# The system on the left shown here is called an Atwood's machine. m$_1$ and m$_2$ are suspended at the ends of a string of negligible mass passing over a friction-less pulley as shown here. The system on the right just has a hanging mass m. What should m be in terms of m$_1$ and m$_2$ so that the tension in the top string is the same in both cases? Assume that the two systems have equal downward acceleration (a').

Answer: $m = \large\frac{4m_1m_2}{m_1+m_2}$
In the system on the right, we have $mg - T = ma' \rightarrow T = m (g-a')$
Net force in the direction of motion of $m_1$ is $F_1 = m_1g - T/2$ which is equal to $m_1(a'-a)$ as per Newton's second law.
Net force in the direction of motion of $m_1$ is $F_2 = m_2g - T/2$ which is equal to $m_2(a'+a)$ as per Newton's second law.
Here $a$ represents the acceleration of $m_1$ and $m_2$.
Let's define a new variable $g'$ such that $g' = g-a$. We can rewrite the equations as follows:
$(i)\;\; \quad mg' = T$
$(ii)\; \quad m_1g' = T/2 - m_1a$
$(iii) \quad m_2g' = T/2 +m_2a$
Eliminating $a$ from $(ii)$ and $(iii)$ and solving for $T$, gives us:
$\Rightarrow T = \large\frac{4m_1m_2}{m_1+m_2}$$g'$
Substituting in $(i)$, we get: $m = \large\frac{4m_1m_2}{m_1+m_2}$
The two-mass system in the left may be equivalently treated as a mass m, given by this equation above, as far as the upper string is concerned
edited Aug 19, 2014