton’s 1st Law:
 net force = zero at constant velocity; so
ForceN = ForceW and ForceE = ForceF
4. Solution
FE = 200 N
FN = 12,000N
 Draw a free body
diagram.
 There is no net force
upward, so the weight of
the car is an equal
FF = -200 N
downward force of
−12,000 N.
 The forward engine
force balances the
friction force so the
friction force is −2,000 N.
FW = -12,000N
Solving Problems
Chapter Six: Laws of Motion
6.1 Newton’s First Law
6.2 Newton’s Second Law
6.3 Newton’s Third Law and
Momentum
Chapter 6.2 Learning Goals
Define Newton’s second law by
relating force, mass, and acceleration.
Apply Newton’s second law
quantitatively.
Describe the relationship between net
force and acceleration.
Investigation 6A
Newton’s First and Second Laws
Key Question:
 What is the relationship between force
and motion?
6.2 Newton’s second law
 Newton’s first law tells us that
motion cannot change without a
net force.
 According to Newton’s second law,
the amount of acceleration
depends on both the force and the
mass.
6.2 The newton
 The S.I. unit of
force (newton) is
defined by the
second law.
 A newton is the
amount of force
needed to
accelerate a 1 kg
object by 1m/s.
6.2 Newton’s second law
 There are three main ideas related
to Newton’s Second Law:
1. Acceleration is the result of
unbalanced forces.
2. A larger force makes a
proportionally larger acceleration.
3. Acceleration is inversely
proportional to mass.
6.2 Newton’s second law
 Unbalanced forces cause changes in
speed, direction, or both.
6.2 Acceleration and force
The second law says
that acceleration is
proportional to force.
If force is increased
or decreased,
acceleration will be
increased or
decreased by the
same factor.
6.2 Acceleration and direction
Another important factor of the second law
is that the acceleration is always in the
same direction as the net force.
6.2 Acceleration and mass
The greater the mass, the smaller the
acceleration for a given force.
This means acceleration is inversely
proportional to mass.
6.2 Acceleration, force and mass
The acceleration caused by a force is
proportional to force and inversely
proportional to mass.
The stronger the
force on an object,
the greater its
acceleration.
 Force is directly
proportional to
acceleration.
 If twice the force
is applied, the
acceleration is
twice as great.
The greater the
mass, the smaller
the acceleration
for a given force.
 Mass is
inversely
related to force.
 An object with
twice the mass
will have half
the acceleration
if the same
force is applied.

6.2 Applying the second law
Keep the following
important ideas in mind:
1. The net force is what
causes acceleration.
2. If there is no acceleration,
the net force must be
zero.
3. If there is acceleration,
there must also be a net
force.
4. The force unit of newtons
is based on kilograms,
meters, and seconds.
Solving Problems
 A car has a mass of 1,000 kilograms.
If a net force of 2,000 N is exerted
on the car, what is its acceleration?
1. Looking for:
 …car’s acceleration
2. Given
 …mass = 1,000 kg; net force = 2,000 N
3. Relationships:
 a=F/m
4. Solution:
 2, 000 N ÷ 1,000 kg = 2 N/kg = 2 m/s2
Chapter Six: Laws of Motion
6.1 Newton’s First Law
6.2 Newton’s Second Law
6.3 Newton’s Third Law and
Momentum
Chapter 6.3 Learning Goals
Describe action-reaction force pairs.
Explain what happens when objects
collide in terms of Newton’s third law.
Apply the law of conservation of
momentum when describing the
motion of colliding objects.
Investigation 6B
Newton’s Third Law
Key Question:
What happens when equal and opposite forces
are exerted on a pair of Energy Cars?
6.3 Newton’s Third Law
 Newton’s Third
Law (actionreaction) applies
when a force is
placed on any
object, such as a
basketball.
6.3 The Third Law: Action/Reaction
 Newton’s Third Law
states that every action
force creates a reaction
force that is equal in
strength and opposite
in direction.
 There can never be a
single force, alone,
without its actionreaction partner.
6.3 The Third Law: Action/Reaction
 It doesn’t matter
which force you call
the action and which
the reaction.
 The forces do not
cancel because we
One force acts on the
ball, and the other force
can only cancel
acts on the hand.
forces acting on the
same object.
6.3 Action and reaction
 When sorting out
action and
reaction forces it
is helpful to
examine or draw
diagrams.
Here the action force is on the ________________, and
the reaction force is on the _______________.
Solving Problems
A woman with a
weight of 500
newtons is sitting
on a chair.
Describe one
action-reaction pair
of forces in this
situation.
Solving Problems
1.

2.

3.

4.



Looking for:
Fc = 500 N
…pair of action-reaction forces
Given
…girl’s forceW = -500 N (down)
Fw = -500 N
Relationships:
Action-reaction forces are equal and opposite and act on
different objects.
Solution
Draw a free body diagram
The downward force of 500 N exerted by the woman on
the chair is an action.
Therefore, the chair acting on the woman provides an
upward force of 500 N and is the reaction.
6.3 Collisions
 Newton’s third law tells us that any time
two objects hit each other, they exert
equal and opposite forces on each other.
 The effect of the force is not always the
same.
6.3 Momentum
Momentum is the mass of a object
times its velocity.
The units for momentum are
kilogram-meter per second (kg·m/s).
6.3 Momentum
The law of
conservation of
momentum states that
as long as the
interacting objects are
not influenced by
outside forces (like
friction) the total
amount of momentum
is constant or does
not change.
6.3 Momentum
We use positive and
negative numbers to show
opposite directions.
The result of a
skateboarder
throwing a 1-kg ball at
a speed of -20 m/sec
is that he and the
skateboard with a
total mass of 40 kg
move backward at a
speed of +0.5 m/sec
(if you ignore friction).
6.3 Collisions
When a large truck
hits a small car, the
forces are equal.
The small car
experiences a much
greater change in
velocity much more
rapidly than the big
truck.
Which vehicle ends up
with more damage?
Solving Problems
If an astronaut in
space were to release
a 2-kilogram wrench
at a speed of 10 m/s,
the astronaut would
move backward at
what speed?
The astronaut’s mass
is 100 kilograms.
Solving Problems
1. Looking for:
 … the velocity of the astronaut (backward)
2. Given
 …velocity1 = 10 m/s; mass1= 2 kg;
 ...mass2 = 100 kg;
3. Relationships:
 m1v1 = m2v2
4. Solution
 Draw a free body diagram.
Investigation 6C
Collisions
Key Question:
Why do things bounce back when they collide?
Forensic Engineering
Human bodies are not
designed to handle the
impact of crashing into a
stationary object after
traveling through space at
the speed of a car.
The study of how vehicles move before,
during, and after a collision is called
vehicular kinematics.