# Hydrostatic Pressure

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## Hydrostatic Pressure

Liquids and gases are called fluids, as they can change their shapes easily. A fluid is a substance that can flow. In the figure above, we notice that the water from the right container is going further than the water from the left container. The only difference between the two container is the height of the water column. Hence, we know that the amount of pressure inside a body of fluids increases with its depth (or height of fluid column).

### Pressure Due To Fluid Column

Pressure due to a fluid column (p) = height of column (h) X density of the fluid ($\rho$) X gravitational field strength (g). (The derivation of this formula can be found at the bottom of this post.)

$$p = \rho g h$$

In a fluid, if two points are separated by a vertical height, their difference in pressure is:

$$\Delta p = \rho g \Delta h$$

From the formula, $p = \rho g h$, we can see that the pressure depends on the depth and density of the liquid and NOT on the cross-sectional area or volume of the liquid. Hence, the water pressure at the bottom of the container in the figure above will be the same throughout the container as the water level is the same! The shape of the container does NOT matter in the computation of the water pressure!

Actual pressure at base of column of fluid  $p = h \rho g$ + atmospheric pressure at liquid surface

### Properties (Summary) Of Hydrostatic Pressure

• Pressure is transmitted throughout the liquid
• Pressure acts in all directions
• All points at the same depth in a fluid are at the same pressure
• Pressure increases with depth
• Pressure is dependent not on the shape of the container but on its depth.

## Pressure At Different Depths

A object immersed in a uniform liquid will experience a pressure which depends only on the height of the liquid above the object.

Hence, consider the picture below.

$$\text{Pressure at point A} = P_{0} + \rho g h$$

, where $P_{0}$ is the atmospheric pressure

## Upthrust

Upthrust or buoyancy force

• An upward force on a body produced by the surrounding fluid (i.e., a liquid or a gas) in which it is fully or partially immersed, due to the pressure difference of the fluid between the top and bottom of the object
• Net upward buoyancy force is equal to the magnitude of the weight of fluid displaced by the body
•  Upthrust = Vρg Or Upthrust = mass of liquid displaced x g = mg

## Archimedes’ Principle

Archimedes’ Principle

• States that the upthrust experienced by an object partially or entirely immersed in a fluid is equal to the weight of the fluid displaced by the object
• Can be applied to objects of all shapes and sizes
• An object floats in equilibrium if the upthrust = weight of object
• An object sinks if upthrust < weight of object

Upthrust In Equilibrium Scenario

Upthrust (in equilibrium)

= weight of object supported by water/fluid
= weight of fluid displaced
= volume of object submerged in water X density of fluid X g

## Simple Derivation Of The Formula For Hydrostatic Pressure ($p = \rho g h$):

Consider a column of water occupying a total volume V and a base surface area of A.

The weight of all the water is:

$$W = mg$$

Recall that the mass of the water is just the density of the water X volume of water: $m = \rho V$. Hence,

$$W = \rho V g$$

We note that volume is a product of surface area and height: $V = Ah$

$$W = \rho A h g$$

We recall that formula for pressure: $p = \frac{F}{A}$. Sub. in W into F to give:

\begin{aligned} p &= \frac{\rho A h g}{A} \\ &= \rho h g \end{aligned}

Done.

## Worked Examples

### Example 1

A string supports a solid iron object of mass 0.180 kg totally immersed in a liquid of density 800 kg m-3. If the density of iron is 8000 kg m-3, calculate:
(a) the upthrust acting on the object;
(b) the tension in the string

(a)

\begin{aligned} \text{Upthrust} &= V\rho g \\ &= \left(\frac{0.180}{8000}\right)(800)(9.81) \\ &= 0.177 \, \text{N} \end{aligned}

(b)
\begin{aligned} \text{Tension} &= \text{Weight of object}-\text{Upthrust on object} \\ &= (0.180)(9.81)-0.177 \\ &= 1.59 \, \text{N} \end{aligned}

### Example 2

An ice cube of sides 2.0cm floats in a cup of tea. One of its faces is 0.20cm above the surface of the tea in the cup. Calculate the density of the tea if the density of ice is 920 kg m-3.

\begin{aligned} \text{Weight of ice} &= \text{Weight of liquid displaced} \\ V_{\text{ice}}\rho_{\text{ice}}g &= V_{\text{tea}}\rho_{\text{tea}}g \\ (0.020)^{3}(920) &= (0.020)^{2}(0.018)\rho_{\text{tea}} \\ \rho_{\text{tea}} &= 1.02 \times 10^3 \, \text{ kg m}^{-3} \end{aligned}

### Example 3

A 50cm3 block of lead and a 50cm3 block of copper are completely under water. Each is suspended by a thread so that they both are at the same height in an aquarium filled with water. The volume of the aquarium is 300 cm3.Which of the following is true?

1. The buoyant force is greater on the lead than on the copper
2. The buoyant force is greater on the copper than on the lead
3. The buoyant force is the same on both blocks

Buoyant force or upthrust = volume of water displaced x density of water x g.
Since the volume of water displaced is the same in each case, the upthrust is the same.

### Example 4

A submarine is in equilibrium in a fully submerged position. What causes the upthrust on the submarine?

1. The air exerts a greater upward force on the submarine than the weight of the steel.
2. The air in the submarine is less dense than the sea water.
3. There is a difference in water pressure acting on the top and bottom of the submarine.
4. The submarine displaces its own volume of water.

Upthrust occurs because the pressure in a fluid increases with increasing depth.