1. Newton’s Laws
El Paso Independent High School

2. Newton’s First Law
Newton’s First Law: An object at rest or an
object in motion at constant speed will
remain at rest or at constant speed in the
absence of a resultant force.
A glass is placed on a board and the board is
jerked quickly to the right. The glass tends to
remain at rest while the board is removed.

3. Newton’s First Law (Cont.)
Newton’s First Law: An object at rest or an
object in motion at constant speed will
remain at rest or at constant speed in the
absence of a resultant force.
Assume glass and board move together at
constant speed. If the board stops suddenly,
the glass tends to maintain its constant speed.

4. Inertia
• Newton's first law is often referred to as the
law of inertia.
• Inertia is the resistance or unwillingness of an
object to accelerate (speed up, slow down, or
change directions).
• The more mass an object has, the harder it is
to accelerate. Thus, more mass equals more
inertia.

5. Understanding the First Law
(a) The driver is forced to move forward. An
object at rest tends to remain at rest.
Discuss what the driver
experiences when a car
accelerates from rest and
then applies the brakes.
(b) Driver must resist the forward motion as
brakes are applied. A moving object tends
to remain in motion.

6. Newton’s Second Law
• Second Law: Whenever a resultant force
acts on an object, it produces an
acceleration that is directly proportional to
the force and inversely proportional to the
mass.
m
F
amaF
Force Units: Newton’s or N

7. Acceleration and Force
Pushing a cart with twice the force
produces twice the acceleration. Three
times the force triples the acceleration.

8. Newton: The Unit of Force
One newton is the force required to give an
acceleration of 1 m/s2 to a mass of 1 kg.
F (N) = m (kg) a (m/s2)
What resultant force will give a 3 kg mass an
acceleration of 4 m/s2?
F = 12 N
F = ?
a = 4 m/s2
3 kg )/4()3( 2
smkgmaF

9. Example A 40 N resultant force causes a block to
accelerate at 5 m/s2. What is the mass?
F = 40 Nm=?
a = 5 m/s2
kg
sm
N
a
F
m
maF
8
/5
40
2

10. Example A net force of 4.2 x 104 N acts on a 3.2 x 104
kg airplane during takeoff. What is the force on the
plane’s 75-kg pilot?
F = 4.2 x 104 N
m = 3.2 x 104 kg
+
F = ma
a = 1.31 m/s2
To find the force on the pilot, assume same
acceleration:
First we find the
acceleration of the
plane.
kgx
Nx
m
F
a 4
4
102.3
102.4
NsmkgmaF 4.98)/31.1)(75( 2

11. Example A 54-g tennis ball is in contact with the racket for a
distance of 40 cm as it leaves with a velocity of 48 m/s. What
is the average force on the ball?
Given: vo = 0; vf = 48 m/s x = 0.40 m;
m = 0.0540 kg; a = ?
First, draw sketch and list
given quantities:
Given: vo = 0; vf = 48 m/s
x = 40 cm; m = 54 g
a = ?
Consistent units require converting grams to
kilograms and centimeters to meters:
Cont. . .

12. Example (Cont). A 54-g tennis ball is in contact with the
racket for a distance of 40 cm as it leaves with a
velocity of 48 m/s. What is the average force on the
ball?
F= (0.054 kg)(2880 m/s2) F = 156 N
2
2
fv
a
x
Knowing that F = m a, we need
first to find acceleration a:
2
2222
22
/2880
)4(.2
048
2
2
sm
x
vv
a
axvv
of
of

13. Weight and Mass
• Weight is the force an object applies as a
result of gravity pulling it downward. It is
directed downward and it varies with gravity.
• Mass is a universal constant which is a
measure of the matter that makes up an
object. It is always constant regardless of
location
mgW
maFBecause weight is a force created
by the downward acceleration of
gravity, we can make this
substitution.

14. Weight and Mass: Examples
What is the weight of a 10-kg block?
9.8 m/s2 W
m10 kg
W = mg = (10 kg)(9.8 m/s2)
W = 98 N
The weight of an object is the
force it causes as a result of
gravity.
The weight of an object is also referred to as the force of gravity

15. Always Remember!!
In Physics, the use of Newton’s second law
and many other applications makes it
absolutely necessary to distinguish between
mass and weight. Use the correct units!
Metric SI units: Mass is in kg; weight is in N.
Always give preference to the SI units.
Pounds should never be used!!!

16. Example A resultant force of 40 N gives a block an
acceleration of 8 m/s2. What is the weight of the
block near the surface of the Earth?
W=?
F = 40 Na
8 m/s2
To find weight, we must first
find the mass of the block:
Now find weight of a
5-kg mass on earth.
W = mg
= (5 kg)(9.8 m/s2)
W = 49.0 N
a
F
mmaF ;
kg
sm
N
a
F
m 5
/8
40
2

17. Newton’s Third Law
• Third Law: For every action force, there
must be an equal and opposite reaction
force. Forces occur in pairs.
Action
Reaction ActionReaction

18. Action and Reaction Forces
Use the words by and on to study
action/reaction forces below as they
relate to the hand and the bar:
The action force is exerted by
the _____ on the _____.
The reaction force is exerted
by the _____ on the _____.bar
hands bar
hands
Action
Reaction

19. m= 1 kg
Forces
If a bar of gold that has a mass of 1 kg is sitting at
rest on a table, what forces are acting on it and to
what magnitude?
FN= 9.8 N
The normal force of
an object pushes up
with a force equal to
the bar’s downward
force. Because both
forces are present,
the bar remains at
equilibrium.
Gravity acts on the
object in the
downward direction.
However, because the
bar is not moving
downward, there must
be another force
counteracting it.
Fg= 9.8 N

20. Free-body Diagrams
• A free body diagram is a diagram that shows
all of the forces present on a given object.
• When drawing a free body diagram, objects
are always represented as points or dots and
forces are drawn as arrows pointing away
from the object.
• For now the only forces we will consider are
gravity, tension, normal, applied, and friction.

21. Free-body Diagrams
• Gravitational Force: This is the only force that
will always be present and will always point
straight down.
• Tension Force: This type of force is caused by
a rope, chain, string, etc. and goes in the
direction of the rope, chain, string, etc.
• Normal Force: This type of force is present
when the object is in contact with a surface.
It always points perpendicular to the surface.

22. Free-body Diagrams
• Applied Force: This type of force is used when
an outside agent is acting on the object such
as a person.
• Frictional Force: This type of force is the result
of two surfaces being drug across each other,
creating friction. It always points in the
direction opposite of the movement.
The size of the arrow should be proportional to the
size of the force. Larger forces should have larger
arrows!!!

23. Examples
Fa
FN
FF
Fg
Fa
FT
Fg
FT
Friction acts
parallel to the
surface but
opposite the
direction of travel.
There is no normal
force present
because the object
is not resting on a
surface.

24. Example A cart and driver have a mass of 120 kg. What
force F is required to give an acceleration of 6 m/s2
with no friction?
1. Read problem and draw a sketch.
2. Draw a free body diagram and label forces.
Diagram for Cart:
FN
Fg
Fa
3. Choose x-axis along motion and indicate the
right direction as positive (+).
x
+

25. Example What force F is required to give an
acceleration of 6 m/s2?
Fy = 0; FN - Fg = 0
The normal force FN
is equal to weight Fg
Fx = max; Fa = ma
Fa = (120 kg)(6 m/s2)
Fa = 720 N
m = 120 kg
4. Write Newton's Law equation for both axes.
ay = 0
Diagram for Cart:
FN
Fg
Fa

26. Example What is the tension FT in the rope below if
the block accelerates upward at 4 m/s2? (Draw
sketch and free-body)
10 kg
a = +4 m/s2
FT a
FT
Fg
+
Fx = m ax = 0
Fy = m ay = m a
FT- Fg = m a
Fg=mg = (10 kg)(9.8 m/s) = 98 N
m a= (10 kg)(4 m/s) = 40 N
FT - 98 N = 40 N FT = 138 N

27. Example Find the acceleration of the blocks if there is no
friction on the surfaces.
2 kg 4 kg
12 N
First apply F = ma to entire system (both masses).
12 N
FN
Fg=(m2 + m4)g
Fx = (m2 + m4) a
12 N = (6 kg) a
2
/2
6
12
sm
kg
N
a
Because the blocks are tied
together and move together,
we can treat them as one
object

28. Now find the tension force in the rope
connecting the two blocks.
2 kg 4 kg
12 N
Now find tension T
in connecting cord.
Apply F = m a to the 2 kg mass where a = 2 m/s2.
FT
FN
m2 g
Fx = m2 a
FT = (2 kg)(2 m/s2)
FT= 4 N

29. Example (Cont.) The two-body problem.
2 kg 4 kg
12 N Same answer for FT
results from focusing
on 4-kg by itself.
Apply F = m a to the 4 kg mass where a = 2 m/s2.
Fx = m4 a
12 N - FT = (4 kg)(2 m/s2)
T = 4 N
Fa
FT
m4g
FT