Does Depth Affect Buoyancy Force on a Spherical Object?

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

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For this problem,

The solution is,

I agree with their solution from Archimedes principle, however, if I look at the equation $PA = F$, where $A = 4\pi r^2$ and $F$ is the force over the entire apple assuming it is spherical. I thought the pressure will increase with depth and therefore I think that therefore so should the buoyancy force.

However, then I realize that the buoyancy force is due to the difference in pressure between the top and bottom of the apple assuming that the hemisphere area is the same. Therefore, since the pressure increases by the same amount assuming constant density, then the buoyancy force will be constant.

Is my alternative reasoning please correct?

Many thanks!

 

[jbriggs444]

However, then I realize that the buoyancy force is due to the difference in pressure between the top and bottom of the apple assuming that the hemisphere area is the same. Therefore, since the pressure increases by the same amount assuming constant density, then the buoyancy force will be constant.

Is my alternative reasoning please correct?

Yes, your alternative reasoning quoted above is correct. Buoyancy is about pressure differences, not absolute pressure.

 

[kuruman]

I agree with their solution from Archimedes principle, however, if I look at the equation $PA = F$, where $A = 4\pi r^2$ and $F$ is the force over the entire apple assuming it is spherical.

This is faulty reasoning. The total force exerted on the spherical apple by the water is most certainly not $F=PA=4\pi r^2P.$ You correctly abandoned your expression when you realized that it assumes that the pressure is constant over the surface of the apple. Setting that aside, there is another reason that makes the expression incorrect. Do you see what it is? Hint: Force is a vector.

 

[ChiralSuperfields]

This is faulty reasoning. The total force exerted on the spherical apple by the water is most certainly not $F=PA=4\pi r^2P.$ You correctly abandoned your expression when you realized that it assumes that the pressure is constant over the surface of the apple. Setting that aside, there is another reason that makes the expression incorrect. Do you see what it is? Hint: Force is a vector.

Thank you for your replies @jbriggs444 and @kuruman!

@kuruman, I think I was meant to say that the force was inward. I see why my expression is incorrect I think. Since pressure is not constant along the surface of the apple I think we need to do some kind of integral.

Many thanks!

 

[kuruman]

@kuruman, I think I was meant to say that the force was inward. I see why my expression is incorrect I think. Since pressure is not constant along the surface of the apple I think we need to do some kind of integral.

There is a reason in addition to the pressure not being the same at different part on the apple. What is that reason? I gave you a hint.

Yes, we need to do some kind of integral. That integral has a special name. Do you know what it is?

 

[ChiralSuperfields]

There is a reason in addition to the pressure not being the same at different part on the apple. What is that reason? I gave you a hint.

Yes, we need to do some kind of integral. That integral has a special name. Do you know what it is?

Thank you for your reply @kuruman!

It looks like it is a surface integral. Is the other reason that the pressure is not the same at different points on an apple due to my assumption of the apple being spherical invalid?

Many thanks!

 

[kuruman]

It looks like it is a surface integral. Is the other reason that the pressure is not the same at different points on an apple due to my assumption of the apple being spherical invalid?

That the pressure varies on the skin of the apple from one point to the other is valid and not an assumption. The hydrostatic pressure is $~p=\rho_{\text{water}}g~d~$ where $d$ is the depth at which the pressure is considered. Any apple, spherical or not, that has points at different depths experiences a force that varies from one point to another. The other reason is not related to that.

Here is a picture of a potato submerged under water.

The arrows represent the force acting at specific points. The integral that you propose must add these arrows somehow. The other reason is related to how exactly you are going to add these arrows. "Surface integral" is just a name. What is the procedure that you are going to follow to find a number for this so-called surface integral?

Moreover, my related additional question is what is this surface integral also known as?

 

[ChiralSuperfields]

That the pressure varies on the skin of the apple from one point to the other is valid and not an assumption. The hydrostatic pressure is $~p=\rho_{\text{water}}g~d~$ where $d$ is the depth at which the pressure is considered. Any apple, spherical or not, that has points at different depths experiences a force that varies from one point to another. The other reason is not related to that.

Here is a picture of a potato submerged under water.
View attachment 325772
The arrows represent the force acting at specific points. The integral that you propose must add these arrows somehow. The other reason is related to how exactly you are going to add these arrows. "Surface integral" is just a name. What is the procedure that you are going to follow to find a number for this so-called surface integral?

Moreover, my related additional question is what is this surface integral also known as?

Thank you for your reply @kuruman!

Sorry I have not done multivariate calculus yet, but the other term for the surface integral is a double integral. Maybe I should come back to this at a maximum time frame of a year where I should have finished Stewart's.

Many thanks!

 

[jbriggs444]

I believe that the point that @kuruman is trying to make is not so much about integrals as it is about vectors.

Way back up in your original post, you spoke about getting the "total force" on a spherical object. You were going to calculate this total force in a simple way: average pressure times total surface area.

In effect, you were going to calculate "total force" by adding up all of the smaller forces on all of the little bits of surface area. The problems is that forces are vectors. You cannot add up a bunch of non-parallel vectors by adding up their magnitudes.

If you wanted total force on a flat surface under a constant pressure then $\vec{F} = P\vec{A}$ would be correct. That would be the distributive law at work. $\sum_i ( p_i \times a_i ) = p_\text{constant} \times \sum_i a_i$

But it turns out that the individual area elements here are not scalars. They are vectors. Directed areas. The magnitude of a directed area is just the surface area of the area element. The direction is that of a vector normal to a particular side of the surface. The multiplication of a directed area by a pressure is not technically the multiplication of two scalars yielding a scalar. Technically, it is the multiplication of a tensor (the Cauchy stress tensor) by the directed area vector yielding a force vector.

Do not panic. You'll likely get out of first year physics without ever learning about the Cauchy stress tensor. Probably without learning about tensors at all. However, I do not want to leave you hanging. Nothing here is beyond the grasp of a willing student...

In component form, the Cauchy stress tensor is a 3 x 3 matrix with direction (x, y and z) on one axis and force components (x, y and z) on the other axis. Each of the nine terms gives a component of stress (force per unit area): "how much force in this direction exists across a unit of directed area in that direction".

In a fluid only the three terms on the main diagonal will be non-zero. Fluids cannot maintain shear forces. Those main diagonal terms will all be equal to one another and will be equal to the pressure within the fluid. So all you have is an identity matrix scaled up by a factor of pressure.

When you matrix-multiply an area element by this tensor, it is just like multiplication of a vector by a scalar. You get a force in the direction normal to the surface of the area element. Just as one would expect for the force resulting from a particular pressure on a particular area element.

If we bring this back to the problem at hand, we are trying to compute "total force" on an object. We assume that pressure is approximately constant and are trying to assert that:$$F = \sum_i \vec{a_i} \times P_\text{constant} = P_\text{constant} \times \sum_i \vec{a_i} = P_\text{constant} \times 4\pi r^2$$The difficulty is the same as for average speed versus average velocity:$$0 = |\sum_i \vec{a_i}| \neq \sum_i |\vec{a_i}| = 4 \pi r^2$$The magnitude of the sum is not the same as the sum of the magnitudes.

[One can easily see that the sum of directed areas for a sphere is zero. For every directed area on the one side of a sphere, there is an equal and opposite directed area diametrically opposed. The sum has to be zero. With some vigorous handwaving, one can assert that the same holds for any two-dimensional surface that encloses a volume.]

 

[kuruman]

I believe that the point that @kuruman is trying to make is not so much about integrals as it is about vectors.

Your belief is correct.
A (relatively) simple way for doing the integral is posted by @Chestermiller here.
However, the name of what this integral represents still eludes us $~\dots$

 

[jbriggs444]

However, the name of what this integral represents still eludes us $~\dots$

It is eluding me as well. Possibly you are trying to point out that this vector sum of forces from external fluid pressure on an object is called "buoyancy".^(*)

Possibly you would go on to assert some sort of theorem that equates this surface integral of force over area with a corresponding integral of force over volume: "The buoyant force on a submerged object is equal to the weight of the displaced fluid".

(*) Back when I was young and stupid, I wondered whether "buoyancy" was just a name for the net effect of fluid pressure difference or whether it was a separate effect in its own right. Never got an answer. It seemed obvious that the two would be equal in value regardless of shape, but my instructors never explicitly identified an underlying cause for buoyancy. I eventually realized that there was just the one thing. "Buoyancy" is just a word for the net effect of [static] fluid pressure difference.

 

[kuruman]

I am trying to point out that the vector sum of the elemental force vectors $d\vec F=-(p~dA)\hat n$ ($\hat n$ is the outward normal) is the net force exerted by the fluid on the immersed object, more commonly known as "the buoyant force".

There are only two entities exerting forces on the object, Earth and fluid. The force exerted by the Earth is already accounted for by $mg$. That leaves the net force exerted by the (ideal) fluid on the immersed object which has been given the name "buoyant force."

 

[ChiralSuperfields]

Thank you for your replies @jbriggs444 and @kuruman!

I will look into this in greater detail when I have more time in a few months.

Many thanks!