question..
Would this iron bar float, for example
in water?
No, right?
However, by conveniently changing the shape,
of this bar
yes it would  yes it floats....How is that possible??
Well, the Archimedean Principle explains it
and we will see it next.
Good
to analyze the principle
Archimedes
What I have done is a drawing where
we have a container that contains
has a liquid, for example water
but in case the liquid tends to rise
the level of that liquid
that liquid that would have risen
is going to evict
if this hole were not present, the liquid level will rise
and we collect it here in a container
Now, what is the Archimedean Principle?
if we introduce a body
in the liquid
that liquid
tends to rise, in the container
but as the hole is, it will drain
and if we weighed
that liquid
the weight of that liquid
is what we call push
the weight of that liquid is a force from below
upwards
which will be equal to the weight of the body
and therefore they will balance
whether this body is totally or partially introcuba
let's analyze, then, the definition
of what Archimedes says
One body, total
or partially submerged in a fluid
and a fluid is a liquid or a gas
What example is there of a gas fluid with this principle?
a floating balloon
when the air inside
it is less dense than the outside atmosphere
will float
then this is true for liquids and gases
in a liquid at rest
because it is not moving, it is not flowing
that's why this topic is inside hydrostatic
this principle
experiences an upward force
of equal value to the weight of
liquid dislodged
I also clarify here that
that the liquid dislodged
has the same volume as the part
submerged body
that is to say, if you were to totally submerge
the body in the liquid
the dislodged volume has the same volume
than the volume of the body, but if a
part of the body, the volume of the liquid evacuated
would have a volume equal to the part
submerged body
half, for example, a quarter, etc.
if it were an iceberg for example
a piece of ice in salty water, submerged
eighty-neve percent of
volume
and the dislodged volume would be eighty-nine
the body's volume
good
we had said, that ..
how could this iron bar
could float in water, in this case
and it is possible because if
we give it a proper shape
how is that proper shape?
if we could stretch this massive iron bar
in such a way to form a container
iron, in this case
but in such a way that it's so big
big enough
that container
what we do with the iron bar
with thin walls of course
such that it contains a quantity of water inside
that container, made with this iron bar
contain a quantity of water
have a weight
same
to the body
to this bar, which has approximately two hundred grams
if we made the container
and inside a quantity could enter
of water, whose weight is equal to the weight
of the bar
this floating bar
but in the shape of a container, a box
this is what happens on ships
they are usually iron
or alliances of iron, forged steel, whatever
they are designed in such a way that
the volume they dislodge
you wonder where the ships dispose of the water?
well, the sea level rises very little
one thousandth of a millimeter, but
multiplied by the sea surface, it dislodges
or would evict a huge amount
of water, whose weight exceeds
to the weight of the boat, rather to the part
to the weight of the entered part of the boat
therefore ships float
for example the SS Faith
which is made of reinforced concrete
that is, it is made of concrete and yet it floats
due to the Archimedean Principle
who says that
a body, totally or partially submended
in a liquid
Resting
experiences an upward force of equal value
to the weight of the evicted liquid, but ..
the evicted liquid has the same volume
that the submerged part of the body
the formulas that we are going to apply
says the specific weight is the same
Specific weight is equal to weight over volume
where the specific weight can be obtained
multiplying density by gravity
density is the amount of mass per unit volume
and then the specific weight is
the amount of weight per unit volume
if we wanted to clear the weight, we passed
the volume multiplying and we have
weight equals specific weight
by volume
which follows from this
Now let's see
one, two, three exercises, I see ..
but two examples for sure we will see
and I think they are going to be quite interesting
Here we have the first exercise
which is a cork immersed in water
the data is that the volume of the cork
immersed in water, is the fourth
I experienced this
put a cork, here it is
in water and approximately by eye, I saw that
a quarter of the cork sank
the part of the cork that was under the water was a quarter
"to eye"
then I saw a table on the Internet that gave
what the density of the cork will give me
that is, it was well estimated
just by chance, because it is very difficult to see how much it sinks in the water
I was lucky ;)
and the density of water which is one grams per
cubic centimeters
or passing it to kilograms per cubic meter
are a thousand
remember there is a topic "passage of units" you can know how to pass
and the density of the cork is requested
maybe also calculate what is the thrust
and I'm going to see what is the apparent weight of the cork in the water
but but is the difference between the weight
of the body in the air
the actual body weight minus the thrust
before starting to do this exercise
I saw even in high school and college
that there were three types of situations that could arise
between thrust and body weight
at least that's what the books apparently said
one that the thrust is equal to the weight of the body, another that
the thrust is less and another that the thrust is greater
and that is impossible, it cannot be the thrust greater than the weight of the body, if not
I would be levitating the body, I would be going up, impossible
what can happen is that
what the books could be meaning
is that the thrust is greater than the weight
but from the part
of the body introduced into the water or liquid
and then yes, the thrust would be greater but to the weight
of the part of the body introduced into the liquid
the apparent weight is going to be zero when
is floating
when will it not be zero?
when the body goes to the bottom
and the bottom provides the part of the force
to compensate for body weight
so there the body will be weighing less
But it's when the thrust can't hold it
and the apparent weight
going to be less than the weight in the air
but when it's floating, the thrust is going to be equal to the weight of the body
always
therefore the apparent weight is zero
so let's start, we have to ...
the push
remember that..
the push is a weight
This formula starts here, of the specific weight
igaul to step on volume
that passing the weight over there
specific weight by volume
now, the weight of the liquid that is dislodged
is the push
so instead of putting "P" I write "E"
igaul to the specific weight of the water
which are a thousand kilograms per cubic meter
by neve coma eight
this would be density by gravity
nine point eight, meters per second squared
by volume
evicted, it's not going to be the volume
full of cork if not
the fourth part
I can calculate it later
whatever the volume of the cork
evicts an amount of liquid that is a quarter
cork volume
the weight of the cork is going to be
the density
what do i want to find out
by nine point eight meters per second squared
density by gravity
by the full volume of the cork
so
the weight of the cork
is going to be equal to the push
I have left that, the density of the cork
by gravity by volume
equals a thousand
kilograms per cubic meter by gravity
I am not going to put nine commas eight, because later they are simplified
by volume over four
I simplify the gravity, the volumes
this four is dividing everything, here is a one
here too, one by one by four
it's four, it's divided by four
then the density of the cork would be a thousand
kilograms per cubic meter
divided four, which gives two hundred fifty
kilograms per cubic meter
then go online
I noticed and this was the density
at least one cork
typical
Now I am going to calculate what is the
what is the weight of the cork
this cork has the measurements of
diameter
two centimeters
and I had high
if I remember correctly, four point five
centimeters
then the volume is
remember that there is a video about areas and volume of geometric bodies
serious base surface
pi times the radius squared
by the height of the body
three point fourteen times
half the diameter (the radius) by
four point five centimeters
this gives me
3.14 per
4.5
fourteen point thirteen
cubic centimeters
this would be the volume
to take the weight off the body
I will multiply
at the density that I have calculated
by gravity
by volume, obtained recently
ah, this volume as it is in cubic centimeters I can't use it like that
because there are meters and cubic meters here
then I will pass it to cubic meters
running comma six places
are two steps
cubic centimeters to cubic meters
are two steps or six places the comma
then here
the comma is here, one two
three four five and six here
then now I have cubic meters
same same
body weight
then two hundred and fifty
by nine point eight
by zero point four zeros
and fourteen thirteen
and this results, zero point zero
thirty-five
Newton
How much would give them the push?
equal ... if it is floating, it doesn't matter the weight
however we will calculate it
thrust is the density of water
by gravity
and multiplied by the volume
but no longer from the cork, because the water dislodged
had a volume that is a quarter of the body's volume
the volume of the cork is this
divided four
then give
thousand for
nine point eight
by zero point zero zero zero zero fourteen thirteen
my students' notes seem ..;)
divided four
and it turns out the same
zero point zero thirty five newton
therefore, the apparent weight
is
body weight less
the push gives zero
remember that it will not give zero when
the body goes to the bottom, touch the floor of the container
from the sea to the river, whatever
and in that case, it will be relieved by the push
but it is not zero, because you need
the force that makes the base of the container
couple holding the body
well now let's see another exercise
before starting exercise two
I show you the last part of
exercise one because it seems that
it didn't look good (complete)
the apparent weight
cork
in the water it was zero, because
the weight of the cork in the air minus the thrust
they are equal
gives zero
and now yes, let's ...
exercise two
I'm going to do only two exercises, ok?
but if you want more, write me, leave me comments
and they ask me for some exercise that they want to solve
make them difficult ..
because I love this topic
me during the race, fourth or fifth year
I almost decided to specialize in
hydraulic, but after
but then continue just civil engineer
difficult but not impossible huh;;)
we have an ice, which is going to be
the iceberg times
at sea
and what i want to find out is the icerberg part
and what I want to find out is the part
of the ice that is below the sea
the part that is entered
the data is ..
a two by two meter base
and a height of 8 meters
I already calculated the volume
is the surface of the base by the height
two by two four
square meters, for eight, thirty-two cubic meters
the density of the ice, the ice ...
this the icerberg
is formed with water
sweet, with rain water
it is not with sea water
is formed in the polar caps
or on glaciers, due to rainfall
freezes, breaks free from there and enters the sea
but it is not sea water
fresh water has a density of
thousand kilograms per cubic meter, but as
it is less dense, the ice
ice has more volume
ice is one of the few substances that
behaves contrary to most elements
at the beginning, behaves the same as the others
that is, it descends
the volume as it cools
but when it reaches four degrees above zero
molecular structure breaks down
of the water
and
takes up more volume. It is like these leaves that
they are arranged like this, they occupy a certain volume
but if I wrinkle them, I fold them and then I put them
one on the other, they occupy more volume
because there is more empty space between the sheets
then at four centigrade grads
the molecular structure is broken
and the volume
increases and therefore mass over volume which is density
to equal more but greater volume
density decreases
and on average
and on average an iceberg has
nine hundred seventeen kilograms per cubic meter
and the sea water as it is denser
than fresh water, because it has salt
on average it is one thousand twenty five
entonces, veamos
the thrust, which would be the weight of the
liquid dislodged
specific weight
the evicted water is the sea water
then one thousand twenty-five
kilograms per cubic meter, by gravity
and multiplied by the volume
of the evicted water
I don't know her
but if I know it will have a base
but if I know it will have a way
which will have a base equal to the one above
the base of the body
then it is, four cubic meters
squares, by height
what is the height that I want to calculate
this part down here
from here to below
the surface of the base, by the height
that's the volume evicted
and the weight of the ice
is nineteen seventeen
kilograms per cubic meter
by gravity and by thirty-two meters
cubic, what if the weight of the body
includes all volume
not just the dislodged volume
then the push
must be equal to the weight of the ice
one-thousand twenty-five
kilograms per cubic meter by gravity
for the height, which is the unknown
equal to nine hundred and seventeen
kilograms per cubic meter by gravity
for thirty two cubic meters
I simplify gravity with gravity
I multiply these two ..
nine hundred
seventeen seven
for thirty two and it gives me
twenty nine thousand three hundred forty four
kilograms and I divide it by one thousand twenty-five
something is missing here ..
I lacked to put the four square meters
I lacked to put the four square meters
It was here after the twenty-fifth, I needed to put it on
then one thousand twenty five
kilograms
per cubic meter, per four square meters
instead of multiplied to ca, I pass them both
this division gives me
all this goes with two of here
and I have one left here
give me
and I get as a result
seven point sixteen meters
that is the answer
what is the percentage of? ..
of the volume of the
ice introduced into the sea
you can see later on the Internet that
on average the percentage
of the volume of an iceberg introduced into the sea
is he
eighty-nine percent and eleven percent,
above the sea
so let's calculate
seven like sixteen
divided eight
and we multiply that by a hundred
this gives us
zero point eighty nine five
and for a hundred
it's eighty-nine like five percent
that's the percentage of the iceberg left
introduced into the water
all right
that's it, this is ..
the Archimedean Principle
fascinating subject, if any
