Lecture 9
Physics I
Chapter 7
Newton’s Third Law
Course website:
http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Today we are going to discuss:
Chapter 7:





Some leftover (Ch.6)
Interacting Objects: Section 7.1
Analyzing Interacting Objects: Section 7.2 (skip Propulsion)
Newton’s Third Law: Section 7.3
Ropes and Pulleys: Section 7.4
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
ConcepTest
A box of weight 100 N is at
rest on a floor where s = 0.5.
A rope is attached to the box
and pulled horizontally with
tension T = 30 N. Which way
does the box move?
Will It move?
A) moves to the left
B) moves to the right
C) the box does not move
D) moves down
E) moves up
Ffr   S N  S mg  0.5 100  50N
The static friction force has a
maximum of sN = 50 N. The
tension in the rope is only 30 N.
So the pulling force is not big
enough to overcome friction.
Follow-up: What happens if the tension is 55 N?
m T=30N
Demonstration
Two Interleaved Books
Simply lay the pages of two phone books on top of each other one by one before
attempting to pull them apart.
It increases
max friction
FA
Fperp
http://www.youtube.com/watch?v=AHq82D78Igg
When we apply FA force, we create two
components Fpull and Fperp.
Since there is no motion, we have a static
friction force Ffr equal to the Fpull.
As we increase Fpull, Fperp gets larger increasing
(Ffr )max , which means that our goal moves
farther away. So the more we pull the larger
goal becomes  But luckily Fpul grows faster
and with two tanks it can be done.
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Fpull
Ffr
Ffr
Fpull
It pulls books apart
NxFfr
ConcepTest. Going Sledding
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
A) pushing her from behind
B) pulling her from the front
C) both are equivalent
easiest way to
D) it is impossible to move the sled
accomplish this?
E) tell her to get out and walk
In case 1, the force F is pushing down
(in addition to mg), so the normal
force is larger. In case 2, the force F
1
is pulling up, against gravity, so the
normal force is lessened. Recall that
the frictional force is proportional to
the normal force.
2
Newton’s
rd
3
Law
Forces come from other objects
 Chapter 7.
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Newton’s Third Law of Motion
Whenever one object exerts a force on a second object, the second
exerts an equal force in the opposite direction on the first.
=
“For every action force, there is an equal
and opposite reaction force”
NOTE: Action and Reaction forces act on
different objects!
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Let’s try to justify N.3rd law.
When a hammer hits a nail, it exerts a forward
force on the nail.
Is it true that a nail also exerts a
force on a hammer (in return)?
When a hammer hits a nail, the hammer stops very quickly
0 /
10 /
i.e. the hammer changes velocity, i.e. it accelerates (a)
If the hammer accelerates (decelerates), there must be a force acting on the
hammer from the poor nail, according to the N. 2nd law.
If you still don’t believe me, hit the nail with a glass hammer.
It’s the force of the nail on the hammer that would cause the glass to shatter!
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Me
Action-Reaction Pair of Forces
Let’s come up with a convention how to denote these pairs of forces
FB on A
Or simply
FBA
FA on B
A
B
FAB
Or simply
=
If object A exerts a force on object B, then object B exerts a force on object A.
- The pair of forces, as shown, is called an action/reaction pair.
Helpful notation:
the second subscript is the object that the force is being exerted on; the first is the source.
A key to the correct application of the third law is that
the forces are exerted on different objects.
Make sure you don’t use them as if they were acting on
the same object.


FWS   FSW
PHYS.1410 Lecture 9 Danylov
- i.e. If you want to describe a motion of a skater,
you have to use a force FWS acting from the wall on the skater
- And if you want to describe the wall, use FSW .
Department of Physics and Applied Physics
Newton’s Third Law of Motion
Rocket propulsion can also be explained using Newton’s third law: hot
gases from combustion spew out of the tail of the rocket at high speeds.
The reaction force is what propels the rocket.
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Acceleration Constraints
 If two objects A and B move together, their accelerations are
constrained to be equal: aA = aB
 This equation is called an acceleration constraint.
 Consider a car being towed by a truck.
 In this case, the acceleration
constraint is aCx = aTx = ax .
 Because the accelerations
of both objects are equal,
we can drop the subscripts
C and T and call both of
them ax .
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Acceleration Constraints
 Sometimes the
acceleration of A and B
may have different signs.
 Consider the blocks A and
B in the figure.
 The string constrains
the two objects to
accelerate together.
 But, as A moves to the right in the +x direction, B moves
down in the −y direction.
 In this case, the acceleration constraint is aAx = −aBy .
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Internal forces cancel each other!!!!!!
Cancel due to Newton’s Third Law
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Two blocks problem
We can now forget about the internal forces
Given: F, m1, m2
y
F
Find:
m=m1+m2
x
a
(common acceleration)
Treat, m=m1+m2, as the system (one big block)
Apply N. 2nd law to m
x component of
N. 2nd law


 F  ma
F
x
 max
F  ma  (m1  m2 )a
F
a
m1  m2
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
ConcepTest
Crazy Mosquito
A) The magnitude of the mosquito’s acceleration is
larger than that of the truck.
A mosquito runs head-on
into a truck. Which is true B) The magnitude of the truck’s acceleration is larger
during the collision?
than that of the mosquito.
C) The magnitude of the mosquito’s acceleration is the
same as that of the truck.
D) The truck accelerates but the mosquito does not.
E) The mosquito accelerates but the truck does not.
FMT FTM
Newton’s 3rd law:
FMT=FTM=F
Newton’s 2nd law:
aM >> aT
Don’t confuse cause and effect! The same force can have very different effects.
The same idea can be applied to an interaction of an apple and the Earth in the next slide.
But you don’t have to read the next slide. Only if you want.
(read if you want)
Falling ball exerts force on Earth
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Ropes and Pulleys
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Tension
If a flexible cord pulls an object, the cord is said to be under
TENSION
Let’s assume that the cord is a described object and apply N 2nd law
T1
0
m
Often in problems the mass of the string or rope is much
less than the masses of the objects that it connects. m=0
T2
massless string approximation: Tension is the same at any point of the rope
,
For problems in this book, you can assume that any strings or ropes are massless unless it explicitly states otherwise.
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Example: Two boxes and a pulley.
Two boxes are connected by a cord running over a pulley. The coefficient of
kinetic friction between box A and the table is 0.20. We wish to find the
acceleration, a, of the system. As box B moves down, box A moves to the right.
FN
T
Ffr
T
Fg=mA g
Fg=mB g
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Thank you
See you on Wednesday
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Example: A ramp, a pulley,
and two boxes.
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Example: A ramp, a pulley, and two boxes
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics
Two blocks problem: y-equation
Box is a described
object
FN
y
m
x
Fg=mg
F
y
 may
FN  mg  ma y
ay = 0 (no motion in y direction)
FN  mg
Y equation gives a normal force
PHYS.1410 Lecture 9 Danylov
Department of Physics and Applied Physics