# Simplifying/Re-arranging Troubles

1. Nov 3, 2005

### caddyguy109

In the middle of a statics problem full of unknowns, and am trying to simplify a sum of moments down to the point where I have: a= everthing else (a as a function of several other terms).

Keep getting close, but then always have an extra variable or two that won't go anywhere.

Here's the equation (the k is just noting a unit vector, so ignore):
http://www.ihostphotos.com/show.php?id=192455" [Broken]

The first thing I did was multiply through by a factor of sqrt(a^2 + 4r^2) to get rid of that factor in two of the denominators and the sqrt sign on the other, but then it just started getting REALLY messy.

Can anyone figure out what this equation (in the pic) should be, as "a=...." only??

The other variables in that are: P, L, r, and Nb.

Last edited by a moderator: May 2, 2017
2. Nov 3, 2005

### caddyguy109

Anybody?

It's just algebra, but good lord is it confusing me!

Last edited: Nov 3, 2005
3. Nov 3, 2005

### Gale

once you get rid of the sq rts, you're left with a quadratic with respect to a. so, group together your a^2, a^1 and a^0 terms, and then plug in the coefficients into the quadratic formula eh?

4. Nov 3, 2005

### caddyguy109

Hmm...after multiplying through by sqrt(a^2 + 4r^2), I got:

[2PLr - Nb(a^2 + 4r^2) + WLa/2 = 0

then I multiplied through by 2 to get rid of the division by 2 in the last term, and got:

[4PLr - 2Nb(a^2 + 4r^2) + WLa = 0

But how do I apply that to the quadratic formula, if I want to get "a" all alone on one side??

5. Nov 3, 2005

### Gale

you can't get a all alone on one side... just like you can't get x on one side if you have x^2 + 2x+3=0. you don't even need to multiply by the two. distribute the N_B and get your equation into ax^2+ bx+c form. then apply the quadratic formula. you'll get two possible answers for a, you'll have to decide which one makes more sense.

6. Nov 3, 2005

### caddyguy109

And that would still end up as "a" as a function of the other terms?

Hmm...going back to my math skills from many years ago...

7. Nov 3, 2005

### caddyguy109

Okay, got it--thanks!

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