Hi, and welcome to Module 11 of an Introduction to Engineering Mechanics.

Today's learning outcome is to calculate the moment of a force about a line or an

axis. Here's the definition of the moment of the force about a line or an axis. And

this figure is taken again from the McGill, King text, which I referred to in

Module 6. So, the moment about a line or an axis is equal to a position vector from

the line or axis to the line of action of the force itself crossed with the force in

vector form. And then, dotted with a unit vector in the direction of the line or

axis. That result is going to give us a scaler in the overall direction of the

moment, or the about the line or axis is going to be in that u direction. And so

the, the easiest way to solve this, or learn this theory is to go ahead and solve

a problem. And so, here again, is a problem from McGill and King. We have a

chain that's carrying 20 pounds of tension and, and it holds open this trap door. And

we want to find the moment of the chain force exerted at point A about this

x-axis. And over here on the right, I've included my theory with the, the, the line

about which we're finding the moment. The line of action of the force and the

equation that we're going to use. So, first of all, what I want you to do is,

we're going to need the force in vector form. So, using what you learned in Module

4, I want you to express the 3D force in the vector form exerted by the chain at

point A. And once you've done that, I want you to go ahead and come on back.

Okay. Here's the result you should have gotten.

You have a force vector acting at point A along the line of action of the chain.

This is the force expressed in vector form. And so, now, I've got the moment

about the x-axis is going to be equal to the position vector crossed with this

force that we just found in the u direction, dotted in with the u direction

vector and it's all, overall going to be in the u direction. And so the first thing

we need to, the next thing we need to find is the position vector from the axis or

the line about which we find in the moment. Any point on that line or access

to a point on the line of action of the force. Here's the line of action of the

force. And so, I have several points here. I've got point O defined here's point B up

here. This is point A down here. Let's call this point C at this corner. And we

have r, vector r is directed from any point on the x-axis, in this case, to any

point on the line of action, I'll abbreviate LOA, of the force. And I want

to emphasize any point so I'll underline those in red. So in this case we have

several options. We could take a, the point O which is on the, the x-axis, to

point A. Or we can take a position vector from O to B, or we can take a position

vector from C to A. C being on the x-axis, A being on the line of action of the

force, or C to B. So, let's write all those down. So, we can have r from O to A,

r from C to A, r from O to B, or r from C to B, and any one of those would work. In

fact, I'm going to choose r from O to A, and I would encourage you to redo the

problem on your own using these other position vectors and see that it comes out

to be the same result. And so, we'll choose rOA. And rOA, in this case, we have

to go 4 feet in the x direction and 3 feet in the j direction. So that's 4 i plus 3 j

feet. Okay. So, recapping here, we have our force in

vector form. This is the equation for finding the moment. We've chosen r is 4 i

plus 3 j feet. So, we have this term, we have this term. All we need now is the u

vector. And the u vector is the unit vector in the direction of the line or

axis. So, the u is the vector in the direction of the line or axis, in this

case, the x-axis. And so, this one is real simple. We'll just say, therefore, again

the, the symbol for therefore this three dots. u, in this case, is just i because i

is a unit vector along the x-axis. And if you had a more complicated line of action

of the line, you should be able to easily find it from earlie r modules, the unit

vector along the direction of that line. Okay. So now, we're getting towards the

end of our problem. We now have the force vector f. We have the position vector from

O to A. We have the unit vector u, and so all I have to do is substitute in. And so,

I've got the moment about the x-axis equals, r from O to A is 4 i plus 3 j.

We're going to cross that with a vector f. Vector f is minus 8.82 i minus 3.31 j plus

17.6 k. And, we're going to dot that cross product result with the unit vector in the

u direction, which is i, in this case. I could write u or, that's also i. And then,

that whole thing. This whole thing in brackets gives us a scale result and the

direction it's going to be in the u direction or the i direction, in this

case. And so, I get the moment about the x-axis. IF I do that math, is equal to

52.8 in the i direction, and the units are pounds feet. Feet for r sub OA, and pounds

for the force, so it's pounds foot, or foot pounds. And, that's my answer. So

now, you have a technique for finding the force about a line, or the moment due to a

force about a line or an axis. And I'll see you next module.