# Pulley Problem: Keeping boxes stationary on another box

• Jazz
In summary: I'll have to push on ##M## with a force such that its acceleration is the same to that of ##m_2##.And I'll need to push two masses: ##M## and ##m_1##, so ##F = (M+m_1)\frac{m_1g}{m_2}##.If ##m_2## accelerates, then the force on this box is ##m_1g## and its acceleration is ##\frac{m_1g}{m_2}##.OK, but what is your reasoning behind this?I took ##m_1## out of
Jazz

##F_{net} = ma##

## The Attempt at a Solution

If ##m_2## accelerates, then the force on this box is ##m_1g## and its acceleration is ##\frac{m_1g}{m_2}##. If I disregard the presence of ##m_1## for a minute, I'll have to push on ##M## with a force such that its acceleration is the same to that of ##m_2##. And I'll need to push two masses: ##M## and ##m_1##, so ##F = (M+m_1)\frac{m_1g}{m_2}##.

I'm not sure whether this conclusion is correct. ##m_1## has a downward acceleration and a horizontal force ##F## cannot prevent it from falling; unless there is friction between ##M## and ##m_1##. On the other hand, if a push on ##M##, by Newton's First Law ##m_2## will attempt to remain at rest, which is the same as exerting a force ##F## on ##m_2## in the opposite direction and ##M## remaining at rest. Since I want the smaller boxes to remain stationary, my problem is reduced to one where I need to counteract the force making them move, hence ##F## would have to be ##m_1g##.

What is the correct approach?

Jazz said:
If ##m_2## accelerates, then the force on this box is ##m_1g## and its acceleration is ##\frac{m_1g}{m_2}##.
OK, but what is your reasoning behind this?

Jazz said:
If I disregard the presence of ##m_1## for a minute, I'll have to push on ##M## with a force such that its acceleration is the same to that of ##m_2##. And I'll need to push two masses: ##M## and ##m_1##, so ##F = (M+m_1)\frac{m_1g}{m_2}##.
Why do you only have to push two masses?

Jazz said:
I'm not sure whether this conclusion is correct. ##m_1## has a downward acceleration and a horizontal force ##F## cannot prevent it from falling; unless there is friction between ##M## and ##m_1##.
There's a rope attached to ##m_1##.

Jazz said:
On the other hand, if a push on ##M##, by Newton's First Law ##m_2## will attempt to remain at rest, which is the same as exerting a force ##F## on ##m_2## in the opposite direction and ##M## remaining at rest.
I don't understand that statement.

Jazz said:
What is the correct approach?

Doc Al said:
OK, but what is your reasoning behind this?

The force making the smaller boxes move is ##m_1g##. Mmm, I think both boxes will move with the same acceleration, so it should be ##\frac{m_1g}{m_1 + m_2}##.
Doc Al said:
I don't understand that statement.

I took ##m_1## out of the picture and imagined just ##M## and ##m_2##. If I push ##M## with a force ##F##, ##m_2## will remain at rest and will eventually fall to the ground. It's the same as pushing on ##m_2## with a force ##F## in the opposite direction making it fall from the top of ##M##. If I go back to the original problem, ##m_1g## is the force I need to cancel out. I could do it by pushing on ##m_2## with a force of ##m_1g## to the left, or by pushing on ##M## with the same magnitude and the opposite direction (to the right). At least, that was my reasoning there.

Doc Al said:
Why do you only have to push two masses?

There's a rope attached to ##m_1##.

Well, I see that my system is the three masses together.

If I'm not wrong again, the acceleration I need to provide to my system is ##\frac{m_1g}{m_1 + m_2}##. The force would have to be:

##(M + m_1 + m_2)\frac{m_1g}{m_1 + m_2}##

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Jazz said:
If I'm not wrong again, the acceleration I need to provide to my system is ##\frac{m_1g}{m_1 + m_2}##. The force would have to be:

##(M + m_1 + m_2)\frac{m_1g}{m_1 + m_2}##

If that's the acceleration, then what is the tension in the string?

Jazz said:
If I'm not wrong again, the acceleration I need to provide to my system is ##\frac{m_1g}{m_1 + m_2}##.
You had the acceleration right the first time. But I want to know your reasoning.

As PeroK suggested, start with ##m_1##. What forces must act on it so that it doesn't fall? All else comes from that.

PeroK said:
If that's the acceleration, then what is the tension in the string?

If ##\frac{m_1g}{m_1+m_2}## is the acceleration, then the tension in the string would be ##\small{T = m_1(g - \frac{m_1g}{m_1+m_2})}## or ##\small{T = m_2\frac{m_1g}{m_1+m_2}}##.

Doc Al said:
You had the acceleration right the first time. But I want to know your reasoning.

As PeroK suggested, start with ##m_1##. What forces must act on it so that it doesn't fall? All else comes from that.

Let ##a## be the latter acceleration I wrote.

I went through the reasoning behind it and I came to the conclusion that my first acceleration wasn't right. In that case I just considered a force ##m_1g## acting on ##m_2## alone, which gives a different acceleration than that on ##m_1##, which in turn would mean that one moves faster than the other, in a given time. The string prevents that from happening, isn't it? I thought it's more like putting ##m_1## right behind ##m_2## and then pushing both to the right with a force magnitude ##m_1g##. So the acceleration would be ##a##.

My new attempt:

If I think on ##m_1##, it will move downward with an acceleration ##a##. A force ##m_1a## that needs to be counteracted. One can prevent ##m_1## from falling by placing a hand underneath of it or by exerting that force either along the string (opposite to its motion) or on ##m_2## itself. So I would solve it in this way:

- If I think just in the pulley system, I would say that a force ##m_1a## to the left (on ##m_2##) is enough to prevent ##m_1## from falling. In this case, the acceleration of ##m_2## would be ##\frac{m_1a}{m_2}##.
- I can't push on ##m_2## to the left but on ##M## to the right instead.
- If I'm able to supply to my whole system an acceleration ##\frac{m_1a}{m_2}## to the right, by Newton's First Law ##m_2## will try to stay motionless relative to my frame of reference (knowing that the surfaces are frictionless), which is the same as accelerating ##m_2## at ##\frac{m_1a}{m_2}## to the left in the ##M##'s frame of reference. That's is just what I needed. Then, the force required is ##(M + m_1 + m_2)\frac{m_1a}{m_2}##.

Last edited:
Jazz said:
If ##\frac{m_1g}{m_1+m_2}## is the acceleration, then the tension in the string would be ##\small{T = m_1(g - \frac{m_1g}{m_1+m_2})}## or ##\small{T = m_2\frac{m_1g}{m_1+m_2}}##.
Two problems. One, do assume any value for the acceleration. Two, why would a horizontal acceleration affect the tension in the string?

Jazz said:
My new attempt:

If I think on ##m_1##, it will move downward with an acceleration ##a##.
Since we want ##m_1## not to move downward, what must be its vertical acceleration?

Jazz said:
A force ##m_1a## that needs to be counteracted. One can prevent ##m_1## from falling by placing a hand underneath of it or by exerting that force either along the string (opposite to its motion) or on ##m_2## itself. So I would solve it in this way:

- If I think just in the pulley system, I would say that a force ##m_1a## to the left (on ##m_2##) is enough to prevent ##m_1## from falling. In this case, the acceleration of ##m_2## would be ##\frac{m_1a}{m_2}##.
- I can't push on ##m_2## to the left but on ##M## to the right instead.
- If I'm able to supply to my whole system an acceleration ##\frac{m_1a}{m_2}## to the right, by Newton's First Law ##m_2## will try to stay motionless relative to my frame of reference (knowing that the surfaces are frictionless), which is the same as accelerating ##m_2## at ##\frac{m_1a}{m_2}## to the left in the ##M##'s frame of reference. That's is just what I needed. Then, the force required is ##(M + m_1 + m_2)\frac{m_1a}{m_2}##.
You're making things way too complicated, which hides what's going on.

Answer these questions, one by one: What is the vertical acceleration of ##m_1##? What is the net vertical force on ##m_1##? What is the tension in the string? What is the acceleration of ##m_2##?

Doc Al said:
Two problems. One, do assume any value for the acceleration.

I did not understand /:

Doc Al said:
Two, why would a horizontal acceleration affect the tension in the string?

I can affect the tension in the string by applying an acceleration in the vertical section of it or in the horizontal one. I'm asked to find a horizontal force ##F##, so I need to apply a horizontal acceleration that allow me to counteract the ##m_1##'s motion. I'm relying in the fact that ##m_2## can slip, so I've been trying to argue that an acceleration on ##m_2## to the left or and acceleration on ##M## to the right will increase the tension in the string and prevent ##m_1## from falling. The opposite happen if an acceleration on ##m_2## to the right or on ##M## to the left is applied.

Is this assumption wrong?

Doc Al said:
What is the vertical acceleration of ##m_1##?

##g##

Doc Al said:
What is the net vertical force on ##m_1##?

My new attempt [: :

If there is no motion of ##m_1##, ##F_{net} = 0##, and tension has the same magnitude that ##m_1##'s weight. Here that doesn't happen, so:

##F_{net} = m_1g - T##

Doc Al said:
What is the tension in the string?

The force is ##m_1g## and both boxes will move with the same acceleration ##\frac{m_1g}{m_1 + m_2}##. One way in which I view tension is as the pulling force exerted to move the object I want to drag. ##m_2## is what I want to pull with the string, so ##T = m_2\frac{m_1g}{m_1 + m_2}##.

Doc Al said:
What is the acceleration of ##m_2##?

##\frac{m_1g}{m1 + m_2}##

-------------##F_{net} = m_1g - T##

##F_{net} = m_1g - m_2\frac{m_1g}{m_1 + m_2}##

This would be the net force making ##m_1## move downwards. An extra tension with this magnitude upwards or horizontal (and to the left) will bring ##m_1## to rest. And that extra tension should be provided by ##F##.

Sorry if I'm not getting the idea yet :)

Last edited:
The vertical acceleration of ##m_1## is 0, not g!
Jazz said:
My new attempt [: :

If there is no motion of ##m_1##, ##F_{net} = 0##, and tension has the same magnitude that ##m_1##'s weight. Here that doesn't happen, so:

##F_{net} = m_1g - T##
Good! (Remember: This tells you what T is.)

Jazz said:
The force is ##m_1g##
Good. Don't forget this.

Jazz said:
and both boxes will move with the same acceleration ##\frac{m_1g}{m_1 + m_2}##.
Again, do not assume any value for the acceleration! The reasoning we are going through will allow us to determine the acceleration, which is needed to find the force F. (If we could just assume the acceleration, then we would be done: F = total mass X acceleration. But we don't know the acceleration yet.)

Jazz said:
One way in which I view tension is as the pulling force exerted to move the object I want to drag. ##m_2## is what I want to pull with the string, so ##T = m_2\frac{m_1g}{m_1 + m_2}##.
You already figured out the tension! Go back and use the value that you already found for T to solve for the acceleration of ##m_2##.

There is much an easier way to do that-

You need to assume that the big F = a(M+m1+m2)

Now if you know that the "Ideal" situation that you are looking for - Masses m1 and m2 are stationary - write down the free-body diagram for each block IF you know that they are Not moving.
once you figure this one out - all you will have to do is to plug it in in the equation of F.

LiorSh said:
There is much an easier way to do that-

You need to assume that the big F = a(M+m1+m2)
That's the final step. (If you look through the thread, you'll see that it's been used several times already.) It's just the application of Newton's 2nd law to the entire system. But before taking that last step, one needs to know the acceleration.

Doc Al said:
You already figured out the tension! Go back and use the value that you already found for T to solve for the acceleration of ##m_2##.

I'm somewhat unsure but here I go:

##T = m_2\frac{m_1g}{m_1 + m_2}##

##F_{net} = m_1g - T##
##F_{net} = m_1g - m_2\frac{m_1g}{m_1 + m_2}##

This is the net force on ##m_2##, so:

##m_2a = m_1g - m_2\frac{m_1g}{m_1 + m_2}##

Then:

##a = \left(m_1g - m_2\frac{m_1g}{m_1 + m_2}\right)\frac{1}{m_2}##

With this acceleration ##m_2## is moving to the right. I'm going to apply my moth-eaten argument again: Making the whole system move with the same acceleration of ##m_2## will keep the boxes stationary:

##F = \small{(M+m_1+m_2)}\left(m_1g - m_2\frac{m_1g}{m_1 + m_2}\right)\frac{1}{m_2}##

You've gone off in the wrong direction. You're missing the key point about ##m_1## not accelerating in the vertical direction. This gives you then tension in the string. Because the pulley is frictionless, you must have the same tension pulling ##m_2##.

The thing that seems to be putting you off is the idea that the force ##m_1g## is also accelerating ##m_1##. It's not. It is an unusual situation, so you need to get this straight.

What I thought was that ##m_1## had indeed a vertical acceleration ##g##, then I concluded that this was not its net acceleration, because of the string and ##m_2## attached to it, but that acceleration I've written before.

It seems that I just want to obtain an answer but I actually want to know where I started to miss the point. Could it be that I've been just thinking in accelerations? (it's like I've been trying to equate acceleration instead of forces /: ).

I need more tension to prevent ##m_1## from falling. That extra tension must be supplied by ##F##, right? Since the ##F_{net}## I found above is the tension I'm lacking, shouldn't it be the force applied to the system?

Jazz said:
I need more tension to prevent ##m_1## from falling. That extra tension must be supplied by ##F##, right? Since the ##F_{net}## I found above is the tension I'm lacking, shouldn't it be the force applied to the system?

This is where you are missing the point. You have two vertical forces on ##m_1##. As m_1 is not accelerating in this direction they are balanced. So, you know the tension.

Now, turn your attention to the forces on ##m_2##. How many horizontal forces are acting on ##m_2##?

Jazz said:
I'm somewhat unsure but here I go:
Jazz said:
##T = m_2\frac{m_1g}{m_1 + m_2}##
Let's skip this one.

Jazz said:
##F_{net} = m_1g - T##
This is the one you want, but don't complicate things. What does ##F_{net}## equal? (We've gone over this before, and you had it right, but for some reason you keep dropping it.)

Using just this last equation, tell me what the tension must equal? Nothing complicated. (Again, you had solved for the tension long ago, but dropped it.)

PeroK said:
This is where you are missing the point. You have two vertical forces on ##m_1##. As m_1 is not accelerating in this direction they are balanced. So, you know the tension.

The tension I want, right: ##m_1g##?

PeroK said:
Now, turn your attention to the forces on ##m_2##. How many horizontal forces are acting on ##m_2##?

Since the force of gravity due to ##m_1## is transmitted undiminished to ##m_2## through the string, there is only one: ##m_1g##.

Doc Al said:
Let's skip this one.

Ok. |:

Doc Al said:
This is the one you want, but don't complicate things. What does ##F_{net}## equal? (We've gone over this before, and you had it right, but for some reason you keep dropping it.)

The ##F_{net}## I want has to be zero.

Doc Al said:
Using just this last equation, tell me what the tension must equal? Nothing complicated. (Again, you had solved for the tension long ago, but dropped it.)

Tension has to be equal to ##m_1g##.

|:

Jazz said:
The ##F_{net}## I want has to be zero.
Tension has to be equal to ##m_1g##.

Good. Now that you have the rope tension, it is time to analyze the forces on ##m_2## and apply Newton's 2nd law.

Do that.

On ##m_2##, the vertical forces are in balanced. There is no friction between the surfaces so only the tension, pulling it to the right, is my ##F_{net}## there:

##F_{net} = T##

##m_2a = m_1g##

My acceleration has to be: ##\frac{m_1g}{m_2}##.

Jazz said:
On ##m_2##, the vertical forces are in balanced. There is no friction between the surfaces so only the tension, pulling it to the right, is my ##F_{net}## there:

##F_{net} = T##

##m_2a = m_1g##

My acceleration has to be: ##\frac{m_1g}{m_2}##.
Perfect. Now you have the acceleration. Now you can find F.

Doc Al said:
Perfect. Now you have the acceleration. Now you can find F.

Since my system are the three boxes (and there is no friction on the floor either), my force ##F## has to be:

##F = \small{(M + m_1 + m_2)}\frac{m_1g}{m_2}##

It doesn't look that bad, does it?

Jazz said:
Since my system are the three boxes (and there is no friction on the floor either), my force ##F## has to be:

##F = \small{(M + m_1 + m_2)}\frac{m_1g}{m_2}##

It doesn't look that bad, does it?
Perfecto! :)

Doc Al said:
Perfecto! :)

Now I think I can print the thread and stick it to the refrigerator :) . I almost lost my mind with this problem.

I would like to understand why the reasoning I went through, after having skipped the first acceleration I got, is actually wrong.

I wanted to know the acceleration of ##m_2## in order to exert and equal acceleration on the whole system (the first time I wrongly applied it to just two boxes). So I assumed that if there to be a net acceleration to the right, it had to be ##\frac{m_1g}{m_1+m_2}##. What is the problem with this assumption?

Last edited:
Jazz said:
So I assumed that if there to be a net acceleration to the right, it had to be ##\frac{m_1g}{m_1+m_2}##. What is the problem with this assumption?
What possible basis is there for this assumption?

Doc Al said:
What possible basis is there for this assumption?

There is no basis, honestly. Since the only force acting on ##m_2## is ##T##, the acceleration is ##\frac{T}{m_2}##. I can't argue against that, but I still can't throw out the other idea from my head either.

But what if I don't applied the force ##F##? what would the acceleration of ##m_2## and ##m_1## be? would It be that I've written a couple of times before?

The picture I had in mind is that that I wrote before: both boxes on ##M##, touching each other with a force ##m_1g## to the right.

Maybe this was making me miss the fact that the horizontal force on ##m_2## does not affect the vertical force on ##m_1## (under the absence of friction or any other opposing force). Then, if I don't exert the force ##F##, ##m_1## will move downward at ##9.8\ m/s^2## and ##m_2## to the right at ##\frac{m_1g}{m_2}##.

Somehow I was thinking that having to pull ##m_2## was making ##m_1## to move downward at an acceleration less than ##g##, as my mental picture was suggesting.

Have I discovered where I've been getting trapped all the time? I think so.

Last edited:

## 1. How do pulleys help keep boxes stationary on another box?

Pulleys use the principle of mechanical advantage to distribute the weight of the boxes evenly and reduce the force needed to keep them stationary on top of each other. This allows for a more efficient and stable stacking of boxes.

## 2. What is the best way to set up pulleys for this type of problem?

The best way to set up pulleys for keeping boxes stationary on another box is to use a multi-pulley system. This involves attaching multiple pulleys to a fixed point and threading the rope through each one to create a mechanical advantage. This will distribute the weight more evenly and make it easier to keep the boxes stationary.

## 3. Can I use any type of rope or cord for this problem?

It is important to use a strong and durable rope or cord for this type of problem. Nylon ropes are a popular choice as they are strong and have a low stretch, which helps maintain tension on the pulleys. It is recommended to use a rope specifically designed for use with pulleys.

## 4. How do I calculate the amount of weight a pulley system can hold?

The amount of weight a pulley system can hold is determined by the mechanical advantage of the system. This is calculated by dividing the weight of the load by the force needed to lift it. For example, a 4:1 mechanical advantage means that the system can support four times the weight of the load.

## 5. Are there any safety precautions I should take when using pulleys for this problem?

It is important to always follow proper safety precautions when using pulleys, such as wearing gloves to protect your hands and ensuring the pulleys are securely attached to a stable and strong anchor point. It is also recommended to regularly inspect the pulleys and ropes for any signs of wear and tear to prevent accidents from occurring.

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