WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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Today we are going to talk about fluids -- starting a new unit, specifically about density and buoyancy.
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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.
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As we start this new topic of fluids, let us talk about what a fluid is.
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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.
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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.
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So to begin, let us get into density.
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Density, which gets the symbol, Greek letter ρ, is the ratio of an objects mass to the volume it occupies.
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Less dense fluid flow on top of more dense fluids and less dense solids will float on top of more dense fluids.
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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.
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Let us take a look at an example with a density of water.
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A single kilogram of water fills a cube of length, 0.1 meter. What is the density of the water?
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Well we have our cube of water and the length of each side is 0.1 meter. What is its density?
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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.
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That is probably a good one to remember -- density of fresh water is 1,000 kg/m³.
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Let us take a look at the volume of gold. It has a density of 19,320 kg/m³. It is very dense.
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How much volume does a single kilogram of gold occupy?
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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.
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Let us take a look at an example of things that are floating.
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Fresh water has a density of 1,000 kg/m³, we just calculated that.
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Which of the following materials will float on water?
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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.
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It is less dense than the water.
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Magnesium has a density of 1740 kg/m³. It is more dense than water, so it is going to sink.
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Cork, of course, 250 kg/m³, is going to float. That is why we make bobbers out of cork when we go fishing.
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Glycerol is 1260 kg/m³ is more dense so it is going to sink.
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So those two will float because they are less dense.
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Let us talk a little bit about buoyancy now.
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Buoyancy is a force exerted by a fluid on an object and it opposes the objects weight when it is in that fluid.
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The buoyant force, typically written as Fb, is determined using what is known as Archimedes' principle.
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The buoyant force is equal to the density of fluid times the volume of the fluid displaced times the acceleration due to gravity.
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Let us spell these out because they are easy to mix up.
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(G) of course is the acceleration due to gravity.
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The volume is going to be the volume of the fluid displaced by your object.
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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.
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And of course, Fb is the buoyant force.
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So let us do an example with the buoyant force.
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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.
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Well the buoyant force, Fb, is equal to the density of the fluid, ρ, times the volume (v) of the fluid displaced times (g).
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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.
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Let us get exciting. Let us talk about a shark tank.
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A steel cable holds a 120 kg shark tank 3 m below the surface of salt water.
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Salt water is slightly more dense than freshwater at 1025 kg/m³.
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If the volume of water displaced by the shark tank is 0.1 m³, what is the tension in the cable?
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Well, let us start with our free body diagram (FBD).
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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).
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We want to find the tension in the cable.
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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.
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We could write Newton's Second Law for the y direction -- Fnety equals tension plus the buoyant force minus the weight and all of that must be equal to 0.
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If we want tension, that must be equal to the weight minus the buoyant force.
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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).
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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).
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Therefore, our tension must equal about 175N.
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Let us go to a favorite project in physics -- building a concrete boat that floats.
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A rectangular boat made out of concrete with a mass of 3,000 kg floats on a freshwater lake.
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The density of freshwater, again, is 1,000 kg/m³.
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If the bottom area of the boat is 6 ² meters -- it is a pretty big boat -- how much of the boat is submerged?
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Well, let us start with a FBD.
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We have the weight of the boat down (mg) and we have the buoyant force holding it up.
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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).
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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).
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Therefore, we could say then that density of the fluid times the volume times (g) equals (mg).
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I can see right away that there is a simplification that we can make -- we can divide (g) out of both sides.
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We could also say then that the volume of our boat is going to be its area times its depth.
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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).
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Therefore the density of our fluid -- (ρ) fluid -- times our volume (ad) must equal (m).
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We rearrange this to find (d), the depth of the boat submerged.
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(D) is going to be equal to the mass over density of the fluid times the area or 3,000 kg...
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...about 3 tons divided by density of our fluid (1,000) and the area (6 ²m) or about 0.5 m.
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About 1/2 a meter is going to be under the water, submerged.
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All right, let us take a look at a problem of apparent mass.
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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.
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What is their mass on dry land?
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Well, what does that mean? Let us think about this for a second.
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If we were to go make a scale and on it we are going to put a bunch of bricks -- there we go.
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And our scale has a reading here.
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It has an apparent mass of 2400 kg when it is submerged.
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The scale reads like it is 2400 kg there, so what would its actual mass be on dry land?
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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?
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I am going to start off with a FBD.
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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.
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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.
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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...
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...so 24,000 plus the buoyant force -- density of our fluid, times the volume of the fluid displaced, times (g) minus (mg) equals 0.
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This implies then that 24,000 plus density of our fluid (1025) times the volume of our fluid displaced...
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...which was 1 times g (10) - 10 m must equal 0.
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Therefore, 24,000 + 1025 × 1 × 10 = 10 m.
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Therefore (m) must equal -- Well divide both of these by 10 -- 2400 + 1025 = 3,425 kg is the actual mass.
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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.
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Let us take a look at the volume of a submerged cube.
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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.
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If the force of gravity on the cube is 10N what is the tension in the string?
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Let us see if we cannot draw this out a little bit first -- feeble attempt at drawing a glass.
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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.
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All right it is time for FBD again.
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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.
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So net force in the y direction, which is the buoyant force minus (mg) minus (t) must equal 0.
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Solving for the tension then -- tension equals the buoyant force minus (mg) which implies then that the tension must be the buoyant force...
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...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...
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...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.
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Therefore, the tension comes out to be 1,000 × 0.002 × 10 -- 20 - 10 = 10N.
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Let us try one more practice problem here -- determining density.
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This one is a little bit more involved.
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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.
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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.
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I am going to start with a FBD.
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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.
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When it is in water, our FBD is going to look similar, but a little bit different.
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We still have the weight down, but now we have the buoyant force up along with the force of the scale (Fw).
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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.
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And we also know that the force of air is equal to (mg), so I could rewrite this as (Fb) + (Fw) = (Fa).
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But that buoyant force is equal to ρ fluid (v)(g).
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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).
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Now we have to take another step that is maybe not quite so obvious.
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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.
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Therefore the volume of our object is equal to the mass of the object over the density of the object.
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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.
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We have mass of the object over density of the object times (g).
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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)...
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...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.
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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...
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Well we have (Fa) times the density of our fluid divided by the scale reading in air minus the scale reading in water.
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Hopefully that gets you a good start on density and buoyancy as we start this new section on fluids.
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I appreciate your time and make it a great day everyone.