Three forces are applied to an object, as indicated in the drawing. Force 1 has a magnitude of 33.0 newtons (33.0 N) and is directed 30.0° to the left of the +y axis. Force 2 has a magnitude of 26.0 N and points along the +x axis. What must be the magnitude and direction (specified by the angle θ in the drawing) of the third force 3 such that the vector sum of the three forces is 0 N?
There is a drawing included, that I can post if it's wanted. Otherwise, Force 3 extends left along the x axis as well as left along the y (making it negative?)
I'm guessing just Ax = A cos theta and Ay = A sin theta. Oh and tan^-1 = Ay/Ax
The Attempt at a Solution
First, Force 1 = A, Force 2 = B and Force 3 = C.
So, I found the combinant vectors for A and B.
Ax = 33 cos 30 = -28.57 (negative because it is heading left along the x-axis)
Ay = 33 sin 30 = 16.5
Since B simply extends right along the x-axis, Bx = 26 and By = 0
So, I found that Rx (resultant vector of A+B) was -2.57, and Ry = 16.5
So, basically, Cx would have to be 2.57 and Cy would have to be -16.5.
Is this right so far?
However, it asks me for the total magnitude, and without the angle, how do I find that?
Then, how do I find the angle? I could use the formulas above but it appears, to me, that I'm missing a piece of information.
Any help is much appreciated :)