4
$\begingroup$

In Wikipedia, the Archimedes' principle is stated like this:

"The upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces"

In one section of our Physics textbook, they showed the proof of Archimedes' principle like this:

Lets say a cylinder is submerged completely in some fluid.Lets say the height if the cylinder is $h$ and the cross-sectional area of the cylinder is $A$ .Let us imagine that the cylinder is submerged in the fluid in such a way that the depth of the upper surface is $h_1$ and the depth of the lower surface is $h_2$.

We said a lot of times to you that the pressure in fluid / Gas does not act in a particular direction. The pressure acts on all direction .

So, the downward pressure acting on the cylinders upper surface is,

$$P_1=h_1\rho g$$

And, the upward pressure acting on the cylinders downward surface is,

$$P_2=h_2\rho g$$

So, the downward force acting on the cylinders upper surface and the upward force acting on the cylinders downward surface is ,

$$F_1 = AP_1 = Ah_1\rho g$$ $$F_2 = AP_2 = Ah_2\rho g$$

We don't have to think about the force acting around the cylinder .because if a force acts on one side of the cylinder , then another opposite force cancels out the original force.Because $h_2$ is greater than $h_1$ , $F_2$ has to be greater than $F_1$ .So the net force will be pointing upwards and its magnitude will be,

$$F = F_2 -F_1 = A(h_2 - h_1)\rho g = Ah\rho g$$

At this point , the proof is done.

There is some places where i felt some doubt myself.For example , They first say that pressure has no direction . But then they start saying things like "upward and downward pressure" in the proof.Also , Why that use the equation $P = h\rho g$ to calculate the upward pressure that the fluid puts on the bottom surface of the cylinder? ...etc.

At this point I have two question,

  1. Is this proof valid?

  2. How can I write a proof with any general solid? (not just cylinder)

Edit : For now,the meaning of the word ‘immersed’ will be ‘fully surrounded by a liquid’. It is true that it will make Phenomenons like 'boats floating' out of scope , but I have did this to reduce too much complexity.

$\endgroup$

2 Answers 2

4
$\begingroup$
  1. Is this proof valid ?

Yes, this proof for the cylindrical body is valid.

But the author should better say "upward and downward pressure force", instead of "upward and downward pressure", because pressure has no direction (as you correctly pointed out).

  1. How can I write a proof with any general solid ? (not just cylinder)

Archimedes' principle for an arbitrarily shaped body can most easily be proved with Gauss' gradient theorem. This theorem relates an integral over a closed surface area $\partial V$ to an integral over the enclosed volume $V$. $$\oint_{\partial V} p(\vec{r})\ d\vec{A} = \int_V \vec{\nabla} p(\vec{r})\ dV \tag{1}$$ where $p(\vec{r})$ is any position-dependent function, and $\vec{\nabla}$ is the gradient operator.

Now, as the position-dependent function we choose the pressure $$p(\vec{r})=p_0-\rho gz \tag{2}$$ where $z$ is the vertical position coordinate and $p_0$ is the pressure at zero-level ($z=0$). We need a minus sign here, because pressure increases when going down in the liquid (i.e. in negative $z$-direction).

Then the gradient of (2) is $$\vec{\nabla}p(\vec{r})=-\rho g\hat{z} \tag{3}$$ where $\hat{z}$ is the unit-vector in $z$-direction (i.e. upwards).

Inserting (3) into (1) we get $$\oint_{\partial V} p(\vec{r}) d\vec{A} = \int_V (-\rho g\hat{z})\ dV. $$

Now on the left side $p\ d\vec{A}$ obviously is the pressure force acting on the surface area element $d\vec{A}$ (except for a minus sign, because the force element $d\vec{F}$ points inside the body, while the area element $d\vec{A}$ points outside). And on the right side, the constants $(-\rho g\hat{z})$ can be factored out. So we get $$-\oint_{\partial V} d\vec{F}=-\rho g \hat{z} \int_V dV$$ or finally $$\vec{F}=\rho g \hat{z} V.$$ This is just Archimedes' principle (Buoyant force is pointing upwards and equal to the weight of the displaced liquid).

$\endgroup$
2
$\begingroup$

The proof is valid for a cylinder. You are right, though, that it is incorrect to talk about 'upward pressure' and 'downward pressure'. It is fine, of course, to talk about the upward force and downward force due to fluid pressure.

There are two well known ways to derive A's Principle for a general shape of solid. The first is a generalisation of the method you have quoted for a cylinder: we consider the sum (or integral) of vertical components of the forces due to hydrostatic pressure on the variously-angled elements of the solid's surface. The second method is this...

Consider the chunk, C, of fluid that used to occupy the space now occupied by the solid. C will have been in equilibrium (if the fluid is stationary), so the net force from the surrounding fluid on C must be equal and opposite to the weight of C. Now that C is replaced by the solid with exactly the same shape of surface as C, the resultant hydrostatic upthrust will be the same. Hence Archimedes' Principle.

$\endgroup$
3
  • $\begingroup$ I found this interesting line of text in a paper : " Archimedes developed rigorous mathematical proofs for most his ideas . However, the derivation of the exact force exerted by an in-homogeneous fluid on an arbitrarily-shaped body immersed in it, as will be shown here, demands the knowledge of the divergence theorem, a mathematical tool that was out of reach for the ancients. Therefore, the validity of the Archimedes propositions for this more general case was not formally proved on his original work." $\endgroup$ Nov 18, 2020 at 14:57
  • $\begingroup$ That rings true. I didn't, of course, consider an inhomogeneous fluid, though I can't, at the moment, see why the method I offered in my last paragraph wouldn't still apply. $\endgroup$ Nov 18, 2020 at 15:02
  • $\begingroup$ So I guess it means , proving AP for homogeneous (in-compressible) fluid = easy enough . Proving AP for in-homogeneous fluid = hard ? $\endgroup$ Nov 18, 2020 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.