00:01
So we have a picture given to us of the force is being applied to an object.
00:04
And we have this first one.
00:06
We say it's f1, and we know that the magnitude of this is 40 newtons, and it's at an angle of 30 degrees.
00:14
And the other force vector is a little longer vector.
00:19
It is 60 newtons of force.
00:21
This is f2, and that's at negative 45 degrees.
00:25
So why don't we put them in the ij unit vector form? so we know that first vector f1 is equal to the vector if we take, well, i'm not going to use that notation.
00:37
Let me erase that.
00:39
If we take the cosine of 30 degrees times its magnitude times i, that will give us our x component of that vector.
00:53
So we're finding this x component by getting that vector.
01:00
Plus, now let's get the y component of that vector, and that's going to be 40 times, i'll go back to black, 40 times the sign of 30 degrees.
01:11
All right, and for a 30, 6090 triangle, we know for a 3060 90 triangle that this leg would be one half on the triangle, and this leg would be square root of 3 over 2.
01:21
So the cosine of that, we get 40 times square root of 3 over 2 i plus, and the sign of 30 degrees is 1 half.
01:31
Oops, i never got to put the little j there...