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### Course: Class 11 Physics (India) > Unit 14

Lesson 2: Buoyant force and Archimedes' principle# Buoyant force example problems

A couple of problems involving Archimedes' principle and buoyant forces. Created by Sal Khan.

## Want to join the conversation?

- Could one walk on water by wearing shoes on their feet that are far less dense than water? Or would the entire body's volume and density contribute in determining whether the person with very low-density shoes on their feet remains afloat?(29 votes)
- An object will float if its average density is less than the average density of the fluid it is immersed in. But you might ask, how can a cruise ship, which is constructed of metals weighing thousands of kilograms, float on water if all those dense metals have a higher average density than water?

Well, air has a lower average density than water and cruise ship's have air submerged beneath the water (ie, between the ship's lower decks) which lowers the ship's average density to be less than that of water.

So if you built shoes that lowered your average density to be less than water than yes, you would be able to walk on water.(17 votes)

- What will be the buoyant force if the gravity is zero?(12 votes)
- If there is no gravity, there will be no buoyancy. Buoyancy results from pressure differentials caused by gravity.(25 votes)

- He states a cubic meter = 27 "square" feet.

A cubic meter = 35.3147 "cubic" feet.

A square meter = 10.7639 "square" feet.

A cubic yard = 27 "cubic" feet.

Am I missing something?(17 votes)- When he said "square" feet at4:50, he actually meant to say "cubic" feet(7 votes)

- Can someone explain, why water doesn't rise even the ice melts?

I am facing a doubt ,if ice floating on water melts more water will be added to existing water so the water level should rise.Why isn't so ?(7 votes)- see when ice floats on water 11th part out of its remaining 12 parts remain in the water and only one part floats above the water level, hence when the ice melts its fills the gap created by it during its ice form, thus the water level does not rise .But when in the polar caps the ice melts as the ice is collected above the land mass hence it does not create any gap in the surface of the water hence when it melt it forms extra water , thus increasing the water level of the waterbody it falls on.(7 votes)

- If W = m * g and the mass of the object and the force of gravity don't change when the object is submerged, then how can the weight of the object change when it's put into water?(5 votes)
- The gravitational weight doesn't change, but the apparent weight (F_gravity - F_buoyancy) will change since the buoyant force will counteract the gravitational weight.(5 votes)

- the cube in the begining..will it go downward?? coz force down is 10 N while up is 8N..(5 votes)
- Yes, as the (net) weight force acting on the cube is 2N (you can see from 10N - 8N, but Sal also mentions that it's 2N in the video)(5 votes)

- If you're standing on the bottom of a swimming pool how is a buoyant force exerted on you from below? Does the normal force become the buoyant force or is no buoyant force exerted on you from directly below?

Obviously there will still be buoyant force on your arms, etc. but does that combined with the normal force constitute the total buoyant force?(4 votes)- If you form a solid seal with the floor, like a suction cup, there will be less upwards pressure. Otherwise, as long as water is capable of getting under your feet, it will push up with the same pressure as usual. The normal force is separate from the bouyant force.(2 votes)

- If two objects have same mass but different volume, do they have same buoyant force?(2 votes)
- No, the buoyant force is the weight of the displaced fluid.

Consider 1 kg block of solid of iron and 1 kg block of solid styrofoam, the iron will sink but the styrofoam will float.(5 votes)

- At5:24, Sal introduces a new question does some calculation.

Can't we straight away calculate the specific gravity of balsa wood and multiply it by 100% ?(3 votes)- Yes, but this process shows you why the specific gravity 'formula' works when finding the percentage submerged. Also, it would be 100, not 100%, as 100% is just equal to 1.(4 votes)

- i thought buoyant force = weight of water immersed then how come it is 8N while weight of object is 10N? Shouldn't buoyant force also equal 10N? "Archimedes' principle indicates that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces."(1 vote)
- Buoyant force is equal to the weight of the water displaced. The weight of the water displaced has nothing to do with the weight of the object (10N), it has to do with the volume of the object.(4 votes)

## Video transcript

Let's say that I have some
object, and when it's outside of water, its weight
is 10 newtons. When I submerge it in water-- I
put it on a weighing machine in water-- its weight
is 2 newtons. What must be going on here? The water must be exerting some
type of upward force to counteract at least
8 newtons of the object's original weight. That difference is the
buoyant force. So the way to think about is
that once you put the object in the water-- it could be a
cube, or it could be anything. We know that we have a downward
weight that is 10 newtons, but we know that once
it's in the water, the net weight is 2 newtons, so there
must be some force acting upwards on the object
of 8 newtons. That's the buoyant force that
we learned about in the previous video, in the video
about Archimedes' principle. This is the buoyant force. So the buoyant force is equal
to 10 minus 2 is equal to 8. That's how much the water's
pushing up. And what does that
also equal to? That equals the weight of the
water displaced, so 8 newtons is equal to weight of
water displaced. What is the weight of
the water displaced? That's the volume of the water
displaced times the density of water times gravity. What is the volume of
water displaced? It's just the volume of water,
divide 8 newtons by the density of water,
which is 1,000 kilograms per meter cubed. A newton is 1 kilogram meter
per second squared. Then, what's gravity? It's 9.8 meters per
second squared. If we look at all the units,
they actually do turn out with you just ending up having
just meters cubed, but let's do the math. We get 8 divided by 1,000
divided by 9.8 is equal to 8.2 times 10 to the negative 4. V equals 8.2 times 10 to the
minus 4 cubic meters. Just knowing the difference in
the weight of an object-- the difference when I put
it in water-- I can figure out the volume. This could be a fun game
to do next time your friends come over. Weigh yourself outside of water,
then get some type of spring or waterproof weighing
machine, put it at the bottom of your pool, stand on it, and
figure out what your weight is, assuming that you're dense
enough to go all the way into the water. You could figure out somehow
your weight in water, and then you would know your volume. There's other ways. You could just figure out how
much the surface of the water increases, and take
that water away. This was interesting. Just knowing how much the
buoyant force of the water was or how much lighter we are when
the object goes into the water, we can figure out the
volume of the object. This might seem like a very
small volume, but just keep in mind in a meter cubed, you
have 27 square feet. If we multiply that number
times 27, it equals 0.02 square feet roughly. In 0.02 square feet, how many--
in a square foot, there's actually-- 12 to the
third power times 12 times 12 is equal to 1,728 times 0.02. So this is actually
34 square inches. The object isn't as small as you
may have thought it to be. It's actually maybe a little bit
bigger than 3 inches by 3 inches by 3 inches, so it's
a reasonably sized object. Anyway, let's do another
problem. Let's say I have some balsa
wood, and I know that the density of balsa wood is 130
kilograms per meter cubed. I have some big cube of balsa
wood, and what I want to know is if I put that-- let
me draw the water. I have some big cube of balsa
wood, which I'll do in brown. So I have a big cube of balsa
wood and the water should go on top of it, just so
that you see it's submerged in the water. I want to know what percentage
of the cube goes below the surface of the water? Interesting question. So how do we do that? For the object to be at rest,
for this big cube to be at rest, there must be zero net
forces on this object. In that situation, the buoyant
force must completely equal the weight or the force
of gravity. What's the force of gravity
going to be? The force of gravity is just the
weight of the object, and that's the volume of the balsa
wood times the density of the balsa wood times gravity. What's the buoyant force? The buoyant force is equal to
the volume of the displaced water, but that's also the
volume of the displaced water and it's the volume of the cube
that's been submerged. The part of the cube that's
submerged, that's volume. That's also equal
to the amount of volume of water displaced. We could say that's the volume
of the block submerged, which is the same thing, remember,
as the volume of the water displaced times the density
of water times gravity. Remember, this is density
of water. So remember, the buoyant force
is just equal to the weight of the water displaced and that's
just the volume of the water displaced times the density
of water times gravity. Of course, the volume of the
water displaced is the exact same thing as the volume
of the block that's actually submerged. Since the block is stationary,
it's not accelerating upwards or downwards, we know that
these two quantities must equal each other. So V, the volume of the wood,
the entire volume, not just the amount that's submerged,
times the density of the wood times gravity must equal the
volume of the wood submerged, which is equal to the volume of
the water displaced times the density of water
times gravity. We have the acceleration
of gravity. We have that on both sides,
so we can cross it out. Let me switch colors to
ease the monotony. What happens if we divide both
sides by the volume of the balsa wood? I'm just rearranging
this equation. I think you'll figure it out. We divide both sides by that,
and you get the volume submerged divided by the volume
of the balsa wood-- I just divided both sides by VB
and switched sides-- is equal to the density of the balsa
wood divided by the density of water. Does that make sense? I just did a couple of quick
algebraic operations, but hopefully that got rid of
the g, and that should make sense to you. Now we're ready to solve
our problem. My original question is
what percentage of the object is submerged? That's exactly this number. If we say this is the volume
submerged over the total volume, this is the
percent submerged. That equals the density of
balsa wood, which is 130 kilograms per meter cubed,
divided by the density of water, which is 1,000 kilograms
per meter cubed, so 130 divided by 1,000 is 0.13. Vs over VB is equal to
0.13, which is the same thing as 13%. So, exactly 13% percent of this
balsa wood block will be submerged in the water. That's pretty neat to me. It actually didn't have
to be a block. It could have been shaped
like a horse. I'll see you in the
next video.