In the figure below, two long barges are moving in the same direction in still w
ID: 3894667 • Letter: I
Question
In the figure below, two long barges are moving in the same direction in still water, one with a speed of 10 km/h and the other with a 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 1150 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. (faster barge)
Explanation / Answer
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).
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).
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