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# Three box system problem

In this video David explains how to easily find the acceleration of a three box system by treating it as a single mass. Created by David SantoPietro.

## Want to join the conversation?

- Wait, isn't the friction force an internal force ?, i thought internal forces were not included in the easy way strategy.(0 votes)
- The table is applying the force of friction and the table is external to the system of the three boxes.(41 votes)

- The force of friction is shown to go to the left but does it not also go to the right?

It is being pulled from both sides.(7 votes)- Friction opposes motion, thus it is to the left.(12 votes)

- What would happen if the total external forces on the system are negative due to friction?(7 votes)
- Nikolay's point on the reactive nature of friction to motion is right

but if your point of asking is about the situation in which a maximum friction happens to be large enough to offset the applied force (yes it is definitely possible and most non-moving bodies on a surface are doing exactly this), the answer is no motion at all as you may expect

for this specific case with the same masses of boxes, the smallest friction coefficient would be around 0.167 which makes the boxes not move. any bigger than this would require larger mass of the right box or smaller mass of the left one or the upper one on the table itself to make them move

hope this fill your curiosity a bit(1 vote)

- How do we find out the direction of the acceleration of the whole system?(3 votes)
- The direction of acceleration(if there is an acceleration ) of the system is the direction of the larger mass hanging.(3 votes)

- since there are two strings shouldn't there be two different accelerations?(1 vote)
- aren't the strings going to move together?(6 votes)

- Why do we have two different tensions?As we are using the same rope don't we have T on left side=T on right side.(3 votes)
- Two different ropes results in two different tensions. The tensions given by the weight of each mass hanging from each rope.(2 votes)

- Wouldn't the mass of the 12 kg object affect the acceleration of the system, and in turn, the force? Because, just like friction force, since the mass of the 2nd object is significantly larger than the other object, wouldn't it slow down or stop the acceleration and the force of the 5 kg object and basically the entire system?(3 votes)
- in fact, the reality is the opposite (friction could make the system faster)

1. if the friction is large enough to cancel out the net external force, the system simply don't move

2. if not, it lets them move as we saw in the example above

3. then, the friction applied to the system would change to friction of kinetic (sliding) than of static. and as we learned from previous videos, F_k <= F_s and in many cases you can say F_k < F_s. if that's the case of ours, the whole system would move faster toward left and downward than at the moment of starting to move

in short, friction plays a (kind of) role of a switch or a threshold. it turns on motion, if there's enough net external force applied. if not, it turns off and the system doesn't move. but once it starts to move, the friction could be smaller than when it began moving cause F_k <= F_s thus let the system move faster

one more thing, if your real concern is the effect of its sheer bigger mass of 12kg than 5kg of the right object, the friction itself is the effect. other than that? nope. that's why we need to only care the simple equation of Normal_force*Friction_coefficient to get an impact of this object and even the acceleration of the entire system(2 votes)

- Could we treat this example as a system even if there is no acceleration?(2 votes)
- Go on try it out yourself put
**a=0**in the equation and see what happens.(2 votes)

- Why is the horizontal frictional force counted as well? Is it because they are in a single system?(2 votes)
- The horizontal frictional force is an external force that affect the acceleration of the system.(2 votes)

- I can't crack how to get the tension from each side of the rope, does the 12kg box facto into this?

Because if I find the tension from the 3kg box its, 30.576N, and from the tension of the 5 box its 50.96N. I think I might be missing something.(2 votes)- Solving for the force of tension is just using the fact that net force = ma.

Finding the tension in the left rope as m(g + a), you get 30.576 N (which you got). The tension in the other rope should be m(g - a) = 47.04 N (you added instead of subtracting). What do you feel like you're missing?(2 votes)

## Video transcript

- So, check out this problem. You've got a 12 kilogram
mass sitting on a table, and on the left hand
side it's tied to a rope that passes over a pulley and that rope gets tied
to a three kilogram mass. And then on the right side
of this 12 kilogram box, you've got another rope and that rope passes over
another pulley on the right and is tied to the five
kilogram box over here. The question is, what's the acceleration of the 12 kilogram box? Let's make it even harder. Let's say there's a
coefficient of kinetic friction between this 12 kilogram
box and the table of 0.1. Now you're looking at
a really hard problem if you try to solve this the hard way. And by the hard way, I mean
using Newton's second law for each box individually
and then trying to solve what'd you end up with is at least three equations and three unknowns because you're gonna have
three different accelerations. For each of these, you'll
have two different tensions cause this left rope is under a different tension from the right rope now. It's only when you have
a single rope can you say that it's the same tension. This problem's gonna be hard. There's gonna be tons of
Algebra mistakes potentially. And so to avoid that, we
can solve this the easy way. And if you remember, the
easy way is just by saying, well, let's treat all of these boxes as if they're a single object. And we can do that cause
they're all gonna have the same magnitude of acceleration that I'm just calling a system. That's gonna be the magnitude
of acceleration of our system. All these boxes will accelerate
with the same magnitude. Some may have negative accelerations, some may have positive
accelerations like these, but they're gonna all have the same magnitude of acceleration
cause we're gonna assume that these ropes don't break. And if they broke, then
they'll be different magnitudes or if they stretch, but we're
assuming that doesn't happen. And the way we can find
this is by just saying, well, if this is just a single object, I don't have to worry
about any internal forces, now these tensions become internal forces. And those don't make a system accelerate, only external forces are gonna
make a system accelerate. So, all I have to do is find out what are all the external forces that try to make this system go, try to accelerate it, and ones that try to prevent acceleration. I'll call this F external and then I divide by the total mass because this is just
simply Newton's second law as if this were one big object. So, what are my external forces? Well, the force that makes it go is gonna be this five
kilogram's force of gravity so I'm gonna have a force
of gravity over here. That tries to propel the system forward. This is the one that's
gonna be driving the system. If I let go of these boxes, it's gonna start shifting
in this direction because this five kilogram mass has a larger force of gravity
than this three kilogram mass. So, I'm gonna include that as a positive. I'm just gonna define direction of motion as positive, 'cause it's easy. You could do it differently
if you wanted to. You could find the other way as positive. So five times 9.8 meters
per second squared is how you find this force of gravity. Are there any other forces that
are propelling this forward? No, no external forces are. So, are there any forces that are trying to reduce the acceleration? Yeah, there's this force
of gravity over here. This force of gravity on
the three kilogram mass is trying to prevent
the acceleration because it's pointing opposite
the direction of motion. The motion of this system
is upright and down across this direction but this force is pointing opposite that direction. This force of gravity right here. So, I'm gonna have to
subtract three kilograms times 9.8 meters per second squared. Am I gonna have any other forces that try to prevent the system from moving? You might think the force of gravity on this 12 kilogram box, but look, that doesn't
really, in and of itself, prevent the system from
moving or not moving. That's perpendicular to this direction. I've called the direction of motion, this positive direction. If it were a force this way, if it were a force this
way or a force that way it'd try to cause
acceleration of the system. This force of gravity just gets negated by the normal force, so I don't even have to worry about that force. So, are there any forces associated with the 12 kilogram box
that try to prevent motion? It turns out there is. There is going to be a force of friction between the table because there's this coefficient of kinetic friction. So, I've got a force this way, this kinetic frictional force, that's gonna be, have a size of Mu K times f n. That's how you find the normal force and so this is gonna be
minus, the Mu K is 0.1 and the normal force
will be the normal force for this 12 kilogram mass. So, I'll use 12 kilograms times 9.8 meters per second squared. You might object, you might say, "Hey, hold on, 12 times 9.8,
that's the force of gravity. "Why are you using this force? "I thought you said we didn't use it?" Well, we don't use this force by itself, but it turns out this force of friction depends on this force. So, we're really using a horizontal force, a force that tries to prevent motion, which is why we've got
this negative sign here, but it's a horizontal force. It just so happens that
this horizontal force depends on a vertical force,
which is the normal force. And so that's why we're
multiplying by this .1 that turns this vertical force, which is not propelling the system, or trying to stop it,
into a horizontal force which is trying to reduce the
acceleration of the system. That's why I subtracted and
then I divide by the total mass and my total mass is
gonna be three plus 12 plus five is gonna be 20 kilograms. Now, I can just solve. If I solve this, I'll
get that the acceleration of this system is gonna be
0.392 meters per second squared. So, this is a very fast way. Look it, this is basically a one-liner. If you could put this together right, it's a one-liner. There's much less chance for
error than when you're trying to solve three equations
with three unknowns. This is beautiful. When you apply this though, be careful. The acceleration of the
five kilogram mass would be negative 0.392 because
it's accelerating downward. The acceleration of the 12,
we'd call positive 0.392 because it's accelerating to the right and we typically call rightward
accelerations positive. And then the three kilogram
mass also would have positive .392 because
it's accelerating upward.