[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.57,0:00:04.68,Default,,0000,0000,0000,,So I have got this block of wood here that has a mass of 5 kilograms
Dialogue: 0,0:00:04.68,0:00:08.83,Default,,0000,0000,0000,,and it is sitting on some dirt and we are near the surface of the earth
Dialogue: 0,0:00:08.83,0:00:16.72,Default,,0000,0000,0000,,and the coefficient of static friction between this type of wood and this type of dirt is 0.60
Dialogue: 0,0:00:16.72,0:00:23.03,Default,,0000,0000,0000,,and the coefficient of kinetic friction between this type of wood and this type of dirt is 0.55
Dialogue: 0,0:00:23.03,0:00:25.63,Default,,0000,0000,0000,,This was measured by someone else long ago
Dialogue: 0,0:00:25.63,0:00:28.06,Default,,0000,0000,0000,,or you found it in some type of a book someplace
Dialogue: 0,0:00:28.06,0:00:32.30,Default,,0000,0000,0000,,And let's say we push on this side of the block with a force of a 100 N
Dialogue: 0,0:00:32.30,0:00:34.44,Default,,0000,0000,0000,,What is going to happen?
Dialogue: 0,0:00:34.44,0:00:37.25,Default,,0000,0000,0000,,So the first thing you might realize is if there is no friction
Dialogue: 0,0:00:37.25,0:00:39.31,Default,,0000,0000,0000,,if this was a completely frictionless boundary and there is
Dialogue: 0,0:00:39.31,0:00:42.64,Default,,0000,0000,0000,,no air resistance, we are assuming that there is no air resistance in this example
Dialogue: 0,0:00:42.64,0:00:45.32,Default,,0000,0000,0000,,That in this dimension, in the horizontal dimension
Dialogue: 0,0:00:45.32,0:00:48.22,Default,,0000,0000,0000,,there would only be one force here, this 100 N force
Dialogue: 0,0:00:48.22,0:00:51.81,Default,,0000,0000,0000,,It would be completely unbalanced and that would be the net force
Dialogue: 0,0:00:51.81,0:00:56.35,Default,,0000,0000,0000,,and so you would have a force going in that direction of a 100 N on a mass of 5 kilograms
Dialogue: 0,0:00:56.35,0:00:59.54,Default,,0000,0000,0000,,Force = Mass times acceleration
Dialogue: 0,0:00:59.54,0:01:01.58,Default,,0000,0000,0000,,acceleration and force are vector quantities
Dialogue: 0,0:01:01.58,0:01:04.11,Default,,0000,0000,0000,,So you would have the force divided by the mass
Dialogue: 0,0:01:04.11,0:01:08.64,Default,,0000,0000,0000,,would give you 20 meters per second of acceleration in the rightward direction
Dialogue: 0,0:01:08.64,0:01:10.95,Default,,0000,0000,0000,,That is if there were no friction
Dialogue: 0,0:01:10.95,0:01:13.04,Default,,0000,0000,0000,,but there is friction in this situation
Dialogue: 0,0:01:13.04,0:01:15.16,Default,,0000,0000,0000,,So let's think about how we'll deal with it
Dialogue: 0,0:01:15.16,0:01:17.95,Default,,0000,0000,0000,,So the coefficient of friction tells us
Dialogue: 0,0:01:17.95,0:01:22.74,Default,,0000,0000,0000,,So this right here is the ratio between the magnitude of the force
Dialogue: 0,0:01:22.74,0:01:24.46,Default,,0000,0000,0000,,that I have called the budging force
Dialogue: 0,0:01:24.46,0:01:28.82,Default,,0000,0000,0000,,The amount of force you need to apply to get this thing to budge
Dialogue: 0,0:01:28.82,0:01:32.67,Default,,0000,0000,0000,,to get this thing to start moving. So we can start using the coefficient of kinetic friction
Dialogue: 0,0:01:32.67,0:01:38.20,Default,,0000,0000,0000,,It's the ratio between that and the magnitude of the force of contact
Dialogue: 0,0:01:38.20,0:01:41.10,Default,,0000,0000,0000,,between this block and the floor or ground here
Dialogue: 0,0:01:41.10,0:01:44.04,Default,,0000,0000,0000,,And the magnitude of that force of contact is the same thing
Dialogue: 0,0:01:44.04,0:01:48.41,Default,,0000,0000,0000,,as the normal force that the ground is applying on the block
Dialogue: 0,0:01:48.41,0:01:52.81,Default,,0000,0000,0000,,the magnitude of the normal force the ground is applying on the block
Dialogue: 0,0:01:52.81,0:01:54.15,Default,,0000,0000,0000,,Then once its moving
Dialogue: 0,0:01:54.15,0:01:58.65,Default,,0000,0000,0000,,then we can say that this is going to be--this will then be equal to
Dialogue: 0,0:01:58.65,0:02:01.71,Default,,0000,0000,0000,,this over here will be equal to the force of friction
Dialogue: 0,0:02:01.71,0:02:05.14,Default,,0000,0000,0000,,So this is the force that really overcome friction
Dialogue: 0,0:02:05.14,0:02:08.36,Default,,0000,0000,0000,,and this over here will be equal to the force of friction
Dialogue: 0,0:02:08.36,0:02:13.28,Default,,0000,0000,0000,,The magnitude of the force of friction over the force of contact
Dialogue: 0,0:02:13.28,0:02:16.72,Default,,0000,0000,0000,,the contact force between those two, so over the normal force
Dialogue: 0,0:02:16.72,0:02:18.14,Default,,0000,0000,0000,,and it makes sense
Dialogue: 0,0:02:18.14,0:02:21.03,Default,,0000,0000,0000,,that the larger the contact force
Dialogue: 0,0:02:21.03,0:02:23.05,Default,,0000,0000,0000,,the more that these are being pressed together
Dialogue: 0,0:02:23.05,0:02:26.21,Default,,0000,0000,0000,,the little at the atomic level, they kind of really get into each others grooves
Dialogue: 0,0:02:26.21,0:02:28.43,Default,,0000,0000,0000,,the more budging force you would need
Dialogue: 0,0:02:28.43,0:02:31.62,Default,,0000,0000,0000,,or the more friction force would go against your motion
Dialogue: 0,0:02:31.62,0:02:33.65,Default,,0000,0000,0000,,And in either situation
Dialogue: 0,0:02:33.65,0:02:35.92,Default,,0000,0000,0000,,the force of friction is going against your motion
Dialogue: 0,0:02:35.92,0:02:37.32,Default,,0000,0000,0000,,So even if you push it in that way
Dialogue: 0,0:02:37.32,0:02:40.42,Default,,0000,0000,0000,,sounds like force of friction is all of a sudden going to help you
Dialogue: 0,0:02:40.42,0:02:44.16,Default,,0000,0000,0000,,So let's think about what the necessary force will we need
Dialogue: 0,0:02:44.16,0:02:47.50,Default,,0000,0000,0000,,to overcome the force of friction right here in the static situation
Dialogue: 0,0:02:47.50,0:02:51.96,Default,,0000,0000,0000,,So the force of gravity on this block
Dialogue: 0,0:02:51.96,0:02:57.76,Default,,0000,0000,0000,,is going to be the gravitational field which is 9.8 m/s^2 times 5 kilograms
Dialogue: 0,0:02:57.76,0:03:05.55,Default,,0000,0000,0000,,9.8 m/s times 5 kilograms gives 49 kilogram meters per second or 49 newtons down
Dialogue: 0,0:03:05.55,0:03:08.39,Default,,0000,0000,0000,,This is the force, the magnitude of the force due to gravity
Dialogue: 0,0:03:08.39,0:03:11.54,Default,,0000,0000,0000,,the direction is straight down towards the center of the earth
Dialogue: 0,0:03:11.54,0:03:17.14,Default,,0000,0000,0000,,The normal force, and that force is there because this block is not accelerating downwards
Dialogue: 0,0:03:17.14,0:03:21.39,Default,,0000,0000,0000,,So there must be some force that completely balances off the force of gravity
Dialogue: 0,0:03:21.39,0:03:26.03,Default,,0000,0000,0000,,And in this example, it is the normal force
Dialogue: 0,0:03:26.03,0:03:30.10,Default,,0000,0000,0000,,So it is acting 49 newtons upward
Dialogue: 0,0:03:30.10,0:03:35.06,Default,,0000,0000,0000,,and so these net out. And that's why this block does not accelerate upwards or downwards
Dialogue: 0,0:03:35.06,0:03:36.97,Default,,0000,0000,0000,,So what we have is the budge the
Dialogue: 0,0:03:36.97,0:03:41.05,Default,,0000,0000,0000,,magnitude of the budging force, needs to be equal to, over the magnitude of the normal force
Dialogue: 0,0:03:41.05,0:03:43.62,Default,,0000,0000,0000,,well this thing right over here is going to be 49 newtons
Dialogue: 0,0:03:43.62,0:03:46.81,Default,,0000,0000,0000,,Is equal to 0.60
Dialogue: 0,0:03:46.81,0:03:52.37,Default,,0000,0000,0000,,Or we could say that the magnitude of the budging force
Dialogue: 0,0:03:52.37,0:03:56.70,Default,,0000,0000,0000,,is equal to 49 newtons times the coefficient of static fiction
Dialogue: 0,0:03:56.70,0:04:06.71,Default,,0000,0000,0000,,Or that's 49 newtons times 0.60
Dialogue: 0,0:04:06.71,0:04:09.91,Default,,0000,0000,0000,,And remember coefficient of friction are unitless
Dialogue: 0,0:04:09.91,0:04:13.04,Default,,0000,0000,0000,,So the units here are still going to be in newtons
Dialogue: 0,0:04:13.04,0:04:22.58,Default,,0000,0000,0000,,So this 49 times .6 gives us 29.4 newtons
Dialogue: 0,0:04:22.58,0:04:30.55,Default,,0000,0000,0000,,This is equal to 29.4 newtons
Dialogue: 0,0:04:30.55,0:04:34.52,Default,,0000,0000,0000,,So that's the force that's started to overcome static friction
Dialogue: 0,0:04:34.52,0:04:36.79,Default,,0000,0000,0000,,which we are applying more than enough of
Dialogue: 0,0:04:36.79,0:04:39.92,Default,,0000,0000,0000,,so with a 100 newtons, we would just start to budge it
Dialogue: 0,0:04:39.92,0:04:42.10,Default,,0000,0000,0000,,and right when we are in just in that moment
Dialogue: 0,0:04:42.10,0:04:44.19,Default,,0000,0000,0000,,where that thing is just starting to move
Dialogue: 0,0:04:44.19,0:04:45.58,Default,,0000,0000,0000,,the net force--
Dialogue: 0,0:04:45.58,0:04:47.67,Default,,0000,0000,0000,,so we have a 100 newtons going in that direction
Dialogue: 0,0:04:47.67,0:04:51.09,Default,,0000,0000,0000,,and the force of static friction is going to go in this direction--
Dialogue: 0,0:04:51.09,0:04:53.58,Default,,0000,0000,0000,,maybe I could draw it down here to show it's coming from right over here
Dialogue: 0,0:04:53.58,0:04:57.97,Default,,0000,0000,0000,,The force of static friction is going to be 29.4 newtons that way
Dialogue: 0,0:04:57.97,0:05:00.65,Default,,0000,0000,0000,,and so right when I am just starting to budge this
Dialogue: 0,0:05:00.65,0:05:03.03,Default,,0000,0000,0000,,just when that little movement--
Dialogue: 0,0:05:03.07,0:05:04.93,Default,,0000,0000,0000,,because once I do that, then all of a sudden it's moving
Dialogue: 0,0:05:05.03,0:05:09.18,Default,,0000,0000,0000,,and then kinetic friction starts to matter, but just for that moment
Dialogue: 0,0:05:09.65,0:05:15.93,Default,,0000,0000,0000,,just for that moment I'll have a net force of 100 - 29.4
Dialogue: 0,0:05:16.36,0:05:28.66,Default,,0000,0000,0000,,to the right, so I have a net force of 70.6 N
Dialogue: 0,0:05:28.66,0:05:32.82,Default,,0000,0000,0000,,for just a moment while I budge it
Dialogue: 0,0:05:32.82,0:05:35.25,Default,,0000,0000,0000,,So just exactly while I'm budging it
Dialogue: 0,0:05:35.25,0:05:42.38,Default,,0000,0000,0000,,While we're overcoming the static friction, we have a 70.6 N net force in the right direction
Dialogue: 0,0:05:42.69,0:05:47.98,Default,,0000,0000,0000,,And so just for that moment, you divide it by 5 kg mass
Dialogue: 0,0:05:48.00,0:05:52.43,Default,,0000,0000,0000,,So just for that moment, it will be accelerating at 14.12 m/s^2
Dialogue: 0,0:05:53.05,0:06:00.28,Default,,0000,0000,0000,,So you'll have an acceleration of 14.1 m/s^2 to the right
Dialogue: 0,0:06:00.30,0:06:04.25,Default,,0000,0000,0000,,but that will just be for that absolute moment, because once I budge it
Dialogue: 0,0:06:04.36,0:06:07.42,Default,,0000,0000,0000,,all of a sudden the block will start to be moving
Dialogue: 0,0:06:07.44,0:06:10.88,Default,,0000,0000,0000,,And once it's moving, the coefficient of kinetic friction starts to matter
Dialogue: 0,0:06:10.88,0:06:12.88,Default,,0000,0000,0000,,We've got the things out of their little grooves
Dialogue: 0,0:06:12.90,0:06:17.16,Default,,0000,0000,0000,,and so they're kind of gliding past each other on the top, although there still is resistant
Dialogue: 0,0:06:17.20,0:06:20.55,Default,,0000,0000,0000,,So once we budge it, we'll have that acceleration for just a moment
Dialogue: 0,0:06:20.56,0:06:23.55,Default,,0000,0000,0000,,Now all of a sudden, the coefficient of kinetic friction comes to play
Dialogue: 0,0:06:23.93,0:06:27.08,Default,,0000,0000,0000,,And the force of friction, assuming we're moving
Dialogue: 0,0:06:27.41,0:06:31.12,Default,,0000,0000,0000,,the magnitude of the force of friction will always go against our movement
Dialogue: 0,0:06:31.55,0:06:35.46,Default,,0000,0000,0000,,is going to be--remember, our normal force is 49 N
Dialogue: 0,0:06:35.70,0:06:38.85,Default,,0000,0000,0000,,So we can multiply both sides of this times 49
Dialogue: 0,0:06:39.00,0:06:57.78,Default,,0000,0000,0000,,We get 49 N times 0.55 which is equal to 26.95 N
Dialogue: 0,0:06:57.80,0:07:00.40,Default,,0000,0000,0000,,This is the force of friction; this is the magnitude
Dialogue: 0,0:07:00.52,0:07:02.40,Default,,0000,0000,0000,,and it's going to go against our motions
Dialogue: 0,0:07:02.41,0:07:05.69,Default,,0000,0000,0000,,So as soon as we start to move in that direction, the force of friction
Dialogue: 0,0:07:05.89,0:07:09.48,Default,,0000,0000,0000,,is going to be going in that direction
Dialogue: 0,0:07:09.74,0:07:15.05,Default,,0000,0000,0000,,So once we start moving, assuming that I'm continuing to apply this 100 newtons of force
Dialogue: 0,0:07:15.13,0:07:18.28,Default,,0000,0000,0000,,what is the net force? So I have 100 N going that way
Dialogue: 0,0:07:18.38,0:07:23.18,Default,,0000,0000,0000,,and I have 26.95 going that way
Dialogue: 0,0:07:23.20,0:07:25.08,Default,,0000,0000,0000,,Remember, with vectors, I don't have to draw them here
Dialogue: 0,0:07:25.08,0:07:28.89,Default,,0000,0000,0000,,I can draw all of their tails start at the center of mass of the
Dialogue: 0,0:07:28.89,0:07:32.81,Default,,0000,0000,0000,,object. I can draw them whatever, but remember this is acting on the object
Dialogue: 0,0:07:32.81,0:07:37.14,Default,,0000,0000,0000,,If we want to be precise, we can show it on the center of mass because
Dialogue: 0,0:07:37.14,0:07:39.94,Default,,0000,0000,0000,,we can view all of these atoms as one collective object
Dialogue: 0,0:07:39.95,0:07:42.15,Default,,0000,0000,0000,,But anyway, what is the net force now?
Dialogue: 0,0:07:42.16,0:07:46.42,Default,,0000,0000,0000,,We have 100 N to the right; we have 26.95 to the left
Dialogue: 0,0:07:47.08,0:07:51.75,Default,,0000,0000,0000,,100 minus 26.95
Dialogue: 0,0:07:51.75,0:07:53.60,Default,,0000,0000,0000,,100 N that I'm applying to the right
Dialogue: 0,0:07:53.60,0:07:59.26,Default,,0000,0000,0000,,- 26.95 N which is the force of friction to the left always acting against us
Dialogue: 0,0:07:59.41,0:08:03.41,Default,,0000,0000,0000,,means that there's a net force to the right of 73.05
Dialogue: 0,0:08:03.43,0:08:10.84,Default,,0000,0000,0000,,So once we're moving, we have a net force to the right of 73.05 N
Dialogue: 0,0:08:10.84,0:08:18.07,Default,,0000,0000,0000,,This is the net force and it's acting to the right
Dialogue: 0,0:08:18.26,0:08:22.85,Default,,0000,0000,0000,,Right after we budge it, how quickly will this accelerate?
Dialogue: 0,0:08:22.85,0:08:32.70,Default,,0000,0000,0000,,Well, 73.05 divided by the mass, divided by 5 kg, gives us 14.61
Dialogue: 0,0:08:33.22,0:08:40.43,Default,,0000,0000,0000,,So the acceleration once we're moving is going to be 14.61 m/s squared
Dialogue: 0,0:08:40.63,0:08:44.45,Default,,0000,0000,0000,,to the right
Dialogue: 0,0:08:44.49,0:08:47.19,Default,,0000,0000,0000,,So I really want to make sure you understand what's happening here
Dialogue: 0,0:08:47.22,0:08:51.27,Default,,0000,0000,0000,,We always have enough force to start budging it
Dialogue: 0,0:08:51.56,0:08:53.51,Default,,0000,0000,0000,,but right when we budged it
Dialogue: 0,0:08:53.51,0:08:56.43,Default,,0000,0000,0000,,we overcome the static friction for just a moment
Dialogue: 0,0:08:56.43,0:08:59.81,Default,,0000,0000,0000,,our acceleration was slower
Dialogue: 0,0:08:59.82,0:09:01.90,Default,,0000,0000,0000,,because we're overcoming that static friction
Dialogue: 0,0:09:01.90,0:09:04.90,Default,,0000,0000,0000,,but once we budged it and once it's moving
Dialogue: 0,0:09:04.92,0:09:08.60,Default,,0000,0000,0000,,and assuming that we're continuing to apply a constant force over here
Dialogue: 0,0:09:08.79,0:09:10.48,Default,,0000,0000,0000,,then all of a sudden, the force of friction since
Dialogue: 0,0:09:10.48,0:09:14.40,Default,,0000,0000,0000,,we're kind of bump it along the top now and not stuck in their grooves
Dialogue: 0,0:09:14.92,0:09:17.57,Default,,0000,0000,0000,,we're now using the coefficient of kinetic friction
Dialogue: 0,0:09:17.59,0:09:22.78,Default,,0000,0000,0000,,And so once it's moving, the net force becomes greater in the rightward direction because
Dialogue: 0,0:09:22.93,0:09:27.59,Default,,0000,0000,0000,,you can kind of view that force of friction will become less once it starts moving
Dialogue: 0,0:09:27.79,0:09:31.82,Default,,0000,0000,0000,,And so now the force of friction went down a little bit to 26.95 N
Dialogue: 0,0:09:31.87,0:09:38.63,Default,,0000,0000,0000,,And so now we're accelerating to right at a slightly faster rate 14.61 m/s^2
Dialogue: 0,0:09:38.64,0:09:41.86,Default,,0000,0000,0000,,So right when you budge it, it accelerates at 14.1 m/s^2
Dialogue: 0,0:09:41.86,0:09:44.87,Default,,0000,0000,0000,,but just for a moment, almost unnoticeable moment once it starts moving
Dialogue: 0,0:09:44.92,0:09:48.78,Default,,0000,0000,0000,,Then you're going to be going to the right with this constant acceleration