00:01
So in this problem, we need to find the resultant of these two forces.
00:04
So we have our cartesian coordinate system.
00:08
Other in the picture, it's a bit more like that.
00:12
And we have the pole in a, which i guess pretty hard to draw perspective.
00:19
Do the best.
00:21
And we have our two forces, f2, and f1.
00:30
So the first thing we want to do is find the point of this end of the pole here, which was point a.
00:37
And if we look at the dimensions given, we see that in the x, it's zero because it is on the yz plane.
00:42
If we have x, y, and z.
00:46
So the first coordinate here is zero.
00:48
Then the next one, in the y, we get directly given that it is four meters by looking at the top.
00:54
And in the z, also directly given that it is six meters.
00:59
So that's the first thing we need to do is find this point.
01:02
And we want to do this so we can describe the lines that f2 and f1 are on.
01:06
So if we extend f2 and f1, they reach somewhere down to the x, y plane.
01:19
And then we can say they are at some location, and we want to find each of these points.
01:27
The point to where f1 goes is negative 2.
01:32
We shouldn't do this in blue.
01:35
Negative 2, because we see it's on the other side of the y, so negative 2 and the x.
01:40
8 in the y, because we add 4 and 1 and 3.
01:43
And then in the z, 0, because it's on the x, y plane.
01:48
And for force 2, it's looking at the point of 4, 5, and 0.
01:56
Now what we're going to do this for is to set up some similar triangles.
01:59
So let's look at 2d.
02:00
This works the same way in 3d, though, of a force and its components, fx, and we're going to draw fy here, even though it's not the proper place, but we see it makes a triangle, which is how we work with forces a lot.
02:16
And some line that is on the radius and it's x and y components.
02:22
Well, these triangles are similar because the ratios are the same.
02:27
And so what that means is we can set up a ratio, like, for example, fx over f is equal to x over r.
02:39
And so if we want to find the opponent of force in the x direction, which we're going to need to do to find the resultant and get it in the i.
02:46
J and k notation.
02:47
We can say fx is equal to fx or f times x.
02:52
So keep that straight times the radius.
02:59
So we have the vector.
03:00
We have the points of these, but we need to find the distance of a to the point where force two is going, which is b and then where force one is going c.
03:13
So to do this, we first need to find what that vector is.
03:15
And we want to subtract the second point from the first point of the coordinates, and now we'll give us our radius.
03:24
So for the radius of 1, if we look at force 1, we have a negative 2, and we subtract 0.
03:33
And so we see that radius vector goes from 0 to negative 2, and so its total x displacement is negative 2, because it's kind of what this is, the displacement vector.
03:42
In the y, we have our 8 minus 4, so 4, and in the z we had 6 minus 0 to be negative.
03:52
For 0 minus 6 to be negative 6.
03:57
And we can find the magnitude of that by taking the sum of the squares and taking the square root.
04:08
So we see that radius to be 2 root 14.
04:13
A lot of times it's easy to just leave this in radicals...
