Remember:
• Force is a push or a pull.
• The unit of force is Newton (N)
• Net Force exerted on an object may cause
change in motion (acceleration).
Chapter 3: Forces
Section 1: Newton’s Second Law
Chapter 3: Forces
Section 1: Newton’s Second Law
Newton’s second law of motion describes
how the force exerted on an object, its
mass, and its acceleration are related
The direction of the net force in the above figure
is to the left.
The rope will move to left.
The direction of acceleration is to the left.
The direction of the net force in the above figure is
to the right.
The rope will move to right.
The direction of acceleration is to the right.
Conclusion: Always the direction of acceleration is in
the same direction of the net force
Figure 1
Figure 2
The sled in figure 2 will move with greater velocity when the dogs
start pulling because the net force is greater.
The sled in figure 2 will have greater acceleration because of
greater net force
In other words, acceleration is directly proportional to the force
Figure 1
Figure 2
The sled in figure 2 will move with greater velocity when the dog
starts pulling because the mass is smaller.
The sled in figure 2 will have greater acceleration because of
smaller mass
In other words, acceleration is inversely proportional to the mass
Newton’s second law of motion states that
The acceleration of an object is in the
same direction of the net force on the object,
and that the acceleration can be calculated
from the following equation
a =
Fnet
m
Newton’s second law of motion equation
could be put as:
Fnet
m a
So we can write:
a=
Fnet
m
or
Fnet = m a
or
m=
Fnet
a
Examples:
1- If the mass of a plane is 2000 Kg and the net
force on it is 16000 N, what is the plane’s
acceleration?
Solution:
Data: m = 2000 kg
Fnet = 16000 N
a=?
a=
Fnet
m
=
16000
2000
= 8 m/s2
Examples:
2- what is the net force on a truck with a mass
of 800 kg if its acceleration is 3 m/s2?
Solution:
Data: m = 800 kg
a = 3 m/s2
Fnet = ?
Fnet = m a
Fnet = (800)(3)
Fnet = 2400 N
Examples:
3- A wagon is being pulled by a horse. What is
the wagon’s mass if the net force on the
wagon is 1500 N and it has an acceleration
of 2 m/s2?
Solution:
Data: Fnet = 1500 N
a = 2 m/s2
m=?
m=
Fnet
a
=
1500
2
= 750 kg
Friction
Friction is the force that opposes the sliding
motion of two surfaces that are touching
each other.
What causes friction?
Friction is caused by microwelds that form
where the surfaces are in contact.
microwelds
Microwelds are formed when two surfaces are
in contact and the bumps touch each other
Types of friction
There are three kinds of friction:
1- Static friction.
2- Sliding friction.
3- Rolling friction.
Static friction
Static friction is the frictional
Force that prevents two surfaces
From sliding past each other.
In this case, the applied force is
not large enough to break the
Microwelds.
The box is not moving
Sliding friction
Sliding friction is the force
that opposes the motion of
two surfaces sliding past each
other.
In this case, the microwelds
constantly breaks and then
forms again as the two surfaces
slides along each other
The box is moving
Sliding friction is always less than static friction
Rolling friction
Rolling friction is the
frictional force between
a rolling object and the
surface it rolls on.
Rolling friction is always
Less than sliding friction
Air resistance
Air resistance is a friction-like
force that opposes the motion
of objects that move through
the air.
The amount of air resistance
on an object depends on its:
1. speed
2. size
3. shape
Net force
Remember that:
• Net force is the combination of forces on one object.
• If the object is under the effect of one force then the
net force is the same force.
• If the object is under the effect of two forces then:
* If the two forces are in the same direction
F
then:
Fnet = F1 +F2
F
1
2
* If the two forces are in opposite directions
then:
Fnet = F1 - F2
F
2
F1
More examples:
m
1. A 1500-kg car moves with
a 2
an acceleration of 2 m/s .
If the car’s engine force is
Fengine 4000 N, find the amount
of friction force on the car.
Engine
force
Ffriction= ?
Solution:
Fnet = m a = (1500)(2) = 3000 N
Fnet = Fengine - Ffriction
3000 = 4000 – Ffriction
Ffriction = 1000 N
Friction
force
More examples:
m
1. A 9000-kg train moves with
a = 0 constant velocity. If the
Engine
friction force on the train force
Ffriction is 4500 N, find the force
of the train’s engine. Fengine = ?
Solution:
Fnet = m a = (9000)(0) = 0 N
Fnet = Fengine - Ffriction
0 = Fengine – 4500
Fengine = 4500 N
Friction
force