WEBVTT
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This is Giancoli Answers
with Mr. Dychko.
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This picture on the left
shows the forces on each of the blocks
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and the block experiences
a tension force upwards
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and that tension
is constant along the rope,
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and so, the right
hand block experiences
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the same tension force up
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and they experience different
forces of gravity downwards
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depending on their masses.
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Now, the pulley experiences
two tension forces downwards,
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one on each side.
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On this edge there's a single
tension force pulling down.
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And because the rope is not slipping.
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And likewise on this edge
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there's a tension force
pulling down on the pulley
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because this arrow here
shows the tension force upwards
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on the block.
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And the Newton's third law
counterpart to this force
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is the force applied
on the rope due to the block,
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and that eventually is really applied
up here on the pulley as well.
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So, there's the picture
for the pulley,
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two times the tension force down
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and then the force
of the cable connecting it
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to the ceiling upwards,
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and this is ultimately
what we have to find.
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But in order to do that we need
to determine the tension force.
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And so, I've written down
Newton's second law
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for each of the blocks.
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Block one has tension force
up minus m1g downwards,
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the gravity down.
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And that equals
the mass times its acceleration.
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And for the second block
it's the same tension force up
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minus m2g equals m2a2.
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These have different accelerations
sort of,
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the same magnitudes
but they're opposite directions
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because they're all part
of the same system,
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the rate at which
they accelerate will be the same
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but one will go up
and the other one will go down.
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So, one is the negative
of the other.
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So, let's rewrite this Newton's
second law for the first block.
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Substituting with negative
a2 in place of a1,
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otherwise it's a copy.
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Copied this part here.
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And then we'll solve that for a2
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and then make a substitution
into the second equation.
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So, we'll move this a2
to the right hand side,
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or sorry, left hand side.
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And...
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I'm sorry. Yeah, well.
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And I also multiply everything
by negative 1.
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It's nice to have
this unknown on the left,
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making it positive is sort
of what I'm doing here.
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So, multiply everything by negative 1.
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So, the sign of everything changed.
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And then flip the sides around.
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So, the m1gis positive
and the FT is negative
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and then this thing is positive.
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Sign of everything changed
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and put the m on the left
and then divide by m1
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so, acceleration 2 is m1g
minus force tension divided by m1.
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And that is what we'll substitute
into this expression.
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And instead of a2
will now write all of this.
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So, it's FT minus m2g,
just copied,
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equals m2, also copied,
times this substitution for a2.
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And we'll multiply everything by m1
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to get rid of this denominator.
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So, that's FT times m1
minus m1m2g
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equals m1m2g minus m2FT.
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So, the m1 cancelled
on the right hand side
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but th+en I also distributed
the m2 into the brackets.
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And then move this to the left,
makes it positive FTm2.
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And move this to the right
and it's the same term,
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they're like terms,
so, that's 2m1m2g.
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And then the next line,
factor out the common factor FT.
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So, it's Ft
times m1 plus m2
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equals 2m1m2g.
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And the tension force then is
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when you divide
by this bracket here.
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The tension force is
2 times m1m2g
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divided by m1 plus m2.
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And so, the force
on the cable
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connecting the pulley to the ceiling
has two times that tension force.
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So, it's 2 times 1.2 kilograms
times 3.2 to kilograms
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times 9.8 Newtons per kilogram.
all divided by the sum of the masses,
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and that gives 34 Newtons.