Is the Work Done by Static Friction Always Zero in a Round Trip?

[Nikhil Rajagopalan]

Friction is said to be a non conservative force. And I see various sources state that work done by a non conservative force on a round trip is always non zero. But is that always true?
In a case where a coin is placed on a turn table, and the table is rotated, and the coin does not move during one complete spin of the table, form a point of reference fixed outside the turn table, the coin has completed one round trip, with a force of static friction acting on it as the centripetal force which is always perpendicular to the circular path, hence making the work done, zero. Is this analysis correct or is anything technically wrong there.

 

[anorlunda]

In the reference frame of the turn table, the coin doesn't move at all. No trip, no energy.

 

[weirdoguy]

But is that always true?

Work done by static friction is AFAIK always zero, because it appears in "static situations".

 

[Nikhil Rajagopalan]

Thank you.

In the reference frame of the turn table, the coin doesn't move at all. No trip, no energy.

But, i believe that work done is dependent on the frame of reference that we choose. In that case, if we choose a frame of reference that is outside the turn table, there is a round trip and the work done is still zero. Is that consideration possible, can i say that work done is zero in the frame of reference that i chose.

 

[Khashishi]

Work done by static friction is AFAIK always zero, because it appears in "static situations".

That's totally wrong. If you put a box in the bed of a pickup truck and start driving (slowly enough such that the box doesn't slip off), the friction does work.

Friction is said to be a non conservative force.

I suppose only kinetic friction is nonconservative. Static friction is conservative.

 

[weirdoguy]

If you put a box in the bed of a pickup truck and start driving (slowly enough such that the box doesn't slip off), the friction does work.

Gosh, of course, my comment was stupid

 

[Nidum]

That's totally wrong. If you put a box in the bed of a pickup truck and start driving (slowly enough such that the box doesn't slip off), the friction does work.

So what exactly is the mechanics of the static friction doing work in that situation ?

 

[Khashishi]

The box was accelerated by the friction. Draw a free body diagram if you wish.

 

[Nikhil Rajagopalan]

The box was accelerated by the friction. Draw a free body diagram if you wish.

I totally agree with that and hence,depending on the frame of reference that we choose.Static friction can do positive negative or zero work. Static friction is said to be non-dissipative, and in the frame of reference where the surface is stationary it performs no net work. But now the question boils down to the nature of Static friction,if it is conservative or non conservative.

 

[anorlunda]

If the coin was doing work on the turntable because of friction, the coin and the table would both get warmer. Do you think that would happen in your scenario.

If the coin does work and if energy is conserved, where does the energy go?

 

[jbriggs444]

I totally agree with that and hence,depending on the frame of reference that we choose.Static friction can do positive negative or zero work. Static friction is said to be non-dissipative, and in the frame of reference where the surface is stationary it performs no net work. But now the question boils down to the nature of Static friction,if it is conservative or non conservative.

But what, exactly, do you mean by a "conservative" force?

Certainly, the force of static friction acting on a box on a flatbed truck need not do zero work as the truck completes a lap around a track.

 

[sophiecentaur]

But what, exactly, do you mean by a "conservative" force?

Certainly, the force of static friction acting on a box on a flatbed truck need not do zero work as the truck completes a lap around a track.

The problem with that approach is that it also suggests that all the intermolecular forces of the block and truck could also be 'doing work'. That doesn't take us anywhere useful does it?

 

[jbriggs444]

The problem with that approach is that it also suggests that all the intermolecular forces of the block and truck could also be 'doing work'. That doesn't take us anywhere useful does it?

If you want to talk about a conservative force, you have to know what "conservative" means. One meaning is that the force forms a conservative vector field. I do not know any other meaning. So there we are.

And yes, intermolecular forces do work.

 

[Cutter Ketch]

Status friction does not do work. Forget friction for a minute. Suppose you have a mass being pulled by a rope. Would you say that the rope is doing the work? The rope applies a force, the mass moves a distance, why isn't the rope doing work? Work is being done on the mass to be sure. But if you want to know if the rope is doing the work you have to add up the forces on it. There is the reaction force at the mass end, but there is an equal and opposite force doing the pulling at the other end. The rope isn't applying the force. It didn't gain or lose any energy in this process, and it didn't do any work. The same is true for static friction. It doesn't do work.

Regarding work being done on a turntable: no work is done. Consider a planet in orbit around the sun. If it required energy to keep the planets in orbit they would have decayed into the sun long ago. There is a force and there is displacement, why isn't there work? The force is always perpendicular to the motion. Work is force times displacement only for that portion of the displacement which is in the direction of the force. You can also see that the energy of the orbiting body doesn't change: same velocity, same gravitational potential energy. No work is being done. It is the same for the penny on the turn-table.

The box on a truck is a more apt example. The truck accelerated accelerating the box. Work was done on the box. Work was done through static friction, but static friction did not do the work. The truck did the work. (Technically the chemical energy in the gasoline and air did the work)

Static friction is not dissipative. It does not cost energy. However, that is NOT the same as saying that it is conservative. A conservative force is a force where the net work done by the force depends only on the net displacement. This description has no bearing in regards to a reaction force like static friction.

 

[jbriggs444]

Status friction does not do work

Static friction acting across a boundary from outside a system to inside a system can do work on that system. This is not open for debate.

The "static" means that the two obects are not in relative motion at the point of contact, not that this point is stationary.

 

[Cutter Ketch]

Static friction acting across a boundary from outside a system to inside a system can do work on that system. This is not open for debate.

That is true, but also irrelevant. That is a sufficient answer for how much work is done on a closed system but no answer at all for what is doing the work. This is exactly analogous to the rope and mass. When you want to discuss the closed system it is sufficient to add up the forces and say how much work was done. In that context it is correct to say x amount of work was done by the force from the rope. However it is incorrect to say the rope did x amount of work. The rope did not do the work. If you then want to discuss the nature of ropes and state that ropes do work you are 100% wrong. (Ok, some smartalec will come up with some wet rope getting shorter as it dries scenario, but you know what I mean)

Here we are discussing the nature of the force. Static friction is always a reaction force and it is always an agency for some other force. It never in itself imparts energy. It never does work. It can never be described as a scalar field and should never be called conservative.

 

[jbriggs444]

That is true, but also irrelevant. That is a sufficient answer for how much work is done on a closed system but no answer at all for what is doing the work.

It is true, full stop. If someone is asking whether the force of static friction does work, a correct answer to that question is very much relevant. The force of static friction can do work on a system. That's pretty much the definition of work in introductory physics: a force acting over a distance.

This is exactly analogous to the rope and mass. When you want to discuss the closed system it is sufficient to add up the forces and say how much work was done. In that context it is correct to say x amount of work was done by the force from the rope. However it is incorrect to say the rope did x amount of work.

You are correct that if one draws the lines around the closed system so that the rope is completely contained therein, then a [taut, inextensible, massless] rope can do no net work on the system. Any work done by one end will be undone by the other. However, we are not restricted to drawing the system boundaries in such a manner. A rope pulling a sled up a hill most certainly does work on the sled. Similarly, a hand grasping a rope can most certainly do work on the rope based on the static friction between hand and rope.

Here we are discussing the nature of the force. Static friction is always a reaction force and it is always an agency for some other force.

There is no difference between an action and a reaction force. In any case, the third law partner of static friction is... static friction.

 

[Cutter Ketch]

The "static" means that the two obects are not in relative motion at the point of contact, not that this point is stationary.

I have reread my post carefully and cannot guess to what this comment refers. Obviously I agree, anyway.

 

[jbriggs444]

I have reread my post carefully and cannot guess to what this comment refers. Obviously I agree, anyway.

You had claimed that the force of static friction does no work. One possible misconception that could lead to such a claim would be the idea that the point of application of a "static" force would necessarily be static.

 

[sophiecentaur]

The force of static friction can do work on a system.

I think this is turning into one of those PF discussions. If we are dealing with truly STATIC friction then there is no slipping so how is this 'static' force any different from all the other non-slipping forces (bonding) in the system? So they can all do work - if you are looking at the point at which the force (any of them) is acting. Is this conclusion of any use at all? How many actual angels are there on this particular pin head?

 

[Cutter Ketch]

You had claimed that the force of static friction does no work. One possible misconception that could lead to such a claim would be the idea that the point of application of a "static" force would necessarily be static.

There is no such implication.

Lest silence be taken for agreement, I will also say one more time that there is no misconception. No work can be derived from static friction. It does not do work. You cannot make a static friction engine. Noting that work is done on a system via a force of constraint is entirely different than ascribing agency to the force of constraint and claiming it did the work. Work is done on the system through the rope or through static friction, but it is logically incorrect to say either does the work. You can add up the forces on the system without knowing what is pulling on the other end of the rope, but it is always true that something is pulling on the other end of the rope. The rope never pulls all by itself. There is no available energy there.

Further, you cannot move against forces of constraint by definition. Static friction IS static in reference to the objects in contact. They cannot move with respect to each other against the force of static friction and so the force of static friction does no work. It can translate force from one object to another. It can be used to impart energy from one object to another, but it neither dissipates nor creates energy. The forces, accelerations, energy, and work are always imparted from elsewhere.

Now I will not state the same thing again after this, but please don't think that means I have changed my mind.

 

[sophiecentaur]

@Cutter Ketch I think your word "Agency" is the nub of this argument. I cannot see how you can possibly be wrong in this. Anyone who is disagreeing with you is clearly working with a different set of definitions.

 

[anorlunda]

The force of static friction can do work on a system. That's pretty much the definition of work in introductory physics: a force acting over a distance.

That's self contradictory. If it is static, there is no distance.

Take @sophiecentaur 's hint. If the attachment were by a bolt rather than friction, does it do work then? Do the separate parts of a rigid body do work independent of the body a whole?

I'll repeat my challenge from an earlier post. If the coin does work, and energy is conserved, where does the energy go?

 

[jbriggs444]

That's self contradictory. If it is static, there is no distance.

It is not contradictory. It is correct.

The "static" in static friction refers to the relative motion of two objects at their point of contact. It does not refer to a lack of motion of the point of contact.

I'll repeat my challenge from an earlier post. If the coin does work, and energy is conserved, where does the energy go?

In the rotating frame of reference in which the turntable is stationary, the coin does no work. The point of contact does not move.

In the inertial frame of reference in which the turntable rotates [at a uniform rate] around a stationary axis, the coin does no work. The point of contact has no component of motion in the direction of the applied force.

In pretty much every other frame of reference the coin does work at a non-zero rate throughout the rotation, but the total [for a uniformly rotating turntable] is zero.

If the turntable is decelerating as it turns and one adopts the inertial frame of reference in which the center of the turntable is at rest, the coin does positive work throughout the cycle.

 

[jbriggs444]

No work can be derived from static friction. It does not do work.

You are absolutely wrong about this and are using an incorrect definition of work.

 

[sophiecentaur]

It is not contradictory. It is correct.

The "static" in static friction refers to the relative motion of two objects at their point of contact. It does not refer to a lack of motion of the point of contact.

So what is the difference between a contact with static friction and a nut and bolt? You can say that every part of the chain is doing work on the next part but how is that useful? Or at least why does the friction situation have some extra significance?

 

[jbriggs444]

So what is the difference between a contact with static friction and a nut and bolt? You can say that every part of the chain is doing work on the next part but how is that useful? Or at least why does the friction situation have some extra significance?

Force is force. The dot product of force and distance gives work. Yes, every link in a chain does work on the next (and extracts work from the previous). The friction situation changes nothing. It's just a force.

As for the question of usefulness, it is useful because it is a simple consistent definition that always works the same way. You can draw the boundaries of a system where you like, tot up the work being done across the various interfaces and wind up with an energy balance equation.

 

[sophiecentaur]

The friction situation changes nothing. It's just a force.

That's fine - so there is no special significance and, as you say, each piece of the chain can be said to be doing work on the next. But the OP was about Friction being non conservative. That's pretty obvious and, when there is slippage, there is something significant about the transmission of power when friction is involved.

 

[jbriggs444]

That's fine - so there is no special significance and, as you say, each piece of the chain can be said to be doing work on the next. But the OP was about Friction being non conservative. That's pretty obvious and, when there is slippage, there is something significant about the transmission of power when friction is involved.

But that's not about being "conservative". That's about not being "dissipative". The two words address different concepts.

Again, I plead with you: Define your terms.

 

[anorlunda]

If there is no slippage, then we treat the turntable-coin system as a rigid body. We analyze rigid bodies as a whole with it's CG and moments of inertia. We can do work on the system.

I suppose one could analyze it one atom at a time with all the interatomic forces, but what's the point?

 

[jbriggs444]

If there is no slippage, then we treat the turntable-coin system as a rigid body. We analyze rigid bodies as a whole with it's CG and moments of inertia. We can do work on the system.

I suppose one could analyze it one atom at a time with all the interatomic forces, but what's the point?

You already split the problem into a coin and a turntable. One can draw a boundary there and ask about the work done across the boundary. No need to worry about molecules.

If you are going to ask a question about work done across an interface, do not object when someone answers your question.

 

[A.T.]

If it is static, there is no distance.

Distance is frame dependent, and so is the work done by static friction. You are misunderstanding what "static" means here.

 

[A.T.]

Static friction IS static in reference to the objects in contact.

Or rather in the rest frame of the contact patches. But in other frames static friction can do work.

 

[anorlunda]

In the reference frame of the interface between two surfaces, static friction does no work, because there is never displacement between the surfaces. In the same reference frame, kinetic friction is always in the direction opposite the motion, and does negative work.[64] However, friction can do positive work in certain frames of reference. One can see this by placing a heavy box on a rug, then pulling on the rug quickly. In this case, the box slides backwards relative to the rug, but moves forward relative to the frame of reference in which the floor is stationary. Thus, the kinetic friction between the box and rug accelerates the box in the same direction that the box moves, doing positive work.[65]

Are you guys saying that if the coin is attached by welding rather than friction the coin still does work in certain reference frames?

Are you rejecting the model of the coin-turntable system as a single rigid body?

 

[anorlunda]

You already split the problem into a coin and a turntable. One can draw a boundary there and ask about the work done across the boundary. No need to worry about molecules.

Distance is frame dependent, and so is the work done by static friction. You are misunderstanding what "static" means here.

I'm feeling dense, not understanding what you're saying.

With static friction, there is no motion between the two objects. Relative motion is the only thing that differentiates static friction from kinetic friction.

The coin moves zero displacement relative to the turntable. Zero distance remains zero in all frames of reference.