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

physics playlist, we only just
figured out the magnitude of

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

torque is actually
a vector and it's

direction can be found.

And that is because torque is
defined as the cross product

between the radial distance from
your axis of rotation and

the rotational force
being applied.

So these are both vectors.

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

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

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

Except now with the cross
product, besides just the

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

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

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

I don't know how intuitive
it really is.

But what did I teach you
before about torque?

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

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

So it would rotate around
this object.

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

Let's say that distance is 10.

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

equal to 10.

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

and F I will do in yellow.

I apply some force F.

Let me draw it straight.

I apply some force
F at some angle.

That's my force F.

It's also a vector.

It has magnitude
and direction.

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

apply a force of 7 newtons.

Let me make it more
interesting.

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

And I just threw that out there
because I think the

numbers will all work out.

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

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

radians this time.

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

that's 60 degrees.

pi over 3 radians is
equal to theta.

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

torque, what is the torque
around this pivot?

Or how much torque is being
applied by this force?

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

really the only hard part about
these problems is that

you don't just multiply the
entire rotational force times

the distance from the
axis of rotation.

You have to multiply the
component of that force that

is actually doing the rotation,
or the component of

the force that is perpendicular
to this rotating

arm, or perpendicular
to this moment arm.

So how do we figure that out?

Well, the component of this
force that is perpendicular to

this arm-- I can visually
draw it here.

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

I could draw it there.

I could also draw
it here, right?

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

that is perpendicular to this
rotating arm, and the

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

care about that.

That's not contributing
to the rotation.

The only thing that is
contributing to the rotation

is this component
of the force.

And what is the magnitude of
this vector right here?

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

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

triangle down here.

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

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

I know it's hard to read.

What is this length
right here?

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

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

can think about it.

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

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

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

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

root of 3 over 2.

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

3, so that equals 3/2.

So the magnitude of this force
vector that is perpendicular,

the component that is
perpendicular to the arm, is

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

the torque.

It's 3/2 newtons times
10 meters.

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

little bit more careful with
my notation right now to

remind you that torque actually
is a vector, or you

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

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

So what is the magnitude
of the torque vector?

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

just drew this vector
here just to

show you the component.

I could just shift the vector
here because this is actually

where the force is
being applied.

You could draw that same vector
here because you can

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

that makes it a little
bit clearer.

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

your pivot arm, so times
10 meters, and so

that is equal to what?

15 newton meters.

So the magnitude of the torque
is 15 newton meters.

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

bit familiar.

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

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

the torque.

But what if we wanted to
know the direction?

And that's where the cross
product comes in.

So what was the definition
of the cross product?

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

times the magnitude of F times
sine of the smallest angle

between them times some vector
that is perpendicular to both.

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

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

So these don't specify
the direction.

The direction is completely
specified by this unit vector,

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

pointing in some direction.

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

part that just gives us
magnitudes, we just calculated

that using what we knew
before of torques.

The magnitude of our force
vector times sine of theta,

that gave us the component of
the force vector that is

perpendicular to the arm.

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

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

We can leave out the newton
meters for now.

15, and then its direction
is this vector that we

specified by n.

We can call it the
normal vector.

And what do we know
about this vector?

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

perpendicular to F.

And the only way that I
can visualize in our

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

both this and this is
if it pops in or out

of this page, right?

Because both of these vectors
are in the plane that are

defined by our video.

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

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

perpendicular to both
of these vectors.

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

into the page?

We use the right hand
rule, right?

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

F is our middle finger, and
whichever direction our thumb

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

the cross product.

So let's draw it.

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

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

hand sitting on top
of this screen.

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

right hand.

Remember, it only works
with your right hand.

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

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

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

you to draw this.

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

know what this is.

So this is the nail on
my index finger.

This is the nail on
my middle finger.

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

My thumb is going to
be popping out.

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

Hopefully, that makes
some sense, right?

That's the palm of my hand.

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

hopefully, that makes
some sense.

This is my index finger.

This is the middle finger.

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

that the torque is actually
pointing out of the page.

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

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

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

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

See you in the next video.