Force = Change of Momentum?

I think what you are doing wrong is assuming that the engines are not affecting the speed of the barges. If the engines were to provide a force then the equation would become F_engine+F_engine=0 which is not the case.f
  • #1
Two long barges are moving in the same direction in still water, one with a speed of 10km/h and the other with a speed of 20km/h. While they are passing each other, coal is shoveled from the slower to the faster one at a rate of 1000 kh/min. How much additional force must be prvided by the driving engines of (a) the faster barge and (b) the slower barge if neither is to change speed? (There is a picture to this but I think the description of the problem is well enough)

Assume that the shoveling is always perfectly sideways and that the frictional forces between barges and the water do not depend on the mass of the barges.

This is a problem from Fundamentals of Physics 8/e Ch9 #79 on the back of the chapter.
(BTW this is not homework, just some randomly popped up question in my head)

The way I can solve this problem, which corresponds to the answer on the back of the book is to find the change of momentum of the coal when they are being transferred from one to the other.
However, the problem that I'm having with is that I cannot seem to find a way to use F=v*dm/dt + m*dv/dt to solve this particular problem?
Is it because F = dp/dt applies only to constant mass or can anyone provide a title for a book that has problems using this particular formula? ( Seems to me that there isn't any such problems in this whole book)

The answers are a) 46 N b) 0 since the engine does not need to provide any force
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  • #2
Rather than use the product rule, why not tackle it directly using F = d(mv)/dt. Figure out the rate at which the momentum of the transferred coal must be increased to maintain the speed of the faster barge. (Every minute a certain mass of coal is transferred. What is that coal's initial momentum? By how much must its momentum be increased?)
  • #3
Like I said, that's how I solved this problem :)
But the thing is I WANT to use the product rule, it's a habit I kinda got used to so I can verify my answers on quizzes and tests by solving the same problem differently.
  • #4
Like I said, that's how I solved this problem :)
D'oh! Sorry about that. :uhh:
  • #5
F=dp/dt always applies. Well, the boat is not to change speed, so you know that dv/dt=0. F_engine + F_vdm/dt = 0 should give the right answer too.
  • #6
F_engine + F_vdm/dt = 0 should give the right answer too.

I'm not gettin how you came up with this formula, can you go deeper into this?
Isn't it F_engine = v*dm/dt?
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  • #7
Right, because you want there to be an equilibrium of no total force. You know that there is a force from the change in momentum by the mass changing, and you want the force from the engine to be opposite that force so that both the force sum to zero. That's your equation says, except not because you are missing a minus sign.
  • #8
Actually F_engine= v*(dm/dt) is equal to F_engine+v*(dm/dt)=0 for the barge losing mass since dm/dt will be a negative quantity.
  • #9
It's been a while but I've been busy and forgot about this thread :p

First let me explain my equations.
My system will be focused on the faster barge hence leaving out the coal that is yet to come, and I will choose my reference relative to the water which is still.

For the momentum of Pf, f for fast, Pf = mf * vf. For the initial condition F_engine = friction since it is at constant velocity. Applying Newton's Second Law F_engine + delta F_engine - friction = vf * dmf/dt for constant velocity. Therefore, delta F_engine = vf * dmf/dt.

The problem here is that vf is measured relative to the water, which gives the wrong answer. This equation will be correct if I plugged in the relative velocities between the barges. Since force should not differ according to different inertial frames there must be something wrong with this reasoning. Also F = ma + vdm/dt, the only thing that would be different is v measuring from different reference frames would result in a different force, this seems strange to me.
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