# Finding acceleration of blocks in a pulley system

1. Jun 20, 2010

### hihowareu

I am a bit confused when doing questions about 2 masses connected by a pulley system.

Sometimes i get the right answer when i use just one of the masses to get the acceleration but other times i have to use both masses to get the correct acceleration.

How do you know when to use just one or both masses in the Fnet equation.

Last edited: Jun 20, 2010
2. Jun 20, 2010

### K^2

Depends on whether you are adding together forces acting on one or all of the masses.

For example. Consider this situation. You have a cart with mass M sitting on the table with a string attached to it. The string is connected via pulley to a suspended mass m. Classical problem. Find acceleration of the cart.

There are two ways to go about solving it.

1) Consider forces acting on both the car and suspended mass.

The car has a string attached to it. Lets say the tension in it is T. That's the only force acting on the cart, so:

Ma = T

Now the suspended mass. It's being pulled down by its own weight, mg, and up by the tension in the string T:

ma = mg - T

Of course, a and T are the same in both of these equations. So we can substitute T = mg - ma into first eqn.

Ma = mg - ma

Ma + ma = mg

(M+m)a = mg

a = g*m/(M+m)

2) Forget the fact that the cart is separate from the suspended weight.

Since they are accelerating at the same rate, we can treat it as a single object. That object's combined mass is M+m. The net force acting on it is just the weight of suspended mass, which is mg. Tension in the string cancels out. (It's not an external force anyways, so it can't be causing acceleration of center of mass. Internal forces will always cancel out.)

So that gives you:

(M+m)a = mg

a = g*m/(M+m)

Exactly the same answer.

So you really don't have to guess. As long as you add all of the forces acting on whatever you call the mass, Newton's Second will still work.

Why consider forces acting on individual components? Well, if you only need total acceleration, it really doesn't matter. But second approach doesn't tell you what tension in the string will be. First approach does. Simply substitute your result for acceleration into one of the equations to find T.

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