Tricky friction problem

[mrbolha]

I've been trying to solve this problem for a while.

Suppose two blocks with mass 1kg each are stacked on a frictionless table. Suppose the coefficients of friction between the blocks are 0.6(static) and 0.5(kinetic).

If I push the upper block with a 7N horizontal force, what should be the final acceleration of each block?

[swansont]

What is the force on the lower block when you push on the upper block?

[mrbolha]

That is what I want to find out

 

I *suppose* the only horizontal force acting on the lower block (since the table is frictionless) is the friction from the upper block.

 

Question is: do the blocks move in relation to each other, or do they "stick"?

[LazerFazer]

Alright, so assuming that the only horizontal force on the lower block is the friction from the upper block, how would YOU go about determining the friction FORCE between the two blocks?

[mrbolha]

Well, let's see:

The Normal force on the upper block is 10N (let's work with g=10m/s^2, ok?)

That would make the maximum static friction force 0.6*10 = 6N, right?

 

Since the force applied on the upper block is 7N, should I assume that the blocks "slide" on each other?

[swansont]

Well' date=' let's see:

The Normal force on the upper block is 10N (let's work with g=10m/s^2, ok?)

That would make the maximum static friction force 0.6*10 = 6N, right?

 

Since the force applied on the upper block is 7N, should I assume that the blocks "slide" on each other?[/quote']

 

Is 7N the actual force on the lower block?

[mrbolha]

No, the force is being applied on the *upper* block.

The lower block will be pushed by the friction force, right?

[swansont]

No' date=' the force is being applied on the *upper* block.

The lower block will be pushed by the friction force, right?[/quote']

 

Right. So you have to figure out what the acceleration of each part of the system is.

[mrbolha]

Well, I asked the very same question to:

* another physics forum, where it was said that the resulting accelerations would be:

7m/s^2 (upper)

5m/s^2 (lower)

* a physics teacher, who said the resulting accelerations would be:

4.5m/s^2 (upper)

2.5m/s^2 (lower)

 

Somehow, I don't think either is right. I think both blocks will move together, with a final acceleration of:

3.5m/s^2 (each)

 

Rationale:

In the first answer, linear momentum in the system grows faster than we can expect from the original force

In the second answer, we find out that the lower block will have a net force of 2.5N. But the only force acting on it is the friction, and it could not be below 6N.

Therefore, I concluded that, in this case, static friction is *not* overcome, and the blocks move along together.

 

Am I wrong?

[mrbolha]

Where I said: "In the second answer, we find out that the lower block will have a net force of 2.5N. But the only force acting on it is the friction, and it could not be below 6N."

I mean "In the second answer, we find out that the lower block will have a net force of 2.5N. But the only force acting on it is the kinetic friction, (since the blocks would be moving in relation to each other) and it could not be below 5N."

[J.C.MacSwell]

Well' date=' I asked the very same question to:

* another physics forum, where it was said that the resulting accelerations would be:

7m/s^2 (upper)

5m/s^2 (lower)

* a physics teacher, who said the resulting accelerations would be:

4.5m/s^2 (upper)

2.5m/s^2 (lower)

 

Somehow, I don't think either is right. I think both blocks will move together, with a final acceleration of:

3.5m/s^2 (each)

 

Rationale:

In the first answer, linear momentum in the system grows faster than we can expect from the original force

In the second answer, we find out that the lower block will have a net force of 2.5N. But the only force acting on it is the friction, and it could not be below 6N.

Therefore, I concluded that, in this case, static friction is *not* overcome, and the blocks move along together.

 

[b']Am I wrong[/b]?

 

If it is not overcome, how do they move? Note that the normal force between the two blocks is only half that of the force between the bottom block and the surface.

 

edit: oops missed the frictionless surface part

 

I think you are correct and the net force on the bottom block will be 3.5 N

[swansont]

Since there is a force on the bottom block, the net force on the upper block will be less than 7N, and you need to know that to determine the acceleration. Think of it like this: if you were blindfolded and exerting this 7N, the block would feel like it had larger than a 1 kg mass; the lower block is coupled to the upper. The question is: how much coupling?

 

The top block exerts some force F on the bottom block, and the bottom block exerts -F on the upper (by Newton's third law) and this is obviously limited (it obviously can't be bigger than 7N, for example, since that would leave you with zero net force and no way to start moving the system); the static limit is when the frictional force is 3.5N, so that each block accelerates at (7N/2kg) = 3.5 m/s^2, and the system looks like a 2 kg block.

 

Since the calculated frictional force (i.e. the threshold where friction would fail to hold them together, 9.8N* mu) will always exceed that, I conclude that the blocks are couples all the times, so no sliding takes place, and the acceleration of each is 3.5 m/s^2

 

Agree or not? Flaws in my reasoning?

[J.C.MacSwell]

Since there is a force on the bottom block' date=' the [b']net[/b] force on the upper block will be less than 7N, and you need to know that to determine the acceleration. Think of it like this: if you were blindfolded and exerting this 7N, the block would feel like it had larger than a 1 kg mass; the lower block is coupled to the upper. The question is: how much coupling?

 

The top block exerts some force F on the bottom block, and the bottom block exerts -F on the upper (by Newton's third law) and this is obviously limited (it obviously can't be bigger than 7N, for example, since that would leave you with zero net force and no way to start moving the system); the static limit is when the frictional force is 3.5N, so that each block accelerates at (7N/2kg) = 3.5 m/s^2, and the system looks like a 2 kg block.

 

Since the calculated frictional force (i.e. the threshold where friction would fail to hold them together, 9.8N* mu) will always exceed that, I conclude that the blocks are couples all the times, so no sliding takes place, and the acceleration of each is 3.5 m/s^2

 

Agree or not? Flaws in my reasoning?

 

I think this is correct. The net force on each is 3.5 N and they accelerate at 3.5 m/s^2.

[mrbolha]

It should be fairly easy to design a lab experimento to try this out.

I imagined a setup with two blocks on a very slippery table. The lower block is initially stuck to the table (with some sticky tape or something), while a rope with a pulley is attached to the upper block.

Then, one could gradually increase the charge on the other end of the pulley rope, till the upper block barely starts to slide. Then, we release the lower block so it can slide freely on the table, and watch if the blocks move together or if they slip.

That is the setting: the force is just high enough to break the static friction if the lower block was fixed, but low enough that, if you assume sliding, the kinetc friction force would be higher than half of the applied force (assuming both blocks have the same weight).

[danny8522003]

If the table is frictionless, then any mu value greater than 0 between the blocks would mean they would both accelerate at 3.5m/s. Surely?

[swansont]

If the table is frictionless, then any mu value greater than 0 between the blocks would mean they would both accelerate at 3.5m/s. Surely?

 

No, I disagree. Look at the limiting behavior will help. The lower block has to be able to exert a sufficient frictional force on the upper, and feel the resulting reaction force, in order to accelerate at 3.5 m/s^2. If mu is zero, the upper block "wants" to accerate at 7 m/s^2, and a blind person pushing would only feel like it was a 1 kg block. As mu increases, the system "feels" more massive, until you reach a mu where they always stick, and beyond that it's effectively a 2 kg mass. Until that point, the lower block has a smaller acceleration.

 

Left as an exercise: at what value of mu will the blocks actually both accelerate at 3.5 m/s^2? (assume kinetic and sliding are equal for simplicity). I think an analysis of that problem will highlight what I said here and in the previous post.

[danny8522003]

There is no force acting to prevent the movement of the lower block because of the frictionless table though. Any force greater than 0 will give rise to an acceleration in the lower block. In order to prevent the lower block from accelerating at 3.5m/s wouldnt there need to be a force opposing this acceleration in the form of friction?

 

Any force on the upper block will result in a force on the lower block, but how can it exert a force back if there is nothing opposing acceleration but it's own inertia?

[swansont]

There is no force acting to prevent the movement of the lower block because of the frictionless table though. Any force greater than 0 will give rise to an acceleration in the lower block. In order to prevent the lower block from accelerating at 3.5m/s wouldnt there need to be a force opposing this acceleration in the form of friction?

 

Any force on the upper block will result in a force on the lower block' date=' but how can it exert a force back if there is nothing opposing acceleration but it's own inertia?[/quote']

 

No. An equivalent situation would be if you were hanging on to a rocket in outer space. The engine kicks in. If you hold on with your pinky finger, will you go along with the rocket, or be left behind after having received only a small impulse?

 

The lower block will feel an acceleration of F_friction/m, maxing out at 3.5. But it can have a smaller acceleration, meaning there will be sliding.

[J.C.MacSwell]

There is no force acting to prevent the movement of the lower block because of the frictionless table though. Any force greater than 0 will give rise to an acceleration in the lower block. In order to prevent the lower block from accelerating at 3.5m/s wouldnt there need to be a force opposing this acceleration in the form of friction?

 

Any force on the upper block will result in a force on the lower block' date=' but how can it exert a force back if there is nothing opposing acceleration [b']but it's own inertia[/b]?

 

I think you just answered your question.

 

3.5m/s is the maximum acceleration for the lower block. Any slippage and it is reduced.

 

3.5m/s is the minimum acceleration for the upper block. Any slippage and it is increased.