Archimedes' Principle & Buoyancy - Ninth Grade Science
Drop a solid steel pebble into a lake, and it sinks immediately to the muddy floor. Yet, a colossal ocean liner constructed from thousands of tons of that exact same steel floats gracefully across the sea, carrying passengers and heavy cargo without sinking. If you attempt to push an inflated basketball beneath the surface of a swimming pool, you encounter an aggressive upward resistance that forces the ball back to the surface the moment you release it. Why do heavy objects feel noticeably lighter when submerged in water? These fascinating phenomena are governed by buoyancy and Archimedes' Principle, discovered in Syracuse over two thousand years ago. In this chapter, you will investigate how fluid pressure generates buoyant forces, derive Archimedes' Principle, examine the laws of flotation, and explore practical marine engineering.
The Concept of Upthrust and Buoyant Force
Whenever an object is partially or wholly submerged in a fluid (a liquid or a gas), the fluid exerts a net upward force on the object. This upward force is called the buoyant force or upthrust ($F_b$ or $U$).
+--------------------------------------------------------------------------+
| Buoyant Force (Upthrust) |
+--------------------------------------------------------------------------+
| Definition: The net upward force exerted by a fluid on a submerged body |
| Origin: Pressure difference between top and bottom of the body |
| Vector: Points strictly vertically upward opposing gravity |
| SI Unit: Newton (N) |
+--------------------------------------------------------------------------+
Why Fluids Exert an Upward Buoyant Force
Water Surface
~~~~~~~~~~~~~~~~~~~~~
|
v Depth h_1 (Lower Pressure P_1 = h_1*rho*g)
+----+----+
| v | Downward Force F_1 = P_1 * A
| |
| Block | Submerged Height = L
| |
| ^ | Upward Force F_2 = P_2 * A
+----+----+
^ Depth h_2 (HIGHER Pressure P_2 = h_2*rho*g)
Net Force = F_2 - F_1 = (P_2 - P_1) * A (UPWARD!)
The Physical Origin of Buoyant Force
Why does a fluid push upward rather than downward? The answer lies in the fundamental nature of hydrostatic pressure, which increases linearly with fluid depth ($P = h \rho g$). Consider a solid rectangular block submerged vertically in water. The lateral pressures acting on the vertical sides cancel out pairwise due to symmetry. However, the top face of the block is at a shallower depth $h_1$, experiencing a downward hydrostatic pressure $P_1$ and a downward force $F_1 = P_1 A$. The bottom face of the block is at a greater depth $h_2$, experiencing a higher hydrostatic pressure $P_2$ and a larger upward force $F_2 = P_2 A$. Because the upward force pushing on the bottom exceeds the downward force pushing on the top ($F_2 > F_1$), the fluid exerts a net upward resultant force on the block. This difference is the buoyant force.
Statement and Derivation of Archimedes' Principle
Around 250 BCE, the Greek polymath Archimedes discovered the quantitative rule that governs buoyancy:
+--------------------------------------------------------------------------+
| Archimedes' Principle |
+--------------------------------------------------------------------------+
| When a body is immersed fully or partially in a fluid, it experiences |
| an upward buoyant force equal to the weight of the fluid displaced |
| by the body. |
+--------------------------------------------------------------------------+
Derivation of Archimedes' Principle
1. Pressure on Top Face (Depth h_1):
P_1 = h_1 * rho_fluid * g ===> F_1 = h_1 * rho_fluid * g * A (Down)
2. Pressure on Bottom Face (Depth h_2 = h_1 + L):
P_2 = h_2 * rho_fluid * g ===> F_2 = h_2 * rho_fluid * g * A (Up)
3. Net Upward Buoyant Force (F_b):
F_b = F_2 - F_1 = (h_2 - h_1) * rho_fluid * g * A
4. Notice that (h_2 - h_1) = L (Height of Block), and L * A = V (Volume):
F_b = V * rho_fluid * g
5. Mass of Displaced Fluid m_fluid = V * rho_fluid:
F_b = m_fluid * g = Weight of Displaced Fluid!
The Buoyancy Equation
Archimedes' Principle can be stated in a single universal equation:
F_b = V_{\text{sub}} \, \rho_{\text{fluid}} \, g
Where:
- $F_b$ is the upward buoyant force (newtons).
- $V_{\text{sub}}$ is the volume of the submerged portion of the object, which equals the volume of fluid displaced ($m^3$).
- $\rho_{\text{fluid}}$ is the density of the surrounding fluid ($kg/m^3$).
- $g$ is the acceleration due to gravity ($9.8\text{ m/s}^2$).
Notice that the buoyant force does not depend on the mass, material, or density of the submerged object; it depends solely on the volume of the object submerged and the density of the displaced fluid. A one-cubic-meter block of solid gold and a one-cubic-meter block of aluminum experience the exact same buoyant force when submerged in water, because both displace exactly one cubic meter of water.
Apparent Weight of Submerged Objects
When an object is immersed in a fluid, the upward buoyant force counteracts a portion of its downward gravitational weight. The reduced weight registered by a spring balance is its apparent weight:
\text{Apparent Weight} = \text{True Weight in Air} - \text{Buoyant Force}
W_{\text{app}} = W_{\text{true}} - F_b
This explains why human bodies feel weightless while floating in a swimming pool and why heavy submerged boulders can be lifted off the riverbed with ease until they break the water surface.
The Principle and Laws of Flotation
When an object is placed in a fluid, two opposing forces act on it simultaneously: its downward gravitational weight ($W = m g = V_{\text{obj}} \rho_{\text{obj}} g$) and the upward buoyant force ($F_b = V_{\text{sub}} \rho_{\text{fluid}} g$). The competition between these two forces dictates whether the object sinks, suspends, or floats.
+-------------------+--------------------+---------------------------------+
| Density Condition | Force Relationship | Resulting State of Motion |
+-------------------+--------------------+---------------------------------+
| rho_obj > rho_fl | W > F_b,max | Object Sinks to the bottom |
| rho_obj = rho_fl | W = F_b,max | Object Suspended anywhere |
| | | in fluid (Neutral buoyancy) |
| rho_obj < rho_fl | W < F_b,max | Object Floats partially |
| | | submerged on the surface |
+-------------------+--------------------+---------------------------------+
Case 1: Sinking Case 2: Floating
(Iron Block) (Hollow Iron Ship)
~~~~~~~~~~~~~ ~~~~~~+------+~~~~~~
+---+ | |
| | +------+
+---+ |
| v Displaces large
v Heavy weight water volume
exceeds small F_b balances Weight!
water displacement!
Sinking ($\rho_{ ext{obj}} > \rho_{ ext{fluid}}$)
If the density of the object exceeds the density of the fluid, the maximum possible buoyant force (when fully submerged) is less than the object's gravitational weight. The net force is downward, and the object accelerates toward the bottom. A solid pebble sinks in water because the density of stone ($\approx 2600 ext{ kg/m}^3$) is much greater than the density of water ($1000 ext{ kg/m}^3$).
Floating ($\rho_{ ext{obj}} < \rho_{ ext{fluid}}$)
If the average density of the object is less than the density of the fluid, the object does not need to submerge completely to displace its own weight in fluid. It sinks into the liquid only until the submerged portion displaces a volume of fluid whose weight exactly matches the object's total weight:
\text{Weight of Floating Body} = \text{Weight of Displaced Fluid}
V_{\text{obj}} \rho_{\text{obj}} g = V_{\text{sub}} \rho_{\text{fluid}} g \implies \frac{V_{\text{sub}}}{V_{\text{obj}}} = \frac{\rho_{\text{obj}}}{\rho_{\text{fluid}}}
The fraction of the object submerged equals the ratio of the object's average density to the fluid's density. For an iceberg with a density of $917 ext{ kg/m}^3$ floating in seawater ($\rho = 1025 ext{ kg/m}^3$), the submerged fraction is $917 / 1025 \approx 0.895$ (nearly 90 percent), meaning nine-tenths of an iceberg lurks invisibly beneath the waves.
Why Steel Ships Float: Average Density
A solid iron bar sinks instantly in water because iron has a density of $7800 ext{ kg/m}^3$. Yet, an ocean freighter constructed from thousands of tons of iron floats easily. A ship is not a solid block of iron; it is shaped into an expansive, hollow hull enclosing vast volumes of empty air. When you divide the total mass of the ship (steel hull plus enclosed air and cargo) by its immense external volume, the average overall density of the vessel is significantly less than the density of water ($1000 ext{ kg/m}^3$). Consequently, the ship displaces its entire weight while submerging only a fraction of its hull.
Chapter Exercises
Exercise 1: A solid aluminum cylinder with a volume of $0.004 ext{ m}^3$ is suspended from a spring balance in air and registers a weight of 108.0 newtons. The cylinder is then completely immersed in fresh water of density $1000 ext{ kg/m}^3$. Taking $g = 9.8 ext{ m/s}^2$, calculate the upward buoyant force exerted on the cylinder by the water, and determine the apparent weight of the cylinder recorded by the spring balance while submerged.
Exercise 2: A rectangular wooden block with dimensions 0.20 meters by 0.10 meters by 0.10 meters has a mass of 1.4 kilograms. It is placed into a tank of fresh water ($\rho = 1000 ext{ kg/m}^3$). Determine whether the block floats or sinks. If it floats, calculate the volume of water displaced by the block and calculate the percentage of the block's total volume that remains submerged beneath the water surface.
Exercise 3: Submarines are designed to navigate both on the ocean surface and at deep underwater depths. Explain the physical mechanism that allows a submarine to submerge, cruise at a constant depth, and surface, referencing Archimedes' Principle and ballast tanks.
Solutions and Explanations
Solution 1: The buoyant force depends on the volume of fluid displaced, which equals the total volume of the fully submerged cylinder: $V = 0.004 ext{ m}^3$. Using Archimedes' Principle: $F_b = V \rho_{ ext{water}} g = 0.004 ext{ m}^3 imes 1000 ext{ kg/m}^3 imes 9.8 ext{ m/s}^2 = 39.2$ newtons upward. The apparent weight of the submerged cylinder is its true weight minus the upward buoyant force: $W_{ ext{app}} = W_{ ext{true}} - F_b = 108.0 - 39.2 = 68.8$ newtons. The spring balance reads 68.8 newtons.
Solution 2: Calculate the total volume of the wooden block: $V_{ ext{obj}} = 0.20 imes 0.10 imes 0.10 = 0.002 ext{ m}^3$. The density of the wood is mass divided by volume: $\rho_{ ext{obj}} = 1.4 ext{ kg} / 0.002 ext{ m}^3 = 700 ext{ kg/m}^3$. Because the density of the wood ($700 ext{ kg/m}^3$) is less than the density of water ($1000 ext{ kg/m}^3$), the block floats. By the law of flotation, the weight of the floating block equals the weight of the water displaced: $W_{ ext{obj}} = W_{ ext{displaced}} \implies m_{ ext{obj}} g = V_{ ext{sub}} \rho_{ ext{water}} g$, which gives $V_{ ext{sub}} = m_{ ext{obj}} / \rho_{ ext{water}} = 1.4 / 1000 = 0.0014 ext{ m}^3$ of water displaced. The submerged fraction is $V_{ ext{sub}} / V_{ ext{obj}} = 0.0014 / 0.002 = 0.70$, meaning exactly 70.0% of the block's volume is submerged beneath the water line.
Solution 3: A submarine controls its depth by manipulating its average density using internal ballast tanks that can be filled alternately with seawater or compressed air. To submerge, the submarine opens flood valves, allowing seawater to rush into the ballast tanks while venting out trapped air; this increases the vessel's total mass and raises its average density above that of surrounding seawater ($\rho_{ ext{sub}} > \rho_{ ext{water}}$), producing a net downward force that causes the vessel to sink. To cruise at a constant depth, the crew adjusts the water ballast so that the submarine's average density matches seawater exactly ($\rho_{ ext{sub}} = \rho_{ ext{water}}$), establishing neutral buoyancy where weight equals the buoyant force. To surface, compressed air is blown into the ballast tanks to expel the seawater; this drastically reduces the vessel's overall mass and drops its average density below that of seawater, creating a net upward buoyant force that carries the submarine to the surface.