WEBVTT

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In all of the torque problems
I've done so far in the

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physics playlist, we only just
figured out the magnitude of

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torque, frankly because that's
what normally matters, but

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torque is actually
a vector and it's

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direction can be found.

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And that is because torque is
defined as the cross product

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between the radial distance from
your axis of rotation and

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the rotational force
being applied.

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So these are both vectors.

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So let's take a look at how I
taught you vectors the first

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time, and then I'll show you
how that's really the same

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thing as what we're doing here
with the cross product.

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Except now with the cross
product, besides just the

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magnitude for torque, we're also
getting the direction.

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But then we'll also see that
direction is a little bit--

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it's just the definition of
the direction of torque.

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I don't know how intuitive
it really is.

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But what did I teach you
before about torque?

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Well, let's say I had some arm,
and let's say this could

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be the hand of a clock or it's
pinned down to the wall there.

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So it would rotate around
this object.

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Let's say it's some distance,
r, from the pivot.

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Let's say that distance is 10.

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This is the same thing as r,
and the magnitude of r is

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equal to 10.

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At some distance 10 from the
pivot, I apply some force F,

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and F I will do in yellow.

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I apply some force F.

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Let me draw it straight.

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I apply some force
F at some angle.

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That's my force F.

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It's also a vector.

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It has magnitude
and direction.

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Let's say that this is 10
meters, and let's say that I

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apply a force of 7 newtons.

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Let me make it more
interesting.

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Let's say I apply a force of
square root of 3 newtons.

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And I just threw that out there
because I think the

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numbers will all work out.

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And let's say that the angle
between my force and the lever

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arm, or the arm that's
rotating-- let's stick to

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radians this time.

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Let's say it's pi over 3, but if
you need to visualize that,

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that's 60 degrees.

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pi over 3 radians is
equal to theta.

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And so just based on what we
already know about moments or

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torque, what is the torque
around this pivot?

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Or how much torque is being
applied by this force?

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And when we learn torque or we
learn moments, we realize

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really the only hard part about
these problems is that

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you don't just multiply the
entire rotational force times

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the distance from the
axis of rotation.

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You have to multiply the
component of that force that

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is actually doing the rotation,
or the component of

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the force that is perpendicular
to this rotating

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arm, or perpendicular
to this moment arm.

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So how do we figure that out?

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Well, the component of this
force that is perpendicular to

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this arm-- I can visually
draw it here.

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Let's see, it would look
something like this.

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I could draw it there.

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I could also draw
it here, right?

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This would be the component, or
this would be the component

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that is perpendicular to this
rotating arm, and the

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component that is parallel would
be this, but we don't

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care about that.

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That's not contributing
to the rotation.

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The only thing that is
contributing to the rotation

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is this component
of the force.

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And what is the magnitude of
this vector right here?

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The component of vector F that
is perpendicular to this arm.

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Well, if this angle--
let me draw a little

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triangle down here.

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If this is square root of 3,
this is pi over 3 radians, or

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60 degrees, and this is a right
angle, it's pi over 3.

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I know it's hard to read.

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What is this length
right here?

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Well, it's a 30-60-90 triangle,
and we know that

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this length here-- I mean,
there's a couple of ways you

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can think about it.

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Now that we know trigonometry,
we know that this is just the

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square root of 3 times the sine
of pi over 3, or the sine

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of 60 degrees, and so that
equals the square root of 3.

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Sine of pi over 3, or sine
of 60 degrees, is square

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root of 3 over 2.

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So the square root of 3 times
the square root of 3 is just

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3, so that equals 3/2.

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So the magnitude of this force
vector that is perpendicular,

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the component that is
perpendicular to the arm, is

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3/2 newtons, and now we can
figure out the magnitude of

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the torque.

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It's 3/2 newtons times
10 meters.

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So we know the magnitude of the
torque, and I'm being a

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little bit more careful with
my notation right now to

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remind you that torque actually
is a vector, or you

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can almost view it as, they use
this term pseudovector,

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because it's kind of a-- well,
anyway, I won't go into that.

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So what is the magnitude
of the torque vector?

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Well, it's 3/2 newtons times the
distance, and remember, I

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just drew this vector
here just to

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show you the component.

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I could just shift the vector
here because this is actually

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where the force is
being applied.

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You could draw that same vector
here because you can

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shift vectors around, so this is
also 3/2 newtons and maybe

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that makes it a little
bit clearer.

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So it's 3/2 newtons times the
distance that you are from

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your pivot arm, so times
10 meters, and so

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that is equal to what?

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15 newton meters.

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So the magnitude of the torque
is 15 newton meters.

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But all we did now-- and
hopefully this looks a little

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bit familiar.

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This is what we learned when we
learned moments and torque,

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but all we did now is we figured
out the magnitude of

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the torque.

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But what if we wanted to
know the direction?

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And that's where the cross
product comes in.

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So what was the definition
of the cross product?

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Cross product: r cross F, that
is equal to magnitude of r

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times the magnitude of F times
sine of the smallest angle

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between them times some vector
that is perpendicular to both.

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And this is really where it's
going to help, because all of

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these right here, these are all
scalar quantities, right?

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So these don't specify
the direction.

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The direction is completely
specified by this unit vector,

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and a unit vector is just a
vector of magnitude 1 that's

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pointing in some direction.

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Well, look, this cross product,
this part of it, the

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part that just gives us
magnitudes, we just calculated

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that using what we knew
before of torques.

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The magnitude of our force
vector times sine of theta,

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that gave us the component of
the force vector that is

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perpendicular to the arm.

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And we just multiply that times
the magnitude of r, and

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we got the magnitude of the
torque vector, which was 15.

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We can leave out the newton
meters for now.

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15, and then its direction
is this vector that we

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specified by n.

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We can call it the
normal vector.

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And what do we know
about this vector?

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It's perpendicular to both r--
this is r, right-- and it's

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perpendicular to F.

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And the only way that I
can visualize in our

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three-dimensional universe, a
vector that's perpendicular to

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both this and this is
if it pops in or out

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of this page, right?

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Because both of these vectors
are in the plane that are

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defined by our video.

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So if I'm a vector that is
perpendicular to your screen,

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whatever you're watching this
on, then it's going to be

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perpendicular to both
of these vectors.

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And how do we figure out if that
vector pops out or pops

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into the page?

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We use the right hand
rule, right?

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In the right hand rule, we
take-- r is our index finger,

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F is our middle finger, and
whichever direction our thumb

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points in tells us whether or
not we are-- the direction of

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the cross product.

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So let's draw it.

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Let me see if I can do a
good job right here.

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So if that is my index finger,
and you could imagine your

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hand sitting on top
of this screen.

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So that's my index finger
representing r, and this is my

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right hand.

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Remember, it only works
with your right hand.

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If you do your left hand, it's
going to be the opposite.

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And then my middle finger is
going to go in the direction

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of F, and then the rest of my
fingers are-- and I encourage

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you to draw this.

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So if I were to draw it-- let
me draw my nails just so you

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know what this is.

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So this is the nail on
my index finger.

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This is the nail on
my middle finger.

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And so in this situation, where
is my thumb going to be?

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My thumb is going to
be popping out.

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I wish I could-- that's
the nail of my thumb.

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Hopefully, that makes
some sense, right?

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That's the palm of my hand.

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That's the other side of my--
and I could keep drawing, but

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hopefully, that makes
some sense.

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This is my index finger.

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This is the middle finger.

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My thumb is pointing out of
the page, so that tells us

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that the torque is actually
pointing out of the page.

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So the direction of this unit
vector n is going to be out of

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the page, and we could signify
that by a circle with a dot.

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And I'm almost at my time limit,
and so there you have

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it: the cross product as it
is applied to torque.

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See you in the next video.