Solving Two Barges moving in the Same Direction

AI Thread Summary
Two barges are moving in the same direction at speeds of 25 km/h and 50 km/h, with coal being transferred from the slower to the faster barge at a rate of 900 kg/min. To maintain their speeds, additional force must be calculated for both barges. The equation F=change in M/change in T is referenced, but it's noted that this approach may not be correct. A suggestion is made to use calculus, specifically the product rule, to derive the force needed as mass varies while velocity remains constant. The discussion emphasizes the need for accurate calculations to solve the problem effectively.
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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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