AP Physics 1 & 2 Density & Buoyancy

Hi everyone and welcome back to Educator.com.0000

Today we are going to talk about fluids -- starting a new unit, specifically about density and buoyancy.0003

Our objectives are going to be to calculate the density of an object, to determine whether an object will float given its average density, and to calculate the forces on a submerged or partially submerged object using Archimedes' principle.0010

As we start this new topic of fluids, let us talk about what a fluid is.0026

Fluid is matter that flows under pressure -- things like liquids -- a great example might be water -- gases, like air; and even plasmas like what you would get from an arc welder.0031

Now fluid mechanics is going to be the study of fluids and how they move -- fluids at rest, fluids in motion, forces applied to fluids and then the forces exerted back by fluids.0044

So to begin, let us get into density.0056

Density, which gets the symbol, Greek letter ρ, is the ratio of an objects mass to the volume it occupies.0059

Less dense fluid flow on top of more dense fluids and less dense solids will float on top of more dense fluids.0065

Now if density is the ratio of an objects mass to volume -- density (ρ), is mass over volume and the units are going to be kilograms per meter cubed.0072

Let us take a look at an example with a density of water.0077

A single kilogram of water fills a cube of length, 0.1 meter. What is the density of the water?0090

Well we have our cube of water and the length of each side is 0.1 meter. What is its density?0096

Well density is mass divided by volume, so that is going to be 1 kg and the volume of the cube (length × width × height) is going to be 0.1 m × 0.1 meter × 0.1 meter for a total of 1,000 kg/m³.0107

That is probably a good one to remember -- density of fresh water is 1,000 kg/m³.0132

Let us take a look at the volume of gold. It has a density of 19,320 kg/m³. It is very dense.0140

How much volume does a single kilogram of gold occupy?0147

Well, if density is mass over volume, then volume will be mass over density or 1 kg/19,320 kg/m³. 0169 Or about 5.18 × 10 ^-5m³.0151

Let us take a look at an example of things that are floating.0184

Fresh water has a density of 1,000 kg/m³, we just calculated that.0188

Which of the following materials will float on water?0193

Ice has a density of 917 kg/m³ and if you have ever had a glass of ice water, you already know the answer -- ice will float on water.0196

It is less dense than the water.0206

Magnesium has a density of 1740 kg/m³. It is more dense than water, so it is going to sink.0209

Cork, of course, 250 kg/m³, is going to float. That is why we make bobbers out of cork when we go fishing.0215

Glycerol is 1260 kg/m³ is more dense so it is going to sink.0224

So those two will float because they are less dense.0229

Let us talk a little bit about buoyancy now.0234

Buoyancy is a force exerted by a fluid on an object and it opposes the objects weight when it is in that fluid.0236

The buoyant force, typically written as F_b, is determined using what is known as Archimedes' principle.0242

The buoyant force is equal to the density of fluid times the volume of the fluid displaced times the acceleration due to gravity.0248

Let us spell these out because they are easy to mix up.0256

(G) of course is the acceleration due to gravity.0259

The volume is going to be the volume of the fluid displaced by your object.0266

Typically, you just see this written as density, but I like to put the fluid after it to remind me that it is the density of the fluid that we need in this calculation and not the density of the object, so that is going to be the density of the displaced fluid.0280

And of course, F_b is the buoyant force.0302

So let us do an example with the buoyant force.0312

What is the buoyant force on a 0.3 m³ box, which is fully submerged in freshwater if it has a density of 1,000 kg/m³.0314

Well the buoyant force, F_b, is equal to the density of the fluid, ρ, times the volume (v) of the fluid displaced times (g).0323

So the density of the fluid displaced is 1,000 kg/m³ and we have a volume of 0.3 m³, and (g) -- we will estimate as 10 m/s² for a force of about 3,000N.0335

Let us get exciting. Let us talk about a shark tank.0356

A steel cable holds a 120 kg shark tank 3 m below the surface of salt water.0359

Salt water is slightly more dense than freshwater at 1025 kg/m³.0365

If the volume of water displaced by the shark tank is 0.1 m³, what is the tension in the cable?0369

Well, let us start with our free body diagram (FBD).0376

There is our object and we have its weight pulling it down, pulling it down, and we have the buoyant force opposing that and we also have a cable on it that has some tension in it (t).0379

We want to find the tension in the cable.0392

If it is just sitting there 3 m below the surface of the saltwater, it is not accelerating, so the net force must be 0.0394

We could write Newton's Second Law for the y direction -- F_nety equals tension plus the buoyant force minus the weight and all of that must be equal to 0.0400

If we want tension, that must be equal to the weight minus the buoyant force.0414

This implies then that the tension is equal to the weight minus -- well the buoyant force is the density of our fluid -- (ρ) fluid -- times the volume of the fluid displaced times (g).0421

So that is going to be mass (120), g (10 m/s²) minus density of our fluid (1025 kg/m³) times the volume of the water displaced by the tank (0.1) times g (10 m/s²).0434

Therefore, our tension must equal about 175N.0454

Let us go to a favorite project in physics -- building a concrete boat that floats.0468

A rectangular boat made out of concrete with a mass of 3,000 kg floats on a freshwater lake.0473

The density of freshwater, again, is 1,000 kg/m³.0480

If the bottom area of the boat is 6 ² meters -- it is a pretty big boat -- how much of the boat is submerged?0484

Well, let us start with a FBD.0490

We have the weight of the boat down (mg) and we have the buoyant force holding it up.0494

Net force in the y direction then, must be the buoyant force minus (mg) and because it is not accelerating up or down, that must be equal to 0, therefore the buoyant force must be equal to (mg).0500

But we also know that the buoyant force is equal to the density of the fluid times the volume of the fluid displaced times (g).0517

Therefore, we could say then that density of the fluid times the volume times (g) equals (mg).0527

I can see right away that there is a simplification that we can make -- we can divide (g) out of both sides.0537

We could also say then that the volume of our boat is going to be its area times its depth.0543

The volume of the fluid displaced is going to be the area of the boat times how much of it is submerged -- that depth submerged (d).0549

Therefore the density of our fluid -- (ρ) fluid -- times our volume (ad) must equal (m).0559

We rearrange this to find (d), the depth of the boat submerged.0568

(D) is going to be equal to the mass over density of the fluid times the area or 3,000 kg...0572

...about 3 tons divided by density of our fluid (1,000) and the area (6 ²m) or about 0.5 m.0582

About 1/2 a meter is going to be under the water, submerged.0597

All right, let us take a look at a problem of apparent mass.0604

A cubic meter of bricks have an apparent mass of about 2400 kg when they are submerged in saltwater with a density of 1025 kg/m³.0610

What is their mass on dry land?0619

Well, what does that mean? Let us think about this for a second.0623

If we were to go make a scale and on it we are going to put a bunch of bricks -- there we go.0627

And our scale has a reading here.0646

It has an apparent mass of 2400 kg when it is submerged.0648

The scale reads like it is 2400 kg there, so what would its actual mass be on dry land?0654

What would its -- well the mass is not actually changing, so what would the scale read on dry land? 0665 What would its weight be on dry land?0660

I am going to start off with a FBD.0668

We have the weight down and while it is submerged, we have the buoyant force up and we have the normal force from the scale and as you know, scales tell you the normal force.0672

We wanted to write Newton's Second Law equation -- normal force plus the buoyant force minus (mg) must equal 0, because acceleration is 0 because the bricks are just sitting there on the scale; they are not accelerating.0683

Therefore, we could write that -- well knowing that the apparent mass is 2400 kg, that means the normal force which must be (mg), it must read that that is normal force of 24,000N...0698

...so 24,000 plus the buoyant force -- density of our fluid, times the volume of the fluid displaced, times (g) minus (mg) equals 0.0716

This implies then that 24,000 plus density of our fluid (1025) times the volume of our fluid displaced...0731

...which was 1 times g (10) - 10 m must equal 0.0742

Therefore, 24,000 + 1025 × 1 × 10 = 10 m.0760

Therefore (m) must equal -- Well divide both of these by 10 -- 2400 + 1025 = 3,425 kg is the actual mass.0773

It appears to have a lower mass when it is in water because the buoyant force is helping lift it on the scale, providing some upward force to counteract that weight.0788

Let us take a look at the volume of a submerged cube.0802

We have a cube of volume 0.002 m³ submerged in a glass of freshwater and attached to the bottom of the glass by a massless string.0805

If the force of gravity on the cube is 10N what is the tension in the string?0814

Let us see if we cannot draw this out a little bit first -- feeble attempt at drawing a glass.0819

There it is and somewhere in the glass we have our cube and it is attached by a string to the bottom there and we have some sort of freshwater in our glass.0826

All right it is time for FBD again.0847

Here we have our object -- we have its weight (mg) down, we have the tension in the string down, and we have the buoyant force up and again because it is just sitting there, it is not accelerating up or down, it must be an equilibrium -- the net force must be 0.0850

So net force in the y direction, which is the buoyant force minus (mg) minus (t) must equal 0.0867

Solving for the tension then -- tension equals the buoyant force minus (mg) which implies then that the tension must be the buoyant force...0876

...density of our fluid times the volume of the fluid displaced times (g) by Archimedes' principle minus (mg) or that is going to be 1,000 kg/m³ since it is freshwater...0889

...volume displaced is 0.002 and (g) is 10 minus mg -- well, it gives us the force of gravity on it which is 10N, which is its weight, which is (mg) minus 10.0904

Therefore, the tension comes out to be 1,000 × 0.002 × 10 -- 20 - 10 = 10N.0919

Let us try one more practice problem here -- determining density.0935

This one is a little bit more involved.0941

The density of an unknown specimen may be determined by hanging the specimen from a scale in air and in water and then comparing the two measurements.0943

If the scale reading in air -- we are going to call (Fa) -- and the scale reading in water is (Fw), let us develop a formula for the density of the specimen in terms of the scale reading in air, in water, and the density of the fluid.0952

I am going to start with a FBD.0967

When we are in air, we have (mg) down and we have (Fa) on the scale up and those will be balanced because it is sitting on the scale at equilibrium.0970

When it is in water, our FBD is going to look similar, but a little bit different.0983

We still have the weight down, but now we have the buoyant force up along with the force of the scale (Fw).0991

Starting with the water, we have (Fb), the buoyant force, plus the force of the scale when it is in water must be equal to its weight because it is the equilibrium.1001

And we also know that the force of air is equal to (mg), so I could rewrite this as (Fb) + (Fw) = (Fa).1011

But that buoyant force is equal to ρ fluid (v)(g).1027

I could rewrite this then as (Fa) - (Fw), with a little bit of rearranging, must equal density of our fluid times our volume displaced times (g).1034

Now we have to take another step that is maybe not quite so obvious.1048

Let us take a look and let us say that the density of our object is equal to the mass of the object divided by the volume of the object.1053

Therefore the volume of our object is equal to the mass of the object over the density of the object.1064

I am going to use that as I rewrite this equation to say that (Fa) - (Fw) = -- we have our density of our fluid, but I am going to replace the volume displaced with my new formula for volume -- mass of the object over the density of the object.1073

We have mass of the object over density of the object times (g).1090

Now it is just a little bit of Algebra to prove that the density of the object is going to be equal to the density of our fluid times the mass of the object times (g) divided by (Fa) - (Fw)...1101

...just a little bit of rearrangement to get the density of the object all by itself and finally one more step, that (Fa) -- we can change that a little bit, we can rearrange things.1126

Let us then say then that the mass of the object times (g) -- that is just its weight in air (Fa) -- so mass of the object times (g) right here -- I am going to replace with (Fa) to write the density of our object is equal to...1137

Well we have (Fa) times the density of our fluid divided by the scale reading in air minus the scale reading in water.1160

Hopefully that gets you a good start on density and buoyancy as we start this new section on fluids.1180

I appreciate your time and make it a great day everyone.1185