Two bars connected by a spring on a floor with friction

[Null_Void]

Homework Statement

Two bars of masses $m1$ and $m2$ are connected by an ideal spring and are kept on a floor with friction. The coefficient of friction is $k$. Find the minimum constant force to be applied to mass $m2$ in order to shift the bar.

Pic attached below.

Relevant Equations

$F - kx - μm_1g = m_1a$

$kx= μm_2g$

$Fx - 1/2kx^2 - μm_1gx = 1/2mv^2$

So I do realise that this problem involves energy, I still wrote equations for forces acting on each block. I will assume coefficient of friction to be $μ$ for convenience.

For block 1:
$F - kx - μm_1g = m_1a$

For block 2:
$kx= μm_2g$

I then wrote an energy equation assuming a velocity $v$:

Say the block moves a distance $x$

$Fx - 1/2kx^2 - μm_1gx = 1/2mv^2$

To solve this equation what value of v am I supposed to assume? Do the force equations play any part in solving
the problem?

 

[PeroK]

Where's the picture?

 

[Null_Void]

Where's the picture

I'm extremely sorry, I've attached the picture now.

 

[PeroK]

The question, as I understand it, is: what is the minimum force $F$ such that $m_1$ moves?

 

[PeroK]

Which is equivalent to: what is the maximum force $F$ such that $m_1$ does not move?

 

[Null_Void]

Which is equivalent to: what is the maximum force $F$ such that $m_1$ does not move?

Yes, I believe so.

 

[erobz]

Do we know the coefficient of kinetic friction acting on block 2 $m_2$? It slides at constant velocity for the smallest such $F$ that would release $m_1$? Or am I missing something, perhaps $\mu_k = \mu_s = \mu$...

Anyhow to the OP. Forces are going to be the key here.

 

[PeroK]

Anyhow to the OP. Forces are going to be the key here.

Forces and energy!

 

[Null_Void]

Do we know the coefficient of kinetic friction acting on block 2 $m_2$? It slides at constant velocity for the smallest such $F$ that would release $m_1$? Or am I missing something, perhaps $\mu_k = \mu_s = \mu$...

Anyhow to the OP. Forces are going to be the key here.

Yes, even though it isn't stated I believe we should take the coefficients for static and kinetic friction to be the same.

But the variable $a$ is the problem here. I should either somehow eliminate it or find another equation, but I can't think of anything. Could you point where I'm lagging?

Also if you don't mind just a side thought:
In such situations, what will be the direction of force acting on the blocks?. In case of a string, I know that it's tendency is to pull. But a spring seems more abstract and complex.

 

[PeroK]

The first thing to think about is what happens if $F$ is large enough to move $m_2$ but not $m_1$?

Acceleration is not much use as it's variable.

 

[erobz]

But the variable $a$ is the problem here. I should either somehow eliminate it or find another equation, but I can't think of anything. Could you point where I'm lagging?

$a$ isn't the problem you think it is. You do a FBD of each block, ask yourself (just beyond very beginning - an instant of its motion) does the front block does it need to accelerate?

In such situations, what will be the direction of force acting on the blocks?. In case of a string, I know that it's tendency is to pull. But a spring seems more abstract and complex.

Free body diagrams of each block and spring should help you work that out.

Anyhow, @PeroK is here so I defer.

 

[PeroK]

$a$ isn't the problem you think it is. You do a FBD of each block, ask yourself (just beyond very beginning - an instant of its motion) does the front block does it need to accelerate?

Free body diagrams of each block and spring should help you work

The problem is a bit too tricky for that!

 

[erobz]

The problem is a bit too tricky for that!

Apparently I've underestimated it. Thats why I'll butt out!

 

[Null_Void]

The first thing to think about is what happens if $F$ is large enough to move $m_2$ but not $m_1$?

Acceleration is not much use as it's variable.

But shouldn't $m_2$ decelarate? There will be an elongation in the spring which increases with time and at one point the block will be in equilibrium? I'm sorry, but could you tell what would happen?

 

[PeroK]

But shouldn't $m_2$ decelarate? There will be an elongation in the spring which increases with time and at one point the block will be in equilibrium? I'm sorry, but could you tell what would happen?

Equal forces means no acceleration, but $m_2$ keeps moving.

 

[Null_Void]

Free body diagrams of each block and spring should help you work that out

But won't it depend on whether the spring is contracting or elongating? How do you know if it will push or pull on a certain block in a double block system?

 

[Null_Void]

Equal forces means no acceleration, but $m_2$ keeps moving.

So it will move with constant velocity and since the spring length in that instant is constant, so $m_1$ will also travel with the same velocity?
The magnitude of velocity at equilibrium depends on the magnitude of F? Therefore for minimum F, velocity should also be minimum?

 

[PeroK]

But won't it depend on whether the spring is contracting or elongating? How do you know if it will push or pull on a certain block in a double block system?

This is what the question is all about!

 

[Null_Void]

This is what the question is all about!

So say if $m_1$ is at rest and $m_2$ elongates by some distance, the force acting on both ends of the spring (on the mass) would be to towards the centre and if $m_1$ is at rest and $m_2$ contracts the spring by some distance,
The force acting on both ends would be away from the centre?
If so why is that? Why can't it push one and pull another?

 

[PeroK]

So say if $m_1$ is at rest and $m_2$ elongates by some distance, the force acting on both ends of the spring (on the mass) would be to towards the centre and if $m_1$ is at rest and $m_2$ contracts the spring by some distance,
The force acting on both ends would be away from the centre?
If so why is that? Why can't it push one and pull another?

That's not a coherent analysis of the scenario.

 

[SammyS]

Homework Statement: Two bars of masses $m1$ and $m2$ are connected by an ideal spring and are kept on a floor with friction. The coefficient of friction is $k$. Find the minimum constant force to be applied to mass $m2$ in order to shift the bar.

Pic attached below.
Relevant Equations: $F - kx - μm_1g = m_1a$

$kx= μm_2g$

$Fx - 1/2kx^2 - μm_1gx = 1/2mv^2$

So I do realise that this problem involves energy, I still wrote equations for forces acting on each block. I will assume coefficient of friction to be $μ$ for convenience.

For block 1:
$F - kx - μm_1g = m_1a$

For block 2:
$kx= μm_2g$

I then wrote an energy equation assuming a velocity $v$:

Say the block moves a distance $x$

$Fx - 1/2kx^2 - μm_1gx = 1/2mv^2$

To solve this equation what value of v am I supposed to assume? Do the force equations play any part in solving
the problem?View attachment 352822

It appears to me that there are some errors and/or typos in the statement of the problem as it is given.

It seems that the intention was that the coefficient of friction is $\mu$ and the spring constant is $k$ .

Then as @PeroK has stated in Post #4 :

The question, as I understand it, is: what is the minimum force $F$ such that $m_1$ moves?

I agree with that, where force, $F$, is applied to the block with mass, $m_2$, as in the figure.

As a hint, I would include the question:
What is the minimum force that the spring must exert on the $m_1$ block so that it moves?

 

[Null_Void]

It appears to me that there are some errors and/or typos in the statement of the problem as it is given.

It seems that the intention was that the coefficient of friction is $\mu$ and the spring constant is $k$ .

Then as @PeroK has stated in Post #4 :

I agree with that, where force, $F$, is applied to the block with mass, $m_2$, as in the figure.

As a hint, I would include the question:
What is the minimum force that the spring must exert on the $m_1$ block so that it moves?

The minimum spring force should be equal to the friction force
$kx = μm_1g$

 

[Null_Void]

That's not a coherent analysis of the scenario.

Well then could you explain what would happen?

 

[PeroK]

Well then could you explain what would happen?

That's not the way homework help works. You are supposed to be capable of tackling the problem yourself.

One more hint: imagine that $m_1$ is very large, so that it doesn't move. You could even imagine $m_1$ is fixed to the floor. Analyse the motion of $m_2$ under a constant force. And by "analyse" I mean both a description of the motion and the relevant equations.

If you can do that, then you have almost solved the problem.

 

[haruspex]

The minimum spring force should be equal to the friction force
$kx = μm_1g$

Right. So the problem reduces to moving $m_2$ a distance $μm_1g/k$.
Now you can think about work.

 

[Null_Void]

That's not the way homework help works. You are supposed to be capable of tackling the problem yourself.

One more hint: imagine that $m_1$ is very large, so that it doesn't move. You could even imagine $m_1$ is fixed to the floor. Analyse the motion of $m_2$ under a constant force. And by "analyse" I mean both a description of the motion and the relevant equations.

If you can do that, then you have almost solved the problem.

I will assume that $m_1$ is fixed to the floor or rather that it is a rigid wall to which the spring and the mass system is connected.
Since the force applied is constant, the block will start decelerating as the spring elongates and at one point of time it will be in equilibrium.

At Equilibrium:
$F = kx + μm_2g$

Using Work-Energy theorem:

$Fx - μm_2gx - 1/2kx^2 = -1/2m_2v^2$

Where $v$ is the initial velocity of the block when it begins to move.
I'm highly doubtful of the sign of the term on the right as it won't give us a proper solution.
But the equation still seems right.

 

[PeroK]

I will assume that $m_1$ is fixed to the floor or rather that it is a rigid wall to which the spring and the mass system is connected.
Since the force applied is constant, the block will start decelerating as the spring elongates and at one point of time it will be in equilibrium.

At Equilibrium:
$F = kx + μm_2g$

Why do you care about the equilibrium point?

 

[Null_Void]

Right. So the problem reduces to moving $m_2$ a distance $μm_1g/k$.
Now you can think about work.

But an extra velocity term $v$ is still present? What value should $v$ take in this case

 

[Null_Void]

Why do you care about the equilibrium point?

Because it is when the acceleration will be $0$ and I thought it might be useful later on but I'm not sure.

 

[PeroK]

Because it is when the acceleration will be $0$ and I thought it might be useful later on but I'm not sure.

You need to think about what happens after the equilibrium point.

 

[Null_Void]

You need to think about what happens after the equilibrium point.

Is this the equation we should be concerned with?
$Fx - μm_2gx - 1/2kx^2 = -1/2m_2v^2$

 

[PeroK]

Is this the equation we should be concerned with?
$Fx - μm_2gx - 1/2kx^2 = -1/2m_2v^2$

No. That's no use, as you yourself have repeatedly complained about the $v^2$ term being a nuisance.

 

[Null_Void]

No. That's no use, as you yourself have repeatedly complained about the $v^2$ term being a nuisance.

Will it perform simple harmonic motion about the equilibrium point?

 

[PeroK]

Will it perform simple harmonic motion about the equilibrium point?

No.

 

[haruspex]

But an extra velocity term $v$ is still present? What value should $v$ take in this case

Suppose there is still a velocity term at that point. What would happen if you were to make F just a tiny bit less?