[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.27,0:00:02.07,Default,,0000,0000,0000,,- [Instructor] We talk about\Nfluid pressure all the time,
Dialogue: 0,0:00:02.07,0:00:05.97,Default,,0000,0000,0000,,for example, blood pressure\Nor pressure inside our tires,
Dialogue: 0,0:00:05.97,0:00:08.04,Default,,0000,0000,0000,,but what exactly are these numbers?
Dialogue: 0,0:00:08.04,0:00:09.39,Default,,0000,0000,0000,,What does 120 mean over here?
Dialogue: 0,0:00:09.39,0:00:11.91,Default,,0000,0000,0000,,Or what does 40 mean over here?
Dialogue: 0,0:00:11.91,0:00:15.90,Default,,0000,0000,0000,,So the big question over here\Nis what exactly is pressure?
Dialogue: 0,0:00:15.90,0:00:17.70,Default,,0000,0000,0000,,That's what we're gonna\Nfind out in this video,
Dialogue: 0,0:00:17.70,0:00:18.69,Default,,0000,0000,0000,,so let's begin.
Dialogue: 0,0:00:18.69,0:00:19.92,Default,,0000,0000,0000,,To make sense of this,
Dialogue: 0,0:00:19.92,0:00:22.56,Default,,0000,0000,0000,,let's hang a perfectly cubicle wooden box
Dialogue: 0,0:00:22.56,0:00:24.30,Default,,0000,0000,0000,,and think about all the forces.
Dialogue: 0,0:00:24.30,0:00:26.43,Default,,0000,0000,0000,,We know that there's gravitational\Nforce acting downwards,
Dialogue: 0,0:00:26.43,0:00:29.16,Default,,0000,0000,0000,,which is perfectly balanced\Nby the tension force,
Dialogue: 0,0:00:29.16,0:00:32.40,Default,,0000,0000,0000,,but there is another\Nset of force over here.
Dialogue: 0,0:00:32.40,0:00:34.92,Default,,0000,0000,0000,,Remember that there are\Nair molecules in our room,
Dialogue: 0,0:00:34.92,0:00:37.65,Default,,0000,0000,0000,,and so these air molecules\Nare constantly bumping
Dialogue: 0,0:00:37.65,0:00:40.26,Default,,0000,0000,0000,,and exerting tiny forces.
Dialogue: 0,0:00:40.26,0:00:42.30,Default,,0000,0000,0000,,But how do we model these forces?
Dialogue: 0,0:00:42.30,0:00:44.88,Default,,0000,0000,0000,,There are billions of air molecules around
Dialogue: 0,0:00:44.88,0:00:47.49,Default,,0000,0000,0000,,and they're constantly bumping,\Nso how do we model this?
Dialogue: 0,0:00:47.49,0:00:49.92,Default,,0000,0000,0000,,Well, we can model them to be continuous,
Dialogue: 0,0:00:49.92,0:00:52.32,Default,,0000,0000,0000,,that simplifies things,\Nbut more importantly,
Dialogue: 0,0:00:52.32,0:00:54.27,Default,,0000,0000,0000,,if you think about the surface area,
Dialogue: 0,0:00:54.27,0:00:55.65,Default,,0000,0000,0000,,since this is the perfect cube,
Dialogue: 0,0:00:55.65,0:00:57.78,Default,,0000,0000,0000,,and the surface areas\Nare exactly the same,
Dialogue: 0,0:00:57.78,0:01:01.11,Default,,0000,0000,0000,,that means the number of\Nmolecules bumping per second
Dialogue: 0,0:01:01.11,0:01:04.23,Default,,0000,0000,0000,,on each surface is pretty much the same,
Dialogue: 0,0:01:04.23,0:01:06.09,Default,,0000,0000,0000,,and therefore, we could model these forces
Dialogue: 0,0:01:06.09,0:01:09.48,Default,,0000,0000,0000,,to be pretty much the same\Nfrom all the directions,
Dialogue: 0,0:01:09.48,0:01:13.35,Default,,0000,0000,0000,,and that's why these\Nforces together cancel out
Dialogue: 0,0:01:13.35,0:01:16.47,Default,,0000,0000,0000,,and they do not accelerate\Nthe box or anything like that.
Dialogue: 0,0:01:16.47,0:01:20.52,Default,,0000,0000,0000,,But these forces, look,\Nare pressing on the box
Dialogue: 0,0:01:20.52,0:01:22.23,Default,,0000,0000,0000,,from all the direction,
Dialogue: 0,0:01:22.23,0:01:26.22,Default,,0000,0000,0000,,so it's these forces that\Nare related to pressure.
Dialogue: 0,0:01:26.22,0:01:28.14,Default,,0000,0000,0000,,But how exactly?
Dialogue: 0,0:01:28.14,0:01:30.69,Default,,0000,0000,0000,,Well, to answer that question,
Dialogue: 0,0:01:30.69,0:01:32.79,Default,,0000,0000,0000,,let's think of a bigger box.
Dialogue: 0,0:01:32.79,0:01:35.07,Default,,0000,0000,0000,,Let's imagine that the\Nsurface area over here
Dialogue: 0,0:01:35.07,0:01:39.72,Default,,0000,0000,0000,,was twice as much as the\Nsurface area over here.
Dialogue: 0,0:01:39.72,0:01:43.05,Default,,0000,0000,0000,,Then what would be the amount of force
Dialogue: 0,0:01:43.05,0:01:44.91,Default,,0000,0000,0000,,that the air molecules\Nwould be putting here
Dialogue: 0,0:01:44.91,0:01:46.47,Default,,0000,0000,0000,,compared to over here?
Dialogue: 0,0:01:46.47,0:01:49.11,Default,,0000,0000,0000,,Pause the video and think about this.
Dialogue: 0,0:01:49.11,0:01:52.08,Default,,0000,0000,0000,,Okay, since we have twice the area,
Dialogue: 0,0:01:52.08,0:01:53.04,Default,,0000,0000,0000,,if you take any surface,
Dialogue: 0,0:01:53.04,0:01:54.90,Default,,0000,0000,0000,,if you consider, for\Nexample, the top surface,
Dialogue: 0,0:01:54.90,0:01:57.54,Default,,0000,0000,0000,,the number of molecules bumping per second
Dialogue: 0,0:01:57.54,0:01:59.70,Default,,0000,0000,0000,,would be twice as much\Ncompared to over here,
Dialogue: 0,0:01:59.70,0:02:01.11,Default,,0000,0000,0000,,because you have twice the area,
Dialogue: 0,0:02:01.11,0:02:03.12,Default,,0000,0000,0000,,and there are air molecules everywhere.
Dialogue: 0,0:02:03.12,0:02:06.30,Default,,0000,0000,0000,,So can you see just from that logic,
Dialogue: 0,0:02:06.30,0:02:07.68,Default,,0000,0000,0000,,the amount of force over here
Dialogue: 0,0:02:07.68,0:02:09.96,Default,,0000,0000,0000,,must be twice as much as over here.
Dialogue: 0,0:02:09.96,0:02:12.21,Default,,0000,0000,0000,,If the surface area was\Nthree times as much,
Dialogue: 0,0:02:12.21,0:02:14.85,Default,,0000,0000,0000,,the amount of force\Nmust be thrice as much.
Dialogue: 0,0:02:14.85,0:02:16.20,Default,,0000,0000,0000,,In other words, you can see
Dialogue: 0,0:02:16.20,0:02:19.20,Default,,0000,0000,0000,,the force that the air molecules\Nare exerting on the box
Dialogue: 0,0:02:19.20,0:02:22.05,Default,,0000,0000,0000,,is proportional to the surface area.
Dialogue: 0,0:02:22.05,0:02:27.05,Default,,0000,0000,0000,,Or in other words, the force\Nper area is a constant.
Dialogue: 0,0:02:27.15,0:02:31.32,Default,,0000,0000,0000,,That is the key characteristics\Nof the force exerted by air
Dialogue: 0,0:02:31.32,0:02:33.15,Default,,0000,0000,0000,,or, in general, any fluid.
Dialogue: 0,0:02:34.02,0:02:38.10,Default,,0000,0000,0000,,And that ratio, force per area,
Dialogue: 0,0:02:38.10,0:02:42.33,Default,,0000,0000,0000,,is what we call, in general, stress, okay?
Dialogue: 0,0:02:42.33,0:02:44.01,Default,,0000,0000,0000,,But what exactly is pressure?
Dialogue: 0,0:02:44.01,0:02:47.07,Default,,0000,0000,0000,,Well, in our example, notice that
Dialogue: 0,0:02:47.07,0:02:48.81,Default,,0000,0000,0000,,the forces are not in\Nsome random direction.
Dialogue: 0,0:02:48.81,0:02:52.53,Default,,0000,0000,0000,,All the forces are perpendicular\Nto the surface area,
Dialogue: 0,0:02:52.53,0:02:54.69,Default,,0000,0000,0000,,but how can we be so sure, you ask?
Dialogue: 0,0:02:54.69,0:02:57.57,Default,,0000,0000,0000,,I mean, these are random\Nmolecules bumping into it, right?
Dialogue: 0,0:02:57.57,0:03:00.12,Default,,0000,0000,0000,,Well, if you zoom in, what do you notice?
Dialogue: 0,0:03:00.12,0:03:01.86,Default,,0000,0000,0000,,You notice that when the molecules bump
Dialogue: 0,0:03:01.86,0:03:03.57,Default,,0000,0000,0000,,and they sort of collide,
Dialogue: 0,0:03:03.57,0:03:07.59,Default,,0000,0000,0000,,and they reflect off of the\Nwalls of our cubicle box,
Dialogue: 0,0:03:07.59,0:03:09.48,Default,,0000,0000,0000,,what you notice is the acceleration
Dialogue: 0,0:03:09.48,0:03:11.55,Default,,0000,0000,0000,,is always perpendicular to the box,
Dialogue: 0,0:03:11.55,0:03:14.25,Default,,0000,0000,0000,,and therefore, the force that\Nthe box is exerting on them
Dialogue: 0,0:03:14.25,0:03:16.65,Default,,0000,0000,0000,,is also perpendicular to the box.
Dialogue: 0,0:03:16.65,0:03:18.24,Default,,0000,0000,0000,,And therefore, from Newton's Third Law,
Dialogue: 0,0:03:18.24,0:03:21.45,Default,,0000,0000,0000,,the force that the molecules\Nare exerting on the box
Dialogue: 0,0:03:21.45,0:03:24.96,Default,,0000,0000,0000,,will also be equal and opposite\Nperpendicular to the box.
Dialogue: 0,0:03:24.96,0:03:27.81,Default,,0000,0000,0000,,And that's why the forces over here
Dialogue: 0,0:03:27.81,0:03:29.37,Default,,0000,0000,0000,,are always perpendicular.
Dialogue: 0,0:03:29.37,0:03:33.15,Default,,0000,0000,0000,,And so look, there's a special\Nkind of stress over here.
Dialogue: 0,0:03:33.15,0:03:34.56,Default,,0000,0000,0000,,It's not just any force,
Dialogue: 0,0:03:34.56,0:03:36.27,Default,,0000,0000,0000,,the force there are always perpendicular,
Dialogue: 0,0:03:36.27,0:03:39.03,Default,,0000,0000,0000,,and that particular special kind of stress
Dialogue: 0,0:03:39.03,0:03:42.33,Default,,0000,0000,0000,,where the forces are perpendicular,\Nthat is called pressure.
Dialogue: 0,0:03:42.33,0:03:45.18,Default,,0000,0000,0000,,So you can think of pressure\Nas a special kind of stress
Dialogue: 0,0:03:45.18,0:03:48.09,Default,,0000,0000,0000,,where the forces are\Nperpendicular to the area.
Dialogue: 0,0:03:48.09,0:03:50.04,Default,,0000,0000,0000,,But wait, this raises another question.
Dialogue: 0,0:03:50.04,0:03:52.92,Default,,0000,0000,0000,,Is it possible for fluids to exert forces
Dialogue: 0,0:03:52.92,0:03:55.50,Default,,0000,0000,0000,,which are paddle to the area as well?
Dialogue: 0,0:03:55.50,0:03:56.91,Default,,0000,0000,0000,,Yes.
Dialogue: 0,0:03:56.91,0:03:59.82,Default,,0000,0000,0000,,Consider the air molecules\Nagain, they're all moving,
Dialogue: 0,0:03:59.82,0:04:03.21,Default,,0000,0000,0000,,but they're all moving in\Nrandom direction, isn't it?
Dialogue: 0,0:04:03.21,0:04:07.05,Default,,0000,0000,0000,,But now, imagine that\Nthere was some kind of wind
Dialogue: 0,0:04:07.05,0:04:09.15,Default,,0000,0000,0000,,that could happen if the\Nblock itself is moving down,
Dialogue: 0,0:04:09.15,0:04:11.58,Default,,0000,0000,0000,,or there is wind blowing\Nupwards, whatever that is,
Dialogue: 0,0:04:11.58,0:04:14.07,Default,,0000,0000,0000,,there's some kind of a relative motion.
Dialogue: 0,0:04:14.07,0:04:15.51,Default,,0000,0000,0000,,Now, if there is a wind,
Dialogue: 0,0:04:15.51,0:04:17.85,Default,,0000,0000,0000,,let's say the air molecules\Nare moving upwards
Dialogue: 0,0:04:17.85,0:04:18.96,Default,,0000,0000,0000,,along with the random motion
Dialogue: 0,0:04:18.96,0:04:20.85,Default,,0000,0000,0000,,that they're doing along with that,
Dialogue: 0,0:04:20.85,0:04:23.88,Default,,0000,0000,0000,,then because these molecules also interact
Dialogue: 0,0:04:23.88,0:04:25.29,Default,,0000,0000,0000,,with the molecules of the box,
Dialogue: 0,0:04:25.29,0:04:28.56,Default,,0000,0000,0000,,they will also exert force on them,
Dialogue: 0,0:04:28.56,0:04:31.20,Default,,0000,0000,0000,,pulling them upwards.
Dialogue: 0,0:04:31.20,0:04:33.69,Default,,0000,0000,0000,,This force is called the viscous force,
Dialogue: 0,0:04:33.69,0:04:35.46,Default,,0000,0000,0000,,and at the heart of it comes from the fact
Dialogue: 0,0:04:35.46,0:04:37.50,Default,,0000,0000,0000,,that molecules can\Ninteract with each other
Dialogue: 0,0:04:37.50,0:04:39.27,Default,,0000,0000,0000,,and as a result, look,
Dialogue: 0,0:04:39.27,0:04:42.84,Default,,0000,0000,0000,,this total viscous force is parallel
Dialogue: 0,0:04:42.84,0:04:45.42,Default,,0000,0000,0000,,to the surface area,
Dialogue: 0,0:04:45.42,0:04:49.23,Default,,0000,0000,0000,,and this force tends to shear our cube.
Dialogue: 0,0:04:49.23,0:04:50.79,Default,,0000,0000,0000,,It's kind of like this deck of cards.
Dialogue: 0,0:04:50.79,0:04:52.83,Default,,0000,0000,0000,,If you press it\Nperpendicular to the surface,
Dialogue: 0,0:04:52.83,0:04:55.05,Default,,0000,0000,0000,,then the stress is just the pressure.
Dialogue: 0,0:04:55.05,0:04:57.81,Default,,0000,0000,0000,,But if you press at an angle,\Nthen there's a pedal component
Dialogue: 0,0:04:57.81,0:04:59.82,Default,,0000,0000,0000,,which shears the deck of cards.
Dialogue: 0,0:04:59.82,0:05:02.25,Default,,0000,0000,0000,,And shearing stresses\Ncan be quite complicated,
Dialogue: 0,0:05:02.25,0:05:03.24,Default,,0000,0000,0000,,but we don't have to worry about it
Dialogue: 0,0:05:03.24,0:05:06.93,Default,,0000,0000,0000,,because in our model, we\Nare considering ideal fluids
Dialogue: 0,0:05:06.93,0:05:09.75,Default,,0000,0000,0000,,and ideal fluids do\Nnot have any viscosity.
Dialogue: 0,0:05:09.75,0:05:11.40,Default,,0000,0000,0000,,And we're also going to\Nassume that we are dealing
Dialogue: 0,0:05:11.40,0:05:14.37,Default,,0000,0000,0000,,with static fluids, which\Nmeans no viscus forces,
Dialogue: 0,0:05:14.37,0:05:16.53,Default,,0000,0000,0000,,no relative motion, and therefore,
Dialogue: 0,0:05:16.53,0:05:19.11,Default,,0000,0000,0000,,we can completely ignore shearing stress.
Dialogue: 0,0:05:19.11,0:05:23.07,Default,,0000,0000,0000,,And therefore, in our model,\Nwe only have pressure.
Dialogue: 0,0:05:23.07,0:05:25.32,Default,,0000,0000,0000,,Forces will always be only perpendicular
Dialogue: 0,0:05:25.32,0:05:26.67,Default,,0000,0000,0000,,to the surface area.
Dialogue: 0,0:05:26.67,0:05:28.41,Default,,0000,0000,0000,,All right, so let's try\Nto understand pressure
Dialogue: 0,0:05:28.41,0:05:29.37,Default,,0000,0000,0000,,a little bit better.
Dialogue: 0,0:05:29.37,0:05:30.63,Default,,0000,0000,0000,,What about its units?
Dialogue: 0,0:05:30.63,0:05:32.82,Default,,0000,0000,0000,,Well, because it's force per area,
Dialogue: 0,0:05:32.82,0:05:33.93,Default,,0000,0000,0000,,the unit of force is Newtons
Dialogue: 0,0:05:33.93,0:05:36.12,Default,,0000,0000,0000,,and the unit of area is meters squared,
Dialogue: 0,0:05:36.12,0:05:37.17,Default,,0000,0000,0000,,so the unit of pressure,
Dialogue: 0,0:05:37.17,0:05:38.22,Default,,0000,0000,0000,,at least a standard unit of pressure,
Dialogue: 0,0:05:38.22,0:05:39.87,Default,,0000,0000,0000,,becomes Newtons per meters squared,
Dialogue: 0,0:05:39.87,0:05:42.12,Default,,0000,0000,0000,,which we also call pascals.
Dialogue: 0,0:05:42.12,0:05:43.68,Default,,0000,0000,0000,,So over an area of one meter squared,
Dialogue: 0,0:05:43.68,0:05:46.47,Default,,0000,0000,0000,,if there's one Newton of force\Nacting perpendicular to it,
Dialogue: 0,0:05:46.47,0:05:49.44,Default,,0000,0000,0000,,then we would say that the\Npressure is one pascal.
Dialogue: 0,0:05:49.44,0:05:50.91,Default,,0000,0000,0000,,And that is a very tiny pressure.
Dialogue: 0,0:05:50.91,0:05:52.65,Default,,0000,0000,0000,,One Newton exerted over one meter squared
Dialogue: 0,0:05:52.65,0:05:54.27,Default,,0000,0000,0000,,is incredibly tiny.
Dialogue: 0,0:05:54.27,0:05:56.28,Default,,0000,0000,0000,,So the big question is,\Nwhat is the pressure
Dialogue: 0,0:05:56.28,0:05:57.84,Default,,0000,0000,0000,,over here in a room?
Dialogue: 0,0:05:57.84,0:05:59.19,Default,,0000,0000,0000,,What is the atmospheric pressure?
Dialogue: 0,0:05:59.19,0:06:01.65,Default,,0000,0000,0000,,Pressure that the air\Nmolecules are pushing
Dialogue: 0,0:06:01.65,0:06:03.69,Default,,0000,0000,0000,,on the sides of this cube with?
Dialogue: 0,0:06:03.69,0:06:05.49,Default,,0000,0000,0000,,Well, turns out that pressure
Dialogue: 0,0:06:05.49,0:06:08.46,Default,,0000,0000,0000,,is about 10 to the power of five pascals.
Dialogue: 0,0:06:08.46,0:06:12.12,Default,,0000,0000,0000,,In other words, that is a 100,000 pascals.
Dialogue: 0,0:06:12.12,0:06:14.76,Default,,0000,0000,0000,,That is insanely huge.
Dialogue: 0,0:06:14.76,0:06:18.45,Default,,0000,0000,0000,,A 100,000 Newtons of force is exerted
Dialogue: 0,0:06:18.45,0:06:21.39,Default,,0000,0000,0000,,by the air molecules per square meter.
Dialogue: 0,0:06:21.39,0:06:24.30,Default,,0000,0000,0000,,That is insanely high. I\Nwouldn't have expected.
Dialogue: 0,0:06:24.30,0:06:27.30,Default,,0000,0000,0000,,That is the amount of\Npressure we are all feeling
Dialogue: 0,0:06:27.30,0:06:28.68,Default,,0000,0000,0000,,just sitting in our room
Dialogue: 0,0:06:28.68,0:06:30.42,Default,,0000,0000,0000,,due to the air molecules over there.
Dialogue: 0,0:06:30.42,0:06:32.22,Default,,0000,0000,0000,,But that was an important question.
Dialogue: 0,0:06:32.22,0:06:35.67,Default,,0000,0000,0000,,Why don't things get\Ncrushed under that pressure?
Dialogue: 0,0:06:35.67,0:06:37.65,Default,,0000,0000,0000,,I mean, sure, this cube\Nis not getting crushed
Dialogue: 0,0:06:37.65,0:06:40.26,Default,,0000,0000,0000,,because the internal forces\Nare able to balance that out.
Dialogue: 0,0:06:40.26,0:06:42.72,Default,,0000,0000,0000,,But what if I take a plastic bag
Dialogue: 0,0:06:42.72,0:06:44.76,Default,,0000,0000,0000,,and there's nothing inside it,
Dialogue: 0,0:06:44.76,0:06:47.25,Default,,0000,0000,0000,,then because there's so much air pressure
Dialogue: 0,0:06:47.25,0:06:48.63,Default,,0000,0000,0000,,over from the outside,
Dialogue: 0,0:06:48.63,0:06:50.70,Default,,0000,0000,0000,,shouldn't the plastic bag just get crushed
Dialogue: 0,0:06:50.70,0:06:52.23,Default,,0000,0000,0000,,due to the air pressure?
Dialogue: 0,0:06:52.23,0:06:53.51,Default,,0000,0000,0000,,Well, the reason it doesn't get crushed
Dialogue: 0,0:06:53.51,0:06:55.47,Default,,0000,0000,0000,,is because there's air inside as well.
Dialogue: 0,0:06:55.47,0:06:58.53,Default,,0000,0000,0000,,And that air also has\Nthe exact same pressure,
Dialogue: 0,0:06:58.53,0:07:01.05,Default,,0000,0000,0000,,which means it is able to\Nbalance out the pressure
Dialogue: 0,0:07:01.05,0:07:01.95,Default,,0000,0000,0000,,from the outside.
Dialogue: 0,0:07:01.95,0:07:04.56,Default,,0000,0000,0000,,But what if you could\Nsomehow suck that air out?
Dialogue: 0,0:07:04.56,0:07:06.09,Default,,0000,0000,0000,,Ooh, then the balance will be lost
Dialogue: 0,0:07:06.09,0:07:07.17,Default,,0000,0000,0000,,because the pressure drops
Dialogue: 0,0:07:07.17,0:07:11.85,Default,,0000,0000,0000,,and then we would see the\Nplastic bag collapsing,
Dialogue: 0,0:07:11.85,0:07:14.73,Default,,0000,0000,0000,,getting crushed under the\Npressure from the outside.
Dialogue: 0,0:07:14.73,0:07:16.32,Default,,0000,0000,0000,,That all makes sense, right?
Dialogue: 0,0:07:16.32,0:07:18.06,Default,,0000,0000,0000,,Okay, before moving forward,\Nlet's also quickly talk
Dialogue: 0,0:07:18.06,0:07:19.50,Default,,0000,0000,0000,,about a couple of other units of pressure
Dialogue: 0,0:07:19.50,0:07:21.27,Default,,0000,0000,0000,,that we usually use in our daily life.
Dialogue: 0,0:07:21.27,0:07:23.67,Default,,0000,0000,0000,,For example, when it\Ncomes to tire pressure,
Dialogue: 0,0:07:23.67,0:07:28.23,Default,,0000,0000,0000,,we usually talk in terms\Nof pounds per square inch.
Dialogue: 0,0:07:28.23,0:07:31.02,Default,,0000,0000,0000,,And just to give some feeling for numbers,
Dialogue: 0,0:07:31.02,0:07:33.06,Default,,0000,0000,0000,,10 to the power of five\Npascals happens to be close
Dialogue: 0,0:07:33.06,0:07:36.36,Default,,0000,0000,0000,,to 14.7 pounds per square inch.
Dialogue: 0,0:07:36.36,0:07:38.13,Default,,0000,0000,0000,,So that is the atmospheric pressure
Dialogue: 0,0:07:38.13,0:07:39.54,Default,,0000,0000,0000,,in pounds per square inch.
Dialogue: 0,0:07:39.54,0:07:41.91,Default,,0000,0000,0000,,Another unit is millimeters of mercury,
Dialogue: 0,0:07:41.91,0:07:44.52,Default,,0000,0000,0000,,and 10 to the power of five\Npascals happens to be close
Dialogue: 0,0:07:44.52,0:07:46.56,Default,,0000,0000,0000,,to 760 millimeters of mercury.
Dialogue: 0,0:07:46.56,0:07:49.68,Default,,0000,0000,0000,,In other words, the atmospheric\Npressure can pull up
Dialogue: 0,0:07:49.68,0:07:54.12,Default,,0000,0000,0000,,the mercury up to 760\Nmillimeters in a column.
Dialogue: 0,0:07:54.12,0:07:56.40,Default,,0000,0000,0000,,Anyways, let's focus on Pascals for now.
Dialogue: 0,0:07:56.40,0:07:59.37,Default,,0000,0000,0000,,And the big question\Nnow, is pressure a scalar
Dialogue: 0,0:07:59.37,0:08:01.89,Default,,0000,0000,0000,,or a vector quantity, what do you think?
Dialogue: 0,0:08:01.89,0:08:03.54,Default,,0000,0000,0000,,Well, my intuition says it's vector
Dialogue: 0,0:08:03.54,0:08:06.03,Default,,0000,0000,0000,,because there's force in one over here,
Dialogue: 0,0:08:06.03,0:08:07.26,Default,,0000,0000,0000,,but let's think a little\Nbit more about it.
Dialogue: 0,0:08:07.26,0:08:11.97,Default,,0000,0000,0000,,In fact, think about\Npressure at a specific point.
Dialogue: 0,0:08:11.97,0:08:13.29,Default,,0000,0000,0000,,How do we do that?
Dialogue: 0,0:08:13.29,0:08:16.23,Default,,0000,0000,0000,,Well, one way to do that\Nis you take this box
Dialogue: 0,0:08:16.23,0:08:17.91,Default,,0000,0000,0000,,and shrink it down.
Dialogue: 0,0:08:17.91,0:08:20.64,Default,,0000,0000,0000,,Let's say we shrink the size of the box.
Dialogue: 0,0:08:20.64,0:08:23.55,Default,,0000,0000,0000,,Well, now, the area has\Nbecome, let's say half,
Dialogue: 0,0:08:23.55,0:08:27.36,Default,,0000,0000,0000,,but the number of air molecules\Nwill also become half,
Dialogue: 0,0:08:27.36,0:08:29.85,Default,,0000,0000,0000,,and therefore, the\Nforce also becomes half,
Dialogue: 0,0:08:29.85,0:08:31.98,Default,,0000,0000,0000,,making sure force per area stays the same.
Dialogue: 0,0:08:31.98,0:08:33.93,Default,,0000,0000,0000,,So the pressure stays the same.
Dialogue: 0,0:08:33.93,0:08:35.55,Default,,0000,0000,0000,,We'll keep shrinking it, keep shrinking,
Dialogue: 0,0:08:35.55,0:08:36.84,Default,,0000,0000,0000,,and keep shrinking it.
Dialogue: 0,0:08:36.84,0:08:41.19,Default,,0000,0000,0000,,Now shrink it all the way\Nto an extremely tiny point.
Dialogue: 0,0:08:41.19,0:08:43.35,Default,,0000,0000,0000,,We call that as an infinitesimal.
Dialogue: 0,0:08:43.35,0:08:45.09,Default,,0000,0000,0000,,Now, the areas are incredibly tiny,
Dialogue: 0,0:08:45.09,0:08:47.04,Default,,0000,0000,0000,,the forces are incredibly tiny,
Dialogue: 0,0:08:47.04,0:08:50.04,Default,,0000,0000,0000,,yet the force per area stays the same.
Dialogue: 0,0:08:50.04,0:08:52.80,Default,,0000,0000,0000,,That is the pressure\Nat a particular point.
Dialogue: 0,0:08:52.80,0:08:55.77,Default,,0000,0000,0000,,And now the question is,\Nshould we assign a direction
Dialogue: 0,0:08:55.77,0:08:58.44,Default,,0000,0000,0000,,to this number, to this pressure?
Dialogue: 0,0:08:58.44,0:09:01.59,Default,,0000,0000,0000,,Well, not really, because\Nall I need is a number,
Dialogue: 0,0:09:01.59,0:09:04.02,Default,,0000,0000,0000,,because what this number is\Nsaying is that if you zoom in
Dialogue: 0,0:09:04.02,0:09:06.09,Default,,0000,0000,0000,,and if you have any area,
Dialogue: 0,0:09:06.09,0:09:09.36,Default,,0000,0000,0000,,then there will always be\Nforces perpendicular to the area
Dialogue: 0,0:09:09.36,0:09:12.69,Default,,0000,0000,0000,,that is exerted by the fluid,\Nand that force per area
Dialogue: 0,0:09:12.69,0:09:14.58,Default,,0000,0000,0000,,will be 10 to the power of five pascals.
Dialogue: 0,0:09:14.58,0:09:16.50,Default,,0000,0000,0000,,And that is valid from any direction.
Dialogue: 0,0:09:16.50,0:09:18.54,Default,,0000,0000,0000,,It doesn't matter how\Nyour area is oriented,
Dialogue: 0,0:09:18.54,0:09:21.30,Default,,0000,0000,0000,,that will always be the case.
Dialogue: 0,0:09:21.30,0:09:24.42,Default,,0000,0000,0000,,Which means, look, all I need is a number.
Dialogue: 0,0:09:24.42,0:09:26.67,Default,,0000,0000,0000,,I don't need a direction
Dialogue: 0,0:09:26.67,0:09:28.56,Default,,0000,0000,0000,,to communicate the idea of pressure,
Dialogue: 0,0:09:28.56,0:09:32.37,Default,,0000,0000,0000,,and therefore, pressure\Nis a scalar quantity.
Dialogue: 0,0:09:32.37,0:09:33.84,Default,,0000,0000,0000,,And because the number of air molecules
Dialogue: 0,0:09:33.84,0:09:36.30,Default,,0000,0000,0000,,bumping per square meter is\Npretty much the same everywhere
Dialogue: 0,0:09:36.30,0:09:39.03,Default,,0000,0000,0000,,and their speeds are pretty\Nmuch the same anywhere you take,
Dialogue: 0,0:09:39.03,0:09:43.44,Default,,0000,0000,0000,,therefore, the pressure\Nnow is the same everywhere.
Dialogue: 0,0:09:43.44,0:09:46.92,Default,,0000,0000,0000,,But if we zoom out and look\Nat the entire atmosphere,
Dialogue: 0,0:09:46.92,0:09:49.14,Default,,0000,0000,0000,,for example, that's not the case.
Dialogue: 0,0:09:49.14,0:09:50.91,Default,,0000,0000,0000,,In fact, that 10 to the\Npower of five we said
Dialogue: 0,0:09:50.91,0:09:53.16,Default,,0000,0000,0000,,is the pressure close to the sea level,
Dialogue: 0,0:09:53.16,0:09:54.63,Default,,0000,0000,0000,,but if you were to go\Na little bit above it,
Dialogue: 0,0:09:54.63,0:09:56.86,Default,,0000,0000,0000,,say at 10 kilometers, which\Nis the cruising altitude
Dialogue: 0,0:09:56.86,0:09:59.67,Default,,0000,0000,0000,,of commercial airplanes,\Nyou would now find
Dialogue: 0,0:09:59.67,0:10:01.56,Default,,0000,0000,0000,,that the pressure is about one fourth
Dialogue: 0,0:10:01.56,0:10:03.03,Default,,0000,0000,0000,,is what we'd find over here.
Dialogue: 0,0:10:03.03,0:10:04.62,Default,,0000,0000,0000,,Why is the pressure different over here
Dialogue: 0,0:10:04.62,0:10:05.91,Default,,0000,0000,0000,,compared to over here?
Dialogue: 0,0:10:05.91,0:10:07.44,Default,,0000,0000,0000,,Well, that's because\Nthe molecules over here
Dialogue: 0,0:10:07.44,0:10:09.87,Default,,0000,0000,0000,,are carrying the weight\Nof the entire atmosphere
Dialogue: 0,0:10:09.87,0:10:10.70,Default,,0000,0000,0000,,on top of it.
Dialogue: 0,0:10:10.70,0:10:12.78,Default,,0000,0000,0000,,That entire weight is pushing down,
Dialogue: 0,0:10:12.78,0:10:14.70,Default,,0000,0000,0000,,pressing the molecules over here.
Dialogue: 0,0:10:14.70,0:10:17.34,Default,,0000,0000,0000,,However, if you consider\Nthe molecules at this level,
Dialogue: 0,0:10:17.34,0:10:18.78,Default,,0000,0000,0000,,they're not carrying the entire weight,
Dialogue: 0,0:10:18.78,0:10:20.55,Default,,0000,0000,0000,,they're carrying the\Nweight only on top of them.
Dialogue: 0,0:10:20.55,0:10:22.26,Default,,0000,0000,0000,,They're not carrying the\Nweight of this amount
Dialogue: 0,0:10:22.26,0:10:23.09,Default,,0000,0000,0000,,of air molecules.
Dialogue: 0,0:10:23.09,0:10:25.29,Default,,0000,0000,0000,,And that's why the pressure\Nover here is slightly lower,
Dialogue: 0,0:10:25.29,0:10:27.12,Default,,0000,0000,0000,,which means the pressure depends
Dialogue: 0,0:10:27.12,0:10:29.07,Default,,0000,0000,0000,,on the height if you zoom out.
Dialogue: 0,0:10:29.07,0:10:31.50,Default,,0000,0000,0000,,And so now, the next\Nobvious question would be,
Dialogue: 0,0:10:31.50,0:10:33.75,Default,,0000,0000,0000,,is there a relationship between\Nthe pressure and the height?
Dialogue: 0,0:10:33.75,0:10:35.64,Default,,0000,0000,0000,,Well, there is, and it's\Nharder to think about that
Dialogue: 0,0:10:35.64,0:10:37.77,Default,,0000,0000,0000,,for air molecules because air molecules
Dialogue: 0,0:10:37.77,0:10:40.26,Default,,0000,0000,0000,,are very compressible, so\Nit's a little hard over there.
Dialogue: 0,0:10:40.26,0:10:42.66,Default,,0000,0000,0000,,But let's consider\Nnon-compressible fluids,
Dialogue: 0,0:10:42.66,0:10:45.18,Default,,0000,0000,0000,,like water, for example.
Dialogue: 0,0:10:45.18,0:10:46.77,Default,,0000,0000,0000,,So here's a specific question.
Dialogue: 0,0:10:46.77,0:10:49.20,Default,,0000,0000,0000,,If we know the pressure\Nat some level over here,
Dialogue: 0,0:10:49.20,0:10:53.79,Default,,0000,0000,0000,,what is the pressure at\Nsome depth, say h, below?
Dialogue: 0,0:10:53.79,0:10:55.56,Default,,0000,0000,0000,,So let's say the pressure at the top is PT
Dialogue: 0,0:10:55.56,0:10:56.86,Default,,0000,0000,0000,,and the pressure at the bottom is PB.
Dialogue: 0,0:10:56.86,0:10:58.92,Default,,0000,0000,0000,,Well, we know that the pressure\Nat the bottom is higher
Dialogue: 0,0:10:58.92,0:10:59.76,Default,,0000,0000,0000,,than the pressure at the top.
Dialogue: 0,0:10:59.76,0:11:00.84,Default,,0000,0000,0000,,So we could say pressure at the bottom
Dialogue: 0,0:11:00.84,0:11:04.53,Default,,0000,0000,0000,,equals pressure at the top,\Nplus some additional pressure
Dialogue: 0,0:11:04.53,0:11:06.24,Default,,0000,0000,0000,,due to this weight of the water.
Dialogue: 0,0:11:06.24,0:11:07.83,Default,,0000,0000,0000,,But how do we figure that out?
Dialogue: 0,0:11:07.83,0:11:10.83,Default,,0000,0000,0000,,Well, for that, let's just\Ndraw a cuboidal surface,
Dialogue: 0,0:11:10.83,0:11:12.60,Default,,0000,0000,0000,,having the surface area A.
Dialogue: 0,0:11:12.60,0:11:14.91,Default,,0000,0000,0000,,Now, the additional pressure\Nthat we are getting over here
Dialogue: 0,0:11:14.91,0:11:17.19,Default,,0000,0000,0000,,is due to the weight of\Nthis cuboidal column.
Dialogue: 0,0:11:17.19,0:11:20.70,Default,,0000,0000,0000,,To be precise, it's going\Nto be the weight per area.
Dialogue: 0,0:11:20.70,0:11:23.55,Default,,0000,0000,0000,,So this term over here is\Ngonna be weight per area.
Dialogue: 0,0:11:23.55,0:11:26.70,Default,,0000,0000,0000,,But what exactly is the\Nweight of this cubital column?
Dialogue: 0,0:11:26.70,0:11:28.11,Default,,0000,0000,0000,,Well, weight is just the force of gravity,
Dialogue: 0,0:11:28.11,0:11:30.90,Default,,0000,0000,0000,,so it's gonna be MG, where\NM is the mass of the water
Dialogue: 0,0:11:30.90,0:11:34.29,Default,,0000,0000,0000,,in this column divided\Nby the area, which is A,
Dialogue: 0,0:11:34.29,0:11:36.06,Default,,0000,0000,0000,,but how do we figure out what is the mass
Dialogue: 0,0:11:36.06,0:11:38.43,Default,,0000,0000,0000,,of the column of this water?
Dialogue: 0,0:11:38.43,0:11:41.88,Default,,0000,0000,0000,,Well, we know that density\Nis mass per volume.
Dialogue: 0,0:11:41.88,0:11:44.67,Default,,0000,0000,0000,,So mass can written as\Ndensity times volume,
Dialogue: 0,0:11:44.67,0:11:47.22,Default,,0000,0000,0000,,and that's because we know\Nthe density of a fluid.
Dialogue: 0,0:11:47.22,0:11:51.72,Default,,0000,0000,0000,,So we can write this as\Ndensity of the fluid,
Dialogue: 0,0:11:51.72,0:11:53.94,Default,,0000,0000,0000,,density of water, times\Nthe volume of the column
Dialogue: 0,0:11:53.94,0:11:55.71,Default,,0000,0000,0000,,times GD divided by A.
Dialogue: 0,0:11:55.71,0:11:58.20,Default,,0000,0000,0000,,But the final question is what\Nis the volume of this column?
Dialogue: 0,0:11:58.20,0:11:59.97,Default,,0000,0000,0000,,Hey, we know the volume of the column.
Dialogue: 0,0:11:59.97,0:12:01.20,Default,,0000,0000,0000,,Volume of this cuboidal column
Dialogue: 0,0:12:01.20,0:12:03.54,Default,,0000,0000,0000,,is just going to be area times the height.
Dialogue: 0,0:12:03.54,0:12:05.97,Default,,0000,0000,0000,,And so we can plug that in over here,
Dialogue: 0,0:12:05.97,0:12:07.92,Default,,0000,0000,0000,,and if we cancel out the areas,
Dialogue: 0,0:12:07.92,0:12:10.23,Default,,0000,0000,0000,,we finally get our expression.
Dialogue: 0,0:12:10.23,0:12:11.64,Default,,0000,0000,0000,,The pressure at the bottom will equal
Dialogue: 0,0:12:11.64,0:12:15.75,Default,,0000,0000,0000,,the pressure at the top plus\Nthis additional pressure
Dialogue: 0,0:12:15.75,0:12:18.03,Default,,0000,0000,0000,,due to the weight of this column.
Dialogue: 0,0:12:18.03,0:12:20.10,Default,,0000,0000,0000,,And this equation will work anywhere
Dialogue: 0,0:12:20.10,0:12:22.59,Default,,0000,0000,0000,,as long as you're dealing\Nwith a non-compressible fluid
Dialogue: 0,0:12:22.59,0:12:25.17,Default,,0000,0000,0000,,because we are assuming\Nthe density to be the same.
Dialogue: 0,0:12:25.17,0:12:27.09,Default,,0000,0000,0000,,If you consider a\Ncompressible fluid, like air,
Dialogue: 0,0:12:27.09,0:12:30.12,Default,,0000,0000,0000,,the density varies and this\Ncalculation becomes harder,
Dialogue: 0,0:12:30.12,0:12:34.35,Default,,0000,0000,0000,,and so you'll get a considerably\Ndifferent expression.
Dialogue: 0,0:12:34.35,0:12:36.63,Default,,0000,0000,0000,,But what I find really\Nsurprising about this
Dialogue: 0,0:12:36.63,0:12:38.82,Default,,0000,0000,0000,,is that for a given\Nnon-compressible fluid,
Dialogue: 0,0:12:38.82,0:12:40.89,Default,,0000,0000,0000,,which means it has a specific density,
Dialogue: 0,0:12:40.89,0:12:42.90,Default,,0000,0000,0000,,the pressure difference between two points
Dialogue: 0,0:12:42.90,0:12:46.44,Default,,0000,0000,0000,,only depends on their\Nheight, nothing else.
Dialogue: 0,0:12:46.44,0:12:48.03,Default,,0000,0000,0000,,In other words, this means\Nthe pressure difference
Dialogue: 0,0:12:48.03,0:12:50.28,Default,,0000,0000,0000,,between two points, say\N10 centimeter apart,
Dialogue: 0,0:12:50.28,0:12:52.77,Default,,0000,0000,0000,,whether you consider that in an ocean
Dialogue: 0,0:12:52.77,0:12:56.22,Default,,0000,0000,0000,,or a tiny test tube, it's the same.
Dialogue: 0,0:12:56.22,0:12:59.22,Default,,0000,0000,0000,,It doesn't matter how much\Nwater you're dealing with,
Dialogue: 0,0:12:59.22,0:13:02.40,Default,,0000,0000,0000,,it's just the height that matters.
Dialogue: 0,0:13:02.40,0:13:04.74,Default,,0000,0000,0000,,Anyways, now we can introduce\Ntwo kinds of pressure.
Dialogue: 0,0:13:04.74,0:13:07.05,Default,,0000,0000,0000,,The pressure that we have\Nover here, these two,
Dialogue: 0,0:13:07.05,0:13:09.18,Default,,0000,0000,0000,,they're called absolute pressures.
Dialogue: 0,0:13:09.18,0:13:11.46,Default,,0000,0000,0000,,For example, if this was\Nthe atmospheric pressure,
Dialogue: 0,0:13:11.46,0:13:13.59,Default,,0000,0000,0000,,then that is the absolute pressure.
Dialogue: 0,0:13:13.59,0:13:15.87,Default,,0000,0000,0000,,The absolute pressure of the atmosphere
Dialogue: 0,0:13:15.87,0:13:16.77,Default,,0000,0000,0000,,close to the sea level
Dialogue: 0,0:13:16.77,0:13:20.28,Default,,0000,0000,0000,,is about 10 to the power of five pascals.
Dialogue: 0,0:13:20.28,0:13:21.81,Default,,0000,0000,0000,,But now, look at this term.
Dialogue: 0,0:13:21.81,0:13:23.40,Default,,0000,0000,0000,,What does that term represent?
Dialogue: 0,0:13:23.40,0:13:26.13,Default,,0000,0000,0000,,That represents the extra pressure
Dialogue: 0,0:13:26.13,0:13:27.90,Default,,0000,0000,0000,,that you have at this point
Dialogue: 0,0:13:27.90,0:13:29.91,Default,,0000,0000,0000,,over and above the atmospheric pressure.
Dialogue: 0,0:13:29.91,0:13:34.11,Default,,0000,0000,0000,,That extra pressure is what\Nwe call the gauge pressure.
Dialogue: 0,0:13:34.11,0:13:36.15,Default,,0000,0000,0000,,And most of the time when\Nwe're talking about pressure
Dialogue: 0,0:13:36.15,0:13:37.29,Default,,0000,0000,0000,,in our day-today life,
Dialogue: 0,0:13:37.29,0:13:39.36,Default,,0000,0000,0000,,we are not talking about\Nthe absolute pressure,
Dialogue: 0,0:13:39.36,0:13:41.07,Default,,0000,0000,0000,,we're talking about the gauge pressure.
Dialogue: 0,0:13:41.07,0:13:43.53,Default,,0000,0000,0000,,So for example, when we talk\Nabout the blood pressure,
Dialogue: 0,0:13:43.53,0:13:46.02,Default,,0000,0000,0000,,we say it's 120 millimeters of mercury.
Dialogue: 0,0:13:46.02,0:13:47.49,Default,,0000,0000,0000,,What does that even mean?
Dialogue: 0,0:13:47.49,0:13:49.77,Default,,0000,0000,0000,,Well, remember that the\Natmospheric pressure
Dialogue: 0,0:13:49.77,0:13:52.80,Default,,0000,0000,0000,,is 760 millimeters of mercury.
Dialogue: 0,0:13:52.80,0:13:55.53,Default,,0000,0000,0000,,This is the pressure over and above that.
Dialogue: 0,0:13:55.53,0:13:59.19,Default,,0000,0000,0000,,So the pressure in the arteries, or veins,
Dialogue: 0,0:13:59.19,0:14:00.75,Default,,0000,0000,0000,,during a cysto for example,
Dialogue: 0,0:14:00.75,0:14:05.43,Default,,0000,0000,0000,,is 760 millimeters plus\N120 millimeters of mercury.
Dialogue: 0,0:14:05.43,0:14:06.48,Default,,0000,0000,0000,,That's what it really means.
Dialogue: 0,0:14:06.48,0:14:08.40,Default,,0000,0000,0000,,So this is the additional pressure
Dialogue: 0,0:14:08.40,0:14:10.29,Default,,0000,0000,0000,,above the atmospheric pressure,
Dialogue: 0,0:14:10.29,0:14:12.09,Default,,0000,0000,0000,,so this is the gauge pressure.
Dialogue: 0,0:14:12.09,0:14:14.07,Default,,0000,0000,0000,,The same as the case with our tires.
Dialogue: 0,0:14:14.07,0:14:16.14,Default,,0000,0000,0000,,For example, if you look at\Nthe pressure inside the tire,
Dialogue: 0,0:14:16.14,0:14:19.44,Default,,0000,0000,0000,,you can see it's about 40 PSI,
Dialogue: 0,0:14:19.44,0:14:21.42,Default,,0000,0000,0000,,but that is a gauge pressure,
Dialogue: 0,0:14:21.42,0:14:23.46,Default,,0000,0000,0000,,meaning it's over and above\Nthe atmospheric pressure.
Dialogue: 0,0:14:23.46,0:14:26.34,Default,,0000,0000,0000,,Remember, the atmospheric\Npressure is 14.7 PSI.
Dialogue: 0,0:14:26.34,0:14:30.15,Default,,0000,0000,0000,,So the pressure in the tire is 40 PSI
Dialogue: 0,0:14:30.15,0:14:32.40,Default,,0000,0000,0000,,above the atmospheric pressure.
Dialogue: 0,0:14:32.40,0:14:35.52,Default,,0000,0000,0000,,So most of the time, we're\Ndealing with gauge pressures.
Dialogue: 0,0:14:35.52,0:14:36.84,Default,,0000,0000,0000,,Okay, finally. coming back over here,
Dialogue: 0,0:14:36.84,0:14:39.36,Default,,0000,0000,0000,,suppose we were to draw a\Ngraph of the gauge pressure
Dialogue: 0,0:14:39.36,0:14:40.77,Default,,0000,0000,0000,,versus the depth.
Dialogue: 0,0:14:40.77,0:14:42.63,Default,,0000,0000,0000,,Okay, what do you think the\Ngraph would look like for,
Dialogue: 0,0:14:42.63,0:14:44.82,Default,,0000,0000,0000,,say, a lake and for the ocean?
Dialogue: 0,0:14:44.82,0:14:46.44,Default,,0000,0000,0000,,Why don't you pause it and\Nhave a think about this?
Dialogue: 0,0:14:46.44,0:14:47.79,Default,,0000,0000,0000,,Okay, let's consider the lake.
Dialogue: 0,0:14:47.79,0:14:49.86,Default,,0000,0000,0000,,Right at the surface, the\Ngauge pressure is zero
Dialogue: 0,0:14:49.86,0:14:51.63,Default,,0000,0000,0000,,because the pressure\Nover there is the same
Dialogue: 0,0:14:51.63,0:14:53.85,Default,,0000,0000,0000,,as the atmospheric pressure,\Nso we start from zero,
Dialogue: 0,0:14:53.85,0:14:56.46,Default,,0000,0000,0000,,and then you can see that the\Ngauge pressure is proportional
Dialogue: 0,0:14:56.46,0:14:58.02,Default,,0000,0000,0000,,to the height, it's proportional to it,
Dialogue: 0,0:14:58.02,0:14:59.28,Default,,0000,0000,0000,,so we get a straight line.
Dialogue: 0,0:14:59.28,0:15:02.88,Default,,0000,0000,0000,,And so we'd expect the\Npressure to increase linearly.
Dialogue: 0,0:15:02.88,0:15:05.58,Default,,0000,0000,0000,,That's what we would get for\Nthe lake. What about the ocean?
Dialogue: 0,0:15:05.58,0:15:08.01,Default,,0000,0000,0000,,Well, ocean is also water,\Nso it has the same density,
Dialogue: 0,0:15:08.01,0:15:09.06,Default,,0000,0000,0000,,or does it?
Dialogue: 0,0:15:09.06,0:15:10.89,Default,,0000,0000,0000,,Remember, ocean has salt water,
Dialogue: 0,0:15:10.89,0:15:12.75,Default,,0000,0000,0000,,so the density is slightly higher.
Dialogue: 0,0:15:12.75,0:15:14.61,Default,,0000,0000,0000,,So for the ocean, we expect\Nthe line to be steeper,
Dialogue: 0,0:15:14.61,0:15:16.65,Default,,0000,0000,0000,,having slightly higher slope.
Dialogue: 0,0:15:16.65,0:15:18.33,Default,,0000,0000,0000,,Finally, before wrapping up the video,
Dialogue: 0,0:15:18.33,0:15:20.46,Default,,0000,0000,0000,,if you were to submerge\Na cube inside water,
Dialogue: 0,0:15:20.46,0:15:22.17,Default,,0000,0000,0000,,earlier, we said that the\Npressure is gonna be the same
Dialogue: 0,0:15:22.17,0:15:23.49,Default,,0000,0000,0000,,from all directions,
Dialogue: 0,0:15:23.49,0:15:25.08,Default,,0000,0000,0000,,but now we know that the\Npressure on the bottom
Dialogue: 0,0:15:25.08,0:15:27.84,Default,,0000,0000,0000,,is slightly higher than\Nthe pressure on the top,
Dialogue: 0,0:15:27.84,0:15:29.64,Default,,0000,0000,0000,,which means the forces on the bottom
Dialogue: 0,0:15:29.64,0:15:31.53,Default,,0000,0000,0000,,would be slightly higher\Nthan the force on the top
Dialogue: 0,0:15:31.53,0:15:33.03,Default,,0000,0000,0000,,because the area is the same.
Dialogue: 0,0:15:33.03,0:15:36.90,Default,,0000,0000,0000,,So wouldn't that produce\Na net upward force?
Dialogue: 0,0:15:36.90,0:15:40.29,Default,,0000,0000,0000,,Yes, it would, and that's\Ncalled the buoyant force,
Dialogue: 0,0:15:40.29,0:15:43.08,Default,,0000,0000,0000,,which is responsible for\Nmaking certain things float.
Dialogue: 0,0:15:43.08,0:15:45.88,Default,,0000,0000,0000,,And that's something we'll\Ntalk about in a future video.