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# Treating systems (the hard way)

Created by David SantoPietro.

## Want to join the conversation?

- At0:16how can you take the acceleration of the (3kg) body to be downwards?? (5kg) has greater mass than (3kg) so there would not be any acceleration in the downward direction the masses will just hang in there without any movement!(21 votes)
- While that might seem to be the case, that is actually incorrect for this situation. The force of gravity is pulling down on the 3kg block. That force is then transferred through the rope and is applied to the 5kg block. This force is acting upon the 5kg block in the horizontal direction, and since there is no friction on the table, it is the only force acting on the 5kg block in the horizontal direction. This will cause there to be a net force on the 5kg block in the horizontal direction, which will cause it to accelerate to the right.(61 votes)

- I dint get why the acelerations have oposite signs in10:30.

Vectors with oposite signs means oposite directions and thats not the case(10 votes)- This comes about as a result of 2 things:

1. The way he has defined his coordinate system; and

2. The way pulleys work

First of all, he has defined motion in the upwards direction to be positive (thus downwards motion is negative), and motion in the rightwards direction to also be positive (thus leftwards motion is negative). This is pretty standard convention when teaching and learning elementary physics, but really, as long as you're consistent, you can choose which direction is positive and negative.

Secondly, there's the way pulleys work. As you can see from the way the picture is drawn, if the mass on the table (the 5kg mass) was to move leftwards, it would actually pull the 3kg mass up. Likewise, if the 5kg mass were moved to the right, the 3kg mass would have to move down. Remember from point 1 (above) that upwards and rightwards motion is positive, whilst leftwards and downwards motion is negative. So to rephrase what I just wrote, if the 5kg mass were moved to the right (positive), the 3kg mass would have to move down (negative). To reiterate what was already said in the video, the pulley will translate horizontal motion into vertical, and vice versa.(16 votes)

- since there is no friction nor air ressistence the mass on the table could be 10000kg and the acceleration would still be the same, right?

And why did we assume that we had an acceleration?(9 votes)- If you have something floating in space, the fact that it's floating in space doesn't mean you need no force to move it, only that the force you need to move it is less than in circumstances when you have other forces interfering. In this case the force of friction is negligible but there is still a tension between the two boxes that is pulling the 3kg box upwards. If i'm not mistaken, the magnitude of the force of tension depends on the masses of both boxes and, if you think about it, if the box on the table was say 100 grams, it would accelerate a lot faster. It makes sense that the opposite is also true. Anyway, you can calculate for a box of 10000kg if you want with the ecuation he used almost at the end: 29.4N=(5kg)(a5x)+(3kg)(a5x) you replace the 5kg with 1000kg and you get a5x=29.4/1003=0.03m*s^-2 So you can see that, no matterwhat the mass of the box on the table is, because it has no friction it will allways accelerate, but not allways at the same rate: as it becomes bigger the acceleration will aproach 0 but never reach it. That's why we asume that there is an acceleration.(12 votes)

- Should not the tension equal the weight of the 3kg since there is no friction?

(The 5kg should not even be able to hold the 3kg since no force of friction or any force is in the -x directions of the 5kg. So there is no tension force pulling up the 3kg. Right?)

Also there should not be any upward tension for the 3kg box, (since nothing is pulling it up) right?(14 votes)- If we assume T up at 3kg box is 30N, so a=0 for the 3kg box.

So T pulling the 5 kg box will be 30N, thus a= F/m=6ms-2

The magnitude of acceleration at two boxes is different! If this happens, tension will not retain, so in this situation, this can't happen!

From the video, the assumption made is that the magnitude of acceleration of two boxes are the same so tension will keep constant(3 votes)

- Shouldn't the acceleration just be 9.81m/s since there is no friction and thus there is no force that is acting against the weight? I thought tension would not exist in this case because the 5kg block is not creating any force against the rope?(5 votes)
- You are forgetting about inertia. To move the 5 kg block, the rope has to pull on it, even in the absence of friction. Remember how Newton's first law tells us that a body at rest will remain at rest as long as there is no net force acting on it? Also, the tension results because there is another block of mass 3 kg attached at the other end of the rope and pulling on it.

As the rope pulls on the 5 kg block, the block pulls back on the rope with an equal amount of force and in the opposite direction. Similarly, the 3 kg block is being pulled up by the rope while gravity is pulling it down. So the 3 kg block is*not*in free fall and therefore it's acceleration will be less than 9.81 m/s^2.

If there were friction, the acceleration would have been even smaller!(15 votes)

- Shouldn't the 5kg mass not move as it is heavier than the other mass (3kg)?

Please tell me!(4 votes)- Since there is no friction between the table and the 5kg mass, it will move wherever a force pushes or pulls it. The acceleration on the 3kg mass accelerates the 3kg mass down, which pulls the rope, which pulls on the 5kg mass (to the right).(10 votes)

- wait, wouldn't it be incorrect to assume that the acceleration of the two objects are equal like David did at10:50, because those two objects have different masses, and thus it cant be the same acceleration!(3 votes)
- The two masses are bound together by the string, which means that the 3kg box will pull the 5kg box, and the 5kg box will slow the 3kg box so that they both accelerate the same. The weight force on the 3kg box provides the acceleration for the total system.(6 votes)

- Would't the forces be different because of different mass? I.e he says that T is the same for both the 5kg box and the 3kg box.

But with both boxes having the same Acceleration= 3,68m/s^2. The forces will be different by Newtons 2. law.

F5x = m5* a5x -> F5x = 5* 3,68 = 18,4N.

F3y = m3* a3y -> F3y = 3* 3,68 = 11N

So how can T- tension be the same when the forces exerted is different because of mass!?(2 votes)- This same thing is true if you pull 2 objects of different masses in a straight line. How would the force of tension be multiplied along a straight piece of string?

It helps if you see the next video and understand both objects as part of one system:

Force of tension on 5kg object(and 3 kg object as we will see):

5*3.675=18.375newtons

Force on both objects due to gravity and gravity acting through tension in the rope:

3*9.8=29.4newtons

Tension on the 3kg object

29.4-(3*3.675)=18.375newtons of tension

The total force available for acceleration is only 3kg*gravity or 29.4N. Because acceleration is only 3.675m/s^2 the remaining force is being absorbed by tension. Otherwise we would see the 3kg have a larger acceleration and deliver more force on impact with the ground.(6 votes)

- why can't we just do A5x=A3y?(2 votes)
- a3y is in the DOWNWARD direction, and since acceleration is a vector, its important to count the direction. usually we call left and down negative. If down was positive, than your answer for a3y would be positive, and that would work aswell.(2 votes)

- At5:55, I wonder why he uses minus instead of plus, since the Fg is negative, and he will in fact adds T and Fg together if he uses minus. In my opinion, it should be T + Fg. Am I right?(2 votes)
- Remember, ropes can't push so that tension can only pull on an object. That means rope affects the object upward and we assume it as the positive direction. On the other hand, Fg is downward toward the center of the Earth and we assume it as the negative direction. These two forces affect opposite directions. We must add these two magnitudes vectorially. This means F=T+(-Fg)=T-Fg(2 votes)

## 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.