# Pulley Problem: Keeping boxes stationary on another box

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1. Dec 18, 2014

### Jazz

1. The problem statement, all variables and given/known data

2. Relevant equations

$F_{net} = ma$

3. 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?

2. Dec 18, 2014

### Staff: Mentor

OK, but what is your reasoning behind this?

Why do you only have to push two masses?

There's a rope attached to $m_1$.

I don't understand that statement.

3. Dec 18, 2014

### PeroK

4. Dec 18, 2014

### Jazz

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 $\frac{m_1g}{m_1 + m_2}$.

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.

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:

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

Last edited: Dec 18, 2014
5. Dec 18, 2014

### PeroK

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

6. Dec 18, 2014

### Staff: Mentor

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.

7. Dec 19, 2014

### Jazz

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 $\small{T = m_2\frac{m_1g}{m_1+m_2}}$.

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: Dec 19, 2014
8. Dec 19, 2014

### Staff: Mentor

Two problems. One, do assume any value for the acceleration. Two, why would a horizontal acceleration affect the tension in the string?

Since we want $m_1$ not to move downward, what must be its vertical acceleration?

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 $m_2$?

9. Dec 19, 2014

### Jazz

I did not understand /:

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?

$g$

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:

$F_{net} = m_1g - T$

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 $T = m_2\frac{m_1g}{m_1 + m_2}$.

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

-------------

$F_{net} = m_1g - T$

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

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: Dec 19, 2014
10. Dec 19, 2014

### Staff: Mentor

The vertical acceleration of $m_1$ is 0, not g!
Good! (Remember: This tells you what T is.)

Good. Don't forget this.

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.)

You already figured out the tension! Go back and use the value that you already found for T to solve for the acceleration of $m_2$.

11. Dec 19, 2014

### LiorSh

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.

12. Dec 20, 2014

### Staff: Mentor

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.

13. Dec 22, 2014

### Jazz

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:

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

14. Dec 22, 2014

### PeroK

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.

15. Dec 22, 2014

### Jazz

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?

16. Dec 22, 2014

### PeroK

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 $m_2$?

17. Dec 22, 2014

### Staff: Mentor

Let's skip this one.

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.)

18. Dec 22, 2014

### Jazz

The tension I want, right: $m_1g$?

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

Ok. |:

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

Tension has to be equal to $m_1g$.

|:

19. Dec 22, 2014

### Staff: Mentor

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.

20. Dec 22, 2014

### Jazz

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: $\frac{m_1g}{m_2}$.