# If you have two masses connected by a system of pullies, will the acceleration be different?

1. Oct 4, 2014

### Hereformore

So if you have two masses connected by a massless string in a multiple pulley system as shown below, the acceleration will be different right?

I ask because when doing normal pulley problems where two masses are connected by a single thread, you assume the acceleration is equal, though the net force would not be, to solve the problems.

But in this case, the acceleration would not be equal, right?
I understand it if i look at the lengths, as if you pull the left string down one meter the right will be shortened by an equivalent amount so between the two sections of that string, on the right each will shorten by .5. So the second mass should shorten/rise at a rate 1/2 of the mass on the left (which is falling).

But i feel im missing a crucil distinction here. Rather than recognizing the type of problem id like to understand what makes the cases unique in considering the accelerations. Any help?

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2. Oct 4, 2014

### Orodruin

Staff Emeritus
You are correct, the right mass will move only half the distance of the left and in the opposite direction. This is a geometrical contstraint and the way to arrive at it is essentially what you hinted at. It is different because you need to shorten both ends of the string to lift the right mass and the total shortening of those strings must equal how much longer the left string becomes. This is not the case for a single string pulley.

3. Oct 4, 2014

### Hereformore

I see so when it comes to pulleys, there is a constraint on accelerations given by thd ropes themselves because the masses cannot acclerate more than what is supplied (or denied) them by tension.

So in determining the relationship, ill always look first at how the objects would move if they were to move. And also, if there is more than one section of rope attached to a pulley, whether each section pulls individually to determine whether to double triple etc the tension force on the final object.

Are there any other considerations i should make/factors i could be missing?

4. Oct 4, 2014

### Orodruin

Staff Emeritus
Yes, as you are saying, everything is based on the total fixed length of the rope. The rope length will not change and thus provides this geometric constraint. The remaining equations you need should be provided by making free body diagrams of each pulley.

5. Oct 4, 2014

### Hereformore

Ah okay great thank you that clears up a lot.

I had one more tangent question regarding pulleys and tension.

So normally the angle at which you pull a rope for a single pulley system does not change the force applied because the rope perfectly transmits the force of the pull over the entire rope across both sides of the pulley.

But for a hanging pulley (this was a workbook problem of mine), where you have the rope attached as shown below:

you need to consider the angle since it is the Y component of the 2 tensions that cancels out the weight of the mass downward.

Imagine if the entire right wall and pulley were erased so it was just a rope being pulled on the right side, at the same angle as before, but with no pulley. Here you would still consider the angle when determining the tension force required to maintain static equilibrium (assuming youre keeping the angle constant).

Now that I'm writing this out, I feel i'v lost what i felt confused about before (i think the imagine i used to parallel the problem in my book helped clear things up), but my question remains what the difference is between the yellow image pulley where you need to consider the angle and a scenario where it's just one pulley with one rope where the angle at which you pull does not matter.

Is it becuase youre keeping the height of the pulley constant in one (when the angle does not matter)?

6. Oct 4, 2014

### Orodruin

Staff Emeritus
In this case you again must consider the geometry of the problem in order to relate the accelerations. The angle will be part of that geometry and you would need to consider how much the middle block would lift if you pulled the right block down. Sorry that I am not answering in more detail, but it is quite late here and I am falling asleep on top of my keyboard (rather under my iPad, but you get the idea).

7. Oct 4, 2014

### Hereformore

Ah no problem I completely understand. Thanks for answering, that makes sense and it is probably the best way to go about it. Cheers