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coefficient of friction between the 2 blocks is U and the surface below the bigger block is smooth . Find the min and max force that can be applied in order to keep the smaller blocks at rest with respect to the bigger block.

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|dw:1356949501640:dw| is that right ?
sorry.. this is right! missed the pseudo on second small block..

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Other answers:

This problem was already solved on OS. Will try to find the reference.
eqns of motion would be Umg + m(F/M) = T and T + Um(F/M) = mg => Umg + m(F/M) + Um(F/M) = mg the value of F that am getting is not right ! ;)
Sorry that case was when it tends to move to the right
Now, |dw:1356962249332:dw|
Fr1= umg Fr2 =umF/(M+2m)
F' = mF/(M+2m) W=mg
From W-Fr2-F'-Fr1=0 we get F=g(M+2m)(1+u)(1-u) From F'-mg-Fr2-Fr1=0 we get F =g(M+2m)(1+u)/(1-u)
i see!! thanks a lot! that helped !
i understand it now..

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