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# Fluids (part 6): A couple of problems involving Archimedes' principle and buoyant forces.

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- 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.

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