Learn about Cross Product and Torque Video

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Learn about Cross Product and Torque

All of the torque problems I have done so far on the Physics playlist. We only just figure out the magnitude of torque but - well frankly because that is what normally matters. But torque is actually a vector and its 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 of vectors the first time. And then I will show you how that is really the same thing as what we are doing here with the cross-product except now at the cross-product. Besides just the magnitude for torque, we are also getting the direction but then we will also see that direction is a little bit. It is just kind of the definition of the direction of torque. I don’t how intuitive it really is, but what I teach you before about torque? Well I said, if I had some arm and let’s say it could be the hand of a clock or it’s pinned down to the wall there. So it rotate around this object and then -- let’s say at some distance are 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 apply some force F at some angle. That is my force F, it is also a vector it has magnitude and direction and let us say this is 10m. 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 drew that out there because I think the numbers are all will work out. And let us say that the angle between my force and the lever arm or the arm that is rotating. Let’s stick to radiance this time, let us say it’s π/3 but if you need to visualize that that is 60°. π/3 radiance it says it equal to theta and so it just basically what we are 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 and then we learn moments, you guys the hard part about this problem is that. You don’t just multiply the entire rotational force times the distance from the axis 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, that component of this, of this force that is perpendicular to this arm. I can visually draw it here and it would be like look something like this, l can draw out there and I could also draw it here. This would be the component or this would be the component to that is perpendicular to this rotating arm. And the component that is parallel would be this but we don’t care about that. That is kind 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 three - this is π/3 radiance or 60°. Then this is a right angle, it is π/3 and it hard to read. What is this line right here? Well it is 369 triangle and we know that this length here. I mean there is couple of ways you could think about it. Now we know Trigonometry, we know now this is just a square root three times the sine of π/3 or the sine of 60° and so that equals to the square root of three. Sine of π/3 or sine of 60° is square root of 3/2, so the square root of 3 times the square root of three is just three. So that equals 3/2, so the magnitude of this force vector that is perpendicular are there any component that is perpendicular to the arm is 3/2 Newtons. And now we can figure out the magnitude of the torque, its 3/2 Newtons (10m), so we know the magnitude of the torque. And I am being a little bit more careful with my notation right now, that are reminding you the torque actually is a vector. Or you can almost -- they use this term pseudo-vector because- I will, won’t go into that. So what is the magnitude of the torque vector? Well its 3/2 Newtons times the distance and remember I just drew this vector here just to show you the component. I couldn’t just shift the vector here because this is actually where the force is being applied - you could draw that some vector here. Because you can shift vectors around, so this is also 3/2 Newtons, maybe that makes it a little bit clear. So its 3/2 Newtons times the distance that you are from your pivot arms, so the times 10m 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 learn when we learn moments and torque. But all we did now is we figure out the magnitude of torque, but what if we wanted to know the direction. And that is where the cross- product comes in, so what is 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 is where is going help, because all of these right here - these are all scale or quantities right? So this won’t specify the directions, the direction is completely specified by this unit vector. Now you know vectors is just a vector of magnitude one that is pointing in some direction. Well look at 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 could leave out the Newtons meters for now and then its direction is this vector that we specified by N. We can call the normal vector and what do we know about this vector? It is perpendicular to both R and it is perpendicular F and the only way that I can visualize in our three dimensional universe. A vector that is perpendicular to both this and this if it pops in or out of this page. Because both of these vectors are in the plane that are defined by our video, so if I have a vector that is perpendicular to your screen whatever you are watching on. Then this is 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, the right hand rule, we take R is our index finger. F is our middle finger and then which ever 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, that is my index finger and you kind of imagine like your hand sitting on top of this screen. So that is my index finger, representing R and this is my right hand, remember it only works with your right hand. If you do with your left hand is going to be the opposite and then my middle finger is going to the direction of F and then the rest of my fingers. And I encourage you to draw this, so if I were to draw - let me draw my nails so you just you know what this is. So this is the nail on my index finger that is the nail on my middle finger. And so in this situation, where is my thumb is going to be? My thumb is going to be popping out, I wish I could. And that is the nail of my thumb, hopefully that makes some sense that is the palm --I could drawing. But hopefully that make 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 that 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 am almost at my time limit and so there you have it. Torque as applied to or the cross-product as it is applied to torque. See you in the next video.

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All of the torque problems I have done so far on the Physics playlist. We only just figure out the magnitude of torque but - well frankly because that is what normally matters. But torque is actually a vector and its 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 of vectors the first time. And then I will show you how that is really the same thing as what we are doing here with the cross-product except now at the cross-product. Besides just the magnitude for torque, we are also getting the direction but then we will also see that direction is a little bit. It is just kind of the definition of the direction of torque. I don’t how intuitive it really is, but what I teach you before about torque? Well I said, if I had some arm and let’s say it could be the hand of a clock or it’s pinned down to the wall there. So it rotate around this object and then -- let’s say at some distance are 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 apply some force F at some angle. That is my force F, it is also a vector it has magnitude and direction and let us say this is 10m. 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 drew that out there because I think the numbers are all will work out. And let us say that the angle between my force and the lever arm or the arm that is rotating. Let’s stick to radiance this time, let us say it’s π/3 but if you need to visualize that that is 60°. π/3 radiance it says it equal to theta and so it just basically what we are 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 and then we learn moments, you guys the hard part about this problem is that. You don’t just multiply the entire rotational force times the distance from the axis 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, that component of this, of this force that is perpendicular to this arm. I can visually draw it here and it would be like look something like this, l can draw out there and I could also draw it here. This would be the component or this would be the component to that is perpendicular to this rotating arm.

And the component that is parallel would be this but we don’t care about that. That is kind 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 three - this is π/3 radiance or 60°. Then this is a right angle, it is π/3 and it hard to read. What is this line right here? Well it is 369 triangle and we know that this length here. I mean there is couple of ways you could think about it. Now we know Trigonometry, we know now this is just a square root three times the sine of π/3 or the sine of 60° and so that equals to the square root of three.

Sine of π/3 or sine of 60° is square root of 3/2, so the square root of 3 times the square root of three is just three. So that equals 3/2, so the magnitude of this force vector that is perpendicular are there any component that is perpendicular to the arm is 3/2 Newtons. And now we can figure out the magnitude of the torque, its 3/2 Newtons (10m), so we know the magnitude of the torque. And I am being a little bit more careful with my notation right now, that are reminding you the torque actually is a vector. Or you can almost -- they use this term pseudo-vector because- I will, won’t go into that.

So what is the magnitude of the torque vector? Well its 3/2 Newtons times the distance and remember I just drew this vector here just to show you the component. I couldn’t just shift the vector here because this is actually where the force is being applied - you could draw that some vector here. Because you can shift vectors around, so this is also 3/2 Newtons, maybe that makes it a little bit clear. So its 3/2 Newtons times the distance that you are from your pivot arms, so the times 10m 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 learn when we learn moments and torque.

But all we did now is we figure out the magnitude of torque, but what if we wanted to know the direction. And that is where the cross- product comes in, so what is 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 is where is going help, because all of these right here - these are all scale or quantities right? So this won’t specify the directions, the direction is completely specified by this unit vector. Now you know vectors is just a vector of magnitude one that is pointing in some direction. Well look at 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 could leave out the Newtons meters for now and then its direction is this vector that we specified by N. We can call the normal vector and what do we know about this vector? It is perpendicular to both R and it is perpendicular F and the only way that I can visualize in our three dimensional universe. A vector that is perpendicular to both this and this if it pops in or out of this page. Because both of these vectors are in the plane that are defined by our video, so if I have a vector that is perpendicular to your screen whatever you are watching on.

Then this is 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, the right hand rule, we take R is our index finger. F is our middle finger and then which ever 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, that is my index finger and you kind of imagine like your hand sitting on top of this screen. So that is my index finger, representing R and this is my right hand, remember it only works with your right hand. If you do with your left hand is going to be the opposite and then my middle finger is going to the direction of F and then the rest of my fingers. And I encourage you to draw this, so if I were to draw - let me draw my nails so you just you know what this is.

So this is the nail on my index finger that is the nail on my middle finger. And so in this situation, where is my thumb is going to be? My thumb is going to be popping out, I wish I could. And that is the nail of my thumb, hopefully that makes some sense that is the palm --I could drawing. But hopefully that make 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 that 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 am almost at my time limit and so there you have it. Torque as applied to or the cross-product as it is applied to torque. See you in the next video.

Transcription by: Scribe4you Transcription Services