Transcript Newton`s Third - HRSBSTAFF Home Page

Newton’s Third Law
Physics 11
Humour again..
Newton’s Third Law:

For every action, there is an equal and opposite
reaction

Forces come in Pairs!
Newton’s Third Law:

If object A exerts a force on object B, object B will
exert a force that is equal and opposite on A
How many force pairs?
(Assume no air)
Push pair
Weight pair
Friction Pair
Two cars crash.
What is true?
They hit with the same
force.
Neither hits harder.
A man pushes his daughter!
Newton’s 3rd Law:
Action = girl rolls left.
Reaction = dad rolls right.
Compare the forces!
Newton’s 3rd Law:
The forces are equal
and in opposite
directions!
Who will have a
greater change in
speed after pushing?
Skaters!
Small mass
Large change
in speed!
Skaters!
A big
mass
means…
A small acceleration
Crash site = Action/Reaction
The forces
are…
Opposite and Equal!
The forces are
opposite and
equal
but…who is
hurt most by
this crash?
Why?
Applying Newton’s Third Law:

Typically when we consider Newton’s Third Law, we
will investigate one of two types of systems
One object applying a force to another
 A system of connected objects and the force each
applies to the other

Acceleration of the Earth


When you are in free fall, the Earth exerts a
gravitational force on you that is equal to your
weight. As a result, you accelerate towards the
Earth at 9.81m/s2
According to Newton’s Third Law, you exert an
equal and opposite force on the Earth.
Determine the acceleration of the Earth
Acceleration of the Earth


Fg  mg


FgEarth   mg


FgEarth  M E a


 mg  M E a

 75kg(9.81m / s )  5.98 x10 kga

 22
2
a  1.2 x10 m / s
2
24
Multiple Objects

Two masses are connected (12.5kg and 10.0kg)
and are moved with a constant velocity across
the floor. If the coefficient of kinetic friction
between the objects and the floor is 0.35,
determine:
a.
b.

The total applied force
If the force is applied to the 10.0kg mass, the
force the 10.0kg mass applies to the 12.5kg mass
and vice versa!
Draw diagram!
FBD (system):
FN
Ff
Fa
Fgtotal
The total applied force


Fnet  mtotala  0



Fnet  F f  Fa

0  (0.35)( 22.5kg)( 9.81m / s )  Fa

Fa  77 N
2
FBD (12.5 kg box)
FN
Ff
Fa10
Fg12.5
The force the 10.0kg mass applies
to the 12.5kg mass


Fnet12.5  m12.5 a  0



Fnet  F f  Fa10

2
0  (0.35)( 12.5kg)( 9.81m / s )  Fa10

Fa10  42 N
FBD
FN
F12.5
Ff
Fa
Fg10
The force the 10.0kg mass applies
to the 12.5kg mass


Fnet  m10 a  0


 
Fnet  F f  Fa  Fa12.5

0  (0.35)( 10.0kg)( 9.81m / s )  77 N  Fa12.5

Fa12.5  42 N
2
Practice:

Page 182 # 18, 19(a), 20(a)

Answers:
18. a) 1554N
19. a) 21.2N
20. a) 73.5 N
b) 909N