# Easy kinetics friction/force

1. Oct 6, 2009

### jun9008

1. The problem statement, all variables and given/known data
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 $$\mu$$s=.2, $$\mu$$k=.1 respectively. let g=10m/s^2

2. Relevant equations

3. 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.

2. Oct 6, 2009

### turin

The force is applied to the LARGE (20 kg) block, not the SMALL (10 kg) block. You need to consider inertia.

3. Oct 6, 2009

### jun9008

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.