Pulley and particles problem

[Dumbledore211]

Homework Statement

Two particles of masses m and 2m lie together on a smooth horizontal table. A string which joins them hangs over the edge and supports a pulley carrying a mass 3m. prove that the acceleration of the latter mass is 9g/17

Homework Equations

f=m1g/m1+m2

The Attempt at a Solution

Here, I think by the latter mass they mean is the one carrying 3m. The system is moving down with an acceleration f due to the mass 3m. Will the two particles m and 2m have different acceleration in the same direction?? I am really at a sheer loss as to where I should with this problem. Any help would be greatly appreciated...

 

[paisiello2]

It's not clear which one they are talking about but more strangely if there is a rope tying everything together then everything must be accelerating the same.

 

[dauto]

No, everything is not accelerating the same. Each particle has a different acceleration. The problem is clear. They want the acceleration of the third particle - the one with mass = 3m (latter here means the last one).

 

[Dumbledore211]

@dauto So, How should I set up the equations for each mass? Are they all acceleration in the same direction? Moreover, what should I take as the starting hypothesis since little information is presented in the problem?

 

[paisiello2]

No, everything is not accelerating the same. Each particle has a different acceleration. The problem is clear. They want the acceleration of the third particle - the one with mass = 3m (latter here means the last one).

If each particle has a different acceleration then the string connecting must be stretching or compressing relative to this motion. This would then change the physics of the problem so I can't see how this is possible.

Also I think the problem would have too many unknowns to solve if this were the case.

Is it possible we need to include the mass of the pulley itself?

 

[dauto]

If each particle has a different acceleration then the string connecting must be stretching or compressing relative to this motion. This would then change the physics of the problem so I can't see how this is possible.

That would be the case if the pulley was fixed. This is a moving pulley we have here

Also I think the problem would have too many unknowns to solve if this were the case.

The moving pulley still provides a constraint relating the three accelerations. That constraint is different than the one provided by a fixed pulley (more general)

Is it possible we need to include the mass of the pulley itself?

That would just make the problem harder and since the mass of the pulley was not provided, we must assume it's massless.

 

[dauto]

@dauto So, How should I set up the equations for each mass? Are they all acceleration in the same direction? Moreover, what should I take as the starting hypothesis since little information is presented in the problem?

The problem provides enough information. Start by trying to find the constraint relating the three different accelerations due to the fact that they are all connected to a single pulley (two masses are connected to a string that rolls around the pulleys edge while the third is attached to the center.)

 

[paisiello2]

That would be the case if the pulley was fixed. This is a moving pulley we have here

The moving pulley still provides a constraint relating the three accelerations. That constraint is different than the one provided by a fixed pulley (more general)


That would just make the problem harder and since the mass of the pulley was not provided, we must assume it's massless.

You are probably right in your interpretation but from the verbal description it doesn't seem to be clear to me at all. For one thing, if the pulley is free to move then the string must be fixed to something on the other end of the pulley. In that case the acceleration of mass 3m would be half of the acceleration of m and 2m.

 

[dauto]

You are probably right in your interpretation but from the verbal description it doesn't seem to be clear to me at all. For one thing, if the pulley is free to move then the string must be fixed to something on the other end of the pulley. In that case the acceleration of mass 3m would be half of the acceleration of m and 2m.

m and 2m are at opposite ends of a string that passes through the pulley (as described in the problem). They have different accelerations.

 

[paisiello2]

You are probably right but the problem didnt say the string passes through the pulley, it only says it supports a pulley. That could mean a number of different things. I suspect the OP left out an accompanying diagram.

 

[dauto]

You are probably right but the problem didnt say the string passes through the pulley, it only says it supports a pulley. That could mean a number of different things. I suspect the OP left out an accompanying diagram.

The problem seems pretty clear to me. There are only two ways to attach a string to a pulley. Either a string goes through the pulley, leaving two lose ends to attach elsewhere, or it connects to the center, leaving only one lose end. The configuration I described is the only one that makes sense. The problem has an additional clue, it gives you the correct answer. I solved the problem as I described and got that same answer, so my interpretation is correct. A diagram would've been helpful, but the problem is solvable as stated.

 

[paisiello2]

You must be right then but it wasn't clear to me and I don't think it was clear to the OP. If "support" can mean "pass through" then so can "tied".

The problem as interpreted by you is also 3 dimensional. You then have to make a number of assumptions about the geometry. The problem could have been better reformulated as a two dimensional problem and still get the same concepts across.

 

[Dumbledore211]

Does the problem say the third mass is attached to the center of the pulley or the other end of the string that holds the pulley? I am still in the dark as to how I should set up the tension and accelerations for each masses? The pulley is mass less and the string is inelastic. So, It definitely rules out the possibility of the extension and compression of the string...

 

[paisiello2]

According to dauto he gets the right answer by taking the 3m mass attached to the center of the pulley.

First step is to set up the acceleration relationship between each mass.

2nd step is to do the free body diagram for each mass.

 

[Dumbledore211]

The equation for the latter mass in this case is 3mg- T= 3mf3
The equation for both the masses which is 2m in this case is T=2mf1+ mf2
Correct me if I am wrong. If there are three accelerations for each particle then we need three equations to solve for the three different variables but we only have two

 

[paisiello2]

What are f1, f2, and f3?

Also T is different for each portion.

 

[Dumbledore211]

Are m and 2m are attached to the two opposite ends of the string according to dauto? Sorry, f1, f2 and f3 are actually a1, 12 and a3

 

[paisiello2]

Yes, so the string is tied to m, goes over the table edge and down, loops through the pulley that carries 3m, comes back up and over the table edge again, and then finally ties back to 2m.

So we should have a total of 5 unknowns: a1, a2, a3, T1, and T2.

We can get 3 equations from ∑F=Ma for each mass. And we can get two more from relating a3 to a1 and a2 for the pulley system.

 

[haruspex]

Yes, so the string is tied to m, goes over the table edge and down, loops through the pulley that carries 3m, comes back up and over the table edge again, and then finally ties back to 2m.

So we should have a total of 5 unknowns: a1, a2, a3, T1, and T2.

We can get 3 equations from ∑F=Ma for each mass. And we can get two more from relating a3 to a1 and a2 for the pulley system.

The tensions must be the same, and there's only one equation relating the three accelerations.

 

[paisiello2]

OK, I agree we can set the tensions to be the same since the pulley is massless and I agree there is only one kinematic relationship between all 3 accelerations.

 

[Dumbledore211]

That means the whole system is in motion. The mass with 3m is fixed at the center of the pulley or the string and the two masses m and 2m are constantly changing positions in a loop I guess. So the equation for the 3m mass is 3mg - T2= 3ma2. The equation for m is T1=ma1 and for 2m T2= ma2? Still don't get it...

 

[Dumbledore211]

@haruspex how can one equation relate the three different accelerations if the tension is same for all of them?
Can you please explain?

 

[paisiello2]

You almost got the 3 equations right.

The first one should be a3 not a2. Also we should consider the pulley and 3m mass as a single system so we should have T1 and T2 acting both acting instead of just T2.

Also you left out the "2" for the 2m mass in the last equation.

 

[haruspex]

@haruspex how can one equation relate the three different accelerations if the tension is same for all of them?
Can you please explain?

The three accelerations are related by the fact that the string has constant length.

 

[Dumbledore211]

@Paisiello2 So the equation for 3m should include both T1 and T2 like this 3mg-(T1+T2)=3ma3. The other two equations for the other two masses T1=ma1 and T2=2ma2. We need two more equations to solve for a1, a2 and a3 and what are they going to be like?

 

[paisiello2]

Ok, one more equation can find the trivial relationship between T1 and T2: take a free body diagram of the pulley itself and sum moments about it's center.

 

[dauto]

@Paisiello2 So the equation for 3m should include both T1 and T2 like this 3mg-(T1+T2)=3ma3. The other two equations for the other two masses T1=ma1 and T2=2ma2. We need two more equations to solve for a1, a2 and a3 and what are they going to be like?

That's still not right. The masses m and 2m are attached to the same string so they are under the action of the same tension T1. The third mass 3m is attached to a separate string connecting to the center of the pulley so it is under the action of a different tension T2. Draw four free body diagrams. One for each mass and one for the pulley. You also must find the constraint relating a1, a2, and a3.

 

[paisiello2]

I think he's on the right track. You're assuming T1 = T2 which turns out to be true for this problem, but this is not because they are the same string; rather it's because the pulley is assumed to have negligible mass.

 

[Dumbledore211]

@dauto There is only one string attaching all the three masses.. I guess I won't be able to solve it on my own but thanks anyway

 

[paisiello2]

Don't give up yet! You are almost there!

 

[dauto]

@dauto There is only one string attaching all the three masses.. I guess I won't be able to solve it on my own but thanks anyway

Don't give up just yet. The second string doesn't have to be a string. It might be a rod or something. The point is that the third mass must be connected to the axis of the pulley

 

[dauto]

I think he's on the right track. You're assuming T1 = T2 which turns out to be true for this problem, but this is not because they are the same string; rather it's because the pulley is assumed to have negligible mass.

Well, duh...

 

[paisiello2]

Well, duh...

So you admit you were wrong then.

 

[dauto]

So you admit you were wrong then.

No I wasn't wrong. I said the two tensions at opposite ends of the string in the problem are identical. That's correct.

 

[paisiello2]

Yeah, but you said it was because it was part of the same string. This isn't the real reason. And in fact the forces are not identical but rather the difference is considered negligible like the mass of the pulley.