00:01
So in this problem, we're looking at moments, and when we look at the definition of the moment first, where a moment is a force times a distance, if these are perpendicular.
00:12
We're going to use the vector notation here in a bit, but if we start with this basic one, if the distance goes to zero, then so does the moment.
00:19
So in the case of force two, we're not drawing on the coordinate system here because we see we're not going to care about this in a second.
00:27
If we want to find its moment about point a, well, force two acts through point a, so it's perfect.
00:31
Perpendicular distance is zero.
00:33
And so the moment due to force two, we call it moment two, is equal to zero.
00:39
So that's really nice in this problem.
00:41
We can just stop caring about force two and only look at force one.
00:47
So in our coordinate system, we have force two somewhere out in space here.
00:56
And we're given the direction cosines of it.
00:59
Let's do this also in blue, under 20 degrees to the x.
01:07
We have x, y, and z, 60 degrees to the y, and 45 degrees to the z.
01:19
So with this 120 here, what we can tell is that force two is actually pointing into the page.
01:24
It's pretty hard to see, though.
01:26
So that 120 degrees does tell you that.
01:29
And what we need to do is find the distance.
01:32
And how do you do that? well, we have point a over here.
01:36
And we want to draw the vector.
01:38
We want to find what the vector is from point a to any point along the line of action of force f.
01:44
So the easiest way to do it is just use the point where force f is applied, and this is actually force 1.
01:54
The location of it is correct.
01:56
So we want to find this radius, and when we do that, we can then use the more general definition of a moment, that the moment vector is equal to the radius cross -product with the force.
02:11
So we need to find both of these vectors.
02:15
Well, let's look at what point a is at.
02:21
We see that we have a 5 foot line that's at 30 degrees.
02:28
So what would be the distance on the y -axis to a? would be 5 times a sine of 30 because we see sign 30 would give us the opposite side of the triangle, and that is the corresponding 1 to a.
02:44
So if we do five, make that five a bit more like a five.
02:53
We see that's directly 2 .5, so that's pretty nice.
02:57
So if you look at point a, we see it's at zero.
03:02
It's at, on the x, it's zero because we see it's on the y -axis.
03:11
In the y, it's negative 2 .5, like we found from the 5 sine 30.
03:15
And then on the z, it's also zero.
03:17
So that is point a.
03:18
And the point b, where force f is, we have the similar kind of thing, 30 degrees and 4 feet.
03:36
So if we look at the sign of 4 sine 30, we saw with 5 or it was half, and so that's what sign 30 is, is 1ā2.
03:48
So 4 times 1ā2 would be 2, which is our side here, and that's the x component of our point b.
04:02
And so point b is at 2 in the x.
04:07
In the y, it's going to be 4 cosine 30, which is root 3 over 2.
04:15
But actually that is the cosine of 30 by itself.
04:18
When we multiply it by 4, we're going to end up with our 2 root 30, or 2 root 3.
04:28
And in the z, we're directly given that it is at 6 units.
04:32
So the radius vector is going to be, our radius vector is basically going to be the point b, minus the point a for each component...