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No integration is required to answer this problem. It is not a calculus integration problem..Scott said: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.
Shouldn't the answer involve the squares of the radii?.Scott said:In that case: (2R1+R2)/(2R2+R1)
Since we're doing ratios, I will drop the H and Pi terms.Chestermiller said:Shouldn't the answer involve the squares of the radii?
.Scott said:In that case: (2R1+R2)/(2R2+R1)
Huh?zul8tr said: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.
It seems to me @.Scott had it nailed in post #10.Charles Link said: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.
OK.zul8tr said:"Which has a greater absolute vertical force only on the conical wall and by what relative amount?"
Chestermiller said:Huh?
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.Chestermiller said:It seems to me @.Scott had it nailed in post #10.
Yes. This, I agree with.zul8tr said: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.
So you have the formula for the volume:Charles Link said:@.Scott and @Chestermiller Please see the edited part of post 17.
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)..Scott said: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.
.Scott said: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.
It's Latex.zul8tr said:How do you get the lower case subscripts on R1 and R2?
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":.Scott said:(I added a space so that it wouldn't convert it).
Hydrostatic force is the pressure exerted by a fluid at rest due to the weight of the fluid above it.
Hydrostatic force is calculated by multiplying the density of the fluid by the gravitational acceleration and the height of the fluid column above the point where the force is being measured.
The main factors that affect hydrostatic force include the density of the fluid, the gravitational acceleration, and the height of the fluid column. Additionally, the shape and size of the container can also affect the distribution of the force.
Using identical containers filled with equal volumes of water allows for a controlled experiment, where the only variable is the height of the fluid column. This allows for a more accurate calculation and comparison of the hydrostatic force.
Hydrostatic force is used in various real-world applications, such as in hydraulic systems, dams, and water towers. It is also an important factor in understanding the stability of ships and other floating structures.