6 newton law and gravitation

Upload
galihsuryono 
Category
Education

view
318 
download
2
Transcript of 6 newton law and gravitation
Physics 111: Lecture 4, Pg 1
Physics 111: Lecture 4Physics 111: Lecture 4
Today’s AgendaToday’s Agenda
Recap of centripetal acceleration Newton’s 3 laws
How and why do objects move? DynamicsDynamics
Physics 111: Lecture 4, Pg 2
Review: Centripetal AccelerationReview: Centripetal Acceleration
UCM results in acceleration:Magnitude: a = v2 / R = R Direction:  rr (toward center of circle)
R aa
^̂
Useful stuff:Useful stuff:
f = rotations / sec
T = 1 / f
ω = 2 / T = 2 f = rad/sec
v = R
Physics 111: Lecture 4, Pg 3
Isaac Newton (1643  1727) published Principia Mathematica in 1687
Physics 111: Lecture 4, Pg 4
DynamicsDynamics
In this work, he proposed three “laws” of motion:
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 reaction.)
Physics 111: Lecture 4, Pg 5
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.
The following statements 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!
1. Dishes2. Monkey
Physics 111: Lecture 4, Pg 6
Is Urbana a good IRF?Is Urbana a good IRF? Is Urbana accelerating? YES!
Urbana is on the Earth.The Earth is rotating.
What is the centripetal acceleration of Urbana? 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. Urbana is a pretty good IRF.
av
RR
TRU
2 22
2
Icepuck
Physics 111: Lecture 4, Pg 7
Newton’s Second LawNewton’s Second Law
For any object, FFNET = FF = maa.
The acceleration aa of an object is proportional to the net force FFNET acting on it.
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 thatobject, and is independent of external influences.
Force has units of [M]x[L / T2] = kg m/s2 = N (Newton)
Physics 111: Lecture 4, Pg 8
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.
FF1 FF2
aaFF1
FF2
aa
FFNET
FFNET = maa
Physics 111: Lecture 4, Pg 9
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 aX and apply
the things we learned about kinematics over the last few weeks:
tavv
ta21
tvxx
xx0x
2xx00
+=
++=
Physics 111: Lecture 4, Pg 10
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 50 N in the ii direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10 m?
FF
v = 0
m a
ii
Physics 111: Lecture 4, Pg 11
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 50 N in the ii direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ?
d
FF
v
m a
ii
Physics 111: Lecture 4, Pg 12
Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...
Start with F = ma.a = F / m.Recall that v2  v0
2 = 2a(x  x0 ) (Lecture 1)
So v2 = 2Fd / m vFd
m
2
d
FF
v
m a
ii
Physics 111: Lecture 4, Pg 13
Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...
Plug in F = 50 N, d = 10 m, m = 100 kg:Find v = 3.2 m/s.
d
FF
v
m a
ii
vFd
m
2
Physics 111: Lecture 4, Pg 14
Lecture 4, Lecture 4, Act 1Act 1Force and accelerationForce and acceleration
A force F acting on a mass m1 results in an acceleration a1.The same force acting on a different mass m2 results in an acceleration a2 = 2a1.
If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration?
(a)(a) 2/3 a1 (b)(b) 3/2 a1 (c)(c) 3/4 a1
F a1
m1 F a2 = 2a1
m2
F a = ? m1 m2
Physics 111: Lecture 4, Pg 15
Lecture 4, Lecture 4, Act 1Act 1Force and accelerationForce and acceleration
Since a2 = 2a1 for the same applied force, m2 = (1/2)m1 !
m1+ m2 = 3m1 /2
(a)(a) 2/3 a1 (b)(b) 3/2 a1 (c)(c) 3/4 a1
F a = F / (m1+ m2)m1 m2
So a = (2/3)F / m1 but F/m1 = a1
a = 2/3 a1
Physics 111: Lecture 4, Pg 16
ForcesForces
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:» Gravity» Electricity
Physics 111: Lecture 4, Pg 17
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
Physics 111: Lecture 4, Pg 18
Action at a distanceAction at a distance
Gravity:
Physics 111: Lecture 4, Pg 19
GravitationGravitation(Courtesy of Newton)(Courtesy of Newton)
Newton found that amoon / g = 0.000278 and noticed that RE
2 / R2 = 0.000273
This inspired him to propose the Universal Law of Gravitation:Universal Law of Gravitation: FMm = GMm / R2
R RE
amoong
where G = 6.67 x 10 11 m3 kg1 s2
Physics 111: Lecture 4, Pg 20
Gravity...Gravity...
The magnitude of the gravitational force FF12 exerted on an object having mass m1 by another object having mass m2 a distance R12 away is:
The direction of FF12 is attractive, and lies along the line connecting the centers of the masses.
F Gm m
R12
1 2
122
R12
m1 m2FF12 FF21
Physics 111: Lecture 4, Pg 21
Gravity...Gravity...
Near the Earth’s surface:R12 = RE
» Won’t change much if we stay near the Earth's surface.
» i.e. since RE >> h, RE + h ~ RE.
RE
m
M
h2E
Eg R
mMGF FFg
Physics 111: Lecture 4, Pg 22
Gravity...Gravity...
Near the Earth’s surface...
2
E
E2E
Eg R
MGm
RmM
GF
So Fg = mg = ma
a = g
All objects accelerate with acceleration g, regardless of their mass!
22E
E s/m81.9RM
Gg Where:
=g
Leaky Cup
Physics 111: Lecture 4, Pg 23
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
FFg The force that gravity exerts on any object is
called its Weight
W = 539 N
Physics 111: Lecture 4, Pg 24
Lecture 4, Lecture 4, Act 2Act 2Force and accelerationForce and acceleration
Suppose you are standing on a bathroom scale in 141 Loomis and it says that your weight is W. What will the same scale say your weight is on the surface of the mysterious Planet X ? You are told that RX ~ 20 REarth and MX ~ 300 MEarth.
(a)(a) 00.75.75 W (b)(b) 1.5 W
(c)(c) 2.25 W
E
X
Physics 111: Lecture 4, Pg 25
Lecture 4, Lecture 4, Act 2Act 2SolutionSolution
The gravitational force on a person of mass m by another object (for instance a planet) having mass M is given by:
F GMm
R
2
WW
FF
X
E
X
E
Ratio of weights = ratio of forces:
G
M m
R
GM m
R
X
X
E
E
2
2
MM
RR
X
E
E
X
2
WW
X
E
300120
752
.
Physics 111: Lecture 4, Pg 26
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 “reaction”.
We have already seen this in the case of gravity:
F F121 2
122 21 G
m m
R
R12
m1m2
FF12 FF21
Newton’sSailboard
Physics 111: Lecture 4, Pg 27
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
2 Skateboards
Physics 111: Lecture 4, Pg 28
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
Physics 111: Lecture 4, Pg 29
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
Physics 111: Lecture 4, Pg 30
Lecture 4, Lecture 4, Act 3Act 3Newton’s 3rd LawNewton’s 3rd Law
Two blocks are stacked on the ground. How many actionreaction pairs of forces are present in this system?
(a) 2
(b) 3
(c) 4
a
b
Physics 111: Lecture 4, Pg 31
Lecture 4, Lecture 4, Act 3Act 3Solution:Solution:
(c) 4
FFE,aE,a
FFa,Ea,E
a
b
FFE,bE,b
a
bFFb,Eb,E
FFb,ab,a
FFa,ba,b
a
b
FFg,bg,b
FFb,gb,g
a
b
Physics 111: Lecture 4, Pg 32
Recap of today’s lectureRecap of today’s lecture
Newton’s 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.
ExtinguisherCart