# Physics Tutorial: Liquid Pressure. Pascal's Principle

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In this Physics tutorial, you will learn:

• The meaning of liquid pressure
• The factors affecting liquid pressure
• The other name used for water pressure
• How can we use U-shaped tubes to find the density of unknown liquids
• What does Pascal's Principle say?
• How can we use the Pascal's Principle in practice?

## Introduction

Why do you feel a light pain in the chest when you go for the first time at beach?

Why people cannot dive so deep in water even though they may have oxygen cylinders with them?

Do you think it is important the position of an object inside water if the depth doesn't change?

Do you think the situation changes when you replace water with another liquid in the above examples?

All the above questions relate to the concept of liquid pressure, which will be explained in this tutorial.

## Liquid Pressure

By definition (and as indicated by the name), liquid pressure is the pressure exerted by liquids on all objects immersed in them.

Obviously, the unit of liquid pressure is Pascal [Pa].

We can still use the equation

P = F/A

to calculate the liquid pressure as we did for solid pressure in the previous tutorial. However, giving that the lower base of containers is not always regular, we can look for any alternative formula in order to ease the calculation of liquid pressure. For this, let's consider the following situation.

Suppose you have a rectangular container in the form of a cuboid and you have poured some liquid in it. Then, you put inside the liquid a heavy plate with negligible height. Obviously, the plate will sink and it will rest at bottom of container as shown in the figure below.

From the definition of pressure, it is obvious that only the liquid column above the plate exerts a downward force on it, due to gravity. Therefore, we must consider only the liquid column that lies above the plate, which forms a small cuboid of base A (equal to the plate's area) and height h (equal to the liquid's depth), as shown below.

Thus, the force exerted on the plate is equal to the weight of the liquid column. From the equation of pressure

P = F/A

we obtain for the liquid pressure

Pliquid = Fliquid/Aplate
= Wliquid/Aplate
= mliquid × g/Abase of liquid column
= ρliquid × Vliquid × g/Abase of liquid column
= ρliquid × Abase of liquid column × hliquid × g/Abase of liquid column
= ρliquid × hliquid × g

In this way, we obtained an alternative formula for liquid's pressure

Pliquid = ρliquid × hliquid × g

besides the standard formula

Pliquid = Fliquid/Abase

we knew before.

The new formula of liquid pressure, offers great advantages compared to the standard one. Some of them include:

1. No area is involved in the calculations. This means we don't have to remember all geometric formulae for area of figures or to find the area of irregular shapes by integration.
2. The shape of container is not important for the same reason.
3. It is not necessary to find the weight of liquid. This advantage is especially important when dealing with objects immersed in seawater as we don't need to calculate the mass of the entire seawater.
4. It is enough if we know the density of liquid and the depth of the immersed object from the surface to calculate the liquid's pressure. Hence, it is enough a small sample of liquid to calculate its density experimentally. As for the liquid's depth, we can use sensors that use sound echo to calculate it.

In this way, we obtain an important property of liquid pressure:

"Liquid's pressure depends only by depth if the density of liquid is known."

In other words, liquid pressure is the same everywhere at a certain depth, regardless its position within the container, as shown in the figure below.

Remarks!

1. The quantity h in the formula Pliquid = ρliquid × g × h shows the depth from the surface, not the height from the ground as in Kinematics.
2. The shape and size of the base area do not affect the magnitude of liquid pressure as long as the depth is the same. Look at the figure below.

If the liquid in all four containers is the same, we have PA = PB = PC = PD for the pressure at the bottom of each container, because the height of liquid is the same in all containers.

### Example 1

A swimming pool filled with water has three levels as shown in the figure below.

A girl who is bathing in this pool wears a watch which can hold up to 150 000 Pa more than the normal air pressure. Which level/s of the pool she can use without taking off her watch? Take g = 9.81 m/s2 and ρwater = 1000 kg/m3.

### Solution 1

We must calculate which depth does a 150 000 Pa water pressure correspond. Thus, from the equation of liquid pressure

Pliquid = ρliquid × g × h

we work out the maximum depth the watch can hold. Thus,

h = Pwater/ρwater × g
= 150 000 Pa/1000 kg/m3 × 9.81 m/s2
= 1.53 m
= 153 cm

This means the girl can use the first level of the pool (70 cm), the second level (70 cm + 70 cm = 140 cm) but not the third level as it is 70 cm + 70 cm + 70 cm = 210 cm > 153 cm deep, because if she puts her hand at bottom of the third level, her watch may be damaged.

The same thing occurs when we immerse our body in water. We feel a slight pain on the chest due to the additional pressure exerted by water on our body. Thus, water pressure causes a pushing force throughout the surface of our body. As a result, when we breathe, we feel some pain, especially in the first day of swimming season.

Remark! Water pressure is otherwise known as "gauge pressure" as we often use a pressure gauge to measure it.

We can use the formula of liquid pressure to calculate the density of an unknown liquid by using a U-shaped tube as shown in the figure below.

First, we pour a known liquid in the tube (liquid 2) and then, an unknown liquid above it (liquid 1). Obviously, this method works for liquids that do not mix with each other. At the position where the two liquid meet, there is the same pressure in both sides (P1 = P2). This means the pressure above this level in both sides of the tube is equal as well. Using the equation of liquid pressure, we can write

P1 = P2
ρ1 × g × h1 = ρ2 × g × h2

where ρ1 and ρ2 are the densities of liquid 1 and 2 respectively, h1 and h2 are their heights above the meeting level and g is the gravity.

Simplifying g from both sides, we obtain

ρ1 × h1 = ρ2 × h2

Thus, we can write for the density of the unknown liquid ρ1

ρ1 = ρ2 × h2/h1

### Example 2

What is the density of the unknown liquid in the U-shaped tube shown in the figure below if the other liquid is mercury? Take the density of mercury equal to 13.6 g/cm3.

### Solution 2

We have the following clues in this problem:

ρ1 = 13.6 g/cm3

h1 = 4.8 cm

h2 = 6.0 cm

ρ2 = ?

Thus, applying the equation

ρ1 × h1 = ρ2 × h2

we obtain for the density of the unknown liquid, ρ2:

ρ2 = ρ1 × h1/h2
= 13.6 g/cm3 × 4.8 cm/6.0 cm
= 10.88 g/cm3

## Pascal's Principle

The way in which pressure acts inside a liquid, was first discovered and outlined by the famous French scientist Blaise Pascal. For this reason, the unit of pressure bears his name, i.e. Pascal. Thus, pressure in liquids obeys to a universal rule, known as "Pascal's Principle" which says:

"In a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container."

This principle is particularly useful in communicating vessels, in which we can use a small force in a small area to transmit pressure in a larger area and in this way, to increase the value of force. Look at the figure below.

Thus, since liquid pressure is transmitted equally in all positions of the communicating vessel, we obtain

P1 = P2

or

F1/A1 = F2/A2

Pascal's principle is the operating principle of all hydraulic machines, i.e. in mechanisms that use the water pressure to lift heavy weights or in other words, to apply large forces as an output by applying a very small force at input, as the one shown in the above figure. As examples in this regard we can mention hydraulic lifts, hydraulic press machines, hydraulic steering wheels, etc.

### Example 3

A hydraulic lift has a 4 cm2 area at the button side and it is used to lift a maximum load of 6 tons as shown in the figure.

What is the maximum force the user can apply on the button (left side) of this hydraulic lift if the wider area is 8 m2? For convenience, take g = 10 m/s2.

### Solution 3

First, let's convert the units into the standard ones. We have:

A1 = 4 cm2 = 0.0004 m2 = 4 × 10-4 m2

A2 = 8 m2

F2 = m × g = 6 t × 10 m/s2 = 6000 kg 10 m/s2 = 60 000 N = 6 × 104 N

F1 = ?

Given that pressure is transmitted equally in all directions of a communicating system (based on Pascal's Principle), we can write for pressure at the positions where the two forces act:

P1 = P2

and therefore,

F1/A1 = F2/A2

Substituting the known values, we obtain

F1/4 × 10-4 = 6 × 104/8
8 × F1 = 4 × 10-4 × 6 × 104
8 × F1 = 24
F1 = 3 N

Thus, by only 3 N of applied force, we can lift a 6 t object using this kind of hydraulic lift.

## Summary

Liquid pressure is the pressure exerted by liquids on all objects immersed in them.

The unit of liquid pressure is Pascal [Pa].

Besides the standard formula of pressure

Pliquid = Fliquid/Abase

we can use the alternative formula

Pliquid = ρliquid × hliquid × g

for calculating the liquid pressure. This formula offers some advantages compared to the standard formula of pressure such as

1. No area is involved in the calculations. This means we don't have to remember all geometric formulae for area of figures or to find the area of irregular shapes by integration.
2. The shape of container is not important for the same reason.
3. It is not necessary to find the weight of liquid. This advantage is especially important when dealing with objects immersed in seawater as we don't need to calculate the mass of the entire seawater.
4. It is enough if we know the density of liquid and the depth of the immersed object from the surface to calculate the liquid's pressure. Hence, it is enough a small sample of liquid to calculate its density experimentally. As for the liquid's depth, we can use sensors that use sound echo to calculate it.

In this way, we obtain an important property of liquid pressure:

"Liquid's pressure depends only by depth if the density of liquid is known."

In other words, liquid pressure is the same everywhere at a certain depth, regardless its position within the container.

It must me noted that:

1. The quantity in the formula Pliquid = ρliquid × g × h shows the depth from the surface, not the height from the ground as in Kinematics.
2. The shape and size of the base area do not affect the magnitude of liquid pressure as long as the depth is the same. Look at the figure below.

Water pressure is otherwise known as "gauge pressure" as we often use a pressure gauge to measure it.

We can use the formula of liquid pressure to calculate the density of an unknown liquid by using a U-shaped tube in which we pour two non-mixable liquids. We measure the levels starting from the meeting position and then, use the equation

ρ1 × h1 = ρ2 × h2

to calculate the density of an unknown liquid.

Pressure in liquids obeys to a universal rule, known as "Pascal's Principle" which says:

"In a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container."

This principle is particularly useful in communicating vessels, in which we can use a small force in a small area to transmit pressure in a larger area and in this way, to increase the value of force.

The equation that represents mathematically the Pascal's Principle is

F1/A1 = F2/A2

Pascal's principle is the operating principle of all hydraulic machines, i.e. in mechanisms that use the water pressure to lift heavy weights or in other words, to apply large forces as an output by applying a very small force at input. Some examples in this regard include hydraulic lifts, hydraulic press machines, hydraulic steering wheels, etc.

## Liquid Pressure. Pascal's Principle Revision Questions

1. Humpback anglerfish is a deep-sea creature that can bear up to 30.3 million Pascal of water pressure.

Given that the density of sea-water is 1030 Pa and gravity g = 9.81 m/s2, calculate the maximum depth a humpback anglerfish can immerse.

1. 30 m
2. 300 m
3. 3000 m
4. 30 000 m

2. Calculate the density (in g/cm3) of the unknown liquid X shown in the figure below. Take density of water equal to 1 g/cm3.

1. 750 g/cm3
2. 0.75 g/cm3
3. 1333 g/cm3
4. 1.33 g/cm3

3. How many 80-kg people can we lift by using the hydraulic lift shown in the figure if the apply a 4 N force on the narrow side? Take g ≈ 10 m/s2.

1. 3
2. 6
3. 30
4. 60