Emptying a Freight Car

[ThEmptyTree]

Homework Statement:

A freight car of mass $m_c$ contains sand of mass $m_s$ . At $t = 0$ a constant horizontal force of magnitude $F$ is applied in the direction of rolling and at the same time a port in the bottom is opened to let the sand flow out at the constant rate $b = \frac{dm_s}{dt}$. Find the speed of the freight car when all the sand is gone (Figure 12.6). Assume that the freight car is at rest at $t = 0$ .

Relevant Equations:

$$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$$

$$\overrightarrow{p_i}=\overrightarrow{0},~\overrightarrow{p_f}=m_c\overrightarrow{v_f}$$

$$\overrightarrow{p_f}-\overrightarrow{p_i}=\int\limits_{t_i}^{t_f}\overrightarrow{F}dt$$

$$t_i=0,~t_f=\frac{m_s}{b}$$

$$m_c\overrightarrow{v_f}=\overrightarrow{F}\frac{m_s}{b}\Rightarrow v_f=\frac{Fm_s}{bm_c}$$

However, their solution uses differential analysis for states $t$ and $t+\Delta{t}$, yielding

$$v(t)=\frac{F}{b}ln\Big(\frac{m_c+m_s}{m_c+m_s-bt}\Big)$$

For $t=t_f=\frac{m_s}{b}$ according to their solution

$$v_f=\frac{F}{b}ln\Big(1+\frac{m_s}{m_c}\Big)$$

Notice that using approximation $ln(1+x)\approx x$ for small $x$ their solution becomes equivalent to mine.

What am I doing wrong? I am confused. What am I skipping when trying to analyze only the initial and final state?

 

[BvU]

What am I doing wrong?

The sand that falls out also has momentum ... how does your solution account for that ?

$\$

 

[ThEmptyTree]

The sand that falls out also has momentum ... how does your solution account for that ?

$\$

I agree that the sand has speed, but I am only analyzing the cart.

Also, in their solution $\Delta{m_c}=-\Delta{m_s}$ and the sand simplifies.

 

[BvU]

but I am only analyzing the cart

$$\overrightarrow{p_f}-\overrightarrow{p_i}=\int\limits_{t_i}^{t_f}\overrightarrow{F}dt$$ has a righthand side that only applies to the car ?

 

[ThEmptyTree]

$$\overrightarrow{p_f}-\overrightarrow{p_i}=\int\limits_{t_i}^{t_f}\overrightarrow{F}dt$$ has a righthand side that only applies to the car ?

That's what came to my mind, that $\overrightarrow{F}$ acts over the cart + fallen sand, not only the cart. But how is that, if $\overrightarrow{F}$ originates from friction?

 

[BvU]

Also, in their solution $\Delta{m_c}=-\Delta{m_s}$ and the sand simplifies.

$m_s$ and $m_c$ are fixed numbers in their solution.

(but not in $b = \frac{dm_s}{dt}$, so I grant you that their use of the symbols is misleading...)

 

[BvU]

$\vec F$ is the externally applied force (e.g. from a locomotive)

 

[ThEmptyTree]

$\vec F$ is the externally applied force (e.g. from a locomotive)

Still, the locomotive tracts the cart, not the sand, right ?

 

[BvU]

The locomotive is pulling on cart + remaining sand !

$$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$$this is one of the (rare) cases where both $m$ and $\vec p$ change ...

$\$

 

[ThEmptyTree]

The locomotive is pulling on cart + remaining sand !

I agree with the remaining sand. The problem is that in their solution they consider the system of cart + fallen sand, and when they write $\overrightarrow{F}=\frac{d\overrightarrow{p_{sys}}}{dt}$ $\overrightarrow{F}$ appears to act on both cart + fallen sand at $t+\Delta{t}$.

When I said I only analyze the cart, that means the cart + the sand left in it.

At $t=0$ even if there is sand in the cart the momentum is null, and at $t=t_f$ there is no sand left.

 

[ThEmptyTree]

@BvU I figured it out. The problem is that $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ only works for constant mass. In my approach I chose the system of cart + remaining sand, but that is not constant mass because sand is constantly falling. I think this is what you were trying to explain, but you did not emphasize it, or I couldn't understand. Thanks for your time!

Case solved.

 

[bob012345]

@BvU I figured it out. The problem is that $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ only works for constant mass. In my approach I chose the system of cart + remaining sand, but that is not constant mass because sand is constantly falling. I think this is what you were trying to explain, but you did not emphasize it, or I couldn't understand. Thanks for your time!

Case solve

Force is the time rate of change of momentum and both mass and velocity can change but applying $\vec F = \large\frac {d \vec P}{dt}$ correctly can be a bit tricky. In general $\large\frac {d \vec P}{dt} = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$. In a rocket, the second term would be the generator of the force. In this case the force is fixed and external and it produces an increasing acceleration as the total mass of the car decreases.

Starting with the variables in the diagram of post $10$ we can write the system momentum at times $t$ and $t + Δt$;

$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and

$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

Then we can write $\vec F = \large \frac{Δ\vec P}{Δt} = \large \frac{\vec P(t + Δt) - \vec P(t )}{Δt}$

as $$\vec F Δt = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c) - (m_c + m_s -bt)\vec v_c(t)$$

Then $$\vec F Δt = -bΔt\vec v_c(t) +(m_c + m_s -bt)Δ\vec v_c - bΔtΔ\vec v_c) + Δm_s\vec v_c(t) + Δm_sΔ\vec v_c)$$ Then getting rid of double differentials and noticing that $Δm_s = bΔt$ we get

$$\vec F Δt = (m_c + m_s -bt)Δ\vec v_c~~~ or ~~~ \frac {d\vec v_c}{dt} = \frac { \vec F}{(m_c + m_s -bt)}~~~ as Δt →0$$.

 

[ThEmptyTree]

@bob012345 I appreciate doing all the calculations, but my enigma was why we could not just analyze the initial and final state only.

No, I believe that's not accurate.

I still believe the reason is because during that time interval, the mass of the object (whether it is a rocket, a freight cart, a snowplow gathering snow) changes, and Newton's generalized 2nd law (in the way I applied it) only works for constant mass, either for a single object or a system of objects.

 

[bob012345]

I was just reacting to your statement ;

The problem is that $F=\large \frac{dp}{dt}$ only works for constant mass.

It works for changing mass too in general. You are correct that what doesn't work is not taking changing mass into account not that Newton's formulation only works for constant mass. If you assume constant mass in this problem all you get is that $v(t) = \frac{F}{m_c + m_s} t$ which is just $v(t) = at$.


But I did just use the initial and final states to derive the equation that needed to be solved to show it works. The simplest way to do this problem however is to just recognize that $\vec a = \large\frac{\vec F}{m_c+m_s -bt}$ and integrate to find $v(t)$.

 

[ThEmptyTree]

I was just reacting to your statement ;

The problem is that $F=\large \frac{dp}{dt}$ only works for constant mass.

It works for changing mass too in general.

What I said there is indeed misleading.

Now after meditating on it for a while I still don't think I understand what I did wrong in my first approach.

I did not take into account that $$d\overrightarrow{p}=md\overrightarrow{v}+dm\overrightarrow{v}$$
but
$$\int\limits_{t_i}^{t_f} d\overrightarrow{p}=\int\limits_{t_i}^{t_f} \overrightarrow{F}dt$$
should still work right?

 

[haruspex]

It works for changing mass too in general.

$\vec F=\frac{d\vec p}{dt}$ applies to a closed system, and in closed systems mass does not change.
If the mass is changing then you need to consider what momentum the gained (lost) mass is adding (subtracting). To make the equation work you need to treat that as an applied force.
In the example of sand being dropped vertically from a static hopper into a moving cart, you get away with it because the sand arrives with no horizontal momentum. But if sand leaks from a moving cart, it carries its momentum away with it. Overlooking that leads to the conclusion that the cart should accelerate.

 

[bob012345]

$\vec F=\frac{d\vec p}{dt}$ applies to a closed system, and in closed systems mass does not change.
If the mass is changing then you need to consider what momentum the gained (lost) mass is adding (subtracting). To make the equation work you need to treat that as an applied force.
In the example of sand being dropped vertically from a static hopper into a moving cart, you get away with it because the sand arrives with no horizontal momentum. But if sand leaks from a moving cart, it carries its momentum away with it. Overlooking that leads to the conclusion that the cart should accelerate.

In my understanding it is more general than $F = ma$. One just has to be careful how to apply it.

 

[haruspex]

In my understanding it is more general than $F = ma$. One just has to be careful how to apply it.

For clarity, please illustrate its use in the leaking sand model.

 

[bob012345]

For clarity, please illustrate its use in the leaking sand model.

I thought I did that in post #12.

 

[haruspex]

I thought I did that in post #12.

That post was rather confusing.
You wrote
$\vec F = \large\frac {d \vec P}{dt}$
and
$\large\frac {d \vec P}{dt} = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$
from which one would conclude
$\vec F = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$.
but followed with

In a rocket, the second term would be the generator of the force.

If the force is $\large\frac { \vec V dm}{dt}$, what does $\vec F$ mean in that equation?
And what is V now, the velocity of the rocket, of the exhaust, or the difference between the two?

It seems to me you did not actually use $\vec F = \large\frac {d \vec P}{dt}$.

 

[bob012345]

That post was rather confusing.
You wrote
$\vec F = \large\frac {d \vec P}{dt}$
and
$\large\frac {d \vec P}{dt} = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$
from which one would conclude
$\vec F = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$.
but followed with

If the force is $\large\frac { \vec V dm}{dt}$, what does $\vec F$ mean in that equation?
And what is V now, the velocity of the rocket, of the exhaust, or the difference between the two?

It seems to me you did not actually use $\vec F = \large\frac {d \vec P}{dt}$.

Sorry if I confused you. I indeed fell victim to a common error of assuming $\large\frac {d \vec P}{dt} = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$
which assumes momentum is always simply $\vec P = m\vec V$ when there can be different contributions. However, that was only in my general remarks and I did not use that directly in the working of the problem. There I used the differential form of $\vec F = \large\frac {Δ \vec P}{Δt}$ to set up the OP problem as recommended in my text An Introduction to Mechanics by Kleppner and Kolenkow to arrive at the result;

$$ \frac {d\vec v_c}{dt} = \frac { \vec F}{(m_c + m_s -bt)}$$.

Which will give $\vec v_c(t)$ when integrated.


.

 

[haruspex]

which assumes momentum is always simply $\vec P = m\vec V$

Which, for a defined assembly, it is.

we can write the system momentum at times $t$ and $t + Δt$;
$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

The second equation does not follow from the first. In fact, if, as you defined, P is the momentum of the cart and its remaining sand then the second equation is not true.
Rather, you would get
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) $$.
It's no good countering that the extra term you added comes from the momentum lost in the loss of sand. That term did not appear in the initial equation. You cannot go adding terms just because you know you will otherwise get the wrong answer.

 

[bob012345]

Which, for a defined assembly, it is.

The second equation does not follow from the first. In fact, if, as you defined, P is the momentum of the cart and its remaining sand then the second equation is not true.
Rather, you would get
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) $$.
It's no good countering that the extra term you added comes from the momentum lost in the loss of sand. That term did not appear in the initial equation. You cannot go adding terms just because you know you will otherwise get the wrong answer.

I wrote those equations for the system, not just the cart. That is exactly what we are talking about, the system not one part. They seem correct to me. They led to the correct expression for velocity. That extra term is the sand. I used the figure from post #10 as reference.

 

[haruspex]

I wrote those equations for the system, not just the cart.

No, your first equation was

$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$

which is clearly just for the cart and its remaining sand. If your next equation is for the cart plus all the sand then you changed the meaning of P(t) . The two equations are clearly not consistent algebraically.

Even if we do redefine P that way for the second equation it is wrong:

$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

Presumably the idea is
"Momentum of whole system at time t+dt = momentum of cart with remaining sand at t+dt + momentum of all lost sand at time t+dt."
Unfortunately the second term on the right is not the momentum of all lost sand at time t+dt; it is only the momentum of the sand lost during time dt. This makes it hard to understand how P(t) is being defined.

 

[bob012345]

No, your first equation was
which is clearly just for the cart and its remaining sand. If your next equation is for the cart plus all the sand then you changed the meaning of P(t) . The two equations are clearly not consistent algebraically.

The cart and remaining sand is the system for the first equation. All previous fallen sand has zero momentum. The second equation is for the cart with $Δm_s = bΔt$ less sand than the first and the same amount of sand moving with $v_c + Δv_c$. We are only concerned with each element of sand just as it falls and looking at one is all we need to do. We are not accounting for all the fallen sand from $t=0$. We start the problem at some arbitrary time.

Even if we do redefine P that way for the second equation it is wrong:

Presumably the idea is
"Momentum of whole system at time t+dt = momentum of cart with remaining sand at t+dt + momentum of all lost sand at time t+dt."
Unfortunately the second term on the right is not the momentum of all lost sand at time t+dt; it is only the momentum of the sand lost during time dt. This makes it hard to understand how P(t) is being defined.

I invite you then to do the problem yourself and see if you get the right answer.

 

[haruspex]

The cart and remaining sand is the system for the first equation.

Which, as I wrote, is what I assumed in the first place:

as you defined, P is the momentum of the cart and its remaining sand

So why did you respond with:

I wrote those equations for the system, not just the cart

So, now you are saying that in
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
$\vec P(t + Δt)$ is the momentum of the cart+remaining sand, yes?
Clearly that is wrong.
$\vec P(t + Δt)$ = momentum of cart plus remaining sand at time (t + Δt)
= [mass of cart plus remaining sand at time (t + Δt)][velocity of cart plus remaining sand at time (t + Δt)]
=$(m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$

All previous fallen sand has zero momentum.

We do not know the fate of the sand, but it certainly had momentum when it left the cart.

see if you get the right answer.

I have no problem getting the right answer, and whether you got the right answer is not the question I'm posing.
The issue is, what exactly is your method, did it really use $F=dp/dt$ with p being the momentum of a variable mass subsystem, with those variables defined how, and is your method correct?

 

[ThEmptyTree]

I am confused. I don't know what to believe, does $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ work or not for changing mass? If it works, I still can't explain what I did wrong. If it doesn't work, then it contradicts physics?

 

[haruspex]

I am confused. I don't know what to believe, does $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ work or not for changing mass? If it works, I still can't explain what I did wrong. If it doesn't work, then it contradicts physics?

$\frac{d\vec p}{dt}=m\frac{d\vec v}{dt}+\vec v\frac{dm}{dt}$ is certainly valid - that's just algebra. The question is, when can we write $\vec F=\frac{d\vec p}{dt}$, F being the force applied to a system and m being its mass?
The short answer is, only when no mass change is adding momentum to or removing it from the system.
With the current problem, the leaking sand is taking momentum away from the cart+remaining sand system.
One way to deal with that is to treat the resulting change in momentum of the chosen system as another applied force. Since the sand leaves at the current velocity of the cart, the rate at which it adds momentum is $v\dot m$ ($\dot m$ being negative). This makes the force equation (I'll drop the vectors) $F+v\dot m=\dot p$, which reduces to $F=m\dot v$.

 

[jbriggs444]

I am confused. I don't know what to believe, does $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ work or not for changing mass? If it works, I still can't explain what I did wrong. If it doesn't work, then it contradicts physics?

It can work. But it is fraught with opportunities to confuse yourself. If you are contemplating a system with variable mass then you are contemplating a (hopefully continuous) mass flow across the system boundary. That mass flow can be treated as a momentum flow -- i.e. as a force.

It is ever so tempting to differentiate $\frac{dp}{dt}=\frac{d(m(t)v(t))}{dt}$ and obtain $m\frac{dv}{dt} + v\frac{dm}{dt} = ma + v\frac{dm}{dt}$. And it is even correct -- as far as it goes.

But it is easy to forget that this applies to a scenario where the mass being added or removed from the system comes in already moving at the same velocity as the system or leaves still moving at the same velocity as the system.

If you have a stream of sand that is at rest and is added to a moving train, it does not carry in a momentum flow that matches $v\frac{dm}{dt}$. It carries in zero momentum instead.

If you have a rocket that is streaming out an exhaust at relative velocity $v_e$, it does not carry out a momentum flow that matches $v\frac{dm}{dt}$. It carries out a momentum flow given by $(v-v_e)\frac{dm}{dt}$ instead.

In the case at hand, we have a freight car streaming sand out of the hopper at a rate of $\frac{dm}{dt}$ The dropped sand is indeed initially moving at the speed of the car, $v$. The egress momentum flow is indeed given by $v\frac{dm}{dt}$. (One should be aware that $\frac{dm}{dt}$ will be negative).

 

[ThEmptyTree]

If you have a stream of sand that is at rest and
In the case at hand, we have a freight car streaming sand out of the hopper at a rate of $\frac{dm}{dt}$ The dropped sand is indeed initially moving at the speed of the car, $v$. The egress momentum flow is indeed given by $v\frac{dm}{dt}$. (One should be aware that $\frac{dm}{dt}$ will be negative).

So I applied it correctly in the first place but got the wrong answer?

 

[haruspex]

So I applied it correctly in the first place but got the wrong answer?

You overlooked that some of the work done by the force goes into accelerating sand that is later lost.

 

[ThEmptyTree]

You overlooked that some of the work done by the force goes into accelerating sand that is later lost.

OK, so that's what you mean by "fallen sand should be counted as applied force". So instead of $\overrightarrow{F}$ I should have $\overrightarrow{F}+\frac{dm}{dt}\overrightarrow{v}$ as you said in post #28. I think it's clear now.

One thing I didn't understand from the beginning is that $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ is actually the definition of the force, that means total physical force that acts over the system as observed by the chosen frame. I was thinking of $\overrightarrow{F}$ as just the traction force.

Now that I understand the precautions when analyzing the initial and final state only, I consider it's safer to analyze problems at $t$ and $t+\Delta{t}$ since the system has constant mass.

I think now it's case solved.

 

[bob012345]

I am confused. I don't know what to believe, does $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ work or not for changing mass? If it works, I still can't explain what I did wrong. If it doesn't work, then it contradicts physics?

You mentioned you knew the right answer. Was it from a textbook and did they show some steps for the problem?

 

[ThEmptyTree]

You mentioned you knew the right answer. Was it from a textbook and did they show some steps for the problem?

I am currently studying from an MIT OpenCourseWare. In the introductory lecture of chapter Momentum and Impulse they told that $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ is the most general form for Newton's 2nd law, which works even for systems with a change in mass over time. However for changing mass I realized that it contradicts $\overrightarrow{F}=m\overrightarrow{a}$ so I was confused. But after some research I found out $\overrightarrow{F}=m\overrightarrow{a}$ is actually the case of $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ for constant mass.

What was most confusing for me was that people on the internet were saying $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ only works for constant mass, which would have easily explained the mistake in my approach. However the mistake was that I was counting $\overrightarrow{F}$ as traction force, not total force.

I had the right answer with full steps, that was not the problem. The exercise is from the course e-book. My dilemma was why my method was not right.

 

[bob012345]

So, now you are saying that in
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
$\vec P(t + Δt)$ is the momentum of the cart+remaining sand, yes?
Clearly that is wrong.

I said it is the momentum of cart + remaining sand in the cart plus the sand that just left the cart at that instant.

$\vec P(t + Δt)$ = momentum of cart plus remaining sand at time (t + Δt)
= [mass of cart plus remaining sand at time (t + Δt)][velocity of cart plus remaining sand at time (t + Δt)]
=$(m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$


We do not know the fate of the sand, but it certainly had momentum when it left the cart.

And I accounted for that.

I have no problem getting the right answer, and whether you got the right answer is not the question I'm posing.

Please just show your version of the first and second equations so I can see how you want to do it.

The issue is, what exactly is your method, did it really use $F=dp/dt$ with p being the momentum of a variable mass subsystem, with those variables defined how, and is your method correct?

My method was using the figure in post #10 and writing the momentum before and after the mass $Δm_s$ is released. Then I subtracted those two momentum equations and divided by $Δt$ and took the limit at $Δt →0$ to get a differential equation that can be solved for $v_c(t)$.