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I have some conceptual questions and I'd be very pleased if a physicist solves them.

First, let's consider two stacked blocks in rest. Suddenly, the bottom block starts to move right with constant acceleration (MRUV) while the top block starts moving along with it because of static friction as expected.

After some time the stacked blocks reach a certain speed and from there they continue to move right with the same speed. During this proccess does the static friction force on the top block above decrease to 0? I'm not sure...

Third, how can I predict when the block above will slide? And what does it depend on? PLEASE, HELP ME!!!

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  • $\begingroup$ I don't understand your first question: "Then, why is it that the block above continuously tends to go the opposite way (left) if there's no force pulling it that way?" In this situation, it sounds like there is a force pulling the bottom block to the right. Therefore, a static friction force is pulling the top block to the right as well. And the top block is moving to the right. What do you mean "continuously tends to...". It's moving to the right, not the left! $\endgroup$ – levitopher Aug 28 '17 at 17:01
  • $\begingroup$ I have submitted an edit, dealing only with the post in its present form, (paragraphs, tags and a title change). If this edit is approved, could you please check it and ensure any further edits preserve these changes, thanks. $\endgroup$ – user167453 Aug 28 '17 at 17:17
  • $\begingroup$ When no external forces are applied the bottom block is still going to decelerate due to the friction with the top block. The only time the block maintains constant speed is when friction is zero. $\endgroup$ – ja72 Aug 28 '17 at 19:13
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The maximum friction force between the blocks is $F=\mu mg$ where $m$ is the mass of the upper block and $\mu$ is the coefficient of static friction. This is the maximum force which accelerates the upper block. The maximum acceleration of the upper block is therefore $F/m=\mu g$.

If the acceleration $a$ of the lower block is greater than $\mu g$ then the upper block will slide against the lower block, because there is not enough friction to keep it accelerating at the same rate $a$.

The friction force between the blocks becomes zero when the lower block stops accelerating. This is because the friction force is required to accelerate the upper block, but no force is required to keep it moving at constant velocity (Newton's 1st law).

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The block on top tends to have a leftward pull because its inertia causes it to resist the acceleration the bottom block is subjecting it to which happens because there surfaces are weekly bonded (what we know as static friction) and the bottom block is being accelerated rightwards. So your 'why' wrt an inertial frame is that it's inertia imparts it a tendency to go left. From a non inertial frame (the bottom block), the net force acting on the block will be non zero (the friction, acting rightwards), but it's acceleration will be zero, which will require us to assume the presence of a pseudoforce which can be attributed to it's "inertia", acting leftwards.

Yes, when the system goes on to being in constant linear motion, the frictional force equals zero and hence the transition will be from F-> 0. The free body diagram will include the normal force the lower block exerts on the upper one and it's weight.

You can predict when the third block will slide by calculating the maximum value of their shared static friction, for which you need to know or determine the coefficient of friction (static) between them. Once you have the max value, you can plug it into the equations you get by applying D'Alembert's Principle on both of the bodies and doing the necessary. Remember, the "when" this will help you answer won't be a time but a certain minimum instantaneous acceleration or the instantaneous force applied to the lower block going above which will cause the block to slide relative to the lower block, because the frictional force will no longer suffice to keep the blocks static wrt each other. I hope this helps and you get along in your physics course well. ;)

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  • $\begingroup$ Awesome! so after all it is impossible to calculate a time? $\endgroup$ – Mr. Mister Aug 28 '17 at 17:49
  • $\begingroup$ The time does not matter, relative sliding only occurs when the instantaneous limit for the interactional force is crossed. An analogy: think about something of mass x tied to a ceiling with a string who's maximum tension is 5N, will the time play a role in when the string breaks? Or will the point the string breaks only depend on the mass x? (Mass x corresponds to a downward pull of gx N). $\endgroup$ – Dan Aug 28 '17 at 17:56
  • $\begingroup$ I get it, so if the two blocks are moving together and reach let's say 250 km/h, the top block won't never slide off right unless a force breaks the limit of static friction right? it has nothing to do with velocity right? $\endgroup$ – Mr. Mister Aug 28 '17 at 18:26
  • $\begingroup$ Yep, exactly. I hope you enjoy your physics course. :) $\endgroup$ – Dan Aug 28 '17 at 18:28
  • $\begingroup$ I'm still at school sooo it's a good intro. Thanks! $\endgroup$ – Mr. Mister Aug 28 '17 at 18:32
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You can see my comment above which illustrates my confusion, but let me try to answer your question by carefully discussing what happens. (I have answered for an inertial frame of reference because in the noninertial case it depends which specific frame you pick, and I think a more universal answer will be more helpful)

First, the two blocks are at rest. The only forces acting on them are gravity and the contact force between them.

Then, I push on the bottom block. Assuming it slides, it begins to move to the right according to Newton's second law. Two possible things can happen to the top block:

  • If the coefficient of static friction is large enough, the top block will be dragged along to the right by the force of friction from the bottom block.
  • If the coefficient of static friction is not large enough, the top block will "slide off" the bottom block.

In the second case, kinetic friction would take over. The bottom block would move to the right, dragging the top block with it but since the force of kinetic friction is less than the maximum force of static friction, the force on it would be smaller, it's acceleration is less, and the top block would eventually fall off.

For your second question, let's assume the blocks reach a certain speed without the top block falling off, and then continue to move at a constant speed. Then, what is the acceleration of the bottom block? Zero, obviously. So, what is the acceleration of the top block? Also zero. Therefore there is not net force in the horizontal direction, and the force of static friction must be zero.

Remember to use Newton's second law "in both directions". The forces can tell you something about the motion of objects, but if you know the acceleration, you can often tell something about the forces on the objects as well.

EDIT: Two additional questions asked in the comments below:

1) In the second case, if the bottom block starts moving and the top block slides off means that from earth (inertial) the top block wont move and will stay where it was initially right?

That would happen if the static friction coefficient was small enough, and if you pulled hard enough (this would be the "pulling the tablecloth from under the china" trick). But the "realistic" answer is that you can't pull suddenly with enough force to avoid the frictional force all together. If you have pull 50 N to break static friction, you will likely have to "ramp up" from 0 N to 50 N, and sometime in that time period you will pass through engaging static friction and moving the top block.

2) If the two blocks are moving together and reach let's say 250 km/h, the top block won't never slide off right? it has nothing to do with velocity right? Then what will cause the top block to slide off?

This would depend on the force you applied. If you apply a very small force, you could eventually reach 250 km/h without the block falling. Once you hit that speed, if you stopped applying a force, the top block will never fall off. If you applied a force which was too large, you would quickly get to 250 km/h, but the top block would probably fall.

So "it has nothing to do with speed" is correct - it only has to do with the applied force. We should probably note that we are assuming air resistance is negligible, which is certainly not true at 250 km/h - the top block would essentially "blow off".

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  • $\begingroup$ Thanks! In the second case, if the bottom block starts moving and the top block slides off means that from earth (inertial) the top block wont move and will stay where it was initially right? Another thing, if the two blocks are moving together and reach let's say 250 km/h, the top block won't never slide off right? it has nothing to do with velocity right? Then what will cause the top block to slide off? $\endgroup$ – Mr. Mister Aug 28 '17 at 18:16
  • $\begingroup$ Both of those questions have to do with the details of the problem. I've added answers in my original answer. $\endgroup$ – levitopher Aug 28 '17 at 19:02
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If the acceleration of the bottom brick is greater then A>u.g then the upper block will slide back with the speed of V=t.u.g which is still moving righ slower than lwer block. And time is t = V/x2-x1, two arbitary points on the top suface of lower block.

This acceleration however will decrease once part of the upper block has passed the edge of bottom block because of reduction in friction and spending part of friction energy to angular momentum of rolling action of the upper block till it falls off!

Calculating the time is possible, having the dimensions and mass and angular moment of inertia of the blocks!

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During this proccess does the static friction force on the top block above decrease to 0?

Yes. There are no other sideways forces, so if friction was present it would be unbalanced and would cause acceleration (Newton's 2nd law). Since it doesn't accelerate, we know that there is no friction anymore.

Third, how can I predict when the block above will slide? And what does it depend on?

We have a formula for that:

$$f_s\leq \mu_s n$$

Static friction always obeys this. It cannot be larger than $\mu_s n$. So, if the box below accelerates so fast that the necessary static friction to pull the top one along is higher than $\mu_s n$, then the static friction simply can't pull it along. Then sliding starts.

So, from this formula we see that it depends on $\mu_s$, which is the roughness and "stickiness" etc, as well as $n$, which is how much the surfaces are being pressed together. And that's all.

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First, let's consider two stacked blocks in rest. Suddenly, the bottom block starts to move right with constant acceleration (MRUV) while the top block starts moving along with it because of static friction as expected.

Assume no friction between lower block and supporting surface. Assume the external force is applied to the right on the lower block. Since both blocks have the same acceleration to the right, and no slipping is occurring, the external applied force must be less than the maximum static friction force.

After some time the stacked blocks reach a certain speed and from there they continue to move right with the same speed. During this process does the static friction force on the top block above decrease to 0? I'm not sure...

If they are both moving to the right at the same speed relative to the ground, then there are no net external forces to the right operating on either block. This means at some point the external force applied to the lower block was removed and the blocks continued on with the same speed they had just prior to removing the force. With no external force being applied, there would be no friction force opposing it, so yes the static friction force would be zero.

Third, how can I predict when the block above will slide? And what does it depend on? PLEASE, HELP ME!!!

You will need to know the coefficient of static friction between the blocks, and the mass of the upper block. Then slippage will occur when the force on the lower block is increased to equal the maximum static friction force, or: $$F=μ_smg$$ Where $F$ is the force to the right applied to the lower block, $μ_s$ is the coefficient of static friction, $m$ is the mass of the upper block and $g$ is the acceleration due to gravity.

Hope this helps.

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