Theory for acceleration of a cart: lab experiment

[Ascrapper]

Homework Statement

In a lab experiment in my introductory physics class, we are asked to verify Newton's Second Law by taking data from an experiment and then comparing that data to a theory. We are given a cart of unknown mass m$_{}1$ is put on a horizontal track with a string attached over a pulley to a hanging mass m$_{}2$. It is given that both masses are moving as one system and therefore have the same acceleration. My problem lies with the theory part of this experiment: how am I supposed to solve for the acceleration with known values? Let the acceleration of the system be a, the rolling frictional force f$_{}r$, the coefficient of rolling friction μ$_{}r$, the normal force F$_{}N$, the tension in the string T, and the acceleration due to gravity g.
This picture is similar to what our experiment looks like: http://www.physicssource.ca/images/cart_forcesensor_track.gif

Homework Equations

Newton's Second Law, F=ma

The Attempt at a Solution

Perhaps there is something intuitive about the mass of the cart and the coefficient of rolling friction that I am not seeing but I just can't figure it out. Here is what I have so far:

First I set up a free body diagram of the mass of the cart m$_{}1$ to show that:
ƩF$_{}x$=f$_{}r$-T=m$_{}1$*a (Equation 1)
ƩF$_{}y$=F$_{}N$-m$_{}1$*g=0, So F$_{}N$=m$_{}1$*g
Also, we know that f$_{}r$=μ$_{}r$*F$_{}N$, So f$_{}r$=μ$_{}r$*m$_{}1$*g

Now I set up a free body diagram of the hanging mass to show:
ƩF$_{}x$=0
ƩF$_{}y$=T-m$_{}2$*g, so T=m$_{}2$*(g+a)

Substituting all back into Equation 1:
μ$_{}r$*m$_{}1$*g-m$_{}2$*(g+a)=m$_{}1$*a
μ$_{}r$*m$_{}1$*g-m$_{}2$*g-m$_{}2$*a=m$_{}1$*a
g*(μ$_{}r$*m$_{}1$-m$_{}2$)=a*(m$_{}1$+m$_{}2$)
(g*(μ$_{}r$*m$_{}1$+m$_{}2$))/(m$_{}1$+m$_{}2$)=a

This is where I am stuck: How am I to get rid of or solve for these two unknown values μ$_{}r$ and m$_{}1$?

 

[Spinnor]

Except for a minus sign I got what you have. I'm guessing you measure acceleration and are supposed to figure out m1? Without knowing μ_r you have done as much as you can.