# Pressure

## Pressure Revision

**Pressure**

**Pressure** is the **force per unit area**. Fluids have **pressure** because the particles exert a **force** on whatever container the fluid is in.

**Pressure at the Surface of a Liquid**

A **fluid** is a substance that can flow and has no fixed shape, because the particles can move past each other. **Liquids** and **gases** are fluids.

Particles in fluid move around and collide with the surfaces of the container, and also with each other. These collisions exert a **force** on the surface of the container. **Pressure** is the **force per unit area **of a surface, so the particles are exerting a **pressure** on the container.

Pressure in a fluid causes a force at **right angles** (normal) to any surface. The **pressure** at a certain surface can be calculated using:

\textcolor{aa57ff}{p = \dfrac{F}{A}}

- \textcolor{aa57ff}{p} is the
**pressure**in**Pascals**\left(\text{Pa}\right) - \textcolor{aa57ff}{F} is the
**force**normal to the surface in**Newtons**\left(\text{N}\right) - \textcolor{aa57ff}{A} is the
**area of the surface**in**meters squared**\left(\text{m}^2\right).

**Pressure in a Column of Liquid**

**Density** is a measure of how close the particles are in a substance. Density is uniform in liquids, meaning that it is the same everywhere in the liquid. However this is not true for gases.

We can relate density to **pressure**. If a substance is more dense, then there are more particles in the volume of space, and so there are more collisions between the particles. Collisions give rise to **pressure**, so a higher density means a higher pressure.

As the** depth** of a liquid increases, there is more particles above this depth. That means that the weight of these particles is on top of the liquid. This weight adds to the pressure in the liquid. Therefore as **depth** increases, so does the **pressure**.

To show this, let’s look at a cylinder filled with water. We have marked two points, A and B, on the cylinder. See how there is more liquid above point B than point A. Therefore there is more particles and more mass above point B than point A, adding to the pressure of the liquid. So the **pressure** at B is higher than at A.

Pressure at a certain depth can be calculated using the following equation:

\textcolor{f21cc2}{p = h \rho g}

- \textcolor{f21cc2}{p} is the
**pressure**in**Pascals**\left(\text{Pa}\right) - \textcolor{f21cc2}{h} is the
**height**of the column in**metres**\left(\text{m}\right) - \textcolor{f21cc2}{\rho} is the
**density**of the liquid in**kilogram per metres cubed**\left(\text{kg/m}^3\right) - \textcolor{f21cc2}{g} is the
**gravitational field strength**in**newtons per kilogram**\left(\text{N/kg}\right) - Remember that \textcolor{f21cc2}{g} on earth is 9.8 \text{N/kg}.

**Upthrust**

- When an object is submerged in a liquid, it experiences
**pressure**due to the pressure in the liquid. - Because
**pressure**increases with**depth**, the bottom surface of this object will experience more pressure than the top surface. - Because this bottom pressure is larger, there is a
**resultant force**on the object upwards (remember that pressure and force are related). This resultant force upwards is called**upthrust**. - The
**upthrust**is equal to the weight of the water displaced by the object. This is just the weight of the water with the same volume as the object.

The balance of the forces will determine whether and object** sinks** or **floats**.

- If the
**upthrust**is equal to or more than the weight of the object, the object**floats**because the forces are balanced. - If the
**upthrust**is less than the objects weight, the object**sinks**.

This is because floating or sinking depends on the **density** of the object. Upthrust is the weight of the displaced water. So if the density of this water is less than the object, then the weight of the water is less than the object as well. This means that the** upthrust** is less than the weight of the object. Hence the object would sink.

**Atmospheric Pressure**

There is a thin layer of air around the earth called the** atmosphere**. The atmosphere gets less dense as the **height** above earth (altitude) increases.

Air molecules collide with the surface of the **atmosphere**, creating **atmospheric pressure**. This atmospheric pressure decreases with increasing altitude. This is because as the altitude increases, the atmosphere is less dense. Therefore there are less particles to collide with the surface, so the pressure decreases.

Also, as the height increases, there is less air particles on top of the surface. This means less weight of air on top of the surface contributing to the pressure. So this contribution to the pressure also decreases with altitude.

**Example: Calculating differences in pressure**

Calculate the change in pressure when an object is plunged from \textcolor{2730e9}{45 \: \text{m}} below the surface of the water to \textcolor{d11149}{100 \: \text{m}} below the surface of the water. The density of the water is \textcolor{f21cc2}{1000 \: \text{kg/m}^3}. The gravitational field strength is \textcolor{aa57ff}{9.8 \: \text{N/kg}}.

**[3 marks]**

At 45 \: \text{m}:

p = h \rho g

p = \textcolor{2730e9}{45} \times \textcolor{f21cc2}{1000} \times \textcolor{aa57ff}{9.8} = 441000 \: \text{Pa}

At 100 \: \text{m}:

p = h \rho g

p = \textcolor{d11149}{100} \times \textcolor{f21cc2}{1000} \times \textcolor{aa57ff}{9.8} = 980000 \: \text{Pa}

**Change in pressure** = 980000 \: \text{Pa} - 441000 \: \text{Pa} = \bold{539000} \: \textbf{Pa}

## Pressure Example Questions

**Question 1**: A potato is submerged into a bucket of water. The potato is more dense than water. State whether the potato will sink or float, and give a reason for your answer.

**[3 marks]**

The potato will **sink**.

This is because the density of the potato is more than water, so the **weight** **of the potato will be** **more than** **the weight of the displaced water**.

The force of the **weight overcomes the upthrust **and the potato sinks.

**Question 2**: Calculate the pressure at the surface of a liquid when the force is 0.4\: \text{N} and the area of the surface is 0.02 \: \text{m}^2.

**[2 marks]**

\begin{aligned}p &= \dfrac{F}{A} \\ p &= \dfrac{0.4\: \text{N}}{0.02 \: \text{m}^2} \\ \boldsymbol{p} &\boldsymbol{= 20} \: \textbf{Pa} \end{aligned}

**Question 3**: Calculate the change in pressure when an object is raised from 73 \: \text{m} below the surface of the water to 4 \: \text{m} below the surface of the water. The density of the water is 1000 \: \text{kg/m}^3. The gravitational field strength is 9.8 \: \text{N/kg}.

**[3 marks]**

At 73 \: \text{m}:

p = h \rho g

p = 73 \times 1000 \times 9.8 = 715400 \: \text{Pa}

At 4 \: \text{m}:

p = h \rho g

p = 4 \times 1000 \times 9.8 = 39200 \: \text{Pa}

**Change in pressure** = 715400\: \text{Pa} - 39200 \: \text{Pa} = \bold{676200} \: \textbf{Pa}

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