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Newton’s Third Law
And Apparent Weight
Physics 11
Newton’s Third Law

For every action, there is an equal and
opposite reaction
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
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:


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
FBD
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
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  m10a  0


 
Fnet  F f  Fa  Fa12.5

0  (0.35)( 10.0kg)( 9.81m / s )  77 N  Fa12.5

Fa12.5  42 N
2
Weight vs. Apparent Weight


As we have discussed, you don’t
experience weight (force of gravity) but
rather the normal force
So, if you feel heavier or lighter it is
because there is more or less force
being applied to you
Elevator Problem

A person with a mass
of 85kg steps onto an
elevator



What would a scale
read?
If the elevator
accelerates up at
2.5m/s2, what is the
new scale reading?
If the elevator slows at
-1.5m/s2, what is the
new scale reading?
Scale Reading


Fg  mg

2
Fg  85kg(9.81m / s )

Fg  830 N



Fnet  Fg  FN


FN   Fg

FN  830 N
Scale Reading when accelerating
at 2.5m/s2

Fg  830 N


Fnet  ma

Fnet  85kg(2.5m / s 2 )

Fnet  210 N



Fnet  Fg  FN

210 N  830 N  FN

FN  1.0 x103 N
Scale Reading when accelerating
at -1.5m/s2

Fg  830 N


Fnet  ma

Fnet  85kg(1.5m / s 2 )

Fnet  130 N



Fnet  Fg  FN

 130 N  830 N  FN

FN  710 N
Free Fall



http://www.youtube.com/watch?v=r7fE
HYkGxd0
http://www.youtube.com/watch?v=Njcd
wY41jJE
http://www.youtube.com/watch?v=BXn
hEDMUJt8
Free Fall


While in free fall, all objects are
experiencing the force of gravity
(assuming zero air resistance) and
therefore all accelerate at the same rate
Therefore, objects appear “weightless”
because there is no normal force to
allow read from a scale or to perceive
as weight
Practice Problems

Page 182


18-20
Page 186

21-23