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Have you ever wondered why the sharp edge of a knife can cut through a fruit seamlessly whereas if you used the blunt edge it would require a lot more effort? This phenomenon can be explained by understanding the concept of pressure. In this case, information about the magnitude of the force that is exerted by the knife is not…

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Pressure

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Jetzt kostenlos anmeldenHave you ever wondered why the sharp edge of a knife can cut through a fruit seamlessly whereas if you used the blunt edge it would require a lot more effort? This phenomenon can be explained by understanding the concept of pressure. In this case, information about the magnitude of the force that is exerted by the knife is not enough to know whether or not cutting fruit is easy or not. We need another measure that quantifies how much force is exerted over a particular area. At the end of this article, you will have learned what pressure is, learned how to use the pressure equation, and worked on a few examples to understand the importance of pressure in our daily lives.

Before we introduce the formula for pressure and look at some worked examples, let us first define what we mean by pressure in physics.

Pressure is defined as the force exerted per unit area of a surface.

The amount of force exerted by a knife is not enough information to know whether or not we will be able to cut into a fruit. Pressure takes into account how **concentrated** the force is over a particular area.

So how do we calculate the concentration of force in a given area? Simple, we know that pressure is nothing but the force acting per unit area which in mathematical terms give

$P=\frac{F}{A}$

or in words

id="2827891" role="math" $\mathrm{Pressure}=\frac{\mathrm{Force}\mathrm{normal}\mathrm{to}\mathrm{the}\mathrm{surface}}{\mathrm{Area}}$

Pressure is expressed in Pascals ($\mathrm{Pa}$),$F$is the force in$\mathrm{N}$, and the area is measured in units of${\mathrm{m}}^{2}$**.**

A pressure of$1\mathrm{Pa}$can be defined as a force of$1\mathrm{N}$acting on an area of$1{\mathrm{m}}^{2}$(unit area). The above formula is only valid when the force acts at **right angles** to the surface. As you can see from the equation. Pressure is directly proportional to the force applied and inversely proportional to the area on which it acts. This means that to **increase** the pressure we can either

**Increase**the**force**keeping the area constant or**Decrease**the**area**keeping the force at a constant.

The SI unit of pressure is$\mathrm{Pa}\mathrm{or}\mathrm{N}/{\mathrm{mm}}^{2}$1 Pascal =$1\mathrm{N}/{\mathrm{mm}}^{2}$.^{ }For large quantities of pressure, we can use $1\mathrm{kPa}={10}^{3}\mathrm{Pa}$and$1\mathrm{MPa}={10}^{6}\mathrm{Pa}$. You may also see other units of pressure such as torr**, $\mathrm{barr},\mathrm{atm},\mathrm{psi}\mathrm{and}\mathrm{mm}\mathrm{Hg}.$ **

The atmospheric pressure is a commonly used unit ($\mathrm{atm}$). The pressure exerted by the Earth's atmosphere on the surface of the earth is what is known as atmospheric pressure. The atmospheric pressure in pascals is equal to$101325\mathrm{Pa}$. It is also sometimes referred to as standard atmospheric pressure.

We can categorise types of pressure in terms of the states of matter that are exerting the pressure. In this section we'll have a look at each of the types of pressure and some examples of each type too.

Pressure can be exerted by solids, liquids, and gases. **Solids** exert pressure through their point of contact. **Liquids and gases** exert pressure on a solid due to the collision of their particles with the solid.

- A nail being hammered into a wall is an example of pressure acting from one solid onto another solid.
- We are constantly facing the force of the atmosphere on our bodies. which is nothing more than the pressure exerted by the gases in the atmosphere.

The pressure exerted by liquids is mainly due to the **weight** of the liquid. Imagine taking a dive at your local swimming pool. Let's say you dive into the deep end. Now due to the effects of gravity, all of the weight of the water above you will act to press the water molecules down onto your body and any other object submerged in the water. This pressing is what we mean by the pressure exerted due to a liquid. The number of water molecules that are present above you will increase as you keep going deeper. This is why the pressure exerted by liquids increases with increasing depth and it's why the pressure you feel increases as you dive deeper into a body of water. Another factor that comes into play is the density of the liquid. Since density measures the mass per unit volume of a liquid. Liquids with higher density will exert a larger pressure at the same depths due to their larger weights.

Lets now consider how to calculate the pressure exerted by a liquid at a particular depth. We consider a rectangular column of water with a base area$A$and height$h$.

We already know that the formula for pressure is:

$P=\frac{F}{A}$

The weight of the liquid situated directly above the base area of consideration$A$is given by the following equation:

$F=mg$

Or in words

$\mathrm{Weight}=\mathrm{mass}\times \mathrm{gravitational}\mathrm{field}\mathrm{strength}$

The next step in our calculation was considered to consider the density of the liquid in order to calculate the mass of the column of water:

$\rho =\frac{m}{V}$,

or in words,

$\mathrm{Density}=\frac{\mathrm{mass}}{\mathrm{volume}}$.

Rearranging for mass we get:

$m=\rho V$.

The volume of a cuboid can be written as the product of its base area$A$and its height$h$:

$V=Ah$,

or in words,

$\mathrm{Volume}=\mathrm{Base}\mathrm{area}\times \mathrm{height}$.

We Substitute the formula for weight in terms of mass before substituting the formula for mass in terms of density and volume. Finally, we substitute the formula for volume in terms of base area and height into the equation for pressure:

$P=\frac{mg}{A}=\frac{\rho \times V\times g}{A}=\frac{\rho \times A\times h\times g}{A}$

After further rearrangement, we arrive at the formula for the pressure exerted by a column of liquid pressure in terms of the density of the liquid$\rho $, the height of the column of liquid$h$, and the base area of the column of liquid$A$:

$P=\rho hg$

$\mathrm{pressure}=\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{column}\times \mathrm{density}\mathrm{of}\mathrm{the}\mathrm{liquid}\times \mathrm{gravitational}\mathrm{field}\mathrm{strength}$

Where pressure is measured in Pascals ($\mathrm{Pa}$), the depth or height of column$h$in meters ($\mathrm{m}$) and density$\rho $in$\mathrm{kg}/{\mathrm{m}}^{3}$and gravitational field strength,$g$in$\mathrm{N}/\mathrm{kg}$.

Next, let us look at a few examples where we use these formulas to calculate the pressure.

Atmospheric pressure is due to the air molecules in the atmosphere colliding with the earth or any other object contained within. The atmospheric pressure reduces as the altitude increases because the density of the air decreases at higher altitudes. At higher altitudes, there is less air, therefore the weight of the air pressing down on an object is reduced. This is the reason why some people experience ear pain during air travel that occurs due to quick changes in air pressure. The atmospheric pressure can be calculated using the same formula as the pressure due to liquids.

A fluid has a weight of$300\mathrm{N}$which is exerted over a surface area of$8.2{\mathrm{m}}^{2}$. Calculate the pressure acting on the surface.

$F=300\mathrm{N},A=8.2{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}P=\frac{300\mathrm{N}}{8.2{\mathrm{m}}^{2}}=36.585\mathrm{Pa}$

Calculate the pressure exerted on the ground by a person weighing$60\mathrm{kg}$wearing

- A flat shoe with a contact area of$100{\mathrm{cm}}^{2}$
- A heeled shoe with a total contact area of$26{\mathrm{cm}}^{2}$

What can you infer from the results?

Unit conversion

$1\mathrm{cm}=0.01\mathrm{m}\phantom{\rule{0ex}{0ex}}1{\mathrm{cm}}^{2}=0.0001{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}100{\mathrm{cm}}^{2}=0.01{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}26{\mathrm{cm}}^{2}=0.0026{\mathrm{m}}^{2}$

Pressure on the ground due to flat shoes

$m=60\mathrm{kg},\mathrm{A}=0.01{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}P=\frac{F}{A}\phantom{\rule{0ex}{0ex}}F=mg=60\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}=588\mathrm{N}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}P=\frac{F}{A}=\frac{588\mathrm{N}}{0.01{\mathrm{m}}^{2}}=58,800Pa=58.8\mathrm{kPa}$

Pressure on the ground due to heels

__$m=60\mathrm{kg},A=0.0026{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}P=\frac{F}{A}\phantom{\rule{0ex}{0ex}}F=mg=60\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}=588\mathrm{N}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}P=\frac{F}{A}=\frac{588\mathrm{N}}{0.0026\mathrm{m}/{\mathrm{s}}^{2}}=226153.846Pa=226.15\mathrm{kPa}$__

The pressure exerted by the heels is almost **5 times** that of the flat sole shoes. This is the reason why heeled shoes are uncomfortable to wear over long periods as they exert a large amount of pressure on the ground and equally on the persons who's wearing the shoe's feet.

Calculate the pressure in the water well at a depth of$50\mathrm{m}$. Take the density of water to be$1000\mathrm{kg}/{\mathrm{m}}^{3}.$

Step 1: List the given quantities

$\rho =1000\mathrm{kg}/{\mathrm{m}}^{3},h=50\mathrm{m},g=9.8\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}$

Step 2: Calculate the pressure using the appropriate equation

$P=\rho hg$

$P=1000\mathrm{kg}/{\mathrm{m}}^{3}\times 50\mathrm{m}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}=490,000\mathrm{Pa}=490\mathrm{kPa}$

Calculate the density of air at an altitude of$5\mathrm{km}$where the pressure of the surrounding atmosphere is$12.5\mathrm{kPa}.$

Step 1: List the given quantities

$h=5\mathrm{km}=5000\mathrm{m},g=9.8\mathrm{m}/{\mathrm{s}}^{2},P=12.5\mathrm{kPa}$

Step 2: Calculate the pressure using the appropriate equation

$P=\rho hg\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\rho =\frac{P}{h\times g}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\rho =\frac{12.5\times 1000\mathrm{Pa}}{5000\mathrm{m}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\rho =0.25\mathrm{Kg}/{\mathrm{m}}^{3}$

- Pressure is defined as the force exerted per unit area of a surface.
- Pressure can be calculated by dividing the force by the area across which it is acting:
$P=\frac{F}{A}$ or in words, $\mathrm{Pressure}=\frac{\mathrm{Force}\mathrm{normal}\mathrm{to}\mathrm{surface}}{\mathrm{Area}}$.

There are different types of pressure depending on the type of matter exerting the pressure.

The pressure exerted by a liquid is given by $P=h\rho g$.

The pressure exerted by a liquid increases as the depth of the liquid increases.

Gases exert pressure on the objects they come into contact with.

Atmospheric pressure is due to the weight of the air molecules on top of the earth.

Atmospheric pressure reduces as the altitude increases.

Pressure is defined as the force per unit area.

The formula for pressure is P=F/A

The formula for pressure is P=F/A

A knife cutting a slab of butter, A nail being hammered into a wall are examples of pressure

The types of pressure are

- Absolute pressure.
- Gauge pressure.
- Differential pressure.
- Sealed pressure or vacuum pressure.

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