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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â