Solving Two Barges moving in the Same Direction

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SUMMARY

The discussion centers on calculating the additional force required by two barges moving at 25 km/h and 50 km/h, respectively, while coal is transferred from the slower barge to the faster one at a rate of 900 kg/min. The key equation to use is F = d/dt(m*v), where m is the mass of coal being transferred and v is the constant speed of the faster barge. The conclusion reached is that the additional force needed to maintain speed is 0 N for the faster barge, as the momentum change due to the coal transfer does not affect its velocity.

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Homework Statement


Two long barges are moving in the same direction in still water, one with a speed of 25 km/h and the other with a speed of 50 km/h. While they are passing each other, coal is shoveled from the slower to the faster one at a rate of 900 kg/min. How much additional force must be provided by the driving engines of each barge if neither is to change speed? Assume that the shoveling is always perfectly sideways and that the frictional forces between the barges and the water do not depend on the weight of the barges.



Homework Equations




I tried using the equation F=change in M/change in T
and then multiplying that by (Va-Vb)


The Attempt at a Solution


I know that the second part of the question is 0 N I only need the first part.
 
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"F=change in M/change in T" isn't quite right.
Do you know calculus? If so, try using F = d/dt of (m*v).
Tricky units in this question.
 


Wouldn't F=(d/dt) (mv) just equal F=ma
 


It would if m was constant and v a variable. But in this case, m varies and v is constant. Using the product rule, right?
 

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