Need help understanding some concepts about friction

[al4n]

We have 2 objects, m₁ and m[SUPlB]2[/SUB]

Friction is present between the two objects but not between m₁ and the floor. A force is exerted on the bottom object which causes it to accelerate parallel to the floor. The thing I'm wondering for while now is, how do I prove that the acceleration of the top object is equal to the acceleration of the bottom object when the magnitude of the applied force is within some range?

 

[PeroK]

We have 2 objects, m₁ and m[SUPlB]2[/SUB]View attachment 328111
Friction is present between the two objects but not between m₁ and the floor. A force is exerted on the bottom object which causes it to accelerate parallel to the floor. The thing I'm wondering for while now is, how do I prove that the acceleration of the top object is equal to the acceleration of the bottom object when the magnitude of the applied force is within some range?

By using Newton's laws of motion.

 

[phinds]

Draw a force diagram and then, as @PeroK said, apply Newton's Laws.

 

[A.T.]

We have 2 objects, m₁ and m[SUPlB]2[/SUB]View attachment 328111
Friction is present between the two objects but not between m₁ and the floor. A force is exerted on the bottom object which causes it to accelerate parallel to the floor. The thing I'm wondering for while now is, how do I prove that the acceleration of the top object is equal to the acceleration of the bottom object when the magnitude of the applied force is within some range?

Look up static friction coefficient.

 

[Lnewqban]

Welcome, @al4n !

Please, see:
https://en.wikipedia.org/wiki/Friction#Static_friction

https://en.wikipedia.org/wiki/Friction#Kinetic_friction

When the mass is not moving, the object experiences static friction. The friction increases as the applied force increases until the block moves. After the block moves, it experiences kinetic friction, which is less than the maximum static friction.

 

[al4n]

Draw a force diagram and then, as @PeroK said, apply Newton's Laws.

Hi. I tried this but I cant seem to make it work.

Lets say F_a is the force applied on the lower object. It will experience a frictional force say F_f1 opposite and with equal value to F_a.The object on top will, to my understanding will also experience a force opposite and with equal magnitude to F_f1. We can call it say F_r1 for force of reaction.
Then the bottom object will experience an acceleration (lets call it a₁) equal to a₁ = (F_a- F_f1)/m₁ which is 0. The upper object however will have an acceleration of a = F_r1/m₂ which is not 0. The accelerations of the objects are not equal but Im trying to prove that they are.

I tried this a 2nd time but letting the upper object also experience friction. The frictional force (Im calling it F_f2) felt is equal in magnitude ( I think ) to the reaction force F_r1. Like with my first attempt, this frictional force will create a reaction force this time on the lower object (calling it F_r2) also with equal magnitude. For this case, the acceleration experienced by the lower object is now a₁ = (F_a- F_f1 + F_r2)/m₁
which is just F_a/m₁.
The acceleration of the upper object however is now a₂ = (F_r1-F_f2)/m₂ which is 0. Again the accelerations of the objects are not equal.

I think Im missing something but I dont know what.

 

[PeroK]

I think Im missing something but I dont know what.

You can approach this in two ways, I would say.

First, from a physical perpsective, we can describe what happens when a force is applied to $m_1$. As there is no friction with the floor, any force will accelerate the system according to $F = (m_1 + m_2)a$. This is because static friction holds the objects $m_1$ and $m_2$ together for low acceleration.

However, eventually the force increases to the point where static friction is not sufficient to hold $m_2$ to $m_1$ and $m_2$ slips backwards (relative to $m_1$). In the first approach, we try to calculate the size of the force needed for this.

This counts as a "proof" in physics, because although we make an assumption, we eventually justify that assumption.

The second approach is what you attempted by considering all forces on the system and trying to balance your equations somehow. What you are missing, perhaps, is that static friction is a variable reaction force. That means it varies from zero to some maximum possible value to prevent relative motion between surfaces. (Once the surfaces begin to slide across each other, we have a constant kinetic friction, which acts against the motion.)

The important point is that static friction is variable with a maximum possible value. In a sense, it adjusts to what is needed to prevent motion - which is, perhaps, the assumption/knowledge you are missing.

 

[al4n]

This is because static friction holds the objects $m_1$ and $m_2$ together for low acceleration.

This statement right here is what Im trying to prove. I just want a neat series of statements that start from say the laws of motions to ideally: a₁ = a₂

 

[PeroK]

This statement right here is what Im trying to prove. I just want a neat series of statements that start from say the laws of motions to ideally: a₁ = a₂

Physics isn't mathematics. The nature of static friction does not follow directly from Newton's Laws or any fundamental postulates of physics (*). You can't prove directly from Newton's laws that there is even such a thing as friction. The basic properties of friction (static and kinetic) must be added to the fundamental laws.

(*) If you have a model for intermolecular behaviour, you may be able to derive the properties of friction from that. But, that is beyond the scope of basic mechanics.

 

[phinds]

Hi. I tried this but I cant seem to make it work.

Tried WHAT? I don't see any force diagram.

 

[nasu]

Hi. I tried this but I cant seem to make it work.

Lets say F_a is the force applied on the lower object. It will experience a frictional force say F_f1 opposite and with equal value to F_a.
I think Im missing something but I dont know what.

There is no reason for the friction force to be equal in magnitude to the force applied to the bottom object. They are not a Newton third law pair. In your ideal case, with no friction with the ground, no matter how small the applied force is, the system will accelerate. Or at least, the bottom body will accelerate. And so, the friction at the top will be allways less than the force applied.

 

[A.T.]

This statement right here is what Im trying to prove. I just want a neat series of statements that start from say the laws of motions to ideally: a₁ = a₂

I assume you mean only the horizontal accelerations here.

So you don't demand a proof that there is no vertical relative motion between the two bodies, because the body on top cannot simply fall through the body below. How is that fundamentally different from assuming sufficient static friction in the horizontal direction?

 

[Lnewqban]

This statement right here is what Im trying to prove. I just want a neat series of statements that start from say the laws of motions to ideally: a₁ = a₂

Can you imagine that both masses are sliding on the frictionless surface, just joined by a string?
Static friction behaves like our imaginary string, which has a limited resistance to tension.

As long as that horizontal force on the pulling object, which accelerates parallel to the floor, induces a tension in our imaginary string that is not greater than its resistance to pulling, it holds.
Therefore, both objects slide with the same acceleration and instantaneous velocities.

If greater force is applied, the string reaches its limit and breaks (our static friction becomes dynamic).
Force is still applied to the pulling object after that, and it continues accelerating at a higher rate.
The pulled object stops accelerating because the string tension disappears (not the case of the dynamic friction, which still applies some lesser pulling force that makes the top object continue accelerating at a lower rate).