Solving a Physics Problem Using Archimedes Principle

[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,

 

[Tanya Sharma]

Should I think of the portion of ball on the left in isolation ? I mean should I consider the forces acting on the ball(truncated sphere) on the left side?

If yes , then

Force due to fluid on left+Force due to wall +Force exerted by the portion of ball towards right on the left portion = 0 .

 

[TSny]

Let's say you want the force that the fluid on the left side of the wall exerts on the ball in your picture in #34. This is the same as the force that the fluid in my picture exerts on the curved portion of the truncated sphere if Po = 3p. But in my picture you can relate this force to the force that the fluid exerts on the flat portion of the truncated sphere.

 

[Tanya Sharma]

When I first came across this problem , I got the answer right . But on second thoughts my reasoning seemed too muddled up.

I have understood your post#35 . In your picture the force due to fluid on left would be equal and opposite to that of force exerted on the flat surface = $3pπR^2$ .

But I am sorry I am not able to relate your picture with the setup I posted in post#34 .

But in my picture you can relate this force to the force that the fluid exerts on the flat portion of the truncated sphere.

Which fluid is exerting force on the flat surface of truncated sphere (on the left), the one with pressure P or 3P ? There is also the wall exerting force on the flat surface . In addition, there would be force from the remaining part of the sphere (the one on the right) .

 

[TSny]

In post #35, the purpose was to see that the fluid exerts a force on the curved surface of $3p\pi R^2$ by comparing with the force on the flat portion. Now you can forget the flat portion and conclude that whenever the curved portion is exposed to a fluid of uniform pressure $3p$, the force on the curved portion will be $3p\pi R^2$. In your problem, the curved portion of the sphere on the left of the wall is exposed to a pressure of $3p$. So, the force on that portion will be $3p\pi R^2$. I think that's a valid argument unless I am overlooking something.

 

[Tanya Sharma]

Thank you very much