System of Equations with Two Unknowns, Algebra Help

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kwixson
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Homework Statement



I have two equations with two unknowns. I know m, F, R, r and I. I need to find a and f.

m = 2
F = 28
R = 0.25
I = 0.0625
r = 0.1875

I know the ultimate answers are [itex]a = 16\frac{1}{3}[/itex] and [itex]f = 4\frac{2}{3}[/itex]

Homework Equations



(1) [itex]Fr-fR=I\frac{a}{R}[/itex]

(2) F+f=ma

The Attempt at a Solution



[itex]Fr-I\frac{a}{R}=fR[/itex]

[itex]f=\frac{Fr}{R}-I\frac{a}{R^2}[/itex]

[itex]I(1+\frac{r}{R})=a(m+\frac{I}{R^2})[/itex]

[itex]a=F(\frac{1+\frac{r}{R}}{m+\frac{I}{R^2}})[/itex]

[itex]28(\frac{1+\frac{0.1875}{0.25}}{2+\frac{0.0625}{0.25^2}})[/itex] = ?

16.333 (Yay!)

I can't get back to f from here. :( Help!
 
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Ok, fair enough. But the answer guide said you could arrive at the following equation for f and I want to know how they got it:

[itex]f=\frac{\frac{r}{R}-\frac{I}{mR^2}}{1+\frac{I}{mR^2}}[/itex]
 
kwixson said:
Ok, fair enough. But the answer guide said you could arrive at the following equation for f and I want to know how they got it:

[itex]f=\frac{\frac{r}{R}-\frac{I}{mR^2}}{1+\frac{I}{mR^2}}[/itex]

F is missing from the expression for f. It should be

[itex]f=F\frac{\frac{r}{R}-\frac{I}{mR^2}}{1+\frac{I}{mR^2}}[/itex]

Plug in your equation [itex]a=F(\frac{1+\frac{r}{R}}{m+\frac{I}{R^2}})[/itex] for a into the equation f+F=ma.

ehild
 
kwixson said:

Homework Statement



I have two equations with two unknowns. I know m, F, R, r and I. I need to find a and f.

m = 2
F = 28
R = 0.25
I = 0.0625
r = 0.1875

I know the ultimate answers are [itex]a = 16\frac{1}{3}[/itex] and [itex]f = 4\frac{2}{3}[/itex]

Homework Equations



(1) [itex]Fr-fR=I\frac{a}{R}[/itex]

(2) F+f=ma
In matrix form, your system of equations is
$$\begin{pmatrix} R & \frac{I}{R} \\ -1 & m \end{pmatrix}\begin{pmatrix} f \\ a \end{pmatrix} = \begin{pmatrix} Fr \\ F \end{pmatrix}$$ A good technique in cases like this one is to use Cramer's rule. It'll get you the solution with minimal algebra.

http://en.wikipedia.org/wiki/Cramer's_rule#Applications