Pressure and Hydrostatic Pressure for the ESAT

Understanding Pressure in Solids

When a force is applied to a surface, the pressure exerted is defined by the relationship between that force and the area over which it is distributed. The formula is:

$pressure = \frac{force}{area}$

In this context, area refers to the specific area of contact where the force is applied. Therefore, pressure is effectively the force per unit area acting on that surface.

If the force is measured in newtons (N) and the area is measured in square metres ($m^{2}$), the resulting unit for pressure is the pascal (Pa). By definition, $1 \text{ Pa} = 1 \text{ N m}^{-2}$.

When an object is placed on a horizontal surface, the force it exerts on that surface is equal to the object's weight. For example, consider a box with a mass of 12 kg and a base area of 0.50 $m^{2}$ resting on the floor. The pressure exerted by the box is calculated as follows:

$pressure = \frac{weight}{area} = \frac{mass \times gravitational field strength}{area}$

Using $g = 10 \text{ N kg}^{-1}$:

$pressure = \frac{12 \times 10}{0.50} = 240 \text{ Pa}$

Controlling Pressure through Force and Area

As the mathematical relationship indicates, pressure can be manipulated in two ways. To increase pressure, one can either increase the applied force or decrease the contact area. Conversely, to decrease pressure, one must either decrease the force or increase the area.

Practical applications of high pressure include tools like sewing needles and knives. These are designed with extremely small contact areas at their points or edges. This small area maximises the pressure generated from the user's force, allowing the needle to pierce cloth or the knife to slice through food easily.

Applications of low pressure include snowshoes and skis. These items feature a large contact area with the snow. By spreading the wearer's weight over a greater area, the pressure is minimised, preventing the person from sinking into the soft snow.

A drawing pin (thumbtack) illustrates both principles. It has a broad head at one end and a sharp point at the other. When you push the pin into a board, the force applied by your thumb at the head is equal to the force delivered at the sharp point. However, because the area of the point is so small, the pressure at the tip is high enough to penetrate the board, while the pressure at the thumb end remains low enough to avoid causing pain.

It is important to distinguish between different sources of pressure. The total pressure on a surface is the sum of the pressure from the object (the solid) and the pressure from the surrounding gas, such as the atmosphere. The term 'pressure due to the solid' specifically refers to the contribution from the object's weight alone.

Worked Example: Table Pressure

A table with a weight of 250 N has four legs. Each leg has a square cross-section with a width of 5.0 cm in contact with the ground. To find the pressure exerted on the ground:

1. Calculate the contact area of one leg in square metres: $0.050 \text{ m} \times 0.050 \text{ m} = 0.0025 \text{ m}^{2}$.
2. Calculate the total contact area for all four legs: $4 \times 0.0025 = 0.010 \text{ m}^{2}$.
3. Apply the pressure formula: $pressure = \frac{250}{0.010} = 25000 \text{ Pa}$.

Hydrostatic Pressure in Fluids

Fluids include both liquids and gases, which are substances capable of flowing. Pressure in a fluid acts in all directions and increases with depth. The pressure at any point in a fluid is caused by the weight of the fluid situated above that point. This hydrostatic pressure depends on the depth ($h$), the density of the fluid ($\rho$), and the gravitational field strength ($g$):

$hydrostatic pressure = h \rho g$

When $h$ is in metres, $\rho$ is in $kg \text{ m}^{-3}$, and $g$ is in $N \text{ kg}^{-1}$, the pressure is calculated in pascals. Crucially, hydrostatic pressure depends only on these variables and is independent of the shape of the container. If multiple containers hold the same liquid, the pressure at the same depth $h$ will be identical in all of them, regardless of the container's width or profile.

Deriving the Hydrostatic Pressure Formula

We can derive this formula by imagining a cuboid of liquid within a larger body of liquid. The cuboid has a cross-sectional area $A$ and a height $h$, with its top surface at the liquid's surface.

The pressure at the base of this cuboid is due to the weight of the liquid column above it:

1. The volume of the liquid cuboid is $V = A h$.
2. The mass of the liquid is $m = V \rho = A h \rho$.
3. The weight of the liquid (the force) is $W = m g = A h \rho g$.
4. Using $P = F/A$, the pressure at the base is $P = \frac{A h \rho g}{A} = h \rho g$.

Just as with solids, the total pressure at a depth in a liquid is the sum of the hydrostatic pressure and the atmospheric pressure pressing down on the liquid's surface.

Worked Example: Sea Floor Depth

At a specific point on the sea floor, the total pressure is $3.16 \times 10^{6} \text{ Pa}$. Given the density of seawater is $1020 \text{ kg m}^{-3}$, atmospheric pressure is $1.0 \times 10^{5} \text{ Pa}$, and $g = 10 \text{ N kg}^{-1}$, calculate the depth.

1. Find the pressure due to the water alone (hydrostatic pressure): $3.16 \times 10^{6} - 1.0 \times 10^{5} = 3.06 \times 10^{6} \text{ Pa}$.
2. Rearrange the formula to solve for height: $h = \frac{P}{\rho g}$.
3. Substitute the values: $h = \frac{3.06 \times 10^{6}}{1020 \times 10} = \frac{3060000}{10200} = 300 \text{ m}$.