A rock, as you know, sinks like a rock. Wood floats on the surface of a lake. A penny with a mass of a few grams sinks, but a massive steel aircraft carrier floats. How can we understand these diverse phenomena?

An air mattress floats effortlessly on the surface of a swimming pool. But if you’ve ever tried to push an air mattress underwater, you know it is nearly impossible. As you push down, the water pushes up. This net upward force of a fluid is called the buoyant force.

The basic reason for the buoyant force is easy to understand. FIGURE 14.17 shows a cylinder submerged in a liquid. The pressure in the liquid increases with depth, so the pressure at the bottom of the cylinder is larger than at the top. Both cylinder ends have equal area, so force $\vec{F}{}_{\text{up}}$ is larger than force $\vec{F}{}_{\text{down}}$. (Remember that pressure forces push in all directions.) Consequently, the pressure in the liquid exerts a net upward force on the cylinder of magnitude $F_{\text{net}}=F_{\text{up}}-F_{\text{down}}$. This is the buoyant force.

FIGURE 14.17 The buoyant force arises because the fluid pressure at the bottom of the cylinder is larger than at the top.

The submerged cylinder illustrates the idea in a simple way, but the result is not limited to cylinders or to liquids. Suppose we isolate a parcel of fluid of arbitrary shape and volume by drawing an imaginary boundary around it, as shown in FIGURE 14.18a. This parcel is in hydrostatic equilibrium; consequently, the gravitational force pulling down on the parcel must be balanced by an upward force. The upward force, which is exerted on this parcel of fluid by the surrounding fluid, is the buoyant force $\vec{F}{}_{\text{B}}$. The buoyant force matches the weight of the fluid: $F_{\text{B}}=mg$.

FIGURE 14.18 The buoyant force.

Full Alternative Text

Imagine that we could somehow remove this parcel of fluid and instantaneously replace it with an object of exactly the same shape and size, as shown in FIGURE 14.18b. Because the buoyant force is exerted by the surrounding fluid, and the surrounding fluid hasn’t changed, the buoyant force on this new object is exactly the same as the buoyant force on the parcel of fluid that we removed.

When an object (or a portion of an object) is immersed in a fluid, it displaces fluid that would otherwise fill that region of space. This fluid is called the displaced fluid. The displaced fluid’s volume is exactly the volume of the portion of the object that is immersed in the fluid. Figure 14.18 leads us to conclude that the magnitude of the upward buoyant force matches the weight of this displaced fluid.

This idea was first recognized by the ancient Greek mathematician and scientist Archimedes, perhaps the greatest scientist of antiquity, and today we know it as Archimedes’ principle.

Suppose the fluid has density ${\rho }_{\text{f}}$ and the object displaces volume $V_{\text{f}}$ of fluid. The mass of the displaced fluid is $m_{\text{f}}={\rho }_{\text{f}}{V}_{\text{f}}$ and so its weight is $m_{\text{f}}g={\rho }_{\text{f}}{V}_{\text{f}}g$. Thus Archimedes’ principle in equation form is

EXAMPLE 14.8 Holding a block of wood underwater

A $10\text{cm}\times 10\text{cm}\times 10\text{cm}$ block of wood with density $700\text{kg}\text{/}{\text{m}}^3$ is held underwater by a string tied to the bottom of the container. What is the tension in the string?

MODEL The buoyant force is given by Archimedes’ principle.

VISUALIZE FIGURE 14.19 shows the forces acting on the wood.

FIGURE 14.19 The forces acting on the submerged wood.

FIGURE 14.19 Full Alternative Text

SOLVE The block is in equilibrium, so

$$\sum F_y=F_{\text{B}}-T-m_{\text{o}}g=0$$

Thus the tension is $T=F_{\text{B}}-m_{\text{o}}g$. The mass of the block is $m_{\text{o}}={\rho }_{\text{o}}{V}_{\text{o}}$, and the buoyant force, given by Equation 14.12, is $F_{\text{B}}={\rho }_{\text{f}}{V}_{\text{f}}g$. Thus

$$T={\rho }_{\text{f}}{V}_{\text{f}}g-{\rho }_{\text{o}}{V}_{\text{o}}g=({\rho }_{\text{f}}-{\rho }_{\text{o}})V_{\text{o}}g$$

where we’ve used the fact that $V_{\text{f}}=V_{\text{o}}$for a completely submerged object. The volume is $V_{\text{o}}=1000{\text{cm}}^3=1.0\times {10}^{-3}{\text{m}}^3$, and hence the tension in the string is

$$\begin{array}{ll}T=\hfill & \left(\left(1000\text{ kg}\text{/}{\text{m}}^3\right)-\left(700\text{kg}\text{/}{\text{m}}^3\right)\right)\hfill \\ \hfill & \times (1.0\times {10}^{-3}{\text{ m}}^3)(9.8{\text{ m/s}}^2)=2.9\text{ N}\hfill \end{array}$$

REVIEW The tension depends on the difference in densities. The tension would vanish if the wood density matched the water density.