Animated Solution for Physics - Laws of Motion: Two masses $m_1=5$ kg and $m_2=4.8$ kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move ? ($g=9.8{\text{ ms}}^{-2}$)

The Atwood machine is one of the most elegant and fundamental setups in classical mechanics. It beautifully demonstrates the interplay between gravity, tension, and Newton's Second Law of Motion. In this problem, we are given two masses, $m_1=5\text{ kg}$ and $m_2=4.8\text{ kg}$, hanging over a light, frictionless pulley. Our goal is to find the acceleration of the system when it is released.

Let's dive into the physics and see how these two masses dictate each other's motion!

Imagine the scenario: we have a heavier mass ($m_1$) and a slightly lighter mass ($m_2$) connected by a string. Because $m_1$ is heavier, it will naturally want to fall downwards, pulling the lighter $m_2$ upwards.

Since the string is inextensible (it doesn't stretch), both masses must move together. If $m_1$ accelerates downwards at a rate $a$, then $m_2$ must accelerate upwards at the exact same rate $a$. This shared acceleration is the key to unlocking the problem.

To find this acceleration, we need to apply Newton's Second Law ($F_{\text{net}}=ma$) to each mass individually. Let's draw a free body diagram for both.

For the heavier mass ($m_1$): Gravity pulls it downwards with a force of $m_{1}g$, while the tension $T$ in the string pulls it upwards. Since it is accelerating downwards, the net force is in the downward direction.

For the lighter mass ($m_2$): Gravity pulls it downwards with a force of $m_{2}g$, and the tension $T$ pulls it upwards. Since it is accelerating upwards, the tension must be greater than gravity.

We now have a beautiful system of two equations. Notice how tension $T$ acts as an internal force—it's negative in the first equation and positive in the second. If we add the two equations together, the tension cancels out completely!

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

This formula makes perfect intuitive sense. The driving force of the system is the difference in their weights $(m_1-m_2)g$, while the total inertia resisting the motion is the sum of their masses $(m_1+m_2)$.

Now, all that's left is to substitute our given values into the master formula. We know $m_1=5\text{ kg}$, $m_2=4.8\text{ kg}$, and $g=9.8{\text{ ms}}^{-2}$.

Let's simplify the numerator and the denominator:

Look at how perfectly the numbers are set up! The $9.8$ in the denominator perfectly cancels out with the value of $g$ outside the bracket.

And there we have it! The system accelerates at a gentle $0.2{\text{ ms}}^{-2}$. Because the masses are so close in value, the net driving force is small, resulting in a slow and steady acceleration.