Hi, welcome to module 8 of an Introduction to Engineering Mechanics. Today we're going to review the math operation of a cross product. And we're going to visualize the direction of a vector resulting from a cross product, using the right hand rule. We'll use the clock method and the determinate method for performing the cross product operation. We'll define what a moment or a torque is, and we'll define the calculation of a moment or a torque using the a scaler method and a vector method. As a math review, here's the definition of a cross product. If I want to take the vector A and I want to cross it with vector B to find a resulting vector C, I take the magnitude of the vector A times the magnitude of the vector B times the sine of the angle between them in a normal direction to the plane of vector A and B. And so we use the, what we call the right hand rule for direction. Here's a, here's an example of a, a cross product. And so I'm going to cross vector r with f. And so, using the right hand rule, I have a position vector r and a force vector f, and an angle between them. And to find the normal direction, what I so is, I point my fingers in the direction of the r. Vector. I cross it with the F vector. And where my thumb points in my right hand is the normal direction out of the plane for the, the cross product direction. So let's go ahead and do a cross product calculation. We'll use the example of the definition for a moment, r cross F. So I can have a position vector that has an x, y and a z component, and a force vector that has an x, y and a z component. both expressed in Cartesian coordinates. And the first method we could use is the clock method. The clock method says that if I cross I with J and a, I get K. So if, so if I turn in a clockwise direction, right here, I get a positive value. So i cross j is k, j cross k is i, k cross i is j. If I go in the opposite direction, counter clockwise, I get a negative result. So i cross k is minus j. K cross J is minus I, and J cross I is minus K. So let's use that rule, and we'll take the moment about point P. Okay. I cross i is 0. i cross j is k. So we have r x Fy times k. Then I have i cross k. Well, i cross k is minus j, so I have rxFz times minus k minus, excuse me, minus j. So I've got minus rx fz and j, okay? And I, let's take the next term. j cross i is minus k, so I'm going to have minus ry, fx, minus ry. Fx, and j cross i, I said, was minus k. And then j cross j is 0. J cross k is i, so I have plus ry Fz. I and then finally k cross i is j so that's plus rz Fx j. K cross j is minus i so that's minus rz. F y i. And then finally k cross k is, is 0. So that's my, my result. Another method for doing this is, is by finding the determinant of a matrix where I put i j k in the first row, the components of r in the second row, and the components of F in the third row. , to, to take that determinant, there's a number of ways for a 3 by 3, an easy way to do it is to add 2 more columns with the first 2 columns inside the matrix. So I'm going to use i, j, r, x, r, y, f, x, Fy. Sometimes folks call this the, the basket weave method. And so, what we do is we multiply this diagonal, and add it to this diagonal, and add it to this diagonal, and then subtract this diagonal, and this diagonal, and this diagonal. So let's try it. We've got this diagonal is. Ry, fz i, so if I look up at my answer, ry fz i is that term. And then I add rz fx j. And so if I look up again, rz fx j is that term. And then I have rx Fy k. And there's rx Fy k. And now I subtract this diagonal, Fx ry k, which is that term, and then minus rz Fy i. Which is that term. And then minus rxFzj which is that term. And so you see you get the same result. So either way and you can also use other methods to find the determinant and whatever works for you from your math review. Okay. Let's now define what a moment or a torque is in your own words. So take a minute. Write it down on a piece of paper. And then come back and we'll, we'll discuss it. Okay. Now that you've answered that question, a moment or a torque, or a force. Let's take this Wrench here, okay? It's a tendency of a force to cause a rotation about a point of axis. So, if I take a point or an axis down here at this end, and I push with a force here, I'm getting a torque, or a moment., due to that force. And so that's, that's the definition of a torque or a moment, the tendency of a force to cause a rotation about a point or an axis. So let's look at that definition in 2 ways. First of all is a scalar method. The magnitude of a moment is equal to the line of action. Or the force, the magnitude of the force * a perpendicular distance from the point about which you're rotating. To the line of action of the force so d perpendicular defined is the perpendicular distance from the point of rotation. Here this is point P, to the line of action of the force causing the rotation. So, the longer this distance perpendicular distance is the larger the magnitude of the moment will be, or the larger the force is, the large the, magnitude of the moment will be. And so let's look at a demonstration here. So If we take my joint here as being point p, and I have a barbell out here that weighs 150 pounds (no, it doesn't actually weigh 150 pounds, it's only eight pounds); but if I have my arm fully extended, then the perpendicular distance Is longer than if I have it down here. So I get more of a moment. And you can try this with a heavy object at home, or wherever you're at. Take a heavy object in your, in your hand, and put it out there. And so, the line of the, the line of action of the force is always straight down due to gravity. And the perpendicular distance is less the lower we go. And so you get more magnitude of moment up here than we do down here. So that gives you a good physical feel for the moment causing rotation about a point or an axis. We can also find a moment using the vector method. And the definition is, The position vector from the point about which we're rotati ng P to any point on the line of action of the force and this, place, on this picture I've designated as B. And you cross that with the force F and we know how to do the cross product now and that'll give you. In a moment, and we'll do some examples of both methods in the moment due to a force. And so I'll see you next time.