Solving a Physics Problem Using Archimedes Principle

[Tanya Sharma]

Homework Statement

Homework Equations

The Attempt at a Solution

On integrating the force due to fluid pressure on the cap , I have arrived at the correct result $ρgπR^2(H-\frac{2}{3}R)$ . This in turn would be the force with which the cap presses the bottom.

But I would like to solve this problem with less maths i.e may be by using Archimedes principle .

The weight of the volume of fluid displaced would be $ρgπ\frac{2}{3}R^3$ . Shouldn't this be the force exerted by fluid on the cap ?

I would be grateful if somebody could help me with the problem .

 

[ShayanJ]

You can't use Archimedes principle here because the force is not due to buoyancy but pressure.
Anyway, I get a different result than yours. Are you sure you didn't make any mistake?

EDIT: Bad wording! Of course buoyancy is due to pressure too, I just meant Archimedes principle is for the upward force.

 

[TSny]

The weight of the fluid displaced by a submerged hemisphere would be equal to the upward buoyant force acting on the hemisphere. Note that this upward force is due to fluid pressure on the curved surface as well as the flat surface of the hemisphere. You only want the force on the curved surface. So, you have little more work to do.

 

[Tanya Sharma]

Hello TSny

Note that this upward force is due to fluid pressure on the curved surface as well as the flat surface of the hemisphere.

But force on the flat surface is not due to the fluid . It is due to the floor on which the vessel is resting .

 

[TSny]

In this problem, the force on the flat surface is due to atmospheric pressure since the flat surface is exposed to the atmosphere because of the hole.

Archimedes principle assumes that the object is completely surrounded by fluid. [EDIT: Or the object is partially submerged. Here, the bottom of the hemisphere is not exposed to the fluid since it is covering a hole. So, the weight of fluid displaced by the hemisphere does not equal the net force of the fluid on the hemisphere. So, in this sense, Archimedes' principle does not apply.]

 

[ShayanJ]

Using Pascal's law, the pressure at a depth z under the water is $P=\rho g z$. We want the pressure on the surface of the hemisphere, so z should be calculated accordingly. If we consider a spherical coordinate system with the origin at the centre of the hemisphere, we can write $z=H-R\sin\theta$. So the downward force at each surface element of the hemisphere is $\rho g (H-R\sin\theta)R^2\sin^2\theta d\theta d\varphi$. Integrating it over the surface of the hemisphere gives $\pi \rho g R^2 (\pi H-\frac{8R}{3})$. I don't see how Archimedes principle can help.

 

[Tanya Sharma]

In this problem, the force on the flat surface is due to atmospheric pressure since the flat surface is exposed to the atmosphere because of the hole.

Archimedes principle assumes that the object is completely surrounded by fluid.

I think I am misunderstanding some things .

Isn't the vessel resting on a horizontal surface ? Hasn't the cap plugged the hole ? How is atmospheric pressure exerting force on the flat surface of the hemisphere ?

 

[Tanya Sharma]

Using Pascal's law, the pressure at a depth z under the water is $P=\rho g z$. We want the pressure on the surface of the hemisphere, so z should be calculated accordingly. If we consider a spherical coordinate system with the origin at the centre of the hemisphere, we can write $z=H-R\sin\theta$. So the downward force at each surface element of the hemisphere is $\rho g (H-R\sin\theta)R^2\sin^2\theta d\theta d\varphi$. Integrating it over the surface of the hemisphere gives $\pi \rho g R^2 (\pi H-\frac{8R}{3})$. I don't see how Archimedes principle can help.

You may be right . I am quite prone to making mistakes . May be TSny could verify the result.

I am just looking at a more intuitive solution . By no means I am saying that this problem could be solved by Archimedes Principal . But it might very well be .

 

[Chestermiller]

I get the same answer that Tanya got. I don't see an easier way of doing the problem.

Chet

 

[Tanya Sharma]

Hi Chet

Thanks for the confirmation . Do you agree that Archimedes Principle is not applicable in this situation ? If not , what is the reason ? Is it because the cap rests on a surface i.e does not have fluid under it .

 

[Chestermiller]

Hi Chet

Thanks for the confirmation . Do you agree that Archimedes Principle is not applicable in this situation ? If not , what is the reason ? Is it because the cap rests on a surface i.e does not have fluid under it .

Yes. See TSny post #5.

Chet

 

[TSny]

You can use Archimedes' principle as a way to get to the answer. You just have to modify the weight of the fluid displaced to take into account that you only want the force due to pressure on the curved surface alone. The weight of the fluid displaced by the volume of hemisphere will give you the net upward force due to fluid pressure acting on all surfaces of the hemisphere if the hemisphere were completed surrounded by fluid. Since your hemisphere is not completely surrounded by fluid, you will need to modify what Archimedes' principle gives you with a simple calculation.

 

[Tanya Sharma]

Since your hemisphere is not completely surrounded by fluid, you will need to modify what Archimedes' principle gives you with a simple calculation.

No headway . How to do what you are suggesting ?

 

[TSny]

Suppose the hemisphere is completely submerged in the fluid so that the hemisphere is completely surrounded by fluid. So, right now there is no hole in the container and the hemisphere is not at the bottom of the container.

Archimedes' principle gives you the buoyant force. But the buoyant force is just the net upward force due to pressure of the fluid acting on the curved and flat surfaces of the hemisphere. What would you have to do this buoyant force to get only the net force acting on the curved surface alone? Assume the hemisphere is oriented as in your problem with the flat surface horizontal.

 

[Tanya Sharma]

Force due to liquid on curved surface + Force due to liquid on flat surface = $ρgπ\frac{2}{3}R^3$ .

Force due to liquid on curved surface = $ρgπ\frac{2}{3}R^3$ - Force due to liquid on flat surface .

Is this what you are suggesting ?

 

[TSny]

Yes, just be careful with signs due to force on curved surface being downward while force on flat surface is upward.

 

[Tanya Sharma]

Now if liquid was present , force due to liquid on flat surface would have been = $ρgπR^2H$ .

So, Force due to liquid on curved surface = $ρgπ\frac{2}{3}R^3 - ρgπR^2H$

But then ,this expression has a negative sign as opposed to what I got in the OP .

 

[TSny]

In OP you calculated the downward force on the curved surface. In #17 you calculated the upward force on the curved surface.

 

[TSny]

Assuming that the system is surrounded by the atmosphere, you should make sure that you have taken atmospheric pressure into account. For example, in your OP the force that you calculated by integrating the pressure over the curved surface should include an additional term due to atmospheric pressure. But would this affect the answer for the force that the hemispherical cap presses on the bottom of the container?

 

[Tanya Sharma]

Assuming that the system is surrounded by the atmosphere, you should make sure that you have taken atmospheric pressure into account. For example, in your OP the force that you calculated by integrating the pressure over the curved surface should include an additional term due to atmospheric pressure

Ok . So the force should be $P_0πR^2+ρgπR^2(H-\frac{2}{3}R)$ . Right ??

 

[TSny]

Ok . So the force should be $P_0πR^2+ρgπR^2(H-\frac{2}{3}R)$ . Right ??

Yes, that would be the actual force of the fluid on the curved surface of the hemisphere.

 

[Tanya Sharma]

But would this affect the answer for the force that the hemispherical cap presses on the bottom of the container?

No ,it wouldn't .

We need to deduct an extra term $P_0πR^2$ when applying Archimedes principle .

So, if liquid was present , force due to liquid on flat surface would have been = $ρgπR^2H-P_0πR^2$

So, Upward force due to liquid on curved surface = $ρgπ\frac{2}{3}R^3−ρgπR^2H-P_0πR^2$ .

Do you agree ??

 

[Chestermiller]

Yes, that would be the actual force of the fluid on the curved surface of the hemisphere.

What about the effect of the air pressure above the upper water surface?

Chet

 

[Tanya Sharma]

What about the effect of the air pressure above the upper water surface?

Chet

Hasn't that been taken care by including $P_0$ in post#20 ?

 

[Chestermiller]

Hasn't that been taken care by including $P_0$ in post#20 ?

Yes. Sorry. But that should be canceled by the pressure of the air below the dome.

Chet

 

[Tanya Sharma]

Yes. Sorry. But that should be canceled by the pressure of the air below the dome.

Chet

There is no air pressure below the dome . It is resting on a surface.

Edit : I was misinterpreting the setup . There is air pressure below the dome.

 

[Chestermiller]

There is no air pressure below the dome . It is resting on a surface.

Well what's inside, vacuum?

Chet

 

[Tanya Sharma]

Well what's inside, vacuum?

Chet

Sorry . You are right . There is atmospheric pressure acting on the inside surface .

 

[TSny]

No ,it wouldn't .

We need to deduct an extra term $P_0πR^2$ when applying Archimedes principle .

So, if liquid was present , force due to liquid on flat surface would have been = $ρgπR^2H-P_0πR^2$

So, Upward force due to liquid on curved surface = $ρgπ\frac{2}{3}R^3−ρgπR^2H-P_0πR^2$ .

Do you agree ??

Yes. [EDIT: I think you have a wrong sign in the expression for the force on the flat surface, but your expression for the upward force on the curved surface is correct.]

Now back to the original problem. You know that the force pushing down on the curved surface is $ρgπR^2H - ρgπ\frac{2}{3}R^3+P_0πR^2$. How do you use this to get the force that the hemisphere presses down on the bottom of the container with the hole?

 

[Tanya Sharma]

Now back to the original problem. You know that the force pushing down on the curved surface is $ρgπR^2H - ρgπ\frac{2}{3}R^3+P_0πR^2$. How do you use this to get the force that the hemisphere presses down on the bottom of the container with the hole?

Assuming cap is massless , shouldn't this be the force with which the cap presses onto the bottom .There are only two forces acting on the cap ,one due to fluid , other normal reaction force due to the bottom .Both equal and opposite to each other .

Isn't it ??

 

[TSny]

If the container is surrounded by the atmosphere, then there would be three forces acting on the cap. The cap covers the hole which is open to the atmosphere. So, the atmosphere is in contact with the inner surface of the hemispherical cap. (As I interpret the diagram in the problem, the hemispherical cap does not contain a flat bottom. It is just a thin hemispherical shell.)

 

[Tanya Sharma]

If the container is surrounded by the atmosphere, then there would be three forces acting on the cap. The cap covers the hole which is open to the atmosphere. So, the atmosphere is in contact with the inner surface of the hemispherical cap. (As I interpret the diagram in the problem, the hemispherical cap does not contain a flat bottom. It is just a thin hemispherical shell.)

Ok .

Considering upward positive and normal reaction force due to floor be N ,the sum of the three forces should be zero .

Force due to liquid on curved surface + Force due to atmospheric pressure on the inner surface of shell + Normal reaction = 0 .

$(ρgπ\frac{2}{3}R^3−ρgπR^2H-P_0πR^2) + (P_0πR^2)+ N= 0$

Hence , $N = ρgπR^2H-ρgπ\frac{2}{3}R^3$ .

 

[TSny]

Ok .

Considering upward positive and normal reaction force due to floor be N ,the sum of the three forces should be zero .

Force due to liquid on curved surface + Force due to atmospheric pressure on the inner surface of shell + Normal reaction = 0 .

$(ρgπ\frac{2}{3}R^3−ρgπR^2H-P_0πR^2) + (P_0πR^2)+ N= 0$

Hence , $N = ρgπR^2H-ρgπ\frac{2}{3}R^3$ .

Yes, I believe that's it.

I think Chet was pointing out earlier that the extra atmospheric force term on the curved surface is canceled by the atmospheric force acting on the underside of the hemisphere. You see that in the calculation above.

 

[Tanya Sharma]

Thank you very much.

Yes, Chet was right . I was misinterpreting the setup.

Please have a look at the attached image . Kindly help me in understanding why is force due to liquid on right side on the ball = $pπR^2$ and that due to fluid on left side is $3pπR^2$ .

 

[TSny]

Think of a truncated sphere immersed in a region of uniform pressure P_o. Assume that the only force acting on the truncated sphere is due to this surrounding pressure. If it makes sense to you that the sphere will not be pushed in any direction by the fluid pressure, then you conclude that the net force due to the pressure must be zero. So, you can compare the force on the curved surface with the force on the flat portion,