What does it mean if a = g in this situation? I basically solved this problem, but I am unsure what the final equation actually means.

blocks on table http://img560.imageshack.us/img560/9804/83508925.jpg
Write an expression for solving the mass of block C if mass B moves to the right with an acceleration of $a$. Note that the coefficient of friction between the table and mass B is $\mu$ and assume the mass of the string and pulley are massless.
Hint: $m_C > m_A + m_B$

Not sure how to include my Free-Body Diagrams, but my final answer looks like
$m_C = \frac{1}{g-a}\left (a\left ( m_A + m_B \right ) + g \left ( m_A + \mu m_b \right )  \right )$
Here is my question what happens if $a \to g$? It certainly can't mean mass C will become infinite. How would that even work? I also thought it might mean that the string might break but wouldn't that mean all strings tied to any block accelerating at g would break?
Thanks
 A: If the system is accelerating to the right with an acceleration of $g$, this means that the upward tension on mass $m_{c}$ is negligible compared to the weight of the mass $m_{c}$.  Since this tension ultimately depends on some combination of the masses $m_{b}$ and $m_{a}$ and the frictional force between $m_{b}$ and the table, what this is telling you is that, for the acceleration to approach $g$, it is necessary that $m_{c} \gg m_{b}$ and $m_{c} \gg m_{a}$.  Hence, the apparent divergence.  For any finite $m_{c}$, the acceleration will be less than $g$.  
In other words, it's best to think of $m_{c}$ as the independent variable and $a$ as the dependent variable, rather than how you've solved it, which is the other way around.
A: If B moves with acceleration $g$, then so does C, which means C is in free-fall, which means $m_A=m_B=0$, so there is no problem here.
A: Let $T$ be the tension pulling mass C, and $m_c$ be the mass of C.
Since $m_c$ accelerates downward, with a magnitude of $a$, by Newton's Second Law
$$m_cg - T = m_ca$$
Multiplying both sides by $1/m$ 
$$g - T/m = a$$
Adding $T/m$ to both sides
$$g = a + T/m$$
Since $T > 0$ and $m > 0$, it follows that $g > a$. Therefore, $a$ does not equal $g$ unless $m_c$ is extremely large (as noted above by Jerry).
