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One of our classic
problems to analyze

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using Newton's second
law is the motion

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of two blocks with a rope
that's wrapped around a pulley.

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So imagine we have a pulley, P,
and we hang from that block 1

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and we hang from that block 2.

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The pulley is suspended
by another rope,

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and it's attached to a wall.

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And let's now try to
use Newton's second law

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to analyze this motion.

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Now, going back to our
methodology, the first thing

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we want to ask
ourselves is to identify

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what are all the moving pieces?

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So I'm going to make my
first assumption here,

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that the rope/pulley a
surface so that's right

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here is frictionless.

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Now, what that assumption
means is that the pulley

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will remain at rest.

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And therefore, when we want
to break this problem down,

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our first question is to
identify all the moving

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objects.

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And so we really have three.

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We have mass, which we
label by 1, 2, and the rope.

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So our goal now is to identify
all the moving objects

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and to draw free body force
diagrams for each moving

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object.

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Now, one of the things
that I'm especially

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interested in talking about
is to keep in mind to identify

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action-reaction pairs.

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So as I go through and draw
these free body diagrams,

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we want to ask
ourselves, what forces

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form Newton's third law
action-reaction pairs?

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So what I'll do is I'll
start with object 1.

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Now, I'm object 1 has a
gravitational force M 1 g.

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And the rope is
pulling object 1 up

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with the tension at the
end of the rope T rope 1.

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So the action-
reaction pair to M 1 g

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is the Earth, is the force
of this mass on the Earth,

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which we're not considering.

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What about the rope?

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So now, let's draw a
picture of our rope.

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And mass, this object here
is pulling the rope down,

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and so we have, on the rope,
object 1 pulling it down.

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And this is our
action-reaction pair.

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What are the other
forces on the rope?

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Well over here, object 2, t
2, is pulling the rope down

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with the tension at
the end of that rope.

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Now, also, the pulley
is exerting a force

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on the rope upwards,
because we have that force.

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And so these are the force
diagrams on the rope.

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Now, what about the
action-reaction pair

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to T 2 rope?

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Well, let's continue and draw 2.

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We have, again,
gravitational force on 2,

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the Earth is the
action-reaction pair.

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The force of 2 on
the Earth upwards

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is equal to m 2g of
the Earth downwards.

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We're not drawing the
Earth in this picture,

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so we don't show it.

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Now, here is the force
of the rope on 2.

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And there is our other
action-reaction pair.

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So these are Newton's third
law pairs in this object.

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Now, now that we've
identified all the forces,

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we want to apply
Newton's second law

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to each of these
objects that are moving.

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So let's begin by remembering
that if our rope--

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our next assumption
here, it's assumed

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that the mass of the
rope is very light.

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And then the tension in
the rope is constant.

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And the implication of
that is T 2 r equals t 1 r.

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And that's why we can
now identify this as T

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and that as T.

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Little bit later on, when
we talk about pulleys

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that have mass with
friction between them,

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the tension on the two
sides of the pulley

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will not be the same.

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But for the moment, we've
made these assumptions,

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and that holds.

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And now, we just really have
objects 1 and object 2 analyze.

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And so one of the
things that helps a lot

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is, we need to choose
unit vectors in order

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to write down the vector
equations for Newton's

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second law.

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So let's just here
assume for ourselves

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that M 1 is greater than M 2.

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This gives us some feeling
for how the system moves.

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I like to do this
because it gives me

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a way to choose my
unit vectors to make

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all the accelerations positive.

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So when M 1 is bigger than
M 2, I expect M 1 to go down

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and 2 go up.

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And that's how I'm going to
choose unit vectors j hat 1.

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Now, here is the very
interesting thing.

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We choose a separate coordinate
system for each object.

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So I'm going to write that down.

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We want to choose separate
unit vectors for each object.

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And our accelerations
will be with respect

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to those unit vectors.

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This is where a lot of
people get tripped up.

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So here, I'm choosing j hat 2.

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And when I choose
that, I expect a 2

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to be positive
because 2 is going up

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in the direction of j hat.

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And here, I expect a
1 to be positive also,

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because 1 is going down.

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Now I can draw Newton's
second law on 1.

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So I have F 1 equals M 1 a 1.

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And now I analyze my forces,
M 1 g is in the positive j hat

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direction, t is in the
negative j hat direction,

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so I have M 1 g minus
t equals M 1 a 1,

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which I'm expecting
to be positive.

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And in the same way, on
2, F 2 equals M 2 a 2.

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I look at my force diagram.

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Now, notice, j hat is
up, so t is positive

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and 2 g is minus my direction,
and 2 g equals M 2 a 2.

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So I now have two equations, but
when I look at these equations,

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I see that I have
three unknowns.

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I have t, a 1, and a 2.

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But I have a constraint,
because these objects are moving

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together and the way I've
chosen the coordinate systems,

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a 1 is equal to plus a 2,
both are positive with respect

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to my choice of
coordinate systems.

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And so these two
as are the same.

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And now I can solve my
equations for t and a.

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And let's quickly do that.

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M 1 g minus t equals M 1 a.

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And over here, let's
solve for the tension,

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for the accelerations here.

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T is equal to m 2 a plus m 2 g.

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And when I substitute
the t into that equation,

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we'll find some space for that.

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we'll write that as, if I put in
the M 2 minus M 2 a minus M 2 g

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and bring the M 2 a
to the other side,

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then it's a little
bit of algebra,

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but I think you'll trust me
that this is M 1 minus M 2

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g over M 1 plus M 2.

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Now, we're almost done.

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We need to check our answers.

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First off, does it have the
dimensions of acceleration?

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Answer, yes, mass
divided by mass,

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g has the dimensions
of acceleration.

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So that's my first check
always with my algebra.

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Second check, what if
M 1 is equal to M 2?

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Then the acceleration is 0,
I expect them to be balanced.

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I've said that M 1
is greater than M 2.

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Positive sign, my A is positive.

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That's how I set up
my coordinate system,

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so I expect that too.

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By the way, our symbol
a was equal to those.

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And so I'm pretty confident
that my result is correct.

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And that's how we apply
Newton's second law

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to the pulley with
several assumptions,

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frictionless surface,
massless rope.

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And think about how we
chose our unit vectors.

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Each object gets its
separate coordinate system.

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I'm not using the
same coordinate system

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for each object.

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I could've, but
then there would be

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a subtlety to this
condition, the accelerations

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would be opposite in sign.