Density, Pressure and Upthrust

A substance that can flow is called fluid. Unlike solids, liquids and gases can flow and hence they are fluids. Liquids are said to be nearly incompressible while gases are easily compressible.

Density

The density is the measure of how much dense a material is. If a body of mass $m$ and volume $V$ is given, its density is its mass per unit volume. Therefore, the density of the body denoted by $\rho$ is

\[\rho = \frac{m}{V} \tag{1}\]

The SI unit of density is $kg/{m^3}$. The density of a body may not be uniform throughout its volume. One example is the Earth whose density varies throughout its volume. In such a case we calculate the average density. There is another term called specific gravity which is the ratio of the density of a material to the density of water at ${4^ \circ }C$. Therefore, the specific gravity is a dimensionless quantity without unit. The name "specific gravity" is weird as it has nothing to do with gravity but you still have to say the same name.

Pressure in a Fluid

When you immerse a body in a fluid such as water, the fluid exerts normal force on the surface of the body. In Figure 1 a body is immersed in a fluid and the fluid exerts the force perpendicular to the surface of the body. The force exerted by the fluid on the body at a point on its surface is always perpendicular to the surface at that point regardless of the body's shape.

Figure 1 A fluid exerts force perpendicular to the surface of an immersed body.

If an infinitesimally small normal force of magnitude $d{F_n}$ acts over an infinitesimally small surface area $dA$, the pressure is the force $d{F_n}$ divided by $dA$. Therefore, the pressure $P$ over the small area is

\[P = \frac{d{F_n}}{{dA}} \tag{2} \label{2}\]

But if the normal force $F_n$ is uniform over an area $A$, the pressure due to the force on the area is $P = F_n/A$ and therefore the pressure on the surface is the normal force per unit area of the surface. If the normal force is not uniform, you should use Eq. \eqref{2} for infinitesimally small normal force $dF_n$ acting over an infinitesimally small area $dA$. The SI unit of pressure is $N/{m^2}$. You can also call the SI unit of pressure as $\text{Pa}$ from the original name Pascal so, $1{\rm{Pa}} = {\rm{1}}N/{m^2}$. Note that pressure is not a vector quantity even if the force is a vector. The air in the atmosphere exerts normal force on the surface of the Earth and also on every object on its surface. Note that the force exerted by a fluid on a body is always perpendicular to the surface of the body. The normal atmospheric pressure denoted by $\text{atm}$ at sea level is $1{\rm{atm}} = 101.3{\rm{kPa}}$ or $1{\rm{atm}} = {\rm{1}}{\rm{.013}} \times {\rm{1}}{{\rm{0}}^5}{\rm{Pa}}$.

The fluid exerts force on the immersed body which in turn results pressure. We derive an expression for the pressure of a fluid at a particular depth here. We consider a fluid of uniform density $\rho$ in a vessel as shown in Figure 2. We consider an imaginary fluid element as a box having imaginary boundary. The imaginary fluid element is at rest that is the fluid element is in hydrostatic equilibrium. If the net force on a particular portion of the fluid is zero, the portion of the fluid is said to be in hydrostatic equilibrium.

Figure 2 The imaginary fluid portion has an imaginary boundary of a box. Since the fluid element is in hydrostatic equilibrium, the net force on the element should be zero.
Figure 3 The pressure at depth $h$ of the fluid in the vessel is $P = P_0 + \rho gh$. The pressure on the top surface of the fluid is $P_0$.

We distinguish the force acting on the top and bottom surfaces of the fluid element as vertical force while the force acting on the other sides as horizontal force. Only the vertical forces are shown in Figure 2. Since the fluid element is in hydrostatic equilibrium, the net force acting on the element is zero. Which concludes that the horizontal forces should cancel each other and the net vertical force should also be zero. Let the area of the top or bottom surface of the fluid element be $A$. Suppose pressure on the top surface of the element is $P_1$ and the pressure on the bottom surface is $P_2$. Note that we choose the positive y-axis to be upwards and therefore the upward force is positive while the downward force is negative. The downward force on the top surface due to the pressure $P_1$ is $-P_1A$ while the upward force on the bottom surface is $P_2A$. Since the weight of the fluid element is also vertically downwards, the pressure on the bottom surface of the fluid element must be greater than the pressure on the top surface. The height of the fluid element is $(y_2-y_1)$ where $y_1$ and $y_2$ are the heights of the bottom and top surfaces of the fluid element from the bottom of the vessel respectively. If $V$ is the volume of the fluid element, its mass is $m = \rho V = \rho A(y_2 - y_1)$. The fluid element is at rest so the sum of all vertical forces must be zero:

\[\begin{align*} - P_1A + P_2A - \rho A(y_2-y_1){\kern 1pt} g &= 0\\ {\rm{or,}}\quad P_2 - P_1 &= \rho g{\kern 1pt} (y_2 - y_1) \tag{3} \label{3} \end{align*}\]

As in Figure 3 let ${P_0}$ be the pressure on the top surface of the fluid in the vessel at height $y_2$ and $P$ be the pressure at height $y_1$, the above equation can be rewritten as

\[P - {P_0} = \rho g({y_2} - {y_1})\]

Let the depth of any point where the pressure is $P$ be $h = y_2 - y_1$. And hence the pressure due to the fluid at any depth $h$ from the top surface is given by

\[P = P_0 + \rho gh \tag{4} \label{4}\]

Suppose the vessel in Figure 3 is open to the atmosphere and the pressure on the top surface of the fluid is the atmospheric pressure. Then the total pressure at the depth $h$ from the top surface is the sum of the atmospheric pressure and the quantity $\rho gh$ as given by Eq. \eqref{4}. The excess pressure greater than the atmospheric pressure is called gauge pressure and the total pressure is called absolute pressure. Therefore, if $P_0$ in Eq. \eqref{4} is the atmospheric pressure, the gauge pressure is $P - P_0 = \rho gh$ and $P$ is the absolute pressure.

Pascal's Law

Pascal's law is based on the transmission of pressure uniformly. It states that the pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid and to the walls of the container. If you increase the pressure $P_0$ on the surface of the fluid in Figure 3, the pressure at any depth $h$ should increase by the same amount and also to the walls of the container.

Figure 4 The applied force $F_1$ on a smaller area $A_1$ is magnified to the greater force $F_2$ on a relatively larger area $A_2$.

One example of Pascal's law is shown in Figure 3. The pressure applied to the piston of small area $A_1$ is transmitted undiminished to every point in the fluid and also to the piston of large area $A_2$. Note that the same pressure is transmitted to the walls of the container as well. The pressure $P_1$ on the piston of area $A_1$ due to the force $F_1$ is $P_1 = F_1/A_1$ and the pressure $P_2$ on the piston of area $A_2$ due to the force $F_2$ is $P_2 = F_2/A_2$. Since both pistons are in the same level, the pressure in both pistons should be the same that is, $F_1/A_1 = F_2/A_2$. And the force on the piston of the larger area $A_2$ is

\[{F_2} = \frac{{{A_2}}}{{{A_1}}}{F_1}\]

We know $A_2>A_1$ and the force $F_2$ is greater than $F_1$. It means the force $F_1$ is magnified to $F_2$ and became able to support the car as in Figure 4. Note that by Newton's third law the car also exerts the downward force of magnitude $F_2$.

Upthrust

If a body is immersed in a fluid, the fluid exerts an upward force on the body called upthrust or buoyancy. You may have noticed while diving into water or going inside water that the water exerts upward force on you. There is a principle known as Archemedes's principle which gives the idea of the upward force exerted by the water. It states that when a body is immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body.

Figure 5 An imaginary fluid element.
Figure 6 The imaginary fluid element replaced by a solid body of the same shape and size.

Consider a fluid in a vessel as shown in Figure 5 where a fluid element at rest is separated by an imaginary boundary. The fluid element isn't moving so the net force on the element should be zero- the total upward force exerted by the surrounding fluid must balance the weight of the fluid element. Now replace the fluid element by a solid body of the same shape and size (see Figure 6). Since the solid body has the same shape and size as that of the fluid element, the upward force on the body by the surrounding fluid is the same as that exerted on the fluid element. And hence the upward force on the solid body is equal to the weight of the fluid displaced by the body. If the weight of the body is greater than the weight of the fluid displaced, the body goes downwards and if the weight of the body is lesser than weight of the fluid displaced, the body goes upwards. Therefore, any body having lower density than the density of fluid floats in the fluid and those bodies having greater density than the density of the fluid sink.

Mechanics

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