Normal force of real pulley: Which direction is it?

[Amin2014]

Consider a pulley fixed to the ceiling. A mass-less string is wrapped around it, with each side of the string hanging down either side of the pulley. Since the pulley has friction with the string, tension along the string will vary. Let's say the string is attempting to move clockwise, so the friction between the pulley and string is counter-clockwise. Which direction is the net normal force exerted on the string by the pulley?

Is my assertion correct that the net normal force would be tilted towards the right (the side with greater tension in the string)?

Also, today I suddenly found my self hung up over this seemingly simple notion: Why do we treat tension in the string as a scalar? For instance, when integrating over an element of the string, we treat the elements of normal force as vectors and the tension as scalar, why? I would appreciate if someone could give a really deep insight into this as I have an appetite for details!

 

[Doc Al]

A mass-less string is wrapped around it, with each side of the string hanging down either side of the pulley. Since the pulley has friction with the string, tension along the string will vary.

If the pulley is massless, why would the tension vary?

 

[Chestermiller]

I think you are thinking of a situation in which the pulley is not allowed to rotate so that the pulley can be replaced by a non-rotating cylinder. (In textile mechanics, be refer to such a cylinder as a pin.) Is this correct?

Chet

 

[Amin2014]

If the pulley is massless, why would the tension vary?

It varies because the string has friction with the pulley. Friction will cause a drop in tension along the string, much like friction can cause drop in potential in electric circuits.
If you write Newton's second law in the tangential direction for each element of the string, you will have to account for three forces, two tensions on either side of the element and friction. Therefore it is the net of these three forces that equals ma equals zero, not the net of the two tensions.

Integrating over the whole string would yield the following relation between T2 and T1 (taking μ as the coefficient of friction):
T2 = T1* e^ (μπ)

 

[Amin2014]

I think you are thinking of a situation in which the pulley is not allowed to rotate so that the pulley can be replaced by a non-rotating cylinder. (In textile mechanics, be refer to such a cylinder as a pin.) Is this correct?

Chet

Good question. In reality, we will have friction either way. The equation we are looking for is this: T2 = T1* e^ (μπ)

What type of pulley do you think that would imply? The solution involves taking μdN as the friction on an element of the string, acting opposite to the direction of motion of string. We can certainly use this type of equation when the pulley is fixed and the string is "skidding" on it, but what about a rotating pulley? How do you think that would change our equations?

 

[sophiecentaur]

I think the easiest argument would be based on Moments about the support point in the ceiling, which will not be equal if there is friction and the pulley spindle will experience a force which will move it about the 'vertical link' to the ceiling when T1 and T2 are not equal.

 

[Chestermiller]

Good question. In reality, we will have friction either way. The equation we are looking for is this: T2 = T1* e^ (μπ)

What type of pulley do you think that would imply? The solution involves taking μdN as the friction on an element of the string, acting opposite to the direction of motion of string. We can certainly use this type of equation when the pulley is fixed and the string is "skidding" on it, but what about a rotating pulley? How do you think that would change our equations?

This equation for the tension variation only applies if the string is slipping relative to the pulley; in that case, mu is the coefficient of kinetic friction. If you have static friction, this equation is incorrect, because the frictional force per unit length is less than the coefficient of static friction times the normal force per unit length. In the case of a rotating pulley, where the string is not slipping relative to the pulley surface (within the contact region), static friction prevails. It is certainly possible to have slippage of a yarn (or string) over the surface of a pulley, and even to have the yarn sticking over part of the pulley and slipping over another part of the pulley (if the yarn is extensible). In such cases, one would have to take into account the deformational strain-tension behavior of the yarn.

Chet

 

[Amin2014]

This equation for the tension variation only applies if the string is slipping relative to the pulley; in that case, mu is the coefficient of kinetic friction. If you have static friction, this equation is incorrect, because the frictional force per unit length is less than the coefficient of static friction times the normal force per unit length. In the case of a rotating pulley, where the string is not slipping relative to the pulley surface (within the contact region), static friction prevails. It is certainly possible to have slippage of a yarn (or string) over the surface of a pulley, and even to have the yarn sticking over part of the pulley and slipping over another part of the pulley (if the yarn is extensible). In such cases, one would have to take into account the deformational strain-tension behavior of the yarn.

Chet

I agree with you analysis, except for one thing: Why can't we have relative motion, and thus kinetic friction with rotating pulley? Can't the string and pulley both rotate, while the string rotates faster? In that case the equation I gave would still hold.

 

[sophiecentaur]

I agree with you analysis, except for one thing: Why can't we have relative motion, and thus kinetic friction with rotating pulley? Can't the string and pulley both rotate, while the string rotates faster? In that case the equation I gave would still hold.

That's what happens in most real circumstances any belt on a motor car that is not actually toothed, for example.

 

[Doc Al]

It varies because the string has friction with the pulley. Friction will cause a drop in tension along the string, much like friction can cause drop in potential in electric circuits.

I mistakenly assumed a massless pulley to go along with the massless string. My bad.

 

[Chestermiller]

I agree with you analysis, except for one thing: Why can't we have relative motion, and thus kinetic friction with rotating pulley? Can't the string and pulley both rotate, while the string rotates faster? In that case the equation I gave would still hold.

You can have this if the imposed tension difference across the pulley is large enough, but in continuous operation like a pulley on a car, the belt would wear out very quickly (thus, I respectfully disagree with Sophiecentaur). The extreme case of this is when a bearing on the pulley seizes up. I have had this happen with a car air conditioning compressor, and first you hear squealing and see smoke pouring out from under the hood, and then the belt breaks.

Chet

 

[Amin2014]

Thank you everyone for your input. Considering the pulley to be non-rotating, what's the answer to the OP? The question was to determine the direction of the normal force exerted on the string by the pulley.

 

[Chestermiller]

Thank you everyone for your input. Considering the pulley to be non-rotating, what's the answer to the OP? The question was to determine the direction of the normal force exerted on the string by the pulley.

At 12 o'clock, the force per unit length is up and to the left.

Chet

 

[Amin2014]

At 12 o'clock, the force per unit length is up and to the left.

Chet

I don't understand. Do you mean to the right? Referring to my picture, tension is greater in the right side of the string (T2>T1). How do you calculate the angle of the normal force with the vertical line?

 

[Chestermiller]

I don't understand. Do you mean to the right? Referring to my picture, tension is greater in the right side of the string (T2>T1). How do you calculate the angle of the normal force with the vertical line?

If the tension is greater to the right, the pulley must be applying a drag force to the left on the string so that T2 = T1 + drag force. At 12 o'clock, the normal component of force on the string is (geometrically) in the vertical direction. So the combined normal and drag force is up and to the left.

Chet

 

[Amin2014]

If the tension is greater to the right, the pulley must be applying a drag force to the left on the string so that T2 = T1 + drag force. At 12 o'clock, the normal component of force on the string is (geometrically) in the vertical direction. So the combined normal and drag force is up and to the left.

Chet

First of all, I asked about normal force, not the combination of normal force and friction. Secondly, if we break up the string into elements and write Newton's second law in the tangential and normal directions for each of those elements , like we would to derive the relation T2 = T1* e^ (μπ), we would get the following picture for the normal force acting on each element of the string (see attachment). According to this picture the net normal force is to the right. Note that we are taking a vector sum over the components to get the net N.

 

[Amin2014]

In the case of constant tension, each of those components would be equal in magnitude and their horizontal components would cancel out once we integrate. Therefore the net normal force in this case would be vertical, i.e. pointing to the ceiling. Remember that N = TdФ where Ф is the angle at the center of the pulley with which we are "counting" our system for integration.

 

[jbriggs444]

We already know that the net drag+normal force must be zero in the horizontal direction -- the total must be purely vertical and equal in magnitude to T2 + T1. If the net horizontal component of the drag is leftward (as it must be for any reasonable distribution of force around the pulley), it follows that the net normal force must be rightward.

It is conceivable that the force distribution of drag around the pulley is pathological. With static friction, the scenario is under-determined and there are multiple distributions of drag that match the givens of the problem. The bulk of the cord might actually be dragged rightward by the pulley except for tiny sections near either end. In that case the net normal force would be leftward.

 

[Chestermiller]

First of all, I asked about normal force, not the combination of normal force and friction. Secondly, if we break up the string into elements and write Newton's second law in the tangential and normal directions for each of those elements , like we would to derive the relation T2 = T1* e^ (μπ), we would get the following picture for the normal force acting on each element of the string (see attachment). According to this picture the net normal force is to the right. Note that we are taking a vector sum over the components to get the net N.

For good reason, I misinterpreted what you were asking for. I assumed you were talking about the local normal force at each location around the pulley, not the cumulative resultant of the normal components of force acting on the yarn along the pulley surface. It isn't clear to me why you would be interested in this. If you are interested in it for some reason, I can show you how to calculate its value precisely. In any event, it causes great confusion to call this resultant force a "net normal force," since it is not oriented perpendicular to the surface of the pulley at any location that has relevance. Note also that, if you are trying to find the total resultant force of the yarn on the pulley (or of the pulley on the yarn), you need to also include the frictional component of force, which also varies with angular location around the pulley.

Chet

 

[Chestermiller]

We already know that the net drag+normal force must be zero in the horizontal direction -- the total must be purely vertical and equal in magnitude to T2 + T1.

Why do you feel that the net force in the horizontal direction on the pulley (or the yarn) must be zero. The spindle of the pulley is capable of exerting a horizontal force on the pulley rotor.

Chet

 

[jbriggs444]

The momentum of the cord is unchanging horizontally. So the net horizontal force must be zero. The net force has contributions from T1 and T2 (both zero), from normal force and from tangential (aka drag) force. So the horizontal component of normal and drag force must sum to zero.

 

[Chestermiller]

The momentum of the cord is unchanging horizontally. So the net horizontal force must be zero. The net force has contributions from T1 and T2 (both zero), from normal force and from tangential (aka drag) force. So the horizontal component of normal and drag force must sum to zero.

The problem statement indicates that the string is mass-less, so its momentum is irrelevant. If we integrate both the horizontal and vertical components of force acting on the string over the surface of the pulley, I contend that we will find that there is a net horizontal component.

Well, on second thought, I'm not so sure. The yarn is still a body, and, if it's massless, the net horizontal force on it must be zero. So probably, integrating both forces over the pulley will result in no net horizontal force. I think you're probably right. I'm going to integrate the forces and see what I get.

Chet

 

[Chestermiller]

Jbriggs444! Good call. I did the integration, and, of course you were exactly right. The integration gave a zero net horizontal force of the pulley on the yarn, and a net vertical force of (T1+T2). Thanks very much for calling that to our attention.

Chet