And now I'm in position to talk about what
is the maximum force.
If I push F harder, as if I push F and the blocks go together,
the accelerations are the same.
I push F harder.
Accelerations are the same.
I push F harder, and I push F so hard,
that the static friction no longer reaches
its maximum value.
And if I push F harder than that,
I will not have-- the static friction
can't get bigger, and the blocks 1 and 2 will start
to slip relative to each other.
So my no slipping condition is that I want two things.
a1 equals a2.
We'll call that a.
And the maximum force condition is that F static
is equal to maximum value.
Now, what is the normal force between that we refer to
in our fiction law?
There's two normal forces-- the ground
and the normal force between the surfaces.
But the static friction that we're talking about
is between the surfaces.
So that's why we use N here and not N ground.
And now I can solve for this F max.
By the way, we also have the condition that fk is mu k.
What normal force are we talking about?
N, which we've called N ground 1.
And now I look at my equations, and my goal
is to solve for F. I know Ng1 from this equation.
It's just equal to M1g plus N. I Know N from that equation.
So Ng1 is just the sum of the masses times g.
So I know this.
I have f, which is m2a.
And I can now apply my result.
So what we'll do is we'll solve for the a's, a1 equals a2,
in terms of F max.
So over here we have that a is f over m2.
That's from this equation.
And now I'll substitute a1 is equal to that.
I'll substitute that there.
And I get that F Max is going to be
equal to fk plus f static max.
That's that one.
Plus m1 times a1, which is f static max divided by m2.
And so I get that F max equals-- now
I'm going to substitute in all of these values.
It's going to look a little complicated.
And so I'd like to have a little space here for that,
to get everything in here.
And we'll see that it's equal to mu k m1 plus m2g, fk, mu k,
plus m1 2mg plus f static max times 1 plus and m1 over m2.
But f static max is mu N, and N is m2g.
So I get m2g times 1 plus m1 over m2.
And there is, if I push any harder than that,
block 2 will slip with respect to block 1.
Again, all of our terms have the dimensions of acceleration.
This is dimensionless-- 1, dimensions of acceleration.
And we did miss one little thing.
We missed the coefficient of static friction.
And there we have it, a tricky problem.