Have you ever tried to push a beach ball under water? This is extremely difficult to do because of the large upward force exerted by the water on the ball. The upward force exerted by water on any immersed object is called a buoyant force. We can determine the magnitude of a buoyant force by applying some logic and Newton’s second law. Imagine that, instead of air, the beach ball is filled with water. If you were standing on land, it would be difficult to hold the water-filled ball in your arms. If you held the ball while standing neck deep in a pool, however, the force you would need to hold it would almost disappear. In fact, the required force would be zero if we were to ignore the thin layer of plastic of which the beach ball is made. Because the water-filled ball is in equilibrium while it is submerged, the magnitude of the upward buoyant force must equal its weight.

If the submerged ball were filled with air rather than water, then the upward buoyant force exerted by the surrounding water would still be present. However, because the weight of the water is now replaced by the much smaller weight of that volume of air, the net force is upward and quite great; as a result, the ball is pushed to the surface.

Archimedes (c. 287–212 B.C.) Archimedes, a Greek mathematician, physicist, and engineer, was perhaps the greatest scientist of antiquity. He was the first to compute accurately the ratio of a circle’s circumference to its diameter, and he showed how to calculate the volume and surface area of spheres, cylinders, and other geometric shapes. He is well known for discovering the nature of the buoyant force. Archimedes was also a gifted inventor. One of his practical inventions, still in use today, is Archimedes’s screw–an inclined, rotating, coiled tube originally used to lift water from the holds of ships. He also invented the catapult and devised systems of levers, pulleys, and weights for raising heavy loads. Such inventions were successfully used to defend his native city Syracuse during a two-year siege by the Romans.

The manner in which buoyant forces act is summarized by Archimedes’s principle, which states that the magnitude of the buoyant force always equals the weight of the fluid displaced by the object. The buoyant force acts vertically upward through the point that was the center of gravity of the displaced fluid.

Note that Archimedes’s principle does not refer to the makeup of the object experiencing the buoyant force. The object’s composition is not a factor in the buoyant force. We can verify this in the following manner: Suppose we focus our attention on the indicated cube of liquid in the container illustrated in Figure 15.9. This cube is in equilibrium as it is acted on by two forces. One of these forces is the gravitational force F_{g} . What cancels this downward force? Apparently, the rest of the liquid in the container is holding the cube in equilibrium. Thus, the magnitude of the buoyant force B exerted on the cube is exactly equal to the magnitude of F_{g} , which is the weight of the liquid inside the cube:

Now imagine that the cube of liquid is replaced by a cube of steel of the same dimensions. What is the buoyant force acting on the steel? The liquid surrounding a cube behaves in the same way no matter what the cube is made of. Therefore, the buoyant force acting on the steel cube is the same as the buoyant force acting on a cube of liquid of the same dimensions. In other words, the magnitude of the buoyant force is the same as the weight of the liquid cube, not the steel cube. Although mathematically more complicated, this same principle applies to submerged objects of any shape, size, or density.

Although we have described the magnitude and direction of the buoyant force, we still do not know its origin. Why would a fluid exert such a strange force, almost as if the fluid were trying to expel a foreign body? To understand why, look again at Figure 15.9. The pressure at the bottom of the cube is greater than the pressure at the top by an amount ρgh, where h is the length of any side of the cube. The pressure difference ΔP between the bottom and top faces of the cube is equal to the buoyant force per unit area of those faces—that is, ΔP = B/A. Therefore, B = (ΔP)A = (ρgh)A = ρgV, where V is the volume of the cube. Because the mass of the fluid in the cube M = ρV, is we see that

where Mg is the weight of the fluid in the cube. Thus, the buoyant force is a result of the pressure differential on a submerged or partly submerged object.

Figure 15.9 The external forces acting on the cube of liquid are the force of gravity F_{g} and the buoyant force B. Under equilibrium conditions, B = F_{g} .

Before we proceed with a few examples, it is instructive for us to compare the forces acting on a totally submerged object with those acting on a floating (partly submerged) object.

Case 1: Totally Submerged ObjectWhen an object is totally submerged in a fluid of density ρ_{f} , the magnitude of the upward buoyant force is B = ρ_{f}V_{o}g where V_{o} is the volume of the object. If the object has a mass M and density ρ_{o} , its weight is equal to F_{g} = Mg = ρ_{o}V_{o}g, and the net force on it is B - F_{g} = (ρ_{f} - ρ_{o})V_{o}g. Hence, if the density of the object is less than the density of the fluid, then the downward force of gravity is less than the buoyant force, and the unconstrained object accelerates upward (Fig. 15.10a). If the density of the object is greater than the density of the fluid, then the upward buoyant force is less than the downward force of gravity, and the unsupported object sinks (Fig. 15.10b).

Case 2: Floating Object Now consider an object of volume V_{o} in static equilibrium floating on a fluid—that is, an object that is only partially submerged. In this case, the upward buoyant force is balanced by the downward gravitational force acting on the object. If V_{f} is the volume of the fluid displaced by the object (this volume is the same as the volume of that part of the object that is beneath the fluid level), the buoyant force has a magnitude B = ρ_{f}V_{f} g. Because the weight of the object is F_{g} = Mg = ρ_{o}V_{o}g, and because F_{g} = B, we see that ρ_{f}V_{f} g = ρ_{o}V_{o}g, or

Under normal conditions, the average density of a fish is slightly greater than the density of water. It follows that the fish would sink if it did not have some mechanism for adjusting its density. The fish accomplishes this by internally regulating the size of its air-filled swim bladder to balance the change in the magnitude of the buoyant force acting on it. In this manner, fish are able to swim to various
depths. Unlike a fish, a scuba diver cannot achieve neutral buoyancy (at which the buoyant force just balances the weight) by adjusting the magnitude of the buoyant force B. Instead, the diver adjusts F_{g} by manipulating lead weights.

Hot-air balloons. Because hot air is less dense than cold air, a net upward force acts on the balloons.

A glass of water contains a single floating ice cube (Fig. 15.11). When the ice melts, does the water level go up, go down, or remain the same?