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We want to look at
this pulley system.

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We want to find out
what this force here

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is, for example, with which
this block is being pulled.

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Now we have two massless pulleys
here and two moving parts.

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And one key component
of this problem

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is to derive the
acceleration constraint.

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How are we going to do that?

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Well, we have to look
at this string here.

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First of all, it's a
fixed string length.

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And that will help us.

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Ultimately, this is
just a distance that

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connects all of these objects.

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And if we eventually
differentiate that twice,

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we get to an acceleration.

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So the first thing
we're going to do

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is to figure out what the
string length is to then derive

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the acceleration condition.

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Let's begin by first
identifying the fixed

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length in this problem.

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So we have a little
distance here.

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We call it sb.

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And we have a distance
here that's fixed, sa.

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And we furthermore know
that this block here

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is fixed to the ground.

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And we can choose a
coordinate system origin.

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Let's say we do that here.

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We know that we have
the distance d here

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to this little block.

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The moving parts,
block 1 and block 2--

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so we have to assign
position functions.

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And actually this
one here is x1.

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It goes up to here
because we want

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to measure the string
length, ultimately,

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and we know this portion here.

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So we can subtract it.

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And then of course we need also
x1, so that goes here-- x2.

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So now we can tally
up the length.

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Let's start with
this portion here.

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That is d minus sa minus x2.

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d minus sa minus x2.

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And then we have a
half circle here, pi r.

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And we know actually that there
is another half circle that's

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going to come there, so we can
just write 2 pi r immediately.

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And then we have this part here.

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That one is x1 minus sb,
and then minus sa minus x2.

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And then finally we
have this portion,

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and that is x1
minus sb minus x2.

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OK, so the next
step is that we need

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to simplify this a little bit.

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So we have one x1 and another
one here, 2x1 plus x2, x2,

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x2-- actually 3-- minus 3x2.

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And then we have all
sorts of consonants.

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We have d, we have
sa, sb, the 2 pi r.

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But, as you will see, if we
differentiate this out here,

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that will actually
all fall away.

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So we're going to make
our life easy and just

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add a constant here.

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And so we want to now do
the second derivative here

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of our string length
with respect to time,

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because these all
position functions.

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So we can differentiate those.

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What's important of course
here is this string length

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is not changing with time.

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So actually we know that
that derivative will be 0.

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And we can just
write this up here

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because x differentiated
twice is a.

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So 2a1 minus 3a2.

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And we immediately see from
that that a1 equals 3/2 a2.

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So this is our
constraint condition

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that we will need later.

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For now, we need to continue
with setting up free body

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diagrams of all four objects.

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Let's start that with object 1.

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What's acting on object 1?

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Well, we have F here.

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Why don't we just
write it as magnitude.

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We have F, and then
we have here a tension

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that goes to the
pulley B. That one

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is different from the
tension in the string.

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So we're going to call this TB.

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And then we have object 2.

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Oh, and of course, i hat
goes in this direction

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because it follows the
motion of the object.

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We have here kind
of the reverse.

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Now we have a tension
of this string here

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that's attached to
pulley A, but it's

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different from this
string tension-- that's

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a specific one-- TA.

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And we also have a T
from the string here.

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So we'll add a T. And then
if we look at pulley A,

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we have two string
tensions, T and T.

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And here we again have a TA.

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And pulley B, TB,
and two T over here.

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So that means we can write
down our equations of motions

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using Newton's Second
Law, F equals ma.

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And since this is just going
in the i hat direction,

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we can just write down the four
equations following the four

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free body diagrams here.

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So we have F minus
TB equals m1 a1.

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We have T plus TA equals m2 a2.

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Then we have 2T
minus TA equals 0.

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That is 0 because we're dealing
with two massless pulleys

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there.

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That means m is 0 and so
our acceleration term is 0.

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So the m is 0 here.

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And finally, we
have TB minus 2T.

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No, not 2 pi-- 2T.

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And that one is also 0 because
it is a massless pulley.

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So here we have our
four equations of motion

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that fully govern
this pulley system.

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And we can use it
then, for example,

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to find this pulling force here.

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We know what TB is from the
equation down here-- so 2T.

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We know what TA is.

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It's also 2T.

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And we can then
solve this for F.

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The only sticky part is that
we have this a1 and a2 in here.

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But for that, we derived this
constraint condition here.

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So with that one, we
can fully solve this.

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Otherwise, we would have
one too many unknowns.

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Alternatively, if one were
to be interested in one

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of these accelerations,
then this equation system

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can also be solved for a1 or a2.