11.6

Archimedes’ Principle

Anyone who has tried to push a beach ball under the water has felt how the water pushes back with a strong upward force. This upward force is called the buoyant force, and all fluids apply such a force to objects that are immersed in them. The buoyant force exists because fluid pressure is larger at greater depths.
In Figure 11-17 a cylinder of height h is being held under the surface of a liquid. The pressure P1 on the top face generates the downward force P1A, where A is the area of the face. Similarly, the pressure P2 on the bottom face generates the upward force P2A. Since the pressure is greater at greater depths, the upward force exceeds the downward force. Consequently, the liquid applies to the cylinder a net upward force, or buoyant force, whose magnitude FB is

We have substituted from Equation 11.4 into this result. In so doing, we find that the buoyant force equals . The quantity hA is the volume of liquid that the cylinder moves aside or displaces in being submerged, and denotes the density of the liquid, not the density of the material from which the cylinder is made. Therefore, gives the mass m of the displaced fluid, so that the buoyant force equals mg, the weight of the displaced fluid. The phrase “weight of the displaced fluid” refers to the weight of the fluid that would spill out, if the container were filled to the brim before the cylinder is inserted into the liquid. The “buoyant force” is not a new type of force. It is just the name given to the net upward force exerted by the fluid on the object.

Figure 11-17 The fluid applies a downward force P1A to the top face of the submerged cylinder and an upward force P2A to the bottom face.

The shape of the object in Figure 11-17 is not important. No matter what its shape, the buoyant force pushes it upward in accord with Archimedes’ principle. It was an impressive accomplishment that the Greek scientist Archimedes (ca. 287–212 B.C.) discovered the essence of this principle so long ago.

| Archimedes’ Principle
| Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces:

| (11.6)

The effect that the buoyant force has depends on its strength compared with the strengths of the other forces that are acting. For example, if the buoyant force is strong enough to balance the force of gravity, an object will float in a fluid. Figure 11-18 explores this possibility. In part a, a block that weighs 100 N displaces some liquid, and the liquid applies a buoyant force FB to the block, according to Archimedes’ principle. Nevertheless, if the block were released, it would fall further into the liquid because the buoyant force is not sufficiently strong to balance the weight of the block. In part b, however, enough of the block is submerged to provide a buoyant force that can balance the 100-N weight, so the block is in equilibrium and floats when released. If the buoyant force were not large enough to balance the weight, even with the block completely submerged, the block would sink. Even if an object sinks, there is still a buoyant force acting on it; it’s just that the buoyant force is not large enough to balance the weight. Example 9 provides additional insight into what determines whether an object floats or sinks in a fluid.

Figure 11-18 (a) An object of weight 100 N is being immersed in a liquid. The deeper the object is, the more liquid it displaces, and the stronger the buoyant force is. (b) The buoyant force matches the 100-N weight, so the object floats.

| Example 9 | | | A Swimming Raft

A solid, square pinewood raft measures 4.0 m on a side and is 0.30 m thick. (a) Determine whether the raft floats in water, and (b) if so, how much of the raft is beneath the surface (see the distance h in Figure 11-19).

Figure 11-19 A raft floating with a distance h beneath the water.

Reasoning To determine whether the raft floats, we will compare the weight of the raft to the maximum possible buoyant force and see whether there could be enough buoyant force to balance the weight. If so, then the value of the distance h can be obtained by utilizing the fact that the floating raft is in equilibrium, with the magnitude of the buoyant force equaling the raft’s weight.
Solution

a. | The weight of the raft is (Equation 4.5), where mpine is the mass of the raft and can be calculated as (Equation 11.1). The pinewood’s density is (Table 11-1), and its volume is . Thus, we find the weight of the raft to be

The maximum possible buoyant force occurs when the entire raft is under the surface, displacing a volume of water that is . According to Archimedes’ principle, the weight of this volume of water is the maximum buoyant force . It can be obtained using the density of water:

Since the maximum possible buoyant force exceeds the 26 000-N weight of the raft, the raft will float only partially submerged at a distance h beneath the water.
b. | We now find the value of h. The buoyant force balances the raft’s weight, so . But according to Equation 11.6, the magnitude of the buoyant force is also the weight of the displaced water, so . Using the density of water, we can also express the weight of the displaced water as , where the water volume is . As a result,

Problem solving insight When using Archimedes’ principle to find the buoyant force FB that acts on an object, be sure to use the density of the displaced fluid, not the density of the object.

In order to decide whether the raft will float in part (a) of Example 9, we compared the weight of the raft to the maximum possible buoyant force . The comparison depends only on the densities and . The take-home message is this: Any object that is solid throughout will float in a liquid if the density of the object is less than or equal to the density of the liquid. For instance, at 0 °C ice has a density of 917 kg/m3, whereas water has a density of 1000 kg/m3. Therefore, ice floats in water.
Although a solid piece of a high-density material like steel will sink in water, such materials can, nonetheless, be used to make floating objects. A supertanker, for example, floats because it is not solid metal. It contains enormous amounts of empty space and, because of its shape, displaces enough water to balance its own large weight. Conceptual Example 10 focuses on the reason ships float.

| Conceptual Example 10 | | | How Much Water Is Needed to Float a Ship?

A ship floating in the ocean is a familiar sight. But is all that water really necessary? Can an ocean vessel float in the amount of water that a swimming pool contains, for instance? Reasoning and Solution In principle, a ship can float in much less than the amount of water in a swimming pool. To see why, look at Figure 11-20. Part a shows the ship floating in the ocean because it contains empty space within its hull and displaces enough water to balance its own weight. Part b shows the water that the ship displaces, which, according to Archimedes’ principle, has a weight that equals the ship’s weight. Although this wedge-shaped portion of water represents the water displaced by the ship, it is not the amount that must be present to float the ship, as part c illustrates. This part of the drawing shows a canal, the cross section of which matches the shape in part b. All that is needed, in principle, is a thin section of water that separates the hull of the floating ship from the sides of the canal. This thin section of water could have a very small volume indeed.

Figure 11-20 (a) A ship floating in the ocean. (b) This is the water that the ship displaces. (c) The ship floats here in a canal that has a cross section similar in shape to that in part b.

The physics of a state-of-charge battery indicator. Archimedes’ principle is used in some car batteries to alert the owner that recharging is necessary, via a state-of-charge indicator, such as the one illustrated in Figure 11-21. The battery includes a viewing port that looks down through a plastic rod, which extends into the battery acid. Attached to the end of this rod is a “cage,” containing a green ball. The cage has holes in it that allow the acid to enter. When the battery is charged, the density of the acid is great enough that its buoyant force makes the ball rise to the top of the cage, to just beneath the plastic rod. The viewing port shows a green dot. As the battery discharges, the density of the acid decreases. Since the buoyant force is the weight of the acid displaced by the ball, the buoyant force also decreases. As a result, the ball sinks into one of the two chambers oriented at an angle beneath it (see Figure 11-21). With the ball no longer visible, the viewing port shows a dark or black dot, warning that the battery charge is low.

Figure 11-21 A state-of-charge indicator for a car battery.

Need more practice?

Interactive LearningWare11.1
A spring has a spring constant of 578 N/m. When used to suspend an object in air, the spring stretches by 0.0640 m. When used to suspend the same object in water (), the spring stretches by 0.0520 m. What is the volume of the part of the object that is under water?
Related Homework: Problem 94
Archimedes’ principle has allowed us to determine how an object can float in a liquid. This principle also applies to gases, as the next example illustrates.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 11 | | | A Goodyear Airship

Normally, a Goodyear airship, such as that in Figure 11-22, contains about of helium (He) whose density is 0.179 kg/m3. Find the weight WL of the load that the airship can carry in equilibrium at an altitude where the density of air is 1.20 kg/m3.

Figure 11-22 (a) A helium-filled Goodyear airship. (b) The free-body diagram of the airship, including the load weight WL, the weight WHe of the helium, and the buoyant force FB. (a, Aaron Vandersommers/Goodyear)

The physics of a Goodyear airship.

Reasoning The airship and its load are in equilibrium. Thus, the buoyant force FB applied to the airship by the surrounding air balances the weight WHe of the helium and the weight WL of the load, including the solid parts of the airship. The free-body diagram in Figure 11-22b shows these forces. The buoyant force is the weight of the air that the ship displaces, according to Archimedes’ principle. The weight of air or helium is given as (Equation 4.5). Using the density and volume V, we can express the mass m as (Equation 11.1).
Knowns and Unknowns The following table summarizes the data:

Description | Symbol | Value
Volume of helium in airship | VHe |
Density of helium |  | 0.179 kg/m3
Density of air |  | 1.20 kg/m3
Unknown Variable |  |
Weight of load | WL | ?

Modeling the Problem

| Step 1 Equilibrium
| Because the forces in the free-body diagram (see Figure 11-22b) balance at equilibrium, we have

Rearranging this result gives Equation 1 at the right. Neither the buoyant force FB nor the weight WHe of the helium is known. We will determine them in Steps 2 and 3.

| (1)

| Step 2 Weight and Density
| According to Equation 4.5, the weight is given by . On the other hand, the density is defined as the mass m divided by the volume V (see Equation 11.1), so we know that . Thus, the weight of the helium in the airship is

This result can be substituted into Equation 1, as shown at the right. We turn now to Step 3 to evaluate the buoyant force.

| Step 3 Archimedes’ Principle
| The buoyant force is given by Archimedes’ principle as the weight of the air displaced by the airship. Thus, following the approach in Step 2, we can write the buoyant force as follows:

In this result, Vair is very nearly the same as VHe, since the volume occupied by the materials of the ship’s outer structure is negligible compared to VHe. Assuming that , we see that the expression for the buoyant force becomes

Substitution of this value for the buoyant force into Equation 1 is shown at the right.

Solution Combining the results of each step algebraically, we find that
The weight of the load that the airship can carry at an altitude where is, then,

Related Homework: Problems 46 and 79

Check Your Understanding 2

A glass is filled to the brim with water and has an ice cube floating in it. When the ice cube melts, what happens? (The answer is given at the end of the book.)

a. | Water spills out of the glass.
b. | The water level in the glass drops.
c. | The water level in the glass does not change.

Background: Answering this question depends first on understanding what forces act on an object floating in a liquid. Then, Archimedes’ principle comes into play. Finally, the concept of density as mass divided by volume is needed.
For similar questions (including calculational counterparts), consult Self-Assessment Test 11.1. The test is described next.

Test your understanding of the central ideas presented in Sections 11.1–11.6:

· | Mass Density
· | Pressure and Depth in a Static Fluid
· | Pascal’s Principle
· | Archimedes’ Principle

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