l  LON-CAPA #3 and Mastering Physics due next
Tuesday
◆  help room hours (Strosacker Help Room, 1248
BPS):
M: 5-8 PM
W: 5-8 PM
F: 2-6 PM
l  Register for Mastering Physics
◆
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l  First iclicker question from Tuesday will not be graded
because of iclicker problems
!
!
l  First exam: Feb 6 in Life Sciences A133
Resolving vectors
l Just as we can add
any two vectors to
get a resultant vector,
we can also resolve
any vector into two
components
perpendicular to each
other
◆
as for example, the
vertical and horizontal
components of a
thrown ball
Note that there is a change in the
velocity of the ball. It is accelerating
in the downward direction.
!
!
Forces
l  A girl hangs from a rope
l  What are the forces involved?
l  Which end of the rope is more
likely to break?
l  The force of gravity is acting
downward
l  …so there must be an equal
and opposite reaction force
l  That force can be divided into
vector components along the
two ropes
l  The larger force is on the rope
segment to the right…so that
is where the rope is more
likely to break
!
!
Friction
l  Consider the forces acting on a
sled initially moving with a
velocity v
l  Gravity is pulling downward with a
force F
l  There’s an equal and opposite
normal force N applied from the
sled to the ground
l  There’s a frictional force
proportional to the normal force
€
◆  fk=µkN
◆  where µk is called the
coefficient of kinetic friction
l  So there’s no net force in the
vertical direction
l  There is a net force in the
horizontal direction
l  So what happens to the sled

N

N
€
!
!

F
Static friction
l  What about when there’s no
motion
l  Is there still friction?
l  Yes
l  There’s a gravitational force
acting downward
l  It can be resolved into two
components
◆
◆
one perpendicular to the surface,
balanced by the normal force
one along the surface
l  There’s a frictional force also
along the surface, proportional to
the normal force
◆
fs=µsN
l  If the horizontal forces are
balanced, the frog stays put
l  What happens if I start tilting the
frog?
!
!
Motion of a car
l Consider all of the
forces acting in this
situation
air resistance
l  As the tire moves over the road,
the tire flattens. The force exerted
by the road on the tire acts to
slow the tire’s rotation.
l  Rolling resistance
l  Less pronounced for steel wheels
normal force
gravity
!
!
Gravitation
l  Remember the story
about Isaac Newton and
the apple
l  Maybe it didn’t happen in
reality, but the reasoning
is roughly correct
l  Newton had the insight to
realize that the force
between the Earth and
the apple is the same as
the force between the
Earth and the Moon
!
!
Gravitation
l  The apple falls to the ground
with an acceleration of 9.8 m/
s2
l  The Moon falls away from the
straight line path it would
follow if there were no forces
acting on it
◆
remember Newton’s first law
l  Isaac Newton compared the
measurements for the Moon’s
acceleration to his
calculations
◆
◆
◆
and found they didn’t agree
so he put his theory away for
20 years, and worked on
optics
…and alchemy
!
!
Gravitation
l Inspired by the
appearance of
Halley’s comet in
1682, he returned to
the problem and
found errors in the
experimental data
l Now the data agreed
with his predictions
l He then published
the universal law of
gravitation
!
!
What did he find?
l  He found the acceleration of
the moon to be ~2.7X10-3 m/s2
l  Radius of Earth =6.37X106 m
l  Average Earth-Moon distance
is 3.84X108 m
l  So the Moon is about 60 times
as far away from the center of
the Earth as is someone
standing on the surface of the
Earth
9.83m /s2
2
≈
3600
=
60
2.7X10 −3 m /s2
l  The acceleration due to
gravity decreases as the
square of the distance from
€ the center of the Earth
!
!
video
Universal law of gravitation
l  Incredibly simple and
beautiful relation
mass1 Xmass2
distance 2
m1m2
Force ∝ 2
d
Force ∝
€
l  The gravitational force
between two masses is
proportional to the €
product of the two
masses and is inversely
proportional to the
square of the distance
between them
l  Before we dealt with
mass in the equation
below, technically known
as the inertial mass
F = ma
l  Here we are dealing with
the gravitational mass
l  It is experimentally
established that the two
are equal and in fact the
equivalence of the two is
an integral part of the
general! theory of
!
relativity
Universal law of gravitation
l  We need a real equation,
which means we need a
constant of proportionality
m1m2
F=G 2
d
€
l  Newton didn’t know the value
of G by itself, only the product
of G times the mass of the
Earth
l  The measurement of G was
big news in 1798
l  Knowing G meant that the
mass of the Earth could be
calculated
◆  6X1024kg
G was determined by
measuring the attraction of two
masses; a difficult measurement
since the force is very small
G=6.67X10-11 Nm2/kg2
!
!
How small of a force?
m1m2
F=G 2
d
F = 6.67X10
−11
m1=1kg
m2=5000kg
d=1m
(1kg)(5000kg)
Nm /kg2
(1m) 2
2
F = 3.34 X10 −7 N
G was determined by
measuring the attraction of two
masses; a difficult measurement
since the force is very small
The small value of G is an indication of how
weak the gravitational force is.
G=6.67X10-11 Nm2/kg2
!
!
Weight
l  Let’s consider another
force, your weight, i.e.
the force the Earth exerts
on you
l  Suppose you weigh 60
kg
l  m1=mEarth=6X1024 kg
l  m2=myou=60 kg
l  d=Rearth=6.37X106 m
F=G
m1m2
d2
F = 6.67X10
F = 591N
−11
(6X10 24 kg)(60kg)
Nm /kg
(6.37X10 6 m) 2
2
2
!
!
Clicker question
l  What is the gravitational force
that you exert on the Earth?
◆
◆
◆
◆
◆
F=G
A) 0 N
B) 591 N
C) 6.67X10-11 N
D) 6X1024 N
E) cannot be determined with the
information given
m1m2
d2
F = 6.67X10
−11
(6X10 24 kg)(60kg)
Nm /kg
(6.37X10 6 m) 2
2
2
F = 591N
!
!
€
Clicker question
l  What is the gravitational force
that you exert on the Earth?
◆
◆
◆
◆
◆
F=G
A) 0 N
B) 591 N
C) 6.67X10-11 N
D) 6X1024 N
E) cannot be determined with the
information given
m1m2
d2
F = 6.67X10
−11
(6X10 24 kg)(60kg)
Nm /kg
(6.37X10 6 m) 2
2
2
F = 591N
!
!
€
Acceleration
l  What is your acceleration due
to the gravitational force of the
Earth?
mEarth myou
G
2
F
d
a=
=
myou
myou
a=G
€
mEarth
= 9.8N /kg = 9.8m /s2 = g
2
d
l  That is, your acceleration is
the same as your friend who
has twice the mass that you
do (and weighs twice as
much)
!
!
m1m2
F=G 2
d
The gravitational force grows weaker as the square
of the distance. Inverse square law.
€
!
!
Inverse square law
l  The further away from
the center of the Earth
that you travel, the
smaller your weight will
be
l  If you weigh 600 N at sea
level, you’ll weigh 598 N
on top of Mt Everest
l  The gravitational force
falls off to zero as the
distance from the center
of the Earth goes to
infinity
!
!
Weightlessness
l  Suppose that you’re in an
elevator that’s stationary (or
moving with a constant velocity)
◆  you stand on a scale and it
gives your normal weight
l  Suppose the elevator accelerates
upward
◆  the scale registers a larger
weight; you feel heavier
l  The elevator accelerates
downward
◆  the scale registers a lower
weight; you feel lighter
l  The elevator cable snaps
◆  the scale registers zero
weight; you feel weightless
!
!
Weightlessness
l If you have enough
money, you book a
flight on a ‘vomit
comet’ that lets you
experience
weightlessness for a
brief period of time
l Or you can spend
some time on the
International Space
Station
!
!
Space elevator
l  One of the problems refers to
a space elevator
l  This would be a convenient
way of transporting material
into a geosynchronous orbit
without the use of any rockets
l  One end of the elevator is
attached to the surface of the
Earth, the other to a
counterweight
l  It’s stable
l  But we don’t currently have
the technology to build a
strong enough tower
l  Carbon nanotubes are close;
they would work for an
elevator on the Moon or on
Mars
!
!
!
!
http://www.pbs.org/wgbh/nova/space/space-elevator.html
Central force
l  Suppose I have a mass m1 and I want
to calculate the gravitational force on
this mass from the Earth
l  What is the direction of the force on
the mass?
◆  towards the center of the Earth
l  What is the direction of the force on
the Earth?
◆  along the line joining m1 and the
center of the Earth, pointing
towards m1
l  What is the magnitude of the force?
m1
d
m1m2
F=G 2
d
€
l  The Earth has a radius of 6.37X106
m, but from a gravitational point of
view, it acts like all of the mass is
concentrated in a point at the center
of the Earth
How do we! know that? Isaac Newton had
to develop !calculus to prove it.
Roundness
l  Why is the Earth round?
l  Because every part of
the Earth attracts every
other part and so the
Earth is pulled together
as tightly as possible
◆  a sphere
l  The Moon is round for a
similar reason
l  The asteroids, except for
the largest ones, are not
round, because they
don’t have sufficient
mass
!
!
Oblateness of the Earth
l  We discussed that gravity is
responsible for the planets
being round
l  But they’re not exactly round
due to the rotational motion of
the Earth
l  The Earth tends to be
somewhat thicker at the
equator than at the poles
l  Not by much
◆  12756 km at equator
◆  12714 km at the poles
l  So a round sphere is still a
very good representation of
the shape of the Earth
How different is the force of gravity
at the North Pole and at the Equator?
Take a mass of 50 kg.
m1m2
F=G 2
d
!
!
Oblateness of the Earth
l  The Earth tends to be
somewhat thicker at the
equator than at the poles
l  Not by much
◆  12756 km at equator
◆  12714 km at the poles
Fequator = W equator = G
Fequator = (6.67X10
−11
mmEarth
d2
(50kg)(6X10 24 kg)
Nm /kg )
(6.378X10 6 m) 2
2
2
Fequator = 491.9N
24
Fpole = (6.67X10 −11 Nm 2 /kg 2 )
(50kg)(6X10 kg)
(6.357X10 6 m) 2
How different is the force of gravity
at the North Pole and at the Equator?
Take a mass of 50 kg.
Fpole = 495.2N
How much did the mass change?
m1m2
F=G 2
d
!
!
Jupiter
l  Jupiter is even more oblate
since
◆  it’s much larger than the
Earth
◆  it rotates much faster (one
day = 10 hours)
◆  it’s composed mostly of
fluid
l  How much would this 50 kg
person weigh on the equator
of the surface of Jupiter?
l  Jupiter is 300 times as
massive as the Earth
l  Why isn’t the weight 300 times
as much?
€
F =W =G
mmJupiter
d2
F = (6.67X10
F = 1246N
−11
!
!
(50kg)(1.90X10 27 kg)
Nm / kg )
(7.13X10 7 m) 2
2
2
Neutron star
l  A neutron star has a mass of
4 X 1030 kg (about twice the
sun’s mass) and a radius of
10 km (about 1/70000th that of
the sun)
l  Remember that most of an
atom is empty space; in a
neutron star all of the atoms
€
collapse down to the size of
their nuclei
l  What would be the weight of a
50 kg person on the surface of
this neutron star?
F=G
mmneutronstar
d2
F = (6.67X10
−11
(50kg)(4 X10 30 kg)
Nm /kg )
(1X10 4 m) 2
2
2
F = 1.33X1014 N
l Or about 300 billion
times as much as on
the surface of the
Earth
l A neutron star has
the same density as
!
the nucleus
of an
!
atom