What is the free-body diagram for a massless pulley in free fall?

[dyn]

Hi.
First of all i have a question regarding a simple Atwood's machine. The pulley has no mass and the string has no mass and is inextensible. If one pulley is suspended from a fixed support and it has 2 masses ; one at either end of the string. If i construct a free-body diagram of the pulley it has 2 forces directed upwards , both being the tension T. So as the pulley does not move , that means there must be a force of 2T acting downwards which is the force that the pulley exerts on the fixed support. If the pulley is suspended by a string , shouldn't the tension 2T be acting upwards ?

If the string suspending the pulley is now cut and the system goes into free-fall ; what does the free-body diagram look like for the massless pulley ?I have that the tension T is now zero so there are no forces acting upwards , but the pulley has no mass so has no weight. So , there are no forces on it , yet it is accelerating downwards ?

Thanks

 

[haruspex]

shouldn't the tension 2T be acting upwards ?

Tension, and likewise compression, is best thought of as a pair of equal and opposite forces. Tension T in each section of the lower string will act upwards on its attached weight and downwards on the pulley. The tension 2T in the top string acts upwards on the pulley, etc.
More generally, you may have a string with mass, so gravity or some other acceleration may lead to changes in tension along the string.

If the string suspending the pulley is now cut and the system goes into free-fall ; what does the free-body diagram look like for the massless pulley ?I have that the tension T is now zero so there are no forces acting upwards , but the pulley has no mass so has no weight. So , there are no forces on it , yet it is accelerating downwards ?

Idealisations, like massless and inextensible, are only valid if viewed as the limiting case as some parameter tends to zero or infinity. If the behaviour converges then we can take this as the behaviour of the ideal limit.
In the situation you describe, whatever the small mass you ascribe to the pulley it has acceleration g downwards, so we can take that as the massless behaviour.

It does happen that problem setters overlook this requirement for idealisations and set questions where their answer does not match the limiting behaviour.

 

[Steve4Physics]

Hi. Tension is not a simple force. If you stretch a rubber band between 2 hands, tension pulls your left hand to the right and your right hand to the left. Keep this mental image!

Tension acts inwardly. Look carefully at all the tensions on this diagram
https://i.stack.imgur.com/i4yek.png
(from https://physics.stackexchange.com/questions/194801/pulley-rope-tension-question)

The tensionc at the bottom of each vertical section of string acts upwards; the tension at the top acts downwards.

 

[Lnewqban]

... If one pulley is suspended from a fixed support and it has 2 masses ; one at either end of the string. If i construct a free-body diagram of the pulley it has 2 forces directed upwards downwards, both being the tension T.

 

[Dale]

If i construct a free-body diagram of the pulley it has 2 forces directed upwards ,

This is incorrect. Why do you think that?

 

[kuruman]

Hi.
First of all i have a question regarding a simple Atwood's machine. The pulley has no mass and the string has no mass and is inextensible. If one pulley is suspended from a fixed support and it has 2 masses ; one at either end of the string. If i construct a free-body diagram of the pulley it has 2 forces directed upwards , both being the tension T.

No, if you construct a free diagram of the pulley, the two tensions from the string supporting the masses will be downwards, i.e. away from the pulley as shown in th diagrams posted above by @Lnewqban.

So as the pulley does not move , that means there must be a force of 2T acting downwards which is the force that the pulley exerts on the fixed support. If the pulley is suspended by a string , shouldn't the tension 2T be acting upwards ?

Yes it should. That is consistent with the downward forces exerted by the two tensions as pointed out above.

If the string suspending the pulley is now cut and the system goes into free-fall ; what does the free-body diagram look like for the massless pulley ?I have that the tension T is now zero so there are no forces acting upwards , but the pulley has no mass so has no weight. So , there are no forces on it , yet it is accelerating downwards ?

Thanks

The massless pulley is an idealization so that you can ignore its moment of inertia and hence its angular acceleration. If the moment of inertia of the pulley divided by the square of its radius (a.k.a. its effective mass) is much smaller than either of the hanging masses, the effect of the pulley on the acceleration may be ignored. However, this is an approximation which becomes inappropriate when the upper string is cut because the pulley's mass in free fall will have an acceleration equal to $g$, the same as the larger masses.

 

[dyn]

This is incorrect. Why do you think that?

I was wrong. The diagram in #3 was great and pointed out where i was wrong. Thanks for everyone's replies.

I have one last question ; it specifically refers to Introduction to Classical Mechanics : with Problem and Solutions by David Morin. At the bottom of P58 it states that the tension in the short string connected to m₂ is 2T because there must be zero net force on the massless right pulley otherwise it would have infinite acceleration. But on P59 the acceleration of m₂ is calculated which means that m₂ and thus the right hand pulley is accelerating with a finite acceleration. I'm confused !

 

[kuruman]

I was wrong. The diagram in #3 was great and pointed out where i was wrong. Thanks for everyone's replies.

I have one last question ; it specifically refers to Introduction to Classical Mechanics : with Problem and Solutions by David Morin. At the bottom of P58 it states that the tension in the short string connected to m₂ is 2T because there must be zero net force on the massless right pulley otherwise it would have infinite acceleration. But on P59 the acceleration of m₂ is calculated which means that m₂ and thus the right hand pulley is accelerating with a finite acceleration. I'm confused !

I too am confused because I don' have access to the book you mentioned and I cannot picture what you are talking about. You mention a right hand pulley. Is there a left hand pulley? Is this a not a simple Atwood machine? Please post a drawing or, even better, start a new thread if this is a different question.

 

[dyn]

This is an Atwood's machine with 2 pulleys. I just posted it here because it was my last question on the topic of Atwood's machines

 

[Dale]

But on P59 the acceleration of m2 is calculated which means that m2 and thus the right hand pulley is accelerating with a finite acceleration

As @kuruman said it is impossible to be specific without the details. However, in general massless objects must have zero net force as you said. However, they may have any finite acceleration while still having 0 net force. (How many N are needed for a 0 kg object to accelerate at 10 m/s^2). You have to determine their acceleration from constraints.

 

[Lnewqban]

I was wrong. The diagram in #3 was great and pointed out where i was wrong. Thanks for everyone's replies.

I have one last question ; it specifically refers to Introduction to Classical Mechanics : with Problem and Solutions by David Morin. At the bottom of P58 it states that the tension in the short string connected to m₂ is 2T because there must be zero net force on the massless right pulley otherwise it would have infinite acceleration. But on P59 the acceleration of m₂ is calculated which means that m₂ and thus the right hand pulley is accelerating with a finite acceleration. I'm confused !

If you imaginarily isolate that right hand pulley and consider all the external forces acting on it, you will be creating its free body diagram.
Those external forces may be or not balanced, but a first step is to imagine that all those forces cancel each other.

By doing so, you can see that the weight of mass 2 must be compensated by two tensions of the U-shaped string, having both the same magnitude.
If the magnitude of that tension equals half that weight, the pulley will not be accelerating.
If it doesn't, the pulley will be accelerating upwards or downwards.

The masses and pulleys of the machine are actually moving in an accelerated fashion under the action of gravity on each of its masses.
Mass 1 is accelerating upwards while mass 2 is accelerating downwards at half the rate of 1, while the tension and length of the string remains constant.

 

[dyn]

Thank you everyone

 

[dyn]

Hi
I have just had one minor thought niggling away at me. For a massless object , it must have zero force acting on it otherwise it would have infinite acceleration from Newton's 2nd Law. So accelaration , a = F/m so for m=0 are we saying that with F=10 for example , a=10/0 = ∞ and with F=0 , a= 0/0 = any finite number ? In other words are we ignoring the mathematical problems of dividing by zero ? Can 0/0 be defined as a finite number ?
Thanks

 

[Dale]

F=0 , a= 0/0 = any finite number ? In other words are we ignoring the mathematical problems of dividing by zero ?

Write it as $F=ma$ then there is no division. For $m=0$ you have $F=0$ for any finite $a$. Newton's 2nd law cannot tell you what $a$ is, that is why you need the constraints, but there is no division by zero problem.

 

[kuruman]

Hi
I have just had one minor thought niggling away at me. For a massless object , it must have zero force acting on it otherwise it would have infinite acceleration from Newton's 2nd Law. So accelaration , a = F/m so for m=0 are we saying that with F=10 for example , a=10/0 = ∞ and with F=0 , a= 0/0 = any finite number ? In other words are we ignoring the mathematical problems of dividing by zero ? Can 0/0 be defined as a finite number ?
Thanks

Perhaps this might help. In this two-pulley problem, think of the pulleys as grooved slippery rings such that the rope slides without friction in the grooves (the rings do not rotate). The mass of the left pulley is irrelevant while the mass of the right pulley can be incorporated into $m_2$. In that case, the analysis is exactly the same. The right pulley can be considered massless, nevertheless the rope has to slip around it as $m_2$ accelerates. That is what I think @Dale means by "constraints".

 

[dyn]

Write it as $F=ma$ then there is no division. For $m=0$ you have $F=0$ for any finite $a$. Newton's 2nd law cannot tell you what $a$ is, that is why you need the constraints, but there is no division by zero problem.

That totally makes sense. I was just thinking like a mathematician for a second and considering the division by zero issue

 

[haruspex]

Hi
I have just had one minor thought niggling away at me. For a massless object , it must have zero force acting on it otherwise it would have infinite acceleration from Newton's 2nd Law. So accelaration , a = F/m so for m=0 are we saying that with F=10 for example , a=10/0 = ∞ and with F=0 , a= 0/0 = any finite number ? In other words are we ignoring the mathematical problems of dividing by zero ? Can 0/0 be defined as a finite number ?
Thanks

I refer you back to the second part of my post #2. Masslessness is an idealisation that does not arise in the real world. It should always be interpreted as a limiting case, not an actual case. Solve the problem for a small but nonzero mass, then see what happens as that mass tends to zero. If the answer converges to a limit then you can claim that as the answer to the massless idealisation.

 

[dyn]

Thanks. Does 0/m converge to a limit as m tends to zero ?

 

[haruspex]

Thanks. Does 0/m converge to a limit as m tends to zero ?

Yes. It is 0 for all m>0, so in the limit it is zero. But I don't see the relevance to your preceding questions.

 

[dyn]

Yes. It is 0 for all m>0, so in the limit it is zero. But I don't see the relevance to your preceding questions.

Because acceleration = force/mass so for zero force and mass tending to zero that would give the acceleration but i do not know if that limit even exists. From earlier posts we have zero force with zero mass gives any finite acceleration so that that limit should be finite but not determined

 

[Dale]

Because acceleration = force/mass so for zero force and mass tending to zero that would give the acceleration but i do not know if that limit even exists. From earlier posts we have zero force with zero mass gives any finite acceleration so that that limit should be finite but not determined

You are putting the cart before the horse here. You are using the fact that the mass is 0 to assert that the force is zero. You cannot then go and take a limit as m approaches 0, it’s already 0.

To do it this way you need to work the whole problem with a nonzero m and at the end take the limit as m goes to zero.

Or you can just use the acceleration required by whatever constraints are involved.

 

[dyn]

I think I'm just trying to confuse myself now. I will just take it that massless objects must have zero force acting on them otherwise they would have infinite acceleration and leave it at that, Thanks again everyone

 

[vanhees71]

Hi
I have just had one minor thought niggling away at me. For a massless object , it must have zero force acting on it otherwise it would have infinite acceleration from Newton's 2nd Law. So accelaration , a = F/m so for m=0 are we saying that with F=10 for example , a=10/0 = ∞ and with F=0 , a= 0/0 = any finite number ? In other words are we ignoring the mathematical problems of dividing by zero ? Can 0/0 be defined as a finite number ?
Thanks

The conclusion is that massless particles don't make sense in Newtonian physics.

 

[Lnewqban]

Hi
I have just had one minor thought niggling away at me. For a massless object , it must have zero force acting on it otherwise it would have infinite acceleration from Newton's 2nd Law. So accelaration , a = F/m so for m=0...

I believe that your confusion comes from the "Solution" paragraph in the attached picture.

For solving problems involving pulleys and strings or ropes, we frequently assume that their masses and friction forces are negligible in comparison to the masses and external forces acting on the system of pulleys.
By doing so, we make the calculations of those easier.

In real life, the rope and the pulleys have masses that accelerate and the rope suffers some degree of elongation.
There is also resistive friction everywhere.

A body accelerates in direct proportion to the external forces acting on it.
At the same time, the smaller its mass is, the greater acceleration it experiments under that force.

 

[anorlunda]

In real life, the rope and the pulleys have masses that accelerate and the rope suffers some degree of elongation.
There is also resistive friction everywhere.

Don't forget that in real life it takes force to bend/unbend cables and ropes. Without considering the bending, the sheaves of the pulleys can be arbitrarily small in diameter, but in real life, we make them have large diameters.

 

[dyn]

The conclusion is that massless particles don't make sense in Newtonian physics.

I agree which makes it confusing that massless pulleys and strings are used to teach Newtonian mechanics !

 

[Dale]

They do make sense, you just have to follow the rules. Don’t divide by zero. The net force is 0. The acceleration can be any finite value needed.

If you forget the rules then you can always derive the rules by assuming that it has mass and taking the limit as it’s mass goes to 0.

 

[A.T.]

Can 0/0 be defined as a finite number ?

Yes, because any finite number x satisfies the equation 0 * x = 0.

See also:
https://en.wikipedia.org/wiki/Indeterminate_form

I was just thinking like a mathematician for a second and considering the division by zero issue.

What issue would a mathematician have with the above?

 

[dyn]

Yes, because any finite number x satisfies the equation 0 * x = 0.

See also:
https://en.wikipedia.org/wiki/Indeterminate_formWhat issue would a mathematician have with the above?

Because mathematicians will not divide by zero ; they might take limits but they won't divide by zero.

Incidentally the wikipedia article you quote gives an example of the limit of x/x³ as x tends to zero as infinity

 

[A.T.]

Because mathematicians will not divide by zero

Neither did I.

 

[Lnewqban]

I agree which makes it confusing that mass-less pulleys and strings are used to teach Newtonian mechanics !

Hope we are not confusing you even more going deeper into the subject.

Also, levers, gears, wedges, belts and slopes, and any part of simple machines, are frequently assumed to have negligible mass, for the very same reasons of simplifying calculations and eliminating the effect of their individual accelerations. By doing so, we devote our neurones solely to the effect of the mechanical energy input into the system on the relatively big and important masses.

Please, see:
https://en.m.wikipedia.org/wiki/Simple_machine

Same concept applies to deflection, stretching, friction and wear of those parts.
In order to make the learning process less confusing, we want to imaging that exactly the same amount of energy or work put into the simple machine goes out at the opposite end of it.

In the case of ideal problems involving mechanical advantage (MA), we assume a theoretical efficiency of 100%.
In practical or experimental problems, where the above assumptions can’t be made, there is a practical MA which magnitude is always less than the ideal MA.

 

[Nugatory]

I agree which makes it confusing that massless pulleys and strings are used to teach Newtonian mechanics !

Yes, and the only reason we can get away with it is that they’re always attached to something with non-zero mass and we’re applying a force to the whole thing.