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Hi, welcome to module 8 of an Introduction to Engineering Mechanics. Today we're

Â going to review the math operation of a cross product. And we're going to

Â visualize the direction of a vector resulting from a cross product, using the

Â right hand rule. We'll use the clock method and the determinate method for

Â performing the cross product operation. We'll define what a moment or a torque is,

Â and we'll define the calculation of a moment or a torque using the a scaler

Â method and a vector method. As a math review, here's the definition of a cross

Â product. If I want to take the vector A and I want to cross it with vector B to

Â find a resulting vector C, I take the magnitude of the vector A times the

Â magnitude of the vector B times the sine of the angle between them in a normal

Â direction to the plane of vector A and B. And so we use the, what we call the right

Â hand rule for direction. Here's a, here's an example of a, a cross product. And so

Â I'm going to cross vector r with f. And so, using the right hand rule, I have a

Â position vector r and a force vector f, and an angle between them. And to find the

Â normal direction, what I so is, I point my fingers in the direction of the r. Vector.

Â I cross it with the F vector. And where my thumb points in my right hand is the

Â normal direction out of the plane for the, the cross product direction. So let's go

Â ahead and do a cross product calculation. We'll use the example of the definition

Â for a moment, r cross F. So I can have a position vector that has an x, y and a z

Â component, and a force vector that has an x, y and a z component. both expressed in

Â Cartesian coordinates. And the first method we could use is the clock method.

Â The clock method says that if I cross I with J and a, I get K. So if, so if I turn

Â in a clockwise direction, right here, I get a positive value. So i cross j is k, j

Â cross k is i, k cross i is j. If I go in the opposite direction, counter clockwise,

Â I get a negative result. So i cross k is minus j. K cross J is minus I, and J cross

Â I is minus K. So let's use that rule, and we'll take the moment about point P. Okay.

Â I cross i is 0. i cross j is k. So we have r x Fy times k. Then I have i cross k.

Â Well, i cross k is minus j, so I have rxFz times minus k minus, excuse me, minus j.

Â So I've got minus rx fz and j, okay? And I, let's take the next term. j cross i is

Â minus k, so I'm going to have minus ry, fx, minus ry. Fx, and j cross i, I said,

Â was minus k. And then j cross j is 0. J cross k is i, so I have plus ry Fz. I and

Â then finally k cross i is j so that's plus rz Fx j. K cross j is minus i so that's

Â minus rz. F y i. And then finally k cross k is, is 0. So that's my, my result.

Â Another method for doing this is, is by finding the determinant of a matrix where

Â I put i j k in the first row, the components of r in the second row, and the

Â components of F in the third row. , to, to take that determinant, there's a number of

Â ways for a 3 by 3, an easy way to do it is to add 2 more columns with the first 2

Â columns inside the matrix. So I'm going to use i, j, r, x, r, y, f, x, Fy.

Â Sometimes folks call this the, the basket weave method. And so, what we do is we

Â multiply this diagonal, and add it to this diagonal, and add it to this diagonal, and

Â then subtract this diagonal, and this diagonal, and this diagonal. So let's try

Â it. We've got this diagonal is. Ry, fz i, so if I look up at my answer, ry fz i is

Â that term. And then I add rz fx j. And so if I look up again, rz fx j is that term.

Â And then I have rx Fy k. And there's rx Fy k. And now I subtract this diagonal, Fx ry

Â k, which is that term, and then minus rz Fy i. Which is that term. And then minus

Â rxFzj which is that term. And so you see you get the same result. So either way and

Â you can also use other methods to find the determinant and whatever works for you

Â from your math review. Okay. Let's now define what a moment or a torque

Â is in your own words. So take a minute. Write it down on a piece of paper. And

Â then come back and we'll, we'll discuss it. Okay.

Â Now that you've answered that question, a moment or a torque, or a force. Let's take

Â this Wrench here, okay? It's a tendency of a force to cause a rotation about a point

Â of axis. So, if I take a point or an axis down here at this end, and I push with a

Â force here, I'm getting a torque, or a moment., due to that force. And so that's,

Â that's the definition of a torque or a moment, the tendency of a force to cause a

Â rotation about a point or an axis. So let's look at that definition in 2 ways.

Â First of all is a scalar method. The magnitude of a moment is equal to the line

Â of action. Or the force, the magnitude of the force * a perpendicular distance from

Â the point about which you're rotating. To the line of action of the force so d

Â perpendicular defined is the perpendicular distance from the point of rotation. Here

Â this is point P, to the line of action of the force causing the rotation. So, the

Â longer this distance perpendicular distance is the larger the magnitude of

Â the moment will be, or the larger the force is, the large the, magnitude of the

Â moment will be. And so let's look at a demonstration here. So If we take my joint

Â here as being point p, and I have a barbell out here that weighs 150 pounds

Â (no, it doesn't actually weigh 150 pounds, it's only eight pounds); but if I have my

Â arm fully extended, then the perpendicular distance Is longer than if I have it down

Â here. So I get more of a moment. And you can try this with a heavy object at home,

Â or wherever you're at. Take a heavy object in your, in your hand, and put it out

Â there. And so, the line of the, the line of action of the force is always straight

Â down due to gravity. And the perpendicular distance is less the lower we go. And so

Â you get more magnitude of moment up here than we do down here. So that gives you a

Â good physical feel for the moment causing rotation about a point or an axis. We can

Â also find a moment using the vector method. And the definition is, The

Â position vector from the point about which we're rotati ng P to any point on the line

Â of action of the force and this, place, on this picture I've designated as B. And you

Â cross that with the force F and we know how to do the cross product now and

Â that'll give you. In a moment, and we'll do some examples of both methods in the

Â moment due to a force. And so I'll see you next time.

Â