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How to Solve a Physics Problem Undergrads Usually Get Wrong

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By Rhett Allain , 07.09.15

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:

1 eqs

If I know the tension, then 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?

Here’s the link to the full article:

How to Solve a Physics Problem Undergrads Usually Get Wrong

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.

Half Atwood machine 2

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:

Half Atwood machine 3

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.

Half Atwood machine 4

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.

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