I worked out formulas for the tension in the ropes and acceleration of the masses for the situation presented below. I was then given specific numbers such that $m_{1}+m_2=m_3$. After plugging these numbers into the formulas, I found, contrary to my intuition, that $m_3$ still experiences acceleration. I already worked out and confirmed the formulas/math. Can someone please explain the intuition behind the result using Netwon's laws? Are there perhaps limiting cases that may make the situation clearer? Also, how should I make sense of this formula?

Also $T_A=\frac {4g}{4/m_3+1/m_1+1/m_2} $

## Best Answer

In the case where $m_1=m_2=\frac{m_3}{2}$. If you work out the equations, you should find that $m_3$, in fact, does not accelerate. This is because the two masses remain stationary relative to each other, so the entire gravitational force is not "used" by $m_1$ or $m_2$, and instead "transmitted" to the tension of the string connecting $m_1$ and $m_2$, balancing out the weight of $m_3$.

Now, you can consider the case where $m_1\gg m_2$. In this case, the string connecting $m_1$ and $m_2$ will obviously "slip", and it would be as if $m_1$ is free falling without anything "holding it back". In this case, the gravitational force will be fully "used" by $m_1$, and none of it will be transmitted to the string connecting $m_1$ and $m_2$. Therefore, $m_3$ will experience near free-fall, since there is no tension to balance it.