Box on Box on Box free body diagram

AI Thread Summary
When applying a force to the center box in a stack of three boxes, the top box can slip if the applied force is strong enough to overcome the static friction between the middle and top boxes. The top box will experience some acceleration in the direction of the applied force, but its acceleration will differ from that of the middle box. If the middle box accelerates quickly enough, the top box may appear immobile, as it won't have time to move significantly before the middle box slips away. The discussion emphasizes the dynamics of friction and acceleration in a stacked system. Overall, understanding these forces is crucial for accurately drawing a free body diagram in this scenario.
wil3
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Okay, so say that I stack three boxes, one on top of another. There is friction between all surface, including the bottom box and the ground.

If I apply a force to the center box only, is there any set of circumstances under which the top box will slip? I know that if the applied force is large enough, static friction between the bottom and center box will be overcome and the middle box will slip over the bottom box. Is there a way for the top box to slip out as well? I am trying to draw a free body diagram, but the only x-axis force I find on the top box is friction, which always points in the direction of the force applied to the center box. I know that the applied force can be great enough to overcome the static friction between the top and middle boxes, but it seems to me like the top box will still accelerate some in hte direction of the applied force. Is this accurate?

Think about this in terms of a playing card deck- if I take a deck of cards and set it on a table, and I pull out on random card halfway, if I flick it hard enough it will go flying out of hte deck without disturbing the cards above or below it.
 
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wil3 said:
but it seems to me like the top box will still accelerate some in hte direction of the applied force. Is this accurate?

Yes.

Middle box:

F_{pull} - F_{f \ bottom} - F_{f \ top} = m_{middle} a_{middle}

Top box:

F_{f \ top} = m_{top} a_{top}

Both accelerations are in the same direction but they are not the same magnitude. And if the middle one accelerate fast enough wrt the top box, the top box will seem immobile (it won't have time to travel very far before the middle box is gone). Similarly, for the bottom box:

F_{f \ bottom} + F_{f \ ground} = m_{bottom} a_{bottom}
 
Thank you very much! Of all of the questions I have ever posted on Physics Forums, this is one of the few that I have received a complete, well-worded, and properly explained answer to.
 
wil3 said:
Thank you very much! Of all of the questions I have ever posted on Physics Forums, this is one of the few that I have received a complete, well-worded, and properly explained answer to.

Thank you. I answer as much for myself than for others. It's exercise for the brain.
 
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