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I basically solved this problem, but I am unsure what the final equation actually means.

blocks on table

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

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I fixed the image for you. New users aren't allowed to include images inline until they get some certain amount of reputation. And since you're asking about the interpretation of your answer (which is the kind of question we like here, by the way), rather than asking for help doing the problem, I think it'll be fine without your free-body diagrams. –  David Z Dec 12 '11 at 3:42
    
Thank you for editing my image –  jake Dec 12 '11 at 3:43
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3 Answers 3

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.

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"if the system is accelerating to the right with an acceleration of g, this means that the upward tension on mass mc is negligible compared to the weight of the mass mc" Maybe my english isn't as good, but does that mean the upward tension is "negligble" because it is smaller than the weight of M_c? –  jake Dec 12 '11 at 3:56
    
@Jake: yes. Though I would say not just smaller, but much, much smaller than the weight of $m_{c}$. See my comment to Chris Gerig's answer below. –  Jerry Schirmer Dec 12 '11 at 5:01
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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.

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Wait, if $m_a = m_b = 0$, wouldn't that mean $m_c = 0$? It isn't just B that will move with g, it is the whole system yes? But I still don't understand what it means when a gets close and closer to g? –  jake Dec 12 '11 at 3:53
    
No, it just means you have $m_C$ dangling from a string with no other attached masses (so your system moving to the right at acceleration $g$ means that your first two objects must be massless). My answer and Jerry's were posted simultaneously and say the same exact thing, just in different forms. –  Chris Gerig Dec 12 '11 at 4:08
    
But Jerry said that if the system is accelerating to the right with g, it means the upward tension on mass C is negligble compared to the weight of mass C. Doesn't that mean the tension is 0 then? IF the tension is 0, can we draw the conclusion that "the string breaks"? –  jake Dec 12 '11 at 4:11
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@Jake: all we're saying is that if you solve your equation above for $a$, you find that $a=g\left(\frac{m_{c}-\mu m_{b} - m_{a}}{m_{a}+m_{b}+m_{c}}\right)$. If your masses are positive (and if $m_{c} \geq m_{a}+\mu m_{b}$), this is always less than $g$ and equal to $g$ only in the case where $m_{c} \gg m_{a}+m_{b}$. –  Jerry Schirmer Dec 12 '11 at 5:00
    
But the system actually cannot accelerate with g, only mass c can (I guess mass C IS the system here!). And when Mass C heads with acceleration g, then the tension tied to mass C is zero and this happens when both mass A and B are 0. Which I tested it out with the expression you gave for acceleration and the inequality. –  jake Dec 12 '11 at 5:46
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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).

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