Calculating Water Volume in a Tank Using Pressure Measurements

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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


46bcb6e308d86c8df7f4bbe05df5a44f6c2c2f0d

81eac1e205430d1f40810df36a0edffdc367af36
is the pressure,
545fd099af8541605f7ee55f08225526be88ce57
is the normal force,
7daff47fa58cdfd29dc333def748ff5fa4c923e3
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.
 
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Regla said:
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?
 
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This question is bonkers - is that the actual wording of the problem as given to you ?
 
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haruspex said:
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.
 
Regla said:
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.
 
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haruspex said:
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?
 
Regla said:
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?
 
haruspex said:
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 said:
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)gLh*1/2 . I can put in all the numbers but what about depth?
 
Chestermiller said:
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.
 
haruspex said:
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 said:
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
 
Regla said:
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.
 
haruspex said:
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.
 
haruspex said:
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)
 
Regla said:
bF=(ro)gxL2
No, x is the depth of an arbitrary strip. You mean h, not x.
Regla said:
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?
 
haruspex said:
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.
 
*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 .
 
Nidum said:
*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.
 
Regla said:
∫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.
 
jbriggs444 said:
@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?
 
Regla said:
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
 
jbriggs444 said:
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.
 
jbriggs444 said:
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?
 

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Regla said:
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##.