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Variable mass system - sand falling on flatcar

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Homework Statement


A flatcar of mass ##m_0## starts moving to the right due to a constant horizontal force F. Sand spills on the flatcar from a stationary hopper. The velocity of loading is constant and equal to ##\mu## kg/s. Find the time dependence of the velocity and acceleration of flatcar in the process of loading. The friction is negligibly small.

(Ans: ##v=Ft/(m_0(1+\mu t/m_0))##, ##a=F/(m_0(1+\mu t/m_0)^2)## )

Homework Equations





The Attempt at a Solution


The given question can be easily solved by the impulse momentum equation. The reason I posted the question here was to clarify a doubt about directly applying F=ma.

For the question in the thread "Sand flowing out of the car", it was suggested that it can be solved by using F=M(t)a(t) but when I try to apply it to this problem, it gives me a wrong answer. :confused:

Also, I think I was a bit lucky while applying the Impulse-Momentum equation to the problem. If the falling sand had some velocity in the horizontal direction, the answer would have been different. So how do I know when to apply Impulse-Momentum equation or F=Ma? :confused:

Any help is appreciated. Thanks!
 

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  • #2
Redbelly98
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Impulse-momentum should always work. And when mass M is constant, it is equivalent to F=Ma. So, use F=Ma when the mass is not changing.

Alternate solution: in principle, you could write an expression for the center of mass of the sand-flatcar system, which does have a fixed mass, and apply F=Ma to the CM motion.
 
  • #3
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Impulse-momentum should always work. And when mass M is constant, it is equivalent to F=Ma. So, use F=Ma when the mass is not changing.
But in this case, the mass is not constant and neither it was in the thread I have posted a link to. :confused:
 
  • #4
TSny
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When the sand is falling out of the cart, are there any action/reaction forces between the cart and the sand leaving the cart?
 
  • #5
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When the sand is falling out of the cart, are there any action/reaction forces between the cart and the sand leaving the cart?
When the sand falls out, there is no contact so I guess the answer is no.
 
  • #6
TSny
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When the sand falls out, there is no contact so I guess the answer is no.
That’s what I was thinking when I first solved the problem (of the sand falling out of the cart) by using ##F_{ext} = M(t)a(t)##. I felt like the sand falling out had no effect on the force side of the equation. The only effect of the sand falling out was to determine ##M(t)##. However, after finding the acceleration this way, I still felt a need to check it with the impulse-momentum approach!

When you think about it, any bit of sand does exert a force on the rest of the system before the bit of sand falls out (because the system is accelerating due to the external force). After the bit of sand falls out, it no longer exerts a force on the cart. While the bit of sand is falling out, it’s not clear (to me) what I should take to be the force acting on the cart by the sand falling out. But you can argue that any such force is negligible compared to the external force and therefore can be neglected.

If the sand is continuously ejected by the cart with a sudden finite change in speed ##\Delta v## (like rocket propulsion), then you can think of that as adding another force to the problem that can’t be neglected. If ##dm## of sand is ejected during time ##dt##, then the reaction force back on the cart is

##(dm)\frac{\Delta v}{dt} = (\lambda dt)\frac{\Delta v}{dt} = \lambda \Delta v##. So the force equation becomes

##F_{ext} +\lambda \Delta v = M(t)a(t)##.

( For the case where the sand "falls out", ##\Delta v = 0## and so we get the ## F_{ext} = M(t)a(t)## equation)
 
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  • #7
ehild
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The given question can be easily solved by the impulse momentum equation. The reason I posted the question here was to clarify a doubt about directly applying F=ma.

For the question in the thread "Sand flowing out of the car", it was suggested that it can be solved by using F=M(t)a(t) but when I try to apply it to this problem, it gives me a wrong answer. :confused:

Also, I think I was a bit lucky while applying the Impulse-Momentum equation to the problem. If the falling sand had some velocity in the horizontal direction, the answer would have been different. So how do I know when to apply Impulse-Momentum equation or F=Ma? :confused:

Any help is appreciated. Thanks!


Always apply the momentum equation, taking into account both the momentum of the loaded car (the varying-mass object) and also the momentum of the added mass. In principle, F=ma is valid only for constant mass. It comes out from the momentum equation that you get F=m(t)a when the variable mass object and the added or removed mass have the same velocity. That was valid for the leaking-out sand, but it is not true for this problem. The falling-in sand has zero horizontal velocity, while the car has velocity v. The added mass has to be accelerated to the velocity of the car, it is done by the interaction with the car and the sand in it. The added sand is accelerated by that interaction force, and the opposite force acts on the car. So it is not accelerated by F alone, but by F-μv.

Physics is not about trying equations blindly, and seeing if they give the same result as the book. You have to know at what conditions is a principle or equation valid and applying them accordingly.
If you think you can apply F=ma for this problem you did not understand what we tried to explain to you in the previous thread. Why have you not asked about your doubts?

ehild
 
  • #8
F=ma can be applied, but that will only make the problem more complex. If you apply it to (flatcar + sand on it), you need to find the forces acting on the system. Sand falling on the flatcar exerts a force to it, because friction has to act on the sand to accelerate it until its velocity equals the velocity of the vehicle.
 
  • #9
haruspex
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In principle, F=ma is valid only for constant mass.
Really? I don't see the difficulty using it here.
Relative to the car, the incoming sand is moving with velocity -v. It therefore exerts a force -μv: ##\dot v (m_0+\mu t) = F-\mu v##. Solving that gives the desired answer.
davidchen9568 said:
F=ma can be applied, but that will only make the problem more complex. If you apply it to (flatcar + sand on it), you need to find the forces acting on the system. Sand falling on the flatcar exerts a force to it, because friction has to act on the sand to accelerate it until its velocity equals the velocity of the vehicle.
You seem to be arguing that there will be a small delay in the retardation of the car due to the incoming sand. That may be true, but you can get around that by defining the speed as being that of the CoM of the car and the sand currently in it, whether or not it is still settling. It doesn't really alter the outcome.
 
  • #10
ehild
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Really? I don't see the difficulty using it here.
Relative to the car, the incoming sand is moving with velocity -v. It therefore exerts a force -μv: ##\dot v (m_0+\mu t) = F-\mu v##. Solving that gives the desired answer.
F was meant as the external applied force acting on the car, and Pranav understood that way. The final formula can be written in the form m(t)a(t)= F(applied)+f where f=μv(t) is the term that results from the interaction between the added/removed mass and the car, and it "comes out" from the momentum-impulse equation.

You can consider the car with its load and the falling in sand separately and writing up the equation ma=F(applied) +f(interaction), but then you need to know the interaction force which is some kind of friction. The advantage of the momentum method is that you do not need to know the the interaction force in detail. It is the same as in case of collisions: You get the outcome velocities without knowing the force of interaction between the colliding objects.

Telling a student that he can apply ma=F for a variable mass system, where F was given in the problem as the applied force, is very misleading. You could see it from Pranav's question: he solved the problem with the momentum method, and then asked, why he did not get the same result from F=ma.

ehild
 
  • #11
haruspex
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F was meant as the external applied force acting on the car, and Pranav understood that way. The final formula can be written in the form m(t)a(t)= F(applied)+f where f=μv(t) is the term that results from the interaction between the added/removed mass and the car, and it "comes out" from the momentum-impulse equation.

You can consider the car with its load and the falling in sand separately and writing up the equation ma=F(applied) +f(interaction), but then you need to know the interaction force which is some kind of friction. The advantage of the momentum method is that you do not need to know the the interaction force in detail. It is the same as in case of collisions: You get the outcome velocities without knowing the force of interaction between the colliding objects.

Telling a student that he can apply ma=F for a variable mass system, where F was given in the problem as the applied force, is very misleading. You could see it from Pranav's question: he solved the problem with the momentum method, and then asked, why he did not get the same result from F=ma.

ehild
To me it seems completely analogous to treating a rocket engine's thrust as a force.
 
  • #12
ehild
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The rocket engine's thrust is an internal force between the rocket and the fuel being just ejected, considering them as parts of a system. If there is gravity, it is the external force acting on the whole system - rocket and fuel ejected in Δt time.

In the sand problem, F was the applied external force: It was wrong to use the equation ma=F, m meaning the variable mass of the car and F the given external force.

ehild
 
  • #13
haruspex
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The rocket engine's thrust is an internal force between the rocket and the fuel being just ejected, considering them as parts of a system. If there is gravity, it is the external force acting on the whole system - rocket and fuel ejected in Δt time.
Instead of having the sand fall vertically into the cart, imagine it magically suspended in the air, statically, being scooped up by the cart as it moves along. The system is essentially the same. Now time-reverse it, so the sand is ejected from the cart, propelling the cart along. Apart from the exhaust speed happening to be just so as to leave the sand stationary relative to the ground, it looks exactly like a rocket.
In the sand problem, F was the applied external force: It was wrong to use the equation ma=F, m meaning the variable mass of the car and F the given external force.
It is certainly a mistake to confuse the specific F given in the question with the generic Fnet in Fnet=ma. But realise that here Fnet = F-μv and it's fine.
 
  • #14
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Very sorry for the late reply. I had some connection problems.

Thanks a lot for the explanations. I have read your posts several times, the way you have explained how to use F=ma is great. I don't think I still have any other doubts on this problem now.

Pardon for a short reply but I can't find anything to add to my post. I have understood what I did wrong while applying F=ma to this problem.
 

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