Archimedes' Principle: Equation with Solved Examples

Have you ever wondered why large and massive steel ships do not sink but a small coin does?

The answer is in the Archimedes' principle which is closely related to the buoyant forces. 

To objects in fluids (such as water or even air!), two main forces applied upward buoyant force and downward gravitational force. Competing between these two forces determines whether an object sinks or floats in a fluid. 

Here, we are going to learn about this topic with some basic and important solved examples. 

This fundamental principle which was discovered by a Greek mathematician in the sixth century B.C. states and defines as below:

Any object wholly or partially submerged in a fluid is buoyed up by a force with a magnitude of the weight of the displaced fluid by the object.

When you lift a heavy object in a swimming pool, in fact, you are experiencing Archimedes' principle as water provides partial support for you to overcome the weight of an object placed in it. 

Or using Archimedes' principle, we can explain why hot air balloons ascend in the air.

When a body is placed into a fluid, an upward force is always exerted on it by surrounding fluid which partially or wholly reduces the impact of downward weight force. This upward force, called buoyant force, was explained with solved examples in another tutorial.

Derivation of Archimedes' principle: 

The physical cause of the upward force exerted by fluids on objects into it, is the pressure difference between the upper and lower sides of an object due to being at different depths of the fluid. 

Upon a surface at depth $h$ below the fluid level, the pressure is $P=P_0+\rho gh$ where $P_0$ is the pressure at the surface of the fluid and $\rho$ is the density of the fluid. 

As you can see, the lower side of an object sits at a greater depth so by definition of pressure, $P=FA$, there is a large force upon it. 

Note that there are also horizontal forces exerted on an object in a fluid but since they are located at the same depth so their net is zero.  

In fact, all horizontal forces exerted on an object with any arbitrary shape can be shown to cancel the effect of each other. 

All that remains are the vertical forces applied on the top and bottom sides of the submerged body which their vector summing gives the upward buoyant force $F_b$. 

Now applying Newton's second law and balancing all forces in the vertical direction, we obtain the following formula for Archimedes' principle 

buoyant force = body's weight 

or \[F_b=W\] Where buoyant force is defined as the product of fluid's density, displaced volume of the fluid by object into it and gravitational constant $g=10\,{\rm m/s^2}$ or \[F_b=\rho_{fluid}\times V_{dis}\times g\]

Now is the time to solve some examples to understand Archimedes' principle. 

Example: a block of wood floats in freshwater with two-fifth of its volume V submerged and in oil with 0.75V submerged. Find the density of (a) the wood (b) the oil. 

Solution: since wood floats in water so its weight must be balanced with the buoyancy force. 

(a) In a partially submerged body, the buoyancy force $F_b$ is defined as the density of fluid $\rho_f$ times the displaced volume of fluid $V_{dis}$ times the gravitational acceleration $g$. Thus, using Archimedes' principle equation, which is equating weight and buoyancy force, we get \begin{align*} W&=F_b \\ \\ \rho_{wood} \times V_{wood}\times g &=\rho_{water}\times V_{dis}\times g\\ \\ \rho_{wood}\times V_{wood} \times g&=(1)\left(\frac{2}{5}V_{wood}\right)g \\ \\ \rho_{wood}&=\frac25 \quad{\rm \frac{g}{cm^3}}\end{align*}

(b) Similarly, we can find the oil's density as above \begin{align*} \left(\rho Vg\right)_{wood}&=\left(\rho' Vg\right)_{oil}\\ \\(400)(V)g&=\rho_{oil}\, (0.75V)g\\ \\\Rightarrow \rho_{oil} &=\frac{400}{0.75}\\ \\&=\frac{1600}{3}\quad {\rm kg/m^3}\end{align*}


Example: an iron object of density $7.8\,{\rm g/cm^3}$ appears 200 N lighter in water than in air. 

(a) What is the volume of the object?
(b) How much does it weigh in the air?

Solution: Since the body has become lighter in water so there must be an upward force acting on the object which cancels some of the downward weight force. In fluids, this force is called floating or buoyancy force. 

(a) According to Archimedes' law, $200\,{\rm N}$ is the buoyancy force acting on the body which is obtained by the formula below \begin{align*} F_b &= \rho_{water} \times V_{object} \times g \\ \\ 200&=100\times V_{object}\times 10 \\ \\ \Rightarrow V_{object}&=\frac{2}{100}\quad {\rm m^3}\end{align*}

(b) Body's weight, $W=\rho V g$ in air is calculated as \[W=(7800)\left(\frac{2}{100}\right)(10)=1560\,{\rm N}\] Where $V$ is the actual volume of the body. 


As you can see above, one of the main applications of Archimedes' principle is finding the density of an unknown object.

Example: a wooden rectangular slab with surface area $5.7\,{\rm m^2}$, volume $V=0.6\,{\rm m^3}$ and density $600\,{\rm kg/m^3}$ is placed slowly in freshwater. By what depth $h$ is the slab submerged? 

Solution: according to Archimedes' principle, the water will apply an upward buoyant force on the slab whose magnitude is equal to the weight of the water displaced by the slab. 

Thus, the buoyant force exerted on the slab is $F_b=m_{water}g=\rho_{water}V_{dis}g$ where $V_{dis}$ is the displaced volume of the water or amount of slab's volume which is underwater. Let $h$ be the height of the slab from the bottom side. Thus, $V_{dis}=Ah$ where $A$ is the base area of the slab.

The weight of the slab is also given by $W=\rho_{slab}V_{slab}g$. 

Next, using Archimedes' principle equation as, $F_b=W$, we get \begin{align*}\rho_{water}\times (Ah) \times g&=\rho_{slab}\times V_{slab}\times g\\ \\ \Rightarrow h&=\frac{\rho_{slab}V_{slab}}{\rho_{water}A}\\ \\ &=\frac{600\times 0.6}{1000\times 5.7} \\ \\&=0.0632\quad {\rm m}\end{align*}

Criteria for floating or sinking:

Archimedes' principle simply gives us a rule of thumb to find out whether an object placed into a fluid sinks or floats. According to this principle, if we write all forces applied by a motionless fluid on a body submerged in it as upward buoyant force $F_b$ and downward weight force $W$, then there will be three situations depending on the sign of the net force $F_{net}=F_b \uparrow-W\downarrow$:

(1) Sinking: when happens $F_{net}<0$, in this case, the upward buoyancy force is less than its downward weight force, then the object sinks. \[\underbrace{\rho_{fluid}V_{fluid}g}_{buoyancy}<\underbrace{\rho_{obj}V_{obj}g}_{weight}\] For example, stone is denser than water, so when it is placed in water, it sinks. 

(2) Floating: when occurs $F_{net}>0$, consequently the positive upward buoyant force is balanced with the negative downward force of gravity (weight), then the object floats on the surface of the fluid. \[\underbrace{\rho_{fluid}V_{fluid}g}_{buoyancy}=\underbrace{\rho_{obj}V_{obj}g}_{weight}\] Wood is less dense than water, so it floats.

(3) Neutral buoyancy: there is a third case when $F_{net}=0$. In these situations, the object remains at that point of releasing in the fluid as motionless. This happens when the densities of object and fluid are equal. An example of neutral buoyancy is swimming fishes in the water. Fishes have a swimming bladder which can be filled with air together with their flesh make a composite object with average adjusted such that balances the density of the water and consequently it neither sinks nor floats in the water.

Question: How much fraction of the volume of an iceberg is under the sea level.

Solution: according to Archimedes' principle, since the iceberg floats on the water, so the upward buoyant force equals its weight. The magnitude of the buoyant force is the product of the iceberg's volume underwater, water's density, and gravitational acceleration. 

On the other hand, weight is defined as the product of the iceberg's actual volume, iceberg's density, and gravitational acceleration. 

Applying floating condition to find the fraction of the volume of the iceberg below sea level. \begin{align*}F_b&=W\\ \rho_{SW}V_{in-water}g&=\rho_{IB}Vg\\ \\ \Rightarrow \frac{V_{in-water}}{V}&=\frac{\rho_{IB}}{\rho_{SW}}\\ \\ &=\frac{0.92\times 10^{3}}{1.025\times 10^3} \\ \\ &=0.9\end{align*} Where in above $\rho_{IB}$ and $\rho_{SW}$ are the densities of iceberg and sea water, respectively. As you can see, about 90% of the volume of an iceberg is underwater.

Author: Ali Nemati

Page Created: 1/31/2021