Forces in equilibrium

[jamie_o]

Hello, this isn't technically a homework, but its something I'm trying to figure for myself. As shown in the diagram (in the attachment), there are three forces acting on a particle. The particle is in equilibrium. I have to find the magnitude of S and the angle theta (which ill write as x, because I dont know how to type theta on a computer screen) Now what I would do is solve it horizontally and vertically.
so horizontally: 6 - Scosx = 0
vertically : 2.5 - Ssinx = 0

My problem is where do I go from here? I know I could make it 6 = Scosx and likewise for the other, but it won't help me solve it. Since they are both equal to 0 I tried setting them equal to each other in the hope that the S would cancel and I would get sinx/cosx which is equal to tanx = to a number and then work from there. However I can't get the equation into that form. Any ideas on how I would solve this? Any help is very much appreciated :)

[Doc Al]

Try writing your equations like this:
(1) 6 = Scosx
(2) 2.5 = Ssinx

Now how can you get the S to cancel?

[jamie_o]

Thank you for the reply. I can see a link there and use Ssinx/Scosx to cancel S and leave tanx = 2.5/6, then use inverse tan of 2.5/6 = x. Then working out S would be straightforward from there. Why would I do this though? Why would I divide the horizontal equation into the vertical equation? Is there theory behind doing this? I don't like being 'monkey sees, monkey does' :) Thanks for the help.

[jamie_o]

Oh and if possible, could someone please tell me why setting the two equal to each other does not work? I might have a serious hole in my understanding of it all. Thanks.

[Doc Al]

You're just solving two equations with two unknowns. There are many ways to approach it; here are two:
(1) Solve for S in one equation, then plug that in to the second. That's a standard approach.
(2) Square both equations and add them. Take advantage of sin^2\theta + cos^2\theta = 1.

Your skill in math often depends on picking up various little "tricks of the trade". Try to have as many "tricks" in your bag as possible.

[jamie_o]

Thank you. I should have spotted that since I am much further on in maths now, from when I got this question. I did try to take advantage of http://www.physicsforums.com/latex_images/35/352887-0.png previous to posting the first time. However I came out with a very large sum, what I had before was incorrect I think.