Maximum Force for Block Stability with Kinetic Friction

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SUMMARY

The maximum force F that can be applied to the large block (M2 = 20kg) without causing the small block (M1 = 10kg) to slide off is determined to be 90N. This calculation incorporates the coefficients of static friction (μs = 0.2) and kinetic friction (μk = 0.1) between the blocks and the table. The net force required for the entire system, factoring in inertia and friction, leads to this definitive value. The solution emphasizes the importance of understanding the dynamics of both blocks under applied forces.

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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 [tex]\mu[/tex]s=.2, [tex]\mu[/tex]k=.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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