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Hoping somebody could give me a hand with this. Unfortunately I haven't taken physics yet, my background is in math/computer science.

I'm studying computational science and the problem is essentially asking what is the force needed to maintain the velocity of a train with while its raining? (water is accumulating, mass of train is increasing).

I'm having trouble figuring out how to relate the changing mass.

In computational science I'm working on this problem as a matrix representation: $$ F = ma \quad \rightarrow \quad f-Ku=Mu''$$ Where $u$ is the displacement vector. $M$ is the mass matrix. $f$ is the external force vector and $Ku$ the internal force vector.

Additional information: The rate the mass is changing is $r= \frac{dm}{dt}$

I'm not looking for just the answer; this is driving me nuts, I should be able to figure it out. But its just not been going well.

-------To break it down--------

train $T$ is traveling at a constant velocity $v$. Mass is a function of time and displacement. The force $F = ma = \frac{d}{dt}(m(u,t)\frac{du}{dt})= \frac{dm}{dt}\frac{du}{dt}$

I am looking for where $ma = 0 \text{ and } F \neq 0 $

Anyways, I'm hoping somebody could give me a nudge here, I don't know how I can represent this properly. Thanks

John Rennie
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spun
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    The equation $F_{total}=ma$ is a special case of the more general $F_{total}= \frac{d(mv)}{dt} = m\frac{dv}{dt}+v\frac{dm}{dt}$ – danimal May 08 '15 at 00:21
  • Why is the increasing mass relevant here? The force is only necessary to accelerate the accumulating rain to the speed of the car. The mass of the car doesn't matter. – BowlOfRed May 08 '15 at 00:30
  • @BowlOfRed Mass increases with velocity, a computational model would have diminished accuracy. – spun May 08 '15 at 00:41
  • Assuming a steady velocity and steady rainfall, the rain quickly accumulates to a uniform sheen of water which covers the moving train. Mass of the moving train doesn't increase any further. So, you shouldn't have to deal with dm/dt. – Ernie May 08 '15 at 00:51
  • My comment is not right. Acceleration isn't instantaneous, so you need t to get the train back up to V after the rain covers it. – Ernie May 08 '15 at 01:01
  • @Ernie Good point. I'm also sorry I wasn't being explicit, there's an open bed container, the water is accumulating. Its just a practice scenario, a lot of assumptions. I imagine this is analogous to a car expending gas over time. $-dm/dt$ – spun May 08 '15 at 01:06
  • Imagine a funnel over the train. The funnel catches the rain and accelerates the water to $v$. Because the water dropping into the car is the same speed as the train, no forces accelerate the train and the total mass in the car is not relevant. Because the mass of the funnel does not change, a constant force is sufficient. – BowlOfRed May 08 '15 at 05:10
  • possible duplicate of Conservation of momentum when rain pours into a wagon – Kyle Kanos May 08 '15 at 10:34

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Use the fact that force is equal to the rate of change of momentum. The rain is accelerated to the same velocity as the train, so if you know the mass of rain per second you can calculate the momentum change per second. This is equal to the force required.

John Rennie
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