Now, finally, I'll do this problem again, but now by vector methods, so

in this case, our vector diagram looks like this.

Here is our resultant force, which is passing along this line of action here, A,

B, C, passing through this point here.

And as usual, I can represent that by its horizontal and

vertical components, F1 and F2.

The radius vector I will take

as being the vector from O to A, although again, it wouldn't make any difference,

I could have taken this vector, or this vector, whatever is, is convenient.

But in this case I'll take it as here.

And here is our general expression for the moment, the cross-product of r and F.

So the radius vector is 2 units horizontal and

4 units vertical, in other words, 2i plus 4j.

The force vector in terms of its horizontal and

vertical components from the previous slide is 600 times

i cosine 40 degrees is the horizontal component.

And 600 times minus j because it's in the negative y direction, sine 40 degrees.

So, the cross-product of those two looks like this.

And again, expanding this out in all it's full glory, is equal to that.

But we realize right away that when we expand this out,

we're only going to have a component of the vector in the z direction.

In other words, out of the picture.

So the first two terms here, the i and

j terms, are going to drop out, which we can see intuitively,

but which will also follow when you actually plug in the numbers.

So all we'll have left is the k component, the z component

which is rx Fy minus ry r.

Fx is equal to k, and here's the arithmetic so the components are as shown.

And evaluating that, we end up with negative 2610k Newton meters.

Again, the same answer as it must be.

But in this case, it's given us the, the vector solution directly.