2-dimensional momentum problem

WEBVTT
Kind: captions
Language: en

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Welcome back.

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We will now do a momentum
problem in two dimensions.

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So let's see what
we have here.

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So we have this ball A and we
could maybe even think of it

00:00:10.070 --> 00:00:13.470
as maybe what's going on on the
surface of a pool table.

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We have ball A and it's moving
with its 10 kilograms. So

00:00:16.730 --> 00:00:19.070
these numbers are the
mass of the balls.

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This is a 10 kilogram ball and
it's moving to the right at 3

00:00:22.300 --> 00:00:23.710
meters per second.

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And then it hits this ball, B,
which is a 5 kilogram ball.

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And then we know that ball A,
ball A kind of ricochets off

00:00:31.920 --> 00:00:36.740
of ball B and gets set onto
this new trajectory.

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Now, instead of going due right,
it's going at a 30

00:00:40.010 --> 00:00:43.250
degree angle to, I guess we
could say, horizontal.

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It's going in a 30 degree angle
at 2 meters per second.

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And the question is, what is
the velocity of ball B?

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So velocity is both magnitude
and direction.

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So we need to figure
out essentially,

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what is ball B doing?

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Ball B is going to be going--
We can just think about it.

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If you ever played pool, we
could guess that ball B is

00:01:01.130 --> 00:01:03.055
going to go roughly
in that direction.

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But we need to figure out
exactly what the angle is and

00:01:05.990 --> 00:01:09.410
exactly what its velocity is.

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So let's do this problem.

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So at first you're saying, oh,
Sal, this looks confusing.

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You know, I know momentum
should be

00:01:15.900 --> 00:01:16.880
conserved and all that.

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But now we have these vectors
and there's two dimensions and

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how do I do that?

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And the key here is that there's
just really one more

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step when you're working on it
in two dimensions or really

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three dimensions or an
arbitrary number of

00:01:28.240 --> 00:01:29.340
dimensions.

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When we did one dimension, you
made sure that momentum was

00:01:31.770 --> 00:01:33.590
conserved in that
one dimension.

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So when you do two dimensions,
what you do is you figure out

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the initial momentum in each
of the dimensions.

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So you break it up into the
x and y components.

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And then you say the final
momentum of both objects are

00:01:44.710 --> 00:01:47.680
going to equal the initial x
momentum and are going to

00:01:47.680 --> 00:01:49.570
equal the initial y momentum.

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So let's figure out the
initial x momentum.

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So P for momentum.

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Because the m is for mass.

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So let's say the initial
momentum in the x direction--

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And we don't have to write
initial or final because

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really, the total momentum in
the x direction is always

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going to be the same.

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So let's say what the initial--
Actually, let me

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write initial just so it hits
the point home that initial

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and final don't change.

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So the initial momentum
in the x direction.

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So i for initial, x-- I should
do something better than keep

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writing these subscripts
--is equal to what?

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Well ball B has no
initial velocity,

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so it has no momentum.

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Ball A is 10 kilograms.

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And what is its velocity
in the x direction?

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Well all of its velocity
is in the x direction.

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So it's 3.

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I mean, this is only moving
in the x direction.

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So the momentum in the x
direction is 30 kilogram

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meters per second.

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Mass times velocity, kilogram
meters per second.

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And what's the initial momentum
in the y direction?

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Well B isn't moving at
all, so it has no

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momentum in any direction.

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And A, all of A's movement
is in the x direction.

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It's not moving at an angle
or up at all, so it has no

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momentum in the y direction.

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So we immediately know that
after the collision, the

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combined momentum of both of
these balls in the x direction

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has to be 30, and the combined
momentum of both of these

00:03:17.670 --> 00:03:21.040
balls in the y direction
has to be 0.

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So let's figure out what A's
momentum in the x and y

00:03:24.580 --> 00:03:26.840
directions are.

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So to figure out what A's
momentum is, we just have to

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figure out what A's velocity in
the x and y directions are

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and then multiply that
times the mass.

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Because mass doesn't
have any direction.

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So let's figure out the x and y
components of this velocity.

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Let's do the x component
of the vector first.

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So the x is just this vector.

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Change colors to keep
things interesting.

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The y is this vector.

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That is the y component.

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And so, what are those?

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And this hopefully, is going to
be almost second nature to

00:04:06.090 --> 00:04:08.810
you if you've been watching
all of the other videos on

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Newton's laws.

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This is just our trigonometry
and we can write out our

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SOH-CAH-TOA again.

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And I reassure you, this is
the hardest part of any of

00:04:19.260 --> 00:04:22.019
these multi-dimensional trig
problems-- Multi-dimensional

00:04:22.019 --> 00:04:24.500
physics problems, which really
are just trig problems.

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So if we want to figure out
the x component, so the

00:04:27.870 --> 00:04:31.880
velocity of A in the
x direction,

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what is it equal to?

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Well this is adjacent
to the angle.

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We know the hypotenuse.

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So we know VA sub x or the
velocity of A in the x

00:04:43.592 --> 00:04:47.470
direction over the hypotenuse,
over 2 meters per second, is

00:04:47.470 --> 00:04:48.770
equal to what?

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Adjacent over hypotenuse.

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Cosine.

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Is equal to cosine
of 30 degrees.

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Or the velocity of A in the x
direction is equal to 2 cosine

00:05:02.720 --> 00:05:04.420
of 30 degrees.

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What's cosine of 30 degrees?

00:05:07.300 --> 00:05:09.000
It's square root of 3 over 2.

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This is square root
of 3 over 2.

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And square root of 3 over
2 times 2 is equal to

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square root of 3.

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So this is equal to the square
root of 3 meters per second.

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And what is the velocity of
A in the y direction?

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Well hopefully, this second
nature to you as well.

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But since opposite over
hypotenuse is equal to the

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sine of 30.

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So VA in the y direction is
equal to 2 times the sine of

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30 degrees.

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The sine of 30 degrees is 1/2.

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So this is 1/2.

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1/2 times 2 is equal to
1 meter per second.

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So after the collision,
A is moving at 1

00:05:54.490 --> 00:05:57.950
meter per second up.

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One meter per second in
the upwards direction.

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And it's moving at square root
of 3 meters per second in the

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rightwards direction.

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So what is going to be A's
momentum in each of the

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directions?

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Well, we figured out its
velocity, so we just multiply

00:06:13.810 --> 00:06:16.160
each of the velocities
times the mass.

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So A has a mass of 10 kilograms.
And this is going

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to be the final momentum.

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Momentum of A in the x direction
is going to equal

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square root of 3 times 10.

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Square root of 3 is the
velocity, 10 is the mass.

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So it's 10 square roots of 3
kilogram meters per second.

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And the momentum of A in the y
direction is going to be-- and

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since it's going up, we'll say
its positive --it's 1 meters

00:06:49.450 --> 00:06:51.210
per second is the velocity
times the mass.

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So 10 times 1 is 10 kilogram
meter per second.

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So now let's figure out B.

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Let's do the y direction first
because they add up to 0.

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I'm going to switch colors.

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We know that the momentum
of-- and this

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is after the collision.

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The momentum of A in the y
direction plus momentum of B

00:07:12.840 --> 00:07:14.780
in the y direction have
to equal what?

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What was the initial momentum
in the y direction?

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Right, it was 0.

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There was no movement in the
y direction initially.

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We know the momentum of
A in the y direction.

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It's 10.

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10 kilogram meters per second
plus the momentum of B in the

00:07:29.790 --> 00:07:32.040
y direction is equal to 0.

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So solving for this, just
subtract 10 from both sides.

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So the momentum of B in the
y direction is equal to 10

00:07:41.080 --> 00:07:42.360
kilogram meters per second.

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You know the units.

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So if its momentum is 10 in the
y direction, what is its

00:07:50.900 --> 00:07:53.140
velocity in the y direction?

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Well, momentum is equal to
mass times velocity.

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So we know that 5 times the
velocity in the y direction--

00:08:02.520 --> 00:08:05.660
that's its mass --is
equal to 10.

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10 is its momentum.

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So the velocity of the y
direction of B must be 2

00:08:10.970 --> 00:08:12.570
meters per second.

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So there we go.

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We figured out B's velocity.

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And so let's say this is B's
velocity vector in the y

00:08:19.110 --> 00:08:22.740
direction is-- And this is
a minus because this is

00:08:22.740 --> 00:08:26.935
equal to minus 10.

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So it's going down.

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It was a velocity of positive
1 going up and then the

00:08:32.740 --> 00:08:35.380
minuses carry through and this
is a velocity of minus 2

00:08:35.380 --> 00:08:40.059
meters per second for B
in the y direction.

00:08:40.059 --> 00:08:41.340
So now let's figure
out the velocity

00:08:41.340 --> 00:08:42.808
of B in the x direction.

00:08:42.808 --> 00:08:45.540
And I'm running out of space
and it's getting messy.

00:08:45.540 --> 00:08:49.860
But we just have to remember
that the momentum of B in

00:08:49.860 --> 00:08:54.480
the-- The momentum of A in the
x direction, which is 10

00:08:54.480 --> 00:09:02.810
square root of 3, plus momentum
of B in the x

00:09:02.810 --> 00:09:05.770
direction has to equal what?

00:09:05.770 --> 00:09:08.090
It has to equal the initial
momentum in the x direction,

00:09:08.090 --> 00:09:10.240
which is 30.

00:09:10.240 --> 00:09:12.590
So to figure out the momentum
of B in the x direction, we

00:09:12.590 --> 00:09:15.010
just subtract 10 square
root of 3 from 30.

00:09:15.010 --> 00:09:17.760
And let's do that.

00:09:17.760 --> 00:09:24.010
So let's figure out 3 square
root times 10 equals.

00:09:24.010 --> 00:09:29.630
And then subtract
that from 30.

00:09:29.630 --> 00:09:34.300
And we get let's
just say 12.7.

00:09:34.300 --> 00:09:38.380
So we know that the momentum
of B in the x direction is

00:09:38.380 --> 00:09:43.420
equal to 12.7, 12.7 kilogram
meters per second.

00:09:43.420 --> 00:09:45.680
And we know the momentum, so we
just divide by the mass and

00:09:45.680 --> 00:09:47.030
we get its velocity in
the x direction.

00:09:47.030 --> 00:09:48.950
So 12.7 divided by 5.

00:09:48.950 --> 00:09:51.630
So velocity of B in
the x direction is

00:09:51.630 --> 00:09:54.970
12.7 divided by 5.

00:09:54.970 --> 00:10:04.460
12.7 divided by 5 is equal to
2.54 meters per second.

00:10:04.460 --> 00:10:07.020
So its velocity in the
x direction is

00:10:07.020 --> 00:10:10.080
2.54 meters per second.

00:10:10.080 --> 00:10:14.250
So it's going faster
in both directions.

00:10:14.250 --> 00:10:15.970
I'm not going to figure out the
angle here because I've

00:10:15.970 --> 00:10:17.650
actually run out of time.

00:10:17.650 --> 00:10:20.070
But if you were to add these
two vectors, you'd get an

00:10:20.070 --> 00:10:21.340
angle something like this.

00:10:21.340 --> 00:10:24.700
And you could figure out the
angle by taking the arc tan.

00:10:24.700 --> 00:10:27.080
Well, I won't go into the--
that's a complexity right now.

00:10:27.080 --> 00:10:29.050
Actually, I'll do that in the
next video just so I won't

00:10:29.050 --> 00:10:29.620
leave you hanging.

00:10:29.620 --> 00:10:33.660
But we know what the x and y
components of B's velocity is.

00:10:33.660 --> 00:10:35.380
See you in the next video.


Momentum: Ice skater throws a ball

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:01.440
Welcome back.

00:00:01.440 --> 00:00:03.950
I'll now do a couple of more
momentum problems.

00:00:03.950 --> 00:00:07.060
So this first problem, I have
this ice skater and she's on

00:00:07.060 --> 00:00:08.630
an ice skating rink.

00:00:08.630 --> 00:00:10.360
And what she's doing is
she's holding a ball.

00:00:10.360 --> 00:00:14.740
And this ball-- let me draw
the ball-- this is a 0.15

00:00:14.740 --> 00:00:15.990
kilogram ball.

00:00:18.610 --> 00:00:20.580
And she throws it.

00:00:20.580 --> 00:00:23.640
Let's just say she throws it
directly straight forward in

00:00:23.640 --> 00:00:25.200
front of her, although
she's staring at us.

00:00:25.200 --> 00:00:27.230
She's actually forward
for her body.

00:00:27.230 --> 00:00:32.790
So she throws it exactly
straight forward.

00:00:32.790 --> 00:00:35.075
And I understand it is hard to
throw something straight

00:00:35.075 --> 00:00:38.490
forward, but let's assume
that she can.

00:00:38.490 --> 00:00:41.510
So she throws it exactly
straight forward with a

00:00:41.510 --> 00:00:44.280
speed-- or since we're going to
give the direction as well,

00:00:44.280 --> 00:00:48.000
it's a velocity, right, cause
speed is just a magnitude

00:00:48.000 --> 00:00:51.200
while a velocity is a magnitude
and a direction-- so

00:00:51.200 --> 00:00:58.160
she throws the ball at 35 meters
per second, and this

00:00:58.160 --> 00:01:03.160
ball is 0.15 kilograms.

00:01:03.160 --> 00:01:08.560
Now, what the problem says is
that their combined mass, her

00:01:08.560 --> 00:01:17.520
plus the ball, is 50 kilograms.
So they're both

00:01:17.520 --> 00:01:20.130
stationary before she does
anything, and then she throws

00:01:20.130 --> 00:01:22.990
this ball, and the question is,
after throwing this ball,

00:01:22.990 --> 00:01:25.000
what is her recoil velocity?

00:01:25.000 --> 00:01:28.930
Or essentially, well how much,
by throwing the ball, does she

00:01:28.930 --> 00:01:30.230
push herself backwards?

00:01:30.230 --> 00:01:33.060
So what is her velocity in
the backward direction?

00:01:33.060 --> 00:01:36.340
And if you're not familiar with
the term recoil, it's

00:01:36.340 --> 00:01:39.600
often applied to when someone,
I guess, not that we want to

00:01:39.600 --> 00:01:42.250
think about violent things, but
if you shoot a gun, your

00:01:42.250 --> 00:01:44.830
shoulder recoils back,
because once

00:01:44.830 --> 00:01:45.900
again momentum is conserved.

00:01:45.900 --> 00:01:48.270
So there's a certain amount of
momentum going into that

00:01:48.270 --> 00:01:51.020
bullet, which is very light
and fast going forward.

00:01:51.020 --> 00:01:54.940
But since momentum is conserved,
your shoulder has

00:01:54.940 --> 00:01:55.780
velocity backwards.

00:01:55.780 --> 00:01:57.250
But we'll do another
problem with that.

00:01:57.250 --> 00:01:58.960
So let's get back
to this problem.

00:01:58.960 --> 00:02:02.410
So like I just said, momentum
is conserved.

00:02:02.410 --> 00:02:05.760
So what's the momentum at the
start of the problem, the

00:02:05.760 --> 00:02:08.289
initial momentum?

00:02:08.289 --> 00:02:09.690
Let me do a different color.

00:02:09.690 --> 00:02:11.730
So this is the initial
momentum.

00:02:11.730 --> 00:02:18.060
Initially, the mass is 50
kilograms, right, cause her

00:02:18.060 --> 00:02:22.110
and the ball combined are 50
kilograms, times the velocity.

00:02:22.110 --> 00:02:23.810
Well the velocity is 0.

00:02:23.810 --> 00:02:29.800
So initially, there is 0
velocity in the system.

00:02:29.800 --> 00:02:34.060
So the momentum is 0.

00:02:34.060 --> 00:02:37.430
The P initial is equal to 0.

00:02:37.430 --> 00:02:41.560
And since we start with a net 0
momentum, we have to finish

00:02:41.560 --> 00:02:42.880
with a net 0 momentum.

00:02:42.880 --> 00:02:44.030
So what's momentum later?

00:02:44.030 --> 00:02:47.730
Well we have a ball moving at
35 meters per second and the

00:02:47.730 --> 00:02:58.040
ball has a mass of 0.15
kilograms. I'll ignore the

00:02:58.040 --> 00:02:59.710
units for now just
to save space.

00:02:59.710 --> 00:03:01.930
Times the velocity
of the ball.

00:03:01.930 --> 00:03:05.060
Times 35 meters per second.

00:03:05.060 --> 00:03:08.930
So this is the momentum of the
ball plus the new momentum of

00:03:08.930 --> 00:03:10.020
the figure skater.

00:03:10.020 --> 00:03:12.060
So what's her mass?

00:03:12.060 --> 00:03:14.440
Well her mass is going
to be 50 minus this.

00:03:14.440 --> 00:03:21.550
It actually won't matter a ton,
but let's say it's 49--

00:03:21.550 --> 00:03:25.330
what is that-- 49.85 kilograms,

00:03:25.330 --> 00:03:28.180
times her new velocity.

00:03:28.180 --> 00:03:29.040
Times velocity.

00:03:29.040 --> 00:03:31.410
Let's call that the velocity
of the skater.

00:03:31.410 --> 00:03:34.890
So let me get my trusty
calculator out.

00:03:37.910 --> 00:03:40.640
OK, so let's see.

00:03:40.640 --> 00:03:50.780
0.15 times 35 is
equal to 5.25.

00:03:50.780 --> 00:03:56.260
So that equals 5.25.

00:03:56.260 --> 00:04:02.350
plus 49.85 times the skater's
velocity, the final velocity.

00:04:02.350 --> 00:04:04.550
And of course, this equals
0 because the initial

00:04:04.550 --> 00:04:05.930
velocity was 0.

00:04:05.930 --> 00:04:10.000
So let's, I don't know, subtract
5.25 from both sides

00:04:10.000 --> 00:04:18.200
and then the equation becomes
minus 5.25 is equal to 49.85

00:04:18.200 --> 00:04:20.279
times the velocity
of the skater.

00:04:20.279 --> 00:04:23.480
So we're essentially saying that
the momentum of just the

00:04:23.480 --> 00:04:25.380
ball is 5.25.

00:04:25.380 --> 00:04:29.480
And since the combined system
has to have 0 net momentum,

00:04:29.480 --> 00:04:32.660
we're saying that the momentum
of the skater has to be 5.25

00:04:32.660 --> 00:04:35.960
in the other direction, going
backwards, or has a momentum

00:04:35.960 --> 00:04:39.230
of minus 5.25.

00:04:39.230 --> 00:04:41.480
And to figure out the velocity,
we just divide her

00:04:41.480 --> 00:04:43.780
momentum by her mass.

00:04:43.780 --> 00:04:48.380
And so divide both sides by
49.85 and you get the velocity

00:04:48.380 --> 00:04:49.695
of the skater.

00:04:49.695 --> 00:04:50.725
So let's see.

00:04:50.725 --> 00:05:01.520
Let's make this a negative
number divided by 49.85 equals

00:05:01.520 --> 00:05:05.370
minus 0.105.

00:05:05.370 --> 00:05:15.520
So minus 0.105 meters
per second.

00:05:15.520 --> 00:05:16.270
So that's interesting.

00:05:16.270 --> 00:05:20.370
When she throws this ball out at
35 meters per second, which

00:05:20.370 --> 00:05:24.670
is pretty fast, she will
recoil back at about 10

00:05:24.670 --> 00:05:28.440
centimeters, yeah, roughly 10
centimeters per second.

00:05:28.440 --> 00:05:30.530
So she will recoil a lot
slower, although

00:05:30.530 --> 00:05:31.740
she will move back.

00:05:31.740 --> 00:05:34.350
And if you think about it, this
is a form of propulsion.

00:05:34.350 --> 00:05:35.790
This is how rockets work.

00:05:35.790 --> 00:05:40.120
They eject something that maybe
has less mass, but super

00:05:40.120 --> 00:05:44.500
fast. And that, since we have a
conservation of momentum, it

00:05:44.500 --> 00:05:47.740
makes the rocket move in
the other direction.

00:05:47.740 --> 00:05:51.550
Well anyway, let's see if we
could fit another problem in.

00:05:51.550 --> 00:05:54.600
Actually, it's probably better
to leave this problem done and

00:05:54.600 --> 00:05:56.760
then I'll have more time for the
next problem, which will

00:05:56.760 --> 00:05:58.515
be slightly more difficult.

00:05:58.515 --> 00:05:59.765
See you soon.


Introduction to momentum

WEBVTT
Kind: captions
Language: en

00:00:00.810 --> 00:00:01.580
Welcome back.

00:00:01.580 --> 00:00:05.930
I will now introduce you to
the concept of momentum.

00:00:05.930 --> 00:00:08.800
And the letter for momentum is,
in physics, or at least in

00:00:08.800 --> 00:00:10.230
mechanics, it's the letter P.

00:00:10.230 --> 00:00:11.890
P for momentum.

00:00:11.890 --> 00:00:15.010
And I assume that's because the
letter M has already been

00:00:15.010 --> 00:00:17.200
used for mass, which
is I guess an even

00:00:17.200 --> 00:00:18.400
more fundamental idea.

00:00:18.400 --> 00:00:20.280
So P for momentum.

00:00:20.280 --> 00:00:21.180
So what is momentum?

00:00:21.180 --> 00:00:25.030
Well, you probably have
a general idea of it.

00:00:25.030 --> 00:00:27.740
If you see a big guy running
really fast, they'll say, he

00:00:27.740 --> 00:00:28.990
has a lot of momentum.

00:00:28.990 --> 00:00:31.440
And if there's a big guy running
really fast and a

00:00:31.440 --> 00:00:33.940
small guy running really fast,
most people would say, well,

00:00:33.940 --> 00:00:36.530
the big guy has more momentum.

00:00:36.530 --> 00:00:37.980
Maybe they don't have a
quantitative sense of why

00:00:37.980 --> 00:00:40.040
they're saying that, but
they just feel that

00:00:40.040 --> 00:00:40.910
that must be true.

00:00:40.910 --> 00:00:43.070
And if we look at the definition
of momentum, it'll

00:00:43.070 --> 00:00:43.730
make sense.

00:00:43.730 --> 00:00:50.450
The definition of momentum is
equal to mass times velocity.

00:00:50.450 --> 00:00:53.060
So something with, say, a
medium mass and a huge

00:00:53.060 --> 00:00:55.400
velocity is going to have
a big momentum.

00:00:55.400 --> 00:00:59.340
Or something with maybe a medium
mass, but-- the other

00:00:59.340 --> 00:00:59.770
way around.

00:00:59.770 --> 00:01:00.860
I forgot what I just said.

00:01:00.860 --> 00:01:03.230
So medium mass and big velocity,
huge momentum, or

00:01:03.230 --> 00:01:03.940
the other way around.

00:01:03.940 --> 00:01:06.280
Huge mass, medium velocity
would have maybe the same

00:01:06.280 --> 00:01:08.230
momentum, but it would still
have a big momentum.

00:01:08.230 --> 00:01:11.910
Or another way of doing momentum
is how little you

00:01:11.910 --> 00:01:16.050
would like to be in the way of
that object as it passes by.

00:01:18.850 --> 00:01:21.610
How unpleasant would it be
to be hit by that object?

00:01:21.610 --> 00:01:25.060
That's a good way of thinking
about momentum.

00:01:25.060 --> 00:01:27.603
So momentum is mass
times velocity.

00:01:30.660 --> 00:01:32.480
So how does it relate
to everything we've

00:01:32.480 --> 00:01:33.420
been learning so far?

00:01:33.420 --> 00:01:41.270
So we know that force is equal
to mass times acceleration.

00:01:41.270 --> 00:01:42.030
And what's acceleration?

00:01:42.030 --> 00:01:44.420
Well acceleration is just
change in velocity.

00:01:44.420 --> 00:01:52.580
So we also know that force is
equal to mass times change in

00:01:52.580 --> 00:01:57.850
velocity per unit
of time, right?

00:01:57.850 --> 00:02:00.890
Per change in time.

00:02:00.890 --> 00:02:02.720
T for time.

00:02:02.720 --> 00:02:06.550
So force is also equal
to-- well, mass

00:02:06.550 --> 00:02:07.290
times change in velocity.

00:02:07.290 --> 00:02:10.039
Mass, let's assume that
mass doesn't change.

00:02:10.039 --> 00:02:15.000
So that could also be viewed
as the change in mass times

00:02:15.000 --> 00:02:18.300
velocity in the unit
amount of time.

00:02:18.300 --> 00:02:20.130
And this is a little tricky
here, I said, you know, the

00:02:20.130 --> 00:02:21.960
mass times the change in
velocity, that's the same

00:02:21.960 --> 00:02:24.230
thing as the change in the
mass times the velocity,

00:02:24.230 --> 00:02:26.160
assuming the mass
doesn't change.

00:02:26.160 --> 00:02:28.960
And here we have mass times
velocity, which is momentum.

00:02:28.960 --> 00:02:34.310
So force can also be
viewed as change in

00:02:34.310 --> 00:02:39.026
momentum per unit of time.

00:02:39.026 --> 00:02:40.420
And I'll introduce
you to another

00:02:40.420 --> 00:02:41.810
concept called impulse.

00:02:41.810 --> 00:02:43.990
And impulse kind of means
that you think it means.

00:02:43.990 --> 00:02:46.640
An impulse is defined
as force times time.

00:02:46.640 --> 00:02:48.980
And I just want to introduce
this to you just in case you

00:02:48.980 --> 00:02:51.760
see it on the exam or whatever,
show you it's not a

00:02:51.760 --> 00:02:53.010
difficult concept.

00:02:53.010 --> 00:02:57.120
So force times change in time,
or time, if you assume time

00:02:57.120 --> 00:02:58.670
starts at time 0.

00:02:58.670 --> 00:03:01.805
But force times change in time
is equal to impulse.

00:03:01.805 --> 00:03:04.430
I actually don't know-- I should
look up what letters

00:03:04.430 --> 00:03:05.720
they use for impulse.

00:03:05.720 --> 00:03:08.490
But another way of viewing
impulse is force

00:03:08.490 --> 00:03:09.360
times change in time.

00:03:09.360 --> 00:03:12.960
Well that's the same thing as
change in momentum over change

00:03:12.960 --> 00:03:15.840
in time times change in time.

00:03:15.840 --> 00:03:16.110
Right?

00:03:16.110 --> 00:03:19.160
Because this is just the
same thing as force.

00:03:19.160 --> 00:03:21.210
And that's just change
in momentum, so

00:03:21.210 --> 00:03:22.460
that's impulse as well.

00:03:26.280 --> 00:03:28.940
And the unit of impulse
is the joule.

00:03:28.940 --> 00:03:30.450
And we'll go more into
the joule when we do

00:03:30.450 --> 00:03:31.200
work in all of that.

00:03:31.200 --> 00:03:33.500
And if this confuses you, don't
worry about it too much.

00:03:33.500 --> 00:03:37.050
The main thing about momentum is
that you realize it's mass

00:03:37.050 --> 00:03:39.100
times velocity.

00:03:39.100 --> 00:03:42.700
And since force is change in
momentum per unit of time, if

00:03:42.700 --> 00:03:47.220
you don't have any external
forces on a system or, on say,

00:03:47.220 --> 00:03:51.960
on a set of objects, their
combined, or their net

00:03:51.960 --> 00:03:53.070
momentum won't change.

00:03:53.070 --> 00:03:54.550
And that comes from
Newton's Laws.

00:03:54.550 --> 00:03:58.260
The only way you can get a
combined change in momentum is

00:03:58.260 --> 00:04:02.150
if you have some type of net
force acting on the system.

00:04:02.150 --> 00:04:03.770
So with that in mind,
let's do some

00:04:03.770 --> 00:04:08.340
momentum problems. Whoops.

00:04:08.340 --> 00:04:10.330
Invert colors.

00:04:10.330 --> 00:04:11.450
OK.

00:04:11.450 --> 00:04:14.470
So let's say we have a car.

00:04:14.470 --> 00:04:17.029
Say it's a car.

00:04:17.029 --> 00:04:19.860
Let me do some more interesting
colors.

00:04:19.860 --> 00:04:23.340
A car with a magenta bottom.

00:04:23.340 --> 00:04:25.390
And it is, let's see, what
does this problem say?

00:04:25.390 --> 00:04:31.100
It's 1,000 kilograms. So
a little over a ton.

00:04:31.100 --> 00:04:35.410
And it's moving at 9 meters
per second east. So its

00:04:35.410 --> 00:04:41.550
velocity is equal to 9 meters
per second east, or to the

00:04:41.550 --> 00:04:43.290
right in this example.

00:04:43.290 --> 00:04:47.460
And it strikes a stationary
2, 000 kilogram truck.

00:04:47.460 --> 00:04:48.710
So here's my truck.

00:04:53.420 --> 00:04:58.450
Here's my truck and this is
a 2,000 kilogram truck.

00:04:58.450 --> 00:05:02.580
And it's stationary, so
the velocity is 0.

00:05:02.580 --> 00:05:06.330
And when the car hits the truck,
let's just say that it

00:05:06.330 --> 00:05:08.600
somehow gets stuck in the truck
and they just both keep

00:05:08.600 --> 00:05:09.740
moving together.

00:05:09.740 --> 00:05:12.340
So they get stuck together.

00:05:12.340 --> 00:05:17.640
The question is, what is the
resulting speed of the

00:05:17.640 --> 00:05:22.050
combination truck and car
after the collision?

00:05:22.050 --> 00:05:24.190
Well, all we have to do is
think about what is the

00:05:24.190 --> 00:05:27.120
combined momentum before
the collision?

00:05:27.120 --> 00:05:27.610
Well let's see.

00:05:27.610 --> 00:05:30.800
The momentum of the car is going
to be the mass times the

00:05:30.800 --> 00:05:33.850
car-- mass of the car.

00:05:33.850 --> 00:05:36.530
Well the total momentum is going
to the mass of the car

00:05:36.530 --> 00:05:43.180
times the velocity of the car
plus the mass of the truck

00:05:43.180 --> 00:05:47.250
times the velocity
of the truck.

00:05:47.250 --> 00:05:49.220
And this is before they
hit each other.

00:05:49.220 --> 00:05:50.370
So what's the mass of the car?

00:05:50.370 --> 00:05:52.660
That's 1,000.

00:05:52.660 --> 00:05:53.760
What's the velocity
of the car?

00:05:53.760 --> 00:05:56.370
It's 9 meters per second.

00:05:56.370 --> 00:05:59.110
So as you can imagine, a unit of
momentum would be kilogram

00:05:59.110 --> 00:06:00.130
meters per second.

00:06:00.130 --> 00:06:02.990
So it's 1,000 times 9 kilogram
meters per second, but I won't

00:06:02.990 --> 00:06:05.950
write that right now just to
keep things simple, or so I

00:06:05.950 --> 00:06:06.960
save space.

00:06:06.960 --> 00:06:09.575
And then the mass of
the truck is 2,000.

00:06:09.575 --> 00:06:10.735
And what's its velocity?

00:06:10.735 --> 00:06:11.540
Well, it's 0.

00:06:11.540 --> 00:06:13.480
It's stationary initially.

00:06:13.480 --> 00:06:16.380
So the initial momentum of the
system-- this is 2,000 times

00:06:16.380 --> 00:06:24.080
0-- is 9,000 plus 0, which
equals 9,000 kilogram meters

00:06:24.080 --> 00:06:25.090
per second.

00:06:25.090 --> 00:06:28.440
That's the momentum before
the car hits

00:06:28.440 --> 00:06:29.970
the back of the truck.

00:06:29.970 --> 00:06:32.170
Now what happens after the car
hits the back of the truck?

00:06:32.170 --> 00:06:33.520
So let's go to that situation.

00:06:33.520 --> 00:06:36.230
So we have the truck.

00:06:36.230 --> 00:06:37.860
I'll draw it a little
less neatly.

00:06:37.860 --> 00:06:41.050
And then you have the car and
it's probably a little bit--

00:06:41.050 --> 00:06:44.040
well, I won't go into whether
it's banged up and whether it

00:06:44.040 --> 00:06:45.360
released heat and all of that.

00:06:45.360 --> 00:06:50.060
Let's assume that there was
nothing-- if this is a simple

00:06:50.060 --> 00:06:51.830
problem that we can do.

00:06:51.830 --> 00:06:53.700
So if we assume that,
there would be

00:06:53.700 --> 00:06:54.990
no change in momentum.

00:06:54.990 --> 00:06:57.410
Because we're saying that
there's no net forces acting

00:06:57.410 --> 00:06:58.010
on the system.

00:06:58.010 --> 00:06:59.910
And when I say system, I
mean the combination of

00:06:59.910 --> 00:07:01.410
the car and the truck.

00:07:01.410 --> 00:07:04.580
So what we're saying is, is
this combination, this new

00:07:04.580 --> 00:07:09.100
vehicle called a car truck, its
momentum will have to be

00:07:09.100 --> 00:07:13.070
the same as the car and the
truck's momentum when they

00:07:13.070 --> 00:07:14.600
were separate.

00:07:14.600 --> 00:07:16.780
So what do we know about
this car truck object?

00:07:16.780 --> 00:07:18.330
Well we know its new mass.

00:07:18.330 --> 00:07:21.050
The car truck object,
it will be the

00:07:21.050 --> 00:07:22.290
combined mass of the two.

00:07:22.290 --> 00:07:26.550
So it's 1,000 kilograms plus
2,000 kilograms. So it's 3,000

00:07:26.550 --> 00:07:30.690
kilograms. And now we can use
that information to figure out

00:07:30.690 --> 00:07:31.510
its velocity.

00:07:31.510 --> 00:07:32.190
How?

00:07:32.190 --> 00:07:36.300
Well, its momentum-- this
3,000 kilogram object's

00:07:36.300 --> 00:07:40.440
momentum-- has to be the same
as the momentum of the two

00:07:40.440 --> 00:07:42.510
objects before the collision.

00:07:42.510 --> 00:07:46.230
So it still has to be 9,000
kilogram meters per second.

00:07:46.230 --> 00:07:49.300
So once again, mass
times velocity.

00:07:49.300 --> 00:07:54.120
So mass is 3,000 times
the new velocity.

00:07:54.120 --> 00:07:57.580
So we could call that, I don't
know, new velocity, v sub n.

00:07:57.580 --> 00:08:01.200
That will equal 9,000.

00:08:01.200 --> 00:08:02.870
Because momentum is conserved.

00:08:02.870 --> 00:08:04.390
That's what you always
have to remember.

00:08:04.390 --> 00:08:07.630
Momentum doesn't change unless
there's a net force acting on

00:08:07.630 --> 00:08:08.030
the system.

00:08:08.030 --> 00:08:13.110
Because we saw a force is change
in momentum per time.

00:08:13.110 --> 00:08:14.670
So if you have no force
in it, you have

00:08:14.670 --> 00:08:16.626
no change in momentum.

00:08:16.626 --> 00:08:17.870
So let's just solve.

00:08:17.870 --> 00:08:22.990
Divide both sides of this by
3,000 and you get the new

00:08:22.990 --> 00:08:28.460
velocity is 3 meters
per second.

00:08:28.460 --> 00:08:29.490
And that kind of makes sense.

00:08:29.490 --> 00:08:32.840
You have a relatively light
car moving at 9 meters per

00:08:32.840 --> 00:08:34.080
second and a stationary truck.

00:08:34.080 --> 00:08:36.299
Then it smacks the truck
and they move together.

00:08:36.299 --> 00:08:41.690
The combined object-- and it's
going to be to the east. And

00:08:41.690 --> 00:08:44.970
we'll do more later, but we
assume that a positive

00:08:44.970 --> 00:08:47.310
velocity is east. If somehow we
ended up with a negative,

00:08:47.310 --> 00:08:50.040
it would have been west. But it
makes sense because we have

00:08:50.040 --> 00:08:53.590
a light object and a stationery,
heavy object.

00:08:53.590 --> 00:08:56.590
And when the light object hits
the stationery, heavy object,

00:08:56.590 --> 00:08:59.510
the combined objects still keeps
moving to the right, but

00:08:59.510 --> 00:09:03.010
it moves at a relatively
slower speed.

00:09:03.010 --> 00:09:04.660
So hopefully that gives you a
little bit of intuition for

00:09:04.660 --> 00:09:08.350
momentum, and that was not too
confusing of a problem.

00:09:08.350 --> 00:09:11.170
And in the next couple of
videos, I'll do more momentum

00:09:11.170 --> 00:09:13.430
problems and then I'll introduce
you to momentum

00:09:13.430 --> 00:09:15.350
problems in two dimensions.

00:09:15.350 --> 00:09:17.450
I will see you soon.


Tension in an accelerating system and pie in the face

WEBVTT
Kind: captions
Language: en

00:00:00.680 --> 00:00:01.360
Welcome back.

00:00:01.360 --> 00:00:03.820
We just finished this problem
with the pulleys and the

00:00:03.820 --> 00:00:04.490
inclined plane.

00:00:04.490 --> 00:00:06.890
And I just wanted to do one
final thing on this problem

00:00:06.890 --> 00:00:08.430
just because I think
it's interesting.

00:00:08.430 --> 00:00:11.150
And then we can move onto
what seems like

00:00:11.150 --> 00:00:12.810
a pretty fun problem.

00:00:12.810 --> 00:00:14.780
So the last thing I want to
figure out is, we figured out

00:00:14.780 --> 00:00:17.720
that this 20 kilo-- actually,
the whole system will

00:00:17.720 --> 00:00:20.860
accelerate up and to
the right at 4.13

00:00:20.860 --> 00:00:22.260
meters per second squared.

00:00:22.260 --> 00:00:24.580
And then the second part of this
question is, what is the

00:00:24.580 --> 00:00:28.485
tension in this rope
or this wire?

00:00:28.485 --> 00:00:30.990
And at first you might say,
this is complicated.

00:00:30.990 --> 00:00:32.940
You know, this thing isn't
static anymore.

00:00:32.940 --> 00:00:34.390
The thing is actually
accelerating.

00:00:34.390 --> 00:00:35.370
How do I do it?

00:00:35.370 --> 00:00:36.900
Well this is how you
think about it.

00:00:36.900 --> 00:00:39.350
Just pick one part
of the system.

00:00:39.350 --> 00:00:43.430
Let's say that all we could see
was this 20 kilogram mass.

00:00:43.430 --> 00:00:48.340
So let's say all we could see
was this 20 kilogram mass.

00:00:48.340 --> 00:00:50.760
And we know it's suspended
from a wire.

00:00:50.760 --> 00:00:54.780
And we also know that this
20 kilogram mass is not

00:00:54.780 --> 00:00:56.840
accelerating as fast
as it would if

00:00:56.840 --> 00:00:57.980
the wire wasn't there.

00:00:57.980 --> 00:01:01.210
It's accelerating only at
4.13 meters per second.

00:01:01.210 --> 00:01:04.510
If the wire wasn't there, it'd
be accelerating at 9.8 meters

00:01:04.510 --> 00:01:06.430
per second, the acceleration
of gravity.

00:01:06.430 --> 00:01:09.140
So the wire must be exerting
some upward

00:01:09.140 --> 00:01:10.520
force on the object.

00:01:10.520 --> 00:01:13.306
And that is the force
of tension.

00:01:13.306 --> 00:01:17.060
That is what's slowing-- that's
what's moderating its

00:01:17.060 --> 00:01:21.090
acceleration from being 9.8
meters per second squared to

00:01:21.090 --> 00:01:24.180
being 4.13 meters per
second squared.

00:01:24.180 --> 00:01:27.290
So essentially, what is the
net force on this object?

00:01:27.290 --> 00:01:29.550
On just this object?

00:01:29.550 --> 00:01:32.310
Well the net force is-- and
you can ignore what I said

00:01:32.310 --> 00:01:36.100
before about the net force
in all the other places.

00:01:36.100 --> 00:01:42.050
But we know that the object
is accelerating downwards.

00:01:42.050 --> 00:01:44.400
Well, we know it's 20 kilograms.
So that's its mass.

00:01:44.400 --> 00:01:46.980
And we know that it's
accelerating downwards at 4.13

00:01:46.980 --> 00:01:48.230
meters per second squared.

00:01:51.220 --> 00:02:00.460
So the net force, 20 times--
see, times 20 is 82-- let's

00:02:00.460 --> 00:02:03.120
just say 83 Newtons.

00:02:03.120 --> 00:02:05.400
83 Newtons down.

00:02:05.400 --> 00:02:08.990
We know that the net force
is 83 Newtons down.

00:02:08.990 --> 00:02:16.520
We also know that the tension
force plus the force of

00:02:16.520 --> 00:02:18.200
gravity-- and what's the
force of gravity?

00:02:18.200 --> 00:02:20.780
The force of gravity is just
the weight of the object.

00:02:20.780 --> 00:02:24.680
So the force of tension, which
goes up, plus the weight of --

00:02:24.680 --> 00:02:29.030
the force of gravity is equal
to the net force.

00:02:29.030 --> 00:02:31.010
And the way I set this up,
tension's going to be a

00:02:31.010 --> 00:02:33.270
negative number.

00:02:33.270 --> 00:02:36.530
Just because I'm saying
positive numbers are

00:02:36.530 --> 00:02:39.060
downwards, so a negative number
would be upwards.

00:02:39.060 --> 00:02:46.870
So tension will be what
is 83 minus 196?

00:02:46.870 --> 00:02:54.800
Minus 196 is equal to
minus 113 Newtons.

00:02:54.800 --> 00:02:57.040
And the only reason why I got a
negative number is because I

00:02:57.040 --> 00:02:59.150
used positive numbers
for downwards.

00:02:59.150 --> 00:03:02.520
So minus 113 Newtons downwards,
which is the same

00:03:02.520 --> 00:03:06.160
thing as 113 Newtons upwards.

00:03:06.160 --> 00:03:09.530
And so that is the tension
in the rope.

00:03:09.530 --> 00:03:11.970
And you could have done the same
thing on this side of the

00:03:11.970 --> 00:03:14.120
problem, although it would
have been-- well, yeah.

00:03:14.120 --> 00:03:15.290
You could have done the
exact same thing on

00:03:15.290 --> 00:03:16.150
this side of the problem.

00:03:16.150 --> 00:03:17.750
You would've said, well what
would it have accelerated

00:03:17.750 --> 00:03:20.870
naturally if there wasn't some
force of tension on this rope

00:03:20.870 --> 00:03:22.230
going backwards?

00:03:22.230 --> 00:03:24.090
And then you're saying, oh,
well, we know it would have

00:03:24.090 --> 00:03:26.350
gone in this direction at some
acceleration, but instead it's

00:03:26.350 --> 00:03:27.570
going in the other direction.

00:03:27.570 --> 00:03:29.090
So you use that.

00:03:29.090 --> 00:03:31.460
You figure out the net force,
and then you say the tension

00:03:31.460 --> 00:03:33.840
plus all of these forces have
to equal the net force.

00:03:33.840 --> 00:03:36.200
And then you should solve
for the tension.

00:03:36.200 --> 00:03:38.510
And it would be the
same tension.

00:03:38.510 --> 00:03:44.760
Now we will do a fun and
somewhat simple, but maybe

00:03:44.760 --> 00:03:46.780
instructive problem.

00:03:46.780 --> 00:03:49.860
So I have a pie.

00:03:49.860 --> 00:03:51.110
This is the pie.

00:03:54.140 --> 00:03:55.580
This is parallel.

00:03:55.580 --> 00:03:58.270
And I have my hand.

00:03:58.270 --> 00:04:02.070
You can tell that my destiny
was really to be a great

00:04:02.070 --> 00:04:04.170
artist. This is my hand.

00:04:04.170 --> 00:04:10.100
And I'm holding a pie, and I'm
looking to smash this pie into

00:04:10.100 --> 00:04:12.670
this individual's face.

00:04:19.382 --> 00:04:24.790
I actually was a, I was the
newspaper cartoonist in high

00:04:24.790 --> 00:04:28.360
school, so I have some
minor-- but anyway.

00:04:28.360 --> 00:04:30.140
Let's make it a bald man.

00:04:30.140 --> 00:04:32.950
Well anyway, I shouldn't be
focusing on the drawing.

00:04:36.400 --> 00:04:37.650
He has a moustache.

00:04:41.280 --> 00:04:45.720
Anyway, I'm looking to throw
this pie into this guy's face.

00:04:45.720 --> 00:04:49.090
And the problem is, I need to
figure out how fast do I need

00:04:49.090 --> 00:04:52.250
to accelerate this pie for
it to not fall down?

00:04:52.250 --> 00:04:52.490
Right?

00:04:52.490 --> 00:04:53.190
Because what's happening?

00:04:53.190 --> 00:04:56.290
Well there's the force of
gravity on this pie.

00:04:56.290 --> 00:04:58.500
There's a force of gravity on
this pie and if I don't

00:04:58.500 --> 00:05:01.220
accelerate it fast enough, it's
just going to slide down.

00:05:01.220 --> 00:05:03.000
And I'll never be able
to, It'll never

00:05:03.000 --> 00:05:04.200
reach the guy's face.

00:05:04.200 --> 00:05:06.440
So I don't want this pie
to slide down at all.

00:05:06.440 --> 00:05:09.070
How fast do I have
to push on it?

00:05:09.070 --> 00:05:11.640
Well, we know that the
coefficient of friction-- you

00:05:11.640 --> 00:05:13.400
don't know this, but I know
that the coefficient of

00:05:13.400 --> 00:05:17.700
friction between my hand and
the pie, the coefficient of

00:05:17.700 --> 00:05:23.170
friction is equal to 0.8.

00:05:23.170 --> 00:05:26.290
So given that, how fast do
I have to accelerate it?

00:05:26.290 --> 00:05:28.650
Well let's see what's
happening.

00:05:28.650 --> 00:05:30.720
So we have the force of
gravity pulling down.

00:05:30.720 --> 00:05:35.550
So let's say that the mass
of the pie is m.

00:05:35.550 --> 00:05:38.860
m equals mass.

00:05:38.860 --> 00:05:40.400
So what is the force of gravity

00:05:40.400 --> 00:05:41.630
pulling down on the pie?

00:05:41.630 --> 00:05:47.350
Well the force of gravity is
just equal to m times 9.8.

00:05:47.350 --> 00:05:48.490
Right?

00:05:48.490 --> 00:05:51.420
The force of gravity is
equal to m times 9.8.

00:05:51.420 --> 00:05:55.810
In order for this pie to not
move down, what do we know

00:05:55.810 --> 00:05:57.950
about the net forces
on that pie?

00:05:57.950 --> 00:06:02.330
Well we know the net forces
on that pie have to be 0.

00:06:02.330 --> 00:06:04.010
So what would be the
offsetting force?

00:06:04.010 --> 00:06:05.630
Well, it would be the
force of friction.

00:06:05.630 --> 00:06:08.106
So we would have a force of
friction acting upwards.

00:06:08.106 --> 00:06:08.690
Right?

00:06:08.690 --> 00:06:11.140
Because the force of friction
always acts opposite to the

00:06:11.140 --> 00:06:14.540
direction that the thing
would move otherwise.

00:06:14.540 --> 00:06:20.190
So essentially, our force of
friction has to be greater

00:06:20.190 --> 00:06:23.160
than, roughly, greater
than or equal to.

00:06:23.160 --> 00:06:24.710
Because if it's greater than,
it's not like the pie is going

00:06:24.710 --> 00:06:25.400
to move up.

00:06:25.400 --> 00:06:28.410
Friction by itself will never
move something, it'll just

00:06:28.410 --> 00:06:29.940
keep something from
being moved.

00:06:29.940 --> 00:06:31.250
But let's just figure
out the minimum.

00:06:31.250 --> 00:06:33.210
I won't do the whole
inequalities.

00:06:33.210 --> 00:06:37.950
The force of friction has to
be equal similarly, to 9.8

00:06:37.950 --> 00:06:42.390
times the mass of the pie.

00:06:42.390 --> 00:06:46.680
So if the coefficient of
friction is 0.8, what is the

00:06:46.680 --> 00:06:50.380
force that I have to apply?

00:06:50.380 --> 00:06:52.830
Well, the force I have to apply
in this case is going to

00:06:52.830 --> 00:06:53.970
be the normal force, right?

00:06:53.970 --> 00:06:58.600
That's normal to the
bottom of the pie.

00:06:58.600 --> 00:07:00.170
Right?

00:07:00.170 --> 00:07:03.270
My hand is now like the
surface of the ramp.

00:07:03.270 --> 00:07:05.830
So this is the normal force.

00:07:05.830 --> 00:07:09.030
And we know that the force of
friction is equal to the

00:07:09.030 --> 00:07:11.550
coefficient of friction times
the normal force.

00:07:11.550 --> 00:07:12.960
I'm going to switch colors
because this is getting

00:07:12.960 --> 00:07:15.300
monotonous.

00:07:15.300 --> 00:07:17.120
And the force of friction,
we know has to be

00:07:17.120 --> 00:07:20.030
9.8 times the mass.

00:07:20.030 --> 00:07:23.010
So 9.8 meters per second
times the mass.

00:07:23.010 --> 00:07:25.420
9.8m is the force of friction.

00:07:25.420 --> 00:07:27.830
And that has to equal to
coefficient of friction times

00:07:27.830 --> 00:07:28.865
the normal force.

00:07:28.865 --> 00:07:31.310
And remember, the normal force
is essentially the force that

00:07:31.310 --> 00:07:33.472
I'm pushing the pie with.

00:07:33.472 --> 00:07:38.270
And we know this is 0.8, so we
have 9.8 times the mass--

00:07:38.270 --> 00:07:43.100
that's not meters, that's the
mass-- is equal to 0.8 times

00:07:43.100 --> 00:07:45.720
the normal force.

00:07:45.720 --> 00:07:51.960
So you have the normal force is
equal to 9.8 times the mass

00:07:51.960 --> 00:07:54.500
divided by 0.8.

00:07:54.500 --> 00:07:56.320
What's 9.8 divided by 0.8?

00:07:56.320 --> 00:08:05.200
9.8 divided by 0.8 is
equal to 12.25.

00:08:05.200 --> 00:08:08.580
So the normal force that
I have to apply is

00:08:08.580 --> 00:08:12.890
12.25 times the mass.

00:08:12.890 --> 00:08:14.860
So that's the force
I'm applying.

00:08:14.860 --> 00:08:15.540
It's time the mass.

00:08:15.540 --> 00:08:17.040
We don't know the
mass of the pie.

00:08:17.040 --> 00:08:19.570
So how fast am I accelerating
the pie?

00:08:19.570 --> 00:08:22.590
Well, force is equal to mass
times acceleration.

00:08:22.590 --> 00:08:29.540
This is the force, 12.25m--
that's the force-- is equal to

00:08:29.540 --> 00:08:31.170
the mass times the

00:08:31.170 --> 00:08:32.720
acceleration of the pie, right?

00:08:32.720 --> 00:08:34.799
And it's the same pie that we're
dealing with the whole

00:08:34.799 --> 00:08:35.970
time, so it's still m.

00:08:35.970 --> 00:08:39.049
And you can take out m from both
sides of the equation.

00:08:39.049 --> 00:08:42.380
So the acceleration, the rate at
which I have to change the

00:08:42.380 --> 00:08:45.450
velocity, or the acceleration
that I have to apply to the

00:08:45.450 --> 00:08:53.860
pie is 12.25 meters per
second squared.

00:08:53.860 --> 00:08:58.090
And so actually, I have to apply
more than 1g, right?

00:08:58.090 --> 00:09:01.330
Because g is the force
of gravity.

00:09:01.330 --> 00:09:05.430
And gravity accelerates
something at 9.8 seconds-- 9.8

00:09:05.430 --> 00:09:06.750
meters per second squared.

00:09:06.750 --> 00:09:09.870
So I have to do something at
12-- I have to push and

00:09:09.870 --> 00:09:13.270
accelerate the pie at 12.25
meters per second squared.

00:09:13.270 --> 00:09:16.720
So it's something a little over
1g in order for that pie

00:09:16.720 --> 00:09:19.910
to not fall and in order for my
normal force to provide a

00:09:19.910 --> 00:09:22.630
force of friction so that
the pie can reach

00:09:22.630 --> 00:09:25.360
this bald man's face.

00:09:25.360 --> 00:09:26.720
I will see you in
the next video.


Introduction to tension (part 2)

WEBVTT
Kind: captions
Language: en

00:00:00.790 --> 00:00:01.450
Welcome back.

00:00:01.450 --> 00:00:04.520
We'll now do another tension
problem and this one is just a

00:00:04.520 --> 00:00:06.340
slight increment harder than the
previous one just because

00:00:06.340 --> 00:00:10.390
we have to take out slightly
more sophisticated algebra

00:00:10.390 --> 00:00:11.490
tools than we did
in the last one.

00:00:11.490 --> 00:00:14.000
But it's not really
any harder.

00:00:14.000 --> 00:00:15.990
But you should actually see this
type of problem because

00:00:15.990 --> 00:00:17.810
you'll probably see
it on an exam.

00:00:17.810 --> 00:00:19.440
So let's figure out the
tension in the wire.

00:00:19.440 --> 00:00:22.950
So first of all, we know
that this point

00:00:22.950 --> 00:00:24.380
right here isn't moving.

00:00:24.380 --> 00:00:26.700
So the tension in this little
small wire right here is easy.

00:00:26.700 --> 00:00:28.170
It's trivial.

00:00:28.170 --> 00:00:30.850
The force of gravity is pulling
down at this point

00:00:30.850 --> 00:00:33.170
with 10 Newtons because you
have this weight here.

00:00:33.170 --> 00:00:35.200
And of course, since this
point is stationary, the

00:00:35.200 --> 00:00:37.750
tension in this wire has to
be 10 Newtons upward.

00:00:37.750 --> 00:00:38.300
That's an easy one.

00:00:38.300 --> 00:00:41.600
So let's just figure out the
tension in these two slightly

00:00:41.600 --> 00:00:44.550
more difficult wires to figure
out the tensions of.

00:00:44.550 --> 00:00:47.500
So once again, we know that this
point right here, this

00:00:47.500 --> 00:00:49.760
point is not accelerating
in any direction.

00:00:49.760 --> 00:00:52.900
It's not accelerating in the
x direction, nor is it

00:00:52.900 --> 00:00:55.340
accelerating in the vertical
direction or the y direction.

00:00:55.340 --> 00:00:57.970
So we know that the net forces
in the x direction need to be

00:00:57.970 --> 00:01:00.800
0 on it and we know the
net forces in the y

00:01:00.800 --> 00:01:03.970
direction need to be 0.

00:01:03.970 --> 00:01:06.410
So what are the net forces
in the x direction?

00:01:06.410 --> 00:01:10.120
Well they're going to be the x
components of these two-- of

00:01:10.120 --> 00:01:12.750
the tension vectors of
both of these wires.

00:01:12.750 --> 00:01:15.820
I guess let's draw the tension
vectors of the two wires.

00:01:15.820 --> 00:01:18.065
So this T1, it's pulling.

00:01:18.065 --> 00:01:21.870
The tension vector pulls in
the direction of the wire

00:01:21.870 --> 00:01:23.240
along the same line.

00:01:23.240 --> 00:01:26.740
So let's say that this is the
tension vector of T1.

00:01:26.740 --> 00:01:30.620
If that's the tension vector,
its x component will be this.

00:01:30.620 --> 00:01:33.690
Let me see how good
I can draw this.

00:01:33.690 --> 00:01:35.440
It's intended to be a straight
line, but that

00:01:35.440 --> 00:01:36.920
would be its x component.

00:01:36.920 --> 00:01:42.570
And its x component, let's
see, this is 30 degrees.

00:01:42.570 --> 00:01:44.440
This is 30 degrees right here.

00:01:44.440 --> 00:01:47.070
And hopefully this is a bit
second nature to you.

00:01:47.070 --> 00:01:49.320
If this value up here
is T1, what is the

00:01:49.320 --> 00:01:51.590
value of the x component?

00:01:51.590 --> 00:01:59.860
It's T1 cosine of 30 degrees.

00:01:59.860 --> 00:02:01.200
And you could do your
SOH-CAH-TOA.

00:02:01.200 --> 00:02:04.810
You know, cosine is adjacent
over hypotenuse.

00:02:04.810 --> 00:02:09.430
So the cosine of 30 degrees is
equal to-- This over T1 one is

00:02:09.430 --> 00:02:10.800
equal to the x component
over T1.

00:02:10.800 --> 00:02:14.010
And if you multiply both sides
by T1, you get this.

00:02:14.010 --> 00:02:15.690
This should be a little bit of
second nature right now.

00:02:15.690 --> 00:02:20.050
That the x component is going to
be the cosine of the angle

00:02:20.050 --> 00:02:23.350
between the hypotenuse and
the x component times the

00:02:23.350 --> 00:02:24.440
hypotenuse.

00:02:24.440 --> 00:02:27.485
And similarly, the x component
here-- Let me

00:02:27.485 --> 00:02:28.860
draw this force vector.

00:02:28.860 --> 00:02:34.675
So if this is T2, this would
be its x component.

00:02:38.260 --> 00:02:44.320
And very similarly, this is 60
degrees, so this would be T2

00:02:44.320 --> 00:02:48.130
cosine of 60.

00:02:48.130 --> 00:02:50.920
Now what do we know about
these two vectors?

00:02:50.920 --> 00:02:55.150
We know that their
net force is 0.

00:02:55.150 --> 00:02:57.500
Or that you also know that the
magnitude of these two vectors

00:02:57.500 --> 00:02:59.340
should cancel each other out
or that they're equal.

00:02:59.340 --> 00:03:01.570
I mean, they're pulling in
opposite directions.

00:03:01.570 --> 00:03:04.030
That's pretty obvious.

00:03:04.030 --> 00:03:05.900
And so you know that their
magnitudes need to be equal.

00:03:05.900 --> 00:03:08.950
So we know that T1 cosine
of 30 is going to equal

00:03:08.950 --> 00:03:10.810
T2 cosine of 60.

00:03:10.810 --> 00:03:12.990
So let's write that down.

00:03:12.990 --> 00:03:27.070
T1 cosine of 30 degrees is
equal to T2 cosine of 60.

00:03:27.070 --> 00:03:29.890
And then we could bring the
T2 on to this side.

00:03:29.890 --> 00:03:31.900
And actually, let's also-- I'm
trying to save as much space

00:03:31.900 --> 00:03:33.950
as possible because I'm guessing
this is going to take

00:03:33.950 --> 00:03:35.550
up a lot of room,
this problem.

00:03:35.550 --> 00:03:38.070
What's the cosine
of 30 degrees?

00:03:38.070 --> 00:03:39.410
If you haven't memorized
it already, it's square

00:03:39.410 --> 00:03:40.040
root of 3 over 2.

00:03:40.040 --> 00:03:45.640
So this becomes square root
of 3 over 2 times T1.

00:03:45.640 --> 00:03:47.510
That's the cosine
of 30 degrees.

00:03:47.510 --> 00:03:49.290
And then I'm going to bring
this on to this side.

00:03:49.290 --> 00:03:54.590
So the cosine of 60
is actually 1/2.

00:03:54.590 --> 00:03:56.520
You could use your calculator
if you forgot that.

00:03:56.520 --> 00:03:58.320
So this is 1/2 T2.

00:03:58.320 --> 00:04:01.590
Bring it on this side so
it becomes minus 1/2.

00:04:01.590 --> 00:04:04.940
I'm skipping more steps than
normal just because I don't

00:04:04.940 --> 00:04:07.020
want to waste too much space.

00:04:07.020 --> 00:04:08.160
And this equals 0.

00:04:08.160 --> 00:04:11.370
But if you seen the other
videos, hopefully I'm not

00:04:11.370 --> 00:04:12.410
creating too many gaps.

00:04:12.410 --> 00:04:14.960
And this is relatively
easy to follow.

00:04:14.960 --> 00:04:19.269
So we have the square root of
3 times T1 minus 1/2 T2 is

00:04:19.269 --> 00:04:19.800
equal to 0.

00:04:19.800 --> 00:04:21.700
So that gives us an equation.

00:04:21.700 --> 00:04:24.320
One equation with two unknowns,
so it doesn't help

00:04:24.320 --> 00:04:25.050
us much so far.

00:04:25.050 --> 00:04:29.250
But let's square that away
because I have a feeling this

00:04:29.250 --> 00:04:32.330
will be useful.

00:04:32.330 --> 00:04:34.330
Now what's going to be happening
on the y components?

00:04:34.330 --> 00:04:39.580
So let's say that this is the y
component of T1 and this is

00:04:39.580 --> 00:04:42.230
the y component of T2.

00:04:42.230 --> 00:04:42.860
What do we know?

00:04:42.860 --> 00:04:44.450
What what do we know about
the two y components?

00:04:44.450 --> 00:04:46.890
I could've drawn them here too
and then just shift them over

00:04:46.890 --> 00:04:48.240
to the left and the right.

00:04:48.240 --> 00:04:52.210
We know that their combined
pull upwards, the combined

00:04:52.210 --> 00:04:54.500
pull of the two vertical tension
components has to

00:04:54.500 --> 00:04:57.770
offset the force of gravity
pulling down because this

00:04:57.770 --> 00:04:59.130
point is stationary.

00:04:59.130 --> 00:05:02.310
So we know these two y
components, when you add them

00:05:02.310 --> 00:05:05.210
together, the combined tension
in the vertical direction has

00:05:05.210 --> 00:05:06.680
to be 10 Newtons.

00:05:06.680 --> 00:05:09.240
Because it's offsetting
this force of gravity.

00:05:09.240 --> 00:05:10.990
So what's this y component?

00:05:10.990 --> 00:05:12.970
Well, this was T1
of cosine of 30.

00:05:12.970 --> 00:05:15.140
This should start to become a
little second nature to you

00:05:15.140 --> 00:05:19.180
that this is T1 sine of 30, this
y component right here.

00:05:19.180 --> 00:05:23.570
So T1-- Let me write it here.

00:05:23.570 --> 00:05:34.875
T1 sine of 30 degrees plus this
vector, which is T2 sine

00:05:34.875 --> 00:05:38.820
of 60 degrees.

00:05:38.820 --> 00:05:39.910
You could review your
trigonometry and your

00:05:39.910 --> 00:05:41.650
SOH-CAH-TOA.

00:05:41.650 --> 00:05:44.830
Frankly, I think, just seeing
what people get confused on is

00:05:44.830 --> 00:05:45.680
the trigonometry.

00:05:45.680 --> 00:05:48.340
But you can review the trig
modules and maybe some of the

00:05:48.340 --> 00:05:50.370
earlier force vector modules
that we did.

00:05:50.370 --> 00:05:52.100
And hopefully, these
will make sense.

00:05:52.100 --> 00:05:53.460
I'm skipping a few steps.

00:05:53.460 --> 00:05:58.870
And these will equal
10 Newtons.

00:05:58.870 --> 00:06:02.370
And let's rewrite this up here
where I substitute the values.

00:06:02.370 --> 00:06:03.650
So what's the sine of 30?

00:06:03.650 --> 00:06:06.180
Actually, let me do
it right here.

00:06:06.180 --> 00:06:07.400
What's the sine of 30 degrees?

00:06:07.400 --> 00:06:15.000
The sine of 30 degrees is 1/2 so
we get 1/2 T1 plus the sine

00:06:15.000 --> 00:06:17.420
of 60 degrees, which is square
root of 3 over 2.

00:06:17.420 --> 00:06:22.150
Square root of 3 over
2 T2 is equal to 10.

00:06:22.150 --> 00:06:23.850
And then I don't like
this, all these 2's

00:06:23.850 --> 00:06:25.780
and this 1/2 here.

00:06:25.780 --> 00:06:27.850
So let's multiply this
whole equation by 2.

00:06:27.850 --> 00:06:30.170
So 2 times 1/2, that's 1.

00:06:30.170 --> 00:06:37.940
So you get T1 plus the square
root of 3 T2 is equal to, 2

00:06:37.940 --> 00:06:40.400
times 10 , is 20.

00:06:40.400 --> 00:06:42.330
Similarly, let's take this
equation up here and let's

00:06:42.330 --> 00:06:46.520
multiply this equation by 2
and bring it down here.

00:06:46.520 --> 00:06:47.980
So this is the original
one that we got.

00:06:47.980 --> 00:06:50.630
So if we multiply this whole
thing by 2-- I'll do it in

00:06:50.630 --> 00:06:52.130
this color so that
you know that

00:06:52.130 --> 00:06:54.230
it's a different equation.

00:06:54.230 --> 00:06:56.410
So if you multiply square root
of 3 over 2 times 2-- I'm just

00:06:56.410 --> 00:06:58.350
doing this to get rid of the
2's in the denominator.

00:06:58.350 --> 00:07:10.360
So you get square root of 3 T1
minus T2 is equal to 0 because

00:07:10.360 --> 00:07:12.080
0 times 2 is 0.

00:07:12.080 --> 00:07:13.410
And let's see what
we could do.

00:07:13.410 --> 00:07:15.960
What if we take this top
equation because we want to

00:07:15.960 --> 00:07:18.790
start canceling out some terms.
Let's take this top

00:07:18.790 --> 00:07:24.880
equation and let's multiply
it by-- oh, I don't know.

00:07:24.880 --> 00:07:28.620
Let's multiply it by the
square root of 3.

00:07:28.620 --> 00:07:36.740
So you get the square
root of 3 T1.

00:07:36.740 --> 00:07:37.950
I'm taking this top equation
multiplied by the

00:07:37.950 --> 00:07:38.320
square root of 3.

00:07:38.320 --> 00:07:39.680
This is just a system
of equations

00:07:39.680 --> 00:07:41.100
that I'm solving for.

00:07:41.100 --> 00:07:43.990
And the square root of 3
times this right here.

00:07:43.990 --> 00:07:46.030
Square root of 3 times square
root of 3 is 3.

00:07:46.030 --> 00:07:54.200
So plus 3 T2 is equal to
20 square root of 3.

00:07:54.200 --> 00:07:59.600
And now what I want to do is
let's-- I know I'm doing a lot

00:07:59.600 --> 00:08:01.160
of equation manipulation here.

00:08:01.160 --> 00:08:04.380
But this is just hopefully, a
review of algebra for you.

00:08:04.380 --> 00:08:09.590
Let's subtract this equation
from this equation.

00:08:09.590 --> 00:08:11.960
So you can also view it as
multiplying it by negative 1

00:08:11.960 --> 00:08:13.300
and then adding the 2.

00:08:13.300 --> 00:08:15.930
So when you subtract this from
this, these two terms cancel

00:08:15.930 --> 00:08:17.620
out because they're the same.

00:08:17.620 --> 00:08:22.010
And so then you're left with
minus T2 from here.

00:08:22.010 --> 00:08:31.170
Minus this, minus 3 T2 is equal
to 0 minus 20 square

00:08:31.170 --> 00:08:33.490
roots of 3.

00:08:33.490 --> 00:08:40.460
And so this becomes minus 4 T2
is equal to minus 20 square

00:08:40.460 --> 00:08:42.480
roots of 3.

00:08:42.480 --> 00:08:46.930
And then, divide both sides by
minus 4 and you get T2 is

00:08:46.930 --> 00:08:52.320
equal to 5 square roots
of 3 Newtons.

00:08:52.320 --> 00:08:54.150
So that's the tension
in this wire.

00:08:54.150 --> 00:08:58.090
And now we can substitute
and figure out T1.

00:08:58.090 --> 00:08:59.460
Let's use this formula
right here because it

00:08:59.460 --> 00:09:02.120
looks suitably simple.

00:09:02.120 --> 00:09:08.160
So we have the square root
of 3 times T1 minus T2.

00:09:08.160 --> 00:09:10.960
Well T2 is 5 square
roots of 3.

00:09:10.960 --> 00:09:14.950
5 square roots of
3 is equal to 0.

00:09:14.950 --> 00:09:19.860
So we have the square root of
3 T1 is equal to five square

00:09:19.860 --> 00:09:21.060
roots of 3.

00:09:21.060 --> 00:09:23.370
Divide both sides by square
root of 3 and you get the

00:09:23.370 --> 00:09:28.010
tension in the first wire
is equal to 5 Newtons.

00:09:28.010 --> 00:09:31.880
So this is pulling with a force
or tension of 5 Newtons.

00:09:31.880 --> 00:09:33.150
Or a force.

00:09:33.150 --> 00:09:36.460
And this is pulling-- the second
wire --with a tension

00:09:36.460 --> 00:09:39.990
of 5 square roots
of 3 Newtons.

00:09:39.990 --> 00:09:43.430
So this wire right
here is actually

00:09:43.430 --> 00:09:45.410
doing more of the pulling.

00:09:45.410 --> 00:09:48.670
It's actually more of the force
of gravity is ending up

00:09:48.670 --> 00:09:49.650
on this wire.

00:09:49.650 --> 00:09:51.820
That makes sense because
it's steeper.

00:09:51.820 --> 00:09:54.800
So since it's steeper,
it's contributing

00:09:54.800 --> 00:09:56.970
more to the y component.

00:09:56.970 --> 00:09:58.600
It's good whenever you do these
problems to kind of do a

00:09:58.600 --> 00:10:01.520
reality check just to make sure
your numbers make sense.

00:10:01.520 --> 00:10:05.800
And if you think about it,
their combined tension is

00:10:05.800 --> 00:10:07.490
something more than
10 Newtons.

00:10:07.490 --> 00:10:10.750
And that makes sense because
some of the force that they're

00:10:10.750 --> 00:10:13.940
pulling with is wasted against
pulling each other in the

00:10:13.940 --> 00:10:15.520
horizontal direction.

00:10:15.520 --> 00:10:18.150
Anyway, I'll see you all
in the next video.


Introduction to tension

WEBVTT
Kind: captions
Language: en

00:00:00.780 --> 00:00:04.070
I will now introduce you to
the concept of tension.

00:00:04.070 --> 00:00:07.390
So tension is really just the
force that exists either

00:00:07.390 --> 00:00:10.760
within or applied by
a string or wire.

00:00:10.760 --> 00:00:13.280
It's usually lifting something
or pulling on something.

00:00:13.280 --> 00:00:16.040
So let's say I had a weight.

00:00:16.040 --> 00:00:21.080
Let's say I have
a weight here.

00:00:21.080 --> 00:00:27.450
And let's say it's
100 Newtons.

00:00:27.450 --> 00:00:31.575
And it's suspended from this
wire, which is right here.

00:00:31.575 --> 00:00:34.920
Let's say it's attached to
the ceiling right there.

00:00:34.920 --> 00:00:38.010
Well we already know that the
force-- if we're on this

00:00:38.010 --> 00:00:41.390
planet that this weight is being
pull down by gravity.

00:00:41.390 --> 00:00:44.370
So we already know that there's
a downward force on

00:00:44.370 --> 00:00:47.730
this weight, which is
a force of gravity.

00:00:47.730 --> 00:00:51.130
And that equals 100 Newtons.

00:00:51.130 --> 00:00:54.090
But we also know that this
weight isn't accelerating,

00:00:54.090 --> 00:00:54.850
it's actually stationary.

00:00:54.850 --> 00:00:56.040
It also has no velocity.

00:00:56.040 --> 00:00:59.570
But the important thing is
it's not accelerating.

00:00:59.570 --> 00:01:05.080
But given that, we know that the
net force on it must be 0

00:01:05.080 --> 00:01:07.950
by Newton's laws.

00:01:07.950 --> 00:01:10.820
So what is the counteracting
force?

00:01:10.820 --> 00:01:12.840
You didn't have to know about
tension to say well, the

00:01:12.840 --> 00:01:14.300
string's pulling on it.

00:01:14.300 --> 00:01:17.490
The string is what's keeping
the weight from falling.

00:01:17.490 --> 00:01:20.920
So the force that the string or
this wire applies on this

00:01:20.920 --> 00:01:24.460
weight you can view as
the force of tension.

00:01:24.460 --> 00:01:27.400
Another way to think about it
is that's also the force

00:01:27.400 --> 00:01:30.350
that's within the wire.

00:01:33.020 --> 00:01:37.320
And that is going to exactly
offset the force of gravity on

00:01:37.320 --> 00:01:37.700
this weight.

00:01:37.700 --> 00:01:43.190
And that's what keeps this point
right here stationery

00:01:43.190 --> 00:01:45.800
and keeps it from
accelerating.

00:01:45.800 --> 00:01:47.040
That's pretty straightforward.

00:01:47.040 --> 00:01:49.640
Tension, it's just the
force of a string.

00:01:49.640 --> 00:01:53.270
And just so you can
conceptualize it, on a guitar,

00:01:53.270 --> 00:01:58.830
the more you pull on some of
those higher-- what was it?

00:01:58.830 --> 00:02:01.820
The really thin strings that
sound higher pitched.

00:02:01.820 --> 00:02:04.040
The more you pull on it,
the higher the tension.

00:02:04.040 --> 00:02:06.780
It actually creates a
higher pitched note.

00:02:06.780 --> 00:02:08.210
So you've dealt with
tension a lot.

00:02:08.210 --> 00:02:11.260
I think actually when they sell
wires or strings they'll

00:02:11.260 --> 00:02:13.730
probably tell you the tension
that that wire or string can

00:02:13.730 --> 00:02:15.650
support, which is important if
you're going to build a bridge

00:02:15.650 --> 00:02:16.980
or a swing or something.

00:02:16.980 --> 00:02:20.170
So tension is something that
should be hopefully, a little

00:02:20.170 --> 00:02:21.600
bit intuitive to you.

00:02:21.600 --> 00:02:24.710
So let's, with that fairly
simple example done, let's

00:02:24.710 --> 00:02:27.980
create a slightly more
complicated example.

00:02:27.980 --> 00:02:29.900
So let's take the same weight.

00:02:29.900 --> 00:02:31.720
Instead of making the
ceiling here, let's

00:02:31.720 --> 00:02:34.510
add two more strings.

00:02:34.510 --> 00:02:37.660
Let's add this green string.

00:02:40.770 --> 00:02:43.010
Green string there.

00:02:43.010 --> 00:02:46.610
And it's attached to the
ceiling up here.

00:02:46.610 --> 00:02:48.970
That's the ceiling now.

00:02:48.970 --> 00:02:49.530
And let's see.

00:02:49.530 --> 00:02:52.630
This is the wall.

00:02:52.630 --> 00:02:54.710
And let's say there's another
string right here

00:02:54.710 --> 00:02:57.240
attached to the wall.

00:02:57.240 --> 00:03:00.600
So my question to you is, what
is the tension in these two

00:03:00.600 --> 00:03:08.620
strings So let's call
this T1 and T2.

00:03:08.620 --> 00:03:12.470
Well like the first problem,
this point right here, this

00:03:12.470 --> 00:03:15.430
red point, is stationary.

00:03:15.430 --> 00:03:17.820
It's not accelerating in
either the left/right

00:03:17.820 --> 00:03:20.570
directions and it's not
accelerating in the up/down

00:03:20.570 --> 00:03:21.260
directions.

00:03:21.260 --> 00:03:24.400
So we know that the net forces
in both the x and y

00:03:24.400 --> 00:03:27.560
dimensions must be 0.

00:03:27.560 --> 00:03:30.950
My second question to
you is, what is

00:03:30.950 --> 00:03:31.950
going to be the offset?

00:03:31.950 --> 00:03:34.770
Because we know already that
at this point right here,

00:03:34.770 --> 00:03:37.330
there's going to be a downward
force, which is the force of

00:03:37.330 --> 00:03:39.270
gravity again.

00:03:39.270 --> 00:03:40.220
The weight of this
whole thing.

00:03:40.220 --> 00:03:43.490
We can assume that the wires
have no weight for simplicity.

00:03:43.490 --> 00:03:46.300
So we know that there's going
to be a downward force here,

00:03:46.300 --> 00:03:47.920
this is the force of
gravity, right?

00:03:47.920 --> 00:03:50.560
The whole weight of this entire
object of weight plus

00:03:50.560 --> 00:03:52.200
wire is pulling down.

00:03:52.200 --> 00:03:55.470
So what is going to be the
upward force here?

00:03:55.470 --> 00:03:57.880
Well let's look at each
of the wires.

00:03:57.880 --> 00:04:02.200
This second wire, T2, or we
could call it w2, I guess.

00:04:02.200 --> 00:04:05.070
The second wire is just
pulling to the left.

00:04:05.070 --> 00:04:06.335
It has no y components.

00:04:06.335 --> 00:04:08.680
It's not lifting up at all.

00:04:08.680 --> 00:04:10.590
So it's just pulling
to the left.

00:04:10.590 --> 00:04:13.910
So all of the upward lifting,
all of that's going to occur

00:04:13.910 --> 00:04:17.500
from this first wire, from T1.

00:04:17.500 --> 00:04:22.430
So we know that the y component
of T1, so let's

00:04:22.430 --> 00:04:25.395
call-- so if we say that
this vector here.

00:04:25.395 --> 00:04:28.510
Let me do it in a
different color.

00:04:28.510 --> 00:04:30.470
Because I know when I draw these
diagrams it starts to

00:04:30.470 --> 00:04:31.720
get confusing.

00:04:34.460 --> 00:04:36.940
Let me actually use
the line tool.

00:04:36.940 --> 00:04:39.480
So I have this.

00:04:39.480 --> 00:04:42.920
Let me make a thicker line.

00:04:42.920 --> 00:04:45.120
So we have this vector
here, which is T1.

00:04:49.110 --> 00:04:51.070
And we would need to figure
out what that is.

00:04:51.070 --> 00:04:53.200
And then we have the other
vector, which is its y

00:04:53.200 --> 00:04:55.890
component, and I'll draw
that like here.

00:04:59.640 --> 00:05:00.890
This is its y component.

00:05:05.780 --> 00:05:09.900
We could call this T1 sub y.

00:05:09.900 --> 00:05:11.920
And then of course, it has an
x component too, and I'll do

00:05:11.920 --> 00:05:15.360
that in-- let's see.

00:05:15.360 --> 00:05:19.160
I'll do that in red.

00:05:19.160 --> 00:05:21.170
Once again, this is just
breaking up a force into its

00:05:21.170 --> 00:05:25.250
component vectors like we've--
a vector force into its x and

00:05:25.250 --> 00:05:27.350
y components like we've been
doing in the last several

00:05:27.350 --> 00:05:30.100
problems. And these are just
trigonometry problems, right?

00:05:32.990 --> 00:05:35.920
We could actually now, visually
see that this is T

00:05:35.920 --> 00:05:38.620
sub 1 x and this is
T sub 1 sub y.

00:05:38.620 --> 00:05:41.130
Oh, and I forgot to give you an
important property of this

00:05:41.130 --> 00:05:44.780
problem that you needed to
know before solving it.

00:05:44.780 --> 00:05:47.820
Is that the angle that the
first wire forms with the

00:05:47.820 --> 00:05:51.370
ceiling, this is 30 degrees.

00:05:51.370 --> 00:05:58.470
So if that is 30 degrees, we
also know that this is a

00:05:58.470 --> 00:06:01.240
parallel line to this.

00:06:01.240 --> 00:06:03.880
So if this is 30 degrees,
this is also

00:06:03.880 --> 00:06:07.290
going to be 30 degrees.

00:06:07.290 --> 00:06:11.240
So this angle right here is also
going to be 30 degrees.

00:06:11.240 --> 00:06:13.370
And that's from our-- you know,
we know about parallel

00:06:13.370 --> 00:06:15.690
lines and alternate
interior angles.

00:06:15.690 --> 00:06:17.870
We could have done
it the other way.

00:06:17.870 --> 00:06:21.745
We could have said that if this
angle is 30 degrees, this

00:06:21.745 --> 00:06:22.900
angle is 60 degrees.

00:06:22.900 --> 00:06:24.620
This is a right angle,
so this is also 30.

00:06:24.620 --> 00:06:26.620
But that's just review
of geometry

00:06:26.620 --> 00:06:27.290
that you already know.

00:06:27.290 --> 00:06:30.280
But anyway, we know that this
angle is 30 degrees, so what's

00:06:30.280 --> 00:06:31.600
its y component?

00:06:31.600 --> 00:06:33.460
Well the y component,
let's see.

00:06:33.460 --> 00:06:36.090
What involves the hypotenuse
and the opposite side?

00:06:36.090 --> 00:06:38.640
Let me write soh cah toa at the
top because this is really

00:06:38.640 --> 00:06:39.920
just trigonometry.

00:06:39.920 --> 00:06:42.670
soh cah toa in blood red.

00:06:42.670 --> 00:06:45.730
So what involves the opposite
and the hypotenuse?

00:06:45.730 --> 00:06:47.770
So opposite over hypotenuse.

00:06:47.770 --> 00:06:55.310
So that we know the sine-- let
me switch to the sine of 30

00:06:55.310 --> 00:07:05.630
degrees is equal to T1 sub y
over the tension in the string

00:07:05.630 --> 00:07:07.610
going in this direction.

00:07:07.610 --> 00:07:15.730
So if we solve for T1 sub y we
get T1 sine of 30 degrees is

00:07:15.730 --> 00:07:20.870
equal to T1 sub y.

00:07:20.870 --> 00:07:23.210
And what did we just say
before we kind of

00:07:23.210 --> 00:07:24.990
dived into the math?

00:07:24.990 --> 00:07:30.600
We said all of the lifting on
this point is being done by

00:07:30.600 --> 00:07:32.550
the y component of T1.

00:07:32.550 --> 00:07:36.310
Because T2 is not doing any
lifting up or down, it's only

00:07:36.310 --> 00:07:38.920
pulling to the left.

00:07:38.920 --> 00:07:44.590
So the entire component that's
keeping this object up,

00:07:44.590 --> 00:07:46.770
keeping it from falling
is the y component of

00:07:46.770 --> 00:07:48.040
this tension vector.

00:07:48.040 --> 00:07:51.930
So that has to equal the force
of gravity pulling down.

00:07:51.930 --> 00:07:54.840
This has to equal the
force of gravity.

00:07:54.840 --> 00:07:58.710
That has to equal this
or this point.

00:07:58.710 --> 00:08:01.260
So that's 100 Newtons.

00:08:01.260 --> 00:08:04.185
And I really want to hit this
point home because it might be

00:08:04.185 --> 00:08:06.140
a little confusing to you.

00:08:06.140 --> 00:08:07.810
We just said, this point
is stationery.

00:08:07.810 --> 00:08:09.000
It's not moving up or down.

00:08:09.000 --> 00:08:10.720
It's not accelerating
up or down.

00:08:10.720 --> 00:08:14.540
And so we know that there's a
downward force of 100 Newtons,

00:08:14.540 --> 00:08:17.480
so there must be an upward force
that's being provided by

00:08:17.480 --> 00:08:18.800
these two wires.

00:08:18.800 --> 00:08:21.150
This wire is providing
no upward force.

00:08:21.150 --> 00:08:24.140
So all of the upward force must
be the y component or the

00:08:24.140 --> 00:08:29.350
upward component of this force
vector on the first wire.

00:08:29.350 --> 00:08:33.309
So given that, we can now solve
for the tension in this

00:08:33.309 --> 00:08:39.409
first wire because we have
T1-- what's sine of 30?

00:08:39.409 --> 00:08:43.050
Sine of 30 degrees, in case you
haven't memorized it, sine

00:08:43.050 --> 00:08:44.990
of 30 degrees is 1/2.

00:08:44.990 --> 00:08:52.640
So T1 times 1/2 is equal
to 100 Newtons.

00:08:52.640 --> 00:08:56.200
Divide both sides by 1/2
and you get T1 is

00:08:56.200 --> 00:09:03.350
equal to 200 Newtons.

00:09:03.350 --> 00:09:06.800
So now we've got to figure out
what the tension in this

00:09:06.800 --> 00:09:08.870
second wire is.

00:09:08.870 --> 00:09:10.790
And we also, there's
another clue here.

00:09:10.790 --> 00:09:14.640
This point isn't moving left
or right, it's stationary.

00:09:14.640 --> 00:09:19.970
So we know that whatever the
tension in this wire must be,

00:09:19.970 --> 00:09:24.120
it must be being offset by a
tension or some other force in

00:09:24.120 --> 00:09:25.260
the opposite direction.

00:09:25.260 --> 00:09:29.150
And that force in the opposite
direction is the x component

00:09:29.150 --> 00:09:31.220
of the first wire's tension.

00:09:31.220 --> 00:09:34.140
So it's this.

00:09:34.140 --> 00:09:39.200
So T2 is equal to the
x component of the

00:09:39.200 --> 00:09:40.990
first wire's tension.

00:09:40.990 --> 00:09:42.410
And what's the x component?

00:09:42.410 --> 00:09:45.520
Well, it's going to be the
tension in the first wire, 200

00:09:45.520 --> 00:09:51.250
Newtons times the cosine
of 30 degrees.

00:09:51.250 --> 00:09:53.900
It's adjacent over hypotenuse.

00:09:53.900 --> 00:09:55.320
And that's square root
of 3 over 2.

00:09:55.320 --> 00:10:00.410
So it's 200 times the square
root of 3 over 2, which equals

00:10:00.410 --> 00:10:03.590
100 square root of 3.

00:10:03.590 --> 00:10:07.940
So the tension in this wire is
100 square root of 3, which

00:10:07.940 --> 00:10:12.590
completely offsets to the left
and the x component of this

00:10:12.590 --> 00:10:16.540
wire is 100 square root of
3 Newtons to the right.

00:10:16.540 --> 00:10:17.446
Hopefully I didn't
confuse you.

00:10:17.446 --> 00:10:19.250
See you in the next video.


Newton's Laws

WEBVTT
Kind: captions
Language: en

00:00:00.890 --> 00:00:03.440
Let's do some more problems
involving Newton's laws.

00:00:03.440 --> 00:00:06.770
OK, so this problem that I am
picking from-- I think it's

00:00:06.770 --> 00:00:08.210
from Oregon University.

00:00:08.210 --> 00:00:10.890
It's zebu.uoregan.edu.

00:00:10.890 --> 00:00:13.260
I want to give them credit
for their problem.

00:00:13.260 --> 00:00:16.585
Let's see, let me draw-- this
is going to be the ground.

00:00:20.050 --> 00:00:23.080
And then the problem says
that I have a train--

00:00:23.080 --> 00:00:27.020
so this is a train.

00:00:27.020 --> 00:00:29.730
Try my best to draw a train.

00:00:29.730 --> 00:00:31.460
Something like that.

00:00:31.460 --> 00:00:32.810
And there's smoke coming out.

00:00:32.810 --> 00:00:36.540
And it says that the
train of mass 6.84.

00:00:36.540 --> 00:00:44.910
So let's say the mass of the
train is 6.84 times 10 to the

00:00:44.910 --> 00:00:49.310
sixth kilograms. That's the
mass of this train.

00:00:49.310 --> 00:00:52.650
Is moving at a speed of 80
kilometers per hour.

00:00:52.650 --> 00:00:55.360
So it's velocity, we could say
it's initial velocity because

00:00:55.360 --> 00:00:57.270
I think it's going to ask me
what happens to its velocity.

00:00:57.270 --> 00:01:03.630
So its initial velocity
is equal to 80

00:01:03.630 --> 00:01:08.395
kilometers per hour.

00:01:08.395 --> 00:01:13.860
So it's going this way at
80 kilometers per hour.

00:01:13.860 --> 00:01:17.380
The brakes, which produce a net
backward force of-- so the

00:01:17.380 --> 00:01:21.900
brakes are going to pull
backwards-- a net backward

00:01:21.900 --> 00:01:29.640
force of-- so force is equal
to 1.93 times 10

00:01:29.640 --> 00:01:31.200
to the sixth Newtons.

00:01:31.200 --> 00:01:33.490
That's 1.93 times 10 to
the sixth Newtons.

00:01:33.490 --> 00:01:36.760
I know it looks a little
bit messy.

00:01:36.760 --> 00:01:41.520
The brakes are going to be
applied for 25 seconds.

00:01:41.520 --> 00:01:45.460
I'll say force sub b or the
force of the brakes is 1.93

00:01:45.460 --> 00:01:48.110
times 10 to the sixth Newtons.

00:01:48.110 --> 00:01:53.760
And then, they will be applied
for 25 seconds.

00:01:53.760 --> 00:01:56.330
So the first question to
ask is, what is the new

00:01:56.330 --> 00:01:57.100
speed of the train?

00:01:57.100 --> 00:01:59.025
So what is the speed of the
train after the brakes are

00:01:59.025 --> 00:02:00.820
applied for 25 seconds?

00:02:00.820 --> 00:02:05.440
Well, all we have to figure out
is how fast is the train

00:02:05.440 --> 00:02:06.900
decelerating?

00:02:06.900 --> 00:02:09.060
Or what rate of deceleration
does this

00:02:09.060 --> 00:02:10.090
backward force create?

00:02:10.090 --> 00:02:12.340
So we just go back
to our basics.

00:02:12.340 --> 00:02:15.640
Let me switch colors because
this red's getting a little

00:02:15.640 --> 00:02:16.800
monotonous.

00:02:16.800 --> 00:02:20.480
Force is equal to mass
times acceleration.

00:02:20.480 --> 00:02:23.530
Probably the easiest formula
ever to memorize.

00:02:23.530 --> 00:02:25.300
And what's the backward force?

00:02:25.300 --> 00:02:34.960
The backward force is 1.93
times 10 to the sixth.

00:02:34.960 --> 00:02:38.370
And that equals mass
times acceleration.

00:02:38.370 --> 00:02:39.130
What's the mass?

00:02:39.130 --> 00:02:40.380
Mass is right here.

00:02:43.920 --> 00:02:44.570
That's a decimal.

00:02:44.570 --> 00:02:51.200
6.84 times 10 to the sixth
times the acceleration.

00:02:51.200 --> 00:02:52.520
And acceleration is what we're
trying to figure out.

00:02:52.520 --> 00:02:55.400
And remember, if the force is
pulling back, then of course,

00:02:55.400 --> 00:02:57.090
the acceleration is also
going to be backwards.

00:02:57.090 --> 00:02:59.960
Or it's going to essentially,
slow down the train.

00:02:59.960 --> 00:03:02.830
I know this is very messy, but
let's just divide-- well, we

00:03:02.830 --> 00:03:04.840
could divide both sides by 10 to
the sixth just to get that

00:03:04.840 --> 00:03:05.260
out of the way.

00:03:05.260 --> 00:03:08.550
That simplifies things
a little bit.

00:03:08.550 --> 00:03:10.540
And then we're left with
the acceleration.

00:03:10.540 --> 00:03:12.810
I'm just flipping both sides
of this equation.

00:03:12.810 --> 00:03:20.890
Acceleration is equal to 1.93
divided by 6.84. soon.

00:03:20.890 --> 00:03:25.110
And this is going to be given in
meters per second squared.

00:03:25.110 --> 00:03:27.660
And I just want to make sure
I got my units right.

00:03:27.660 --> 00:03:31.390
Because this number here, the
mass was in kilograms. And

00:03:31.390 --> 00:03:32.820
then the force was
given in units.

00:03:32.820 --> 00:03:35.660
So the acceleration will be in
meters per second squared.

00:03:35.660 --> 00:03:37.640
And let's get the handy
calculator here because this

00:03:37.640 --> 00:03:40.330
looks-- I don't want to waste
your time reviewing how to

00:03:40.330 --> 00:03:41.350
divide decimals.

00:03:41.350 --> 00:03:42.340
Although, that might
not be a bad

00:03:42.340 --> 00:03:43.960
review; most people forget.

00:03:43.960 --> 00:03:50.980
1.93 divided by 6.-- whoops.

00:03:50.980 --> 00:03:52.270
I think I messed up.

00:03:52.270 --> 00:03:52.830
Let me see.

00:03:52.830 --> 00:04:02.850
1.93 divided by 6.84 is 0.282.

00:04:02.850 --> 00:04:09.040
So its acceleration equals-- is
equal to 0.28-- let's just

00:04:09.040 --> 00:04:13.560
say 282 meters per
second squared.

00:04:13.560 --> 00:04:15.070
And so how fast is the
train going to be

00:04:15.070 --> 00:04:17.800
going after 25 seconds?

00:04:17.800 --> 00:04:19.329
So what's the change
in velocity?

00:04:19.329 --> 00:04:23.630
Well the change in velocity
is equal to

00:04:23.630 --> 00:04:26.050
acceleration times time.

00:04:26.050 --> 00:04:28.240
And if we take its velocity as
a positive number, then its

00:04:28.240 --> 00:04:29.760
acceleration will be a negative
number, right?

00:04:29.760 --> 00:04:32.920
Because it's going to be going
in the opposite direction.

00:04:32.920 --> 00:04:34.400
Could have done it the
other way around.

00:04:34.400 --> 00:04:37.950
So the change in velocity is
going to be equal to this

00:04:37.950 --> 00:04:42.080
minus 0.282 meters per
second squared.

00:04:42.080 --> 00:04:43.510
I won't write the units
here because I'm

00:04:43.510 --> 00:04:44.750
running out of space.

00:04:44.750 --> 00:04:51.200
Minus 0.282 meters per second
squared times 25 seconds.

00:04:51.200 --> 00:04:52.950
Let's get the calculator
back here.

00:04:52.950 --> 00:04:53.700
Where'd it go?

00:04:53.700 --> 00:04:54.140
There it is.

00:04:54.140 --> 00:04:55.430
OK.

00:04:55.430 --> 00:04:57.510
So now we want to multiply that
number we just figured

00:04:57.510 --> 00:05:01.970
out times 25.

00:05:01.970 --> 00:05:09.020
Equals 7.05.

00:05:09.020 --> 00:05:13.560
So the change in velocity
is equal to minus

00:05:13.560 --> 00:05:21.560
7.05 meters per second.

00:05:21.560 --> 00:05:24.020
OK, so let's see what's
going on.

00:05:24.020 --> 00:05:29.920
The initial speed of the train
is 80 kilometers per hour.

00:05:29.920 --> 00:05:32.000
But we have the change
in velocity given

00:05:32.000 --> 00:05:34.590
in meters per second.

00:05:34.590 --> 00:05:37.900
So we can either change this
change in velocity to

00:05:37.900 --> 00:05:40.740
kilometers per hour, or we could
go the other way around

00:05:40.740 --> 00:05:44.220
and do the- or change the
kilometers per hour into

00:05:44.220 --> 00:05:46.200
meters per second.

00:05:46.200 --> 00:05:48.810
I don't know, let's change the
meters per second into

00:05:48.810 --> 00:05:50.360
kilometers per hour.

00:05:50.360 --> 00:05:56.790
So there are how many meters--
so it'll be minus 7.05.

00:05:56.790 --> 00:05:59.920
And how many meters are
there in a kilometer?

00:05:59.920 --> 00:06:02.010
Well there are a thousand
meters in a kilometer.

00:06:02.010 --> 00:06:06.450
So it's going to be going that
number divided by a thousand

00:06:06.450 --> 00:06:06.960
kilometers.

00:06:06.960 --> 00:06:08.890
And if this is a little
confusing, I'm skipping a

00:06:08.890 --> 00:06:09.720
couple of steps.

00:06:09.720 --> 00:06:12.500
You might want to review the
video on unit conversion.

00:06:12.500 --> 00:06:14.910
And then, this is meters per
second, but we want to get

00:06:14.910 --> 00:06:17.150
kilometers per hour.

00:06:17.150 --> 00:06:18.260
How may seconds are
there in an hour?

00:06:18.260 --> 00:06:19.390
There are 3,600.

00:06:19.390 --> 00:06:20.120
right?

00:06:20.120 --> 00:06:23.220
So it'll go 3,600 times as
far in an hour as it

00:06:23.220 --> 00:06:24.740
does it in a second.

00:06:24.740 --> 00:06:28.800
Times 3,600.

00:06:28.800 --> 00:06:31.930
So we can get rid of two 0's
here and two 0's here.

00:06:31.930 --> 00:06:41.120
So it's essentially going
to be 7.05 you times 36.

00:06:41.120 --> 00:06:42.540
I just got rid of the 0's.

00:06:42.540 --> 00:06:44.640
I'll write the 0's there, just
so you don't get confused.

00:06:44.640 --> 00:06:47.640
Divided by a thousand.

00:06:47.640 --> 00:06:49.770
1, 2, 3.

00:06:49.770 --> 00:06:55.830
Equals 25.39 kilometers
per hour.

00:06:55.830 --> 00:06:57.690
So I'll just round.

00:06:57.690 --> 00:06:59.730
25 kilometers per hour.

00:06:59.730 --> 00:07:07.870
So the change in velocity is
minus 25 kilometers per hour.

00:07:07.870 --> 00:07:09.330
What's the change velocity?

00:07:09.330 --> 00:07:12.430
If it was starting off at 80
kilometers and then the change

00:07:12.430 --> 00:07:16.090
in velocity is minus 25, what's
its new velocity?

00:07:16.090 --> 00:07:20.120
Well it's going to be 80 minus
25, the initial velocity minus

00:07:20.120 --> 00:07:21.740
the change in velocity.

00:07:21.740 --> 00:07:24.270
Or plus the change in velocity
depending on how you view that

00:07:24.270 --> 00:07:25.290
negative number.

00:07:25.290 --> 00:07:26.660
So 80 minus 25.

00:07:26.660 --> 00:07:32.920
So we could say the final
velocity is going to be 80

00:07:32.920 --> 00:07:40.040
minus 25 is 55 kilometers
per hour.

00:07:40.040 --> 00:07:42.940
Now the second part of this
question is, how far has the

00:07:42.940 --> 00:07:44.280
train traveled in this time?

00:07:44.280 --> 00:07:48.960
So essentially, how long-- how
far does the train go as it

00:07:48.960 --> 00:07:51.660
brakes from 80 kilometers
per hour to 55

00:07:51.660 --> 00:07:53.170
kilometers per hour?

00:07:53.170 --> 00:07:55.300
Well, all we have to figure out
is the average velocity

00:07:55.300 --> 00:07:57.750
and then multiply that
times the time.

00:07:57.750 --> 00:07:59.690
So what's the average
velocity here?

00:08:02.740 --> 00:08:04.720
Well actually, let me just clear
this because I think

00:08:04.720 --> 00:08:07.736
it's getting-- you don't need
a drawing of a train.

00:08:07.736 --> 00:08:11.900
Let's see if I remember my-- so
the initial velocity was 80

00:08:11.900 --> 00:08:14.110
kilometers per hour.

00:08:14.110 --> 00:08:18.950
Final velocity is 55 kilometers
per hour.

00:08:18.950 --> 00:08:25.120
So the average velocity is going
to be 80 plus 55 over 2,

00:08:25.120 --> 00:08:25.710
which is, what?

00:08:25.710 --> 00:08:28.590
67 and 1/2, something
like that.

00:08:28.590 --> 00:08:35.730
67 and 1/2 kilometers
per hour.

00:08:35.730 --> 00:08:37.549
And then if we want to figure
out the total distance it

00:08:37.549 --> 00:08:41.870
travels, distance is equal to
average velocity times time.

00:08:41.870 --> 00:08:42.929
The average velocity?

00:08:42.929 --> 00:08:45.700
Well, we have the time
in seconds, right?

00:08:45.700 --> 00:08:48.280
So we have 25 seconds here.

00:08:48.280 --> 00:08:50.810
But this is given in kilometers
per hour.

00:08:50.810 --> 00:08:54.770
So we want to convert this into,
maybe meters per second.

00:08:54.770 --> 00:08:58.310
So what we do is take
the 67.5, it's going

00:08:58.310 --> 00:08:59.870
to go times a thousand.

00:08:59.870 --> 00:09:00.040
that's.

00:09:00.040 --> 00:09:01.540
How many meters it's
going to go.

00:09:01.540 --> 00:09:04.535
And then in a second it's going
to go 1/3,600 of that.

00:09:04.535 --> 00:09:07.530
So let's see what
that gives us.

00:09:07.530 --> 00:09:17.305
67.5 times 1,000 divided
by 3,600.

00:09:17.305 --> 00:09:23.380
18.75.

00:09:23.380 --> 00:09:25.340
That's the meters per second.

00:09:25.340 --> 00:09:28.180
That's the average velocity
in meters per second.

00:09:28.180 --> 00:09:31.800
And we go for 25 seconds, so
what does that give us?

00:09:31.800 --> 00:09:38.625
Times 25 equals 468 meters.

00:09:38.625 --> 00:09:42.790
468.75 meters.

00:09:42.790 --> 00:09:46.370
That's how far it traveled
just to brake from 80

00:09:46.370 --> 00:09:49.750
kilometers per hour to 55
kilometers per hour.

00:09:49.750 --> 00:09:51.670
And it took 25 seconds.

00:09:51.670 --> 00:09:53.320
Hopefully I haven't confused
you too much.

00:09:53.320 --> 00:09:54.570
See