Download More Newton`s Laws Applications

More Newton’s Laws
Applications
Unit 3 Presentation 2
Revisit Newton’s 3rd Law


“Every action force has an equal in
magnitude but opposite in direction
reaction force”
The action and reaction forces act
on DIFFERENT objects
Newton’s 3rd Law Examples
Action Force
Reaction Force
A gun is fired, propelling a
bullet forward.
The gun fired backwards,
in the opposite direction of
the bullet.
A person pushes on a wall. The wall pushes back on
the person.
A person walks down a
The ground exerts a
street, exerting a frictional normal force against the
force against the ground.
person, propelling them
forward.
Newton’s 3rd Law Example

A block-block system, like the one pictured below, is
pushed by a force of 15 N from the left to the right.
Calculate
(a) The acceleration of the block-block system
(b) The magnitude of the action-reaction force in between the two
blocks
First, break the two blocks into
individual problems and draw
two free-body diagrams:
5 kg
15 N
2 kg
Normal Force
AR
Force
15 N
2 kg
Action/
Reaction
Force
(same on
both
blocks)
Normal Force
AR
Force
15 N
5 kg
Gravity
2 kg block
Gravity
5 kg block
Newton’s 3rd Law Example (cntd)

Now, consider the free-body diagrams. Remembering to
keep the x and y variables separate, set up some
equations.

Do we need to consider the y direction forces? NO! The motion is
only in the x direction in this example.
Now, using Newton’s 2nd Law on each block in the x-direction only:
m1  2kg
m2  5kg
a?
Far  ?
15 N  Far  m1a
Far  m2 a
Notice, we have 2 equations with 2 unknowns. Lets use substitution
and substitute the right-hand equation into the left-hand equation:
15 N  ( m2 a )  m1a
Next, solve for the AR Force:
15  (5a )  2a
15  7 a
Far  5(2.14)
2.14m / s 2  a
Far  10.7 N
Another Newton’s 3rd Law Example
A block-block system, like the one pictured below, is pushed and
accelerates at 3.2 m/s2 from the left to the right. Calculate:
(a) The force on the 8 kg block
(b) The magnitude of the action-reaction force in between the two
blocks
3 kg
First, break the two blocks into
individual problems and draw
two free-body diagrams:
8 kg
3.2 m/s2
F
8 kg
Action/
Reaction
Force
(same on
both
blocks)
Normal Force
AR
Force
Normal Force
AR
Force
F
3 kg
Gravity
8 kg block
Gravity
3 kg block
Another Newton’s 3rd Law Example
(cntd)

Now, consider the free-body diagrams. Remembering to
keep the x and y variables separate, set up some
equations.

Do we need to consider the y direction forces? NO! The motion is
only in the x direction in this example.
Now, using Newton’s 2nd Law on each block in the x-direction only:
m1  8kg
m2  3kg
F  Far  m1 (3.2m / s )
2
Far  m2 (3.2m / s 2 )
Now, solve the right-hand equation to determine the magnitude of
the action-reaction force:
a  3.2m / s 2
Far  3kg(3.2m / s 2 )
Far  ?
Far  9.6 N
F ?
Next, solve for the Force:
F  (3kg  3.2m / s 2 )  8kg(3.2m / s 2 )
F  9.6 N  25.6 N
F  35.2 N
The Force of Friction

Friction is a force that always opposes
motion


Opposite in direction to the motion of the
object
Friction is directly related to the normal
force:
f f  friction
f f    FN
= Coefficient of Friction (pure number between 0 and 1)
=0 No Friction (frictionless surface)
=1 Highest frictional force possible
Types of Friction




Static Friction: The force of friction that
opposes motion from beginning from rest. Also
known as sticking friction.
Kinetic Friction: The force of friction that
opposes motion as it is occurring. This only
occurs after motion has begun.
Example: When pushing a large block, static
friction prevents you from moving the block at all.
When you finally start pushing it, kinetic friction
takes over and slows the motion.
Always…..static friction is stronger
than or equal to kinetic friction in
magnitude.
Coefficients of Friction

s =Coefficient of Static Friction

k =Coefficient of Kinetic Friction

Always….
s  k
Coefficients of Friction for Common
Surfaces
Materials
s value
k value
Steel on Steel
0.74
0.57
Aluminum on Steel
0.61
0.47
Copper on Steel
0.53
0.36
Rubber on Concrete
~1.0
0.8
Wood on Wood
0.25 – 0.50
0.2
Glass on Glass
0.94
0.4
Waxed wood on wet snow
0.14
0.10
Waxed wood on dry snow
0.01
0.04
Metal on Metal (lubricated)
0.15
0.06
Ice on Ice
0.10
0.03
Teflon on Teflon
0.04
0.04
Synovial joints in Humans
0.01
0.003
Frictional Forces Example

Calculate the static and kinetic frictional forces of a 5 kg
block of aluminum on a steel table if a force were applied
to move it to the right.
First, find the coefficients from the table on the previous slide:
 s  0.61
 k  0.47
Normal Force
f
(friction)
F (applied)
m  5kg
g  9.8m / s 2
Gravity
(Weight)
Noting that there is no acceleration in the y direction and
using Newton’s Second Law:
Fn  mg  0
Fn  mg  5kg(9.8m / s 2 )  49 N
Frictional Forces Example (cntd)

Now, lets calculate the static frictional force:
f s   s  Fn  0.61* 49 N  29.9 N

Now, lets calculate the kinetic frictional force:
f s  s  Fn  0.47 * 49 N  23.0 N