Calculating Water Volume in a Tank Using Pressure Measurements

[Regla]

Homework Statement

on a flat desk is an aquarium with a volume of 640 cm³ ,t's not full of water. The pressure to the bottom of the aquarium is 3 times bigger than to one side of it, how much water is in the tank?

Homework Equations

is the pressure,

is the normal force,

is the area of the surface on contact.

The Attempt at a Solution

I think this mainly requires crunching of crude formulas, I don't even know how to calculate the pressure to the side of a tank even if I have more variables. Thanks for even reading so far, any help is appreciated.

 

[haruspex]

The pressure to the bottom of the aquarium is 3 times bigger than to one side of it, how much water is in the tank?

I'm not sure what the question means. Have you stated it exactly as given to you?
The pressure at the side is not uniform on the sides. It could mean the average pressure there.
Or, if, you knew the shape of the aquarium, it could mean the force, not the pressure.

If the water has depth h, what is the average pressure on the side?

 

[Nidum]

This question is bonkers - is that the actual wording of the problem as given to you ?

 

[Regla]

I'm not sure what the question means. Have you stated it exactly as given to you?
The pressure at the side is not uniform on the sides. It could mean the average pressure there.
Or, if, you knew the shape of the aquarium, it could mean the force, not the pressure.

If the water has depth h, what is the average pressure on the side?

Sorry I missed to mention that the tank is an cube, I forgot to mention it since I had to translate the whole question.

 

[haruspex]

Sorry I missed to mention that the tank is an cube, I forgot to mention it since I had to translate the whole question.

Then I would interpret it as saying the force is the same on the bottom as on each side... particularly if it is a translation.
If the depth is h, can you calculate the force on each side?

Edit: correction... Since we know the depth is less than the length and width, the total force on a side must be less than that on the base, so the right interpretation must be that the average pressures are equal.

 

[Regla]

Then I would interpret it as saying the force is the same on the bottom as on each side... particularly if it is a translation.
If the depth is h, can you calculate the force on each side?

does that mean you would use the same formula as for the bottom? (ro)mg?

 

[haruspex]

does that mean you would use the same formula as for the bottom? (ro)mg?

No. The pressure varies according to depth.
Consider a horizontal strip of width dx at depth x. What is the pressure there? What is the total force on that strip?

 

[Regla]

No. The pressure varies according to depth.
Consider a horizontal strip of width dx at depth x. What is the pressure there? What is the total force on that strip?

so any idea how to find out the h? I'm trying to understand other branches of physics, not just the ones I'm fairly good at.

 

[haruspex]

so any idea how to find out the h? I'm trying to understand other branches of physics, not just the ones I'm fairly good at.

We'll get to that. Please try to answer my question.

 

[Regla]

We'll get to that. Please try to answer my question.

ok so using a formula to find the F on a strip I used F=(ro)gL^h*1/2 . I can put in all the numbers but what about depth?

 

[Chestermiller]

ok so using a formula to find the F on a strip I used F=(ro)gL^h*1/2 . I can put in all the numbers but what about depth?

You need to solve for h using the information you have been given. Do you have another relationship involving h?

 

[Regla]

You need to solve for h using the information you have been given. Do you have another relationship involving h?

This is the part that I get stuck at, if I knew the exerted force on the side or bottom I could just rearrange the variables and be done with it, but all I know is that the force to the bottom is 3 times stronger than to one side.

 

[Chestermiller]

This is the part that I get stuck at, if I knew the exerted force on the side or bottom I could just rearrange the variables and be done with it, but all I know is that the force to the bottom is 3 times stronger than to one side.

What is the force to one side as a function of h (algebraically)?

 

[haruspex]

using a formula to find the F on a strip I used F=(ro)gLh*1/2

I specified a strip width dx at depth x, i.e. from depth x to depth x+dx. Where are those in your formula for the force?

 

[Regla]

I specified a strip width dx at depth x, i.e. from depth x to depth x+dx. Where are those in your formula for the force?

let's say the pressure on that strip is dF which is equal to the pressure at that depth P which we multiply by the area of the strip dA
dF=PdA dA=Ldx
dF=PLdx P=(ro)gx
dF=(ro)gxLdx
something like this?

 

[jbriggs444]

Yes. The force on an incremental strip at depth x is $\rho g x L\ dx$

Now if you integrate that incremental force from top strip (x=0) to bottom strip (x=h), what do you get?

 

[Regla]

Yes. The force on an incremental strip at depth x is $\rho g x L\ dx$

Now if you integrate that incremental force from top strip (x=0) to bottom strip (x=h), what do you get?

I get the force applied to the whole wall of the aquarium, but we don't know the x or h

 

[haruspex]

I get the force applied to the whole wall of the aquarium, but we don't know the x or h

x is the variable of integration. After doing the integration and applying the bounds you will have a function of h.
The force on the base of the tank is also a function of h.
Find both functions.

 

[Regla]

x is the variable of integration. After doing the integration and applying the bounds you will have a function of h.
The force on the base of the tank is also a function of h.
Find both functions.

that makes some sense, I'll try that tomorrow. Thanks.

 

[Regla]

x is the variable of integration. After doing the integration and applying the bounds you will have a function of h.
The force on the base of the tank is also a function of h.
Find both functions.

I've got the formula for the base of the tank bF=(ro)gxL²
but I'm quite lost on how to find a function, we worked a bit with them last year, but we didn't find new functions we just solved given ones.(which is stupid to be honest)

 

[haruspex]

bF=(ro)gxL2

No, x is the depth of an arbitrary strip. You mean h, not x.

lost on how to find a function

For the force on the side, yes?
In post #15 you wrote that on the side $dF=\rho gxL.dx$. Can you write that as an integral and integrate it?

 

[Regla]

No, x is the depth of an arbitrary strip. You mean h, not x.

For the force on the side, yes?
In post #15 you wrote that on the side $dF=\rho gxL.dx$. Can you write that as an integral and integrate it?

∫f(x)dx=dF/(ro)gL I feel like I'm doing a lot of mistakes.

 

[Nidum]

*Length of any edge of cubic tank = h
*Depth of water = k.h

Pressure at top surface of water = ?
Pressure at bottom surface of water = ?

*Pressure varies linearly with depth .
Average pressure = ?

Area of side of tank = ?
Pressure force acting on side of tank = ?

Area of bottom of tank = ?
Pressure force acting on bottom of tank = ?

*Pressure force acting on tank side = 1/3 of pressure force acting on bottom of tank .
*Use to find value of k .
k = ?

*Volume of water in full tank = h³ = a given value .
*Use to find value of h .
h = ?

*Actual volume of water in tank = k.h³
What is final answer for volume of water in tank ?

*Note that final answer is really just k times the amount of water in full tank .

 

[Regla]

*Length of any edge of cubic tank = h
*Depth of water = k.h

Pressure at top surface of water = ?
Pressure at bottom surface of water = ?

*Pressure varies linearly with depth .
Average pressure = ?

Area of side of tank = ?
Pressure force acting on side of tank = ?

Area of bottom of tank = ?
Pressure force acting on bottom of tank = ?

*Pressure force acting on tank side = 1/3 of pressure force acting on bottom of tank .
*Use to find value of k .
k = ?

*Volume of water in full tank = h³ = a given value .
*Use to find value of h .
h = ?

*Actual volume of water in tank = k.h³
What is final answer for volume of water in tank ?

*Note that final answer is really just k times the amount of water in full tank .

k is the coefficient? right? I'' try that in a few hours.

 

[jbriggs444]

∫f(x)dx=dF/(ro)gL I feel like I'm doing a lot of mistakes.

@Nidum is trying to lead you past the integration step so that you do not have to evaluate that integral.

The key is that you are being asked to evaluate $\int_0^h \rho g L \ x \ dx$

The $\rho g L$ factors out and all that's left is evaluating $\rho g L\ \int_0^h x \ dx$

That integral should be easy to evaluate. It should give the same result that @Nidum is jumping to: The force on a side is the average pressure on a side multiplied by the [wetted] area of that side.

 

[Regla]

@Nidum is trying to lead you past the integration step so that you do not have to evaluate that integral.

The key is that you are being asked to evaluate $\int_0^h \rho g L \ x \ dx$

The $\rho g L$ factors out and all that's left is evaluating $\rho g L\ \int_0^h x \ dx$

That integral should be easy to evaluate. It should give the same result that @Nidum is jumping to: The force on a side is the average pressure on a side multiplied by the [wetted] area of that side.

thanks for showing me how I was supposed to form the integral, my only question is what do you mean that ρgL factors out?

 

[jbriggs444]

thanks for showing me how I was supposed to form the integral, my only question is what do you mean that ρgL factors out?

The $\rho$, the $g$ and the $L$ are all constant terms for our purposes. They do not depend on x. They are simply a constant multiplier on the value. As long as you have a constant multiplier on the function you are integrating, for example the k in:
$$\int_0^h k\ f(x)\ dx$$
then you are allowed to move that constant outside of the integral and instead write
$$k\ \int_0^h f(x)\ dx$$
Moving the constant out of the integral is what I call "factoring it out". It seems that Wiki even has a page on this principle: https://en.wikipedia.org/wiki/Constant_factor_rule_in_integration

 

[Regla]

The $\rho$, the $g$ and the $L$ are all constant terms for our purposes. They do not depend on x. They are simply a constant multiplier on the value. As long as you have a constant multiplier on the function you are integrating, for example the k in:
$$\int_0^h k\ f(x)\ dx$$
then you are allowed to move that constant outside of the integral and instead write
$$k\ \int_0^h f(x)\ dx$$
Moving the constant out of the integral is what I call "factoring it out". It seems that Wiki even has a page on this principle: https://en.wikipedia.org/wiki/Constant_factor_rule_in_integration

oh that makes sense.

 

[Regla]

The $\rho$, the $g$ and the $L$ are all constant terms for our purposes. They do not depend on x. They are simply a constant multiplier on the value. As long as you have a constant multiplier on the function you are integrating, for example the k in:
$$\int_0^h k\ f(x)\ dx$$
then you are allowed to move that constant outside of the integral and instead write
$$k\ \int_0^h f(x)\ dx$$
Moving the constant out of the integral is what I call "factoring it out". It seems that Wiki even has a page on this principle: https://en.wikipedia.org/wiki/Constant_factor_rule_in_integration

did I enter everything correct?

 

[jbriggs444]

did I enter everything correct?

You are supposed to be integrating from 0 to h, not from 0 to 4. You are supposed to be integrating x, not 4. And what @haruspex is trying to get you do to is to express the result in symbolic form. You should wind up with an expression involving $\rho$, $g$, $h$ and $L$.

 

[Regla]

You are supposed to be integrating from 0 to h, not from 0 to 4. You are supposed to be integrating x, not 4.

the whole problem was stated as a volume of 64 dm³... so I thought that since 4³=64 doesn't that mean h=4?

 

[jbriggs444]

the whole problem was stated as a volume of 64 dm³... so I thought that since 4³=64 doesn't that mean h=4?

You are trying to find h. It will not be equal to 4.

h in this context refers to the height (or depth) of the water in the tank. The tank is not full.

 

[Regla]

You are trying to find h. It will not be equal to 4.

like this?

 

[jbriggs444]

like this?

That appears to be a correct result. However, as has been suggested previously, it would be better to leave it in symbolic form and to avoid replacing $\rho$, $g$ and $L$ with numeric values:

$$F_{side} = \rho g L \frac{h^2}{2}$$

Now can you write down a formula for the total force on the bottom of the tank in terms of $\rho$, $g$, $L$ and $h$?

 

[Regla]

That appears to be a correct result. However, as has been suggested previously, it would be better to leave it in symbolic form and to avoid replacing $\rho$, $g$ and $L$ with numeric values:

$$F_{side} = \rho g L \frac{h^2}{2}$$

Now can you write down a formula for the total force on the bottom of the tank in terms of $\rho$, $g$, $L$ and $h$?

Like this? or does the 4 have to be an x?

 

[jbriggs444]

Like this? or does the 4 have to be an x?

You can include "L" or you can include "4". Using both is nonsense. As I had suggested in #34, the correct equation is:

$$F_{side} = \rho g L \frac{h^2}{2}$$

You had previously come up with a formula for the force on the base:

I've got the formula for the base of the tank bF=(ro)gxL2

When you wrote that you were apparently using the variable "x" to denote depth. Let us use "h" for depth and reformat that result:

$$F_{base} = \rho g L^2 h$$

Now then. We have formulas for the force on the side and for the force on the base. The problem statement tells us the ratio of those two forces. Can you write down an equation expressing that fact?

 

[Regla]

Now then. We have formulas for the force on the side and for the force on the base. The problem statement tells us the ratio of those two forces. Can you write down an equation expressing that fact?

Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h²/2)=3

 

[jbriggs444]

Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h²/2)=3

Yes. You should be able to simplify that significantly.

 

[Regla]

​

Yes. You should be able to simplify that significantly.

so how do we find the volume or height of the water level? since that only tells the ratio.

 

[jbriggs444]

More calculating. Fewer questions.

 

[Regla]

More calculating. Fewer questions.

so what does the integral in post #34? h? the height of the water level? or the number that I have to put in the post #37 formula?
but we still get the h in the integral.

 

[jbriggs444]

In #37 you had an equation:

Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h2/2)=3

Solve that equation for h. Or at least simplify it as I had suggested in #38.

 

[Regla]

More calculating. Fewer questions.

I get h=0.00991988 or 0.01

 

[Regla]

In #37 you had an equation:
Solve that equation for h. Or at least simplify it as I had suggested in #38.

if I try to find the pressure to the bottom and side by putting in h I get always the same thing as 2. (ρgL²h/ρgL(h²/2)) It's a bit counter intuitive since I was stated that that should be 3, is the formula for the side equation right?

 

[Nidum]

ρgL2h/ρgL(h2/2)=3

Let's do a tidy re-arrangement of that equation :

ρgL²h = (3/2)ρgLh²

the ρg's cancel so :

L²h = (3/2)Lh²

Divide both sides by hL :

L = (3/2)h

Can you go from there ?

 

[Regla]

ρgL2h/ρgL(h2/2)=3

Let's do a tidy re-arrangement of that equation :

ρgL²h = (3/2)ρgLh²

the ρg's cancel so :

L²h = (3/2)Lh²

Divide both sides by hL :

L = (3/2)h

Can you go from there ?

that simplifies a lot of things, thanks, but when I calculate h (if L=40) it I get h=26.6667 or 27. the problem is that if I try to calculate the pressures they don't equal to the statement give in the equation (Pbase/Pside=3)
Pbase=(1000*9.8*40²*27)/40*40=264600
Pside=(1000*9.8*40*(27²/2)/40*27=132300
264600/132300=2. It doesn't match with the statement given. I feel like there is something wrong with the side force equation.

 

[Nidum]

L is not 40 . Simple numerical mistake I think ?

 

[Regla]

L is not 40 . Simple numerical mistake I think ?

If I know that the volume of the tank is 64 dm³ 4³=64 since that 4 is dm I change it into cm which is 40cm, isn't that right?

 

[Nidum]

Cube root of 640 is roughly 8.62

8.62cm x 8.62 cm x 8.62 cm = 640 cm³ .

.

 

[Regla]

Cube root of 640 is roughly 8.62

8.62cm x 8.62 cm x 8.62 cm = 640 cm³ .

.

even if I change L with 8.62 and h changes to roughly 5.75 I still get 2 when I calculate the Pressures in the way stated in post #46