System of Equations with Two Unknowns, Algebra Help

In summary, the conversation is discussing a problem with two equations and two unknowns. The equations are (1) Fr-fR=I\frac{a}{R} and (2) F+f=ma. The known values are m, F, R, r, and I, and the ultimate answers for a and f are a = 16\frac{1}{3} and f = 4\frac{2}{3}. Cramer's rule is suggested as a technique to solve the problem with minimal algebra.
  • #1
kwixson
6
0

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!
 
Last edited:
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  • #2
From (2) ##f = ma - F## so plug your numbers in.
 
  • #3
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]
 
  • #4
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
 
  • #5
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
 

1. What is a system of equations with two unknowns?

A system of equations with two unknowns is a set of two equations that contain two variables, typically represented by x and y. The goal is to find the values of x and y that satisfy both equations simultaneously.

2. How do you solve a system of equations with two unknowns?

There are several methods for solving a system of equations with two unknowns, including substitution, elimination, and graphing. The most common method is substitution, where one equation is solved for one variable and then substituted into the other equation to find the value of the remaining variable.

3. Can a system of equations with two unknowns have no solution?

Yes, it is possible for a system of equations with two unknowns to have no solution. This occurs when the two equations are parallel lines and never intersect. Graphically, this would be represented by two lines that are parallel to each other.

4. Can a system of equations with two unknowns have an infinite number of solutions?

Yes, a system of equations with two unknowns can have an infinite number of solutions. This happens when the two equations are equivalent, meaning they are essentially the same line. Graphically, this would be represented by two lines that overlap each other.

5. How do you check the solution to a system of equations with two unknowns?

To check the solution to a system of equations with two unknowns, simply substitute the values of x and y into each equation and see if they make both equations true. If the values satisfy both equations, then they are a valid solution to the system.

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