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A system with varying mass  two long barges 
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#1
Jan2305, 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 ^_^ 


#2
Jan2405, 06:33 PM

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P: 41,570

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? 


#3
Jan2405, 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 ^_^ 


#4
Jan2505, 12:22 PM

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P: 41,570

A system with varying mass  two long barges
Ff = [(25  2) * (5/18)] m/s * [1000 / 60] kg/s 


#5
Jan2505, 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. That's it! Thank you very much for your kindness! ^_^ 


#6
Jan2505, 07:15 PM

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P: 41,570




#7
Jan2605, 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? 


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