Download Newton`s laws Prez - Ms. Gamm

NEWTON’S LAWS
• Force  push or a pull
• Contact Force  physical contact
• Field Forces  No contact: gravity,
magnetic force, etc.
INERTIA
INERTIA
•Newton’s 1st Law
•An object wants to
keep on doing what it
is already doing
•In order to change, it
needs a net force not
equal to zero
å F= Sum of all forces
NEWTON’S
ND
2
LAW: ACCELERATION
•When you do have a net force…
•object will accelerate in direction of the net
force
•This acceleration is proportional to the net
force and inversely proportional to its mass
acceleration µ force
1
acceleration µ
mass
EXPLORING
ND
2
LAW
•Put the two together:
Fnet
a=
mtot
F = ma
•mass units: kg
•acceleration units: m/s2
kg
×
m
•This makes force units:
2
s
•1 kg•m/s2 is called a Newton (N)
MASS IS NOT WEIGHT
•Weight is a common force
•Weight is the force exerted by gravity on an
objects mass
•Weight can change if gravity changes
Mass: 70 kg
Mass: 70 kg
Weight: 700
Newtons
Weight: 112
Newtons
(157 pounds)
(25 pounds)
Mass is
Same!!
MASS
WEIGHT
•Mass is a measure of how
much matter makes up
an object
• Weight is a Force
• Depends on object itself
– NEVER CHANGES
• Mass Is a measure of
inertia
•Mass is measured in:
grams, Kilograms
• Depends on object’s
location
• Weight is measured
in: Newtons (N) or
pounds (lbs)
Mass vs. Weight
You can convert mass into weight using
Newton’s Second Law
F = ma
On Earth, a = 9.8m/s2
Example 1: An object has a mass of 10.0 kg, find its weight
Example 2: A 2500 kg car is pushed with a 250 N force, what
is the acceleration acting on the car due to the force? If the
force is doubled, what will happen to the acceleration?
Example 3: An artillery shell has a mass of 55 kg. The
projectile is fired from the piece and has a velocity of 770 m/s
when it leaves the barrel. The gun barrel is 1.5 m long.
Assuming the force and therefore the acceleration is constant
while the projectile is in the barrel, what is the force that
acted on the projectile?
Equilibrium vs. Static Equilibrium
• If an object is in equilibrium it is not
acceleration (no change in motion)
• If an object is in Static equilibrium it is at
rest (not moving)
• An object is in equilibrium if the sum of
the forces acting on the object is zero
SF = 0
FREE BODY DIAGRAMS
1. Identify all forces acting upon an object
2. Show direction and relative size of forces
3. Vectors originate from the center of mass
Which boxes are in equilibrium?
Only # 3
Which boxes are definitely in static
equilibrium? NONE! b/c it’s possible that
they are all moving
1
2
3
FREE BODY DIAGRAMS
Scenario 1: 10 kg block going upwards
10 kg
SFy = W
W
FREE BODY DIAGRAMS
Scenario 2: 10 kg block pulled across a frictionless
floor by a string
Y Does it fall through
the floor?
FN
Just like projectiles,
we treat x and y
separately
10 kg
W
Normal Force
• perpendicular () to the surface
• equal to the force acting 
on the opposite side of surface
SFy = FN -W = 0
FREE BODY DIAGRAMS
Scenario 2: 10 kg block pulled across a frictionless
floor by a string
X
N
3 Forces
T
10 kg
W
net force only in x direction
S Fy = 0
S Fx = FT
FREE BODY DIAGRAMS
A 1000kg car moving right using its brakes to
Scenario 3:
slow down on a level freeway. Consider
friction.
Does it fall through
Just like projectiles,
the floor?
FN
we treat x and y
separately
Y
10 kg
W
S Fy = FN -W = 0
FREE BODY DIAGRAMS
A 1000kg car moving right, using its
Scenario 3:
brakes to slow down on a level freeway.
Consider friction.
X
N
1000 kg
f
W
WHAT’S UP WITH friction?!
X
f does not actually act on the COM,
but at the surface.
N
1000 kg
f
• Note: f is a
W
force.
• it always acts against the motion
FREE BODY DIAGRAMS
A 1000kg car moving right, using its
Scenario 3:
brakes to slow down on a level freeway.
Consider friction.
X
N
4 Forces
Fapplied
1000 kg
f
net force only in x direction
W
S Fy = 0
S Fx = -Fapplied - f
FREE BODY DIAGRAMS
10 kg block sliding up a ramp at 30° with
Scenario 4:
friction
10 kg
FREE BODY DIAGRAMS
10 kg block sliding up a ramp at 30° with
Scenario 4:
friction
N
There are 3 forces
10 kg
fk
W
FREE BODY DIAGRAMS
10 kg block sliding up a ramp at 30° with
Scenario 4:
friction
N
There are 3 forces
Choose x and y to line
up with the direction
of motion
Weight has both an
x and y component
10 kg
fk
W
FREE BODY DIAGRAMS
q
The angle between
the weight and Fy is
the same as the angle
between the ramp
and the ground
W
Wy= W cosq
W
=
W
sin
q
x
q
FREE BODY DIAGRAMS
Scenario 4: 10 kg block on a frictionless ramp at 30°
N
The y forces cancel
(block does not accelerate in
the y direction)
SFy = N -Wy = 0
the only Net Force is:
SFx= -W sinq - fk
10 kg
fk
W
Fwy
Fwx
Practice!
Drawing a FBD for # 5 from HW 3
35
HW 3: #5
A 34.5 kg block rests on the ramp as shown in the
drawing. What is the tension in the line that is
connected to the top of ramp?
35
Language Objective
Students will be able to explain to each
other what force causes a ball to accelerate
down a hill
Force
Applied
Variable
F1,… F2
Direction
Direction
of push/pull
Direction
of push
Air resistance / Drag
Fair
Opposite to motion
Kinetic Friction
fk
Opposite to motion
Static Friction
fs
Opposite to motion
Weight / gravity
W or Fw
Normal
N or FN
90° to surface
Tension
T or FT
Along the rope/string
Straight down
Practice!
Complete the Free Body Exercises in
your HW packet
ND
2
LAW: EXAMPLE
The Batman, with a mass of 70-kg, rappels down a
rope from his bat-copter with a downward acceleration
of 3.5 m/s2. What vertical force does the rope exert on
Batman?
ND
2
LAW: EXAMPLE
The Batman, with a mass of 70-kg, rappels down a
rope from his bat-copter with a downward acceleration
of 3.5 m/s2. What vertical force does the rope exert on
Batman?
Start all force questions with a diagram
showing the forces
T
Pick the positive direction
(make direction of motion positive)
m1 = 70 kg
anet = 3.5m / s
Identify all givens with symbols
W
2
ND
2
LAW: EXAMPLE
The Batman, with a mass of 70-kg, rappels down a
rope from his bat-copter with a downward acceleration
of 3.5 m/s2. What vertical force does the rope exert on
Batman?
Given: m = 70 kg
anet = 3.5m / s
2
W = mg = (70kg)(9.8m /s )
W = 686 N
Want:
2
T
m1 = 70 kg
anet = 3.5m / s
T
W
2
ND
2
LAW: EXAMPLE
T = 441N
Calculations:
Fnet = mtot anet
SF = W -T = mtot anet
T = W - mtot anet
W
T = (686 N) - (70kg)(3.5m / s )
2
T = 441N
Practice!
Try # 7 from HW 1
AIR RESISTANCE
•Most falling objects don’t accelerate at
9.8 m/s2!
•Because air resistance makes objects
accelerate less 9.8 m/s2!
•Two things determine the magnitude of
air resistance
•surface area (shape)
•speed
TERMINAL VELOCITY
•Terminal velocity is the fastest an object can
fall due to air resistance!
•When FAir resistance = W an object is falling at
terminal velocity and not accelerating
TERMINAL VELOCITY
•When FAir resistance = W the object’s a = zero
10 kg block in freefall traveling at terminal
Scenario 1:
velocity
Fair
10 kg
SFy = 0
Say = 0
W
TIME TO PRACTICE
Start your homework
Turn to pg. 405
Atwood Machine Video
http://www.youtube.com/watch
?v=i2bGTC27OJU
NEWTON’S
RD
3
LAW
•Equal and Opposite
•For every force that one object exerts on a
second object, the second object exerts an
equal (in size) and opposite (in direction)
reaction force on the first object.
NEWTON’S
RD
3
LAW SUBTLETIES
•Forces only come in pairs (you cannot
create a single force)
•Does action cancel out reaction?
•NO! Each force acts on a different object.
•Each force has different effect on their
object!
Action Fore: A bat applies a force to a
baseball.
Action force effect:
•The baseball accelerations a LOT
Reaction force:
•The baseball applies a force to a bat
Reaction force effect:
•The baseball bat (de)accelerates a little
When I am zooming down the highway I
crush small bugs that hit my car
Which exerts the larger force?
a. the bug on my car
b. My car on the bug
•Neither! The forces are equal (but opposite)
How does a rocket work in space if there
is no air to push it?
Lets watch WALL-E
•What happens when the pod reaches
“cruising speed”?
•How does WALL-E move in space when
there is no air or ground for him to push
against?
TIME TO PRACTICE
Check Yourself #11-15
Answers
11.
12.
13.
14.
15.
A
B
B
B
A
http://www.youtube.com/watch?v=8bTdM
mNZm2M
1. Why does the moon orbit the earth?
2. Does the moon put a gravitational pull on the
earth?
3. How does that force compare to the force the
earth puts on the moon?
FRICTION
•Friction is everywhere there is motion on
a surface or within a fluid
•Friction is a resistive force (a force that
opposes motion)
•Friction evidence?
•Heat
•Noise (air resistance)
FRICTION
•Let’s examine sandpaper
40 grit
400 grit
FRICTION
•Even “smooth” sandpaper is rough
40 grit
400 grit
FRICTION
•Slide the surfaces of
two pieces together
•Atoms at the peaks
bond with atoms at
the peaks of the other
•It requires force to
break these bonds
•opposing force is friction
PERPENDICULAR FORCE
•Rub your hands together
•What happens to friction if you press harder?
•So,
f
µ FNormal
NATURE OF SURFACES
•Rub your hands together again
•What would happen to friction if you added
some oil to your hands?
•What would happen to friction if you added
some dry glue?
•This condition of a surface is given as a
quantity known as the coefficient of friction (m)
FORCE OF FRICTION
•Friction between solid surfaces depends on:
1. The perpendicular (Normal) force
between the surfaces in contact
2. The nature of the surfaces in contact
•together, these create a net frictional force
NATURE OF SURFACES
•Each situation has two coefficients of friction
•μs = static friction
•when an object is
being pushed, but
hasn’t moved yet
Surfaces
steel on steel
glass on glass
μs
μk
0.74
0.94
0.57
0.40
tire on dry road 1.0
tire on icy road 0.30
bone joints
0.010
0.80
0.015
0.0030
•μk = kinetic friction
•the friction once the object starts moving
NATURE OF SURFACES
• μs is always bigger
• This means …
It takes more force
to make an object
start moving than it
takes to keep an
object moving
Surfaces
steel on steel
glass on glass
μs
μk
0.74
0.94
0.57
0.40
tire on dry road 1.0
tire on icy road 0.30
bone joints
0.010
0.80
0.015
0.0030
FRICTIONAL FORCE
•Friction is the product of the normal force
and the coefficient of friction
f = m FN
•when μ = 0, surface contact is frictionless
FRICTION EXAMPLE
A 500kg car is going 30 m/s. How long will it
take to come to a stop on a dry road
compared to an icy one?
Givens
m = 500 kg
vo= 30 m/s
vf = 0
Unknown
t=?
adry = ?
aice = ?
Equation
t=
v f - v0
a
FRICTION EXAMPLE
A 500kg car is going 30 m/s. How long will it
take to come to a stop on a dry road
compared to an icy one?
1. Draw a FBD!
2. Identify the + direction
3. Write the Net Force Eqs
N
SFx = - f
SFy = N -W = 0
f
W
FRICTION EXAMPLE
A 500kg car is going 30 m/s. How long will it
take to come to a stop on a dry road
compared to an icy one?
4. Solve for the net
acceleration
SFx = - f = mtot a net
-f
= a net
mtot
N
f
W
FRICTION EXAMPLE
A 500kg car is going 30 m/s. How long will it
take to come to a stop on a dry road
compared to an icy one?
m
fdry = mk FN = (0.8)(500kg)(9.8 2 ) = 3920N
sm
ficy = mk FN = (0.015)(500kg)(9.8 2 ) = 73.5N
s
The difference in
friction is why it takes
so long to stop on ice
N
f
W
FRICTION EXAMPLE
A 500kg car is going 30 m/s. How long will it
take to come to a stop on a dry road
compared to an icy one?
4. Solve for the net
acceleration for each road
- fdry
= a dry
mtot
-3920N
m
= a dry= -7.84 2
s
500kg
N
f
W
FRICTION EXAMPLE
A 500kg car is going 30 m/s. How long will it
take to come to a stop on a dry road
compared to an icy one?
4. Solve for the net
acceleration for each road
- fice
= a ice
mtot
-73.5N
m
= a ice = -0.14 2
s
500kg
N
f
W
FRICTION EXAMPLE
A 500kg car is going 30 m/s. How long will it
take to come to a stop on a dry road
compared to an icy one?
Equation
Givens
m = 500 kg
v f - v0
t=
vo= 30 m/s
a
vf = 0
Unknown
t=?
tdry = 214s
m/s2
adry = -7.84
aice = -0.14 m/s2
tice = 3.83s
Solving Force Problems
1.
2.
3.
4.
Draw a FBD
Identify direction of movement
Write the expression for Fnet = F1+F2-F3…
Set Fnet expression equal to mtotanet
F1+F2-F3…= mtotanet