# Two masses and two pulleys

Masses $M_1$ and $M_2$ are connected to a system of strings and pulleys as shown. The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of $M_1$.

Problem Source : An Introduction to Mechanics

We first look for the relationship between the tension of the two strings. Consider the tension force exert on the lower pulley with mass $m$,

$T_1 - 2T_2 - mg = ma$

Although there is acceleration on the pulley, it’s mass is $0$. So, we can conclude that

$T_1 = 2T_2$

Let the acceleration of $M_1$ be $-a_1$, then the acceleration of the lower pulley will be $a_1$.

If $l$ is the length of the string, we have

$x_1 + x_2 + \pi R = l$

Differentiating twice we have

\begin{aligned} \ddot{x_1} + \ddot{x_2} &= 0 \\ (a_1 - 0) + (a_1 - a_2) &= 0 \\ a_2 &= 2a_1 \end{aligned}

Now, we have all the details needed to solve this problem, with two variables and two equations that study the forces on $M_1$ and $M_2$.

\begin{aligned} 2T - M_1g &= M_1(-a) \\ T - M_2g &= M_2(2a) \end{aligned}

Substitute $T$,

\begin{aligned} \displaystyle M_1g-M_1a &= 2M_2g+4M_2a \\ (M_1+4M_2)a &= (M_1-2M_2)g \\ a &= \bigg( \frac{M_1-2M_2}{4M_2+M_1}\bigg)g \end{aligned}

We can also find the tension force on the string $T$. Substitute $a$ on the second equation,

\begin{aligned} \displaystyle T &= 2M_2a+M_2g \\ &= M_2\bigg( \frac{2M_1-4M_2}{4M_2+M_1} + 1 \bigg)g \\ &= \bigg ( \frac{3M_1M_2}{4M_2+M_1}\bigg )g \end{aligned}