# Direction of Kinetic Friction and Static Friction

• jeff12

#### jeff12

Can someone please explain me the direction of static friction? I know kinetic friction is always going against the motion of an object but for static friction it depends.

Can someone please explain me the direction of static friction? I know kinetic friction is always going against the motion of an object but for static friction it depends.
jeff12, the static friction is a kind of reaction. That means that it is directed against the force with which the body acts on a surface.

Can someone please explain me the direction of static friction? I know kinetic friction is always going against the motion of an object but for static friction it depends.
Static or dynamic friction always act in a direction opposite to the impressed force. If you apply that rule in a totally rigid way, there is really no problem. However, there can be an intuitive problem which makes things look wrong.
The problem often arises when you are considering wheels, driving a vehicle. When you operate the accelerator, there is a force, backwards, against the road surface and a reaction force will be stopping the wheels from slipping (before you have exceeded the limiting static friction). The contact point with the ground will act as a fulcrum (instantaneously) and the torque at the axle will produce a forward accelerating force on the car (as the wheel turns). If there is slipping, the friction force is less and so the accelerating force on the car will be less. When there is slipping, there is a Force times velocity involved and the tyres can get hot and engine power will actually be wasted in producing the heating effect.

Static or dynamic friction always act in a direction opposite to the impressed force.
The dynamic friction does not need in any force. It is really directed in opposit to the velosity. And the static friction can never be without the shearing force and it is always directed in opposit to it. However, you are right - dynamic friction forces can or slow down the movement of the body or accelerate it. But they are always slow down the part of the body that is in dynamical contact with friction surface. For example, "they are trying to slow down" the rotation of the drive wheels and thus accelerate the car.
But TC asked about the static friction force direction. If the non compensated tangential to contact surface force is present the static friction force is directed in opposit to it. And it compensate it until the $k_f \cdot N$ is more or equal to that force.

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The dynamic friction does not need in any impressed force.

If you are talking in terms of the dynamic friction on a 'coasting' object, it is directed against the motion and, likewise, dynamic friction is acting against the motion of a driving wheel over the road surface (and that can involve forward or backward motion, depending on the actual situation). The same rule applies and there is no contradiction. I can't see the OP's problem with the direction of the static friction force as long as we are talking of the forces rather than the motion. Static friction is the odd one out. in as far as it only works 'against' a force and can't be acting in a dynamic situation (slipping). But is this confusing? I don't see how.

As for the dynamic friction force, you can generalise by saying that it will be a cause of energy loss, whether in a slipping drive or a brake. It produces negative (vectorially) acceleration, in both situations, in as far as the reaction (accelerating) force on the road is against the direction that it is applied by the wheel.

But, to be honest, there are far too many posts which are centred on examining how a classification produces confusion during an arm waving discussion. This is one of them. Once you get down to drawing a proper diagram of any situation involving friction, including all forces and velocities, it will give you an equation of motion which can be solved and give the correct prediction. What more can a chap want?

The direction of static friction is decided like this. Imagine that there is no friction, figure out which way the object would slide in such a case. Static friction will be opposite to that direction.

Can someone please explain me the direction of static friction? I know kinetic friction is always going against the motion of an object but for static friction it depends.
Actually static friction is a self adjusting force.The maximum static friction that can act between the surfaces in contact is directly proportional to the normal reaction.The formula is f=μsN.If the applied force is less than the maximum static frictional force then the static frictional force that acts between the surfaces is equal and opposite to the force applied.

n3pix
It is because when my professor he explained it using two blocks, the big one on the bottom and a small on top and he said the static friction was moving in the direction the objects were moving.

jeff12, the static friction is a kind of reaction. That means that it is directed against the force with which the body acts on a surface.
Thats just Newtons 3rd Law, which also applies to dynamic friction.

Can someone please explain me the direction of static friction? I know kinetic friction is always going against the motion of an object but for static friction it depends.
Yes, it depends. You have to look at all other forces on the body to find the static friction required to hold the contact static.

Can you give me an example of it not going opposite of the motion of the object?

Can you give me an example of it not going opposite of the motion of the object?
No one surface moves (relative one to another) when the static friction force acts between them. That force is not "working force" since there is no displacement. Can you give an example where "object motion" is present and it have sense in the static friction context? Or can you explane more detal of your professor example, please? Remember that we are talking about static (non dynamic) friction force.

But, to be honest, there are far too many posts which are centred on examining how a classification produces confusion during an arm waving discussion.
The topic caster spoke about the professor and the two bodies ... Maybe he did not understand the example and then professor of law but we do not know what. But if the meaning is transmitted correctly, the professor also is not to blame. Just when he was a student, another professor showed him this example. The idea of static friction connection with the motion is just of plain bad. But if we do this it must be said that the static friction is directed in oposit to the acceleration of the system and not to its movement (velocity). Application of a non-inertial frame of reference to illustrate the static friction is not justified. This technique is rarely justified.

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Can someone please explain me the direction of static friction? I know kinetic friction is always going against the motion of an object but for static friction it depends.
Just remember that friction opposes slipping between surfaces. For the static case, if you can figure out which way the surfaces would tend to slip (in the absence of friction), then friction will act to oppose that slipping.

It is because when my professor he explained it using two blocks, the big one on the bottom and a small on top and he said the static friction was moving in the direction the objects were moving.
Let's break that down. Say the big block is being pushed to the right by some force and the small block is going along for the ride. So you know that it's static friction accelerating the small block and the static friction on the small block must act to the right. Thinking in terms of surfaces: without friction, the top block would tend to slip to the left with respect to the bottom block, and the static friction must oppose that and act to the right.

Of course, it's the opposite for the bottom block. The static friction acting on the bottom block acts to the left.

Can you give me an example of it not going opposite of the motion of the object?
That very example above (which you gave).

In any case, it's best to think of slipping between surfaces to determine the direction of friction and not the "motion" of the object.

In any case, it's best to think of slipping between surfaces to determine the direction of friction and not the "motion" of the object.
Absolutely agree. The movement is superfluous. I do not understand why that professor has applied a dynamic scheme and even of two bodies. I think the assumption that in this case there is an effect of Sigmund Freud is not sufficiently substantiated.
I think, anyway, that the one body is enough. jeff12, let we have only the one body which is placed on a rough surface in rest. Suppose that at some point in time a very small force begins to act on the body. Let this force is directed to the left (for definiteness). Let the direction of this force does not change, and a module that force is slowly growing until the body begins the slowly moving (to the left). Let denote the magnitude of this force at the start of the movement as a Fmax. Today we have one answer to the question "why the body did not move when the force has a value from zero to Fmax". This answer is: "the force is equal in magnitude and opposite in direction acted on the body" . This answer is the logical consequence of Newton's Law II. Analyzing the bodies that can affect the body, this can be found only one body that can be the source of action and that body is the surface. The force that applied from the surface on the body is a reaction force. This is because this force is always in need of a driving force It occurs as a response to the emergence of the driving force. And it responds to changes in the driving force to make a balance with. The nature of this force is the engagement of micro-roughness of the surface and of the body. As you can see here, there is no any friction, because there is no movement of surfaces relative to each other. However, the engagement of the micro-roughness plays a decisive role in dynamical friction and this force was called the static friction force. Traditions sometimes generate confusion, but all is even worse without them. I believe that spich will help you to find the right direction. :)

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Can you give me an example of it not going opposite of the motion of the object?
If it did, Energy would be manufactured out of nowhere!
As is usual with these problems, considering the Energy situation is a much more reliable approach. If the situation is a static one then using the principle of Virtual Work (From way back in my schooldays), you let things move by an infinitessimal amount and check that the Energy is dropping. If it isn't, then you have assigned the wrong sign to the friction direction.
But you have to make sure which object you are considering. (Is it the portion of the wheel in the footprint or the car that counts, for instance?)

Can you give me an example of it not going opposite of the motion of the object?
What? Static friction? An car accelerating without slippage for example.

What? Static friction? An car accelerating without slippage for example.
I think this is bright example, and at the same time it is a bad example. The trouble is that it is true as far as is true idealization: "without slippage". So, the slightest slip leads to the movement of the wheels against the surface of the friction force. And there is the kinetic (sliding) friction appeared...The movement is a harmful thing to demonstrate the static friction. The static friction is not a friction at all.

Can you give me an example of it not going opposite of the motion of the object?
A brick sitting in the middle of the bed on a pick-up truck rounding a corner. The force of static friction on the brick from the truck is inward toward the corner. The force of static friction on the truck from the brick is outward away from the corner. Neither is in the direction of motion of either truck or brick.

Static friction does not oppose motion. It opposes relative motion -- by just enough to prevent it.

IgorIGP
What? Static friction? An car accelerating without slippage for example.
Hence my caveat about what is the object to consider here. The thing that the friction is stopping is the slipping of the wheel, not the acceleration of the car. N3 rears its head and seems to confuse a lot of people. With or without slippage, the situation is the same and so is the direction of the friction force.

z
in the middle of the bed
Yes! And the pick-up is rounding over the Earth placed on the 3 elephants those with wet feet are standing on a slippery turtle... You are right! If only you know how I understand all that. The complex model with motion and even accelerated motion ( non inertial systems and imaginary forces) is a really road to hell. Going by that road (TC asks an example with a moving object) I would suggest a braking car. Here, the variant of soft braking without sliding is available and static friction force tends to accelerate the braking wheels rather than slow them down. But what does that prove? That an example offered by Professor wrong? I think, the way to inventing of such examples is an example in itself.

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Can you give me an example of it not going opposite of the motion of the object?
If it did, Energy would be manufactured out of nowhere!
What do you mean by that? You can easily construct inertial frames where the same force (dynamic or static friction) is either opposed to motion of the object or in the same direction.

The relative motion of the contact patches is always opposed by dynamic friction, while in the case of static friction there is no such relative motion.

IgorIGP
while in the case of static friction there is no such relative motion
You are absolutely right. But, to be honest, we have not pure reaction in real world. The houses in my street that runs to the sea are moving to the sea with velocity 2sm per year (or two meters per century). What about this... The reaction is always slightly lower than its an active force. This is absolutely fundamental thing. That means that the massive body at a rest on an inclined plane someday will be at the bottom. That means also that the statical friction on driving/braking whils is never purely free of the slip. I am not finding fault. I just want to show that this is no a good way to refute this a wrong model once and for all. This is because this evidence is true up to the idealization and the opponent can always argue that in the real world, his model is correct.

What do you mean by that? You can easily construct inertial frames where the same force (dynamic or static friction) is either opposed to motion of the object or in the same direction.
My point was that, if the (dynamic) friction force is in the same direction as the motion, this would involve getting energy out of the system (Work being force dot displacement). If you ever see a situation where this appears to be happening, you have got the direction of the friction force wrong. Isn't that a very basic requirement, related to thermodynamics? I don't think you can invoke 'symmetry' here.

static friction force tends to accelerate the braking wheels rather than slow them down.
In a complicated system like the drive to car wheels, at every point of contact between the parts, N3 applies and you can choose a force in either direction. The effect on the car is to slow it down (of course) and it is fruitless to spend too much time looking for a paradox; there is not one. More than 20 posts on this thread is evidence that this approach just generates confusion.
I made the point much earlier that everything can be resolved with a diagram (and I haven't seen one on this thread) which includes all the forces with and without slipping. Rather than demanding a resolution to this question, people who have a problem should really sit on their own with a paper and pencil and figure it out independently. Intuition can easily fail you in a case like this
This is not strictly right. But it's really a matter of how you are defining things. Are you referring to acceleration of the CM of the wheel, or to its rotational motion? If there were no friction, the wheels would be traveling forward (rotating or not; that's irrelevant). Friction contact with the ground would, of course, tend to cause peripheral speed (relative to the axle) of the wheel to approach the translational speed of the car. Assume that the wheel is massless. The only forces will be due to the force on the mass of the car, acting through the axle. Take an instantaneous fulcrum, half way between the contact point and the axle and you have equivalent forces slowing the car and forcing the wheel to rotate.
If the wheel has mass (moment of inertia, actually) and not rotating, but moving forward, when dropped onto a surface with finite friction, there will be a backwards force on the contact point until its rotation speed reaches the linear speed. But there will be a transfer of linear KE to rotational KE, which will slow up the wheel. You could say that you have 'sped up' the wheel by making it rotate - but have you?

IgorIGP
You could say that you have 'sped up' the wheel by making it rotate - but have you?
If it in context of the TC question, - yes I have. Because this:
. Are you referring to acceleration of the CM of the wheel, or
is out of context as far as I understand. I have asked TC to describe the professor's example with two bodies more detail, but without an answer. So as I have understood the speech goes about the point of contact moving. Otherwise it can be built arbitrarily complex schemes of arbitrarily complex conjugate bodies and constructive discussion on the verge of it will completely lose the constructivity. I have no to add something and I do not know if it needs because TC keeps silence. Let we wait to him reaction.

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sophiecentaur
My point was that, if the (dynamic) friction force is in the same direction as the motion, this would involve getting energy out of the system (Work being force dot displacement).
Work is frame dependent, and of course you can have inertial frames, where dynamic friction on an object is in the same direction as the motion of that object, thus doing positive work on that object. This doesn't violate energy conservation, because the equal but opposite friction force on the other object is doing more negative work on the other object.

Work is frame dependent,
That's true, in principle but my point is that, if the brakes and the tyres get hot, the procedure cannot be looked upon as reversible. This is what happens with anything other than operating above the limit of static friction. The operation of wheels on the ground can only result in loss of energy, which determines the relevant choice of direction. Hot brakes won't supply Kinetic Energy to either the car or the road.
Does it help to look at a system in a more complicated way (as long as the simple way is not neglecting to include something relevant)?
I see where you're coming from, though.

That's true, in principle but my point is that, if the brakes and the tyres get hot, the procedure cannot be looked upon as reversible.
Its not reversible becuse of entropy, not because of energy conservation.

Hot brakes won't supply Kinetic Energy to either the car or the road.
The force of dynamic friction can supply kinetic energy to them.

The force of dynamic friction can supply kinetic energy to them.
Yes but it doesn't work the other way round, does it? I don't understand why you don't seem to understand.??
Would you say that it is relevant whether the wheels do work on the road or the road does work on the wheels? Either way, there's no work resulting from the heating up effect of the tyres or brakes.

Its not reversible becuse of entropy, not because of energy conservation.
I didn't think I said it was. But the energy deficit, due to the effect of friction only applies ' one way round' and the heating comes from whichever side of the contact is supplying the energy. If you choose the instantaneous point of contact on the turning wheel as your reference frame, it strikes me that you are just making your life more difficult for yourself. It is a pretty established principle to choose the (vastly) more massive body as the reference for this sort of problem. There could be an alternative scenario involving a conveyor belt and an unpowered vehicle but afaics, the assumption was of the more familiar situation.
If the car is accelerating and there is slipping, the heating power is coming from the engine and if it's braking, the heating power is coming from the KE of the car. The friction force (on the road) will be in different directions in the two cases and so will the reaction force (on the contact point of the wheel).

.
The force of dynamic friction can supply kinetic energy to them.
Yes but it doesn't work the other way round, does it?
It does. Dynamic friction can add or remove kinetic energy from an object, by doing positve or negative work on it. It depends on the reference frame.

Would you say that it is relevant whether the wheels do work on the road or the road does work on the wheels?
It's frame dependent.

Its not reversible becuse of entropy, not because of energy conservation.
I didn't think I said it was.
You originally wrote the reversal would manufacture energy out of nowwhere. It wouldn't. But it would decrease entropy.

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You originally wrote the reversal would manufacture energy out of nowwhere. It wouldn't. But it would decrease entropy.
I see what you are getting at now and, of course, I could have stated things better. The reversal I am referring to is not 'that' one. I am talking in terms of Energy flow.
I started writing here because people seemed to be claiming that there was some confusion about the 'direction' to assign the friction force. For some reason, there is an opinion that the directions of Dynamic and Static friction forces could, somehow be different. That has to be either nonsense or due to choosing a different force in each case. The only point I am making is that to resolve this question, it is only necessary to consider that mechanical energy will always be lost due to friction. This principle takes you just one way and will always tell you how to choose a suitable direction to assign the force. There is less work done in accelerating the car, when there is slippage, than the engine is actually doing. I don't think that statement is controversial. is it?
The choice of frame is not relevant here - but I still say that, in the simple case of a car on a road, it would only add further confusion to work in terms of the car's frame - and particularly to work with the lowest point on the tyre. For someone struggling with this stuff, the possibility of using an alternative frame can only be confusing.

There is less work done in accelerating the car, when there is slippage, than the engine is actually doing.
I don't think that statement is controversial. is it?
Possibly not controversial. But not unambiguously correct.

Consider a car accelerating eastward at a red light on the streets of Chicago. It spins its tires briefly and a small cloud of smoke forms. The engine is doing a good amount of work on the drive shaft. But if we adopt an earth-centered inertial frame, the power being applied to the surface of the spinning tires by the pavement during this event is even larger than the power being supplied by the engine. That is because the surfaces of the tires are actually moving eastward at 600 to 700 of [earth-centered-inertial] miles per hour despite the engine's best efforts to get them moving westward at, perhaps, 30 [car-relative] miles per hour.

It would be more correct to say that the work done by the pavement on the surface of the tires plus the work done by the surface of the tires on the pavement adds up to a negative total.

IgorIGP
Possibly not controversial. But not unambiguously correct.

Consider a car accelerating eastward at a red light on the streets of Chicago. It spins its tires briefly and a small cloud of smoke forms. The engine is doing a good amount of work on the drive shaft. But if we adopt an earth-centered inertial frame, the power being applied to the surface of the spinning tires by the pavement during this event is even larger than the power being supplied by the engine. That is because the surfaces of the tires are actually moving eastward at 600 to 700 of [earth-centered-inertial] miles per hour despite the engine's best efforts to get them moving westward at, perhaps, 30 [car-relative] miles per hour.

It would be more correct to say that the work done by the pavement on the surface of the tires plus the work done by the surface of the tires on the pavement adds up to a negative total.
This is all a bit disingenuous, I think. The source of the Power that we are discussing is the fuel in the tank in the car. You can hardly be suggesting that the fuel in the tank is somehow being 'used' by the Earth, in order to accelerate the car (not an inertial frame, because there is acceleration). That is, afaiac, a very relevant point in this discussion. There will, of course, be a finite but very small, increase in the KE of the Earth during the interaction. The source of the energy to achieve this is also in the fuel tank. Do you mean the total power developed by the engine?
But you need to remember that the periphery of the tyres is changing its velocity, constantly and the top of the tyre is actually moving eastwards at 600 - 60 mph. There is no possible reason to use the Earth centre unless you think that the centripetal force involved in the rotation of the Earth is relevant. Could you measure it in practice (in the sort of experiment we are discussing)? We both know that a flat, stationary Earth surface is by far the most sensible frame to be working with.
I would say that is wrong, or at least, unclear. There is a region of tyre and grit upon which the wheel is expending a lot of energy but the effective work done " by the pavement", (being the Reaction Force times the velocity along the road ) is the same as the Work done by the engine, (effective drive force times velocity). Any left over power will be stirring up the grit and the tyre surface - abrading it and getting it hot. If you consider the Earth to me of infinite mass then there is no change in the Earth's KE so it can not be 'supplying' any power (or it would be changing). Momentum remains the same and the velocity change is zero for the earth.

Here's another thing. When the brakes are applied and there is no slippage of the tyres (just on the pad surfaces), is there anything different about the power flow situation compared with the situation where the brakes lock and the slippage is all on the tyre surface (arranged so thet the acceleration is the same in each case)? Again, because the KE of the infinite Earth is not changing, all the energy comes from the KE of the car; where it happens to be dissipated is not relevant.