Atwood’s Machine#

Overview#

Welcome! You’re about to play with an Atwood’s Machine, a classic tool from the world of physics.

Imagine a perfect, frictionless pulley (like a wheel for a rope to slide over). Now, hang a rope over it, and attach a weight to each end. That’s it! You’ve made an Atwood’s Machine. It’s a simple setup used to explore some of the most important rules that govern how things move, especially Newton’s Second Law of Motion.

This simulation lets you be the scientist. You can change the weights and see exactly how they speed up and move, helping you build an intuition for physics in a hands-on way.

How to Use the Simulation#

Let’s run your first experiment. We’ll walk through it together.

  1. Set the Masses Find the sliders for the Blue Block and Red Block. These control how heavy each block is.

    • Set the Blue Block mass to 2.0 kg.

    • Set the Red Block mass to 3.0 kg.

  2. Start the Simulation Press the Start button. Watch what happens! Because the red block is heavier, it should start to fall, pulling the lighter blue block upwards.

  3. Watch the Data As the blocks move, look at the Real-Time Data Display. You’ll see numbers for “Blue block velocity” and “Red block velocity” changing. “Velocity” is just a scientific word for speed in a specific direction. This shows you exactly how fast each block is moving at any moment.

  4. Pause and Reset - You can press Pause at any time to freeze the action and look at the numbers. - When you’re done, press Reset. This puts everything back to the beginning, ready for your next experiment.

Exploring with Graphs The simulation can also draw graphs for you. This is a powerful way to see the patterns in the motion.

  • You can choose which block to track (blue or red).

  • You can plot different things like position (where the block is), velocity (how fast it’s moving), or acceleration (how quickly its speed is changing) against time.

Tip

Try plotting velocity vs. time. For this experiment, you should see a straight, sloped line. This is a visual sign that the blocks are moving with constant acceleration!

The Science Behind the Machine#

So, why does the machine work the way it does? It’s all about a cosmic tug-of-war between forces.

What are Forces? A force is simply a push or a pull. In our simulation, there are two main forces at play:

  1. Gravity: This is the force that pulls everything downwards. The Earth is pulling on both the red and blue blocks. The heavier the block, the stronger gravity’s pull.

  2. Tension: This is the pulling force transmitted through the string. The string pulls up on both blocks equally. Think of it as the rope in a game of tug-of-war.

The Tug-of-War For each block, gravity pulls down and the string’s tension pulls up.

  • If the pull of gravity is stronger than the tension, the block moves down.

  • If the tension is stronger than the pull of gravity, the block moves up.

Since the red block (3.0 kg) is heavier than the blue block (2.0 kg), the force of gravity on the red block wins the overall tug-of-war, and the whole system starts to move. To see exactly how we calculate this, check out the detailed derivations in the next section.

Deriving the Formulas#

Here we’ll walk through the algebra to find the exact formulas for acceleration and tension. We’ll use Newton’s Second Law, ΣF = ma, which says the net force on an object equals its mass times its acceleration.

Let’s call the blue block’s mass m_1 and the red block’s mass m_2. We’ll assume m_2 is heavier.

1. Deriving the Acceleration (a)

First, we look at the forces on each block separately.

  • For the lighter block (m_1): The upward tension T is winning against the downward pull of gravity m_1g. This net force makes it accelerate upwards.

    \[ΣF_1 = T - m_1g = m_1a\]

  • For the heavier block (m_2): The downward pull of gravity m_2g is winning against the upward tension T. This net force makes it accelerate downwards.

    \[ΣF_2 = m_2g - T = m_2a\]

Now we have a system of two equations. The clever trick is to add them together. Watch how the tension T cancels out:

\[(T - m_1g) + (m_2g - T) = m_1a + m_2a\]

The T and -T on the left side cancel, leaving:

\[m_2g - m_1g = m_1a + m_2a\]

Now we just need to solve for a. Let’s factor out g on the left and a on the right:

\[g(m_2 - m_1) = a(m_1 + m_2)\]

Finally, divide both sides by (m_1 + m_2) to get a by itself:

\[a = g \frac{m_2 - m_1}{m_1 + m_2}\]

And that’s our formula for acceleration!

2. Deriving the Tension (T)

To find the tension, we can take our new formula for a and substitute it back into one of our original equations. Let’s use the first one: T - m_1g = m_1a.

First, let’s rearrange it to solve for T:

\[T = m_1a + m_1g\]

Now, substitute the big expression for a into this equation:

\[T = m_1 \left( g \frac{m_2 - m_1}{m_1 + m_2} \right) + m_1g\]

To combine these terms, we need a common denominator, which is (m_1 + m_2).

\[T = \frac{m_1g(m_2 - m_1)}{m_1 + m_2} + \frac{m_1g(m_1 + m_2)}{m_1 + m_2}\]

Now we can combine the numerators over the single denominator:

\[T = \frac{m_1g(m_2 - m_1) + m_1g(m_1 + m_2)}{m_1 + m_2}\]

Let’s distribute the terms in the numerator:

\[T = \frac{(m_1gm_2 - m_1^2g) + (m_1^2g + m_1gm_2)}{m_1 + m_2}\]

The - m_1^2g and + m_1^2g cancel each other out. We are left with two m_1gm_2 terms.

\[T = \frac{2m_1m_2g}{m_1 + m_2}\]

This gives us the final, clean formula for the tension in the string.

3. Deriving the Energy Equation

We can also look at this from an energy perspective. The principle of Conservation of Energy states that energy isn’t created or destroyed, it just changes form. Here, gravitational potential energy (U_g) is converted into kinetic energy (K).

The change in kinetic energy must equal the loss in potential energy: ΔK = -ΔU_g.

  • Kinetic Energy Change (ΔK): The system starts from rest, so the initial kinetic energy is 0. The final kinetic energy is the sum for both blocks.

    \[ΔK = K_{final} - K_{initial} = \left( \frac{1}{2}m_1v^2 + \frac{1}{2}m_2v^2 \right) - 0 = \frac{1}{2}(m_1 + m_2)v^2\]

  • Potential Energy Change (ΔU_g): As the heavier block m_2 falls by a height h, the lighter block m_1 rises by that same height h.

    \[ΔU_g = U_{g,final} - U_{g,initial} = (m_1gh - m_2gh) - 0 = (m_1 - m_2)gh\]

Now, we set ΔK = -ΔU_g:

\[\frac{1}{2}(m_1 + m_2)v^2 = -((m_1 - m_2)gh)\]

Distributing the negative sign on the right gives us our final relationship:

\[\frac{1}{2}(m_1 + m_2)v^2 = (m_2 - m_1)gh\]

What Can You Discover?#

Use the simulation to answer these questions for yourself:

  • What if the masses are equal? Set both the blue and red blocks to the same mass (e.g., 5 kg) and press Start. What happens? Why? (Hint: look at the acceleration formula you just derived).

  • What creates the fastest acceleration? Try to find the combination of masses that makes the blocks move the fastest. Is it a small difference between them, or a very large one?

  • Verify the formula: Choose two masses (e.g., m_1 = 1 kg and m_2 = 4 kg). Use the formula to calculate the theoretical acceleration a. Then, run the simulation and check the “acceleration” value in the data display. Do they match?

Troubleshooting#

Note

If the simulation seems stuck or has been running for a long time, the best first step is to click the Reset button.

If you run into any issues:

  • Make sure both masses are set to a value between 0.1 kg and 10 kg.

  • If the blocks move too fast to see, try using masses that are closer together (e.g., 4 kg and 5 kg instead of 1 kg and 10 kg).

  • If the simulation becomes unresponsive, a quick click on Reset usually fixes it.