The first thing to know is that fluids are not just liquids but include gases as well; anything that freely flows is fluid. These fluids have properties known as density and buoyancy. Essentially, you can think of these two (very informally) as the mass and resistance properties of your fluid. Visually, think of density as the reason a ping pong ball floats in water, and buoyancy as the reason you feel less heavy and can move less agilely under water. Of course, if you swam too deep, you would invoke the next topic, pressure.
Density is the ratio of an object's mass to the volume it occupies.
The density of freshwater is 1000 kilograms per cubic meter.
Buoyancy is a force exerted by a fluid on an object, opposing the object's weight.
The buoyant force is equal to the weight of the fluid displaced by the object.
Density & Buoyancy
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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
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 Fb, 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
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
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
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
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
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