|-------\(adsbygoogle = window.adsbygoogle || []).push({});

| Mf > -> vf

|-------/

|-------\

| Ms > -> vs

|-------/

Two long barges are moving in the same direction in still water, one with a speed of 10 km/h and the other with speed of 20 km/h.

While they are passing each other, coal is shoveled from the slower to the faster one at a rate of 1000 kg/min.

How much additional force must be provided by the driving engines of

(a) the fast barge and (b) the slow 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 mass of the barges.

(Fundamental of Physics 6th ed. pg. 191 #47)

Looking at the problem, I think the fast barge's driving engine must provide additional force because it gets additional mass but the slow barge's driving engine must lose its force because it loses its mass.

Therefore, I solve this by using fact that F = (delta)p / (delta)t = dp / dt.

For the fast barge,

Ff = dp / dt

Ff = { [ (Mf + dm) * vf ] - [ Mf * vf ]} / dt

Ff = vf * ( dm / dt )

Ff = [ 20 * ( 5 / 18 ) ] m/s * [ 1000 / 60 ] kg/s

Ff = 92.60 N

For the slow barge,

Fs = dp / dt

Fs = { [ (Ms - dm) * vs ] - [ Ms * vs ] } / dt

Fs = - vs * ( dm / dt )

Fs = - [ 10 * ( 5 / 18 ) ] m/s * [ 1000 / 60 ] kg/s

Fs = - 46.30 N

Mf = fast barge's mass

Ms = slow barge's mass

vf = fast barge's speed

vs = slow barge's speed

Ff = fast barge's force

Fs = slow barge's force

Thus, my answers are

(a) the fast barge's driving engine must provide additional 92.60 N and (b) the slow barge's driving engine must decrease its force by 46.30 N.

But the answers in the key are (a) 46 N (b) none.

So, where's my fault? Is my reasoning wrong? Isn't the slow barges must lose its force because it loses its mass?

If instead I do the problem using isolated system consists of fast & slow barges,

F external = dp / dt

F external = { [ (Mf + dm) * vf + (Ms - dm) * vs ] - [ (Mf * vf) + (Ms * vs) ] } / dt

F external = ( vf - vs ) * (dm / dt)

F external = [ ( 20 - 10 ) * ( 5/18 ) ] m/s * [ 1000 / 60 ] kg/s

F external = 46.30 N

But this answer doesn't tell me which engine provides this force.

And I still on my previous reasoning that the slow barge's engine must lose some of its force.

Thank you very much for helping me ^_^

**Physics Forums - The Fusion of Science and Community**

# A system with varying mass - two long barges

Have something to add?

- Similar discussions for: A system with varying mass - two long barges

Loading...

**Physics Forums - The Fusion of Science and Community**