Newton’s Third Law
For every force, or action force, there is an equal but opposite force,
or reaction force.

FBA


FAB   FBA
A
B

FAB
If object A exerts a force on object B (an “action”), then
object B exerts a force on body A (a “reaction”).
These two forces have the same magnitude but opposite direction.
Note: these two forces act on different objects.
Example: Two carts are put back-to-back on a track. Cart A has a springloaded piston; cart B, which has twice the mass of cart A, is entirely
passive. When the piston is released, it pushes against cart B, and the
carts move apart. Which of the two forces exerted by the two carts on
each other has a larger magnitude?
It’s a third
law pair!!
1. The force exerted by A.
2. The two forces have equal magnitude.
3. The force exerted by B.
B
A
Example: Gravitation. You attract the Earth!
F g you,Earth = Fg
Earth, you
But the acceleration that this produces on the Earth is
your weight (70 kg)(10 m/s 2 )
22
2
a 


10
m/s
,
24
MEarth
6  10 kg
nothing to be too proud of…
Free Body Diagram
Example: Book on Table – The full story
NBT
WBE
NTE
Action-Reaction Pairs:
Normal force between book and table
NBT = –NTB
Gravitational force between book and earth
WBE = –WEB
Normal force between table and earth
NTE = –NET
WTE
WET
NTB
WEB
NET
Gravitational force between table and earth
WTE = –WET
The book does not accelerate WBE+NBT=0
The table does not accelerate WTE+NTB+NTE=0
Does the earth accelerate?
Example: Two blocks of masses m and 2m
are pushed together along a horizontal,
frictionless surface by a force F.
m
F
A
2m
B
1) What is the magnitude of the net force on block B ?
A.
1/3 F
Entire system:
Fnet,all (  F )  3ma
B.
2/3 F
Block A:
Fnet,A  ma
C.
F
Block B:
Fnet,B  2ma  2
(so ma 
F
)
3
F
3
2) What is the magnitude of the force on block A by block B?
A.
1/3 F
FAB  FBA  f
B.
2/3 F
C.
F
Block A :
Block B :
F  f  ma
f  2ma
NB
NA
FAB
F
A
mg
FBA
B
2mg
2
 f  F
3
Example 1: A box with a mass of 100kg is given an upward
acceleration of 2.2m/s² by a cable. What is the tension in the cable?
T
m=100kg
a=2.2m/s2
T=?
Newton’s equation:
T-mg=ma
T=m(g+a)
m
mg
T=100kg*(9.8 m/s2 +2.2 m/s2 )= 1200 N
Note: this tension is bigger than the box’s weight, w=mg=100 kg* 9.8 m/s2 = 980 N
Example 2: The same box as in example 1 is given an downward
acceleration of 2.2m/s² by a cable. What is the tension in the cable?
m = 100 kg
a = -2.2 m/s2
T=?
T=m(g+a)
T=100 kg*(9.8 m/s2 -2.2 m/s2 )= 760 N
Note: this tension is smaller than the box’s weight, w=mg=100kg* 9.8m/s2 = 980N
Compare examples 1 & 2: the tension depends on acceleration & is independent
from velocity. The tension is equal to the weight if there is now acceleration (a=0).
Example: A fish is being yanked upward out of the water with a line that
can stand a maximum tension of 180 N. The string snaps when the
acceleration of the fish is 8 m/s2. What is the mass of the fish?
A. 8 kg
T
B. 10 kg
C. 18 kg
mg
T  mg  ma
T
180 N
m

 10 kg
2
g  a 10  8m / s
Example: Two boxes hang from the ceiling of an accelerated elevator
T1
a
m1
m1g
T2
m2
m2 g
Upper box:
T1  T2  m1 g  m1 a
Lower box:
T2  m2 g  m2 a
T2  m2 g  a 
T1  T2  m1 g  a   m1  m2 g  a 
Example: John has a mass of 100 kg and standing on a scale in an
elevator which is accelerating upwards from rest at 2 m/s². What will the
scale read?
Non John, by scale
a
What does a scale measure?
The magnitude of the normal force
on the scale by John, |NJS|= |NSJ|
(not part of John’s
free body diagram)
Newton’s 2nd
law on John:
NJS  mJ g  mJa
Won John, by Earth
NJS WJE  mJ a
 mJ  g  a  
John moves with
the elevator
Note: In this course, the word “weight”
refers to mg, not to what a scale reads.
 (100 kg)(9.8 m/s2  2.0 m/s2 )
 1180 N
Non scale, by John
If the scale is in kg, it will read:
NJS
1180 N

 120 kg
g
9.8 m/s2
Example: A hand keeps a 20-kg box from sliding down a frictionless
incline. The plane of the incline makes an angle θ = 30° with the
horizontal. What is the magnitude of the force exerted by hand?
mgsinθ – F = m ax = 0
F = mg sinθ
N – mgcosθ = m ay = 0
N = mg cosθ
y
F = (20 kg)(9.8 m/s2)sin(30°)
= 98 N
N
FB,hand
x
mgcosθ
mgsinθ
Directions:
θ
mg
•Draw the free-body diagram
θ
• Choose axes (draw them!)
• Use Newton’s 2nd law in the
x and y-directions.