0:00:00.000,0:00:03.400
In all of the torque problems[br]I've done so far in the

0:00:03.400,0:00:05.910
physics playlist, we only just[br]figured out the magnitude of

0:00:05.910,0:00:09.610
torque, frankly because that's[br]what normally matters, but

0:00:09.610,0:00:13.500
torque is actually[br]a vector and it's

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direction can be found.

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And that is because torque is[br]defined as the cross product

0:00:21.160,0:00:26.330
between the radial distance from[br]your axis of rotation and

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the rotational force[br]being applied.

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So these are both vectors.

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

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

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

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Except now with the cross[br]product, besides just the

0:00:41.570,0:00:43.590
magnitude for torque, we're also[br]getting the direction.

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

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it's just the definition of[br]the direction of torque.

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I don't know how intuitive[br]it really is.

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But what did I teach you[br]before about torque?

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

0:01:02.410,0:01:05.269
be the hand of a clock or it's[br]pinned down to the wall there.

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So it would rotate around[br]this object.

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Let's say it's some distance,[br]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,[br]and the magnitude of r is

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equal to 10.

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At some distance 10 from the[br]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[br]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[br]and direction.

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Let's say that this is 10[br]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[br]interesting.

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

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And I just threw that out there[br]because I think the

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numbers will all work out.

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

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arm, or the arm that's[br]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[br]you need to visualize that,

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that's 60 degrees.

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pi over 3 radians is[br]equal to theta.

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

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torque, what is the torque[br]around this pivot?

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Or how much torque is being[br]applied by this force?

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And when we learn torque or we[br]learn moments, we realize

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really the only hard part about[br]these problems is that

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you don't just multiply the[br]entire rotational force times

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the distance from the[br]axis of rotation.

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You have to multiply the[br]component of that force that

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is actually doing the rotation,[br]or the component of

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the force that is perpendicular[br]to this rotating

0:03:09.860,0:03:13.090
arm, or perpendicular[br]to this moment arm.

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So how do we figure that out?

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Well, the component of this[br]force that is perpendicular to

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this arm-- I can visually[br]draw it here.

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

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I could draw it there.

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I could also draw[br]it here, right?

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This would be the component, or[br]this would be the component

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that is perpendicular to this[br]rotating arm, and the

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

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care about that.

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That's not contributing[br]to the rotation.

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The only thing that is[br]contributing to the rotation

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is this component[br]of the force.

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And what is the magnitude of[br]this vector right here?

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

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Well, if this angle--[br]let me draw a little

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triangle down here.

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

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60 degrees, and this is a right[br]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[br]right here?

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

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

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can think about it.

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Now that we know trigonometry,[br]we know that this is just the

0:04:28.420,0:04:33.170
square root of 3 times the sine[br]of pi over 3, or the sine

0:04:33.170,0:04:36.020
of 60 degrees, and so that[br]equals the square root of 3.

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

0:04:38.790,0:04:40.820
root of 3 over 2.

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So the square root of 3 times[br]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[br]vector that is perpendicular,

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the component that is[br]perpendicular to the arm, is

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

0:05:01.580,0:05:02.135
the torque.

0:05:02.135,0:05:05.700
It's 3/2 newtons times[br]10 meters.

0:05:05.700,0:05:08.970
So we know the magnitude of the[br]torque, and I'm being a

0:05:08.970,0:05:11.200
little bit more careful with[br]my notation right now to

0:05:11.200,0:05:13.680
remind you that torque actually[br]is a vector, or you

0:05:13.680,0:05:16.520
can almost view it as, they use[br]this term pseudovector,

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

0:05:21.330,0:05:24.060
So what is the magnitude[br]of the torque vector?

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

0:05:31.930,0:05:33.830
just drew this vector[br]here just to

0:05:33.830,0:05:34.450
show you the component.

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I could just shift the vector[br]here because this is actually

0:05:36.400,0:05:37.550
where the force is[br]being applied.

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You could draw that same vector[br]here because you can

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

0:05:43.730,0:05:44.950
that makes it a little[br]bit clearer.

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

0:05:49.600,0:05:54.960
your pivot arm, so times[br]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[br]is 15 newton meters.

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

0:06:07.720,0:06:08.180
bit familiar.

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This is what we learned when we[br]learned moments and torque,

0:06:10.530,0:06:13.690
but all we did now is we figured[br]out the magnitude of

0:06:13.690,0:06:14.010
the torque.

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But what if we wanted to[br]know the direction?

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And that's where the cross[br]product comes in.

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So what was the definition[br]of the cross product?

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

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times the magnitude of F times[br]sine of the smallest angle

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between them times some vector[br]that is perpendicular to both.

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

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these right here, these are all[br]scalar quantities, right?

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So these don't specify[br]the direction.

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The direction is completely[br]specified by this unit vector,

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

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pointing in some direction.

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Well, look, this cross product,[br]this part of it, the

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part that just gives us[br]magnitudes, we just calculated

0:07:02.000,0:07:05.200
that using what we knew[br]before of torques.

0:07:05.200,0:07:08.730
The magnitude of our force[br]vector times sine of theta,

0:07:08.730,0:07:11.110
that gave us the component of[br]the force vector that is

0:07:11.110,0:07:13.110
perpendicular to the arm.

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

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

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We can leave out the newton[br]meters for now.

0:07:23.960,0:07:27.910
15, and then its direction[br]is this vector that we

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specified by n.

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We can call it the[br]normal vector.

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And what do we know[br]about this vector?

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

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perpendicular to F.

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And the only way that I[br]can visualize in our

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

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both this and this is[br]if it pops in or out

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of this page, right?

0:07:48.960,0:07:51.690
Because both of these vectors[br]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[br]perpendicular to your screen,

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

0:08:00.350,0:08:03.050
perpendicular to both[br]of these vectors.

0:08:03.050,0:08:06.150
And how do we figure out if that[br]vector pops out or pops

0:08:06.150,0:08:07.240
into the page?

0:08:07.240,0:08:10.035
We use the right hand[br]rule, right?

0:08:10.035,0:08:14.340
In the right hand rule, we[br]take-- r is our index finger,

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F is our middle finger, and[br]whichever direction our thumb

0:08:17.410,0:08:23.340
points in tells us whether or[br]not we are-- the direction of

0:08:23.340,0:08:24.980
the cross product.

0:08:24.980,0:08:26.930
So let's draw it.

0:08:26.930,0:08:31.350
Let me see if I can do a[br]good job right here.

0:08:31.350,0:08:43.409
So if that is my index finger,[br]and you could imagine your

0:08:43.409,0:08:47.130
hand sitting on top[br]of this screen.

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

0:08:51.160,0:08:51.610
right hand.

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Remember, it only works[br]with your right hand.

0:08:52.850,0:08:56.440
If you do your left hand, it's[br]going to be the opposite.

0:08:56.440,0:09:01.220
And then my middle finger is[br]going to go in the direction

0:09:01.220,0:09:06.460
of F, and then the rest of my[br]fingers are-- and I encourage

0:09:06.460,0:09:07.570
you to draw this.

0:09:07.570,0:09:10.300
So if I were to draw it-- let[br]me draw my nails just so you

0:09:10.300,0:09:11.200
know what this is.

0:09:11.200,0:09:13.470
So this is the nail on[br]my index finger.

0:09:13.470,0:09:15.470
This is the nail on[br]my middle finger.

0:09:15.470,0:09:17.960
And so in this situation, where[br]is my thumb going to be?

0:09:17.960,0:09:19.550
My thumb is going to[br]be popping out.

0:09:19.550,0:09:23.380
I wish I could-- that's[br]the nail of my thumb.

0:09:23.380,0:09:25.400
Hopefully, that makes[br]some sense, right?

0:09:25.400,0:09:26.870
That's the palm of my hand.

0:09:26.870,0:09:29.940
That's the other side of my--[br]and I could keep drawing, but

0:09:29.940,0:09:32.250
hopefully, that makes[br]some sense.

0:09:32.250,0:09:33.220
This is my index finger.

0:09:33.220,0:09:34.080
This is the middle finger.

0:09:34.080,0:09:37.440
My thumb is pointing out of[br]the page, so that tells us

0:09:37.440,0:09:40.280
that the torque is actually[br]pointing out of the page.

0:09:40.280,0:09:43.780
So the direction of this unit[br]vector n is going to be out of

0:09:43.780,0:09:49.080
the page, and we could signify[br]that by a circle with a dot.

0:09:49.080,0:09:51.950
And I'm almost at my time limit,[br]and so there you have

0:09:51.950,0:09:57.980
it: the cross product as it[br]is applied to torque.

0:09:57.980,0:09:59.820
See you in the next video.