# Cant solve these simoltaneous equations

• Apr 22nd 2010, 08:54 AM
fogel1497
Cant solve these simoltaneous equations
Fr = m alpha (L/2) cos(beta)

N - mg = - m alpha (L/2) sin(beta)

Fr cos(beta) - N sin(beta) = 2 Io alpha / L

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The problem is a dynamics problem for an engineering class im in and i've gotten the three equations (which is suppose to be the hard part) that I have to simoltaneously solve. Knowns in the above equation include 'm', 'g', and beta and everything must be in terms of these variables.

Also Io = (mL^2)/12

I'm trying to solve for Fr and N in terms of known quantities but i keep getting stuck with some messy algebra and wrong answers. Any help would be greatly appreciated.
• Apr 22nd 2010, 10:22 AM
running-gag
Hi

The first equation gives Fr = f(alpha)
The second equation gives N = g(alpha)

Substitute into the third equation
f(alpha) cos(beta) - g(alpha) sin(beta) = 2 Io alpha / L

This leads to alpha = 3g sin(beta) / L

Substitute into the first equation Fr = 3mg sin(beta) cos(beta) / 2
and into the second equation
• Apr 26th 2010, 07:03 AM
fogel1497
Quote:

Originally Posted by running-gag
Hi
The first equation gives Fr = f(alpha)

What are you calling f here? Where are the terms L, m, and beta going? I'm just a little confused because in the context here there is no variable defined as 'f'.

Quote:

Originally Posted by running-gag
Hi
The second equation gives N = g(alpha)

When you bring the term 'mg' to the other side your left with:
N = - m alpha (L/2) sin(beta) + mg
I don't see how this condenses into N = g(alpha)

Once again thanks for your help! I'm just trying to understand the algebra thats going on here.
• Apr 26th 2010, 09:09 AM
running-gag
Quote:

Originally Posted by fogel1497
What are you calling f here? Where are the terms L, m, and beta going? I'm just a little confused because in the context here there is no variable defined as 'f'.

I was talking about a function of alpha

$\displaystyle F_r = \alpha \frac{mL}{2}\cos \beta$

$\displaystyle N = mg - \alpha \frac{mL}{2}\sin \beta$

Substitute into the third equation

$\displaystyle \alpha \frac{mL}{2}\cos^2 \beta - mg \sin \beta + \alpha \frac{mL}{2}\sin^2 \beta = \alpha \frac{mL}{6}$

And since $\displaystyle \alpha \frac{mL}{2}\cos^2 \beta + \alpha \frac{mL}{2}\sin^2 \beta = \alpha \frac{mL}{2} (\cos^2 \beta + \sin^2 \beta) = \alpha \frac{mL}{2}$

It comes
$\displaystyle \alpha \frac{mL}{2} - mg \sin \beta = \alpha \frac{mL}{6}$

$\displaystyle \alpha \frac{mL}{3} = mg \sin \beta$

$\displaystyle \alpha = \frac{3g}{L} \sin \beta$