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A system with varying mass - two long barges

by Eus
Tags: barges, mass, varying
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Eus
#1
Jan23-05, 09:03 PM
P: 94
|-------\
| 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 ^_^
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Doc Al
#2
Jan24-05, 06:33 PM
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Quote Quote by Eus
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.
It's not the losing or gaining of mass, but the change in momentum that matters.
Therefore, I solve this by using fact that F = (delta)p / (delta)t = dp / dt.
Right! The force will equal the rate at which each barge is required to change the momentum of the shoveled coal.

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
Your error is using the momentum of the coal where you should use the change in momentum. Realize that the coal shoveled onto the fast barge starts with a speed of 10 km/h, not zero.

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
Again, what is the change in momentum of the shoveled coal? Initial speed is 10 km/h... final speed is 10 km/h... change is zero!

Here's another example to test your thinking: What if both barges traveled side by side at the same speed of 10 km/h. Is any additional force required to maintain the speed of the barges if one shovels coal onto the other? Does the coal change momentum?
Eus
#3
Jan24-05, 10:26 PM
P: 94
Hi Ho! ^_^

Mmmm... I think I got it a little.

About the another example, the answers are
(a) There's no additional force required to maintain the speed of the barges.
Because there's no change in momentum.
(b) The coal doesn't change momentum.
Am I right?

To make me sure that now I've the correct way of reasoning, could you please check this?
Mmmm... two long identical barges are moving in the same direction in still water.
Suppose there are barge A which is empty and barge B which is full of coal.
Barge A's speed is 20 km/h and barge B's speed is 10 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 barge A's speed becomes 25 km/h and barge B's speed becomes 2 km/h?
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.

Now I calculate this according to the change in momentum during dt for system consists of barge A and coal.

Ff = { [ vf2 * ( Mf + dm ) ] - [ vf1 * Mf + vc * dm ] } / dt
Ff = { [ (vf2 - vf1) * Mf ] + [ (vf2 - vc) * dm ] } / dt
Ff = (vf2 - vc) * (dm/dt)
Ff = [(25 - 10) * (5/18)] m/s * [1000 / 60] kg/s
Ff = 69.44 N ... answer to (a). Am I right?

vc = coal's speed
vf2 = barge A's final speed after dt
vf1 = barge A's initial speed
Mf = barge A's mass
dm = coal's mass which is shoveled
Ff = additional force by the driving engine of the fast barge

Then I calculate this according to the change in momentum during dt for system consists of barge B and coal.

Fs = { [ vs2 * ( Ms - dm ) ] - [ vs1 * Ms + vc * dm ] } / dt
Fs = { [ (vs2 - vs1) * Ms ] - [ (vs2 + vc) * dm ] } / dt
Fs = -(vs2 + vc) * (dm/dt)
Fs = -[(2 + 10) * (5/18)] m/s * [1000 / 60] kg/s
Fs = -55.55 N ... answer to (b). Am I right?

vc = coal's speed
vs2 = barge B's final speed after dt
vs1 = barge B's initial speed
Ms = barge B's mass
dm = coal's mass which is shoveled
Fs = additional force by the driving engine of the slow barge

Thank you very much for your help ^_^

Doc Al
#4
Jan25-05, 12:22 PM
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A system with varying mass - two long barges

Quote Quote by Eus
About the another example, the answers are
(a) There's no additional force required to maintain the speed of the barges.
Because there's no change in momentum.
(b) The coal doesn't change momentum.
Am I right?
Yes, you are right.

To make me sure that now I've the correct way of reasoning, could you please check this?
Mmmm... two long identical barges are moving in the same direction in still water.
Suppose there are barge A which is empty and barge B which is full of coal.
Barge A's speed is 20 km/h and barge B's speed is 10 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 barge A's speed becomes 25 km/h and barge B's speed becomes 2 km/h?
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.
I am going to assume that this is exactly the same situation as before, but that the barges just have different speeds. Note that since the speeds of the barges do not change, the masses of the barges do not enter into any calculation. The only thing that needs to be considered is the changing momentum of the coal.

Now I calculate this according to the change in momentum during dt for system consists of barge A and coal.

Ff = { [ vf2 * ( Mf + dm ) ] - [ vf1 * Mf + vc * dm ] } / dt
Ff = { [ (vf2 - vf1) * Mf ] + [ (vf2 - vc) * dm ] } / dt
Ff = (vf2 - vc) * (dm/dt)
Ff = [(25 - 10) * (5/18)] m/s * [1000 / 60] kg/s
Ff = 69.44 N ... answer to (a). Am I right?
No. The fast barge must change the speed of the coal from 2 km/h to 25 km/h, so:
Ff = [(25 - 2) * (5/18)] m/s * [1000 / 60] kg/s

Then I calculate this according to the change in momentum during dt for system consists of barge B and coal.

Fs = { [ vs2 * ( Ms - dm ) ] - [ vs1 * Ms + vc * dm ] } / dt
Fs = { [ (vs2 - vs1) * Ms ] - [ (vs2 + vc) * dm ] } / dt
Fs = -(vs2 + vc) * (dm/dt)
Fs = -[(2 + 10) * (5/18)] m/s * [1000 / 60] kg/s
Fs = -55.55 N ... answer to (b). Am I right?
No. Once again, does the slow barge change the speed of the coal? No! So no additional force needs to be applied to the slow barge.
Eus
#5
Jan25-05, 07:00 PM
P: 94
Aaah, now I see the picture! ^_^
I really got it! ^.^
The things below are the step how I grasped the idea.

Note that since the speeds of the barges do not change, the masses of the barges do not enter into any calculation.
In the last question, in my mind the barges do change in speed during the shoveling process. Barge A accelerates and barge B decelerates.

The only thing that needs to be considered is the changing momentum of the coal.
Yup! In the first thread, my focus was always on the barges and its mass.

That's it!

Thank you very much for your kindness! ^_^
Doc Al
#6
Jan25-05, 07:15 PM
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Quote Quote by Eus
Aaah, now I see the picture! ^_^
I really got it! ^.^
I'm glad you've got it. To test your understanding, solve this problem: Instead of the slow boat (10 km/h) shoveling coal onto the fast boat (20 km/h), have the fast boat shovel the coal onto the slow boat. What changes?
Eus
#7
Jan26-05, 06:18 AM
P: 94
Hi Ho! ^_^

Hehehe... okay! I'll solve it! ^^v

The fast barge doesn't change the speed of the coal.
Thus, Ff = 0 N

The slow barge must change the speed of the coal from 20 km/h to 10 km/h.
Therefore,
Fs = { [ vs * ( Ms + dm ) ] - [ vs * Ms + vc * dm ] } / dt
Fs = (vs - vc) * (dm / dt)
Fs = [ (10 - 20) * (5/18) ] m/s * (1000 / 60) kg/s
Fs = -46.30 N

Am I right?
Doc Al
#8
Jan26-05, 09:55 AM
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Exactly right! What changes is the direction of the force, since the receiving barge must deaccelerate the incoming coal. Good job.


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