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Welcome back to an Introduction to Engineering Mechanics. This is Module 9.

We're going to learn how to use Varignon's Theorem to sum moments today, and we are

going to solve a two-dimensional moment problem. First of all, let's look at the

definition of a moment. And, I've changed it a little bit this time. last time we

said moment about point p. It can be a moment about any point. So, I've just

changed the notation to kind of get you used to doing different notations. So, I

have the sum of the moment about A is equal to r from A to B, where B is a point

along the line of action of the force cross with the force itself. Now,

Varingnon's Theorem says, its also called Principle of Moments, that we can some

take the moment of a force is equal to the algebraic sum of the components of that

force. And so, if I have a force that I can break into two components F1 and F2,

then r cross F is the same as r cross F1 plus r cross F2. And we'll use that in, in

solving the 2D problem today. So, let's look at a, a vector solution of a problem.

Let's take this demonstration over here. I have a, a section of an iBeam, and I'm

going to pull down and to the right on a, on a three on four slope, with a 20 pound

force, and I want to see what the moment is. The tendency to cause rotation about

point O is here in the center of the, the iBeam. And so, that's the set up for the

problem. Okay, let's use the vector method for finding the moment. So, we have the

moment about point O is equal to the position vector from the point about which

we rotated, O, to a point on the line of action of the force. And the most readily

available point on the line of action of force that we know is point A, so we'll go

from r, from O to A, and then cross it with the force vector itself. And so, rOA,

you should be able to easily find on your own. Once you do that, you'll see that rOA

is two units, two inches in the i direction and three inches in the j

direction. So, this is 2 i plus 3 j. Define the force vector, we first ne ed to

find, we know its magnitude, we need to find its, its direction. And so, we have a

position vector along the line of action of the force, AF we'll call it. Goes 4

units in the i direction and minus 3 units in the j direction. So AF is equal to 4 i

minus 3 j. That means that the magnitude of AF is

equal to the square root of the sum of the squares, or 5.

So the magnitude of AF equals 5. And therefore, a unit vector in the

direction of the force is equal to position vector AF divided by its

magnitude, or in this case, 4 divided by 5 is 0.8 in the i direction, and then 3

divided by 5 is minus 0.6 in the j direction. And so, that allows us to write

the force as, in vector form with a magnitude. And its direction. Its

magnitude is 20. Its direction is 0.8 in the i direction minus 0.6 in the j

direction. And so, that's 16 i minus 12 j. And the force units are pounds, in this

case. The position vector units are up here are inches. And so, now I have rOA, I

have F, I can substitute it in and do the cross-product to find the moment about

point O. The moment about point O is then eOA cross F, or 2 i plus 3 j crossed with

16 i minus 12 j. And if you do that cross product, you'll get minus 72 in the k

direction. And the units are, the position vector is inches, the force is pounds, so

this is inch pounds in the negative k direction. If I want to just draw it with

an arrow, the negative k direction is, is back in this direction. So that means you

have a clockwise rotation of my iBeam which makes physical sense. So, I could

also write this as 72 inch pounds, and the direction is clockwise. Excuse me that's

right, I wrote it wrong. It should be clockwise, so it should be the other

direction. Clockwise in the minus k direction. Okay. So, that's the vector

method. Okay, let's redo this problem using the scalar method. And we saw that

by definition, the magnitude of the moment about point O is equal to the magnitude of

the force F times the perpendicular distance to the force F. And so, yo u'll

see here that the line of action for this force to find the, the perpendicular

distance to that line of action would be geometrically difficult. And so, what

we're going to do is we're going to take advantage again, of Varignon's Theorem.

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We're going to split this force into it's x and y components. And so, we've got F of

x, and F of y. And, then we know that this is equal to F sub x times d perpendicular

to the x force plus F of y times d perpendicular to the y force. And, in the,

some of these moments, we're going to have a sine convention for the direction that

these moments are acting. And so, I'll go ahead and arbitrarily choose clockwise is

positive. And so, we've got the moment, and I should have a moment vector symbol

there. So, I've got the magnitude of the moment about O is equal to F of x. Well,

the magnitude of F of x by similar triangles is corresponds to the 4 5th,

component of F or 4 5th times 20. And then, we have to find the, the

perpendicular distance from the line of action of the F sub x force to the point

about which we're rotating. And so, here's the line of action of the force F sub x.

And here's the perpendicular distance, 2.0. And so, we see that, that is equal to

3 inches. And so, we've got 4 5th 20 times that perpendicular distance, which is 3

inches. And, it's going to be positive because that F of x causes a clockwise

rotation about 0.0, and that's positive in accordance with my sign convention. , now

I have F sub y by its similar triangles. It's the 3 5th side. It's also going to

cause a clockwise rotation, so it's going to be positive according to my sine

convention. And like we saw in an earlier module, you could arbitrarily choose an

opposite direction and you'd get the same answer for your sign convention. And so,

we've got plus 3 5th of 20, and we've got to find that perpendicular distance. And

so, here's the line of action of the force F sub y. And so, its perpendicular

distance. The perpendicular distance from O to that line of action of that force is

d perpendicular y equals 2 inches. This is times 2. And if I multiply those out, I

get 72. Since it's positive, it means that the moment is going to be clockwise. So, I

have the moment about point O as a vector is going to be 72. Units are inch pounds,

and the direction is clockwise. Can also put that in vector form. here is my

x-axis, here's my y-axis. If I go clockwise, by the right-hand rule that's

into the board, so that's the negative k direction. So that's equal to minus 72 k

inch pounds. And so, that's the exact same answer that we got using the vector

method. And that's the end of today's module. See you next time.