Archimedes' Principle and Buoyancy

Archimedes' Principle and the buoyant force govern objects submerged in or floating on fluids.

The Buoyant Force

The buoyant force is the upward force exerted by a fluid on any object submerged in it.

Because pressure in a fluid increases with depth ( $p = h \rho g$ ), the upward force pushing on the bottom of a submerged object is greater than the downward force on its top. This net upward force is present whether the object floats, sinks, or remains suspended.

An object's floating or sinking behavior depends on the balance between the buoyant force and its weight:

Sinks: Buoyant force is less than the object's weight ( $F_B < w_{\text{obj}}$ ).
Floats/Suspended: Buoyant force equals the object's weight ( $F_B = w_{\text{obj}}$ ).

Archimedes' Principle

The buoyant force on an object equals the weight of the fluid it displaces:

F_B = w_{\text{fl}} = m_{\text{fl}} g = \rho_{\text{fl}} V_{\text{fl}} g

(1)

Density and Floating

An object's average density ( $\rho_{\text{obj}}$ ) determines whether it floats or sinks relative to the fluid's density ( $\rho_{\text{fl}}$ ):

$\rho_{\text{obj}} < \rho_{\text{fl}}$ : The object floats because displacing its entire volume yields a fluid weight greater than its own weight.
$\rho_{\text{obj}} > \rho_{\text{fl}}$ : The object sinks.

Example: A lump of clay sinks in water, but molding the same lump into the shape of a boat allows it to float. The boat shape displaces a larger volume of water without changing mass, creating a greater buoyant force.

Fraction Submerged

For a floating object in equilibrium, its mass equals the mass of displaced fluid ( $m_{\text{obj}} = m_{\text{fl}}$ ). Substituting $m = \rho V$ yields the fraction of the object's volume submerged:

\text{Fraction Submerged} = \frac{V_{\text{sub}}}{V_{\text{obj}}} = \frac{\rho_{\text{obj}}}{\rho_{\text{fl}}}

(2)

Apparent Weight

When an object is weighed while completely submerged, it suffers an apparent weight loss equal to the weight of the displaced fluid:

w_{\text{apparent}} = w_{\text{obj}} - F_B = w_{\text{obj}} - w_{\text{fl}}

(3)

Measuring an object's true weight in air and its apparent weight submerged allows determination of unknown densities.

Example 1

A $60.0\text{ kg}$ person floats in fresh water ( $\rho_{\text{fl}} = 1000\text{ kg/m}^3$ ) with $97.0\%$ of her volume submerged when her lungs are full of air. What is her average density?

Hint Click to unblur

Rearrange Equation 2 to solve for $\rho_{\text{obj}}$ using the fraction submerged.

Solution Click to unblur

Using Equation 2:

\frac{V_{\text{sub}}}{V_{\text{obj}}} = \frac{\rho_{\text{person}}}{\rho_{\text{fl}}}

\rho_{\text{person}} = \left(\text{Fraction Submerged}\right) \cdot \rho_{\text{fl}} = (0.970)(1000\text{ kg/m}^3) = 970\text{ kg/m}^3

Example 2

A solid metal crown weighs $24.5 \text{ N}$ in air. When completely submerged in fresh water ( $\rho_{\text{fl}} = 1000 \text{ kg/m}^3$ ), its apparent weight is measured to be $22.6 \text{ N}$ . Find the average density of the crown.

Hint Click to unblur

Find the buoyant force using $F_B = w_{\text{obj}} - w_{\text{apparent}}$ , then relate $F_B$ to the displaced fluid volume to calculate $\rho_{\text{obj}}$ .

Solution Click to unblur

First, find the buoyant force from the apparent weight loss (Equation 3):

F_B = w_{\text{obj}} - w_{\text{apparent}} = 24.5 \text{ N} - 22.6 \text{ N} = 1.90 \text{ N}

Since the crown is fully submerged, the volume of displaced water equals the volume of the crown ( $V_{\text{fl}} = V_{\text{obj}}$ ). Use Archimedes' Principle (Equation 1) to find this volume:

F_B = \rho_{\text{fl}} V_{\text{obj}} g \implies V_{\text{obj}} = \frac{F_B}{\rho_{\text{fl}} g} = \frac{1.90 \text{ N}}{(1000 \text{ kg/m}^3)(9.80 \text{ m/s}^2)} \approx 1.939 \cdot 10^{-4} \text{ m}^3

Calculate the mass of the crown from its true weight:

m_{\text{obj}} = \frac{w_{\text{obj}}}{g} = \frac{24.5 \text{ N}}{9.80 \text{ m/s}^2} = 2.50 \text{ kg}

Calculate the crown's average density:

\rho_{\text{obj}} = \frac{m_{\text{obj}}}{V_{\text{obj}}} = \frac{2.50 \text{ kg}}{1.939 \cdot 10^{-4} \text{ m}^3} \approx 1.29 \cdot 10^4 \text{ kg/m}^3