# Question on fluid and basic mechanics

The question given is :

Two vertical cylindrical vessels A and B of horizontal cross–sectional areas S and 2S are connected at their bottoms with a horizontal tube of cross sectional area 0.5S. An amount of water is trapped in the vessels under leak proof pistons, one in each cylindrical vessel. The pistons are connected with a light inextensible thread that passes over an ideal pulley as shown in the figure. The pulley is pulled upwards with a constant velocity v . The vessels are rigidly affixed on the horizontal floor. What will be the following motion of the pistons and with what velocity?

so i had thought both the pistons will move upwards with some velocity but it is given one goes down and other up. i fail to understand how it is possible. then won't the pressure at the bottom of the container be no longer uniform? any insights will be helpful.

then won't the pressure at the bottom of the container be no longer uniform?

The pressure at bottom need not and will not be uniform because this is a fluid dynamic situation and not a fluid static situation . Different points at the same horizontal level will be have a certain velocity and hence will be at different pressures conversely a pressure difference is necessary for fluids to flow . You can even try calculating pressure at some points using Bernoulli's equation .

What will be the following motion of the pistons and with what velocity?

Let $$\vec V_a$$ and$$\vec V_b$$ be the velocity of Piston A and B respectively and $$\vec v$$ be velocity of the pulley .

Volume of the liquid is constant .Thus , volume of liquid rise in A must be equal to volume of liquid fall in the B.

$$\therefore \ {\vec v_a}=2{\vec v_b}$$

The difference in velocities of the Pistons will be the twice velocity of pulley (pulley constraint).

$$\vec v_a- \vec v_b=2\vec v$$

Solve these two equations and you have the velocities.

Both the pistons cannot move up at the same time, since, the fluid is assumed to be of constant density, and the pistons are airtight. If both pistons move up, volume of fluid will increase, and that cannot occur.

It's easier to understand what is happening if we re-arrange the diagram.

In my diagram the pistons are rectangular and 1 unit thick into the page, therefore the volumes of the pistons are proportional to the areas in my drawing.

The right hand piston has area $$2S$$ , and when it moves $$x$$ to the left, it pushes a volume of water $$V=x.2S$$ into the left piston.

The left hand piston has area $$S$$ , and the volume displaced has to be the same as the right hand piston, therefore it has to move twice as far as it has half the area $$V=2x.S$$

Getting back to the original problem , when we pull up on the pulley, the small piston rises $$2x$$ while the large one sinks $$-x$$. the motion of the pulley axle will be the average of these or $$z_p=\frac{(-x + 2x)}{2} = 0.5x$$ , so the small piston moves up at twice the speed of the pulley , the large piston moves down at the same speed as the pulley (but in the opposite direction).

if you release the pulley at some later time, the small piston will descend and the large one rise, until both are at the same level (this only works if the pistons are vertical, if they are horizontal they will stay where you left them). If you release the pulley and invert the apparatus, I think it will be unstable, whichever piston is lower will continue dropping until the other one hits the stop, it's a siphon action.

Let A have an upward velocity $$v_A$$ and B have a downward velocity $$vB$$

By constraint equation $$-2v+v_A-v_V$$. By equation of continuity $$S*v_A=2S*v_B=0.5*v_C$$. Solving these equations you'll get the answer.