- So just to show you how powerful this approach is of
treating multiple objects as if they were a single mass
let's look at this one, this would be a hard one.
We've got a 9kg mass hanging from a rope
that rope passes over a pulley then it's connected
to a 4kg mass sitting on an incline.
And this incline is at 30 degrees,
and let's step it up let's make it hard,
let's say the coefficient of kinetic friction
between the incline and the 4kg mass is 0.2
And that's the coefficient.
so there's going to be friction as well.
If you tried to solve this the hard way it
would be challenging, it's do-able
but you're going to have multiple equations
with multiple unknowns, if you try to analyze each
box separately using Newton's second law.
But because these boxes have to accelerate at the same rate
well at least the same magnitude of acceleration,
then we're just going to be able to find the system's
acceleration, at least the magnitude of it, the size of it.
This 4 kg mass is going to have acceleration in this way
of a certain magnitude, and this 9 kg mass
is going to have acceleration this way
and because our rope is not going to break or stretch,
these accelerations are going to have to be the same.
So we get to use this trick where we treat these
multiple objects as if they are a single mass.
And the acceleration of the single mass
only depends on the external forces on that mass.
So we're only looking at the external forces,
and we're gonna divide by the total mass.
So what would that be?
If we wanted to find the acceleration of this
4 kg mass, let's say what the magnitude of this acceleration
This 9 kg mass is much more massive than
the 4 kg mass and so this whole system
is going to accelerate in that direction,
let's just call that direction positive.
So that's one weird part about treating
multiple objects as if they're a single mass
is defining the direction which is positive
is a little bit sketchy to some people.
We're just saying the direction of motion
this way is what we're calling positive.
And that works just fine, so when I plug in
and go to solve for what is the acceleration
I'm gonna plug in forces which go this way
as positive and forces which go the other way as negative.
What do I plug in up top? What forces make this go?
The force of gravity on this 9 kg mass is driving
this system, this is the force which makes the
whole system move if I were to just let go of these masses
it would start accelerating this way because of
this force of gravity right here.
So that's going to be 9 kg times
9.8 meters per second squared and that's going
to be positive because it's making the system go.
There's no other forces that make this system go.
So now I'm only going to subtract forces that resist
the acceleration, what forces resist the acceleration?
The gravity of this 4 kg mass resists acceleration,
but not all of the gravity.
The gravity of this 4 kg mass points straight
down, but it's only this component this way
which resists the motion of this system in this direction.
What is this component?
This is "m" "g" "sin(theta)" so if that doesn't
make any sense go back and look at the videos about inclines
or the article on inclines and you'll see the component
of gravity that points down an incline
parallel to the surface is equal
to "m" "g" "sin(theta)" so I'm gonna
have to subtract 4 kg times
4 kg times 9.8 which is "g"
times sin of the angle, which is 30 degrees.
We need more room up here because there are more
forces that try to prevent the system
from moving, there's one more force, the force of friction
is going to try to prevent this system
from moving and that force of friction
is gonna also point in this direction.
It's not equal to "m" "g" "sin(theta)"
it's equal to the force of kinetic friction
"mu" "k" times "Fn" and the "mu" "k"
is going to be 0.2, you have to be careful because
the "Fn" is not just equal to "m" "g"
the reason is that on an incline the normal force
points this way so the normal force doesn't
have to counteract all of gravity on an incline
it just has to counteract that component of gravity
that's directed perpendicular to the incline
and that happens to be "m" "g" "cos(theta)"
for an object on an incline and if that makes
no sense go back and look at the video on inclines
or look at the article on inclines and
you'll see that this component of gravity
pointing into the surface is "m" "g" cosine
that means that normal force is "m" "g" cosine.
Because there's no acceleration in this
perpendicular direction and I have to multiply
by 0.2 because I'm not really plugging in the
normal force up here or the force of gravity
in this perpendicular direction.
I'm plugging in the kinetic frictional force
this 0.2 turns this perpendicular force into
this parallel force, so I'm plugging in the
force of kinetic friction and it just
so happens that it depends on the normal force.
That's why I'm plugging that in,
I'm gonna need a negative 0.2 times 4 kg
times 9.8 meters per second squared.
And then I need to multiply by cosine of the angle
in this case the angle is 30 degrees.
Alright, now finally I divide by my
total mass because I have no other forces
trying to propel this system or to make it
stop and my total mass is going to be 13 kg.
You might object and think wait a minute,
there's other forces here like this tension
going this way, why don't we include that?
Well that's internal force and the whole
benefit and appeal of treating this
two-mass system as if it were a single mass
is that we don't have to worry about these
internal forces, it's there but that tension
is also over here and on this side it's
resisting the motion because it's pointing
opposite the directional motion.
On this side it's helping the motion,
it's an internal force the internal force
is canceled that's why we don't care about
them, that's what this trick allows us to do
by treating this two-mass system as a single
object we get to neglect any internal forces
because internal forces always cancel on that object.
So if we just solve this now and calculate,
we get 4.75 meters per second squared
is the acceleration of this system.
So this 4 kg mass will accelerate up the incline
parallel to it with an acceleration
of 4.75 meters per second squared.
This 9 kg mass will accelerate downward
with a magnitude of 4.75 meters per second squared.
Remember if you're going to then go
try to find out what one of these internal
forces are, we neglected them because
we treated this as a single mass.
But you could ask the question,
what is the size of this tension?
Often that's like a part two because
we might want to know what the tension
is in this problem, if we do that now
we can look at the 9 kg mass individually
so I can say for just the 9 kg mass alone,
what is the tension on it and what are the force?
We can find the forces on it simply by
saying the acceleration of the 9 kg mass
is the net force on the 9 kg mass divided
by the mass of the 9 kg mass.
Now this is just for the 9 kg mass since
I'm done treating this as a system.
This trick of treating this two-mass system
as a single object is just a way
to quickly get the magnitude of the acceleration.
Now that I have that and I want to find
an internal force I'm looking at just this 9 kg box.
And I can say that my acceleration is not 4.75
but -4.75 if we want to treat downwards as negative
and upwards as positive then I have to plug
this magnitude of acceleration in as
a negative acceleration since the 9 kg
mass is accelerating downward and that's
going to equal what forces are on the 9 kg mass:
I called downward negative so that
tension upwards is positive,
but minus the force of gravity on
the 9 kg mass which is 9 kg times
9.8 meters per second squared divided by 9 kg.
I don't divide by the whole mass,
because I'm done treating this system
as if it were a single mass and I'm now looking
at an individual mass only so we go back
to our old normal rules for newton's second law
where up is positive and down is negative
and I only look at forces on this 9 kg mass
I don't worry about any of these now because
they are not directly exerted on the 9 kg mass
and at this point I'm only looking at the 9 kg mass.
So if I solve this now I can solve for the tension
and the tension I get is 45.5 newtons
which is less than 9 times 9.8
it's got to be less because this object is
accelerating down so we know the net force
has to point down, that means this tension
has to be less than the force of gravity
on the 9 kg block.
So recapping, treating a system of masses
as if they were a single object is a great
way to quickly get the acceleration
of the masses in that system.
Once you find that acceleration
you can then find any internal force that you want
by using Newton's second law for an individual box.
You're done treating as a system and you just
look at the individual box alone
like we did here and that allows you to find
an internal force like the force of tension.