If you hold an object underwater and then release it, it floats to the surface, sinks, or remains “hanging” in the water. How can we predict which it will do? The net force on the object an instant after you release it is $\vec{F}{}_{\text{net}}=\stackrel{}{(F}{}_{\text{B}}-m_{\text{o}}g\text{,}\text{upward})$. Whether it heads for the surface or the bottom depends on whether the buoyant force $F_{\text{B}}$is larger or smaller than the object’s weight $m_{\text{o}}g$.

The magnitude of the buoyant force is ${\rho }_{\text{f}}{V}_{\text{f}}g$. The weight of a uniform object, such as a block of steel, is simply ${\rho }_{\text{o}}{V}_{\text{o}}g$. But a compound object, such as a scuba diver, may have pieces of varying density. If we define the average density to be ${\rho }_{\text{avg}}=m_{\text{o}}\text{/}{V}_{\text{o}}$, the weight of a compound object is ${\rho }_{\text{avg}}{V}_{\text{o}}g$.

Comparing ${\rho }_{\text{f}}{V}_{\text{f}}g$ to ${\rho }_{\text{avg}}{V}_{\text{o}}g$, and noting that $V_{\text{f}}=V_{\text{o}}$ for an object that is fully submerged, we see that an object floats or sinks depending on whether the fluid density ${\rho }_{\text{f}}$ is larger or smaller than the object’s average density ${\rho }_{\text{avg}}$. If the densities are equal, the object is in equilibrium and hangs motionless. This is called neutral buoyancy. These conditions are summarized in Tactics Box 14.2.

TACTICS BOX 14.2 Finding whether an object floats or sinks

Full Alternative Text

An object sinks if it weighs more than the fluid it displaces—that is, if its average density is greater than the density of the fluid.

Full Alternative Text

An object floats on the surface if it weighs less than the fluid it displaces—that is, if its average density is less than the density of the fluid.

Full Alternative Text

An object hangs motionless if it weighs exactly the same as the fluid it displaces—that is, if its average density equals the density of the fluid.

Workbook Exercises 14–18

If you release a block of wood underwater, the net upward force causes the block to shoot to the surface. Then what? Let’s begin with a uniform object such as the block shown in FIGURE 14.20. This object contains nothing tricky, like indentations or voids. Because it’s floating, it must be the case that ${\rho }_{\text{o}}<{\rho }_{\text{f}}$.

FIGURE 14.20 A floating object is in equilibrium.

Now that the object is floating, it’s in equilibrium. The upward buoyant force, given by Archimedes’ principle, exactly balances the downward weight of the object. That is,

In this case, the volume of the displaced fluid is not the same as the volume of the object. In fact, we can see from Equation 14.13 that the volume of fluid displaced by a floating object of uniform density is

You’ve often heard it said that “90% of an iceberg is underwater.” Equation 14.14 is the basis for that statement. Most icebergs break off glaciers and are fresh-water ice with a density of $917\text{kg}\text{/}{\text{m}}^3$. The density of seawater is $1030\text{kg}\text{/}{\text{m}}^3$. Thus

$V_{\text{f}}$, the displaced water, is the volume of the iceberg that is underwater. You can see that, indeed, 89% of the volume of an iceberg is underwater.

EXAMPLE 14.9 Measuring the density of an unknown liquid

You need to determine the density of an unknown liquid. You notice that a block floats in this liquid with 4.6 cm of the side of the block submerged. When the block is placed in water, it also floats but with 5.8 cm submerged. What is the density of the unknown liquid?

MODEL The block is an object of uniform composition.

VISUALIZE FIGURE 14.21 shows the block and defines the cross-section area $A$ and submerged lengths $h_{\text{u}}$ in the unknown liquid and $h_{\text{w}}$ in water.

FIGURE 14.21 More of the block is submerged in water than in an unknown liquid.

FIGURE 14.21 Full Alternative Text

SOLVE The block is floating, so Equation 14.14 applies. The block displaces volume $V_{\text{u}}=A{h}_{\text{u}}$ of the unknown liquid. Thus

$$V_{\text{u}}=A{h}_{\text{u}}=\frac{{\rho }_{\text{o}}}{{\rho }_{\text{u}}}{V}_{\text{o}}$$

Similarly, the block displaces volume $V_{\text{w}}=A{h}_{\text{w}}$ of the water, leading to

$$V_{\text{w}}=A{h}_{\text{w}}=\frac{{\rho }_{\text{o}}}{{\rho }_{\text{w}}}{V}_{\text{o}}$$

Because there are two fluids, we’ve used subscripts w for water and u for the unknown in place of the fluid subscript f. The product ${\rho }_{\text{o}}{V}_{\text{o}}$ appears in both equations; hence

$${\rho }_{\text{u}}A{h}_{\text{u}}={\rho }_w{}_{}A{h}_{\text{w}}$$

The area $A$ cancels, and the density of the unknown liquid is

$${\rho }_{\text{u}}=\frac{h_{\text{w}}}{h_{\text{u}}}{\rho }_{\text{w}}=\frac{5.8\text{cm}}{4.6\text{cm}}\times 1000{\text{kg/m}}^3=1260{\text{kg/m}}^3$$

REVIEW Comparison with Table 14.1 shows that the unknown liquid is likely to be glycerin.

As an example, steel is denser than water, so a chunk of steel sinks. Oil is less dense than water, so oil floats on water. Fish use swim bladders filled with air and scuba divers use weighted belts to adjust their average density to match the density of water. Both are examples of neutral buoyancy.