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Newton’s Laws of Motion
Chapter 4
Newton’s First
Law of Motion
Force
A force is a push or pull. An object at rest needs
a force to get it moving; a moving object needs
a force to change its velocity.
Force
Force is a vector, having both magnitude and
direction. The magnitude of a force can be
measured using a spring scale.
Newton’s First Law of Motion
This is Newton’s first law, which is often called
the law of inertia:
Every object continues in its state of rest, or of
uniform velocity in a straight line, as long as
no net force acts on it.
Newton’s First Law of Motion
It may seem as though it takes a force to keep an
object moving.
Example: Push your book across a table—when
you stop pushing, it stops moving. Why?
Your book stops because the force of friction
stops it. It doesn’t take a force to keep an object
moving in a straight line—it takes a force to
change its motion.
ConcepTest 4.1b Newton’s First Law II
A hockey puck
slides on ice at
constant velocity.
What is the net
force acting on
the puck?
1) more than its weight
2) equal to its weight
3) less than its weight but more than zero
4) depends on the speed of the puck
5) zero
ConcepTest 4.1b Newton’s First Law II
A hockey puck
slides on ice at
constant velocity.
What is the net
force acting on
the puck?
1) more than its weight
2) equal to its weight
3) less than its weight but more than zero
4) depends on the speed of the puck
5) zero
The puck is moving at a constant velocity, and
therefore it is not accelerating. Thus, there must
be no net force acting on the puck.
ConcepTest 4.1c Newton’s First Law III
You put your book on
the bus seat next to you.
When the bus stops
suddenly, the book
slides forward off the
seat. Why?
1) a net force acted on it
2) no net force acted on it
3) it remained at rest
4) it did not move, but only seemed to
5) gravity briefly stopped acting on it
ConcepTest 4.1c Newton’s First Law III
You put your book on
the bus seat next to you.
When the bus stops
suddenly, the book
slides forward off the
seat. Why?
1) a net force acted on it
2) no net force acted on it
3) it remained at rest
4) it did not move, but only seemed to
5) gravity briefly stopped acting on it
The book was initially moving forward (because it was
on a moving bus). When the bus stopped, the book
continued moving forward, which was its initial state of
motion, and therefore it slid forward off the seat.
Follow-up: What is the force that usually keeps the book on the seat?
ConcepTest 4.4a Off to the Races I
From rest, we step on the gas of our
Ferrari, providing a force F for 4 secs,
speeding it up to a final speed v. If the
applied force were only ½F, how long
would it have to be applied to reach
the same final speed?
1) 16 s
2) 8 s
3) 4 s
4) 2 s
5) 1 s
F
v
ConcepTest 4.4a Off to the Races I
From rest, we step on the gas of our
Ferrari, providing a force F for 4 secs,
speeding it up to a final speed v. If the
applied force were only ½F, how long
would it have to be applied to reach
the same final speed?
1) 16 s
2) 8 s
3) 4 s
4) 2 s
5) 1 s
In the first case, the acceleration
acts over time T = 4 s to give
velocity v = aT. In the second
case, the force is half, therefore
the acceleration is also half, so
to achieve the same final speed,
the time must be doubled.
F
v
Mass
Mass is the measure of inertia of an object. In
the SI system, mass is measured in kilograms.
Mass is not weight!
Mass is a property of an object. Weight is the
force exerted on that object by gravity.
Example: If you go to the Moon, whose
gravitational acceleration is about 1/6 g, you will
weigh much less. Your mass, however, will be the
same.
Newton’s Second
Law of Motion
Newton’s Second Law of Motion
Newton’s second law is the relation between
acceleration and force. Acceleration is
proportional to force and inversely proportional to
mass.
It takes a force to change either the
direction or the speed of an object.
More force means more
acceleration; the same force
exerted on a more massive object
will yield less acceleration.
Newton’s Second Law of Motion
Since force is a vector, so  F  ma is true along
each coordinate axis.
The unit of force in the SI
system is the newton (N).
Note that the pound is a
unit of force, not of mass,
and can therefore be
equated to Newtons but
not to kilograms.
ConcepTest 4.3 Truck on Frozen Lake
A very large truck sits on a
frozen lake. Assume there
is no friction between the
tires and the ice. A fly
suddenly smashes against
the front window. What
will happen to the truck?
1) it is too heavy, so it just sits there
2) it moves backward at constant
speed
3) it accelerates backward
4) it moves forward at constant speed
5) it accelerates forward
ConcepTest 4.3 Truck on Frozen Lake
A very large truck sits on a
frozen lake. Assume there
is no friction between the
tires and the ice. A fly
suddenly smashes against
the front window. What
will happen to the truck?
1) it is too heavy, so it just sits there
2) it moves backward at constant
speed
3) it accelerates backward
4) it moves forward at constant speed
5) it accelerates forward
When the fly hit the truck, it exerted a force on the truck
(only for a fraction of a second). So, in this time period,
the truck accelerated (backward) up to some speed. After
the fly was squashed, it no longer exerted a force, and the
truck simply continued moving at constant speed.
ConcepTest 4.8 On the Moon
An astronaut on Earth kicks
a bowling ball and hurts his
foot. A year later, the same
astronaut kicks a bowling
1) more
2) less
3) the same
ball on the Moon with the
same force. His foot hurts...
Ouch!
ConcepTest 4.8 On the Moon
An astronaut on Earth kicks
a bowling ball and hurts his
foot. A year later, the same
astronaut kicks a bowling
1) more
2) less
3) the same
ball on the Moon with the
same force. His foot hurts...
The masses of both the bowling ball
and the astronaut remain the same, so
his foot feels the same resistance and
hurts the same as before.
Follow-up: What is different about
the bowling ball on the Moon?
Ouch!
ConcepTest 4.5 Force and Mass
A force F acts on mass M for a
time interval T, giving it a final
speed v. If the same force acts
for the same time on a different
mass 2M, what would be the
final speed of the bigger mass?
1) 4v
2) 2v
3) v
4)
1
2
v
5)
1
4
v
ConcepTest 4.5 Force and Mass
A force F acts on mass M for a
time interval T, giving it a final
speed v. If the same force acts
for the same time on a different
mass 2M, what would be the
final speed of the bigger mass?
1) 4v
2) 2v
3) v
4)
1
2
v
5)
1
4
v
In the first case, the acceleration acts over time T to give
velocity v = aT. In the second case, the mass is doubled,
so the acceleration is cut in half; therefore, in the same
time T, the final speed will only be half as much.
Follow-up: What would you have to do to get 2M to reach speed v ?
ConcepTest 4.6 Force and Two Masses
A force F acts on mass m1 giving acceleration a1.
The same force acts on a different mass m2 giving
acceleration a2 = 2a1. If m1 and m2 are glued
together and the same force F acts on this
combination, what is the resulting acceleration?
F
F
F
m1
a1
m2
m2 m1
a2 = 2a1
a
3
3
4
3
2
a1
3)
1
2
a1
4)
4
3
a1
5)
2
3
a1
1)
2)
a1
ConcepTest 4.6 Force and Two Masses
A force F acts on mass m1 giving acceleration a1.
The same force acts on a different mass m2 giving
acceleration a2 = 2a1. If m1 and m2 are glued
together and the same force F acts on this
combination, what is the resulting acceleration?
F
m1
3
4
3
2
a1
3)
1
2
a1
4)
4
3
a1
5)
2
3
a1
1)
2)
a1
a1
F = m1 a1
a2 = 2a1
m2
F
F = m2 a2 = (1/2 m1 )(2a1 )
Mass m2 must be ( 1 m1) because its
2
acceleration was 2a1 with the same
force. Adding the two masses
together gives ( 32 )m1, leading to an
F
m2 m1
a3
acceleration of ( 32 )a1 for the same
applied force.
F = (3/2)m1 a3 => a3 = (2/3) a1
Newton’s Third
Law of Motion
Newton’s Third Law of Motion
Any time a force is exerted on an object, that
force is caused by another object.
Newton’s third law:
Whenever one object exerts a force on a second
object, the second exerts an equal force in the
opposite direction on the first.
Newton’s Third Law of Motion
Rocket propulsion can also be explained using
Newton’s third law: hot gases from combustion
spew out of the tail of the rocket at high speeds.
The reaction force is what propels the rocket.
Note that the rocket does
not need anything to
“push” against.
Newton’s Third Law of Motion
Helpful notation: the first subscript is the object
that the force is being exerted on; the second is
the source.
Newton’s Third Law of Motion
Conceptual Example 4-5: Third law clarification.
Michelangelo’s assistant has been assigned the task of moving a block of
marble using a sled. He says to his boss, “When I exert a forward force
on the sled, the sled exerts an equal and opposite force backward. So
how can I ever start it moving? No matter how hard I pull, the backward
reaction force always equals my forward force, so the net force must be
zero. I’ll never be able to move this load.” Is he correct?
Weight
Weight is the force exerted on an object
by gravity. Close to the surface of the
Earth, where the gravitational force is
nearly constant, the weight of an object
of mass m is:
where
The Normal Force
An object at rest must have no net force on it. If it is
sitting on a table, the force of gravity is still there; what
other force is there?
The force exerted perpendicular to a surface is
called the normal force. It is
exactly as large as needed to
balance the force from the
object. (If the required force gets
too big, something breaks!)
Using Free-Body Diagrams with Newton’s Laws
1. Draw a sketch.
2. For one object, draw a free-body diagram,
showing all the forces acting on the object.
Make the magnitudes and directions as
accurate as you can. Label each force. If
there are multiple objects, draw a separate
diagram for each one.
3. Resolve vectors into components.
4. Apply Newton’s second law to each
component.
5. Solve.
Problem Solving—A General Approach
1. Read the problem carefully; then read it again.
2. Draw a sketch, and then a free-body diagram.
3. Choose a convenient coordinate system.
4. List the known and unknown quantities; find relationships
between the knowns and the unknowns.
5. Estimate the answer.
6. Solve the problem without putting in any numbers
(algebraically); once you are satisfied, put the numbers in.
7. Keep track of dimensions.
8. Make sure your answer is reasonable.
ConcepTest 4.9a Going Up I
A block of mass m rests on the floor
1) N > mg
of an elevator that is moving upward
2) N = mg
at constant speed. What is the
3) N < mg (but not zero)
relationship between the force due
4) N = 0
to gravity and the normal force on
5) depends on the size of the
elevator
the block?
v
m
ConcepTest 4.9a Going Up I
A block of mass m rests on the floor
1) N > mg
of an elevator that is moving upward
2) N = mg
at constant speed. What is the
3) N < mg (but not zero)
relationship between the force due
4) N = 0
to gravity and the normal force on
5) depends on the size of the
elevator
the block?
The block is moving at constant speed, so
it must have no net force on it. The forces
v
on it are N (up) and mg (down), so N = mg,
just like the block at rest on a table.
m
ConcepTest 4.9b Going Up II
A block of mass m rests on the
floor of an elevator that is
accelerating upward. What is the
relationship between the force
due to gravity and the normal
force on the block?
1) N > mg
2) N = mg
3) N < mg (but not zero)
4) N = 0
5) depends on the size of the
elevator
a
m
ConcepTest 4.9b Going Up II
A block of mass m rests on the
floor of an elevator that is
accelerating upward. What is the
relationship between the force
due to gravity and the normal
force on the block?
1) N > mg
2) N = mg
3) N < mg (but not zero)
4) N = 0
5) depends on the size of the
elevator
The block is accelerating upward, so
it must have a net upward force. The
forces on it are N (up) and mg (down),
so N must be greater than mg in order
to give the net upward force!
Follow-up: What is the normal force if
the elevator is in free fall downward?
N
m
a>0
mg
S F = N – mg = ma > 0
\ N > mg
ConcepTest 4.10 Normal Force
Below you see two cases: a
physics student pulling or
pushing a sled with a force
F that is applied at an angle
q. In which case is the
normal force greater?
1) case 1
2) case 2
3) it’s the same for both
4) depends on the magnitude of
the force F
5) depends on the ice surface
Case 1
Case 2
ConcepTest 4.10 Normal Force
Below you see two cases: a
physics student pulling or
pushing a sled with a force
F that is applied at an angle
q. In which case is the
normal force greater?
1) case 1
2) case 2
3) it’s the same for both
4) depends on the magnitude of
the force F
5) depends on the ice surface
Case 1
In case 1, the force F is pushing down (in
addition to mg), so the normal force
needs to be larger. In case 2, the force F
is pulling up, against gravity, so the
normal force is lessened.
Case 2
ConcepTest 4.12 Climbing the Rope
When you climb up a rope,
1) this slows your initial velocity, which
is already upward
the first thing you do is pull
2) you don’t go up, you’re too heavy
down on the rope. How do
3) you’re not really pulling down—it
just seems that way
you manage to go up the
rope by doing that??
4) the rope actually pulls you up
5) you are pulling the ceiling down
ConcepTest 4.12 Climbing the Rope
When you climb up a rope,
1) this slows your initial velocity, which
is already upward
the first thing you do is pull
2) you don’t go up, you’re too heavy
down on the rope. How do
3) you’re not really pulling down—it
just seems that way
you manage to go up the
rope by doing that??
4) the rope actually pulls you up
5) you are pulling the ceiling down
When you pull down on the rope, the rope pulls up on
you!! It is actually this upward force by the rope that
makes you move up! This is the “reaction” force (by the
rope on you) to the force that you exerted on the rope.
And voilá, this is Newton’s Third Law.
ConcepTest 4.16a Tension I
You tie a rope to a tree and
you pull on the rope with a
force of 100 N. What is the
tension in the rope?
1) 0 N
2) 50 N
3) 100 N
4) 150 N
5) 200 N
ConcepTest 4.16a Tension I
You tie a rope to a tree and
you pull on the rope with a
force of 100 N. What is the
tension in the rope?
1) 0 N
2) 50 N
3) 100 N
4) 150 N
5) 200 N
The tension in the rope is the force that the rope
“feels” across any section of it (or that you would
feel if you replaced a piece of the rope). Because
you are pulling with a force of 100 N, that is the
tension in the rope.
ConcepTest 4.16b Tension II
Two tug-of-war opponents each
1) 0 N
pull with a force of 100 N on
2) 50 N
opposite ends of a rope. What is
3) 100 N
the tension in the rope?
4) 150 N
5) 200 N
ConcepTest 4.16b Tension II
Two tug-of-war opponents each
1) 0 N
pull with a force of 100 N on
2) 50 N
opposite ends of a rope. What is
3) 100 N
the tension in the rope?
4) 150 N
5) 200 N
This is literally the identical situation to the
previous question. The tension is not 200 N !!
Whether the other end of the rope is pulled by a
person, or pulled by a tree, the tension in the rope
is still 100 N !!
ConcepTest 4.16c Tension III
You and a friend can
1) you and your friend each pull on
opposite ends of the rope
each pull with a force
2) tie the rope to a tree, and you both
pull from the same end
of 20 N. If you want
to rip a rope in half,
3) it doesn’t matter—both of the above
are equivalent
what is the best way?
4) get a large dog to bite the rope
ConcepTest 4.16c Tension III
You and a friend can
each pull with a force
of 20 N. If you want
to rip a rope in half,
what is the best way?
1) you and your friend each pull on
opposite ends of the rope
2) tie the rope to a tree, and you both
pull from the same end
3) it doesn’t matter—both of the above
are equivalent
4) get a large dog to bite the rope
Take advantage of the fact that the tree can
pull with almost any force (until it falls down,
that is!). You and your friend should team up
on one end, and let the tree make the effort on
the other end.
ConcepTest 4.14a Collision Course I
1) the car
A small car collides
with a large truck.
Which experiences the
greater impact force?
2) the truck
3) both the same
4) it depends on the velocity of each
5) it depends on the mass of each
ConcepTest 4.14a Collision Course I
1) the car
A small car collides
with a large truck.
Which experiences the
greater impact force?
2) the truck
3) both the same
4) it depends on the velocity of each
5) it depends on the mass of each
According to Newton’s Third Law, both vehicles
experience the same magnitude of force.
ConcepTest 4.14b Collision Course II
In the collision
between the car and
the truck, which has
the greater
acceleration?
1) the car
2) the truck
3) both the same
4) it depends on the velocity of each
5) it depends on the mass of each
ConcepTest 4.14b Collision Course II
In the collision
between the car and
the truck, which has
the greater
acceleration?
We have seen that both
vehicles experience the
same magnitude of force.
But the acceleration is
given by F/m so the car
has the larger acceleration,
because it has the smaller
mass.
1) the car
2) the truck
3) both the same
4) it depends on the velocity of each
5) it depends on the mass of each
ConcepTest 4.15a Contact Force I
If you push with force F on either
the heavy box (m1) or the light box
(m2), in which of the two cases
below is the contact force between
the two boxes larger?
1) case A
2) case B
3) same in both cases
A
m2
F
m1
B
m2
m1
F
ConcepTest 4.15a Contact Force I
If you push with force F on either
the heavy box (m1) or the light box
(m2), in which of the two cases
below is the contact force between
the two boxes larger?
1) case A
2) case B
3) same in both cases
The acceleration of both masses together
A
is the same in either case. But the contact
force is the only force that accelerates m1
in case A (or m2 in case B). Because m1 is
m2
F
m1
the larger mass, it requires the larger
B
contact force to achieve the same
acceleration.
m2
m1
F
ConcepTest 4.15b Contact Force II
Two blocks of masses 2m and m
are in contact on a horizontal
frictionless surface. If a force F is
applied to mass 2m, what is the
force on mass m ?
1) 2F
2) F
3)
4)
5)
1
2F
1
F
3
1
F
4
F
2m
m
ConcepTest 4.15b Contact Force II
Two blocks of masses 2m and m
are in contact on a horizontal
frictionless surface. If a force F is
applied to mass 2m, what is the
force on mass m ?
1) 2F
2) F
3)
4)
5)
1
2F
1
F
3
1
F
4
The force F leads to a specific
acceleration of the entire system. In
order for mass m to accelerate at the
same rate, the force on it must be
smaller! How small?? Let’s see...
F
2m
m
ConcepTest 4.11 On an Incline
Consider two identical
blocks, one resting on a flat
surface and the other
resting on an incline. For
which case is the normal
force greater?
1) case A
2) case B
3) both the same (N = mg)
4) both the same (0 < N < mg)
5) both the same (N = 0)
ConcepTest 4.11 On an Incline
Consider two identical
blocks, one resting on a flat
surface and the other
resting on an incline. For
which case is the normal
force greater?
1) case A
2) case B
3) both the same (N = mg)
4) both the same (0 < N < mg)
5) both the same (N = 0)
In case A, we know that N = W.
y
In case B, due to the angle of
the incline, N < W. In fact, we
N
f
can see that N = W cos(q).
q
q
W
Wy
x
Motion with Variable
Acceleration
Motion with Constant Acceleration
If an object is moving with constant acceleration a,
then:
v(t )   adt
v(t )  v0  at
x(t )   v(t )dt
x(t )  x0  v0t  at
1
2
2
Variable Acceleration
If acceleration changes as a function of time, the
previous definitions still apply, but the resulting
equations of motion may look…well…nasty!
Example: Suppose an object has an acceleration
a(t )  Ae .t Find the objects position function.
A t
t
v(t )   a(t )dt  A e dt
v(t )   e  b

A t
x(t )   v(t )dt   ( e  b)dt

A t
x(t )  2 e  bt  c

Velocity-Dependent Forces: Drag
When an object moves through a fluid, it
experiences a drag force that depends on
the velocity of the object.
For small velocities, the force is
approximately proportional to the velocity;
for higher speeds, the force is approximately
proportional to the square of the velocity.
Terminal Velocity
If the drag force on a falling object
is dependent on its velocity, the
object gradually slows until the
drag force and the gravitational
force are equal. Then it falls with
constant velocity, called the
terminal velocity.
Problem 4.17 – Can cars “stop on a dime”?
Calculate the acceleration of a 1400-kg car if it can
stop from 35 km/h on a dime (diameter = 1.7 cm).
How many g’s is this? What is the force felt by the
68-kg driver?
First and foremost, we need to convert all given
quantities to SI units.
35 km/h ≈ 9.7222222 m/s;
1.7 cm = 0.017 m
Problem 4.17 (cont.)
We were not told that the acceleration varies, so
we will assume it is constant. Since we are
assuming constant acceleration, we can actually
use the projectile motion equation to solve the
problem.
Since we are not given time, but rather a distance,
we can use the equation:
v 2  v02  2a( x  x0 )
Solving for acceleration, we get:
v 2  v02
a
2( x  x0 )
Problem 4.17 (cont.)
Now we just substitute our given values, and find
that:
3
2
a  2.8 10 m/s
Dividing that result by 9.80 m/s2, we see that:
a  2.8 10 g ' s
2
So, what is the force felt by the driver? Using
Newton’s second law, we learn that the driver
experienced 1.9105 N.
Question 4.24 – Bear Wants my Lunch
Question 4.24 – Bear Wants my Lunch
Here we are not concerned with the acceleration of the
backpack, but rather how much force does the person
need to apply to keep the backpack at a certain height.
y
T1
T2
q
q
x
w
We are not told where the
back pack is along the
rope, or what angle each
side of the rope makes
with the horizontal, so we
are going to assume the
back pack is midway along
the rope, and that the
angle of the rope is the
same on both sides.
Question 4.24 – Bear Wants my Lunch
We always assume frictionless conditions (unless
otherwise stated), so the branch through which the
men is pulling the rope lets the rope slide freely.
Furthermore, we always assume that the rope does not
stretch, so the magnitude of the tension on the rope
does not change as it bends around the branch.
Therefore:
| T1 || F |
Also, since it is the same rope all through out:
| T2 || T1 || F |
Question 4.24 – Bear Wants my Lunch
Before we apply Newton’s laws to this problem, it is a
good idea to write down all vectors as the sum of their
components.
T1   F cos q ˆi  F sin q ˆj
T2  F cos q ˆi  F sin q ˆj
w  mg ˆj
Now we can apply Newton’s 2nd law to the vertical
components to get our answer:
F
y
 ma
2F sin q  mg  0
mg
F
2 sin q
Question 4.24 – Bear Wants my Lunch
mg
F
2 sin q
Notice that the angle q appears in the denominator. As
the person lifts the backpack higher, the angle q
becomes smaller, which means that sin(q ) becomes
smaller, which in turn makes F larger. Therefore, the
force needed to raise the backpack increases as the
backpack gets higher, and in fact, it is impossible to pull
it hard enough so that it doesn’t sag at all.