Very confused about friction -- Please explain this....

[Prem1998]

Suppose two rigid circular coins are lying on top of each other. A horizontal force on the top coin is applied. So, I think there are two possible scenarios:
1. Both move together with common acceleration (if friction between them is strong)
2. Only the top coin slips over the bottom (if friction between them is weak)
I can understand the second case that when force on the top one is greater that the max value of static friction between them, then the top moves over the bottom with a net force. But I can't understand how the first can happen.
If force on the top one is less than the max value of static friction, then friction between them applies an equal and opposite force to the top coin preventing it from moving. So, if force>max static friction, top slips over and if force<max static friction, net force on top becomes zero and no movement happens. Then, in what case exactly does they move together with common acceleration with friction acting as a connector?
Also, what if unequal horizontal forces are applied on both the coins? In what case will they both move together and in what case one slips over another?
Also, does friction act on the coin on which force is applied or does it act on the other one or does it act on both coins when one tends to move over other?
I need a brief understanding of the concept, please.

 

[andrewkirk]

The first case can occur if the coefficient of static friction for the interface between the two coins is greater than double both the coefficients of static and kinetic friction for the interface between a coin and the horizontal surface. Consider for instance if the two coins were made of rough wood and the surface was ice.

 

[A.T.]

2. Only the top coin slips over the bottom (if friction between them is weak)
I can understand the second case that when force on the top one is greater that the max value of static friction between them

The applied force can also be greater than the force between them, when they accelerate together.

 

[Prem1998]

I still don't understand how can they move together if we apply Newton's laws. I am assuming both coins lie on an ideal frictionless surface on top of each other. So, if horizontal force on the first coin is smaller that the max possible value of static friction between them, then friction is capable of applying an equal and opposite force to the top coin and force on top becomes zero and no movement occurs.
if force on top>max possible static friction, then friction is not able to stop the first coin from moving and it accelerates with a net force=(applied force- kinetic friction) over the bottom coin and the bottom one stays motionless.
So, the case in which they both move as one object becomes impossible, but I know it is possible because friction can act as a connector. Think of when a box is lying on an accelerating car, the force acts on the car but the friction between car floor and box makes the box move together with it. But, I can't explain this by applying the laws.
I am assuming they lie of frictionless surface on top of each other.
Will someone please explain in brief intuitively instead of giving a short answer?

 

[A.T.]

So, if horizontal force on the first coin is smaller that the max possible value of static friction between them, then friction is capable of applying an equal and opposite force

This isn't a 3rd Law pair. There is no reason for them to be equal but opposite.

if force on top>max possible static friction, then friction is not able to stop the first coin from moving and it accelerates with a net force=(applied force- kinetic friction) over the bottom coin...

Why over the bottom coin? They can both accelerate together, if friction = applied force / 2 (for identical coins and frictionless surface). Draw a free body diagram with all the forces.

 

[Prem1998]

This isn't a 3rd Law pair. There is no reason for them to be equal but opposite.Why over the bottom coin? They can both accelerate together, if friction = applied force / 2 (for identical coins and frictionless surface). Draw a free body diagram with all the forces.

Yes, I know that this isn't a third law pair but I have read that if force is applied on one body so that it tends to move over the other, then if the applied force is smaller than the max value of static friction between the surfaces, then the body which tends to move experiences an equal frictional force in the opposite direction to prevent it from moving only until the applied force is greater than the max value of static friction, at that point kinetic friction comes into play and the body attains motion.
So, if the force applied on the truck to make it move is less than the max value of static friction between the surface of the truck and the box lying on it, then the surface of the truck experiences an equal friction force to prevent its motion and it won't move at all but this isn't what happens, is it? In reality the truck moves carrying the box inside it as if the friction between them is acting to connect them instead of preventing the truck's motion.

 

[andrewkirk]

So, if horizontal force on the first coin is smaller that the max possible value of static friction between them, then friction is capable of applying an equal and opposite force to the top coin and force on top becomes zero and no movement occurs.

The equal and opposite force applied by friction is equal and opposite to the force applied to the bottom coin by the top coin, not the force applied to the top coin by the coin-pusher. If the coin-pusher applies a 2N force to the top coin, and the whole lot accelerates at 1/m where m is the mass of a single coin, then the force of the top coin on the bottom coin is only 1N. Thus the net horizontal force on the top coin is 2N-1N=1N, and the net horizontal force on the bottom coin is 1N. Which agrees with them both accelerating at rate 1/m.

THis is not particular to friction. The same set of forces would be in action if the top and bottom coin were bolted together.

 

[Prem1998]

The equal and opposite force applied by friction is equal and opposite to the force applied to the bottom coin by the top coin, not the force applied to the top coin by the coin-pusher. If the coin-pusher applies a 2N force to the top coin, and the whole lot accelerates at 1/m where m is the mass of a single coin, then the force of the top coin on the bottom coin is only 1N. Thus the net horizontal force on the top coin is 2N-1N=1N, and the net horizontal force on the bottom coin is 1N. Which agrees with them both accelerating at rate 1/m.

THis is not particular to friction. The same set of forces would be in action if the top and bottom coin were bolted together.

I am understanding this a bit. So, you're saying that if we apply 2N force on the top coin then only 1N of opposing frictional force acts on the top coin, so force on the top coin is not zero and it moves. So, opposing force is not equal to the applied force. BUT I have read that the force of friction on an object always matches the applied force to prevent its motion until the applied force is less than max static friction. I mean, that's why we are not able to move a block on the ground until we exert some force greater than max value of static friction. This can only be explained if the ground exerts an equal and opposite force to our applied force. Then why in the coin example the applied force is 2N and the friction is only 1N?

 

[andrewkirk]

BUT I have read that the force of friction on an object always matches the applied force to prevent its motion until the applied force is less than max static friction.

That rule applies when the friction is between the object to which the horizontal force is applied and the ground. Since that is by far the simplest and most usual situation used in physics teaching, they often don't bother pointing out that that rule is restricted to such conditions.

The rule is not applicable when the item being pushed is not itself in contact with the ground.

 

[CWatters]

If you glue the coins together does that make it easier to understand? Using glue is just one way to dramatically increase the coefficient of friction.

 

[Prem1998]

That rule applies when the friction is between the object to which the horizontal force is applied and the ground. Since that is by far the simplest and most usual situation used in physics teaching, they often don't bother pointing out that that rule is restricted to such conditions.

The rule is not applicable when the item being pushed is not itself in contact with the ground.

So, is this the generalized definition of friction:
friction is the force with which a surface attracts the other one in contact with it when it tends to move over it so that the other one also attracts it with the same force preventing it's motion only to some extent and this force is not be equal to the force with which the first surface tends to move. But in case the second surface is Earth, then it is equal to the applied force.
But Earth can also be viewed as an object. So, what's different about earth? Why can't the coin and Earth attain a common acceleration? I can't see the case with Earth as an approximation of the general definition when the second surface is a very big one.

 

[andrewkirk]

Why can't the coin and Earth attain a common acceleration?

Because the person pushing the coin is standing on the Earth, so her feet are pushing the Earth backwards exactly the same amount as her finger is pushing it forwards. If the pusher is not connected to the Earth - eg a hovering spaceship using a long pole - then the Earth and the coin do attain the same (incredibly small) acceleration, provided the push is not strong enough to break the frictional lock between the coin and the Earth.

There is nothing special or different about the Earth. To liken the two coins to the Earth, the pusher would need to be standing on the extremely heavy lower coin.

 

[Prem1998]

Because the person pushing the coin is standing on the Earth, so her feet are pushing the Earth backwards exactly the same amount as her finger is pushing it forwards. If the pusher is not connected to the Earth - eg a hovering spaceship using a long pole - then the Earth and the coin do attain the same (incredibly small) acceleration, provided the push is not strong enough to break the frictional lock between the coin and the Earth.

There is nothing special or different about the Earth. To liken the two coins to the Earth, the pusher would need to be standing on the extremely heavy lower coin.

Exactly why would her feet would push the Earth backwards when she pushes the coin forwards? Is it a necessary condition? And even if that's true then she could use a magnet to apply force on the coin
And even if her feet push the Earth backwards, then how does this lead us to the conclusion that the frictional force exerted will be equal to the applied force. I mean can it be proved that if one's feet push the Earth backwards while applying a force on some object, then friction on the object would be equal to the force applied on it?

 

[A.T.]

BUT I have read that the force of friction on an object always matches the applied force to prevent its motion until the applied force is less than max static friction.

Only if the object doesn't accelerate. This is just Newtons 2nd law.

 

[CWatters]

First step is to draw a free body diagram of the lower coin showing the horizontal forces acting on the coin..

The only horizontal force acting on the lower coin is transmitted to it by friction with the upper coin, let's call that friction force F_f. Don't worry about the size of that force because it turns out to be irrelevant to the question "does the lower coin move?". There are no other horizontal forces acting on the lower coin because you assumed it is resting on an ideal frictionless surface. So if there is any friction force between the two coins the net force on the lower coin is non-zero. Newton's laws say that means the lower coin must accelerate so there must be some movement of the lower coin.

Lets assume the coins have the same mass "m". Using Newtons law we can work out that the lower coin will accelerate at ...

a_lower = F_f/m

but since the max value of F_f is umg then the max possible acceleration of the lower coin is given by..

a_lower = umg/m = ug

You can take things further and work out if the lower coin will keep up with the top coin. For example if a very large force was applied to the top coin it might out accelerate the lower coin meaning it will eventually leave the lower coin behind.

To work out how the upper coin moves we draw a free body diagram for the upper coin...

The upper coin has two horizontal forces acting on it, the applied force call it F_a and the friction force with the lower coin F_f

So the net force acting on the top coin is F_a - F_f. Therefore it accelerates at

a_upper = (F_a-F_f)*m

No we can look at the condition under which the coins accelerate together at the same rate...

a_lower = a_upper
or
F_f/m = (F_a-F_f)*m
mass cancels
F_f = F_a-F_f
rearrange
F_a = 2*F_f

So for them to accelerate together the applied force must be no more than twice the friction force between the coins.

We already know that the maximum possible acceleration of the lower coin is μg so the maximum force that can be applied so the coins move/accelerate together at the same rate is given by

F_a = 2mμg

The key is to learn how to draw free body diagrams for the separate objects in the problem. One drawing for each object.

 

[andrewkirk]

Exactly why would her feet would push the Earth backwards when she pushes the coin forwards?

It's Newton's third law. It's the same reason why a passenger in a car can't make it go faster by pushing on the back of the seat in front of them.

A more general way to state the rule of static friction, that doesn't use irrelevant concepts like up, down, Earth, is as follows.

If two objects with flat surfaces are in contact and pushed together by a normal force of N, and additional forces are applied in any way such as to not change the normal force but to cause object A to apply a force of F to object B in direction k, where k is in the plane of the interface (known as a 'shear force'), then object A will move relative to object B in direction k iff $F>N\mu_S$, where $\mu_S$ is the coefficient of static friction for the interface.

 

[CWatters]

Exactly why would her feet would push the Earth backwards when she pushes the coin forwards? Is it a necessary condition?

No it's due to Newton's third law.

And even if that's true then she could use a magnet to apply force on the coin

That makes no difference. Newton's third law means the magnet also experiences an opposite reaction force. If she is holding the magnet the magnet pushes on her and her feet push on the earth.

 

[Prem1998]

No it's due to Newton's third law.

That makes no difference. Newton's third law means the magnet also experiences an opposite reaction force. If she is holding the magnet the magnet pushes on her and her feet push on the earth.

I think she could have made the magnet swing with a thread and connected with a stand, so the reaction force on the magnet would just make it swing and would exert no force on Earth. I mean there must be countless ways to exert a force on something without transmitting the reaction to Earth.

 

[Prem1998]

It's Newton's third law. It's the same reason why a passenger in a car can't make it go faster by pushing on the back of the seat in front of them.

I don't think it works that way. I mean the she could be pushing the coin with her finger and the reaction from the coin could have just caused a compression or slight deformation of her finger. Her body is not completely rigid, so there's no way for such a small reaction force to transmit to the Earth.

 

[A.T.]

I don't think it works that way. I mean the she could be pushing the coin with her finger and the reaction from the coin could have just caused a compression or slight deformation of her finger. Her body is not completely rigid, so there's no way for such a small reaction force to transmit to the Earth.

Momentum does not work this way. It cannot be dissipated or stored in deformation like kinetic energy.

But this whole discussion about the Earth is just confusing you. Remember that static/dynamic friction prevents/opposes the relative motion between the contact surfaces, not the acceleration/movement of the objects in contact in general. Only in special cases with no acceleration you can conclude by Newtons 2nd that friction is equal but opposite to the other force, so they cancel.

 

[Prem1998]

Momentum does not work this way. It cannot be dissipated or stored in deformation like kinetic energy.

But this whole discussion about the Earth is just confusing you. Remember that static/dynamic friction prevents/opposes the relative motion between the contact surfaces, not the acceleration/movement of the objects in contact in general. Only in special cases with no acceleration you can conclude by Newtons 2nd that friction is equal but opposite to the other force, so they cancel.

I now understand that friction opposes only relative motion and it is not equal to the applied force because if it were it would prevent motion. But, I want to know why in the case with Earth, friction always matches the applied force? I have been given the explanation that it is because the pusher lies on Earth so his feet push Earth backwards with the same force as he pushes the object forward. I don't think that explains this. Because even if we consider that that's true then how does that lead us to the conclusion that friction will be equal to the applied force in that case when the pusher pushes Earth backwards?

 

[A.T.]

But, I want to know why in the case with Earth, friction always matches the applied force?

It doesn't, forget such rules. Learn to draw FBDs and apply Newton's Laws.

 

[Prem1998]

It doesn't, forget such rules. Learn to draw FBDs and apply Newton's Laws.

I recently thought of it as an approximation like this:
Suppose that an object lies in contact with Earth and the coefficient of static friction between them is high,
and a force F is applied on the object, then it exerts some force on Earth(between the atoms) and Earth in turn exerts the same force on the object (which is friction),
if M is mass of Earth and m is that of the object then their common acceleration is given by:
a = F/(m+M)
and therefore the force applied by the object on Earth is:
Mass of Earth*a
= MF(m+M)
Since, Earth exerts the same opposing force on the Object (which is friction), so friction on the object is:
Friction = MF(m+M)
since M>>m, so (m + M) = M approx,
so friction = MF/M
= F which is the applied force,
so I think the fact that friction exerts the same opposite force on the object lying on Earth is an approximation of the general fact because of the large mass of the Earth

 

[CWatters]

See my post #15. I showed how to work out what happens in your two coin set up using Free Body Diagrams and Newtons laws without any reference at all to the earth. There is no need to even think about reaction forces with the earth, no need for any approximations etc. You don't need to know how the applied force is produced (person, spring, rocket, magnet, gravity etc) as that's irrelevant to the two coin problem you described.

I have read that the force of friction on an object always matches the applied force to prevent its motion until the applied force is less than max static friction.

That's correct if you change "less than" to "greater than".

The force of static friction between two objects matches the applied force until the applied force exceeds the maximum possible static friction force. The maximum possible static friction force is frequently defined as μ_sN where μ_s is the coefficient of static friction and N is the Normal force.

If the applied force exceeds the maximum possible static friction force then the objects slide relative to each other. Once sliding occurs the friction force is called the Kinetic friction force and is typically given by u_kN where u_k is the coefficient of kinetic friction.