# Force With a Pulley: Theoretical and Experimental

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1. Oct 1, 2015

### Thornak

1. The problem statement, all variables and given/known data

This is not a typical homework problem, so I hope I am posting this in the correct place. I am in 12th grade physics and we are dealing with basic kinematics. However, we recently did a set of problems that I did in a way which makes sense and which our textbook (and the rest of the Internet) says is correct, but my physics teacher claimed otherwise. After a little bit of debate, we ended up recreating the situation in class. The results supported my teacher, but I can't believe the rest of the world is wrong. An example problem is below:

Two objects are connected by a massless string and massless/frictionless pulley. m1 weighs 3 kg and hangs off of the table. m2 weighs 5 kg and sits on a frictionless surface. What is the acceleration of m2?

2. Relevant equations

My way is simple: (m1)*g=(m1+m2)*a, so a=(m1)*g/(m1+m2).

My teacher's way involves drawing free-body diagrams, which you can use in my way and still come to the same answer. However, for the equations of each object, my teacher uses TOTAL mass of the system because "it is all accelerating as one unit." For example, his equation in the y-direction for m1 is Fgravity-Ftension=(m1+m2)*a. And his equation for m2 in the x-direction is Ftension=(m1+m2)*a. When you use these two equations to solve for a, you end up getting Fgravity-(m1+m2)*a=(m1+m2)*a, so a=Fgravity/[2(m1+m2)]=m*a/[2(m1+m2)]. Thus, his answer is always half of mine.

3. The attempt at a solution

With the entire Internet backing me up, I assumed I was correct. But when we recreated the situation in class, the results supported my teacher's answers. The only difference in our experiment was that obviously the massless, frictionless conditions were not met, though my teacher assured the class that they were negligible (we used light pulleys and a toy car on wheels). Also, the setup was slightly different, as shown below:

(Sorry for the crappy diagram. I quickly cooked this up in MS paint. But I think you get the point.)

I don't believe the two pulleys in this experiment matter because they do not create a pulley system. They simply redirect the force twice instead of once. I may be wrong, though.

My question is twofold: First, who is correct and who is not? And second, if I am technically correct, why did the experiment yield results that supported my teacher's claim? This is an issue that will probably reappear as the year goes on, so I need to figure this out. Thank you!

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2. Oct 2, 2015

### ehild

You are right and your teacher is wrong.
In the real situation, you certainly measure less acceleration as the theory indicates. It is not only the moment of inertia of the pulley, but also the friction at the axis of the pulley. And it appears twice in the second set-up. Also the friction/rolling resistance between the toy car and and the table decreases acceleration.
The other thing, how did you measure acceleration.

3. Oct 2, 2015

### Thornak

We measured acceleration by timing the car and then using x=v0*t+(1/2)a*t2, which, because v0=0, simplifies to x=(1/2)a*t2. Thus, a=2x/t2.
I figured the unaccounted frictions/masses were causing the problem, but it was odd how close the calculated acceleration was to half of my theoretical.

4. Oct 2, 2015

### ehild

Wheels on toy cards diminish friction between the mass and the table, but when accelerating, the rotation of the wheels also counts, as if additional mass was given to the mass of the cart. It also can happen that the wheels do not do pure rolling, and then friction does count.
The set-up your teacher used is complicated enough, there are lot of places where friction can appear. Haven't you tried an experience with a single pulley and two hanging masses?

5. Oct 2, 2015

### Staff: Mentor

Yes. I agree with Ehild. Your answer is right and your teacher's answer is wrong. To begin with, your teacher's equation for the tension is wrong. It should just be m2a. Apparently you teacher thinks that he has advanced beyond the point where he no longer needs to use free body diagrams. He's wrong about that too.

My general rule is: always, always, always, always, always use free body diagrams.

Chet