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In this video, we’re going to learn about density and pressure.
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We’ll learn the definition of these two terms, how to calculate them, and how to work with them practically.
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To start out, imagine that as an amateur magician you are getting practice learning your next trick, lying on a bed of nails.
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But as you really look at the bed of nails, you start to wonder, what if when you lie down, the nails actually do go through your skin.
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Suddenly though, an idea comes on how to make lying on this bed of nails safer.
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You fit the bed inside a very large glass aquarium tank and then you fill the tank up to the top with water.
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Then, strapping on a snorkel, you decide you’re ready to start practicing.
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To understand more about why you might not want to lie on a bed of nails and why your solution might help, we need to know a bit about density and pressure.
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When we talk about pressure, we’re talking about a force over a unit area, or a force spread over an area 𝐴.
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One way to understand pressure is to imagine having a board made of wood and a flat piece of metal, almost like a sheet of metal.
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If we put the metal sheet over the board and then try to hammer it into the board, well, you can imagine how well that would work.
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It really wouldn’t work.
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No matter how hard we hammered, we couldn’t drive the sheet of metal into the board.
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However, if we took that same sheet, heated it up, melted it down, and made a nail out of it, then when we put that nail to the board and started to hammer in, it would drive in well.
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The nail goes into the board, while the sheet doesn’t because with the nail we’re able to apply greater pressure.
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Our force through the hammer is the same in both cases.
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But the area of the nail is so much less than the area of the sheet.
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Because pressure is equal to a force divided by an area, its units in terms of SI base units, are newtons per meter squared.
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But there is a special name for a newton per meter squared.
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And that’s a pascal, abbreviated simply pa, a unit which is named after the French polymath Blaise Pascal.
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Given a particular area 𝐴, over which a force acts, there are two different ways that force can be applied.
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In the first case, it can be applied constantly at each point over that area 𝐴.
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In that case, the pressure exerted over that area simply equals that constant force over the area 𝐴.
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Now, the force we apply over this area varies so that, at each infinitesimal area element, we might have a different value for a force.
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That changes our expression for pressure on this area 𝐴.
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Now, that our force varies depending on where on the area 𝐴 it falls, we integrate that force over the entire area and then divide that complete, or net, force over the area 𝐴.
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Once we solve for that over our force though, our equation reduces to what it was before.
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When we consider pressure related to fluids, that’s when we start to connect these two terms of pressure and density.
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If you’ve ever swum down in a deep pool, you know that the farther down you go, the greater the pressure you feel from the water.
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The pressure a height ℎ below the surface of a fluid as we go down into the fluid is equal to that height ℎ multiplied by 𝑔 times the Greek letter 𝜌, which stands for the fluid density.
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The density of an object is equal to its mass divided by the volume, or the space it takes up.
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We often use the Greek letter 𝜌 to symbolize density.
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And its units are kilograms per meter cubed in SI base units.
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With these two expressions one for density and one for pressure, let’s get some practice calculating these terms in an example.
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What is the density of a rock of mass 356 grams that displaces 93.0 cubic centimeter of water?
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We can call the rock density 𝜌.
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And we can recall that 𝜌 is equal to an object’s mass divided by the volume it takes up.
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In our case, the mass of our object, the rock, is 356 grams and the volume it takes up is 93.0 cubic centimeters.
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When we convert this mass to units of kilograms and this volume to units of cubic meters and calculate the fraction, we find that to three significant figures the rock’s density is 3.83 times 10 to the third kilograms per cubic meter.
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That’s the ratio of the rock’s mass to its volume.
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Now, let’s look at an example involving calculating pressure.
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The tip of a nail exerts tremendous pressure when hit by a hammer because it exerts a large force over a small area.
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What magnitude force must be exerted on a nail that has a circular tip with a diameter of 2.50 millimeters in order to produce a pressure with a magnitude of 2.00 times 10 to the ninth newtons per meter squared?
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In this exercise, we want to solve for a force magnitude we can call capital 𝐹.
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When we recall the pressure relationship, pressure equals force divided by area, we see this force is equal to the pressure applied by the tip of the nail multiplied by the area of that tip.
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We’re given the pressure of 2.00 times 10 to the ninth newtons per meter squared.
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And since the tip of the nail is circular, we know it will be equal to 𝜋 times the diameter of the nail in units of meters squared divided by four.
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When we enter these values on our calculator, we find a force of 9.82 times 10 to the third newtons.
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That’s the force that must be applied to the head of the nail, perhaps by way of a hammer, in order to produce this pressure.
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Now, let’s summarize what we’ve learnt about density and pressure.
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We’ve seen that the density of an object is often represented by its mass per unit volume, density being represented by the Greek letter 𝜌.
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And pressure is given by force over area.
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That is an applied force divided by the area over which that force acts, 𝑃 equals 𝐹 over 𝐴.
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And finally, pressure and density are related through fluids.
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The pressure applied by a fluid of density 𝜌 is equal to that density multiplied by 𝑔 the acceleration due to gravity times the depth of the fluid ℎ.