How to Solve a Physics Problem Undergrads Usually Get Wrong 

How to Solve a Physics Problem Undergrads Usually Get Wrong

A cart is connected with a string to a mass hanging over a pulley. In physics, this is called a Half Atwood's Machine. Rhett Allain

This is a classic introductory physics problem. Basically, you have a cart on a frictionless track (call this m₁) with a string that runs over a pulley to another mass hanging below (call this m₂). Here’s a diagram.

Now suppose I want to find the acceleration of the cart after it is let go. The string that attaches the two carts does two things. First, the string makes the magnitude of the acceleration for both carts is the same. Second, the magnitude of the tension on cart 1 and cart 2 has the same value (since it’s the same string). This means I can draw the following two force diagrams for the two masses.

So, how do you find the acceleration of cart 1? It seems clear, right? You just need to find the tension in the string since that’s the only force in the horizontal direction. You could write:

If I know the tension, I can calculate the acceleration. Simple, right? Even simpler, the tension would just be equal to the gravitational force on the hanging mass (m₂). WRONG! This is not the correct way to solve this problem — I actually remember making this exact mistake when I was an undergraduate student. But why is it wrong?

Experimentally Measuring the Tension

If I just solved the problem the correct way, it wouldn’t be that much fun. Instead, I am going to set up an experiment to show that the tension in the string isn’t just the weight of mass 2. You can see the setup in the image above (it’s called a half Atwood machine in case you want to google the solution). Basically, I have a mass on a low friction track with a string connected over a pulley to a hanging mass. Both masses have force sensors on them so that I can measure the tension in the string. Also, I can measure the position, velocity and acceleration of the cart using the Vernier motion encoding system. This is basically just series of lines on the track that the cart uses to determine its position (and thus velocity and acceleration).
With a cart mass of 1207 grams (mass 1) and a hanger mass of 145 grams (mass 2), I get the following data.

Here is what you should look at:

• First, the forces. Yes, there are two force probes — one on each mass. Both of the force sensors give the same force. This means the tension on mass 1 has the same magnitude as the tension on mass 2.
• The first part of the graph shows the two masses at rest (I’m holding it with my hand). In this case, the acceleration is zero m/s² (as seen in the graph showing the acceleration) and the tension in the string is just the weight of mass 2 (0.145 kg)*(9.8 N/kg) = 1.421 N.
• The masses accelerate around a time of t = 1.4 seconds. Ok, this is wrong. They actually accelerate before that, but the first acceleration was caused by me pushing the cart away from the pulley. The cart is accelerating the way you would expect from time of 1.4 seconds to 2.6 seconds.
• During this acceleration, the tension in the string drops from 1.421 Newtons to just 1.285 Newtons.

But why? Why is the tension not the same as the weight of mass 2? The answer is simple — mass 2 is not in equilibrium but instead it is accelerating downward. Since it’s accelerating, the net force is not equal to zero (vector). This means that the tension should be smaller than the weight of mass 2 — which it is.

Solution to the Half-Atwood Machine

The tension in the string depends on the weight of mass 2 as well as the acceleration of mass 2. However, the acceleration of mass 2 is the same as mass 1 — but the acceleration of mass 1 depends on the tension. Does this mean you can’t solve the problem? Of course not, it just means that it’s slightly more complicated.
Let’s say mass 2 is accelerating in the negative y-direction. This means that I can write the following force equation (in the y-direction).

Now I can do a similar thing for mass 1 with its acceleration in the x-direction. Since the magnitudes of these two accelerations are the same, I will use the same variable.

With two equations and two variables (a and T), I can solve for both variables. If I substitute the expression for T for mass 1 into the equation for mass 2, I get:

Instead of completely solving for the acceleration, let me leave it in the form above. Think of the problem like this: suppose you consider the system that consists of both mass 1 and mass 2 and it’s accelerating. What force causes this whole system to accelerate? It’s just the weight of mass 2. So, that is exactly what this equation shows — there is only one force (m₂g) and it accelerates the total mass (m₁ + m₂). From this I can solve for the acceleration.

Using the values of mass 1 = 1.207 kg and mass 2 = 0.145 kg, I get an acceleration of 1.05 m/s². This is pretty close to the experimental value (seen above) at 1.109 m/s². I’m happy.
With the value of the acceleration, I can plug back into the original equation to solve for the tension. With this, I get a tension of 1.267 N. This is fairly close to the experimental value of 1.285 N. Again, I’m happy. It seems physics still works.