Water pressure in a tank

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.

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haruspex
Homework Helper
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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?

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

Regla
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
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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.

Last edited:
Nidum
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
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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
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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.

Regla
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
Mentor
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?
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
Mentor
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
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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
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2019 Award
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
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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
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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
Gold Member
*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
Homework Helper
2019 Award
∫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.