Maximum Force for Block Stability with Kinetic Friction

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The discussion focuses on calculating the maximum force that can be applied to a larger block without causing a smaller block on top to slide off. Given the masses of the blocks and the coefficients of static and kinetic friction, the initial assumption of using static friction to find the force was incorrect. The correct approach involves considering the inertia of both blocks and the net forces acting on the system. Ultimately, the required force to prevent sliding was determined to be 90N. The conversation highlights the importance of understanding the dynamics of the system rather than just applying static friction principles.
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Homework Statement


What is the largest force F that can be applied to the large block such that the small block doesn't slide off.

There are two blocks of M1=10kg, M2=20kg. M1 is on top of M2, M2 is on top of the table.
the coefficients of static and kinetic friction between the two blocks, and also between the large block and the table are \mus=.2, \muk=.1 respectively. let g=10m/s^2


Homework Equations





The Attempt at a Solution



Well, i thought that simply that the force should be applied to where the force of friction would be the max of static friction, so F=-Ff=.2(10kg)(10m/s^2)= 20N

It seems wrong, but I can't figure out what is.
 
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The force is applied to the LARGE (20 kg) block, not the SMALL (10 kg) block. You need to consider inertia.
 
wow shoot me in the head, i am retarded.
so 20N/10kg=2m/s^2, where net force for the whole system is 60n. Then at kinetic friction has to total 30N, therefore the required force is 90N.
And there's the answer. Thanks for the heads-up haha. I seriously got to learn how to read.
 
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