# Newton's Laws

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```					Newton’s Laws
Newton-1: Law of Inertia
• Newton’s First Law
• An object subject to no external forces is
at rest or moves with a constant velocity
if viewed from an inertial reference
frame.
– If no net forces act, there is no acceleration.

F net  0  a  0
Newton 2 Cart Demo

Newton 2 Slide 3
Minimizing Friction

• One way to minimize friction is to float
objects on a cushion of air. You will use an
air track in lab this week.
• Any horizontal force exerted on the cart
is the net force acting on the cart.
Acceleration vs Force
• Experiments show that
Acceleration is proportional to force
F1  a1

2 F1  2a1
Acceleration vs Mass
• Acceleration is inversely proportional to
mass
M  a1

2M  1 a1
2
Newton-2
• Combining these two observations gives
F                                              F
a       and by an appropriate choice of units   a
m                                              m

Or, more familiarly,

Units: Mass has SI units of kg,
and acceleration has SI units of m/s2.
The SI unit of force is:
1 newton = 1 N = 1kg m/s2.
Newton-2 (Second Law of Motion)

Fnet
a
m
The Unit Newton
• Newton’s 2nd law says a = Fnet / m

• So Fnet = ma by algebra
1N  1kg  m / s 2

• 1 Newton of force is the amount of force
necessary to accelerate 1 kg at 1 m/s/s

• This is why 1 kg weighs 9.8 N on Earth,
because the acceleration due to gravity
on earth (g) is 9.8 m/s/s
Example: Accelerated Mass
m1

A net force of 3.0 N produces an acceleration of 2.0
m/s2 on an object of unknown mass.
What is the mass of the object?

F   (3.0 N)
m1  =         2
 1.5 kg
a1 (2.0 m/s )
Newton 2 Practice
The skater’s mass is 25 kg.

Force           Mass       Acceleration
100 N
200 N
10 m/s/s

Complete the table below for a 50 N resistance force.
Force        Acceleration
50 N
100N
200 N
Friction: force that resists motion
• force between the surfaces of two
objects
• Examples: sliding friction, air resistance
• Friction acts in the direction opposite to
motion
Friction Example
• A force of 5 N is used to drag a 1 kg object across
the lecture table at a constant velocity of 1 m/s.
What is the friction force opposing the motion?
– What is the acceleration of the object?
• Velocity constant – acceleration = 0
– What is the net force on the object?
• Acceleration = 0  Fnet = 0
– What is the force of friction opposing the motion?
5N                            FN = 10 N
Ff = 5 N                 5N

Fg = 10 N
Friction Example
• A force of 5 N is used to drag a 1 kg object across
the lecture table at a constant velocity of 1 m/s.
What is the friction force opposing the motion?
– What is the force of friction opposing the
motion?
•5N
• Now a force of 13 N is applied to the object. What
is its acceleration?
FN = 10 N
• Fnet = 13 N – 5 N = 8 N
• a = Fnet/m = 8 N/1 kg = 8 m/s2                 13 N
Ff = 5 N

Fg = 10 N
Kinematics & Dynamics Combo
You are stranded in space, away from your
spaceship. Fortunately, you have a propulsion
unit that provides a constant net force F for
3.0 s. You turn it on, and after 3.0 s you have
moved 2.25 m.
If your mass is 68 kg, find F.

x  x0  v0t  1 at 2  1 at 2
2        2
2 x 2(2.25 m)
a 2                    0.50 m/s 2
t       (3.0 s) 2
a  0.50 m/s 2

Fnet  ma  (68 kg)(0.50 m/s 2 )  34 N
Since there is only one force, we call that direction
positive x and only worry about magnitudes.
In which direction does the object
accelerate?

1. a.
2. b.
3. c.
74%

4. d.
5. e.
19%

6%
0%                    0%

1    2     3     4     5
In which direction does the object
accelerate?

1. a.
2. b.
100%

3. c.
4. d.
5. e.

0%   0%   0%          0%

1    2    3      4     5
Example: Three Forces

Moe, Larry, and Curley push on a 752 kg boat, each exerting a 80.5 N force
parallel to the dock.
(a) What is the acceleration of the boat if they all push in the same direction?
(b) What is the acceleration if Moe pushes in the opposite direction from Larry
and Curley as shown?
Fnet1  FM  FL  FC  3(80.5 N)  241.5 N
a1  Fnet1 / m  (241.5 N) / (752 kg)  0.321 N/kg  0.321 m/s 2

Fnet 2  FM  FL  FC  80.5 N
a2  Fnet 2 / m  (80.5 N) / (752 kg)  0.107 m/s 2
Astronaut
Checkpoint 1

Suppose you are an astronaut in outer space giving a brief push to a
spacecraft whose mass is bigger than your own.
Compare the magnitude of the force you exert on the
spacecraft, Force on spacecraft by astronaut, to the magnitue of the force exerted on
you by the spacecraft, Force on astronaut by spacecraft , while you are pushing.

A. Force on astronaut by spacecraft = Force on spacecraft by astronaut

B. Force on astronaut by spacecraft > Force on spacecraft by astronaut

C. Force on astronaut by spacecraft < Force on spacecraft by astronaut
Astronaut
Checkpoint 2

Suppose you are an astronaut in outer space giving a brief push to a
spacecraft whose mass is bigger than your own.
Compare the magnitude of the force you exert on the
spacecraft, Force on spacecraft by astronaut, to the magnitude of the force exerted
on you by the spacecraft, Force on astronaut by spacecraft , while you are pushing.

A. acceleration of astronaut = acceleration of spacecraft

B. acceleration of astronaut > acceleration of spacecraft

C. acceleration of astronaut < acceleration of spacecraft
Newton’s Third Law of Motion

Forces always come in pairs, acting on
different objects:
If Object 1 exerts a force F on Object 2, then
object 2 exerts a force –F on Object 1.
These forces are called action-reaction pairs.
Newton’s Third Law of Motion
Some action-reaction pairs:
Contact forces:
The force exerted by                          12 N
one box on the other is         3 kg   1 kg
different depending on
which one you push.

Boxes 1 and 2 rest on a
12 N 3 kg
frictionless surface.                1 kg
What is the acceleration
in each case?               Fnet
a
What is the force            m
between the boxes in
12 N
each of the cases?      a        3m s 2

4kg
Contact forces:
The force exerted by one box on
the other is different depending
on which one you push.                                            12 N
Boxes 1 and 2 rest on a frictionless              3 kg     1 kg
surface.
What is the force between the
boxes in each of the cases?

In the first case: find
the net force on mass 1                    12 N
Fnet  ma   3 kg    3 m s          2           3 kg
1 kg

Fnet  9 N this is the force on 1 by 2Fnet  12 N  F2,1
Since there is no friction,

Alternatively, you could analyze
mass 2
3 N  12 N  F2,1
Fnet  ma  1kg   3 m s      2

F2,1  12 N  3N
Fnet  3 N Mass 2 also has the applied force of F2,1  9 N
12 N acting in the opposite direction.
Contact forces:
The force exerted by one box on
the other is different depending
on which one you push.
Boxes 1 and 2 rest on a frictionless          3 kg      1 kg12 N
surface.
What is the force between the
boxes in each of the cases?
In the second case: find net force
on mass 2
Fnet  ma  1kg    3 m s  12 N  2
3 kg      1 kg
Fnet  3 N this is the force on 2 by 1
Since there is no friction,

Fnet  12 N  F1,2
Or as before, mass 1 could be
analyzed .                                        9 N  12 N  F1,2
Fnet  ma   3 kg    3 m s      2
F1,2  12 N  9 N
Fnet  9 N Mass 1 also has the applied force of           F1,2  3 N
12 N acting in the opposite direction.
Freefall
• The ratio of weight
(F) to mass (m) is
the same for all
objects in the same
locality
• Therefore, their
accelerations are
the same in the
absence of air
resistance.
In a vacuum, a coin and a feather fall equally,
side by side. Would it be correct to say that
equal forces of gravity act on both the coin and
the feather when in a vacuum?
77%
1. Yes
2. No

23%

1            2
In a vacuum, a coin and a feather fall equally,
side by side. Would it be correct to say that
equal forces of gravity act on both the coin and
the feather when in a vacuum?

1. Yes
2. No

NO! These objects accelerate
equally not because the forces
of gravity on them are equal, but
because the ratios of their
weights to masses are equal.
Non-Freefall
The Effect of Air Resistance
• Force of air drag on a falling object
depends on two things.
– the frontal area of the falling object—that is,
on the amount of air the object must plow
through as it falls
– the speed of the falling object; the greater
the speed, the greater the force
• As an object falls through air, the force
of air resistance on it increases as its
speed increases
Terminal Speed
v vs t
120

100

80
speed

60

40

20

0
0    10   20          30   40   50
time

• When the force of air resistance is
equal to the force of gravity on an
object, it no longer accelerates. This
speed is called terminal velocity.
Terminal Speed

• The heavier parachutist
must fall faster than the
lighter parachutist for air
resistance to cancel his
greater weight.
Golf Ball & Styrofoam Ball
• A stroboscopic study of a golf ball
(left) and a Styrofoam ball (right)
falling in air. The air resistance is
negligible for the heavier golf ball,
and its acceleration is nearly equal
to g. Air resistance is not
negligible for the lighter
Styrofoam ball, which reaches its
terminal velocity sooner.
A skydiver jumps from a high-flying helicopter.
As she falls faster and faster through the air,
does her acceleration increase, decrease, or
remain the same                         90%

1. It increases
2. It decreases
3. It remains the same
Acceleration decreases because the         7%       3%
net force on her decreases. Net force is
equal to her weight minus her air          1    2    3

resistance, and since air resistance
increases with increasing speed, net
force and hence acceleration decrease.
Newton 2 Slide 34

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