Main content

## Treating systems

Current time:0:00Total duration:15:27

# Treating systems (the hard way)

## Video transcript

- [Voiceover] Alright,
this problem is a classic, you're gonna see this in basically every single physics text book. And the problem is this,
if you've got two masses tied together by a rope, and
that rope passes over a pulley, what's the acceleration of the masses? In other words, what's the
acceleration of the 3 kg mass? And then what's the
acceleration of the 5 kg mass? And if you're wondering
what the heck is a pulley? The pulley is this part right here. This right here is the pulley. So what a pulley does, a pulley
is a little piece of plastic or metal that can rotate. And it's usually got a
groove in it so that a string or a rope can pass over it. What it does, is it rotates
freely so that you can turn what's a horizontal tension on one side, into a vertical tension on the other. Or vice versa. It turns vertical forces
into horizontal forces. It allows you to transfer
a force from one direction to another direction. So that's what these
pulleys are useful for. And if they can spin
freely, and if this pulley has basically no mass,
if there's no resistance to motion at all, then
this tension on this side is gonna be equal to the
tension on this side. This vertical tension gets
transferred fully undiluted, into a horizontal tension
and these tension values will just be the same if
this pulley can spin freely. And if its mass is really small so that there's no inertial reason
why it doesn't wanna spin. So that's the problem. Let's say you wanted to figure this out. What is the acceleration of the 3 kg mass with the acceleration of the 5 kg mass? Now, I've gotta warn you,
there's an easy way to do this, and a hard way to do this. Now, I'm gonna show
you the hard way first. Sorry, no one ever wants to hear that. But, the reason is that, the
easy way won't make any sense unless I show you the hard way first. It won't make any sense why
the easy way works unless I show you the hard way. And for two, the hard way
isn't really all that hard. So I'm calling it the hard way, but it's not really that bad. And for three, sometimes teachers
and professors just wanna see you do it the hard
way, so you should know how to do this. So what do we do? We wanna find acceleration,
well you know how to find acceleration. We're gonna use Newton's second law. So we'll say that the
acceleration in a given direction is gonna equal the net
force in that direction, divided by the mass. Now what do we do? What mass are we gonna choose? We've got a couple masses here. One thing we could do, let's
just pick the 5 kg mass. Just pick one of them. So I'm gonna say that the
acceleration of the 5 kg mass is the net force on the 5
kg mass, divided by the mass of the 5 kg mass. And remember, we should always
pick a direction as well. So do we wanna pick the vertical direction or the horizontal direction? Well, since this box is gonna
be accelerating horizontally, and that's what we're interested
in, I'm gonna put one more sub-script up here, X, to
let us know we're picking the horizontal direction. So I can fill this out
now, I can plug stuff in. The acceleration of the 5
kg mass in the X direction is gonna be equal to. Alright what forces do we
have to figure out what goes up here? You always draw a force diagram. So what forces do I have on the 5 kg mass? I'm gonna have a force
of gravity, I'll draw that straight down. FG, and there's gonna be an equal force, normal force upward. So this normal force up
should be equal to the force of gravity and magnitude
because this box is probably not gonna be accelerating vertically. There's no real reason why it should be if this table is rigid. And there's one more force on this box. There's a force to the right. That's gonna be the force of tension. And if there's no friction on this table, then I have no leftward forces here. I'm ignoring air resistance since we usually ignore air resistance. So that's is, the only horizontal force I've got is T, tension. And I divide by the mass of
the 5 kg box, which is 5 kg. But we got a problem. Look it, we don't know the
acceleration of the 5 kg mass, and we don't know the tension. I can't solve this. Normally what you do in
this case, is you go to the vertical direction, the
other direction in other words. But that's not gonna help me either. That's just gonna tell me
that the normal force is gonna be equal to the force of gravity. And we kind of already knew that. So that doesn't help. So what do we do? Well, you might note, this
is only the equation for the 5 kg mass. And so now I have to do
this for the 3 kg mass. So let's come over here, let's
say that the acceleration of the 3 kg mass is gonna
be equal to the net force on the 3 kg mass, divided by
the mass of the 3 kg mass. And again, which direction should we pick? Well this acceleration over
here is gonna be vertical. So let's solve this for
the vertical direction. I'm gonna add one more
sub-script, Y, to remind myself. And you should do this too so
you remember which direction you're picking. So what forces do I plug in here? You figure that out with a force diagram. I'm gonna have a force of
gravity on this 3 kg mass, and then I'm gonna have
the same size of friction, or sorry, the same as
tension, that I had over here. So the tension on this side
of the rope, it's gonna be the same as the tension on this side. Assuming this pulley offers no
resistance either by its mass or friction. So assuming that its mass is
negligible, there's basically no friction, then I'm
gonna have a tension. That tension is gonna be the same size. So I'll draw that coming upward. But it's not gonna be as big
as the force of gravity is on this 3 kg mass. I've got the force of gravity here. This tension is gonna be
smaller, and the reason is, this 3 kg mass is accelerating downwards. So these forces can't be balanced. The upward force of tension
has gotta be smaller than the force of gravity
on this 3 kg mass. But this tension here should be the same as this tension here. So I'll plug those in. So let's plug this in. A of the 3 kg mass, in the Y
direction is gonna be equal to, I've got two vertical forces. I've got tension up, so
I'll make that positive, 'cause we usually treat up as positive. I've got gravity down, and
so I'm gonna have negative, 'cause it's downward force of
3 kg times the acceleration on Earth, is 9.8 meters
per second squared. Now what do we do? We divide by 3 kg, 'cause that's the mass. But I've still got a problem. I don't know this
acceleration or this tension. So what do I do? You might notice, if you're
clever you'll say wait, I've got my unknown on
this side is acceleration and tension. My unknown on this side is
acceleration and tension. It seems like I've got two
equations, two unknowns, maybe we should combine them. And that's exactly how you do these. So I've got tension in
both of these equations. Let me solve for tension over here, where it's kind of simple. And I'll just get the tension equals 5 kg, times the acceleration of the
5 kg mass in the X direction. So now I know what tension is. Tension is equal to this. And that tension over on
this side is the same as the tension on this side. So I can take this and
I can plug it in for this tension right here. And let's see what we get. We get that the acceleration
of the 3 kg mass vertically, is gonna equal, alright, I'm
gonna have a big mess on top, what am I gonna get? I'm gonna get, so T is the same as 5AX. So I'll plug in 5 kg
times the acceleration of the 5 kg mass in the X direction. And then I get all of
this stuff over here. So I'll get the rest of this right here. I'll just bring that down right there. Alright, now what do I have? I've got 3 kg on the bottom
still, so I have to put that here. Are we any better off? Yeah, we're better, because
now my only unknowns are acceleration. But these are not the same acceleration. Look, this acceleration
here is the acceleration of the 3 kg mass, vertically. This acceleration here is the acceleration of the 5kg mass horizontally. Now here's where I'm gonna
have to make an argument, and some people don't like this. But, it's crucial to
figuring out this problem. And the key idea is this, if
this 3 kg mass moves down, let's say one meter, let's say
it moves downward one meter. Well then this 5 kg mass had
better move forward one meter. Because if it doesn't, then it
didn't provide the one meter of rope that this 3 kg
mass needed to go downward. Which means either the rope
broke, or the rope stretched. And we're gonna assume that
our rope does not break or stretch. That's kind of a lie. All ropes are gonna stretch
a little bit under tension. We're gonna assume that
stretch is negligible. So the argument is that if
this 3 kg mass moves downward a certain amount, this 5
kg mass has to move forward by that same amount in order
to feed that amount of rope for this 3 kg mass to go
downward by that amount. Otherwise, think about it. If this 5 kg mass just sat
here and the 3 kg moved, or the 3 kg moved farther
than the 5 kg mass, then this rope is stretching or breaking. So if you believe that,
if you don't believe it, pause it and think about it. 'Cause you've gotta
convince yourself of that. If you believe that then you
can also convince yourself that, well if the 3 kg
mass was moving downward at a certain speed, let's
say two meters per second. Then the 5 kg mass had
better also be moving forward two meters per second
because otherwise it wouldn't be feeding rope at a
rate that this 3 kg needs to move downward at that rate. And finally, if you believe all that, it's not too much harder
to convince yourself that this 3 kg mass, no matter what its acceleration downward must be, this 5 kg mass had better
have the same magnitude of acceleration forward
so that it's again, feeding the rope so
this rope doesn't break, or snap, or stretch. 'Cause we're gonna assume
the rope doesn't do that. So what I'm saying is that the
acceleration of the 3 kg mass in the Y direction had
better equal the magnitude. So these magnitudes have to be the same. The sign doesn't have to be the same. So this 3 kg mass has
a negative acceleration just 'cause it points
down, and we're assuming up is positive, down is negative. This 5 kg mass has a
positive acceleration 'cause it's pointing to the right,
and we're assuming rightward is the positive horizontal direction. So, they can have different
signs, but the magnitudes had better be the same so
that your feeding this rope at a rate that the other one
needs in order to move. And so we can say that the
magnitudes are the same. In this case, since one
is negative of the other, I can say that the
acceleration of the 3 kg mass vertically downward is gonna
be equal to, let's say negative of the acceleration of the 5
kg mass in the X direction. I could have written it the other way. I could have wrote that A of the 5 kg mass in the X direction is a
negative A of the 3 kg mass in the Y direction. They're just different
by a negative sign is all that's important here. Okay, so this is the link we need. This is it. So this allows us to put
this final equation here in terms of only one variable. 'Cause I know I've got A3Y
on this left hand side. I know A3Y should always be -A5X. If I take this and just plug
it in for A3Y right here, I'm gonna get -A5X =, well
all of this stuff, so I'll just copy this. Save some time. Copy, paste. Just equals all of that. All I did was plug in what I
know A3Y has to be equal to. 'Cause now look, I've got one
equation with one unknown. I just need to solve for what A5X is. It's on both sides. So I'll need to combine
these and then isolate it on one side. So there's gonna be a
little bit of algebra here. Let's just take this, let's
give ourselves some room. Move this up just a little bit. Okay, so what do we do? We're gonna solve for A5X. Let me just get rid of this denominator. Let me multiply both sides by 3 kg. So I'm gonna get -3kg x
A5 in the X direction, if I multiply both sides by 3 kg, and then I get 5 kg x
A5 in the X direction. And I've still got minus,
alright 3 x 9.8 is 29.4 Newtons. So we'll just turn this into
what it's supposed to be. 29.4 Newtons. So let's combine our A terms now. Let's move this negative
3A to the right hand side by adding it to both sides. And let's add this 29.4 to both sides. So I'll get the 29.4 Newtons
over here with a positive, if I add it to both sides. And it'll disappear on
the right hand side. And then I'll add this term to both sides. Add a positive 3 kg x A to both sides. It'll disappear on the
left, and I'll get 5 kg x A5 in the X direction + 3 kg
x A5 in the X direction. Now we're close, look on the
right hand side I can combine these terms because 5A plus
3A is the same thing as 8A. So 29.4 Newtons, .4 = 8 kg x,
I'll put the parenthesis here, times five. 85 in the X direction. Now I can divide both sides
by eight and usually we put the thing we're solving for
on the left, so I'm just gonna put that over here. I'll get 29.4 Newtons
over 8 kg is equal to the acceleration of mass five. The 5 kg mass in the X direction. And if we calculate that, I'll just put that into my calculator. 29.4 divided by eight. I get 3.675. So we'll just round. We'll just say that's 3.68. 3.6 whoops, 3.68 and it's
positive, that's good. We should get a positive
because the 5 kg mass has a positive acceleration. So we get positive 3.68
meters per second squared. But that's just of the 5 kg mass. How do we get the
acceleration of the 3 kg mass? Well that's easy. It's gotta have the same
magnitude of the 5 kg mass. All I have to do is take this number now. I know what A5X is. So if I just plug that in right here, well then I know that A3Y
is just gonna be equal to -3.68 meters per second squared. And I'm done, I did it. We figured out the
acceleration of the 3 kg mass, it's negative. No surprise, 'cause it's
accelerating downward. We figured out the
acceleration of the 5 kg mass, it's positive, not a surprise. It was accelerating to the right. The way we did it, recapping really quick, we did Newton's second
law for the 5 kg mass. That didn't let us solve. We did Newton's second
law for the 3 kg mass, that didn't let us solve. In fact it got really bleak,
because it seemed like we had three unknowns and only two equations, but the link that allowed us
to make it so that we only had one equation with one
unknown, is that we plugged one equation to the other first. We had to then write the
accelerations in terms of each other, that's
because these accelerations are not independent. The accelerations have to
have the same magnitude. And in this case one
had the opposite sign. So when we plugged that
in, we have one equation with one unknown, we
solve, we get the amount of acceleration. So that's the hard way
to do these problems. So in the next video I'll
show you the easy way to do these problems.