Hydrostatic force on identical containers filled with equal volumes of water

[zul8tr]

In the sketch are 2 identical containers filled with equal volumes of water. Which has a greater absolute vertical force only on the conical wall and by what relative amount?
Not a classroom or homework problem.

 

[.Scott]

We are also assuming that the fluid pressure at the top of each tank is zero.
Of course, the vertical force in one case is upward and in the other case downward.

Even though this is not a homework problem, I'm not going to give you the answer outright.
This is a calculus integration problem. There will be more absolute vertical pressure on one were the highest pressure is applied to the greater conical surface area.

Try to set up the equations.

 

[Chestermiller]

We are also assuming that the fluid pressure at the top of each tank is zero.
Of course, the vertical force in one case is upward and in the other case downward.

Even though this is not a homework problem, I'm not going to give you the answer outright.
This is a calculus integration problem. There will be more absolute vertical pressure on one were the highest pressure is applied to the greater conical surface area.

Try to set up the equations.

No integration is required to answer this problem. It is not a calculus integration problem.

 

[zul8tr]

I do not need help, basic laws of hydrostatics apply, think volume weights rather than pressures

 

[zul8tr]

No calculus required just basic hydrostatics some algebra and understanding the vertical forces in this example. Yes there is atmospheric pressure at the open top of A and B and as stated in the sketch both are sealed against water leaking out the bottom and both are secured at the bottom boundary against any movement .

 

[.Scott]

I agree. No calculus is required. Have you found the ratio yet? If not, have you found the volume yet?
Using "gage pressure" eliminates atmospheric effects.

 

[zul8tr]

Yes Answer known before posting for the vertical forces and the ratio

 

[.Scott]

In that case: (2R1+R2)/(2R2+R1)

 

[Chestermiller]

In that case: (2R1+R2)/(2R2+R1)

Shouldn't the answer involve the squares of the radii?

 

[.Scott]

Shouldn't the answer involve the squares of the radii?

Since we're doing ratios, I will drop the H and Pi terms.
The volume is (R2^3-R1^3)/(R2-R1)/3 = (R1^2+R1R2+R2^2)/3
Then we subtract out the pressure against the bottom: Either R1^2 or R2^2.
Again, since we're doing ratio's, I'll multiply through by 3:
We have:
2R1^2-R1R2-R2^2 = (2R1+R2)(R1-R2)
and
-R1^2-R1R2+2R2^2 = (2R2+R1)(R1-R2)

Finally, multiply through by (R1-R2).

No R^2 terms.

 

[zul8tr]

In that case: (2R1+R2)/(2R2+R1)

For now not saying you are correct but answer what was posed in my post based on what you derived:
"Which has a greater absolute vertical force only on the conical wall and by what relative amount?"

In your statement to drop the H and Pi terms would also need to drop the unit weight of water term to be complete statement for the problem.

 

[Chestermiller]

For now not saying you are correct but answer what was posed in my post based on what you derived:
"Which has a greater absolute vertical force only on the conical wall and by what relative amount?"

In your statement to drop the H and Pi terms would also need to drop the unit weight of water term to be complete statement for the problem.

Huh?

 

[Charles Link]

If I understand the diagram correctly, in the first diagram there are basically 3 Forces. The downward force of gravity $Mg=\delta V g$ where $\delta=$ density of water must balance( be equal in magnitude to the resultant of the two pressure forces).(And the volume $V$ has been correctly computed). The first pressure force is an upward $P_{bottom} A_{bottom}$, and the other one is a downward force supplied by the sides of the container, which is basically the unknown in this problem. (Both $P_{bottom}$ and $A_{bottom}$ are very readily computed. I don't think $P_{bottom}$ has been calculated as of yet). $\\$ The second part proceeds similarly, but it would help to solve the first part in its entirety before taking ratios. $\\$ An algebraic expression is needed for the downward force from the sides of the container, that can be obtained by solving the first paragraph above. $\\$ Additional note: I think the atmospheric pressure can be ignored in this problem, because it will not change the force that is computed that needs to be supplied by the structure of the containers. The atmospheric forces will balance out.

 

[Chestermiller]

If I understand the diagram correctly, in the first diagram there are basically 3 Forces. The downward force of gravity $Mg=\delta V g$ where $\delta=$ density of water must balance( be equal in magnitude to the resultant of the two pressure forces).(And the volume $V$ has been correctly computed). The first pressure force is an upward $P_{bottom} A_{bottom}$, and the other one is a downward force supplied by the sides of the container, which is basically the unknown in this problem. (Both $P_{bottom}$ and $A_{bottom}$ are very readily computed. I don't think $P_{bottom}$ has been calculated as of yet). $\\$ The second part proceeds similarly, but it would help to solve the first part in its entirety before taking ratios. $\\$ An algebraic expression is needed for the downward force from the sides of the container, that can be obtained by solving the first paragraph above. $\\$ Additional note: I think the atmospheric pressure can be ignored in this problem, because it will not change the force that is computed that needs to be supplied by the structure of the containers. The atmospheric forces will balance out.

It seems to me @.Scott had it nailed in post #10.

 

[.Scott]

"Which has a greater absolute vertical force only on the conical wall and by what relative amount?"

OK.
Given two cases where:
in the 1st case the diameter at the top is $R_1$ and at the bottom is $R_2$
in the 2nd case the diameter at the top is $R_2$ and at the bottom is $R_1$
Then the ratio of the absolute vertical component of the force on the container walls
for the 1st case over the 2nd case will be:
$(2R_2+R_1)/(2R_1+R_2)$
So when $R_1<R_2$, the absolute vertical force in the first case will be greater.

 

[zul8tr]

Huh?

Not sure what the Huh? relates to. In calculating forces using volumes as I hinted to do the unit weight of the fluid is also needed. Since a relative ratio is needed then H, Pi and unit weight of water drop out.

 

[Charles Link]

It seems to me @.Scott had it nailed in post #10.

I did not get this answer for the ratios. I got two quadratic expressions in $R_1$ and $R_2$, and they don't readily simplify. $\\$ Edit : The pressure at the bottom is computed as $P_{bottom}=\delta g H$. This gets multiplied by $A_{bottom}=\pi R_2^2$ in the first part of the problem. Because of the $(\frac{1}{3})$ factor in the volume in the term $Mg=\delta V g = \delta g (\frac{1}{3}) H \pi (R_1^2+R_1 R_2 +R_2^2)$, there is no simple cancellation.

 

[Charles Link]

@.Scott and @Chestermiller Please see the edited part of post 17.

 

[Chestermiller]

Not sure what the Huh? relates to. In calculating forces using volumes as I hinted to do the unit weight of the fluid is also needed. Since a relative ratio is needed then H, Pi and unit weight of water drop out.

Yes. This, I agree with.

 

[.Scott]

@.Scott and @Chestermiller Please see the edited part of post 17.

So you have the formula for the volume:
$(R_2^3-R_1^3)/(R_2-R_1)/3 = (R_1^2+R_1 R_2+R_2^2)/3$
Since the density of the liquid and the force of gravity are affecting all of these force equations proportionately, we do not have to carry those terms around. We also have a common $\pi H$ terms which we will also drop for the same reason.

The next equations are for the for the force of the fluid onto the bottom of the container:
For case 1: $R_2^2$
For case 2: $R_1^2$

The difference is the vertical force of the side walls:
For case 1: $R_2^2 - (R_1^2+R_1 R_2+R_2^2)/3$
For case 2: $R_1^2 - (R_1^2+R_1 R_2+R_2^2)/3$

Again, we don't need to carry the $1/3$ term:
For case 1: $3R_2^2 - (R_1^2+R_1 R_2+R_2^2) = 2R_2^2 -R_1 R_2 - R_1^2$
For case 2: $3R_1^2 - (R_1^2+R_1 R_2+R_2^2) = 2R_1^2 -R_1 R_2 - R_2^2$

Those are the equations for the (proportional) vertical forces on the sides of the container. Just multiply by $\pi H/3$, gravity, and density.
Also, we don't need to simplify $R_1^2+R_1 R_2+R_2^2$, so we don't care that there is no cancellation of terms.
We only need to deal with the vertical forces on the sides of the container.

 

[Charles Link]

So you have the formula for the volume:
$(R_2^3-R_1^3)/(R_2-R_1)/3 = (R_1^2+R_1 R_2+R_2^2)/3$
Since the density of the liquid and the force of gravity are affecting all of these force equations proportionately, we do not have to carry those terms around. We also have a common $\pi H$ terms which we will also drop for the same reason.

The next equations are for the for the force of the fluid onto the bottom of the container:
For case 1: $R_2^2$
For case 2: $R_1^2$

The difference is the vertical force of the side walls:
For case 1: $R_2^2 - (R_1^2+R_1 R_2+R_2^2)/3$
For case 2: $R_1^2 - (R_1^2+R_1 R_2+R_2^2)/3$

Again, we don't need to carry the $1/3$ term:
For case 1: $3R_2^2 - (R_1^2+R_1 R_2+R_2^2) = 2R_2^2 -R_1 R_2 - R_1^2$
For case 2: $3R_1^2 - (R_1^2+R_1 R_2+R_2^2) = 2R_1^2 -R_1 R_2 - R_2^2$

Those are the equations for the (proportional) vertical forces on the sides of the container. Just multiply by $\pi H/3$, gravity, and density.
Also, we don't need to simplify $R_1^2+R_1 R_2+R_2^2$, so we don't care that there is no cancellation of terms.
We only need to deal with the vertical forces on the sides of the container.

My mistake: Sorry . I see the expressions that I have do indeed factor. And yes, I agree with your answer. $\\$ Editing: And something I originally overlooked: Since for a vertical wall where $R_1=R_2$ there is zero vertical force for both cases, the term $R_2-R_1$ must necessarily factor out of both expressions, ( because if $r$ is a root of a polynomial equation, $x-r$ is a factor).

 

[zul8tr]

OK.
Given two cases where:
in the 1st case the diameter at the top is $R_1$ and at the bottom is $R_2$
in the 2nd case the diameter at the top is $R_2$ and at the bottom is $R_1$
Then the ratio of the absolute vertical component of the force on the container walls
for the 1st case over the 2nd case will be:
$(2R_2+R_1)/(2R_1+R_2)$
So when $R_1<R_2$, the absolute vertical force in the first case will be greater.

Scott you are correct, applause.

Certainly the hinted volume approach is the easier way to go on this problem although there are several other ways to solve it.

Ex:
Realizing that the fluid in the central core has no influence to the vertical force of the conical wall of A or B then all that is left is to calculate the remaining volume weight in each case dealing with only the conical wall.

On hydrostatic problems where the volumes are generated by revolution about an axis the 2nd Therom of Pappus is a handy tool to calculate these volumes. Volume generated by rotation = plane area x path traveled by the centroid of plane area

In case A the vertical up force on the conical wall is the Imaginary volume weight above that wall up to the free surface.

By Pappus:

For case A: FvA = Triangle x 2 Pi x [R1 + 2/3(R2-R1)] x water unit weight

where Triangle is the imaginary plane area above the conical wall of height H and base R2-R1
bounded by the free surface and [...] is the radius to the centroid of plane area from center axis of
the container.

In case B the vertical down force is the real volume weight above the conical surface to the real water surface.

For case B: FvB = Triangle x 2 Pi x [R1 + 1/3(R2-R1)] x water unit weight

where Triangle is the plane area above the conical wall in the fluid to the free surface of height H and
base R2-R1 and [...] is radial distance to its centroid from center axis of container

Triangle is the same area in case A and B, the ratio is:

FvA / FvB = [1 + 2/3(R2 / R1 - 1) ] / [1 + 1/3 (R2 / R1 - 1) ] and this is > 1 for R2 > R1 thus case A has a greater relative vertical force on the conical wall.

Algebra yields your result;

FvA / FvB = (R1 + 2R2) / (2R1 + R2)

Scott how do you get the lower case subscripts on R1 and R2 and superscripts when needed?

 

[.Scott]

How do you get the lower case subscripts on R1 and R2?

It's Latex.
Enter: ## R_1, R_2 # #
(I added a space so that it wouldn't convert it).

 

[jbriggs444]

(I added a space so that it wouldn't convert it).

FYI, instead of neutering the "##" or "$$" delimiters with white space, you can neuter them with color, using the following tip from the PF INFO section, "Latex Primer":

"If for some reason you want to post raw LaTeX code that includes the delimiters, you can prevent the delimiters from being interpreted as delimiters. Select a delimiter (only), click the Text Color icon (the one that looks like a half moon) in the palette at the top of the editing window, and choose black. Repeat for the other delimiter.​

And *voila*: $R_1,R_2$

 

[Charles Link]

I posted it in my edited comments of post 21, but perhaps it is worthwhile also mentioning it in a separate comment: I really was not expecting the quadratic expressions for the answers of the vertical forces from both cases to be factorable, and especially to have a common factor, but since there is zero vertical force in the case of $R_2=R_1$, the factor $R_2-R_1$ is necessarily a factor in both of these expressions, (because if $r$ is a root of the polynomial expression in $x$, then $x-r$ is a factor). $\\$ Kudos to @.Scott for recognizing that the expressions did indeed have a common factor.