Understanding the Problem of F = dp/dt with Sand in a Box

[HAMJOOP]

A box which carries sand is moving with a constant velocity u on a smooth horizontal surface.

The sand is removed during the motion of the box such that the sand has the same horizontal velocity u.

We know that there is no horizontal force exerts on the box.
The box will travels in constant velocity u.

However, the mass of the box is decreasing. Its momentum is decreasing.
So there is a net force acting on the box.


BUT there is no net force. What's wrong ?

 

[HallsofIvy]

What's wrong is that you have described an impossible situation. Yes, F= dp/dx= d(mv)dt= (dm/dt)v+ m(dv/dt). Now, you say, "We know that there is no horizontal force exerts on the box. The box will travels in constant velocity u." Those sentences cannot both be true. With constant velocity, dv/dt= 0 so F= (dm/dt)v. As long as neither v nor dm/dt is 0, F cannot be 0.

 

[Studiot]

What's the problem?


You are just applying conservation of momentum in an inappropriate fashion.

The mass of the box is constant so if its velocity is constant its momentum is constant and no force is applied.

The horizontal velocity of the sand is likewise constant and since its total mass is also constant its momentum is constant and again no horizontal force is applied.

Total horizontal momentum = (momentum of box + momentum of sand) = a constant.

What you cannot do is what you have done, which is start with the momentum of one system (box plus all the sand) and subtract the momentum of a different system (box plus some the sand) to obtain dp the change of momentum.

dp applies to the same system or not at all.

 

[HAMJOOP]

If I choose the system to be the box only, I observe that the momentum of the box is decreasing.

So the box should experience a net force. But there is none.

why?

 

[arildno]

One way to understand this is that "the box" is NOT a system consisting of the SAME material particles over time; "the box" is merely a convenient GEOMETRIC region we wish to analyze the behaviour to, due to the changing forces acting upon the changing particles included within it.
(A similar geometric region might be a bended tube through which fluid flows; it is legitimate to find out what the force acting upon the tube will be, but again, the filled tube is NOT a single, material system.
Some years ago, I made an essay here at PF on such geometric systems, you may read it here:
https://www.physicsforums.com/showthread.php?t=72176

 

[Studiot]

If I choose the system to be the box only, I observe that the momentum of the box is decreasing.

If you chose the box only as the system ( perfectly legitimately) then you'd be wrong in your observation.

I have already shown you the calculation involved, did you not follow it?

 

[Chestermiller]

The change in momentum for the portion of material that has remained in the box (up to any time t) is zero. So it has experienced no net force. The change in momentum of the material that has been jettisoned from the box up to any time t is also zero. So it too has experienced no net force (aside from the force in the perpendicular direction needed to jettison it).

The problem is that the equation d(mv)/dt = f applies only to a closed system in which mass is neither leaving or entering. In your case, mass is leaving.