Extension of spring in a system with two pistons

[Bling Fizikst]

Homework Statement

Refer to the image

Relevant Equations

Refer to the image

Let's say the upper piston goes down by $y_1$ and the lower piston goes down by $y_2$ after the block is suspended \ By volume conservation $s_1 y_1=s_2 y_2$ Let the pressure at the location of the upper piston be $P_c=\frac{ky_1}{s_1}$ Pressure at the lower piston : $P_a=P_c+\rho g(h−y_1+y_2)$ At point of equilibrium , $mg=P_a s_2$ Doing the calculations gives : $y_1=\frac{mg s_1−\rho hgs_1 s_2}{ks_2+\rho g s_1(s_1−s_2)}$ which is the wrong answer . I am not sure where i went wrong .

 

[haruspex]

Let the pressure at the location of the upper piston be $P_c=ky_1s_1$

I don’t understand how you get that. k has dimension$MT^{-2}$, $s_1y_1$ is a volume, giving $ML^3T^{-2}$, not $ML^{-1}T^{-2}$ (pressure).

 

[Bling Fizikst]

I don’t understand how you get that. k has dimension$MT^{-2}$, $s_1y_1$ is a volume, giving $ML^3T^{-2}$, not $ML^{-1}T^{-2}$ (pressure).

sorry that was a typo . i edited it now

 

[haruspex]

sorry that was a typo . i edited it now

I think you are mixing up total pressures with changes in pressure.
If the upper piston descends $y_1$, what is the change in pressure the piston exerts on the water? What is the change in water depth?

 

[Bling Fizikst]

If the upper piston descends by $y_1$ then the corresponding change in pressure should be $\rho g y_1$ ? Change in water depth in the entire system due this change would be $$h-(h-y_1+y_2) = y_1-y_2$$ where $y_2$ can be written in terms of $y_1$ from the volmue conservation equation ?

 

[haruspex]

If the upper piston descends by $y_1$ then the corresponding change in pressure should be $\rho g y_1$ ?

No, I asked what the change in pressure exerted by the upper piston is. You had the expression for $P_c$, but you called it the pressure, not the change in pressure. You also need to consider the sign.

Change in water depth in the entire system due this change would be $$h-(h-y_1+y_2) = y_1-y_2$$

Again, be careful with signs.

 

[Bling Fizikst]

No, I asked what the change in pressure exerted by the upper piston is. You had the expression for $P_c$, but you called it the pressure, not the change in pressure. You also need to consider the sign.

Again, be careful with signs.

So that means the change in pressure exerted by the upper piston should be : $$\frac{k y_1}{s_1} +\rho gy_1$$ And the change in the water depth of the system should be : $$h_f-h_i = (h-y_1+y_2)-h = y_2-y_1$$

 

[haruspex]

So that means the change in pressure exerted by the upper piston should be : $$\frac{k y_1}{s_1} +\rho gy_1$$

Why the second term? It's only the change in the spring length that alters the pressure exerted by the piston. But $y_1$ is an increase in length of the spring, so the pressure it exerts should reduce.

And the change in the water depth of the system should be : $$h_f-h_i = (h-y_1+y_2)-h = y_2-y_1$$

Yes.

 

[Lnewqban]

Doing the calculations gives : $y_1=\frac{mg s_1−\rho hgs_1 s_2}{ks_2+\rho g s_1(s_1−s_2)}$ which is the wrong answer .

Could you show the correct answer?

Because the fluid is incompressible, the new force on the spring will be the addition of the weight of the block plus the unchanged weight of the water.
Our problem is that we don't know the value of the former.

Could you draw four FBD's for the pistons in both conditions?

 

[Bling Fizikst]

I am getting a contradiction even after trying to do this fundamentally

The answer key says : $$\frac{mgS_1}{kS_2-\rho g S_1 (S_1-S_2)}$$

 

[Lnewqban]

Please, consider that for initial conditions:
Pressure at B should be equal to Patm (the bottom piston is at balance).
Pressure at A should be less than pressure at B (with the help of the stretched spring, the top piston is pulling the weight of the water column underneath it).

 

[haruspex]

I am getting a contradiction even after trying to do this fundamentally
View attachment 346105
The answer key says : $$\frac{mgS_1}{kS_2-\rho g S_1 (S_1-S_2)}$$

I cannot follow your reasoning from the various expressions scattered around that picture.

Let's go back to post #7.
I queried the $\rho gy_1$ term in the first equation. What is your answer to that?
I agreed with the second equation.
What do you then get for the pressure change just above the lower piston? What force change does that correspond to?
What equation relates that to mg?

 

[Bling Fizikst]

I cannot follow your reasoning from the various expressions scattered around that picture.

Let's go back to post #7.
I queried the $\rho gy_1$ term in the first equation. What is your answer to that?
I agreed with the second equation.
What do you then get for the pressure change just above the lower piston? What force change does that correspond to?
What equation relates that to mg?

Here is my entire thought process , please do help me with the misconceptions that i have : Let's say initially the atmospheric pressure on the upper piston is $P$ and its the same just below it as well . Now , if the upper piston goes down by $y_1$ then the pressure on the upper piston increases to $P+\rho g y_1$ . This generates a force of $(P+\rho g y_1 - P)s_1$ which is to be balanced by the spring force , $ky_1$ . Giving the equation : $$ ky_1 = \rho g y_1$$ which is contradictory . For the lower piston , the change in pressure times the corresponding corss sectional area should be equal to $mg$ : $$ (P+\rho g h - (P+\rho g (h-y_1+y_2)) = mg $$

 

[haruspex]

Let's say initially the atmospheric pressure on the upper piston is $P$ and its the same just below it as well .

What about the spring? It is not necessarily in its relaxed state. That is why it is easier to work in terms of changes to forces and pressures.

Now , if the upper piston goes down by $y_1$ then the pressure on the upper piston increases to $P+\rho g y_1$ .

Eh? Why? The piston does not become immersed in water.
The spring changes length, though.

 

[Bling Fizikst]

What about the spring? It is not necessarily in its relaxed state. That is why it is easier to work in terms of changes to forces and pressures.

Eh? Why? The piston does not become immersed in water.
The spring changes length, though.

Ahh i see . I redid the calculations now . If the upper piston went down by $y_1$ then the change in pressure exerted by the upperpiston would be $$\frac{ky_1}{s_1}$$ If the lower piston goes down by $y_2$ , then we can write the forces corresponding to change in pressures : $$ \left(\frac{ky_1}{s_1}+\rho g (h-y_1+y_2) - \rho g h\right)s_2 =mg$$ Doing the calculation gives : $$ y_1= \frac{mgs_1}{ks_2+\rho g s_1 (s_1-s_2)}$$ There still seems to be some slight error as there is a '-' in the denominator in the answer key .

 

[haruspex]

Ahh i see . I redid the calculations now . If the upper piston went down by $y_1$ then the change in pressure exerted by the upperpiston would be $$\frac{ky_1}{s_1}$$ If the lower piston goes down by $y_2$ , then we can write the forces corresponding to change in pressures : $$ \left(\frac{ky_1}{s_1}+\rho g (h-y_1+y_2) - \rho g h\right)s_2 =mg$$ Doing the calculation gives : $$ y_1= \frac{mgs_1}{ks_2+\rho g s_1 (s_1-s_2)}$$ There still seems to be some slight error as there is a '-' in the denominator in the answer key .

Two sign errors.
When a spring extends, which way does the force it exerts change?
To balance the added mass at the bottom, which way does the force at the lower piston need to change?

 

[Lnewqban]

Doing the calculation gives : $$ y_1= \frac{mgs_1}{ks_2+\rho g s_1 (s_1-s_2)}$$ There still seems to be some slight error as there is a '-' in the denominator in the answer key .

Let’s assume for a second that the container is perfectly cylindrical and that we have identical cross-section areas for the upper and lower pistons.
The height of the water column h should remain the same, regardless the magnitude of m.

If you make $S_1=S_2=S$;
then, $y_1=y_2=y$
and the equation gets simplified to be:

$$ y= \frac{mg}{k}$$

Because the shape of the vessel in our problem, we have certain mechanical advantage between both pistons.
The upper piston should slide at a slower rate than the lower piston.

Because of the above, $y_1<y_2$
and our equation will need to have a reduced value in the denominator, which will be achieved with “a '-' in the denominator in the answer key.”