Modified Block Problem -- Block on top and on the side

AI Thread Summary
The discussion centers on a modified block problem involving two blocks and their accelerations. The original poster questions their problem definition and how to utilize two derived acceleration equations. Key insights include the need for two acceleration inequalities to prevent sliding of the blocks, with one threshold ensuring m2 remains stationary and another for m1. Additionally, the coefficient of friction is clarified as a crucial factor, emphasizing that it is not a force but a ratio related to the normal force. The conversation concludes with positive feedback on the problem's complexity and solution quality.
rumman18
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<< Thread moved to the schoolwork forums from the technical forums, so no Homework Template is shown >>

The problem is a modified version of two basic block problems. This is my work so far, but I end up with two equations for acceleration from m1 and m2, I don't understand how to use both equations. So my questions are -

1) Did I define the problem correctly to begin with?

2) Which acceleration value do I plug in for the final expression for F? and why?

3) Overall, is my derivation correct?

This is not a homework problem. I just made it up to test my conceptual understanding.

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I think this should be in the Homework section. (it is a "homework style" problem)
I'll take a look at your work, but you are going to need 2 acceleration inequalities. Acceleration will need to be greater than or equal to some threshold value to keep m2 from sliding down. Also, acceleration will need to be less than or equal to some other threshold value so that m1 does not start sliding. So you would have a range of accelerations which would satisfy.

It could be possible that there is only one critical value of acceleration which satisfies both. Or it could work out that the 2 inequalities do not intersect, and there is no solution.
 
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Further to scottdave's reply, there are several errors in the problem statement.
μs.. is not a force, it is a coefficient of friction. If the normal force is N, and there is no slippage, and the actual frictional force is Fs then the formula is |Fs|≤μsN. Note the ≤, not =.
As scottdave writes, this leads to two constraints on the applied force, F, which might or might not allow some range of solutions.
And on a pedantic note, you mean that the two smaller masses do not move relative to M.
 
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I think the total force on M in the x-axis should be ##F-F_{sM}-F_{m_2M}-F_{sm_1}##. Other than that you should replace some equalities with inequalities. for example I think it should be ##F_{sm_1}\geq m_1a##

Nice problem you made (and also the solution is very good , with detailed schemes and equations). It remembered me of a similar problem my teacher made up when I was in high school (about 30 years ago). I gave you a like for this :D.
 
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