- [Voiceover] Let's now tackle part C.
So they tell us block 3 of mass m sub 3,
so that's right over here, is added
to the system as shown below.
There is no friction between block 3 and the table.
Alright, indicate whether the magnitude
of the acceleration of block 2
is now larger, smaller, or the same
as in the original two-block system.
Explain how you arrived at your answer.
So let's just think about the intuition here.
If you think about the net forces
on the system itself, they're the same
as we had before, now the internal forces
are going to be different, you're actually
going to have two different tensions now,
now that you have two different strings
but the net forces, the ones that are
causing this thing to accelerate
in the upward direction on the left-hand side
to the right on the top and then downwards
on the right-hand side, it's still this though
the difference in the weights between
the two blocks but now that difference in those,
inbetween the weights of the two blocks
is moving more mass and we know that force
is equal to mass times acceleration
or acceleration is equal to force
divided by mass and our force, our net force
is being, is the differential between the weights
or the difference between the weights
of the block but now we're going
to be moving more aggregate mass,
this is going to be m1 plus m2 plus m3
and so you're going to have a smaller acceleration.
And so what you could write is acceleration,
acceleration smaller
because same difference,
difference in weights,
in weights,
between m1 and m2
is now accelerating more mass,
accelerating more mass.
And that's the intuitive explanation for it
and if you wanted to dig a little bit deeper
you could actually set up free-body
diagrams for all of these blocks over here
and you would come to that same conclusion.
So let's just do that, just to feel good about ourselves.
So what are, on mass 1
what are going to be the forces?
Well you're going to have the force of gravity,
which is m1g, then you're going to have
the upward tension pulling upwards
and it's going to be larger than the force of gravity,
we'll do that in a different color,
so you're going to have, whoops,
let me do it, alright so you're going
to have this tension, let's call that T1,
you're now going to have two different tensions here
because you have two different strings.
Now the tension there is T1, the tension
over here is also going to be T1
so I'm going to do the same magnitude, T1.
Now since block 2 is a larger weight
than block 1 because it has a larger mass,
we know that the whole system is going
to accelerate, is going to accelerate
on the right-hand side it's going to accelerate down,
on the left-hand side it's going to accelerate up
and on top it's going to accelerate to the right.
And so if the top is accelerating to the right
then the tension in this second string
is going to be larger than the tension in the first string
so we do that in another color.
I'm having trouble drawing straight lines,
alright so that we could call T2,
and if that is T2 then
the tension through,
so then this is going to be T2 as well
because the tension through, the magnitude
of the tension through the entire string
is going to be the same, and then finally
we have the weight of the block,
we have the weight of block 2,
which is going to be larger than this tension
so that is m2g.
Now I've just drawn all of the forces that are
relevant to the magnitude of the acceleration.
If I wanted to make a complete
I guess you could say free-body diagram
where I'm focusing on m1, m3 and m2,
there are some more forces acting on m3.
M3 in the vertical direction,
you have its weight, which we could call m3g
but it's not accelerating downwards
because the table is exerting force
on it on an upwards, it's exerting
an upwards force on it so
of the same magnitude offsetting its weight.
So that's if you wanted to do a more
complete free-body diagram for it
but we care about the things that are
moving in the direction of the accleration
depending on where we are on the table
and so we can just use Newton's second law
like we've used before, saying the net forces
in a given direction are equal to the mass
times the magnitude of the accleration
in that given direction, so the magnitude
on that force is equal to mass
times the magnitude of the acceleration.
And so we can do that first with block 1,
so block 1, actually I'm just going to
do this with specific, so block 1
I'll do it with this orange color.
So block 1, what's the net forces?
Well it is T1 minus m1g,
that's going to be equal to mass times acceleration
so it's going to be m1 times the acceleration.
Now what about block 3?
Well block 3 we're accelerating to the right,
we're going to have T2,
we're going to do that in a different color,
block 3 we are going to have
T2 minus T1,
minus T1 is equal to m
is equal to m3 and the magnitude
of the acceleration is going to be the same.
Here we're accelerating to the right,
here we're accelerating up,
here we're accelerating down,
but the magnitudes are going to be the same,
they're all, I can denote them with this lower-case a.
And then finally we can think about block 3.
We could say that the net forces,
well that's m2g minus T2,
that's going against m2g,
is equal to m2 times its acceleration
and now if we want to solve for acceleration,
and this will be quite convenient,
we can just add up all of the left-hand sides
to get a new left-hand side, and add up
all the right-hand sides to get a new right-hand side,
we can do that algebraically because they're all
this is equal to that, that is equal to that,
that is equal to that so if you add up
these and then you add up those,
well then the sums are going to be equal to each other.
And so what are you going to get?
So if you add up all of this,
this T1 is going to cancel out with the subtracting the T1,
this T2 is going to cancel out with the subtracting the T2,
and you're just going to be left with an m2g,
m2g minus m1g,
minus m1g, m2g minus m1g
is equal to and just for,
well let me just write it out
is equal to m1a plus
m3a plus m2a.
Well we could of course factor the a out
and so let me just write this as
that's equal to a times m1 plus m2
plus m3, and then we could divide
both sides by m1 plus m2 plus m3.
So let's just do that.
So m1 plus m2 plus m3,
m1 plus m2 plus m3,
these cancel out and so this is your,
the magnitude of your acceleration.
And notice you have the same difference in weights
that's providing the net force on the system
but is now accelerating more mass,
so you could even say, hey look,
I have more mass here,
so more mass to accelerate,
more mass,
more mass to accelerate while I have the same
net force acting on the system,
we're not talking about the internal forces,
those all canceled out when I added these equations
and so if you're taking the same net force
and you're dividing it by more mass
you're going to have a smaller,
smaller acceleration.
Hopefully that all made sense to you.