Re: Aberation and the Speed of Gravity

"JM Albuquerque" <jmDOTa2@xxxxxxx> wrote in
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"JM Albuquerque" <jmDOTa2@xxxxxxx> wrote in
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"JM Albuquerque" <jmDOTa2@xxxxxxx> wrote in


To calculate the friction force you need to know the
normal force. You can only calculate the normal force when
you make a force balance based on a coordinate system fixed
to the sliding box.


Incorrect. It is EASIER. It is NOT necessary. To calculate the normal
force all I need to know is the angle BETWEEN gravity and the
direction that the box is moving. I could be standing on my head and
referencing g and the plane to the angles I see from that point of
view. The difference between those directions gives me the angle and
allows me to calculate the normal force.

You are too used to 'doing it the "easy" way' to realize that it is
not the ONLY way.

.....
You cannot calculate it the way you've said.
BTW, whatever the angle you chose that angle reads
90 degrees. So, you only get "0" or "1" out of it.

How about this one?

A box with a mass of 23 gm is at rest on an inclined surface. When the
object is about to start sliding, the inclined plane makes an angle of
13° with the horizontal. What is the static friction coefficient
between the two surfaces?.

Now, tell me why that problem could not be worded and worked in any
coordinate frame I might choose?

And where is a zero or 90 degrees? Yes, they are involved but there are
other angles involved also.


A box sliding on an inclinated surface has nothing to do with
rotations, nor centrifugal force.
There is no fixed point around which rotation takes place.

The next problem could have involved a car following a banked curve.

The car, in that case, represents a box on an inclined plane. The car has
momentum in its current direction, the plane is inclined at a certain
angle, We want to avoid exceeding the static coefficient of friction UP
the plane.

You would include the distance from the center of the circle. I would not
need to bother with that, I just need to know the degree of bank, the
velocity vector that the car needs to follow, the current velocity vector,
and the cars mass.

So, I cannot see the above any valid argument against what
I was saying, nor about the subject of this thread.

Point being that is is NOT necessary to tie our coordinate system to a
particular point to calculate the forces. It may make it easier, but it is
not necessary.

Gravity?
You are saying gravity on a Merry-go-round ?
??????????????????

If you are standing on the rotating platform, there is gravity [and
friction, otherwise you would not be standing there].

I don't call that a merry-go-round, I call that a turn-table or
spinning disk problem.
Nevertheless, the term "merry-go-round" is not used in my
native language, so I assume that I could be responsible for
some confusion on what a "merry-go-round" is for English
native people.

Sorry, I could not say any of this in Portuguese. My wife is fluent in
Spanish and reads French and Italian but I am barely fluent in English.

What you seem to have thought I was talking about is called 'the rotor' in
english.

However, the same box, sliding on a merry-to-round would require
a completely different set of 'fictional' forces to explain its
actions.

The box is held to the platform by gravity. The frictional force of the
rotating platform accelerates the box so that it is traveling at the
same speed as the part of the platform immediately under the box.

The box wants to continue in a straight line, but the merry-go-round is
turning.

The static friction has been overcome so all that we have left
'convincing the box' to travel in the circle is the sliding frictional
forces. So the box begins to slide toward the outside edge of the
platform.

Yes, the box slids toward the outside. Actually it draws a spiral.

Yes.

A spiral is composed by rotation plus an outward motion.
The rotation is due to tangential accelerations

Which is caused by friction between the box and the platform overcoming
the inertia of the box(we assume the box and the platform were not
rotating at the beginning and the platform starts slowly. As it approaches
'operating speed, the box, at first is stationary wrt the platform and
then goes sliding and flying.

, the radial
outward motion is due to centrifugal force - a true force here.

The outward motion is because the friction is insufficient to overcome the
inertia as the box attempts to continue along the tangent (follow a
straight line at its current velocity at each instant in time)

..... big snip....

We were seeing different things, that is all. You missed important
clues. The box certainly could not fly off of the rotating cylinder you
were seeing.

Yep.
But since people are not used to fly out of merry-go-round devices,
for a matter of safety, I never thought that a turn-table could be
regarded as a merry-go-round.

I can calculate the friction force of the box around the inner
surface of the said cylinder in less then 30 seconds.
Since friction doesn't depend on speed, we have:
F_friction = centrifugal force * friction coefficient
= m*r*w^2 * friction coefficient
or else, since v_tangential = w*r,

You calculated the tangential velocity. That is what the box is attempting
to follow at every instant; it's current tangential velocity.

I calculate the tangential velocity at any instant and the delta v is the
acceleration experienced at each moment.

F friction at each instant is mass times delta v at that instant.
I never even need to calculate the next step so the problem is simpler.

centrifugal force = m*v^2 / r

If you want to use any other FoR you will never get the result that
easy (if you get a result at all).

I got them easier.

.....

Since very few things can accelerate without frictional forces.
You could not walk without them. Try walking on ice, where frictional
forces are minimal. Try driving a car or stopping a car without them.

Yes, I know that you know and I should had make such a
comment, but you actually said:
""Frictional forces accelerating it ...""

They do. The platform and the box were stationary. The static friction
between the platform and the box accelerated the box to begin with. As the
box started to slide, there were still frictional forces preventing the
box from following the straight line. THOSE frictional forces continue to
accelerate the box until it flies off the edge of the platform. ....

I agree that centrifugal force, the way most people use the term is
erroneous. In my opinion, the term can and should be avoided to prevent
confusion.

The term is extremelly usefull when applicable.

The problem is that is almost never really necessary and it is confusing.

If you want to re-read all I've said on this thread you
will notice that I never missed that fixed stiff point.

Do you agree with my point?

Of course I agree.
This subject is so easy that I can't see why we don't agree.

It is good to be agreeable, at last. :)
Then there has been enough said about this subject.




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

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