Buoyancy: Archimedes' Principle

1. The Origin of Buoyancy: A Net Pressure Force

Imagine any object submerged in a fluid. From our previous discussions, you know that the pressure exerted by the fluid increases with depth. This means the pressure on the bottom surface of the object is greater than the pressure on its top surface. The fluid also exerts pressure on the sides, but these horizontal forces tend to cancel each other out for a symmetric object. The imbalance in pressure between the top and bottom surfaces results in a net upward force. This net upward force is the buoyant force.

Let's think about this more formally. Consider an arbitrarily shaped object submerged in a fluid. The buoyant force is the vector sum of all the pressure forces acting on every infinitesimal area element of the object's surface.

The beautiful insight, attributed to Archimedes, is that this net upward force is exactly equal to the weight of the fluid that would occupy the volume of the submerged portion of the object. Why? Because if the object weren't there, a volume of fluid identical in shape would be held in equilibrium by the exact same surrounding pressure forces. Therefore, the net pressure force (the buoyant force) must be equal and opposite to the weight of that "displaced" fluid.

Let's see a formal, calculus-based derivation of this result.

The derivation you just read culminates in the mathematical statement of Archimedes' Principle:

where:

• $F_B$ is the magnitude of the buoyant force.
• ${\rho }_{\text{fluid}}$ is the density of the fluid.
• $g$ is the acceleration due to gravity.
• $V_{\text{submerged}}$ is the volume of the portion of the object that is submerged in the fluid.

The term ${\rho }_{\text{fluid}}{V}_{\text{submerged}}$ is the mass of the fluid displaced by the object. Multiplying by $g$ gives the weight of the displaced fluid.

For a more visual and intuitive walkthrough of this derivation and principle, the following video is highly recommended.

2. Applications of Archimedes' Principle

Now that we have the core principle, let's apply it to the two main scenarios: fully submerged objects and floating objects.

Case 1: Fully Submerged Objects and Apparent Weight

When an object is fully submerged, $V_{\text{submerged}}=V_{\text{object}}$. The object is subject to two main vertical forces: its own weight ($W=m_{\text{obj}}g$) acting downwards, and the buoyant force ($F_B$) acting upwards.

The net force on the object determines its acceleration. More practically, if you were to weigh the object while it's submerged (e.g., by hanging it from a spring scale), the scale would not read the true weight. It would read a lower value, known as the apparent weight.

The tension in the string (or the reading on the scale) would be:

This is why, as you've surely experienced, it's much easier to lift a heavy rock underwater than in the air. The buoyant force from the water is helping you.

Let's see this calculated in a concrete example.

Case 2: Floating Objects

An object floats when it is in static equilibrium while partially submerged. For this to happen, the net force on the object must be zero. This means the upward buoyant force must exactly balance the object's downward weight.

Let's expand this using our formulas:

Since $m_{\text{object}}={\rho }_{\text{object}}{V}_{\text{object}}$, we can substitute this in:

Canceling $g$ from both sides gives us a powerful and elegant relationship:

This equation tells us that the fraction of an object's volume that is submerged is equal to the ratio of the object's density to the fluid's density. This immediately explains a few things:

• If ${\rho }_{\text{object}}<{\rho }_{\text{fluid}}$, the fraction is less than 1, and the object floats.
• If ${\rho }_{\text{object}}>{\rho }_{\text{fluid}}$, the required submerged fraction would be greater than 1, which is impossible. The buoyant force, even when fully submerged, is not enough to support the weight, so the object sinks.
• If ${\rho }_{\text{object}}={\rho }_{\text{fluid}}$, the fraction is 1. The object is neutrally buoyant and will remain at any level where it is placed when fully submerged.

The following video explores these concepts with some classic and more advanced examples that are staples of competitive physics exams.

3. A Conceptual Test: The Melting Ice Cube

To truly solidify your understanding, let's consider a classic conceptual problem. An ice cube is floating in a glass of water, filled to the brim. When the ice cube melts, what happens to the water level? Does it rise (and overflow), fall, or stay exactly the same?

Think about this for a moment using the principles we've discussed. The floating ice cube displaces a volume of water whose weight is equal to the ice cube's total weight. When the ice melts, it becomes water. What volume does this new water occupy?

The following reading provides a concise and elegant mathematical proof of the answer.

The conclusion is that the water level remains unchanged! This is a non-trivial result that demonstrates the power of applying these principles correctly.

Conclusion

In this lesson, you've moved from the detailed mechanics of hydrostatic pressure to the elegant and powerful Archimedes' principle. This single principle allows you to solve a vast array of problems involving submerged and floating bodies.

Key Takeaways:

• The buoyant force arises from the pressure difference between the bottom and top of a submerged object.
• Archimedes' Principle states that the buoyant force is equal to the weight of the fluid displaced by the object: $F_B={\rho }_{\text{fluid}}g{V}_{\text{submerged}}$.
• For a fully submerged object, its apparent weight is its true weight minus the buoyant force: $W_{\text{app}}=W_{\text{true}}-F_B$.
• For a floating object, the buoyant force equals the object's weight, which leads to the relationship: $\frac{V_{\text{submerged}}}{V_{\text{object}}}=\frac{{\rho }_{\text{object}}}{{\rho }_{\text{fluid}}}$.

So far, we have only concerned ourselves with the magnitude of the buoyant force. However, just like the hydrostatic forces we studied before, the buoyant force also has a specific line of action. It acts through the centroid of the displaced volume, a point known as the center of buoyancy. In our next lesson, we will investigate the interplay between the center of gravity and the center of buoyancy to analyze the stability of floating bodies—a critical concept in ship design and engineering.