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# Archimedes principle and buoyant force

Introduction to Archimedes' principle and buoyant force. Created by Sal Khan.

Video transcript

Let's say we have
a cup of water. Let me draw the cup. This is one side of the cup,
this is the bottom of the cup, and this is the other
side of the cup. Let me say that it's
some liquid. It doesn't have to be water,
but some arbitrary liquid. It could be water. That's the surface of it. We've already learned that the
pressure at any point within this liquid is dependent
on how deep we go into the liquid. One point I want to make before
we move on, and I touched on this a little bit
before, is that the pressure at some point isn't just acting
downwards, or it isn't just acting in one direction. It's acting in all directions
on that point. So although how far we go down
determines how much pressure there is, the pressure is
actually acting in all directions, including up. The reason why that makes sense
is because I'm assuming that this is a static system,
or that the fluids in this liquid are stationary, or you
even could imagine an object down here, and it's
stationary. The fact that it's stationary
tells us that the pressure in every direction must be equal. Let's think about a
molecule of water. A molecule of water, let's say
it's roughly a sphere. If the pressure were different
in one direction or if the pressure down were greater than
the pressure up, then the object would start accelerating
downwards, because its surface area
pointing upwards is the same as the surface area pointing
downwards, so the force upwards would be more. It would start accelerating
downwards. Even though the pressure is a
function of how far down we go, at that point,
the pressure is acting in every direction. Let's remember that, and now
let's keep that in mind to learn a little bit about
Archimedes' principle. Let's say I submerge a cube into
this liquid, and let's say this cube has dimensions
d, so every side is d. What I want to do is I want to
figure out if there's any force or what is the net
force acting on this cube due to the water? Let's think about what the
pressure on this cube is at different points. At the depths along the side of
the cube, we know that the pressures are equal, because
we know at this depth right here, the pressure is going to
be the same as at that depth, and they're going to offset each
other, and so these are going to be the same. But one thing we do know, just
based on the fact that pressure is a function of depth,
is that at this point the pressure is going to be
higher-- I don't know how much higher-- than at this point,
because this point is deeper into the water. Let's call this P1. Let's call that pressure on top,
PT, and let's call this point down here PD. No, pressure on the
bottom, PB. What's going to be the net
force on this cube? The net force-- let's call that
F sub N-- is going to be equal to the force acting
upwards on this object. What's the force acting
upwards on the object? It's going to be this pressure
at the bottom of the object times the surface area at the
bottom of the object. What's the surface area at
the bottom of the object? That's just d squared. Any surface of a cube is d
squared, so the bottom is going to be d squared minus--
I'm doing this because I actually know that the pressure
down here is higher than the pressure here, so this
is going to be a larger quantity, and that the net force
is actually going to be upwards, so that's why I can
do the minus confidently up here-- the pressure
at the top. What's the force at the top? The force at the top is going to
be the pressure on the top times the surface area of
the top of the cube, right, times d squared. We can even separate out the d
squared already at that point, so the net force is equal to
the pressure of the bottom minus the pressure of the top,
or the difference in pressure times the surface area of either
the top or the bottom or really any of the
sides of the cube. Let's see if we can figure
what these are. Let's say the cube is submerged
h units or h meters into the water. So what's the pressure
at the top? The pressure at the top is
going to be equal to the density of the liquid-- I keep
saying water, but it could be any liquid-- times how
far down we are. So we're h units down, or maybe
h meters, times gravity. And what's the pressure
the bottom? The pressure at the bottom
similarly would be the density of the liquid times the depth,
so what's the depth? It would be this h and then
we're another d down. It's h plus d-- that's our total
depth-- times gravity. Let's just substitute both of
those back into our net force. Let me switch colors to keep
from getting monotonous. I get the net force is equal to
the pressure at the bottom, which is this. Let's just multiply it out, so
we get p times h times g plus d times p times g. I just distributed this out,
multiplied this out. That's the pressure at the
bottom, then minus the pressure at the top, minus phg,
and then we learned it's all of that times d squared. Immediately, we see something
cancels out. phg, phg subtract. It cancels out, so we're
just left with-- what's the net force? The net force is equal to dpg
times d squared, or that equals d cubed times
the density of the liquid times gravity. Let me ask you a question:
What is d cubed? d cubed is the volume
of this cube. And what else is it? It's also the volume of
the water displaced. If I stick this cube into the
water, and the cube isn't shrinking or anything-- you
can even imagine it being empty, but it doesn't have to be
empty-- but that amount of water has to be moved out
of the way in order for that cube to go in. This is the volume of
the water displaced. It's also the volume
of the cube. This is the density-- I keep
saying water, but it could be any liquid-- of the liquid. This is the gravity. So what is this? Volume times density is the mass
of the liquid displaced, so the net force is
also equal to the mass of liquid displaced. Let's just say mass times
gravity, or we could say that the net force acting on this
object is-- what's the mass of the liquid displaced
times gravity? That's just the weight
of liquid displaced. That's a pretty interesting
thing. If I submerge anything, the net
force acting upwards on it, or the amount that I'm
lighter by, is equal to the weight of the water
being displaced. That's actually called
Archimedes' principle. That net upward force due to
the fact that there's more pressure on the bottom than
there is on the top, that's called the buoyant force. That's what makes
things float. I'll leave you there to just to
ponder that, and we'll use this concept in the next couple
of videos to actually solve some problems.
I'll see you soon.