00:01
So in this problem we are calculating resultant force from our three vectors 300 newtons, 200 newtons, and 400 newtons.
00:17
We have our axis x and y.
00:21
Now the angle here is 150 degrees, but we know part of this is 90, and so relative to the x -axis here we have an angle of 60 degrees.
00:32
Now this whole angle is 90 between f2 and f3, which i should probably label.
00:39
I can't do that now because i'm going to have to leave that there.
00:44
This whole angle is 90 degrees, so we have 60 and 30 to get 90, and 90 minus 30 is once again 60 degrees.
00:58
And so that's the angles we're going to be using for our working out because they are all referenced off of our axis.
01:05
Now we can find the resultant vector, and we're told to use a process first of finding an intermediate f2 plus f3.
01:17
And we're going to do this in component form.
01:22
So f prime is equal to f2 plus f3.
01:25
So let's look at the x components of that.
01:28
Well we have a component of f2 adjacent to the angle, so it's 60, cosine 60, as that component.
01:38
So the magnitude cosine 60 there.
01:45
Now f3 is directly in the y direction, so has no component in the x, and this is the i hat direction, and then the j hat direction.
01:54
F3 is all in the y, so it's just negative, oh well we'll keep the same order here.
02:00
F2, we have a component here across from the angle, so that's the sine, and that's positive, so 200 sine 60, and then the f3 is all downwards, so negative 300 j hat.
02:18
And so f prime is 100 i minus 126 .8 j, and leaving in components because it's the convenient form to work with...