Problem with Static Friction -- Two blocks on a platform

[Astr]

Hello, everyone.
This problem is easy if you assume that the friction between the blocks and the platform is the maximum possible. Then the normal force is 0. But how can you show that the normal is 0 even when the friction is less than maximum?
Thank you for your help.

 

[BvU]

Then the normal force is 0

Do you mean: the force of interaction between the blocks ?

What about the acceleration of each of the blocks when the friction between the blocks and the platform is less than maximum ?

if you assume that the friction between the blocks and the platform is the maximum possible. Then the normal force is 0.

This happens when the blocks are at the point of sliding, and beyond that point as well, right ?

$\$

 

[Astr]

Thank you for your questions, BvU.

"Do you mean: the force of interaction between the blocks ?"

Yes, the force of interaction between the blocks, or the normal force between them (i.e. the contact force that is perpendicular to the surface of the blocks).

"What about the acceleration of each of the blocks when the friction between the blocks and the platform is less than maximum ?"

That is the question I can't answer yet. I guess is still 0, but I don't know how to prove it.

"This happens when the blocks are at the point of sliding, and beyond that point as well, right ?"

Yes, that is correct, because this is the equation of the normal force that mass 2 exerted over mass 1:

So, if F_r1=μ_km₁g and if F_r2=μ_km₂g then the normal is 0.$\$

 

[BvU]

That is the question I can't answer yet

In that case none of the blocks is sliding. So their acceleration is exactly equal, and the friction force alone (between blocks and platform) is causing that acceleration. Meaning the net force is the frinction force, ergo the contact force must be zero !

$\$

 

[BvU]

Yes, that is correct, because this is the equation of the normal force that mass 2 exerted over mass 1:

I don't recognize this. Where does it come from ?

$\$

 

[Astr]

I don't recognize this. Where does it come from ?

$\$

That can be deduced from Newton's laws of motion like this:

 

[Astr]

Taking the masses as a whole, this is the Newton's second law for the system:
$$(m_1+m_2)a=Fr_1 + Fr_2$$
(With $Fr_1$ and $Fr_2$ the respectively friction forces)
Which implies:
$$a=\frac{Fr_1+Fr_2}{m_1+m_2}$$
Newton's second law for the For the $m_1$
$$m_1a=Fr_1 -N_{21}$$
Then:
$$N_{21}=Fr_1 -m_1a$$
Replacing $a$, and cancelling similar terms you'll get to:
$$N_{21}=\frac{m_2Fr_1-m_1Fr_2}{m_1+m_2}$$

 

[wrobel]

it is a statically indeterminate problem even for a=0

 

[Astr]

it is a statically indeterminate problem even for a=0

Thank you for your answer, but I don´t think is correct. I get an answer for $a=0$, and for $a=\mu_{est}g$, and the result can be obtained from Newton's second and third laws of motion.

 

[PeroK]

Hello, everyone.
This problem is easy if you assume that the friction between the blocks and the platform is the maximum possible. Then the normal force is 0. But how can you show that the normal is 0 even when the friction is less than maximum?
Thank you for your help.

The question is whether you could have a force between the blocks. And the answer must be yes. A very large coefficient of friction would mean that the blocks are effectively stuck to the surface. It would, therefore, take significant force between them to push them apart (even if the whole system has some horizontal acceleration).

 

[Astr]

The question is whether you could have a force between the blocks. And the answer must be yes. A very large coefficient of friction would mean that the blocks are effectively stuck to the surface. It would, therefore, take significant force between them to push them apart (even if the whole system has some horizontal acceleration).

Thank you for your answer. But, please correct me if I am wrong, are you saying that are friction between the blocks?
If that is what you are saying, this is not the case. The friction is between the platform and the blocks.

 

[PeroK]

Thank you for your answer. But, please correct me if I am wrong, are you saying that are friction between the blocks?
If that is what you are saying, this is not the case. The friction is between the platform and the blocks.

Not friction between the blocks. There could be any force between the blocks. They could, say, be charged and be electromagnetically repelling each other. They could also be attracting each other.

There could be a small compressed spring between them, trying to force them apart.

As @wrobel says, this makes the problem statically indeterminate. You may add internal forces to the system that do not result in motion.

 

[BvU]

But you don't give the meaning of these variables. What are they ?

[edit] this is in reply to #6. It looks as if we are going astray with this thread!
post #7 explains the variables.
It is not very handy to take the two blocks as a whole !
Look at the individual blocks and calculate the acceleration of each one due to the friction with the platform. If they are identical there is no way one block can 'push' the other.

$\$

 

[Astr]

@BvU
"It is not very handy to take the two blocks as a whole"
If you look carefully I analyzed both, the blocks as a whole and the block with mass $m_1$. You can do the same for mass $m_2$, but it is redundant because you can deduce that from Newton's third law.
"Look at the individual blocks and calculate the acceleration of each one due to the friction with the platform"
If you calmly read again the question you can conclude, obviously, that the acceleration of everything -blocks and platform- is the same because, by hypothesis, the acceleration is such that every block is in the static friction regime.
"If they are identical there is no way one block can 'push' the other."

That is the question of the post. How can you prove it?

 

[PeroK]

"If they are identical there is no way one block can 'push' the other."

That is the question of the post. How can you prove it?

It's already been explained several times that this is not the case. You can't prove something that is not true. Especially when it's been shown to be false!

 

[PeroK]

Look at the individual blocks and calculate the acceleration of each one due to the friction with the platform. If they are identical there is no way one block can 'push' the other.

This can't be correct. The blocks may be pushing each other without sufficient force to overcome the friction.

If you are pushing a large block that isn't moving (and neither are you) because of friction, then you have a static scenario yet you do not have zero forces between you and the block. This may not change if the floor starts accelerating: there may still be no motion relative to the floor.

 

[Astr]

Thanks @PeroK.
" There could be any force between the blocks. They could, say, be charged and be electromagnetically repelling each other. They could also be attracting each other."
Suppose there is no such a thing as springs, net electric charge, etc. Only the contact force is present.
Can you prove that the normal force between the blocks is 0 when $Fr<Fr_{max}$.

I can prove that this normal is 0 if friction is proportional to the mass.

 

[PeroK]

Thanks @PeroK.
" There could be any force between the blocks. They could, say, be charged and be electromagnetically repelling each other. They could also be attracting each other."
Suppose there is no such a thing as springs, net electric charge, etc. Only the contact force is present.
Can you prove that the normal force between the blocks is 0 when $Fr<Fr_{max}$.

This makes no sense. There could be a force, so you cannor prove there is not. If you assume there is no contact force, then that isn't a proof.

I can prove that this normal is 0 if friction is proportional to the mass.

If you really could prove that, you could also prove there are no horizontal intermolecular forces within the blocks.

 

[BvU]

I must be missing something important, but I can't find out what. The problem statement mentions

and in my naive book that means $F_{\text{friction, max}}=\mu mg$. So block 1 experiences a friction force
$F\le\mu\, m_1\,g$ and block 2 $F\le \mu \,m_2\, g$. Accelerations are $\mu g$, i.e. the same for both, if $a\ge \mu g$ and $a$ if there is no sliding.

If there is no sliding, there is no contact force between the blocks. If there is sliding, both blocks experience the same acceleration, ergo no contact force between the blocks either.

I acknowledge

it is a statically indeterminate problem even for a=0

but if the exercise composer is devious enough to have an initial nonzero contact force, he/she is more or less obliged to say something about it. I can't believe such a mischievous setup in a textbook.

$\$

 

[jbriggs444]

If there is no sliding, there is no contact force between the blocks.

This assertion is false.

Take, for instance the situation where I am standing in the bed of a pickup truck that is slowly accelerating from a stop. I am facing forward, pushing on a heavy box that is also sitting in the bed.

My shoes are not sliding on the bed. The box is not sliding on the bed. Yet there is a force between us. The situation is statically indeterminate.

 

[PeroK]

If there is no sliding, there is no contact force between the blocks. If there is sliding, both blocks experience the same acceleration, ergo no contact force between the blocks either.

One of the blocks could be leaning on the other. There would be more friction on one block and less on the other, depening on the direction of the external acceleration. And this may apply whenever the friction is less than the maximum.

You may assume that is not the case, but you can't prove those internal forces between the blocks do not exist.

 

[BvU]

I certainly agree, but have a hard time believing the exercise composer wants the student to consider such complications and worry about them. My careful conclusion: not a top quality exercise.

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[PeroK]

I certainly agree, but have a hard time believing the exercise composer wants the student to consider such complications and worry about them. My careful conclusion: not a top quality exercise.

The available answers are all wrong. The answer is indeterminate. It's not that complicated, as the static examples show. It's more a fundamental lack of knowledge on the part of the question setter. Or, just an oversight.

If the question was whether the accelerated motion requires a force between the blocks, then the answer is no.

 

[rudransh verma]

If you are pushing a large block that isn't moving (and neither are you) because of friction, then you have a static scenario yet you do not have zero forces between you and the block. This may not change if the floor starts accelerating: there may still be no motion relative to the floor.

That is not the identical case as the two blocks sitting on a platform. Because you are talking about shifting of center of mass from the stable point. Blocks are stable. They are not leaning on each other. They may have different masses due to different materials but we can see in figure they are exactly identical.

In that case none of the blocks is sliding. So their acceleration is exactly equal, and the friction force alone (between blocks and platform) is causing that acceleration. Meaning the net force is the frinction force, ergo the contact force must be zero !

If we assume they have not been started accelerating and are asked about the force during acceleration then the force between both the blocks will be zero in all cases.