0:00:00.680,0:00:04.610
Let's say that I have a[br]huge, maybe frozen over lake,

0:00:04.610,0:00:06.080
or maybe it's a big pond.

0:00:06.080,0:00:09.750
So I have a huge surface of[br]ice over here-- my best attempt

0:00:09.750,0:00:11.830
to draw a flat surface[br]of ice-- and I'm

0:00:11.830,0:00:14.710
going to put two[br]blocks of ice here.

0:00:14.710,0:00:17.650
So I'm going to put[br]one block of ice

0:00:17.650,0:00:21.814
just like this, one block[br]of ice right over here.

0:00:21.814,0:00:23.980
And then I'm going to put[br]another block of ice right

0:00:23.980,0:00:26.220
over here.

0:00:26.220,0:00:30.490
And then another block[br]of ice right over here.

0:00:30.490,0:00:33.510
And these blocks of[br]ice are identical.

0:00:33.510,0:00:35.230
They're both 5 kilograms.

0:00:35.230,0:00:38.280
They are both 5 kilograms--[br]let me write this down.

0:00:38.280,0:00:42.490
So they are both 5 kilograms.

0:00:42.490,0:00:46.310
Or both of their masses, I[br]should say, are 5 kilograms.

0:00:46.310,0:00:48.700
And the only difference[br]between the two

0:00:48.700,0:00:51.620
is that relative to[br]the pond, this one

0:00:51.620,0:00:56.120
is stationary-- this[br]one is stationary--

0:00:56.120,0:00:59.310
and this one is moving[br]with a constant velocity--

0:00:59.310,0:01:02.745
constant velocity.

0:01:02.745,0:01:06.460
Constant velocity in the[br]right-wards direction.

0:01:06.460,0:01:08.950
And let's say that[br]its constant velocity

0:01:08.950,0:01:13.890
is at 5 meters per second--[br]5 meters per second.

0:01:13.890,0:01:17.164
And the whole reason why I made[br]blocks of ice on top of ice

0:01:17.164,0:01:19.330
is that we're going to[br]assume, at least for the sake

0:01:19.330,0:01:23.540
of this video, that[br]friction is negligible.

0:01:23.540,0:01:26.360
Now what does Newton's[br]First Law of Motion

0:01:26.360,0:01:29.590
tell us about something that[br]is either not in motion--

0:01:29.590,0:01:32.440
or you could view this as[br]a constant velocity of 0--

0:01:32.440,0:01:35.314
or something that has[br]a constant velocity?

0:01:35.314,0:01:36.730
Well Newton's First[br]Law says, well

0:01:36.730,0:01:41.180
look, they're going to keep[br]their constant velocity

0:01:41.180,0:01:43.930
or stay stationary, which is[br]the constant velocity of 0,

0:01:43.930,0:01:47.110
unless there is some[br]unbalance, unless there

0:01:47.110,0:01:49.774
is some net force[br]acting on an object.

0:01:49.774,0:01:51.190
So let's just think[br]about it here.

0:01:51.190,0:01:52.850
In either of these[br]situations, there

0:01:52.850,0:01:56.140
must not be any unbalanced[br]force acting on them.

0:01:56.140,0:01:58.532
Or their must not[br]be any net force.

0:01:58.532,0:01:59.990
But if you think[br]about it, if we're

0:01:59.990,0:02:02.280
assuming that these[br]things are on Earth,

0:02:02.280,0:02:05.830
there is a net force[br]acting on both of them.

0:02:05.830,0:02:08.780
Both of them are at the[br]surface of the Earth,

0:02:08.780,0:02:10.949
and they both have[br]mass, so there

0:02:10.949,0:02:14.460
will be the force of[br]gravity acting downwards

0:02:14.460,0:02:15.900
on both of them.

0:02:15.900,0:02:20.140
There is going to be the[br]downward force of gravity

0:02:20.140,0:02:22.630
on both of these blocks of ice.

0:02:22.630,0:02:25.810
And that downward force of[br]gravity, the force of gravity,

0:02:25.810,0:02:29.540
is going to be equal to[br]the gravitational field

0:02:29.540,0:02:32.910
near the surface of the[br]Earth, times-- which

0:02:32.910,0:02:37.220
is a vector-- times[br]the mass of the object.

0:02:37.220,0:02:41.050
So times 5 kilograms.

0:02:41.050,0:02:47.040
This right over here is 9.8[br]meters per second squared.

0:02:47.040,0:02:49.290
So you multiply that times 5.

0:02:49.290,0:02:53.000
You get 49 kilogram meter[br]per second squared, which

0:02:53.000,0:02:56.080
is the same thing as 49 newtons.

0:02:56.080,0:02:58.670
So this is a little bit[br]of a conundrum here.

0:02:58.670,0:03:02.490
Newton's First Law[br]says, an object at rest

0:03:02.490,0:03:04.250
will stay at rest, or[br]an object in motion

0:03:04.250,0:03:06.610
will stay in motion, unless[br]there is some unbalanced,

0:03:06.610,0:03:08.580
or unless there[br]is some net force.

0:03:08.580,0:03:10.340
But based on what[br]we've drawn right here,

0:03:10.340,0:03:13.320
it looks like there's[br]some type of a net force.

0:03:13.320,0:03:16.780
It looks like I have 49 newtons[br]of force pulling this thing

0:03:16.780,0:03:17.740
downwards.

0:03:17.740,0:03:20.900
But you say, no, no no, Sal.

0:03:20.900,0:03:23.860
Obviously this thing won't[br]start accelerating downwards

0:03:23.860,0:03:25.200
because there's ice here.

0:03:25.200,0:03:32.640
Its resting on a big[br]pool of frozen water.

0:03:32.640,0:03:35.930
And so my answer to you is,[br]well, if that's your answer,

0:03:35.930,0:03:42.350
then what is the resulting[br]force that cancels out

0:03:42.350,0:03:44.990
with gravity to keep[br]these blocks of ice,

0:03:44.990,0:03:47.600
either one of them,[br]from plummeting down

0:03:47.600,0:03:51.320
to the core of the Earth?

0:03:51.320,0:03:53.610
From essentially[br]going into free fall,

0:03:53.610,0:03:56.240
or accelerating towards[br]the center of the Earth?

0:03:56.240,0:04:01.210
And you say, well, I guess if[br]these things would be falling,

0:04:01.210,0:04:04.400
if not for the ice,[br]the ice must be

0:04:04.400,0:04:07.420
providing the[br]counteracting force.

0:04:07.420,0:04:09.630
And you are absolutely correct.

0:04:09.630,0:04:13.390
The ice is providing[br]the counteracting force

0:04:13.390,0:04:15.750
in the opposite direction.

0:04:15.750,0:04:17.839
So the exact magnitude[br]of force, and it

0:04:17.839,0:04:20.220
is in the opposite direction.

0:04:20.220,0:04:23.730
And so if the force of gravity[br]on each of these blocks of ice

0:04:23.730,0:04:27.690
are 49 newtons downwards[br]it is completely

0:04:27.690,0:04:32.150
netted off by the force of[br]the ice on the block upwards.

0:04:32.150,0:04:37.560
And that will be a force 49[br]newtons upwards in either case.

0:04:37.560,0:04:39.320
And now, hopefully,[br]it makes sense

0:04:39.320,0:04:41.650
that Newton's First[br]Law still holds.

0:04:41.650,0:04:45.100
We have no net force on this[br]in the vertical direction,

0:04:45.100,0:04:47.810
actually no net force on[br]this in either direction.

0:04:47.810,0:04:51.740
That's why this guy[br]has a 0 velocity

0:04:51.740,0:04:53.140
in the horizontal direction.

0:04:53.140,0:04:56.010
This guy has a constant velocity[br]in the horizontal direction.

0:04:56.010,0:04:57.570
And neither of them[br]are accelerating

0:04:57.570,0:04:58.970
in the vertical direction.

0:04:58.970,0:05:01.520
Because you have the force[br]of the ice on the block,

0:05:01.520,0:05:03.470
the ice is supporting[br]the block, that's

0:05:03.470,0:05:05.920
completely[br]counteracting gravity.

0:05:05.920,0:05:09.740
And this force, in this example,[br]is called the normal force.

0:05:09.740,0:05:14.570
This is the normal force--[br]it's 49 newtons upwards.

0:05:14.570,0:05:17.040
This right here is[br]the normal force.

0:05:17.040,0:05:20.150
And we'll talk more about the[br]normal force in future videos.

0:05:20.150,0:05:22.610
The normal force[br]is the force, when

0:05:22.610,0:05:24.380
anything is resting[br]on any surface that's

0:05:24.380,0:05:25.630
perpendicular to that surface.

0:05:25.630,0:05:27.463
And it's going to start[br]to matter a lot when

0:05:27.463,0:05:29.692
we start thinking about[br]friction and all the rest.

0:05:29.692,0:05:32.150
So what we'll see in future[br]videos, when you have something

0:05:32.150,0:05:35.220
on an incline, and let's say[br]I have a block on an incline

0:05:35.220,0:05:36.120
like this.

0:05:36.120,0:05:38.440
The normal force[br]from the, I guess

0:05:38.440,0:05:40.080
you could say, this[br]wedge on the block,

0:05:40.080,0:05:44.364
is going to be perpendicular[br]to the surface.

0:05:44.364,0:05:46.530
And if you really think[br]about what's happening here,

0:05:46.530,0:05:48.760
it's fundamentally an[br]electromagnetic force.

0:05:48.760,0:05:56.080
Because if you really zoomed[br]in on the molecules of the ice

0:05:56.080,0:05:58.590
right over here, even better[br]the atoms of the ice here.

0:05:58.590,0:06:01.840
And you really zoomed in on[br]the atoms or the molecules

0:06:01.840,0:06:06.320
of the ice up here, what's[br]keeping this top block of ice

0:06:06.320,0:06:09.030
from falling down[br]is that in order

0:06:09.030,0:06:13.480
for it to go through its[br]molecules would have to kind

0:06:13.480,0:06:16.070
of compress against, or I guess[br]it would have to get closer

0:06:16.070,0:06:19.420
to, the water molecules[br]or the individual atoms

0:06:19.420,0:06:21.530
in this ice down here.

0:06:21.530,0:06:23.720
And the atoms, let me[br]draw it on an atomic level

0:06:23.720,0:06:24.890
right over here.

0:06:24.890,0:06:33.300
So maybe, let me draw one[br]of this guy's molecules.

0:06:33.300,0:06:40.044
So you have an oxygen[br]with 2 hydrogens

0:06:40.044,0:06:41.710
and it forms this big[br]lattice structure.

0:06:41.710,0:06:45.980
And we can talk about more of[br]that in the chemistry playlist.

0:06:45.980,0:06:48.880
And let's talk about this ice[br]as one of these molecules.

0:06:48.880,0:06:53.500
So maybe it looks[br]something like this.

0:06:53.500,0:06:55.810
And it has its 2 hydrogens

0:06:55.810,0:06:58.710
And so what's keeping these guys[br]from getting compressed, what's

0:06:58.710,0:07:02.200
keeping this block of ice[br]from going down further,

0:07:02.200,0:07:05.330
is the repulsion between the[br]electrons in this molecule

0:07:05.330,0:07:06.950
and the electrons[br]in that molecule.

0:07:06.950,0:07:10.570
So on a macro level we view[br]this is kind of a contact force.

0:07:10.570,0:07:13.110
But on a microscopic[br]level, on an atomic level,

0:07:13.110,0:07:17.870
it's really just electromagnetic[br]repulsion at work.