Transcript Dynamics 2

Newton’s Third Law
Sec. 6.3
Interaction Forces
► Objectives
 Explain the meaning of interaction pairs (third
law pairs) of forces and how they are related by
Newton’s third law
 List the four fundamental forces and illustrate
the environment in which each can be
observed.
 Explain the tension in ropes and strings in terms
of Newton’s third law
Newton’s laws of motion
1. The law of inertia. An object in motion
remains in motion with constant velocity if the net
force on the object is zero.
2. Force and acceleration. If the net force acting
on an object of mass m is F, then the acceleration
of the object is a = F/m. Or, F = ma.
3. Action and reaction. For every action there is
an equal but opposite reaction.
Action means force.
Newton’s 3rd Law
For every action force there is
an equal and opposite
reaction force
(You cannot touch without being touched)
Action-reaction Pair
If object A exerts a force on object B, then object B
exerts an equal and opposite force on object A. The pair
of forces (due to one interaction), is called an
action/reaction pair.
NOTA BENE:
The action/reaction pair will never appear in the same free body diagram.
Summary: Interaction forces
(‘Third Law Pairs’)
► Two
forces that are in opposite directions
and have equal magnitude
► They
never act on the same object!
(Therefore never in same FBD!)
► General
Form is:
FA on B = -FB on A
Newton’s third law
For every action there is an
equal but opposite reaction.
Whenever one object exerts a force on
another object, the second object exerts an
equal but opposite force on the first object.
force of A on B
force of B on A
Forces always occur like this, in pairs.
We will see that this is very hard to accept! It is just not
common sense. That is why it took a great genius like
Newton to figure it out.
Action: tire pushes on road
Reaction: road pushes on tire
When you walk or run, what forces occur?
• At constant velocity the horizontal force is 0 and you
continue to move because of inertia.
• To accelerate, you push backward against the floor; the
reaction force, which is a friction force exerted by the
floor on your foot, pushes you forward.
This reaction force may be hard to
visualize, but imagine what would
happen if you were on a frictionless
surface – can’t accelerate!
Recoil force
a = F/
m
Force accelerating bullet
a = F/m
A bug and a car collide. Which
experiences the greater force?
• (a) bug
• (b) car
• (c) neither, they both experience the same
magnitude of force
Consider hitting a baseball with a
bat. If we call the force applied to
the ball by the bat the action force,
identify the reaction force.
(a) the force applied to the bat by the hands
(b) the force applied to the bat by the ball
(c) the force the ball carries with it in flight
(d) the centrifugal force in the swing
Which vehicle exerts a greater force ―
the tow truck or the car?
A puzzle:
The truck pulls to the right. According to Newton’s
third law, the car pulls to the left with an equal
force. So how can they start moving, or accelerate?
Resolution: Consider each part separately, and
don’t forget that other forces are also acting.
Playing catch with
a medicine ball
A
A throws the ball and B catches it.
B
 four forces
When A throws the ball he exerts a force on the ball
(toward the right) and the ball exerts a force on him
so he recoils (toward the left).
► Newton’s third law for the throw
When B catches the ball he exerts a force on the ball
(toward the left to stop it) and the ball exerts a force on
him so he is knocked back (toward the right).
► Newton’s third law for the catch
Example – A collision
The force exerted by the
ball on the toe (reaction) is
equal to the force exerted
by the toe on the ball.
Really hard to accept!
A puzzle : Tug of War
String tension
String tension
Contact force
Contact force
Which team will end up in the puddle?
But aren’t the forces equal but opposite !?
Resolution: Don’t forget that there are other forces acting.
Each team exerts a force on the Earth, so the Earth exerts a
force on the team (3rd law!). The net force on either team is
toward the left.
A puzzle
Horse and Cart
The horse pulls the cart with a force A (to the right).
According to Newton, the cart pulls the horse with a
force –A (to the left).
So how can they start moving, or accelerate?
Resolution: Consider each part separately, and don’t
forget that there are other forces acting.
A small car is pushing
a larger truck that
has a dead battery.
The mass of the truck
is larger than the
mass of the
car. Which of the following statements is true?
A. The truck exerts a larger force on the car than the car exerts
on the truck.
B. The truck exerts a force on the car but the car doesn’t exert a
force on the truck.
C. The car exerts a force on the truck but the truck doesn’t exert
a force on the car.
D. The car exerts a larger force on the truck than the truck
exerts on the car.
E. The car exerts the same amount of force on the truck as the
truck exerts on the car.
A small car is pushing
a larger truck that
has a dead battery.
The mass of the truck
is larger than the
mass of the
car. Which of the following statements is true?
A. The truck exerts a larger force on the car than the car exerts
on the truck.
B. The truck exerts a force on the car but the car doesn’t exert a
force on the truck.
C. The car exerts a force on the truck but the truck doesn’t exert
a force on the car.
D. The car exerts a larger force on the truck than the truck
exerts on the car.
E. The car exerts the same amount of force on the truck as
the truck exerts on the car.
Newton’s Third Law Pairs
Example - analyzing interacting
objects
A person pushes a large
crate across a rough
surface.
• Identify the objects
that are systems of
interest
• Draw free-body
diagrams for each
system of interest.
• Identify all
action/reaction pairs
with a dashed line.
Forces involved in pushing a
crate – FBD of person and crate
Propulsion Force
• The force label fp
shows that the
static friction
force on the
person is acting
as a propulsion
force.
• This is a force
that a system
with an internal
source of energy
uses to drive
itself forward.
Propulsion forces
Free Body Diagrams –
Exercises
Draw a freebody diagram of each object in
the interacting system.
Show action/reaction pair with red/orange
dotted lines.
Draw force vectors in another color.
Label vectors with standard symbols.
Label action/reaction pairs FAonB , FBonA for
example.
A fishing line of negligible mass lifts a fish
upward at constant speed. The line and the fish
are the system, the fishing pole is part of the
environment. What, if anything, is wrong with
the free-body diagrams?
A fishing line of negligible mass lifts a fish
upward at constant speed. The line and the fish
are the system, the fishing pole is part of the
environment. What, if anything, is wrong with
the free-body diagrams?
The gravitational force and the tension
force are incorrectly identified as an
action/reaction pair. The correct
action reaction pair is…?
Action/reaction pairs are never on the
same free body diagram.
Mass of line considered negligible so no
weight force necessary.
Boxes A and B are sliding to the right across a
frictionless table. The hand H is slowing them
down. The mass of A is larger than the mass of B.
Rank in order, from largest to smallest, the
horizontal forces on A, B, and H. Ignore forces on
H from objects not shown in the picture.
Draw a FBD for each
object: A, B, and H.
Only consider horizontal
forces in this problem.
A
FBA
C
B
FHB
FAB
Recognize any third law pairs?
FBH
Boxes A and B are sliding to the right across a
frictionless table. The hand H is slowing them
down. The mass of A is larger than the mass of B.
Rank in order, from largest to smallest, the
horizontal forces on A, B, and H.
FB on H = FH on B > FA on B = FB on A
from Newton’s 2nd and 3rd Laws
Problem-Solving Strategy:
Interacting-Objects Problems
Challenge Problem:
Two strong magnets each weigh 2 N
and are on opposite sides of the
table. The table, by itself, has a
weight of 20 N. The long rangerange attractive force between the
magnets keeps the lower magnet
in place. The magnetic force on
the lower magnet is 3 times its
weight.
a. Draw a FBD for each magnet and
table. Use dashed lines to
connect all action/reaction pairs.
b. Find the magnitude of all forces in
your FBD and list them in a table.
Challenge Problem
Upper
Table
Lower
↓ (Wg) E,U
↓ (Wg) E,T
↓ (Wg) E,L
↑ N T,U
↑ N S,T
↑ LM U,L
↓ LM L,U
↓ N U.T
↓ N T,L
-
↑ N L,T
-
Find Third Law Pairs
Upper
Table
Lower
↓ (Wg) E,U
↓ (Wg) E,T
↓ (Wg) E,L
↑ N T,U
↑ N S,T
↑ LM U,L
↓ LM L,U
↓ N U.T
↓ N T,L
-
↑ N L,T
-
FBDs for EOC 8
Challenge Problem: Answer
Ranking Task – Pushing blocks
Block 1 has a mass of m,
block 2 has a mass of
2m, block 3 has a mass
of 3m. The surface is
frictionless.
Rank these blocks on the
basis of the net force on
each of them, from
greatest to least. If the
net force on each block is
the same, state that
explicitly.
Ranking Task – Pushing blocks
Answer: 3 2 1
Reason: ΣF = ma.
Acceleration is equal
for all blocks.
Ranking Problem Example
Block 1 has a mass of 1 kg, block
2 has a mass of 2 kg, block 3
has a mass of 3 kg. The
surface is frictionless.
a. Draw a fbd for each block. Use
dashed lines to connect all
action/ reaction pairs.
b. How much force does the 2-kg
block exert on the 3-kg block?
c. How much force does the 2-kg
block exert on the 1-kg block?
Ranking Problem Example – Answer:
b. How much force does the 2-kg block exert on the 3-kg
block? – 6N
c. How much force does the 2-kg block exert on the 1-kg
block? – 10N
Newton’s Third Law
Acceleration constraint
• An acceleration
constraint is a welldefined relationship
between the
acceleration of 2 (or
more) objects.
• In the case shown,
we can assume ac
=aT = ax
Is there an acceleration constraint in this situation?
If so, what is it? The pulley is considered to be
massless and frictionless.
Answer: Acceleration constraint is: aA = -aB
The actual signs may not be known until the
problem is solved, but the relationship is known
from the start.
Exercises
17
answers
a2kg = -.5a1kg
Newton’s Third Law
nd
2
Page 163,
Ed (Found in
Chapter 5, 1st ed)
Interacting systems problem
(EOC #35)
A rope attached to a 20
kg wooden sled pulls
the sled up a 200
snow-covered hill. A
10 kg wooden box
rides on top of the
sled. If the tension in
the rope steadily
increases, at what
value of tension will
the box slip?
Interacting systems problem
(EOC #35)
What are the objects of
interest?
What kind of axes for the FBD
for each?
Acceleration constraints?
Draw FBDs, with 3rd law pairs
connected with dashed lines.
Find the max tension in
the rope, so the box
does not slip.
Box: 3 forces
Sled: 6 forces
0.06
0.06
• I suggest starting with the
equations for the sled, since
the unknown of interest is
found there.
• Identify quantities in sled
equations that you can find
by solving box equations.
• Solve box equations.
• Return to sled equations
with newfound booty.
• Plug and chug.
Interacting systems problem
(EOC #35)
A rope attached to a 20
kg wooden sled pulls
the sled up a 200
snow-covered hill. A
10 kg wooden box
rides on top of the
sled. If the tension in
the rope steadily
increases, at what
value of tension will
the box slip?
Answer: 155 N.
The Massless String
Approximation
A horizontal forces only fbd for the string:
TAonS
●
TBonS
ΣF = TBonS – TAonS = ma. If string is accelerating to the right
TBonS = TAonS + ma
The Massless String
Approximation
Often in physics and engineering problems the mass of the
string or rope is much less than the masses of the objects
that it connects. In such cases, we can adopt the following
massless string approximation:
This allows the objects A and B to be analyzed as if
they exert forces directly on each other.
Pulleys
If we assume that the string is massless and the
pulley is both massless and frictionless, no net force
is needed to turn the pulley. TAonB and TBonA act “as
if” they are an action/reaction pair, even though
they are not acting in opposite directions.
Pulleys
• In this case the
Newton’s 3rd law
action/reaction pair
point in the same
direction!
T 100kg on m
Tm on 100kg
All three 50 kg blocks are at rest. Is the
tension in rope 2 greater than, less than,
or equal to the tension in rope 1?
A. Equal to
B. Greater than
C. Less than
All three 50 kg blocks are at rest. Is the
tension in rope 2 greater than, less than,
or equal to the tension in rope 1?
A. Equal to
B. Greater than
C. Less than
In the (moving)
figure to the right,
is the tension in
the string greater
than, less than, or
equal to the
weight of
block B?
A. Equal to
B. Greater than
C. Less than
In the figure to the
right, is the
tension in the
string greater
than, less than, or
equal to the
weight of
block B?
A. Equal to
B. Greater than
C. Less than
Interacting systems problem
(EOC #40)
A 4.0 kg box (m) is on a
frictionless 350
incline. It is
connected via a
massless string over
a massless,
frictionless pulley to
a hanging 2.0 kg
mass (M). When the
box is released:
a. Which way will it go
George?
4.0 kg
350
Interacting systems problem
(EOC #40)
a. Which way will it go?
Even if you have no clue,
follow the plan!
What are the objects of
interest?
What kind of axes for the
FBD for each?
Acceleration constraints??
Draw FBDs, with 3rd law
pairs connected with dashed
lines.
4.0 kg
350
Interacting systems problem
(EOC #40)
How do you figure out
which way the system
will move, once m is
released from rest?
massless string
approx. allows us to
join the tensions as an
“as if” interaction pair
Interacting systems problem
(EOC #40)
Interacting systems problem
(EOC #40)
a = - 0.48 m/s2, T = 21 N.
Which way is the system moving?
How does the tension compare to the
tension in the string while the box
was being held?
Greater than, less than, equal to?
EOC # 33
The coefficient of static
friction is 0.60 between
the two blocks in the
figure. The coefficient of
kinetic friction between the
lower block and the floor is
0.20. Force F causes
both blocks to slide 5
meters, starting from rest.
Determine the minimum
amount of time in which
the motion can be
completed without the
upper block slipping.
EOC # 33
The coefficient of
static friction is
0.60 between the
two blocks in the
figure. The
coefficient of
kinetic friction
between the
lower block and
the floor is 0.20.
Force F causes
both blocks to
slide 5 meters,
starting from rest.
Determine the
minimum amount
of time in which
EOC # 33
amax = 3.27 m/s2
tmin = 1.75 s (time is
important in this one)
EOC # 46
Find an expression for
F, the magnitude of
the horizontal force
for which m1 does not
slide up or down the
wedge. This
expression should be
in terms of m1, m2 , θ,
and any known
constants, such as g.
All surfaces are
frictionless.
EOC # 46
.
EOC # 46
.
F = (m1 + m2) g tan θ
EXAMPLE 7.6 Comparing two
tensions
QUESTION:
EXAMPLE 7.6 Comparing two
tensions
EXAMPLE 7.6 Comparing two
tensions
EXAMPLE 7.6 Comparing two
tensions
EXAMPLE 7.6 Comparing two
tensions
The spring force
---another example of 3rd law
Suppose the spring is stretched beyond its
equilibrium length by a length x.
The force on the mass m1 is F1 = +kx.
(k = Hooke’s constant)

The force on the mass m2 is F2 = -kx.

The forces are equal but opposite.
( + means to the right; - means to the left.)
Example
One end of a spring is attached to a wall.
When the other end is pulled with a force
of 50 N, the spring is stretched by 3 cm.
What force would be required to stretch
the spring by 5.5 cm?
Answer: 91.7 N
Hooke’s law: The strength of a spring force is
proportional to the displacement (extension or
compression). F = k x where k is called Hooke’s
constant for the spring.
Forces obey Newton’s third law.
We’ll consider two examples:
• The force of universal gravitation
• The spring force
Universal Gravitation
--- an example of Newton’s third law
Gm1m2
F1 
nˆ 1
2
r
Gm1m2
F2 
nˆ 2
2
r
The Earth pulls the apple down (“action”).
The apple pulls the Earth up (“reaction”).
The two forces are equal (but opposite).
When does a scientific theory become
accepted as true?
For a laboratory measurement, the
gravitational force is really very weak.
The force on the 1 kg mass is +3.3 x 10-10 N.
The force on the 5 kg mass is –3.3 x 10-10 N.
( + means to the right, i.e., increasing x)
G  6.67  10
-11
3
-2
m s kg
-1
Henry Cavendish, 1798 : first measurement of G
What makes g?
GMm
weight  mg  force of gravity  2
R
GM
9.81
g 2
R
m/s2
Weighing the Earth
Calculate the mass of the Earth.
The force of gravity on m
is, by definition, its
weight,
F  mg
By Newton’s theory of
universal gravitation,
GMm
F 
R2
gR2 9.81  (6.4 E 6) 2
M

kg
G
6.67 E - 11
the mass of the Earth,
24
 6.02 10 kg
relying on the Cavendish
measurement
Quiz Question
The planet is pulled toward the moon (and vice versa).
Calculate the gravitational force on the planet.