# Fluid Pressure

## The Attempt at a Solution

Applying Archimedes priniciple the force exerted by liquid on the upper hemisphere would be ## F = \frac{2}{3}\pi R^3 \rho g## .

Now I am not sure , what does it mean by force due to gauge pressure ?

Any help is appreciated .

Thanks .

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• liquid.PNG
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Chestermiller
Mentor
Applying Archimedes priniciple the force exerted by liquid on the upper hemisphere would be ## F = \frac{2}{3}\pi R^3 \rho g## .

Now I am not sure , what does it mean by force due to gauge pressure ?

Any help is appreciated .

Thanks .
Your calculated force would not be correct unless the bottom of the hemisphere were immersed in fluid, which it is not. The only way I could think of to get the net (downward) force exerted by the fluid on the upper hemisphere was to integrate the pressure (including its vectorial direction of the pressure at the surface) over the upper surface. This gave me one of the four answers in the choices. But it was the force due to the total pressure, not the gauge pressure. I think they meant to say the force due to the total pressure in the problem statement. Otherwise, the P0 shouldn't be in there at all.

Chet

Vibhor
Geofleur
Gold Member
I would just like to add that the gauge pressure is the difference between the absolute pressure and a reference pressure. So, for example, gauge pressure in this problem could be ## P - P_0 ##. I also could not think of another way to solve this problem, by the way - I had to do it the way Chet did!

Your calculated force would not be correct unless the bottom of the hemisphere were immersed in fluid, which it is not.
Chet

Are you suggesting that Archimedes Principle cannot be applied since the sphere rests on the surface at the bottom ?

In the attached figure consider a circular strip of width ##Rdθ## at an angle θ to the horizontal ( the horizontal line at a height R from the bottom) .

The area of the strip is ##2\pi R^2 cos\theta dθ##

Pressure at this strip is ##P_0 + \rho g(3R-Rsin\theta)##

Force due to liquid at this pressure will be ##[P_0 + \rho g(3R-Rsin\theta)][2\pi R^2 cos\theta dθ]## .

But we should only consider the vertical component of this force which would be given by ##[P_0 + \rho g(3R-Rsin\theta)][2\pi R^2 cos\theta sin\theta dθ]##

##\int_0^{\frac{\pi}{2}}[P_0 + \rho g(3R-Rsin\theta)][2\pi R^2 cos\theta sin\theta dθ] = \frac{\pi R^2}{3}[3P_0+7\rho Rg]## .

Is this what you are getting ?

Last edited:
Chestermiller
Mentor

Are you suggesting that Archimedes Principle cannot be applied since the sphere rests on the surface at the bottom ?
No. I'm saying that Archimedes Principle cannot be applied to the upper hemisphere because it is not fully immersed in the surrounding fluid. Its bottom surface is in contact with the lower hemisphere, not surrounding fluid.

Chet

Vibhor
Geofleur
Gold Member
Update: You can check your final answer for the force if you take the limit as ## g \rightarrow 0 ##.

Vibhor and Chestermiller
No. I'm saying that Archimedes Principle cannot be applied to the upper hemisphere because it is not fully immersed in the surrounding fluid. Its bottom surface is in contact with the lower hemisphere, not surrounding fluid.

Chet
Can we calculate force due to liquid on the whole sphere using Archimedes Principle considering the fact that the sphere rests on the bottom surface (the sphere is not completely immersed in water as bottomost point is touching the container)?

Chestermiller
Mentor
Can we calculate force due to liquid on the whole sphere using Archimedes Principle considering the fact that the sphere rests on the bottom surface (the sphere is not completely immersed in water as bottomost point is touching the container)?
The contact is only at a single geometric point, so this doesn't affect the force that the surrounding fluid exerts on the ball.

Chet

Vibhor
Chestermiller
Mentor
In the attached figure consider a circular strip of width ##Rdθ## at an angle θ to the horizontal ( the horizontal line at a height R from the bottom) .

The area of the strip is ##2\pi R^2 cos\theta dθ##

Pressure at this strip is ##P_0 + \rho g(3R-Rsin\theta)##

Force due to liquid at this pressure will be ##[P_0 + \rho g(3R-Rsin\theta)][2\pi R^2 cos\theta dθ]## .

But we should only consider the vertical component of this force which would be given by ##[P_0 + \rho g(3R-Rsin\theta)][2\pi R^2 cos\theta sin\theta dθ]##

##\int_0^{\frac{\pi}{2}}[P_0 + \rho g(3R-Rsin\theta)][2\pi R^2 cos\theta sin\theta dθ] = \frac{\pi R^2}{3}[3P_0+7\rho Rg]## .

Is this what you are getting ?
Yes. Nice job.

Chestermiller
Mentor
Update: You can check your final answer for the force if you take the limit as ## g \rightarrow 0 ##.
Yes. This is a really cute trick. With this trick, you can eliminate 3 of the choices immediately.

Chet

The contact is only at a single geometric point, so this doesn't affect the force that the surrounding fluid exerts on the ball.

Chet
So upward force due to fluid on whole sphere would be ##\frac{4}{3}\pi R^3 \rho g## ?

Yes. This is a really cute trick. With this trick, you can eliminate 3 of the choices immediately.

Chet
Is this a check of what would happen if the container were in a free fall ?

Geofleur
Gold Member
I had in mind to consider the same problem far from any source of gravity, so that ## g = 0 ##. But, according to the equivalence principle of general relativity, a free fall frame locally amounts to the same thing!

Vibhor
Chestermiller
Mentor
I had in mind the case where ρ = 0.

Chet

I had in mind the case where ρ = 0.

Chet
Sorry I did not understand what you are conveying . In addition , could you respond to post#12 ?

Chestermiller
Mentor
So upward force due to fluid on whole sphere would be ##\frac{4}{3}\pi R^3 \rho g## ?
Yes.

Vibhor
Chestermiller
Mentor
Sorry I did not understand what you are conveying . In addition , could you respond to post#12 ?
What I'm saying is that, in the limit as the fluid density approaches zero, the pressure around the ball becomes spatially uniform, and equal to the atmospheric pressure. Under these circumstances, the downward force on the hemisphere is just equal to the atmospheric pressure times the projected area πr2.

Chet

Vibhor