# Understanding Archimedes' Principle and Its Applications

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## Who Was Archimedes?

Archimedes, a notable Greek mathematician, scientist, and astronomer from Syracuse, was born around 287 BC. He made significant contributions during the classical era, laying the groundwork for contemporary calculus, proving numerous geometric theorems, approximating the value of pi, and calculating the surface areas and volumes of three-dimensional shapes.

## What is Archimedes' Principle?

Archimedes' Principle asserts that the upward force, known as the buoyant force, acting on an object immersed in a fluid equals the weight of the fluid displaced by that object. The term "displaced" refers to the fluid that is pushed out of the way; for instance, when stones are dropped into a full container of water, they cause the water level to rise. This principle applies not only to water but to any fluid, including gases like air. For more information about forces, consider exploring my physics tutorial on:

**Force, Mass, Acceleration, and Understanding Newton’s Laws of Motion**

## Experiments to Understand Archimedes' Principle

Let’s conduct some experiments to explore and comprehend Archimedes' Principle.

### Experiment 1: Investigating Negative Buoyancy

**Step 1: Weigh the Object**

Consider an object hanging from a scale, for example, a 6 kg iron weight. We will submerge it in a water tank filled to the brim, ensuring the water level is at the overflow outlet. The weight may either float or sink, but this does not influence our experiment.

**Step 2: Measure the Displaced Water**

As the weight is submerged, it displaces water, which spills over into a container on a second scale. Upon full submersion, the displaced water weighs 2 kg.

**Step 3: Check the Weight on the First Scale**

Revisiting the first scale, we find that the weight now reads only 4 kg.

**Step 4: Conduct Calculations**

By subtracting the new reading from the initial weight, we find that the difference matches the weight of the displaced water:

6 kg - 4 kg = 2 kg

## Principle of Archimedes

Thus, we have confirmed Archimedes' principle: "The buoyant force acting on a body submerged or floating in a fluid equals the weight of the fluid displaced."

Why does the first scale show a lower weight now? This is due to the buoyant force acting upward, making the object appear lighter. The weight of the 6 kg object pulls downward, but the upward force of 2 kg supports it, resulting in a net weight of 4 kg on the scale. However, the mass of the object remains unchanged at 6 kg.

## What are the 3 Types of Buoyancy?

### Negative, Positive, and Neutral Buoyancy

An object in a fluid can exhibit three behaviors:

**Sinking**: This occurs when the object has negative buoyancy.**Floating**: Positive buoyancy is at play when the object stays on the surface. If pushed under the water and released, it will rise back to the surface.**Suspension**: An object can remain submerged at a constant depth without either sinking or floating, known as neutral buoyancy.

### Negative Buoyancy and Sinking Bodies

In our first experiment, the iron weight sank when submerged. The 6 kg weight displaces 2 kg of water, resulting in an upward buoyant force of 2 kg. Since this is less than the weight of the object, it fails to keep the object afloat, demonstrating negative buoyancy. If released from the scale, it would sink.

### Examples of Objects with Negative Buoyancy

- Anchors
- Sinkers for fishing nets

### Experiment 2: Investigating Positive Buoyancy

In this experiment, we will lower a hollow steel ball onto the water's surface.

### Positive Buoyancy and Floating Objects

What happens when a weight floats? When we lower a hollow steel ball weighing 3 kg into the tank, the chain becomes slack as it floats, reading 0 kg on the scale. The buoyant force equals the weight of the displaced water, allowing the ball to remain afloat.

If the ball is pushed under the water, it displaces more liquid, increasing the buoyant force until it rises back to the surface.

### Examples of Objects with Positive Buoyancy

- Lifebuoys
- Marker buoys
- Ships
- Swimmers
- Life jackets
- Fishing line floats
- Flotation devices for salvage operations
- Floating oil rigs and turbines

### Experiment 3: Investigating Neutral Buoyancy

Here, we will examine an object that maintains neutral buoyancy, remaining suspended beneath the water's surface.

Neutral buoyancy occurs when an object's average density matches that of the fluid. It neither sinks nor floats and can remain at any depth until acted upon by another force.

### Examples of Objects with Neutral Buoyancy

- Divers
- Submarines

Submarines manage their buoyancy by filling large tanks with water for negative buoyancy to dive. Once at the desired depth, they balance buoyancy to achieve a neutral state. To ascend, they pump out water and replace it with air, achieving positive buoyancy.

Scuba divers maintain neutral buoyancy by utilizing weight belts to remain at a preferred depth without constant swimming.

## Formula for the Buoyant Force

**Fb = -?gV**

Where:

**?**is the density of the displaced fluid**V**is the volume of the displaced fluid**g**is the acceleration due to gravity

If we consider the downward weight as positive, the negative sign indicates that the buoyant force acts in the opposite direction.

The weight of an object is given by:

**Fg = mg**

Where **m** is the mass of the object.

## Why do Ships Float?

Ships, despite their massive weight, float because they displace a significant amount of water. The extensive internal space of a ship allows it to push a large volume of water aside, generating an upward force that counterbalances its weight.

### Why do Ships Sink?

While positive buoyancy keeps ships afloat, overloading them with heavy cargo can lead to sinking if the total weight surpasses the buoyant force. Additionally, if a hull is compromised, water entering the ship increases weight, causing it to sink.

A ship could also sink if its structure were compressed into a small block, as the reduced volume would not displace sufficient water.

## How Does Liquid Density Affect Buoyancy?

The buoyancy of an object is influenced by the density of the fluid in which it is placed, while still adhering to Archimedes' Principle.

### Average Density of Object

The average density **?** of an object is given by:

**? = m / V**

Where **m** is the object's mass and **V** is its volume.

For instance, a large hollow steel ball has a shell density significantly higher than that of the air inside, resulting in a lower average density that allows it to float in water if this average density is less than that of the water.

### Buoyancy and Average Density

- If an object's average density > fluid density, it will sink (negative buoyancy).
- If an object's average density < fluid density, it will float (positive buoyancy).
- If an object's average density = fluid density, it remains suspended (neutral buoyancy).

For an object to float, its average density must be lower than that of the fluid. For instance, if an object is less dense than water but denser than kerosene, it will float in water but sink in kerosene.

## Why Do Icebergs Float?

Water behaves unusually; most materials become less dense as temperature increases. However, water's density decreases below 4ºC, allowing ice to float since it is less dense than liquid water.

## How Do Helium Balloons Float?

Archimedes' Principle applies to both liquids and gases. Balloons utilize the buoyant force of air to rise.

When a balloon is filled with air, the downward force is the weight of the balloon and the air inside. The upward force is the weight of the displaced air. If the balloon's weight is greater than the buoyant force, it will sink.

When filled with helium, which is less dense than air, the displaced air's weight exceeds that of the balloon and its contents, allowing it to rise.

## Why Do Hot Air Balloons Float?

Hot air balloons float because their internal air is heated, reducing its density compared to the cooler surrounding air. The weight of the displaced air exceeds the weight of the hot air inside, providing lift.

## Worked Examples on Buoyancy

### Question 1:

A hollow steel ball weighing 10 kg and with a diameter of 30 cm is submerged in a pool. Calculate the net upward force acting on the ball.

### Answer:

To find the displaced water volume, we compute the weight and density of the water.

Let **r** be the radius of the sphere. The volume **V** of a sphere is calculated as:

**V = 4/3 ? r³**

Using a diameter of 30 cm (0.3 m), we find:

**r = 0.15 m**

Substituting this into our volume equation gives us the volume of the sphere. Then, using the density of water (1000 kg/m³), we can find the mass of the displaced water.

Thus, we have:

**m = ?V = 1000 × V**

The ball weighs 10 kg, while the displaced water weighs approximately 14.137 kg. Thus, the upward buoyant force is 14.137 kg.

The net force acting on the ball is:

**14.137 kg - 10 kg = 4.137 kg**

The ball demonstrates positive buoyancy and will rise to the surface, balancing its weight with the volume of displaced water.

### Question 2:

How many helium balloons are needed to lift a person weighing 90 kg, assuming each balloon has a diameter of 30 cm and weighs 10 g (including the string)?

### Answer:

Let **d** be the diameter of a balloon (0.3 m), and thus the radius **r** is 0.15 m.

The density of helium is approximately 0.1786 g/L (or 1.786 kg/m³), while the density of air is about 1.225 kg/m³. The volume of one balloon is:

**V = 4/3 ? r³**

Total volume of **N** balloons = **NV**.

The buoyant force must exceed the total weight of the helium, balloon material, and the person.

Setting up the inequality:

**NV(air density) > N(helium density) + N(weight of balloon) + weight of person**

This leads to:

**N > Weight of person / (V(air density) - V(helium density) - weight of balloon)**

Calculating gives approximately 18,765 balloons needed to lift the individual, emphasizing the impracticality of using many small balloons. Larger balloons would provide a better solution as their volume increases cubically compared to surface area.

For larger balloons (3 m diameter), the required number reduces significantly to around 7.

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*This article was originally published on Owlcation on 8th November, 2018*