In this tutorial, the buoyant force formula is explored in a problem-solution based.

Most likely you observed a ship or boat floating on the water. But what is caused some of the objects to float or sink in the fluids?

Let's do an experiment.

Connect a dense body to the end of a spring scale and weigh it. Then, submerge it in the water and again weigh it. You can see that the scale reading in the water is less than that in the air.

Why?

We can conclude that there must be an upward force acting on the body in the fluid (say, water) that opposes the downward gravity force (weight) results in less weight, called apparent weight, in the fluid.

This is a general observation and occurs in all fluids when a body with different density submerged in it.

The force exerted by a fluid on a body completely or partially submerged in it is called buoyant force and its magnitude is equal to the weight of the displaced fluid by the body.

This upward force which is called the buoyant force is not a new kind of force.

The physical origin of this is the *pressure difference between the upper and lower sides of a submerged object into a fluid.*

Consider a body submerged (say, a cube) in a fluid where forces exert on it by the fluid from all sides.

Forces in the horizontal direction, are opposite so cancel out each other (In fact, since they are at the same depth from the fluid level, so the corresponding forces are also the same).

But in the vertical direction, the top and bottom sides are located at the different depths of the fluid.

We know that in the deeper depth the pressure is higher than in the lower one. Thus, using the definition of pressure, $P=FA$, we can realize that the force on the lower side is greater than the upper one which adding them up creates a net upward force which is called buoyant force.

For simplicity, consider a cube of height $h$ with base area $A$ completely submerged into a fluid with a density of $\rho_f$.

In the figure, you can see the vertical forces due to the water pressure at different depths exerting on the cube.

The force $\vec{F}_1$ exerting down due to the fluid pressure on the upper side, similarly there is a force $\vec{F}_2$ on the lower side but in an upward direction.

These are the only vertical forces applied by the fluid on the cube submerged in it.

The horizontal forces are in the opposite directions so cancel each other out.

Vector summing them results in a net upward force which gets the formula of buoyant force as below \begin{align*} F_{net} &= F_2-F_1\\ \\&=P_2A-P_1A\\ \\&=(\rho_f gh_2)A-(\rho_g g h_1)A\\ \\&=\rho_f g(h_2-h_1)A\\ \\&=\rho_f V_{dis}g \end{align*} In above, we used the following facts: pressure at depth $h$ below the fluid surface level is found as $P=P_0+\rho_f gh$ with surface pressure $P_0$ and definition of volume as base $A$ times object's height ($h_2-h_1$).

Thus, in words, buoyant force formula is the product of fluid's density $\rho_f$, displaced volume of the fluid $V_{dis}$ and gravitational constant $g$. \[F_b=\rho_f V_{dis} g\]

We can also use the definition of density as the ratio of mass to volume, $\rho=\frac mV$, to find another useful equation for buoyant force as \[F_b=m_f g\] where, in this version, $m_f$ is the mass of the fluid displaced by the object submerged in it.

**Example: A solid cylinder with base area $A=20\,{\rm cm^2}$ is placed completely underwater as shown in the figure. Calculate the difference between the forces exerted by the water on the upper and lower bases of the cylinder? ($g=10\,{\rm m/s^2}$ and $\rho_{water}=1\,{\rm g/cm^3}$).**

**Solution**: The forces $F_1=P_1 A$ and $F_2=P_2 A$ are exerted on the upper and lower bases of the cylinder by the water. $P=\rho gh$ is also the pressure at depth $h$ of a fluid with density $\rho$.

Upper base is at depth $h_1=0.1\,{\rm m}$ so pressure at that depth is $P_1=1000\times 10\times 0.1=1000\,{\rm Pa}$.

Lower base is at depth $h_2=0.5\,{\rm m}$ so pressure at that depth is $P_2=1000\times 10\times 0.5=5000\,{\rm Pa}$.

Difference in forces is $\Delta F=F_2-F_1=(P_2-P_1)A$ so we have \begin{align*}\Delta F&=F_2-F_1=(P_2-P_1)A\\&=(5000-1000)20\times(0.01)^2 \\&=8\quad {\rm N}\end{align*}

This difference in forces is the upward buoyant force applied on the cylinder by the water.

Thus, we can also find it using buoyant force equation as $F_b=\rho_{water}V_{body}g$ where $V_{body}$ was the amount of the body's volume inside the fluid (here, all volume is completely submerged). \begin{align*}F_b&=\rho_{water}V_{body}g\\ &=(1000)(40\times 20)\times 10^{-5}\times 10\\&=8\quad {\rm N}\end{align*} Where the cylinder's volume is base ($A=20\,{\rm cm^2}$) times the height ($h=40\,{\rm cm}$).

The above equation for buoyancy was obtained for a simple shape, cube, but it can be shown that it holds for any shape.

(1) Mass in the formula is the mass of the displaced fluid, not the mass of the body.

(2) This formula is applicable to any shape since any shape can be approximated by a large number of infinitesimal cubes.

(3) Buoyancy force depends on the surrounding fluid and does not change if the occupied body replaced with a more or less dense one. That is, it does not matter the above cube with what material fills in, say wood or iron. The only thing that matters is the density of the surrounding fluid $\rho_f$.

In the example below, you can see what is meant by displaced volume in the buoyancy force equation.

**Example: A rectangular container of ethanol with density $\rho_{eth}=800\,{\rm kg/m^3}$ floats in the a liquid of density $\rho_{f}=1200\,{\rm kg/m^3}$. Height and base of the container are $H=6\,{\rm cm}$ and $A=20\,{\rm cm^2}$, respectively. $h=4\,{\rm cm}$ of its height is in the fluid. Find the buoyancy force exerted on the container. **

**Solution**: Recall that buoyant force is the weight of the displaced fluid by a submerged body in it $F_b=m_f g$ or the fluid's density times the volume of the object under the fluid surface times $g$.

Here, the amount of total volume under the liquid which displaces it is $V_{dis}=A\times h$. Putting it into the buoyant force formula, we get \begin{align*}F_b&=\rho_{f}V_{dis}g\\&=\rho_f \times (Ah)g\\&=(1200)(20\times 4)\times 10^{-5}\times 10\\&=9.6\quad {\rm N}\end{align*}

The buoyant force is the essential part of the Archimedes' principle which states 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.

**Example: a piece of wood with density $\rho_{wood}=0.6\,{\rm g/cm^3}$ floats on the surface of the water. What is the ratio of the portion of the wood submerged in the water to that of outside the water? ($\rho_{water}=1\,{\rm g/cm^3}$) **

**Solution**: Density of wood is smaller than water so it floats on water. Since the wood is floating so its weight must be balanced by the upward force due to the fluid that is buoyant force. Using buoyant force formula and weight definition, we have, \begin{align*} \text{buoyant force}&=\text{weight}\\ \\ \rho_{fluid}V_{dis}g&=\rho_{body}V_{body}g\\ \\ \Rightarrow \frac{V_{dis}}{V_{body}}&=\frac{\rho_{body}}{\rho_{fluid}}\\ \\ &=\frac{0.6}{1}\\ \\ \Rightarrow V_{fluid}&=0.6\,V_{body}\end{align*} In above, $V_{dis}$ and $V_{body}$ are the volume of the part of the wood beneath the water level and total volume of the piece of wood, respectively.

To find the volume of the part outside of the water, proceed as below \begin{align*} V_{wood}&=V_{wood-in}+V_{wood-out}\\ \\V&=0.6V+V_{wood-out} \\ \\\Rightarrow& V_{wood-out}=0.4V\end{align*}Therefore, we obtained portion of the body inside and outside the water as $V_{wood-in}=0.6V$ and $V_{wood-out}=0.4V$ with their ratio as $\frac 32$.

Buoyant force formula can be used to find the unknown density of a fluid as the example below:

**Example: A piece of wood floats in water so that two-thirds of its volume $V$ is beneath the water level. Repeat that experiment and place the wood in oil. This time, the submerged volume in oil is $0.8V$. Find the density of (a) the wood and (b) the oil. **

**Solution**: The object floats on the fluids (water and oil) so the downward gravitation force and upward buoyant force must balance each other.

Recall that the buoyant force formula is the fluid's density times the volume of the part of the object beneath the fluid level times the gravitational constant.

In the case of water, two-thirds of wood's volume is underwater i.e. $V_{dis}=2/3 V$. Thus, balancing the buoyancy force $F_b=\rho_{water}V_{dis}g$ and weight, gives us the following \begin{align*} F_b&=mg\\ \rho_{water}V_{dis}g&=\rho_{wood}Vg\\(1000)(2/3 V)&=\rho_{wood}\times V\\ \Rightarrow \rho_{wood}&=\frac{2000}{3}\quad{\rm \frac{kg}{m^3}}\end{align*}

Similarly, the density of the oil is also obtained by applying buoyant force equation and weight as below \begin{align*} F_b&=mg\\ \rho_{oil}V_{dis}g&=\rho_{wood}Vg\\ \\\rho_{oil}(0.8 V)&=\underbrace{\rho_{wood}}_{2000/3}\times V\\ \\ \Rightarrow \rho_{oil}&=834 \quad{\rm \frac{kg}{m^3}}\end{align*}

Another application of the buoyant force equation is for finding the apparent weight of objects in fluids as below

**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) Using Archimedes law, 200 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*}Where $V_{object}$ is volume of the water displaced by the iron or explicitly, part of the iron underwater (All of it).

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

**Example: An object with a mass of 150 kg is thrown into the water and displaces a volume of 100 liters of water. Will the object sinks or floats? **

**Solution**: To see what happens after throwing, we must compute the forces applied on the object into the water that is upward buoyant force and its downward weight.

We know that buoyant force is the weight of fluid displaced volume. Here, displaced volume by the object is 100 liters which must be converted into SI units of volume, $m^3$, as $1\,{\rm L}=10^{-3}\,m^3$.

Thus, using buoyant force formula we have \begin{align*} F_b&=\rho_{water}\times V_{dis}\times g\\&=(1000)(100\times 10^{-3})(10)\\&=1000\quad {\rm N}\end{align*} Its downward weight is also given by $W=mg=1500\,{\rm N}$, so applying Newton's second law to find the net force on the object, gives \[F_{net}=1000\uparrow-1500\downarrow=-500\downarrow\,{\rm N}\] Since weight is greater than upward buoyant force, so the object sinks.

In the above, all factors involved in the buoyant force formula taught in detail including the meaning of displaced volume by an object in a fluid, conditions of floating, and immersion.

**Author: Ali Nemati**

**Page Created: 1/31/2021**

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