# Two Barges moving in the same direction (tried multiple times anyone understand first

1. Oct 18, 2009

### patelkey

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. The attempt at a solution
I know that the second part of the question is 0 N I only need the first part.

2. Oct 18, 2009

### Delphi51

Re: Two Barges moving in the same direction (tried multiple times anyone understand f

"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.

3. Oct 18, 2009

### patelkey

Re: Two Barges moving in the same direction (tried multiple times anyone understand f

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

4. Oct 18, 2009

### Delphi51

Re: Two Barges moving in the same direction (tried multiple times anyone understand f

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?