Transcript Newton`s 2nd Law of Motion

Newton’s 1st Law of Motion
Newton’s 1st Law
• Newton’s 1st Law: An object at _______stays
at _______and an object in motion stays in
motion, unless acted upon by an
___________(i.e. net or outside) force
• This motion is in a ___________line, at a
_____________speed
• Often called an objects inertia
What Does Inertia Have To Do With
Rugby?
• Jonah Lomu:
–New Zealand All Blacks
–Mass: 120 kg (translates to
273 lbs)
–Height: 6’ 5’’
–Speed: Runs 100 meters in
10.8 seconds
(Carl Lewis: 100 meters in 9.88
seconds)
What Does Inertia Have To Do With
Rugby?
• Jonah Lomu:
–Very Big (lots of mass – 120 kg)
– Lots of Inertia
–Because of Large Mass (i.e., large amount of
inertia) he is very resistant to changes in motion
•HARD TO STOP WHEN MOVING
•HARD TO START MOVING WHEN AT REST
What Does Inertia Have To Do With
Rugby?
• So when Jonah Lomu
is in motion and
encounters someone
less massive (i.e., with
less INERTIA)…
• HE IS MORE RESISTANT TO THEIR
ATTEMPTS TO CHANGE HIS STATE OF
MOTION
(i.e., bring him to a rest)
What Does Inertia Have To Do With
Rugby?
• So when Jonah
Lomu is at rest and
encounters
someone less
massive (i.e., with
less INERTIA)…
• He is more resistant to their attempts to
change his motion.
What Does Inertia Have To Do With
Rugby?
• Because of his mass, it
is harder to change his
direction when in
motion.
• WHY?
–INERTIA!
•Straight Line Motion
•Inertia resists changes in
motion (in this case the
DIRECTION of the motion)
•The GREATER the MASS, the HARDER it is to CHANGE DIRECTION
Challenge!
• Why does my water bottle roll under my
seat when leaving school!
• Why do you shift left when you turn your
car right?
Newton’s 2nd Law of
Moton
Newton’s 2nd Law
• The acceleration of an object is directly
proportional to the net external force acting
on the object and inversely proportional to
the mass of the object
or:
____________________
Forces
• A push or pull
• The cause of an _________________
• Cause of a change in an object’s state of
motion
–Cause objects to speed up or slow down
–Cause a change of direction
• Unit of force: ________________________
Types of Forces
• Common forces:
–Applied – (FA or F) – Generic name for any force
that is applied
–Tension - (Ft or T) - The force that a “string”
_________on an object
–Normal Force - (FN or N) - Force that a surface
applies to an object (the direction is  to the
surface)
–Weight – (Fg or w) – The force due to
______________(mg)
–Friction - (Ff or f ) - To be defined later
FN, Normal Force (N)
• A force that a surface applies to an
object
• “Normal” means perpendicular
• The direction of the normal force is
perpendicular to the surface
surface
Free Body Diagrams
• Used to analyze the forces
affecting the motion of a single
object
• Shows only the forces acting on
an object
Free Body Diagrams
•One object only
•Shows only forces acting on the
object
•Forces represented as arrows
•Fg down, FN perpendicular to
surface
Examples
Friction
• Force of Friction: Ff , Unit: N – Newton
• μ: ____________of Friction (Unitless)
• Static vs Kinetic
•Static Friction –Two surfaces at _________relative to one another (not
sliding)
•
Always parallel to the surface
• Kinetic Friction – Two surfaces ___________relative to one
another (Sliding)
•
Always ______________to the surface and opposes the motion
Common μ’s
Materials
Oak on oak, dry
μ
0.30
Steel on steel,
dry 0.41
greasy 0.12
Steel on ice
Rubber on asphalt,
0.01
Dry 1.07
wet 0.95
Rubber on ice
0.005
Sample Problems
• It takes 50 N to pull a 6.0 kg object along a desk
at constant speed. What is the coefficient of
friction?
• The coefficient of friction between two
materials is 0.35. A 5.0 kg object made of one
material is being pulled along a table made of
another material. What is the force of friction?
Newton’s 3rd Law
• Every action force has an ______________and
equal reaction force.
–The two forces are called force pairs.
–They act on ________________objects.
–How do we walk by pushing backwards?
–Why do cannons recoil when fired?
Challenge
• Is the force of gravity that the Earth exerts
on the Moon greater than, equal to, or less
than the force the Moon exerts on the Earth?
• How does a hammer push a nail into wood
in light of Newton’s 3rd Law?
Centripetal Force
• ac, __________________acceleration (m/s2)
– “Center seeking”
– Necessary to keep the object traveling in a circle
• Fc, centripetal force (N)
– Newton’s 2nd Law
FNET = ma
• Newton’s 2nd Law for Circles
• NOTE: CENTRIFUGAL FORCE IS AN IMAGINARY
FORCE TO EXPLAIN INERTIA! DOES NOT EXIST!
Centripetal Force
• The centripetal acceleration is caused by
another force such as:
–Friction
–Tension
–Normal Force
–Gravity
Centripetal Force
• Steps to solve centripetal force problems:
–Draw a free body diagram
–Identify the force causing the centripetal force
–Set the causal force = centripetal force eq.
–Solve for unknown
A 0.50 kg box is attached to string on a frictionless horizontal table. The box
revolves in a circle of radius 2.8 m. If the box completes 1 revolution every
2.0 seconds, what is the tension in the string?
FN
r
FT
FT = FC
v2
FT = m
r
2pr 2p (2.8)
v=
=
= 8.80
T
2
8.80
FT = 0.5
2.8
2
FT = 13.8N
Fg
A 1200 kg car approaches a circular curve with a
radius of 45.0 m. If the coefficient of friction
between the tires and road is 1.20, what is the
maximum speed at which the car can negotiate the
curve?
r
A 0.025 kg rubber stopper connected to a string is swung in a
horizontal circle of radius 1.20 m. If the stopper completes 5
revolutions in 2 seconds. Calculate the period of revolution of
the stopper, the magnitude of the velocity of the stopper, the
magnitude of the stopper’s centripetal acceleration and the
tension in the string.
In a popular amusement park ride, a cylinder of radius
3.0 m is set in rotation at a speed of 5.6 m/s.
The floor drops away, leaving the riders suspended
against the wall in a vertical position What minimum
coefficient of friction between a 100.0 kg rider’s
clothing and the wall of the cylinder is needed to
keep the rider from slipping?
Vertical Circle I
Vertical circle 2
Vertical circle 3
A 800 kg car drives down a
hill. If it is going 11 m/s at the
bottom of the hill, what is the
force the road is exerting on
the car (FN)?
r=13 m
FN
FN  mg  Fc
2
Fg
v
FN  m
 mg
r
112
FN  800
 800 * 9.8
13
FN  15286N
A 800 kg car goes up a hill. If
it is going 11 m/s at the top of
the hill, what is the force the
road is exerting on the car
(FN)?
FN
Fg
r=13 m
mg  FN  Fc
FN  mg  Fc
2
11
 394N
FN  800 * 9.8  800
13