Solving Two Barges moving in the Same Direction

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Homework Help Overview

The problem involves two barges moving in the same direction at different speeds, with coal being transferred from the slower barge to the faster one. The objective is to determine the additional force required by the engines of each barge to maintain their speeds despite the transfer of mass.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of force equations, particularly the relationship between force, mass, and velocity. There is an exploration of how to handle the changing mass of the barges due to the coal transfer.

Discussion Status

Some participants have offered alternative perspectives on the equations being used, suggesting a need for a more nuanced understanding of the relationship between mass and velocity in this context. The discussion is ongoing with multiple interpretations being explored.

Contextual Notes

There is mention of specific assumptions, such as the frictional forces being independent of the barges' weight and the nature of the coal transfer. The problem also involves considerations of unit conversions and the implications of varying mass on force calculations.

patelkey
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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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