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One of the big
difficulties students

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have in a problem that
involves many objects is

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to successfully identify
third law interaction pairs.

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So today, I'd like to look
at a problem which shows us

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how to think about that.

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So what we're going to look at
is third law interaction pairs.

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Now, the problem I'd like to
consider is the following.

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Suppose we have a block one
and another block two sitting

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on block one on a surface
and our surface has friction.

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And I'd like to push
block one with a force F.

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And now I'd like to use
Newton's second law to determine

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what is the maximum force I can
push that block two will not

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slip.

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So I'm trying to find F max such
that block two does not slip.

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Now, how do we even begin
to think about this?

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Well, the first thing
that we have to decide

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is when we apply Newton's second
law, our first question will

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be, what is the system
that we'll choose?

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And in this problem, there
are three separate ways

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to think about this.

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Just to show you an example,
so system a will be block one,

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system b will be block
two, and system c

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will be block one and block two.

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And given these
different systems,

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I can address
different questions

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and we'll see that
as we develop this.

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So let's start with
block one and we'll

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start with breaking our
problem down into block one

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and block two.

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And let's draw the free-body
diagrams on those blocks,

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trying to identify
action-reaction pairs.

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So let's begin with block one.

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First off, we know there's
gravitational force

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and the Earth is the other pair
there, which we're not drawing.

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There's friction
between the surfaces.

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So there's a friction between
the ground and block one.

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There is the normal force
of the ground on block one.

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Block two is
sitting on block one

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so there's a normal force
of block two on block one.

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And finally, as
you push block one,

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there's a friction
force between block two

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and block one that's opposing
the fact that block one is

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being pushed forward.

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So we have a friction force
here between blocks two and one.

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Now, this friction
down here is kinetic.

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But if the blocks
are moving together,

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this friction here is static.

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So those are the free-body
force diagrams on one.

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I'll choose unit
vectors in a moment.

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Now, what about two?

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So let's draw two.

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Well, again, m2 g-- the
Earth is the other element

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of the interaction pair.

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And now, block one is
pushing block two up.

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So we have block one pushing
block two up and notice

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our indices make it very easy
to see that our first Newton's

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third law interaction pair is
the normal force of contact

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between the two blocks.

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Now, what else?

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Here's the subtle thing is
that this whole system will

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move to the right.

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What's the force that's making
block two move to the right?

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Well, it's static friction.

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So static friction from
block one and block

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two-- this is the
static friction--

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is causing block two
to move to the right.

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And now we can see, again our
third law interaction pair.

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So in this problem, we have
two third law pairs, this one

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and I'll connect the line there,
and those are the third law

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pairs.

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Now, we know by the third law
that they're equal and opposite

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in magnitude.

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We can identify f.

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We can call this one
N if we wanted just

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to save ourselves the problem
of writing a lot of indices.

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Once we've done that,
we're now ready to apply

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Newton's second law.

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We haven't yet figured
out what the condition is

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that it will just slip.

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We'll get to that.

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But for the moment, we can now
apply vector decomposition.

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So we need to choose
some unit vectors.

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Because I'm pushing the system
this way, it makes sense for me

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to choose my i-hat to the right.

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So here, I'm going to
choose i-hat 1 and j-hat 1.

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Now, over here, I could
choose the same unit vectors,

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even though I'm
thinking about this

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as a completely separate problem
with its own coordinate system.

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And I'll choose
i-hat 2 and j-hat 2.

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But because these unit vectors
are in the same direction,

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they're equal.

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And they're both moving in
the positive i directions

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and so I expect both a1
and a2 to be positive.

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And now I on block one, I can
write down F1 equals m1 a1.

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And because we have two
different directions,

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I'll separate out.

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I like to call this
my "scorecard."

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And now I look at the forces.

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Oh, I missed the pushing force.

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But that's an
interesting exercise.

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When I looked at
this diagram, I saw

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I had two forces going this way.

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I had no force acting that way.

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I went back.

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I checked my free-body
diagram, and recognized

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that I forgot to put
F in there-- always

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a good exercise to double-check
your free-body diagrams

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before you apply Newton's laws.

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So now in the x-direction,
we have the pushing force

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minus the static
friction minus-- we'll

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call this fk-- minus
the kinetic friction

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and that's equal to m1 a1.

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Now in the vertical
direction, we

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have the ground
friction, N ground

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1, minus block two
pushing down on block one

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minus the gravitational force.

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And there is no acceleration
in that direction.

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And I double-check my free-body
diagrams, I check my signs,

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and that looks right to me.

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Now, for block two, I'll
apply the same analysis.

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F2 equals m2 a2.

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Separate out my two
unit directions.

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Notice even though these
unit vectors are the same,

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I'm emphasizing that I'm
talking about block two.

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I could have chosen different
coordinate systems if I wanted.

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Now, I look at my
free-body diagrams.

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On block two.

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I see that its
static friction is

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the only one in the
positive i-hat direction.

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So I have F equals m2 a2.

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And in the vertical
directions, I

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have that my force between
the blocks, the normal force

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between the blocks,
minus gravity is 0.