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## Physics library

# Two masses hanging from a pulley

In this video David explains how to find the acceleration of two masses hanging from a pulley (using the easy method). Created by David SantoPietro.

## Want to join the conversation?

- Is the tension for the 2 boxes is the same?(16 votes)
- Yes. The tension in the rope is equal and upwards on both the masses. This is how he is able to cancel the forces out in the video.(42 votes)

- How did he get 36.75 for tension? He used his calculator, but I wanted to see how he solved it because I got a different answer.(11 votes)
- a=F/m,

-2.45(m/s^2)={T-(gravity)}/5,

-2.45*5=T-mg,

T=-12.25+5*9.8=-12.25+49=36.75.

Therefore, T=36.75(N)

*T=tension, a=accelaration, F=force, m=mass, g=gravitational accelaration(g is approximately 9.8),

a=F/m came from newton's second law.

T-Gravity=F, because, T is towards + direction, and Gravity is towards the opposite direction(-)(14 votes)

- What would happen to the acceleration if the mass and radius of the pulley were taken into account? Such as if mass(pulley)=5kg and radius was 0.25m?(4 votes)
- If you assume the pulley has a non-negligible mass, then you have to factor in what is known as the "moment of inertia" of the pulley. The moment of inertia (symbol I (uppercase i)) is the rotational equivalent of regular inertia to motion. The formula for the m.o.i. of a pulley is 1/2mr^2, where m is the mass and r is the radius. So the m.o.i. of your pulley would be I=1/2*5kg*.25m^2=.156kg*m^2. The product of the m.o.i. and the angular velocity is going to equal the torque (rotational force) on the pulley. The tension in the rope, in turn, is equal to the torque on the pulley multiplied by the radius r. It's late and my answer is getting really drawn out but I only just learned about moment of inertia and its corresponding complications to these types of problems in my last semester, in my third year of university. There's a good sample problem here: http://farside.ph.utexas.edu/teaching/301/lectures/node112.html . For your reference the dot over a symbol means its derivative (this is calculus if you don't know); v dot is equal to accleration, and omega dot (looks like w, represents angular velocity) is equal to angular acceleration. I have no idea what year of school you're in, but as you can maybe see just taking the mass of a pulley into account makes everything significantly harder!(10 votes)

- 7:20, i miss one step how did he got 36.75 N what did he do with the aceleration?(5 votes)
- accelaration of the system is the same as accelaration of the object.

a=-2.45(m/s^2)

*accelaration is - because it is in - direction(downwards.)(1 vote)

- how did you solve for t on the end part? ive been trying to rearrange it but i keep getting a different number(4 votes)
- Kind of random, but, if you are free falling and your acceleration due to gravity is 9.8 m/s^2, and your velocity increases by 9.8 m/s because of this, but your acceleration stays the same. Why is it that we feel more force when we hit the ground from higher up, when it is only our velocity that changes, and not our acceleration?(2 votes)
- You speed increases by 9.8 m/s every second, that is why it is m/s^2. So after the first second you moving at 9.8 m/s and after the second you are going 19.6, 29.4 after 3 seconds, 39.2 m/s after 4 second and 49 m/s after 5 seconds. This is why you hit harder when you fall from further up.(4 votes)

- How would you solve this the "hard way"(2 votes)
- Two free body diagrams, two sets of equations of motions, solved simultaneously.(4 votes)

- Is the Tensions on 3kg and 5kg still the same if we apply a friction on the rope?(5 votes)
- No, because the 5kg block is heaver then the 3kg block, so the 3kg block would go up and the 5kg block will go down so it won't be the same.(0 votes)

- are the tensions the same for both boxes, 3kg and 5kg(1 vote)
- Absolutely the same. if you use the newton's second law in vertical direction of the 3kg mass you have to use the acceleration in the positive direction because it moves upward. so the acc = +2.45m/s^2 and then plug in the force of gravity of 3kg mass to find T and you will find that it has the same force as you find it on the 5kg mass by using acc = -2.45m/s^2.

I hope that help. :)(6 votes)

- if the pulley has a mass, when drawing the free body diagram will it have any other forces or just mg?(3 votes)

## Video transcript

- [Instructor] Let's solve some more of these systems problems. If you remember, there's
a hard way to do this, and an easy way to do this. The hard way is to solve
Newton's second law for each box individually,
and then combine them, and you get two equations
with two unknowns, you try your best to solve the algebra without losing any sins,
but let's be honest, it usually goes wrong. So, the easy way to do this, the way to get the magnitude
of the acceleration of the objects in your system, that is to say, if I
wanna know the magnitude at which this five
kilogram box accelerates, or that this three
kilogram box accelerates, all I need to do is take
the net external force that tries to make my system go, and then I divide by my
total mass of my system. This is a quick way to
get what the magnitude of the acceleration is of
the objects in my system, but it's good to note, it'll only work if the objects in your
system are required to move with the same magnitude of acceleration. And in this case they are, what I have here is a five kilogram mass tied to a rope, and that
rope passes over a pulley, pulls over and connects to
this three kilogram mass so that if this five kilogram mass has some acceleration downward,
this three kilogram mass has to be accelerating
upward at the same rate, otherwise this rope would
break or snap or stretch, and we're assuming that
that doesn't happen. So this rope is the
condition that requires the fact that this rope doesn't break is what allows us to say that the system is just a single, big total mass with external forces exerted on it. So how would we solve this? I'd just say that, well,
what are the external forces? Keep in mind, external forces
are forces that are exerted on the objects in our system from objects outside of our system. So one external force would
just be the force of gravity on this five kilogram mass. So I'm gonna have a force
of gravity this way, and that force of gravity
is just going to be equal to five kilograms times 9.8
meters per second squared, because that's how we
find the force of gravity. Should I make it positive or negative? Well, this five kilogram
is gonna be the one that's pulling downward,
so if the question is, I hold these masses and I let
go, what's the acceleration? This five kilogram mass is
gonna accelerate downward, it's gonna drive the system forward. That's the force making the system go, so I'm gonna make that a positive force. And then I figure out,
are there any other forces making this system go? No, there are not. You might say, well what
about this tension over here? Isn't the tension on
this three kilogram mass? Isn't that tension making this system go? Not really, because
that's an internal force exerted between the objects in our system and internal forces are always opposed by another internal force. This tension will be
pulling the three kilogram, trying to make it move,
but it opposes the motion of the five kilogram mass, and if we think of this
three plus five kilogram mass as a single object, these
end up just canceling on our single object that we're viewing as one big eight kilogram mass. So those are internal forces. We don't include them, they're
not part of this trick. We have to figure out what other forces would try to make this system go or try to prevent it from moving. Another force that tries
to prevent it from moving is the force of gravity on
the three kilogram mass. Or, one force that tries
to prevent the system from moving would be
this force of gravity. How big is that? That's three kilograms times
9.8 meters per second squared. And that's trying to prevent
the system from moving. This five kilogram mass
is accelerating downward, and this force is in the
opposite direction of motion. That trips people out sometimes. They're like, I don't understand, they're both pointing down. Shouldn't they have the same sign? They would when we're
using Newton's second law the way we usually use it, but when we're using this
trick, what we're concerned with are forces in the direction of motion, this is an easy way to figure it out, forces in the direction of
motion we're gonna call positive. And any forces opposite
the direction of motion we're gonna call negative. So, forces that propel the system forward we'll just call that positive direction. Forces that resist the motion, we're just gonna call that
the negative direction. And since this is on this side
of the motion of the system, this system is, everything in
this system is going this way. The three kilogram mass goes up. The string over here goes up. The string up here goes to the right. The string right here goes down. The five kilogram mass goes down. Because all the motion in
the system is this way, we'd find that way's positive, but this force of gravity
on the three kilogram mass is the opposite direction. It's opposing the motion of the system. It's preventing the
system from accelerating as fast as it would have. That's why we subtract it. And now we just divide by the total mass. And the total mass is
just five plus three, is gonna be eight kilograms, and I get the acceleration of my system. So if I just add this up, I get 2.45 meters per second squared. So this is a really fast way
to get what the acceleration of our system is, but
you have to be careful. If the question is,
what's the acceleration of a five kilogram box? Well, technically, that acceleration of the five kilogram box
would be negative 2.45. What we really found here, since we were just finding the magnitude, was the size of the acceleration, since this five kilogram
box is accelerating down, and we usually treat down as negative. You won't wanna forget that negative in putting in that answer the acceleration of the three kilogram box, however, would be positive 2.45
meters per second squared. So when you're applying
this to an individual box, you have to be very careful and make sure that you
apply that acceleration with the correct sign
for that particular box. And if you wanted to find the tension now, now it's easy to find the tension. I could find this tension
right here if I wanted to. If the next step was find
the tension in the string connected to the boxes, now I can just use Newton's second law, but the way we always use it. I'm done with the trick. The trick is just the way to get the magnitude of the acceleration. Now that I have that, I'm
done treating it as a system or a single object. I'll look at this single
five kilogram mass all alone, and I'll say that the acceleration of the five kilogram mass,
which is Newton's second law, is gonna equal the net force
on the five kilogram mass divided by the mass of
the five kilogram mass. I know the acceleration
of the five kilogram mass, but if I'm gonna treat up as positive now, I gotta plug this acceleration
in with a negative sign. So negative 2.45 meters per second squared is gonna equal the net force
on the five kilogram mass. I've got tension up, you might be like, wait, we
said that was an internal force. It was an internal force, and
we didn't include it up here, but we're doing the old rules now. Normal second law in
the vertical direction. So I use vertical forces,
and if they're upward I'm gonna treat them as positive, and if they're downward
like this five times 9.8, I'm gonna treat it as a negative, because it points down. Five times 9.8 meters per second squared, and I divide by the five kilogram mass, 'cause that's the box I'm analyzing. I'm not analyzing the whole system. I'm just analyzing the
five kilogram box now. And I can solve and I can get my tension. The alternate way to do this would be to say, all right, let's
just treat down as positive for this five kilogram mass. I'd then plug my
acceleration in as positive, and I'd plug my force
of gravity in positive, then my tension would be negative. I'd get the same value. Here I'm just solving for the magnitude of the tension anyway. So if I solve this, if I
plug this into the calculator and solve for tension, I'm
gonna get 36.75 Newtons, which is less than the force of gravity, which it has to be, 'cause if
the tension was greater than the force of gravity,
this five kilogram mass would accelerate up. We know that doesn't happen. The tension's gotta be less
than the force of gravity, so that this five kilogram
mass can accelerate downward. So that's a quick way to
solve for the magnitude of the acceleration of
the system by treating it as a single object. We're saying that if it's a single object, or thought of as a single object, which we can do, 'cause these are required to have the same acceleration, or same magnitude of the acceleration, that if we're treating
it like a single object, only external forces matter, and those external forces
that make the system go are going to accelerate the system. And those external forces
that resist the motion are trying to reduce the acceleration, and we divide by the
total mass of the system that we're treating as one
object, we get the acceleration. If that still seems like
mathematical witchcraft, or if you're not sure
about this whole idea, I encourage you to go
back and watch the video. We solved one of these types
of problems the hard way. And you see, you really do end up with the force that tries to make
the system go externally, and the external force
that tries to stop it divided by the total mass
gives you the acceleration. Essentially, what we're saying is that these internal forces cancel if you're thinking of this
system as one single object, 'cause these are applied internally, and they're opposed to each other. One tries to make the system go, one tries to make the system stop.