0
$\begingroup$

Let say there are two pistons, $p_1$ and $p_2$ such that $p1$ has lesser area at its base and $p_2$ has larger area at its base. Then, by pascals law(pressure remains same everywhere on surface bounding fluid) we get more force on $p_2$ than applied upon $p_1$. It should mean that acceleration $(p_1)$ $<$ acceleration $(p_2)$ when mass of both is same. But why this doesn't happen as predicted here?

enter image description here

$\endgroup$

4 Answers 4

1
$\begingroup$

I guess there is a gravity, acting on pistons? If yes, the accelerations of pistons really will be different. Small piston will go down, large piston will go up, not only accelerations will be different, but even directions of accelerations will be different.

It would not be quote difficult to calculate actual accelerations in this case, because the system is not static any more.

If the system is stationary (or everything moves very slowly) there must be some other forces acting on pistons. Total of these forces should result in 0 acceleration of each piston.

$\endgroup$
1
  • $\begingroup$ You can neglect that gravity kind of thing $\endgroup$
    – Spidy
    Sep 27, 2021 at 13:07
1
$\begingroup$

Pascal's Law applies to the initial and final static states of the two pistons. It is not valid when the fluids are accelerating.

The assumption in the law is that the pressure at any level in the fluid is the same in all directions. For the fluid to accelerate in the connecting tube (B in the figure) there needs to be a pressure differential along the tube.

Hope this helps.

$\endgroup$
2
  • $\begingroup$ Would that pressure be still there in moving fluid? I am thinking if this pressure is applying pressure at my brain and creating confusion $\endgroup$
    – Spidy
    Sep 27, 2021 at 13:16
  • $\begingroup$ For the fluid to accelerate, there needs to be pressure difference. $\endgroup$
    – Bob D
    Sep 27, 2021 at 13:22
1
$\begingroup$

Assume the pistons are light enough so that the gravity on them can be neglected.

When they are in equilibrium, we can see that the liquid pressure from bottom is equal to the atomsphere pressure from above.

Now, we want both of them to accelerate under the liquid pressure. This can be done by removing the atomsphere pressure or by increasing the liquid pressure (e.g. press the pistons hard and then release them suddenly).

Then, at the releasing point, the pistons would accelerate according to your description. But the acceleration will stop as long as the liquid is no longer pressed.

Therefore I think the described phenomenon could happen, but not very usual to observe. Because it is hard to make a continuous acceleration.

$\endgroup$
3
  • $\begingroup$ Thanks! I also thought similar to this. F2 is not continuously applied on p2 since the base of p2 looses contact when accelerated more than fluid $\endgroup$
    – Spidy
    Sep 27, 2021 at 13:06
  • $\begingroup$ "Therefore I think the described phenomenon could happen, but not very usual to observe. Because it is hard to make a continuous acceleration" - No, I don't think so. Don't you think it would be a failure of law of conservation of energy. You can assume a small time at which you think the p2 is accelerate and in that time interval we will get a violation of law since the output is more $\endgroup$
    – Spidy
    Sep 27, 2021 at 13:13
  • 1
    $\begingroup$ I'm not sure how do you define the input and output works. If you are meaning the process of pressing p1 to accelerate p2, I would like to notice you that the acceleration of p1 is determined by the difference of external force and liquid pressure. In this case the acceleration of p1 and p2 is not directly related. $\endgroup$
    – Siqi Wu
    Sep 27, 2021 at 13:27
0
$\begingroup$

This is simple $A_1 v_1 = A_2 v_2$

Differentiating $A_1 a_1 = A_2 a_2$

So $a_2\lt a_1$ since $A_2 \gt A_1$

$\endgroup$
2
  • $\begingroup$ Thiss is not the explanation rather it is a derivation using the fact that V•A must remain same. But the question is about why the acceleration or velocity becomes less in terms of Force acting on them $\endgroup$
    – Spidy
    Sep 28, 2021 at 1:35
  • $\begingroup$ Question is " since force on P2 is more than P1 it's acceleration should be more than P1"and I think what I wrote clearly explain why it is not. $\endgroup$
    – Steve
    Sep 28, 2021 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.