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 magnitude of the buoyant force equals the weight of the fluid that the object displaces. This is called Archimedes' Principle. The phrase weight of the displaced fluid refers to the weight of the fluid that would spill out, if a container were filled to the brim before an object 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. The shape of the object is not important. No matter what its shape, the buoyant force pushes it upward in accordance with Archimedes' principle. It was an impressive accomplishment that the Greek scientist Archimedes (ca. 287 - 212 B.C.) discovered this principle so long ago.

Let's try to prove Archimedes' Principle by examining the cylinder of height h being held under the surface of a liquid. The pressure P_{1} on the top face generates the downward force P_{1}A, where A is the area of the face. Similarly, the pressure P_{2} on the bottom face generates the upward force P_{2}A. Since the pressure is more 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 F_{B} is:

F_{B} = P_{2}A - P_{1}A

F_{B} = (P_{2} - P_{1})A

F_{B} = pghA

since V = Ah .... F_{B} = pgV

since m = pV ..... F_{B} = mg (weight of fluid displaced)

Therefore, if the weight of the water displaced (F_{B}) by the submerged or partially submerged object is equal to the downward weight of the object, it will float. If the weight of the water displaced (F_{B}) by the submerged or partially submerged object is less than the downward weight of the object, it will sink. Below are links to two applets that you can experiment with to examine this concept.

Raft Problem: A solid, square pinewood raft measures 4.0 m on a side and is 0.30 m thick. The density of pine is 550 kg/m^{3}. (a) Determine whether the raft floats in water and (b) if so, how much of the raft is beneath the surface?

The Raft Problem

Raft Problem Solution: A solid, square pinewood raft measures 4.0 m on a side and is 0.30 m thick. The density of pine is 550 kg/m^{3}. (a) Determine whether the raft floats in water and (b) if so, how much of the raft is beneath the surface?

Part A: If the raft floats then...

Max. Buoyant Force (F_{B}^{max}) > W_{raft}

F_{B}^{max} = ρ_{water}V_{water}g
F_{B}^{max} = 1000 x 4.8 x 9.81 = 47,088 N

W_{raft} = ρ_{pine}V_{pine}g
W_{raft} = 550 x 4.8 x 9.81 = 25,898 N

Since the maximum buoyant force is greater than the weight of the raft it will clearly float. There is a much easier way to answer this problem. If you notice in our calculations of the Max Buoyant Force and the Weight of the raft above, the only difference is the density of the raft and the density of the water. Therefore if the density of an object is less than the density of the fluid it is in, then the object will float. If the density of an object is greater than the density of the fluid it is in, then the object will sink.

Part B: Since we now know that the raft floats according to Archimedes' Principle, the buoyant force must equal the weight of the raft. Therefore, only 25,898 N of the available maximum buoyant force of 47,088 N must be used, therefore the raft will not have to be completely submerged.

W_{raft} = B_{F}

ρ_{pine}V_{pine}g = ρ_{water}V_{water}g

25,898 N = (1000)(4.0 m x 4.0 m x h)(9.81)

height = 0.165 meters.

There is also an easier way to solve problems like this when we want to find out how much of a floating object is below the surface of the fluid. Let's start again with the below generic equation where the weight of the object equals the buoyant force of the object:

ρ_{object}V_{object}g = ρ_{fluid}V_{fluid}g

this can be simplified to:

ρ_{object} / ρ_{fluid} = V_{fluid} / V_{object}

In this equation the V_{fluid} is the volume of the object that is submerged. Therefore, for any floating object the ratio of the density of the object to the density of the fluid is equal to the ratio of volume of the object that is submerged to the total volume of the object. In other words, the ratio of the density of the object to the density of the fluid times 100 will give you the percentage of the volume of the object that is submerged. In this problem, that would mean that (550/1000) x 100 or 55% of the raft is submerged. Since the volume of the entire raft was 4.8 m^{3} this would mean that the submerged volume would be (4.8)(.55) = 2.64m^{3}. Now we can simply solve for the height.

Archimedes 10 pt. Quiz

Problem #1: If the density of ice is 917 kg/m^{3}, what percentage of an iceberg is below the surface of the water?

Problem #2: If the Goodyear blimp contains 5400 m^{3} of helium, whose density is 0.179 kg/m^{3}. Find how much extra weight the blimp could carry in equilibrium at an altitude where the density of air is 1.20 kg/m^{3}.

Problem #3: 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 in milliliters?

Problem #4: A glass is filled to the brim with water and has an ice cube floating in it. When the ice cube melts, what happens? 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.

Problem #5: A steel beam is suspended completely under water by a cable that is attached to one end of the beam, so it hangs vertically. Another identical beam is also suspended completely under water, but by a cable that is attached to the center of the beam, so it hangs horizontally. Which beam, if either, experiences the greater buoyant force?

Problem #6: A glass beaker, filled to the brim with water, is resting on a scale. A solid block is then placed in the water, causing some of it to spill over. The water that spills is wiped away, and the beaker is still filled to the brim. How do the intial and final readings on the scale compare if the block is made from (a) wood (density less then water) and (b) iron (density greater than water)?

Problem #7: On a distant planet the acceleration due to gravity is less than it is on earth. Would you float more easily in water on this planet than on earth?

Problem #8: As a person dives toward the bottom of a swimming pool, the pressure increases noticeably. Does the buoyant force acting on her also increase?

Problem #9: A pot is partially filled with water, in which a plastic cup is floating. Inside the floating cup is a small block of lead. When the lead block is removed from the cup and placed into the water, it sinks to the bottom. When this happens, what happens to the water level in the pot?

Problem #10: What is the specific gravity of a fluid if it contains a floating object with a density of 234 kg/m^{3} whose volume is 75% submerged?