Can a tension constraint lead to the wrong results?

[Physics freak]

Homework Statement

Find the constraint relation between a1,a2,a3

Relevant Equations

All pulleys are massless.Now for pulley 2 the upward tension T+T so the downward is 2T. And for pulley 1 since upward tension is T the downward ones must be T/2 and T/2 . Now the highlighted string is uniform so constant tension throughout ,but its coming out to be separately T/2 and 2T. Which is only possible when tension is 0, so is it really so or there's some mistake in my approach. The question just asked to find constraint relation between a1,a2,a3. I was trying to use tension constraint here and thus got struck.

 

[haruspex]

Which is only possible when tension is 0,

Quite so.
What does that tell you about the accelerations of the masses? What about the pulleys?

 

[Physics freak]

Quite so.
What does that tell you about the accelerations of the masses? what about the pulleys?

Nothing, but we normally use Σ t.a for such questions.

if the question had simply asked about tension what could we have done then?

 

[haruspex]

Nothing

No, it tells you a great deal about the accelerations of the masses. If all tensions are zero, what forces act on them?

 

[Physics freak]

No, it tells you a great deal about the accelerations of the masses. If all tensions are zero, what forces act on them?

Gravity?

 

[haruspex]

Gravity?

Right. So what are their accelerations?

Edit: if you now find the constraint relations between the five accelerations (three masses, two pulleys) you can figure out the accelerations of the pulleys. Then you may come to understand how it can be that the three masses are in free fall,

Edit 2: the original question is flawed. There are in all five accelerations that involve the strings, three masses and two pulleys. Two strings means two constraint relations. You can use one to eliminate one pulley acceleration, but you are left with the other pulley acceleration. So you cannot find an equation relating only the accelerations of the three masses.

 

[haruspex]

I'm worried that you are still stuck on this. Please respond.

 

[Physics freak]

I'm worried that you are still stuck on this. Please respond.

Hello. I never saw the edit till today. So basically in question, the tension is 0 and all masses and pulleys have acceleration g

 

[haruspex]

Hello. I never saw the edit till today. So basically in question, the tension is 0 and all masses and pulleys have acceleration g

Right. The puzzle is, how is this possible? To understand it, I recommend figuring out the accelerations of the pulleys.

 

[Physics freak]

Right. The puzzle is, how is this possible? To understand it, I recommend figuring out the accelerations of the pulleys.

The pulleys must also have acceleration g

 

[haruspex]

The pulleys must also have acceleration g

No. The strings have fixed lengths.

 

[Physics freak]

No. The strings have fixed lengths.

So they can reach equilibrium if the masses are same

 

[Physics freak]

So they can reach equilibrium if the masses are same

I mean the pulleys which are not fixed

 

[haruspex]

So they can reach equilibrium if the masses are same

I don't know what you mean.
If the masses are all descending at g but the string lengths are constant then it is evident from the diagram that some pulley or pulleys must be accelerating upwards.
To get a full understanding it helps to calculate those accelerations, which you can do by writing equations which represent the constancy of the string lengths.

 

[Physics freak]

I don't know what you mean.
If the masses are all descending at g but the string lengths are constant then it is evident from the diagram that some pulley or pulleys must be accelerating upwards.
To get a full understanding it helps to calculate those accelerations, which you can do by writing equations which represent the constancy of the string lengths.

But why are the pulleys also not accelearting with g

 

[haruspex]

But why are the pulleys also not accelearting with g

Consider the upper string. It connects to the centre of a pulley at one end, a mass at the other end, and supports a pulley in the middle. If the mass and both pulleys were descending at g then the string would be lengthening at 4g. That is not allowed.

 

[Physics freak]

Consider the upper string. It connects to the centre of a pulley at one end, a mass at the other end, and supports a pulley in the middle. If the mass and both pulleys were descending at g then the string would be lengthening at 4g. That is not allowed.

Oh finally i get it. Thanks a lot for help

 

[haruspex]

Oh finally i get it. Thanks a lot for help

Ok, but I still think it helps to finish the calculation.

Let the downward accelerations of the pulleys be a₁, a₂.
For the upper string to have constant length we have a₁+2a₂+g=0.
For the lower string, the lengths of the leftmost and central sections are accelerating at g-a₁, while for the rightmost it is g-a₂.
So 2(g-a₁)+(g-a₂)=0.
From these we get a₁=7g/3, a₂=-5g/3 (i.e. upwards).

I think what makes it appear impossible at first that the masses are in free fall is that the pulleys have to accelerate so much faster than g.