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For a submerged object, buoyant force ($F_b$) is defined as:

$$F_b = V_{\text{submerged}} \times \rho \text{ (density)} \times g \text{ (gravitational constant)}$$

Conceptually, the buoyant force equation says that buoyant force exerted is equal to the weight of the volume of water a given object displaced. Why? I went online to http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Pressure/BouyantForce.html

and found the following explanation, but it seems like non-sequitur to me:

Explanation: When an object is removed, the volume that the object occupied will fill with fluid. This volume of fluid must be supported by the pressure of the surrounding liquid since a fluid can not support itself. When no object is present, the net upward force on this volume of fluid must equal to its weight, i.e. the weight of the fluid displaced. When the object is present, this same upward force will act on the object.

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The argument sounds perfectly reasonable.

Consider arbitrary parcel of fluid in equilibrium. It exerts downward force equal to it's weight on the surrounding fluid, and it does not move. Therefore according to the second law of motion, the downward force must be balanced by upward force of equal magnitude, the buoyant force (otherwise it would start to move, contradicting the equilibrium assumption).

The buoyant force is exerted by the fluid surrounding the parcel. Therefore if we replace the parcel with something else, there is no reason for that force to change.

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  • $\begingroup$ thanks for the help. Just a follow-through: I can understand that the fluid surrounding the parcel exerts the buoyant force. But what causes the fluid surrounding the parcel to exert a buoyant force? Can I draw the following comparison, in which a block rests upon the earth:if the normal force (equal in magnitude to weight and the "counterpart" to the buoyant force) occurs a result of electrostatic repulsion between the block and earth, shouldn't the buoyant force occur as a result of repuslion between parcel and surrounding fluid. Moreover, shouldn't Fb's magnitude = weight of object above it $\endgroup$ – Muno Sep 21 '14 at 2:45
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    $\begingroup$ to make my question more succinct: if buoyant force depends on a difference in pressure, and pressure at a particular depth depends upon the weight above it, why isn't a submerged object's weight factored into pressure? $\endgroup$ – Muno Sep 21 '14 at 10:09
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The heart of that (rather elegant and well-known) argument is that

When the object is present, this same upward force will act on the object.

If you accept this, then everything else is just a straightforward application of Newton's Second Law. A rephrasing of that statement is

The force exerted by the surrounding fluid on a submerged object depends only on what that object's surface looks like and how it is situated in the water.

To see why this is true, recall that at each point $\mathbf x$ in the fluid, we can associate a number $P(\mathbf x)$ which gives the pressure in the fluid at that point. This means that the force exerted on a small directed area element $d\mathbf a$ of a surface is $-P(\mathbf x)d\mathbf a$. The net force on any surface $S$ is obtained by breaking it up into little area elements, and summing up the forces on all of them. In other words, it's obtained by doing an integral; \begin{align} \mathbf F[S] = -\int_S P(\mathbf x)d\mathbf a. \end{align} This means that when you put an object under water, the force exerted on its surface by the water depends only on the shape of its surface, and where it is in the water, not on what it's made of. In particular, it would be the same if it were composed of water or of wood or any other sufficiently rigid material for that matter.

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  • $\begingroup$ if buoyant force depends on a difference in pressure, and pressure at a particular depth depends upon the weight above it, why isn't a submerged object's weight factored into pressure? $\endgroup$ – Muno Sep 22 '14 at 10:28
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    $\begingroup$ @user21945 Pressure at a particular depth arises from the weight of the fluid above that depth, not from the weight of the object. $\endgroup$ – joshphysics Sep 26 '14 at 15:55

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