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### Course: Class 11 Physics (India) > Unit 9

Lesson 11: Treating systems# Treating systems (the easy way)

David shows the easier way to find the acceleration of two masses connected by a rope. Created by David SantoPietro.

## Want to join the conversation?

- Why do we consider only the forces acting on the system which are external because even the internal forces of the system affect the motion?(8 votes)
- the internal forces can effect rotation but not the linear motion.

if you move around inside a space ship, it will continue in the same direction, but you can make it rotate(35 votes)

- Shouldn't the Fg be a negative force as well because it pulls downward?(12 votes)
- now in this method the convention changes the forces responsible for accelerating the system are taken as positive(16 votes)

- whats the tension of the rope(17 votes)
- The next video solves for tension, so definitely don't miss that. For this one, we just use Newton's second law for the 5 kg or the 3kg system, your choice. In other words:

(for 5 kg system)

F = ma

T = (5)(3g/8)

=> T = 15g/8

(for 3 kg system)

T - mg = -ma (don't forget minus sign!)

T = -ma + mg

T = m(g - a)

T = (3)(g - 3g/8)

= (3)(5g/8)

= 15g/8

=>T = 15g/8

As you can see, you can do it both ways. Cheers.(5 votes)

- In the problem described in this video and the previous, why doesn't the pulley contribute to the tension of the rope? Wouldn't the normal force of the rope on the pulley effect S(igma)Forces?(3 votes)
- In real life, the pulley would exert some force (e.g. friction). However, we are told in the beginning that this is a weightless and frictionless pulley - therefore we ignore it. We also ignore other forces such as air resistance.(10 votes)

- At the end of the video he finds the acceleration to be 1.86 or something and in the previous video, it was 3.68 why is that although the two diagrams are exactly the same?(2 votes)
- That's because in this video, he added the force of friction to the system, whereas in the previous video, he assumed that friction was negligible. Hope that helps(10 votes)

- What if I draw another block on the left side and we assume that the middle block has friction. Can I use this method too? Is the force of tension the same? How can i deal with the force of friction?(4 votes)
- the big part that I know is that the tension on the left side is the same through out that side but the right side could have a completely different tension or it could be the same it depends on the mass of the new block(1 vote)

- so when i say m/s squared some people tend to think meters per second per second so that is why i am confused is meters/s squared another way of saying meters per second per second?(2 votes)
- Yes, because in algebra (m/s)/s is m/s^2(2 votes)

- A bird is sitting in a large closed cage which is placed on a spring balance. The spring balance records a weight of 25 N. The bird (mass = 0.5kg) flies upward in the cage

with an acceleration of . The spring balance will now record a weight of

(a) 24 N (b) 25 N (c) 26 N (d) 27 N

Although this question isn't related to tension, it definitely has something to do with systems, internal forces and external forces. The correct answer given is (B) but if the bird isn't sitting on the bottom of the cage anymore, shouldn't the weight recorded by the spring balance reduce?(3 votes)- well, imagine the situation like that: in order for the bird to leave the bottom of the cage, ita has to fly, so it will move its wings, which will produce an air current that will push the cage down, so that the force measured by the spring wouldnt change. Although slightly "unphysical", I believe my explanation illustrates the situation reasonably. Hope this helps !(1 vote)

- At1:18, they say that the 29.4 N must be equal to the force of gravity on the 3kg. However, how is that possible if there was tension pulling it upwards? Thanks!(3 votes)
- Are we combining the horizontal and vertical direction in the numerator? For example, he subtracted friction(horizontal force) from the numerator even though there was a vertical force(gravity) (at8:19). Do you not have to add them vectorially?(2 votes)
- Well, instead of thinking of the forces as "vertical" and "horizontal," you have to think of them as forces that either "resist movement" or "make movement." Also, you must add the forces that make movement and subtract the forces that resist movement to calculate the net force on the entire system. In this case, friction was the force that resisted movement, whereas gravity was the force that made the system move. Does the help?(3 votes)

## Video transcript

- [Voiceover] So in the previous video we solved this problem the hard way. Maybe you watched it, maybe you didn't, maybe you just skipped right
to here and you're like, "I don't even wanna know the hard way. "Just show me the easy way please." Well, that's what we're
gonna talk about now. Turns out there's a trick and the trick is after you
solve this problem the hard way with a five kilogram mass
and a three kilogram mass, when you find the acceleration, what you get is this. That the acceleration of
the five kilogram mass is just 29.4 divided by eight kilograms. But when you do enough of these,
you might start realizing, "Wait a minute. 29.4 Newtons. "That was just the force of gravity "pulling on this three kilogram mass." In other words, the only force that was really propelling this whole entire system forward. Or at least the only external
force propelling it forward. And then eight kilograms down here. You're gonna be like,
"Wait. Eight kilograms? "That's just five kilograms
plus three kilograms. "Is that just a coincidence "or is this telling us
something deep and fundamental?" And it's not a coincidence. Turns out you'll always get this. That what you'll end up with
after solving this hard way, you'll get in the very end, you'll get all the external
forces added up here where forces that make it go like this force of gravity
end up being positive and forces that try to resist the motion. So if there was friction, that
would be an external force that tries to resist
motion, would be up top and then you get the
total mass on the bottom. And this makes sense. The acceleration of this entire system, if we think about it as a single object--- So if you imagine this
was just one single object and you asked yourself, "What's the total acceleration
of this entire system?" Well, it's only gonna depend
on the external forces and in this case, the only
external force making it go was this force of gravity right here. You might object. You might be like, "Wait. "What about this tension right here? "Isn't the tension pulling
on this five kilogram mass "making this system go?" It is but since it's
an internal force now, if we're treating this entire
system as our one object, since this tension is
trying to make it go, you've got another tension over here resisting the motion on this mass, trying to make it stop. That's what internal forces do. There's always equal and opposite on one part of the object than the other so you can't propel yourself forward with an internal force. So these end up cancelling
out essentially. The only force you have in this case was the force of gravity on top, only external forces, and the total mass on the bottom. And that's trick. That's the trick to quickly find the acceleration of some system that might be complicated if you had to do it in multiple equations and multiple unknowns but much, much easier
once you realize this. So the trick, sometimes it's called just "Treating systems
as a single object". Let me just show you really quick. If that made no sense, let me
just show you what this means. If we just get rid of this. So what I'm claiming is this. If you ever have a system where multiple objects
are required to move with the exact same magnitude
of acceleration, right? Because maybe they're
tied together by rope or maybe they're pushing on each other. Maybe there's many boxes in a row and these boxes all have to be pushed at the same acceleration because they can't get
pushed through each other. Right, if there's some
condition where multiple objects must have the same
magnitude of acceleration, then you can simply find the
acceleration of that system as if it were a single object. I'm writing, "SYS" for system. By just using Newton's second law, but instead of looking
at an individual object for a given direction, we're just gonna do all
of the external forces, all of the external forces on our system, treat it as if it were a single object, divide it by the total mass of our system. And so when you plug in
these external forces--- These are forces that are external so external means not
internal to the system. So if I think of this five kilogram box and this three kilogram
box as a single mass, tension would be an internal force because it's applied internally
between these two objects, between objects inside of our system. But the force of gravity
on the three kilogram mass, that's an external force
'cause that's the Earth pulling down on the three kilogram mass and the Earth is not part of our system. Similarly, the normal
force is an external force but it's exactly cancelled
by the gravitational force. So even though those are external, they're not gonna make it in here. I mean, you can put 'em in there but they're just gonna cancel anyway. We only look at forces in
the direction of motion and if it's a force that causes motion, we're gonna make that a positive force. If it's a force in the direction of motion like this force of gravity is, we make those positive forces. So forces will be plugged
in positive into here if they make the system go. And that might seem weird. You might be like, "Wait. "How do I decide if it
makes the system go?" Well, just ask yourself,
"Is that force directed "in the same direction as
the motion of the system?" So, we're just saying the
system is gonna accelerate if there's forces that make it go and we're gonna plug in negative forces, the forces that make the system stop or resist the motion of system. So maybe I should say,
"Resist motion of the system." In this case, for this one down here, I don't have any of those. So resist motion of the system. I don't have any of those. I could have if I had a force of friction. Then there'd be a external
force that resist the motion. I would plug in that
external force as a negative 'cause it resist the motion. So even though this might sound weird, it makes sense if you think about it. The acceleration of our system
treated as a single object is only gonna depend on the forces that try to make the system go and the forces that try
to make the system stop or resist the motion. So if we add those accordingly
with positives and negatives, we divide it by the total mass which gives the total measure
of the inertia of our system, we'll get the acceleration of our system. It makes sense and it works. Turns out it always works and it saves a ridiculous amount of time. For instance, if we
wanted to do this problem, if you just gave me this
problem straight away and you were told, "Do
this however you want.", I would use this trick. And I would say that the
acceleration of this system which is composed of
this five kilogram mass and our three kilogram mass is just gonna be equal to--- I'd ask myself, "What
force makes this system go? "What force drives this system?" And it's this force of gravity
on the three kilogram mass that's driving this system, right? If you took this force away, if you eliminated that force, nothing's gonna happen here. This is the force making the system go so I'd put it in my three
kilograms times nine point eight. And at this point, you might be like, "Well,
okay, that gravity made it go. "Should I include this gravity, too?" But no, that gravity is
perpendicular to the motion for one so this gravity isn't making the system, that's just causing this
mass to sit on the table and for two, it's cancelled
by that normal force. So those cancel anyway, even though they're external forces. This is it, this is the only
one that drives the system. So I put that in here and
I divide by my total mass 'cause that tells me how
much my system resists through inertia, changes in velocity,
and this is what I get. I get the same thing I got before, I get back my three point six eight meters per second squared,
and I get in one line. I mean, this trick is
amazing and it works, and it works in every example where two masses or more masses are forced to move with
the same acceleration. So this is great. This'll
save you a ton of time. This is supposed to be a three here. And to show you how useful it is, let say there was friction, let's say there was a
coefficient of friction of zero point three. Well, now I'd have a frictional force so there'd be an external
frictional force here. It'd be applied this five kilogram mass. I'd have to subtract it up here. So if I get rid of this--- So it's not gonna be three
point six eight anymore. I'm gonna have a force of
friction that I have to subtract. So minus mu K so the force of friction--- I'll just put force of friction. And so to solve for the force of friction, the force of friction
is gonna be equal to--- Well, I know three times
nine point eight is--- Let me just write this in here, 29.4 Newtons minus the force
of friction it's given by. So there's a formula
for force of friction. The force of friction is always mu K FN. So the force of friction
on this five kilogram mass is gonna be mu K which is point three. So it's gonna be zero point
three times the normal force, not the normal force on our entire system. I don't include this three kilogram mass. It's only the normal force
on this five kilogram mass that's contributing to this
force of friction here. So even though we're treating
this system as a whole, we still have to find individual forces on this individual boxes correctly. So it won't be the entire
mass that goes here. The normal force on the five kilogram mass is just gonna be five kilograms
times nine point eight meters per second squared. I divide by my total mass down here because the entire mass is
resisting motion through inertia. And if I solve this from my
acceleration of the system, I get one point eight four
meters per seconds squared. So this is less, less than
our three point six eight and that makes sense. Now, there's a resistive force, a resistive external force, tryna prevent the system from moving. But you have to be careful. What I'm really finding here, I'm really finding the
magnitude of the acceleration. This is just giving me the magnitude. If I'm playing this game where positive forces
are ones that make it go and negative forces are
ones that resist motion, external forces that is, I'm just getting the
magnitude of the acceleration. Individual boxes will have that magnitude of the acceleration but they may have positive
or negative accelerations. In other words, this five kilogram mass
accelerating to the right, gonna have a positive acceleration. In other words, the acceleration
of the five kilogram mass will be positive one point eight four and the acceleration of
the three kilogram mass since it's accelerating downward will be negative one point eight four meters per seconds squared.