Honors Physics , Pg 1
Honors Physics Honors Physics
Today’s AgendaToday’s Agenda
Newton’s 3 laws. How and why do objects move? DynamicsDynamics.
Textbook problems Textbook problems Chapter 4 59-80 You should also try answering 41-58
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The Fundamental Forces of our UniverseThe Fundamental Forces of our Universe
Any object with mass will have an attraction to another object with mass Luckily it is VERY WEAK.This is called the Gravitational Force (due to the large
mass of the earth)
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Electromagnetic ForceElectric and magnetic forces Forces that give objects their strength, their ability to
squeeze, stretch, or shatter Very Large compared to the gravitational force
The Fundamental Forces of our UniverseThe Fundamental Forces of our Universe
• Strong Nuclear Force•Holds the particles in the nucleus together•Strongest force (100 times stronger than electromagnetic)
• Weak Nuclear Force•Radioactive decay of some nuclei (enough said)
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GUTGUTGrand Unified TheoryGrand Unified Theory
At one time in the history of our universe (BIG BANG) all of the forces could not be differentiated due to the nature, temperature, and pressure of the universe. Therefore there was only one force which ruled the universe
Mathematical Models of the BIG BANG theory
Based on some observations between Electromagnetic and the WEAK force yielding the the combined Elecrtroweak Force
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DynamicsDynamics
Issac Newton (1643-1727) published Principia Mathematica in 1687. In this work, he proposed three “laws” of motion:
See text: 5-1 and 5-2
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Newton’s First LawNewton’s First Law An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frameinertial reference frame.
If no forces act, there is no acceleration. For Normal Folks- An object at rest remains at rest and an object in motion remains in motion unless acted upon by an external force. The first statement can be thought of as the definition of inertial reference frames.
An IRF is a reference frame that is not accelerating (or rotating) with respect to the “fixed stars”. If one IRF exists, infinitely many exist since they are related by any arbitrary constant velocity vector!
See text:pge 94
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Is Cincinnati a good IRF?Is Cincinnati a good IRF? Is Cincinnati accelerating? YES!
Cincinnati is on the Earth.The Earth is rotating.
What is the centripetal acceleration of Cincinnati? T = 1 day = 8.64 x 10 4 sec, R ~ RE = 6.4 x 10 6 meters .
Plug this in: aU = .034 m/s2 ( ~ 1/300 g) Close enough to 0 that we will ignore it. Cincinnati is a pretty good IRF.
av
RR
TRU
2 22
2
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Newton’s Second Law...Newton’s Second Law...
What is a force?A Force is a push or a pull.A Force has magnitude & direction (vector).Adding forces is like adding vectors.(next chapter)
FF1 FF2
aaFF1
FF2
aa
FFNET
FFNET = maa
See text: pge 93 and 4.2
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Newton’s Second LawNewton’s Second Law
For any object a= FFNET /m
The acceleration aa of an object is proportional to the net force FFNET acting on it and inversely proportional to the objects mass m
For any object, FFNET = FF = maa. The constant of proportionality is called “mass”, denoted m.
This is the definition of mass.The mass of an object is a constant property of that
object, and is independent of external influences.
Force has units of [M]x[L/T2] = kg m/s2 = N (Newton)
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Newton’s Second Law...Newton’s Second Law...
Components of FF = maa :
FX = maX
FY = maY
FZ = maZ
Suppose we know m and FX , we can solve for a and
apply the things we learned about kinematics over the last few weeks:
2
2
1tatvd
atvv
i
if
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Example: Pushing a Box on Ice.Example: Pushing a Box on Ice.
A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50N in the xx direction. If the box starts at rest, what is it’s speed v after being pushed a distance d=10m ?
FF
v = 0
m a
xx
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Example: Pushing a Box on Ice.Example: Pushing a Box on Ice.
A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50N in the x x direction. If the box starts at rest, what is it’s speed v after being pushed a distance d=10m ?
d
FF
v
m a
xx
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Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...
Start with F = ma.a = F / m.Recall that v2
2 - v12 = 2ad (lecture 1)
So v2 = 2Fd / m vFd
m
2
d
FF
v
m a
xx
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Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...
Plug in F = 50N, d = 10m, m = 100kg:Find v = 3.2 m/s.
d
FF
v
m a
xx
vFd
m
2
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ForcesForces
Units of force (mks): [F] = [m][a] = kg m s2 = N (Newton) We will consider two kinds of forces:
Contact force:» This is the most familiar kind.
I push on the desk. The ground pushes on the chair...
Action at a distance (a bit mysterious):» Gravity» Electromagnetic, strong & weak nuclear forces.
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Contact forces:Contact forces:
Objects in contact exert forces.
Convention: FFa,b means “the force acting on a due to b”.
So FFhead,thumb means “the force on the head due to the thumb”.
FFhead,thumb
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Gravity...Gravity...
Near the earth’s surface...
But we have just learned that: FFg = maaThis must mean that g is the “acceleration due to
gravity” that we already know!
So, the force on a mass m due to gravity near the earth’s surface is FFg = mgg where gg is 9.8m/s2 “down”.
F GM m
Rm G
M
Rmgg
e
e
e
e
2 2
g GM
Re
e
2and
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Example gravity problem:Example gravity problem:
What is the force of gravity exerted by the earth on a typical physics student?
Typical student mass m = 55kgg = 9.8 m/s2.Fg = mg = (55 kg)x(9.8 m/s2 )
Fg = 539 N
The force that gravity exerts on any object is called its Weight
FFg
See text example Mass and Weight.
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Newton’s Third Law:Newton’s Third Law:
Forces occur in pairs: FFA ,B = - FFB ,A.
For every “action” there is an equal and opposite “re-action”.
In the case of gravity:
F F121 2
122 21 G
m m
R
R12
m1m2
FF12 FF21
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Newton’s Third Law...Newton’s Third Law...
FFA ,B = - FFB ,A. is true for contact forces as well:
FFm,w FFw,m
FFm,f
FFf,m
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Example of Bad ThinkingExample of Bad Thinking
Since FFm,b = -FFb,m why isn’t FFnet = 0, and aa = 0 ?
a ??a ??FFm,b FFb,m
ice
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Example of Good ThinkingExample of Good Thinking Consider only the box only the box as the system!
FFon box = maabox = FFb,m
Free Body Diagram (next time).
aaboxbox
FFm,b FFb,m
ice
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The Free Body DiagramThe Free Body Diagram
Newtons 2nd says that for an object FF = maa.
Key phrase here is for an objectfor an object..
So before we can apply FF = maa to any given object we isolate the forces acting on this object:
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ExampleExample
Example dynamics problem:
A box of mass m = 2kg slides on a horizontal frictionless floor. A force Fx = 10N pushes on it in the xx direction. What is the acceleration of the box?
FF = Fx ii aa = ?
m
yy
x x
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Example...Example...
Draw a picture showing all of the forces
FFFFBF
FFFBFFBE
FFEB
y y
x x
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Example...Example...
Draw a picture showing all of the forces. Isolate the forces acting on the block.
FFFFBF
FFFB = mgg
y y
x x
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Example...Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram.
FFFFBF
y y
x x
FFFB = mgg
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Example...Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. Solve Newtons equations for each component.
FX = maX
FBF - mg = maY
FFFFBF
mgg
y y
x x
See strategy: Solving Newton’s Law Problems,
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Example...Example... FX = maX
So aX = FX / m = (10 N)/(2 kg) = 5 m/s 2.
FBF - mg = maY
But aY = 0 So FBF = mg.
The vertical component of the forceof the floor on the object (FBF ) isoften called the Normal Force Normal Force (N).
Since aY = 0 , N = mg in this case.
FX
N
mg
y y
x x
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Problem: ElevatorProblem: Elevator
A student of mass m stands in an elevator accelerating upward with acceleration a. What is her apparent weight?
Apparent weight = the magnitude of the normal force of the floor on her feet.
This is the weight a scale would read if she were standing on one!
See example pge. 99: An Elevator
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Elevator...Elevator...
maa
mggNN
First draw a Free Body Diagram
of the student:
Recall that FFNet = maa.yy
See text: 6-1
See example p 99: An Elevator
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Elevator...Elevator...
Add up the vectors accordingly!
FFNet =
In this case FFNet = NN - mgg.
(note that NN and g g are vectors) Considering the yy (upward) component:
N - mg = ma
N = m (g + a)
maamggNN + =
yy
See text: 6-1
See example p 99: An Elevator
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Elevator...Elevator...
ma
mgN
Interesting limiting cases:
If a = 0, N = mg (ok). Like previous example.
If a = -g, N = 0 (free fall). The vomit comet!
N = m (g + a)
yy
See text: 6-1
See example : An Elevator
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Scales:Scales:
Springs can be calibrated to tell us the applied force. We can calibrate scales to read Newtons, or...Fishing scales usually read
weight in kg or lbs.
02468
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Tools: Pegs & PulleysTools: Pegs & Pulleys
Used to change the direction of forces.
An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:
FF1 ideal peg
or pulley
FF2
| FF1 | = | FF2 |
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Tools: Pegs & PulleysTools: Pegs & Pulleys
Used to change the direction of forces.
An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:
mg
T
m T = mg
FW,S = mg
Honors Physics , Pg 38
Recap of Newton’s 3 laws of motionRecap of Newton’s 3 laws of motion
Newtons 3 laws:
Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an
inertial reference frame.
Law 2: For any object, FFNET = FF = maa
Law 3: Forces occur in pairs: FFA ,B = - FFB ,A. For every “action” there is an equal and opposite “re-action”.
Textbook problemsTextbook problems