## Table of Contents

## Pressure

You can push a drawing pin into a piece of wood – but you cannot push your finger into the wood even if you exert a larger force. The difference in each case is the different area in contact. The pin has a much smaller area in contact than the finger. Hence, exerts a much higher pressure that makes the substance yield to the force.

**Pressure (p)** is defined as the perpendicular force (F) acting on unit area of a surface or the force per unit area (A).

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

- SI unit of pressure is the pascal or newton per metre square. (Pa or $N m^{-2}$)
- It is a scalar quantity.
- Pressure is deemed a scalar quantity as its calculation involves solely the force that acts perpendicular to the contact area. Since the direction of this force remains constant, pressure is determined solely by its magnitude and is not influenced by its direction.

The pressure of one pascal is insufficient for describing pressures encountered in everyday life. Hence, its multiples, such as kilopascal $1 \, \text{kPa} = 10^3 \, \text{Pa}$ and megapascal $1 \, \text{MPa} = 10^6 \, \text{Pa}$, are frequently employed. Another widely utilized unit of pressure is the standard atmosphere $\text{atm}$.

- $1 \times 10^5 \, \text{Pa} = 0.1 \, \text{MPa} = 100 \, \text{kPa}$
- $1 \, \text{atm} = 101 \, 325 \, \text{Pa} = 101.325 \, \text{kPa}$

Analyzing the connection between pressure, force, and area, we observe that:

- An increase in the force’s magnitude results in a higher pressure.
- A reduction in the contact area leads to an increase in pressure.

## Worked Examples

### Example 1

Find the pressure exerted on a person by a few plastic blocks of total mass 42 kg, the bottom of which in contact with the person has an area of 840 $cm^{2}$. State what will happen to the pressure if the area is now decreased to a very small value.

**Click here to show/hide answer**

$$\begin{aligned} \text{Pressure}, \, p &= \frac{F}{A} \\ &= \frac{W}{A} \\ &= \frac{mg}{A} \\ &= \frac{42 \times 10}{840 \times 10^{-4}} \\ &= 5.0 \times 10^{3} \text{ Pa} \end{aligned}$$

In order to find out what will happen to the pressure with a small area, you can replace the area, A in the above equations with a value that is smaller than $840 \text{ cm}^{2}$. You will find that:

**If the area is decreased to a very small value, the pressure would increase to a very large value.**

### Example 2

The girl has a mass of 50 kg. Determine the pressure she exerts on a trampoline in two scenarios:

- When standing upright, with the area of her feet in contact with the trampoline being 500 cm².
- When lying down, with the contact area of her body on the trampoline being 6000 cm².

**Click here to show/hide answer**

Given gravitational field strength $g = 10 \, \text{N/kg}$, the pressure $P$ can be calculated using the formula $P = \frac{\text{Force}}{\text{Area}}$, where $\text{Force} = \text{mass} \times g$.

- When standing upright:

$$ \begin{aligned} P_{\text{upright}} &= \frac{50 \, \text{kg} \times 10 \, \text{N/kg}}{500 \, \text{cm}^2} \\ &= \frac{500}{0.05} \\ &= 10 \times 10^{3} \text{ Pa} \end{aligned} $$ - When lying down:

$$ \begin{aligned} P_{\text{lying down}} &= \frac{50 \, \text{kg} \times 10 \, \text{N/kg}}{6000 \, \text{cm}^2} \\ &= \frac{500}{0.60} \\ &= 833 \text{ Pa} \end{aligned}$$

### Example 3

A wooden plank has dimensions 100 cm by 20 cm by 2 cm and a weight of 40 N. Determine the maximum and minimum pressure exerted by the plank when placed on a flat horizontal table.

**Click here to show/hide answer**

Given the dimensions, the pressure $P$ exerted by the plank can be calculated using the formula $P = \frac{\text{Force}}{\text{Area}}$. The force acting on the table is the weight of the plank $F = 40 \, \text{N}$, and the contact area with the table depends on the orientation of the plank.

**Maximum Pressure:**

Considering the plank in the flat position, the area in contact with the table is the narrow side (20 cm by 2 cm).

$$\begin{aligned} P_{\text{min}} &= \frac{40 \, \text{N}}{20 \, \text{cm} \times 2 \, \text{cm}} \\ &= \frac{40}{4.0 \times 10^{-3}} \\ &= 10 \text{ kPa} \end{aligned}$$

**Minimum Pressure:**

Considering the plank in the upright position, the area in contact with the table is the base (100 cm by 20 cm).

$$

\begin{aligned} P_{\text{max}} &= \frac{40 \, \text{N}}{100 \, \text{cm} \times 20 \, \text{cm}} \\

&= \frac{40}{0.2} \\ &= 200 \, \text{Pa} \end{aligned} $$

### Example 4

A kitchen knife with a length of 15 cm is employed to cut a carrot with fluctuating thickness. The largest cross-sectional diameter of the cut carrot is 3 cm while the smallest cross-sectional diameter of the cut carrot is 1 cm. The cutting edge of the knife has a thickness of 0.1 mm, and a force of 20 N is applied. Determine the minimum pressure exerted by the knife during the carrot cutting process.

**Click here to show/hide answer**

Given the dimensions and force, the pressure $P$ exerted by the knife can be calculated using the formula $P = \frac{\text{Force}}{\text{Area}}$. The force acting on the carrot is $F = 20 \, \text{N}$, and the contact area with the carrot depends on the cross-sectional diameter and the thickness of the cutting edge of the knife.

To find the minimum pressure, consider the scenario where the knife is perpendicular to the carrot’s surface at the thickest part of the carrot, maximizing the contact area.

$$ \begin{aligned} P_{\text{min}} &= \frac{20 \, \text{N}}{3 \text{ cm} \times 0.1 \text{ mm}} \\ &= \frac{20}{3 \times 10^{-6}} \\ &= 6.67 \text{ MPa} \end{aligned}$$

Considerable pressure is applied with a relatively modest force. When engaging in cutting tasks, it proves more manageable to cut through the thinner sections of an object compared to its thicker portions. The rationale behind this lies in the smaller contact area associated with the thinner parts. In instances where the blade is dulled, the contact area expands, necessitating a greater force for effective cutting.

### Example 5

The pressure applied on the ground by an individual wearing roller blades is greater than that when they wear flat sole sneakers. Why?

**Click here to show/hide answer**

This phenomenon can be attributed to the reduced contact area between the roller blades and the surface compared to the flat sole sneakers. The smaller contact area concentrates the force over a smaller region, resulting in higher pressure. The pressure is calculated as the force applied divided by the contact area. Therefore, when wearing roller blades, the force is distributed over a smaller area, leading to increased pressure.