Work done by Gravity on a liquid

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SUMMARY

The discussion focuses on calculating the work done by gravity when liquid levels equalize in two connected cylindrical vessels. The initial heights of the liquid are 1.56 m and 0.854 m, with a liquid density of 1.3 x 10^3 kg/m^3 and a base area of 0.0004 m^2. Participants clarify that the change in potential energy should be calculated using the center of mass of the liquid columns, leading to the formula for work done by gravity as W = (density * g * area * (h1 - h2)²) / 4. The correct answer for the work done is +0.635 J, emphasizing the importance of sign conventions in integral calculations.

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Saladsamurai
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Two identical cylindrical vessels with their bases at the same level each contain the same liquid of density 1.3*10^3 kg/m^3. the area of each base is .0004 m^2.

In one vessel the height if the liquid is 1.56m and in the other 0.854. Find the work done by gravity in equalizing the levels when the vessels are connected.Well I tried to use plain old W=F*d where F=mg=density*V*g and d=the height of the higher liquid minus the equalized level (which I assumed was the average=1.207).

So I used W_g=\rho*Vg\Delta h=\rho*A*h_{higher}g\Delta h

=1.03*10^3(.0004)(1.56)(9.8)(1.56-1.207)

but as you probably know, this is far from correct. What is my conceptual error here?

Casey
 
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Do I need to consider pressures?...I suck at this chapter:smile:
 
Well, have you accounted for all changes in (potential) energy? What about the other cylinder?
 
Well... in the container...hmmm...it looks like some negative work will be done by gravity. Maybe that is my error...let me make some calculations really quick...

Would work done by the lower liquid be m_l*g*\Delta h=\rho*A*h_{lower}*g(\Delta h)

\Rightarrow W_g=1.03*10^3(.0004)(.854)(9.8)(.854-1.207)

...nevermind that is not working either.
 
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This isn't making any sense to me. If it is just work gone by gravity I don't see why it isn't just

W_g+W'_g Where W is the work done by gravity on the liquid going from height 1.56-->1.207
and W' is the work done by gravity on the liquid going from height .854-->1.207... :mad::mad::mad:
 
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Saladsamurai said:
Would work done by the lower liquid be m_l*g*\Delta h=\rho*A*h_{lower}*g(\Delta h)

No no, you don't need to go for Bernoulli.
Note that as liquid is enetring/leaving the cylinder, mass of the liquid in it doesn't remain constant. So, the formula you have used for change in potential energy is wrong.
Instead, use \DeltaPE = (mgh)final - (mgh)initial, for each of the cylinders.
 
saket said:
No no, you don't need to go for Bernoulli.
Note that as liquid is enetring/leaving the cylinder, mass of the liquid in it doesn't remain constant. So, the formula you have used for change in potential energy is wrong.
Instead, use \DeltaPE = (mgh)final - (mgh)initial, for each of the cylinders.

So for the cylinder with initial height 1.56 (mgh)_f-(mgh)_i=? I am confused since m=V*density and V=A*h...so what do I use for m in mgh_f...
 
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Screw this problem.
 
There isn't by any chance an integral involved here? Is there?

Casey
 
  • #10
:mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad:

I am going to sleep:smile:
 
  • #11
No, you won't need integral.
 
  • #12
Hmm.. gone to sleep? Hope, you will solve it in your dreams!
Anyways, if in case you are not able to, to make you start your day happily, here is the soltion: (Both methods are almost same.)
Method I:
Let bottom of the cylinders be our reference line for height. Let h1 = 1.56 m, h2 = 0.854 m
Let density of liquid be d and base of the cylinders be of cross-sectional area A.
Initial PE of first cylinder = mgh = (d*(A*h1))*g*(h1/2).
Note that, centre of mass will be at a height h1/2. {This is where you were going wrong.}
For the second cylinder, PE = (d*(A*h2))*g*(h2/2).

Finally, the equilibrium height will be (h1 + h2)/2 and centre of mass for each of them would be at (h1 + h2)/4.
So, final PE = 2*(d*(A*((h1 + h2)/2))*g*((h1 + h2)/4).
Change in PE = negative of the work done by gravity
Thus, work done by gravity = d*g*A*(h1 - h2)²/4
I hope you will be able to substitute values.
 
  • #13
Thanks Saket! I am still not so sure why I use h/2 in mgh?

Casey
 
  • #14
Saladsamurai said:
Thanks Saket! I am still not so sure why I use h/2 in mgh?

Casey

Consider a stick of length L. Where will its centre of mass be? Mid-point, right? This is at a distance of L/2 from either end.
Similarly in the given problem, we have water column of height, say h. So, centre of mass will be at a height of h/2 .. isn't it? {You have studied centre of mass, isn't it?}
 
  • #15
saket said:
Consider a stick of length L. Where will its centre of mass be? Mid-point, right? This is at a distance of L/2 from either end.
Similarly in the given problem, we have water column of height, say h. So, centre of mass will be at a height of h/2 .. isn't it? {You have studied centre of mass, isn't it?}

Yes, I have studied center of mass, but in recent chapters I have not found any use for it.

Actually, I am still confused as to WHY it is used. I know what and where the center of mass is.

If if the only potential energy here comes from weight and weight is m*g
and m is \rho*V(g) ...I just don't see why center of mass comes into play?

Thanks,
Casey

Also: Unless I am screwing up the calculator, I cannot get your solution to yield the correct answer of .635 J, is that what you got?

and why couldn't you do this as an integral with weight as a funciton of height?
 
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  • #16
Okay. So I went ahead an used an integral. I got the correct numbers, but the wrong sign. I used: W=int F(height) dh

W_{total}=W_{left}+W_{right}

\Rightarrow W_t=\rho gA\int_{1.56}^{1.207}hdh+\rho gA\int_{.854}^{1.207}hdh
=-0.635 J

But the correct answer is +.635 J where is my sign wrong? Do I need to switch my bounds?

Casey
 

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