Animated Solution for Physics - Laws of Motion: A light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (vertically). If the acceleration of the system is $g/8$, then the ratio of the masses is

Finding Mass Ratios from Acceleration

Imagine a classic physics setup: two blocks hanging from a string that passes over a pulley. This elegant arrangement is known as an Atwood machine, and it's a fantastic tool for understanding Newton's laws of motion. In this problem, we are given the acceleration of the system and asked to work backward to find the ratio of the masses. Let's dive into the mechanics of it.

To solve any mechanics problem, our first step is to draw a free body diagram. Let's assume we have two masses, $m_1$ and $m_2$, and let's say $m_2$ is the heavier one ($m_2>m_1$). Because $m_2$ is heavier, it will accelerate downwards, pulling the lighter mass $m_1$ upwards.

Since the string connecting them is inextensible (it doesn't stretch), both masses must move together. This means they share the exact same magnitude of acceleration, which we'll call $a$. Furthermore, because the problem states the pulley is "smooth" (frictionless) and "light" (massless), the tension $T$ in the string is uniform throughout.

Now, let's apply Newton's Second Law ($F_{net}=ma$) to each block individually:

For the lighter block $m_1$, which is accelerating upwards, the upward tension $T$ must be greater than its downward weight $m_{1}g$. The equation is:

For the heavier block $m_2$, which is accelerating downwards, its downward weight $m_{2}g$ must be greater than the upward tension $T$. The equation is:

We have a system of two linear equations. Our goal is to find a relationship involving the acceleration $a$ and the masses, eliminating the unknown tension $T$. The simplest way to do this is to add equation (i) and equation (ii) together. Notice how the $+T$ and $-T$ beautifully cancel each other out!

Rearranging this to solve for acceleration $a$, we get the classic Atwood machine formula:

This equation makes intuitive sense: the net driving force is the difference in weights $(m_2-m_1)g$, and the total mass being accelerated is $(m_1+m_2)$.

The problem gives us a crucial piece of information: the acceleration of the system is $a=\frac{g}{8}$. Let's substitute this value into our master equation:

We can immediately cancel the acceleration due to gravity, $g$, from both sides, leaving us with a clean algebraic relationship:

Now, it's just a matter of simple cross-multiplication to solve for the ratio of the masses:

Let's group the like terms. We'll move all the $m_1$ terms to the left side and all the $m_2$ terms to the right side:

Finally, we rearrange this to find the ratio of the heavier mass to the lighter mass, $\frac{m_2}{m_1}$:

And there we have it! The ratio of the masses is $9:7$.

This problem perfectly illustrates the power of setting up clear free body diagrams and systematically applying Newton's laws. By deriving the general acceleration formula for an Atwood machine, we transformed a physics problem into a straightforward algebra exercise. Always remember to check your assumptions—like the massless string and frictionless pulley—as they are the key to simplifying the equations!