# mrbolha

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## Posts posted by mrbolha

### Tricky friction problem

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).

### Tricky friction problem

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."

### Tricky friction problem

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?

### Tricky friction problem

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

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

### Tricky friction problem

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?

### Tricky friction problem

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"?

### Tricky friction problem

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?

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