00:01
In this problem, we are given a graph containing two forces, f1 and f2.
00:07
We want to determine the magnitude of the resultant force of f1 plus f2 and its direction measured clockwise from the positive u axis.
00:16
So here we have a little bit of an interesting cartesian grid.
00:21
We have the u axis pointing towards the right like so.
00:34
There is no vertical axis but i'm going to call this v, i suppose.
00:47
And so now the goal is to decompose our f1 and f2 vectors in this cartesian system in order to add them up.
01:00
So f1, the vector force, will be equal to 300n times the x component here, well the unit vector containing the x component and the y component.
01:20
I'm saying x and y but really it's u and v, i suppose.
01:24
And this unit vector will correspond to the cost of 30 degrees in this u axis and minus the sine of 30 degrees in this v axis if the v axis is pointed upwards.
01:52
And evaluating this, we obtain a vector force of 259 .81 in this first component and minus 150 in the second component and our forces in newtons.
02:13
Let's do the same thing for f2.
02:16
We can decompose it by taking its magnitude, 500, and multiplying it by the unit vector that describes its direction.
02:26
So how do we find the direction of this vector? so we need to find this angle right here, which is not explicitly given to us but being calculated, it is in fact equal to 180 minus 70 minus 45 degrees, giving us an angle of 65 degrees.
02:51
So we now see that f2 will be equal to 500n times minus the cos of 65 degrees.
03:01
Why minus? because this u component is negative, pointing in the negative u axis...