Note: this section uses some Algebra, and it would help, though isn't necessary, to know how to solve multiple equations with multiple variables. Introduction to Vectors is pretty much required reading before looking at this, though.
OK, you now know what equilibrium is, but, so what? You're going to need some math to use it. What I called "Calculating Equilibrium" in the title involves finding out what all the unknown forces are in the system that you're investigating. As was mentioned in the previous section, you assume that the whole system is in equilibrium. Once you solve the system, you'll know what the forces are going to be. Sometimes, these could be a lot higher than you'd want them to be. That's why we give you a computer simulator; you don't have to clean up messes and it helps reduce the rate of entropy.

In equilibrium, you learned, there is no movement in any direction. Translated into our situation right now, that can mean "all horizontal forces sum to zero" and "all vertical forces sum to zero". To write these two equations, we go back to what we learned about the components of forces (from the Parallelogram Law and Introduction to Vectors). [Example 9] To sum the horizontal forces in the example at right, we just take the horizontal components exerted by each force and add them together. As you may notice, each force has it's unit vector (representing it's direction, remember?) written beside it. To find the horizontal component, we'd multiply the magnitude of the force by the first part of the unit vector (you should remember that!). We only know one of the forces' magnitudes, however. We know that force F3 has a magnitude of 100 Newtons, so it's horizontal component would be .087*100 = 8.7.

OK, now we'll write one of the equations, the one for the horizontal forces. Since the magnitudes of forces F1 and F2 are unknown, we'll represent them by M1 and M2.

-.174*M1 + .866*M2 + 8.7 = 0
For the vertical forces, it's basically the same reasoning.
.985*M1 + .6*M2 + -99.6 = 0
The -99.6 comes from -.996*100, the vertical component of force F3.

If you know how to solve multiple equations with multiple variables, it should be pretty easy to solve the two equations above. If you don't, however, these is a little thing called Cramer's Rule, that, well, really saves the day. You set up your equations into Ax+By=C form (where A, B and C are the coefficients and x and y are the unknowns). You call the first one Ax+By=C and the second, Dx+Ey=F. Then, here's the good part:

x = (C*E-B*F)/(A*E-B*D)
y = (A*F-C*D)/(A*E-B*D)

All you have to do is plug in the values. Now, admittedly, this really isn't so great because I'm not going to (OK, fine, I can't) explain why or how this works. I can tell you that it involves matrices and determinants (and you thought you knew all there was to know in math!) and encourage you to do some investigating on your own. The complete rule (I've only given you part) will work on any size system.

Going back to our example, we'll take the easy way out and use that rule we showed above. Hang on tight, there's a lot of numbers.

The first equation becomes

-.174x + .866y = -8.7 (subtracted 8.7 from both sides)
and the second becomes
.985x + .6y = 99.6 (subtracted -99.6 from both sides)
So,
x=(-8.7*.6-.866*99.6)/(-.174*.6-.866*.985)=95.543
and
y=(-.174*99.6-(-8.7)*.985)/(-.174*.6-.866*.985)=9.150
To wrap everything up, the magnitude of force F1 is 96.386 Newtons (approximately, I rounded a little bit) and that of force F2 is 9.320 (again, it was rounded a little).

You'll notice in our example that there were two equations, one for horizontal forces and one for vertical, and two unknowns (the magnitudes of Forces F1 and F2). This was perfectly solvable. [Example 10] If we had three unknowns, like at right, we'd be stuck with an indeterminate structure, which is covered later in Houston, We Have Indeterminacy. Basically, this means that we can't find the magnitudes of any of the unknown forces. Of course, you can have more than three forces, you just have to know all but two of them . Or have only one unknown, but Cramer's Rule won't work in that case and you'll just have to fall back on good old Algebra. It's a shame, isn't it.

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