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## Treating systems

Current time:0:00Total duration:9:58

# Masses on incline system problem

## Video transcript

- 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.