Understanding Pulleys, Strings, and Tensions: Common Doubts and Clarifications

[Tanya Sharma]

I am having few doubts regarding pulley,strings and tensions .

The doubts are

1)Can a massless pulley rotate ?

My reasoning - Yes .Just like a massless string can accelerate with zero net force .The same way a massless pulley can rotate with zero net torque.

2)Is tension in a massless string always same when it moves over a pulley .The string may be slipping(not sufficient friction) ,not slipping(sufficient friction,so moving along with the rotating pulley).

3)If the pulley is massless and the string is not massless,what are the tensions on the two sides?

My thoughts-Since the pulley is massless,the tensions at the end points of segment of string over the pulley should be same.But the tension in a string having mass is not same at any two points .

4)How does friction between string and the pulley affect the tension on the two sides of the string?

Kindly help me in clearing the concepts.

 

[Doc Al]

1)Can a massless pulley rotate ?

My reasoning - Yes .Just like a massless string can accelerate with zero net force .The same way a massless pulley can rotate with zero net torque.

Sure. Realize that a massless pulley is just an idealization, of course. Real pulleys have mass.

2)Is tension in a massless string always same when it moves over a pulley .The string may be slipping(not sufficient friction) ,not slipping(sufficient friction,so moving along with the rotating pulley).

When discussing friction in a pulley, usually they mean the friction about the axle not the surface of the pulley where it touches the string. So, if the pulley has friction, the strings must exert unequal tensions to overcome that friction.

3)If the pulley is massless and the string is not massless,what are the tensions on the two sides?

My thoughts-Since the pulley is massless,the tensions at the end points of segment of string over the pulley should be same.But the tension in a string having mass is not same at any two points .

Interesting question! Since the segment of the string over the pulley has mass, I'd say that the tensions must be unequal at the endpoints.

4)How does friction between string and the pulley affect the tension on the two sides of the string?

For a massless string over a massless pulley, the tension will be the same. (Friction would be irrelevant.)

I'm not sure I've addressed your concerns. If not, try again. Is there a particular sort of problem that you want to solve?

 

[Tanya Sharma]

Hi Doc Al

Thank you very much for the response .But things are still not clear to me.

I understand what happens when the pulley is considered massless .If the pulley has mass,then we need to consider its inertia as well as rotational inertia.

But the role of friction between pulley and string and how it affects the tension in the string is not clear.

Again role of mass of string in affecting the tensions on the two sides of the string is not clear.

I would be grateful if you could elaborate on these issues .Thanks :)

When discussing friction in a pulley, usually they mean the friction about the axle not the surface of the pulley where it touches the string.

Does that mean,in the problems we encounter the pulley always rotates along with the string ,irrespective of whether the pulley is massless or has mass ?

So, if the pulley has friction, the strings must exert unequal tensions to overcome that friction.

Here you mean friction at the axle ?

Interesting question! Since the segment of the string over the pulley has mass, I'd say that the tensions must be unequal at the endpoints.

Are tensions unequal due to string having mass or because static friction is present between string and pulley ?

For a massless string over a massless pulley, the tension will be the same. (Friction would be irrelevant.)

Again,the tensions are same due to massless string or massless pulley ?

Is there a particular sort of problem that you want to solve?

No. There isn't specific problem in my mind

 

[Buckleymanor]

Sure.

Interesting question! Since the segment of the string over the pulley has mass, I'd say that the tensions must be unequal at the endpoints.

Would that not depend on the length of string over the pulley.Same length over each side, same tensions at the end points.

 

[jbriggs444]

Would that not depend on the length of string over the pulley.Same length over each side, same tensions at the end points.

Negative.

Possibly you are thinking of an arrangement where the string is hanging over the top of a pulley and is either stationary or moving at a fixed speed. Or, similarly, of an arrangement where a pulley is hanging from a loop of string that is either stationary or moving at a fixed speed.

However, if the pulley is being pulled horizontally then gravity can cause a change in tension over the length of the string where it is adjacent to the pulley. Tension at the top will exceed tension at the bottom.

If the string is accelerating then that acceleration will be associated with a change in tension over the length where it is adjacent to the pulley. Otherwise it would not be accelerating.

 

[Tanya Sharma]

I think I have figured out most of the things.But i am still unclear about one thing . When the pulley has mass,with massless string around,having sufficient friction present(i.e no slipping between string and the pulley),the tension on the two sides is unequal .

How is that the net force on the massless string is zero ?

 

[jbriggs444]

I think I have figured out most of the things.But i am still unclear about one thing . When the pulley has mass,with massless string around,having sufficient friction present(i.e no slipping between string and the pulley),the tension on the two sides is unequal .

How is that the net force on the massless string is zero ?

There are four forces on a massless string wrapped around part of a pulley.

1. The tension on the one end.
2. The tension on the other end.
3. The normal force from the pulley applied throughout the arc.
4. The tangential force from the pulley applied throughout the arc.

Number 3 does not contribute to tension.

Number 4 is the interesting one. If the pulley has mass and is accelerating, there must be an unbalanced tangential force between the string and the pulley.

Newton's second law applied to the string says f = ma. But if m is zero, the net force on the string must be zero. If there is unbalanced tangential force contributed by the pulley, that must be balanced by a difference in the tension at the two ends of the string.

If you are a bicycle rider, one way to think about it is to consider the chain leading to your back wheel as if it were a string going to a pulley. The top of the chain is under a great deal of tension. The bottom, not so much.

 

[Tanya Sharma]

Hi jbriggs444

Thanks for the response...Isnt tension in the string throughout the arc ,apart from being at the ends?

 

[Doc Al]

Hi jbriggs444

Thanks for the response...

Yes, thanks for that excellent response. (I have been too busy to respond for the last few days.)

Isnt tension in the string throughout the arc ,apart from being at the ends?

The tension will vary along the arc, but any section will have a net force of zero.

 

[Tanya Sharma]

Hi Doc Al

The tension will vary along the arc, but any section will have a net force of zero.

Okay... And the tension remains same when the string is out of the arc .Is that so?

 

[Doc Al]

Okay... And the tension remains same when the string is out of the arc .Is that so?

Right.

 

[Tanya Sharma]

So the difference between a massless string and a string having mass can be stated as -

Massless string -The tension varies along the arc,remains constant out of the arc .

String having mass -The tension varies throughout the length i.e varies along the arc and also varies out of the arc .

Is that so?

 

[Doc Al]

So the difference between a massless string and a string having mass can be stated as -

Massless string -The tension varies along the arc,remains constant out of the arc .

Yes, for a massless string over a massive pulley.

String having mass -The tension varies throughout the length i.e varies along the arc and also varies out of the arc .

Is that so?

Yes. I think you've got it.

 

[Tanya Sharma]

Thanks Doc Al

Now considering the massive pulley,the net tangential force on the pulley is the frictional force which will be equal to the difference in tensions at the two ends of the arc .

Am I right ?

 

[Doc Al]

Now considering the massive pulley,the net tangential force on the pulley is the frictional force which will be equal to the difference in tensions at the two ends of the arc .

Am I right ?

Yes, that makes sense to me.

 

[Buckleymanor]

Negative.

Possibly you are thinking of an arrangement where the string is hanging over the top of a pulley and is either stationary or moving at a fixed speed. Or, similarly, of an arrangement where a pulley is hanging from a loop of string that is either stationary or moving at a fixed speed.

However, if the pulley is being pulled horizontally then gravity can cause a change in tension over the length of the string where it is adjacent to the pulley. Tension at the top will exceed tension at the bottom.

If the string is accelerating then that acceleration will be associated with a change in tension over the length where it is adjacent to the pulley. Otherwise it would not be accelerating.

No I was thinking about

Would that not depend on the length of string over the pulley.Same length over each side, same tensions at the end points.

No mention of movement and no mention of orientation or speed.
So I reckon that it is too much for anyone to presume it's just a pulley with a stationary piece of string hanging over it with both ends dangling down towards the ground with both ends the same length.
Which I reckon you understood first time round but did not like it's implication or lack of for some reason.

 

[Tanya Sharma]

Thanks jbriggs444 and Doc Al

 

[Tanya Sharma]

For a massless string over a massless pulley, the tension will be the same. (Friction would be irrelevant.)

I didnt quite understand it.

If the pulley is massless ,it requires zero torque to rotate it.This means the tensions at the two ends of the string over the arc are same .But,if there is friction present between string and the pulley,in order to overcome that friction,the tensions need to be unequal.

How would tension be the same and friction becoming irrelevant ?

 

[Doc Al]

I didnt quite understand it.

If the pulley is massless ,it requires zero torque to rotate it.This means the tensions at the two ends of the string over the arc are same .But,if there is friction present between string and the pulley,in order to overcome that friction,the tensions need to be unequal.

How would tension be the same and friction becoming irrelevant ?

How much friction is required to turn a massless pulley?

 

[Tanya Sharma]

How much friction is required to turn a massless pulley?

To turn a massless pulley,zero torque is required.This in turn means,no friction is required.The friction required is equal to the difference in the tensions on the end points of the arc.Hence tension on the two sides is same.

Am I correct ?

If I am correct , does that mean even though sufficient static friction is present ,it doesn't come into picture due to pulley being massless ?

Strangely,I am finding it difficult to comprehend the situation.

 

[Doc Al]

To turn a massless pulley,zero torque is required.This in turn means,no friction is required.The friction required is equal to the difference in the tensions on the end points of the arc.Hence tension on the two sides is same.

Am I correct ?

Right.

If I am correct , does that mean even though sufficient static friction is present ,it doesn't come into picture due to pulley being massless ?

The static friction force is zero. The surfaces may support friction, if needed, but no friction force is needed.

 

[Tanya Sharma]

Thanks Doc Al

Even though I have done quite a few problems with pulleys ,I am managing to confuse myself very nicely :shy: .

Why don't we take static friction into account while drawing FBD of a massive pulley in intro physics problems ?

It doesn't do any work,but why don't we consider it while writing force equations?

 

[Doc Al]

Since the difference in string tension will equal the required friction, that's all you need to worry about.

 

[Tanya Sharma]

Thanks Doc...You are beautifully clearing my doubts :)

If the string has mass ,pulley is massless and frictionless,then the tensions at the end points of the arc are unequal.How is mass of the string creating a difference in the tensions,even though the pulley is massless and frictionless ?

 

[Doc Al]

If the string has mass ,pulley is massless and frictionless,then the tensions at the end points of the arc are unequal.How is mass of the string creating a difference in the tensions,even though the pulley is massless and frictionless ?

The string segments have mass and thus require a net force to accelerate.

 

[Tanya Sharma]

Thanks Doc...You have been absolutely wonderful

 

[Tanya Sharma]

Hi Doc

When discussing friction in a pulley, usually they mean the friction about the axle not the surface of the pulley where it touches the string. So, if the pulley has friction, the strings must exert unequal tensions to overcome that friction.

The friction about the axle is confusing me in case of a massless pulley.If there is friction present at the axle ,we would write

Net Torque =Iα . Now since I=0 ,the net torque required is zero i.e tensions on either side are equal.

But this is contrary to what you have mentioned.May be I am interpreting it incorrectly .Kindly explain.

 

[jbriggs444]

The friction about the axle is confusing me in case of a massless pulley.If there is friction present at the axle ,we would write

Net Torque =Iα . Now since I=0 ,the net torque required is zero i.e tensions on either side are equal.

If there is friction present at the axle then the net torque has three contributions. The torque from the axle, the torque from the tension at one side and the torque from the tension at the other side.

 

[Tanya Sharma]

Oh...I have accidently deleted my post in trying to edit it...I am extremely sorry

If there is friction present at the axle then the net torque has three contributions. The torque from the axle, the torque from the tension at one side and the torque from the tension at the other side.

Thanks jbriggs

Okay so the torque equation should be

(F-f)R = Iα where,

F=Friction present between pulley and the string ,such that F=T1-T2
f=friction at the axle

Is it correct ?

Edit :The deleted post has been restored.Thanks

 

[jbriggs444]

Okay so the torque equation should be

(F-f)R = Iα where,

F=Friction present between pulley and the string ,such that F=T1-T2
f=friction at the axle

The frictional force at the axle would be multiplied by the radius of the axle to get the resulting frictional torque.

So you would have:

Iα = (T2-T1)R - fr​

Where T1 is the one tension, T2 is the other tension, R is the pulley radius, r is the axle radius and f is the frictional force at the axle.

 

[Tanya Sharma]

The frictional force at the axle would be multiplied by the radius of the axle to get the resulting frictional torque.

So you would have:

Iα = (T2-T1)R - fr​

Where T1 is the one tension, T2 is the other tension, R is the pulley radius, r is the axle radius and f is the frictional force at the axle.

Thanks...Things are getting clearer...Now please consider a frictionless pulley with sufficient friction between string and pulley.

Why do we write net torque on the pulley as (T2-T1)R ?

The tangential force acting on the pulley is friction F and not the tension in the string.Tension doesn't act on the pulley.

So,the torque on the pulley is FR.

But,since the friction F is equal to the difference in the tensions at the two ends of the arc we write the torque as (T2-T1)R .

Is my understanding correct ?

 

[jbriggs444]

But,since the friction F is equal to the difference in the tensions at the two ends of the arc we write the torque as (T2-T1)R .

Is my understanding correct ?

Yes.

Note that there is some lack of rigor talking about "friction F" as if it had a well-defined value. Force is a vector. The friction between the string and the pulley is not a single well defined vector whose magnitude is "F". Instead, it is spread out (and may not be spread out evenly) over the surface of the pulley. Fortunately for our purposes, every incremental bit of string that is stretched out over the surface of a circular pulley contributes its frictional force with a moment arm of length R at right angles to the incremental force. So we can safely shrug, treat the net tangential F as if it were a simple number and correctly conclude that its value is T2-T1.

 

[Tanya Sharma]

Thanks jbriggss...The explanation is simply wonderful.

Loosely speaking,this is how I thought that the net friction would be equal to the difference in tensions.

The forces acting on a tiny bit of a string are the net tension difference,say dT and friction f .Now,since the string is massless,the net force on every bit of the string should be zero.So we have dT=f .Summing them along the arc,we get ΔT=F or T2-T1 =F.

Is it correct to think in this manner ?

 

[jbriggs444]

Loosely speaking,this is how I thought that the net friction would be equal to the difference in tensions.

The forces acting on a tiny bit of a string are the net tension difference,say dT and friction f .Now,since the string is massless,the net force on every bit of the string should be zero.So we have dT=f .Summing them along the arc,we get ΔT=F or T2-T1 =F.

Is it correct to think in this manner ?

Yes. This is a good way to come up with a number "F" that is, in some perfectly reasonable sense, the total friction.

The lack of rigor that I had in mind comes up when multiplying this "F" by R to try to compute torque. The result has units of torque and actually gives the right torque in the case at hand, but it would not be a justifiable computation in the case of arbitrary (non-circular) pulleys. There are also complications if the string is wrapped in a slanted or crooked path around the pulley.

 

[jbriggs444]

Loosely speaking,this is how I thought that the net friction would be equal to the difference in tensions.

The forces acting on a tiny bit of a string are the net tension difference,say dT and friction f .Now,since the string is massless,the net force on every bit of the string should be zero.So we have dT=f .Summing them along the arc,we get ΔT=F or T2-T1 =F.

Is it correct to think in this manner ?

Yes. This is a good way to come up with a number "F" that is, in some perfectly reasonable sense, the total friction.

The lack of rigor that I had in mind comes up when multiplying this "F" by R to try to compute torque. The result has units of torque and actually gives the right torque in the case at hand, but it would not be a justifiable computation in the case of arbitrary (non-circular) pulleys. There are also complications if the string is wrapped in a slanted or crooked path around the pulley.