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.