Question

In fig. 9-70, two long barges are moving in the same direction in still water, one with a speed of 10 km/hand the other with a speed of 20 km/hWhile 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 faster barge and (b) the slower barge if neither is to change the speed? Assume that the shoveling is always perfectly side-ways and that the frictional forces between the barges and the water do not depend on the mass of the barges.

Short answer

a) The additional force that must be provided by the driving engines of the faster barge if neither is to change speed is, F = 46 N

b) To maintain a constant speed, the slower barge does not require any additional force.

Step-by-step solution

Listing the given quantities

Speed of the two barges is$v_1=10\mathrm{km}/\mathrm{h},{\mathrm{v}}_2=20\mathrm{km}/\mathrm{h}$

Coal is shoveled from barge 1 to barge 2 with rate$R=1000\mathrm{kg}/\min$

Understanding the concept of law of conservation of momentum

We can find the additional force that must be provided to barge 2 using Newton’s second law which relates force with momentum. From the given information in the problem about frictional force we can guess theadditional force that must be provided by the driving engines of the slower barge (1) if neither is to change speed.

Formula:

$F=\frac{dP}{dt}$

Calculation of speed of two barges

Speed of the two barges is

$$\begin{gathered} v_1=10\mathrm{km}/\mathrm{h} \\ =10\times \frac{5}{18} \\ =2.78\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_2=20\mathrm{km}/\mathrm{h}\times \frac{5}{18} \\ =5.56\mathrm{m}/\mathrm{s} \\ R=1000\mathrm{kg}/\min \\ =\frac{1000}{60} \\ =16.67\mathrm{kg}/\mathrm{s} \end{gathered}$$

To maintain the constant speed after shoveling the coal from the slower barge 1 to the faster barge 2, the additional force should be provided to barge 2.

According to Newton’s second law,

$$\begin{gathered} F=\frac{dP}{dt} \\ F=\frac{d\left(mv\right)}{dt} \\ F=\frac{d\left(mv\right)}{dt} \\ F=\frac{dm}{dt}\left(v_2-v_1\right) \\ =16.67\left(5.56-2.78\right) \\ =46\mathrm{N} \end{gathered}$$

Therefore, the additional force that must be provided by the driving engines of the faster barge (2) if neither is to change speed is 46 N

(b) Explain the reasoning

The coal is shoveled in y direction from barge 1, and there is no frictional force acting on it in x direction.

Therefore, to maintain a constant speed, the slower barge does not require any additional force.