# Case Study on Archimedes’ Law

Archimedes’ law, commonly known as Archimedes’ principle, is a law of physics fundamental to fluid dynamics and buoyancy. The principle was discovered by the ancient Greek mathematician and inventor Archimedes of Syracuse. It states that any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

The concept of buoyancy is essential for the design of ships, submarines, and other maritime vehicles. Moreover, it is also crucial in the field of fluid mechanics and is often used by engineers and scientists.

## Theory

In mathematical terms, the principle can be expressed as:

\[ F_b = \rho \cdot g \cdot V \]

Where,

– \( F_b \) is the buoyant force,

– \( \rho \) is the density of the fluid,

– \( g \) is the acceleration due to gravity, and

– \( V \) is the volume of the fluid displaced.

Archimedes’ law indicates that the buoyant force on an object in a fluid is directed upwards and is equal in magnitude to the weight of the fluid that the object displaces. This principle applies to both floating and submerged objects, explaining not only why ships float but also why objects like stones sink in water.

## Case Study: The Golden Crown

One of the most famous anecdotes relating to Archimedes is the story of how he used his principle to determine whether King Hiero’s crown was made of solid gold, or if a dishonest goldsmith had substituted some of the gold for silver. Archimedes realized that by submerging the crown in water, he could calculate the volume of the crown by the amount of water it displaced. By comparing this to the displacement of an equal weight of pure gold, he could determine its density and thus its purity, as gold has a higher density than silver.

Now, let’s consider some problems and solutions that involve applying Archimedes’ law.

## Problems and Solutions about Archimedes’ Law

### Problem 1:

A cube of gold weighing 19.3 grams is submerged in water. What is the buoyant force acting on the cube if the density of water is \(1 g/cm^3\)?

### Solution 1:

First, find the volume of the gold cube:

\[ Density_{gold} = \frac{Mass}{Volume} \]

\[ Volume = \frac{Mass}{Density_{gold}} = \frac{19.3 g}{19.3 g/cm^3} = 1 cm^3 \]

Using Archimedes’ principle:

\[ F_b = \rho_{water} \cdot g \cdot V_{gold} \]

\[ F_b = 1 g/cm^3 \cdot 9.8 m/s^2 \cdot 1 cm^3 \]

\[ F_b = 9.8 g \cdot m/s^2 \]

\[ F_b = 0.0098 N \] (since \( 1 N = 10^3 g \cdot m/s^2 \))

Therefore, the buoyant force is 0.0098 N.

### Problem 2:

What is the apparent weight of a 500 g steel block (density = \(7.8 g/cm^3\)) when it is fully submerged in water?

### Solution 2:

Calculate the volume of the steel block:

\[ Volume_{steel} = \frac{Mass_{steel}}{Density_{steel}} = \frac{500 g}{7.8 g/cm^3} \approx 64.1 cm^3 \]

The buoyant force is:

\[ F_b = \rho_{water} \cdot g \cdot V_{steel} \]

\[ F_b = 1 g/cm^3 \cdot 9.8 m/s^2 \cdot 64.1 cm^3 \]

\[ F_b = 628.18 g \cdot m/s^2 \]

\[ F_b \approx 0.628 N \]

The apparent weight is the actual weight minus the buoyant force:

\[ Weight_{apparent} = Weight_{actual} – F_b \]

\[ Weight_{apparent} = (500 g \cdot 9.8 m/s^2) – 0.628 N \]

\[ Weight_{apparent} \approx 4.902 N – 0.628 N \]

\[ Weight_{apparent} \approx 4.274 N \]

Therefore, the apparent weight of the block is approximately 4.274 N.

### Problem 3:

A cylindrical object with a radius of 4 cm and a height of 10 cm is floating upright in freshwater with its height transferring from 2 cm above the surface to 3 cm. What is the object’s density?

### Solution 3:

Calculate the volume of the displaced water when the object is floating, which equals the volume of the submerged part of the object.

Volume of the submerged cylinder before the change:

\[ V_1 = \pi r^2 h_1 \]

\[ V_1 = \pi (4 cm)^2 (8 cm) \]

\[ V_1 = 402.12 cm^3 \]

After the change:

\[ V_2 = \pi r^2 h_2 \]

\[ V_2 = \pi (4 cm)^2 (7 cm) \]

\[ V_2 = 351.86 cm^3 \]

Since the object is floating, the weight of the displaced fluid equals the weight of the object. Before and after the transfer, the buoyant force must equal the object’s weight.

\[ F_{b1} = F_{b2} \]

\[ \rho_{water} g V_1 = \rho_{water} g V_2 \]

\[ V_1 = V_2 \]

However, since the weight doesn’t change, and the volume of the water displaced changes, the object’s density must change. Since the volume of the object itself doesn’t change, we can imply that the object consists of multiple material components or is compressible.

This poses conflicting information in the scenario because the shift in height suggests the object’s composition or shape changes, a factor not addressed by Archimedes’ law and not solvable with the provided info. To solve the problem, we need additional data about the structural changes of the object.

### Problem 4:

A swimmer with a volume of 0.07 m³ jumps into a pool and floats with 90% of their body submerged. What is the swimmer’s mass?

### Solution 4:

Since 90% of the swimmer’s volume is submerged, we can find the buoyant force acting on that volume.

\[ F_b = \rho \cdot g \cdot V_{submerged} \]

\[ F_b = 1000 kg/m^3 \cdot 9.8 m/s^2 \cdot 0.9 \cdot 0.07 m^3 \]

\[ F_b \approx 617.4 N \]

Buoyant force is equal to the weight of the swimmer in this scenario:

\[ Weight = Mass \cdot g \]

\[ F_b = Mass \cdot g \]

\[ Mass = \frac{F_b}{g} \]

\[ Mass \approx \frac{617.4 N}{9.8 m/s^2} \]

\[ Mass \approx 63 kg \]

The swimmer’s mass is approximately 63 kg.

### Problem 5:

A rectangular barge 10 m long, 5 m wide, and 1 m deep floats in freshwater. If the barge is loaded with cargo until the water is just at the top edge of the barge, how much cargo in kilograms can it hold? Assume freshwater has a density of \( 1000 kg/m^3 \).

### Solution 5:

To find how much cargo the barge can hold, we need to find the volume of the water that would be displaced when the barge is submerged up to the top edge.

\[ Volume = Length \times Width \times Depth \]

\[ Volume = 10 m \times 5 m \times 1 m \]

\[ Volume = 50 m^3 \]

Given the density of the water, we can find the weight of the water displaced which equals the maximum cargo weight:

\[ Weight = \rho \cdot Volume \]

\[ Weight = 1000 kg/m^3 \cdot 50 m^3 \]

\[ Weight = 50000 kg \]

Therefore, the barge can hold up to 50,000 kg of cargo before it begins to submerge and take in water.

### Problem 6:

A solid iron anchor with a mass of 300 kg is thrown overboard and sinks to the bottom of the sea. Find the buoyant force exerted on the anchor when submerged. The density of seawater is \(1025 kg/m^3\).

### Solution 6:

First, find the volume of the anchor by using the mass and density of iron (\(Density_{iron} = 7874 kg/m^3\)):

\[ Volume = \frac{Mass}{Density_{iron}} \]

\[ Volume = \frac{300 kg}{7874 kg/m^3} \]

\[ Volume \approx 0.0381 m^3 \]

The buoyant force is calculated using Archimedes’ principle:

\[ F_b = \rho_{seawater} g V \]

\[ F_b = 1025 kg/m^3 \times 9.8 m/s^2 \times 0.0381 m^3 \]

\[ F_b \approx 384.073 N \]

The buoyant force acting on the submerged anchor is approximately 384.073 N.

### Problem 7:

A rectangular block of wood with a density of \(600 kg/m^3\) and a volume of 0.002 m^3 is placed in water. How much of the block will be above the water’s surface?

### Solution 7:

First, we calculate the mass of the block:

\[ Mass = Density \times Volume \]

\[ Mass = 600 kg/m^3 \times 0.002 m^3 \]

\[ Mass = 1.2 kg \]

The weight of the block is its mass times the acceleration due to gravity:

\[ Weight = Mass \times g \]

\[ Weight = 1.2 kg \times 9.8 m/s^2 \]

\[ Weight \approx 11.76 N \]

Archimedes’ principle tells us that the buoyant force equals the weight of the displaced water. Since the block is floating, the buoyant force also equals the weight of the block:

\[ F_b = Weight \]

\[ \rho_{water} \times g \times V_{submerged} = Weight \]

\[ 1000 kg/m^3 \times 9.8 m/s^2 \times V_{submerged} = 11.76 N \]

\[ V_{submerged} \approx 0.0012 m^3 \]

The proportion of the block that is submerged can then be found by:

\[ \frac{V_{submerged}}{V_{total}} = \frac{0.0012 m^3}{0.002 m^3} = 0.6 \]

This means that 60% of the block is submerged, and 40% of the block is above the water’s surface.

### Problem 8:

A spherical balloon of radius 10 cm is submerged in glycerin. If the density of glycerin is \(1260 kg/m^3\) and the density of the balloon material (including the air inside it) is \(0.2 kg/m^3\), what is the buoyant force acting on the balloon?

### Solution 8:

First, find the volume of the balloon:

\[ Volume = \frac{4}{3} \pi r^3 \]

\[ Volume = \frac{4}{3} \pi (0.1 m)^3 \]

\[ Volume \approx 0.00419 m^3 \]

Now, calculate the buoyant force using Archimedes’ principle:

\[ F_b = \rho_{glycerin} \cdot g \cdot V \]

\[ F_b = 1260 kg/m^3 \cdot 9.8 m/s^2 \cdot 0.00419 m^3 \]

\[ F_b \approx 52.0 N \]

Therefore, the buoyant force acting on the balloon is approximately 52.0 N.

### Problem 9:

A metal sphere has a mass of 2 kg and is attached to a string and submerged in water. If the sphere displaces 250 cm^3 of water, determine the tension in the string.

### Solution 9:

Calculate the buoyant force acting on the sphere:

\[ F_b = \rho_{water} \cdot g \cdot V \]

\[ F_b = 1 g/cm^3 \times 9.8 m/s^2 \times 250 cm^3 \]

\[ F_b = 2.45 N \]

Find the weight of the sphere:

\[ Weight = Mass \times g \]

\[ Weight = 2 kg \times 9.8 m/s^2 \]

\[ Weight = 19.6 N \]

The tension in the string will be the weight of the sphere minus the buoyant force:

\[ Tension = Weight – F_b \]

\[ Tension = 19.6 N – 2.45 N \]

\[ Tension \approx 17.15 N \]

The tension in the string is approximately 17.15 N.

### Problem 10:

A rectangular prism has a volume of 0.03 m³ and floats in water with one of its 6 m² surfaces parallel to the water surface. If the density of the prism is 800 kg/m³, how deep does it sink into the water?

### Solution 10:

First, calculate the mass of the prism:

\[ Mass = Density \times Volume \]

\[ Mass = 800 kg/m^3 \times 0.03 m^3 \]

\[ Mass = 24 kg \]

Calculate the weight of the prism:

\[ Weight = Mass \times g \]

\[ Weight = 24 kg \times 9.8 m/s^2 \]

\[ Weight = 235.2 N \]

As the prism floats in equilibrium, the buoyant force must be equal to its weight. Thus we can find the volume of water displaced (which will have the same density as freshwater, \(1000 kg/m^3\)):

\[ F_b = \rho_{water} \cdot g \cdot V_{displaced} \]

\[ 235.2 N = 1000 kg/m^3 \times 9.8 m/s^2 \times V_{displaced} \]

\[ V_{displaced} \approx 0.024 m^3 \]

Let’s assume the surface area, A, is one of the sides of the prism:

\[ Volume_{displaced}= A \cdot Depth \]

\[ 0.024 m^3 = 6 m^2 \times Depth \]

\[ Depth \approx 0.004 m \]

The prism sinks 0.004 m or 4 mm into the water.

### Problem 11:

A plastic cube with a side length of 5 cm is floating in oil. If the density of the oil is \(920 kg/m^3\) and the cube floats with one of its faces parallel to the oil surface, how much of the cube’s height is submerged when the density of the plastic is \(850 kg/m^3\)?

### Solution 11:

First, calculate the volume of the cube:

\[ Volume_{cube} = a^3 \]

\[ Volume_{cube} = (0.05 m)^3 \]

\[ Volume_{cube} = 0.000125 m^3 \]

Calculate the mass of the cube:

\[ Mass_{cube} = Density_{plastic} \times Volume_{cube} \]

\[ Mass_{cube} = 850 kg/m^3 \times 0.000125 m^3 \]

\[ Mass_{cube} = 0.10625 kg \]

The weight of the cube is:

\[ Weight_{cube} = Mass_{cube} \times g \]

\[ Weight_{cube} = 0.10625 kg \times 9.8 m/s^2 \]

\[ Weight_{cube} \approx 1.04125 N \]

When the cube is floating, the buoyant force is equal to the weight of the cube:

\[ F_b = Weight_{cube} \]

\[ \rho_{oil} \times g \times V_{submerged} = Weight_{cube} \]

\[ 920 kg/m^3 \times 9.8 m/s^2 \times V_{submerged} = 1.04125 N \]

\[ V_{submerged} = \frac{1.04125 N}{(920 kg/m^3 \times 9.8 m/s^2)} \]

\[ V_{submerged} \approx 0.0001161 m^3 \]

Since the cube is floating with its face parallel to the surface, the submerged volume would be the area of the face times the submerged height:

\[ V_{submerged} = 0.0001161 m^3 = (0.05 m)^2 \times h_{submerged} \]

\[ h_{submerged} = \frac{0.0001161 m^3}{(0.05 m)^2} \]

\[ h_{submerged} = 0.04644 m \]

\[ h_{submerged} = 4.644 cm \]

About 4.644 cm of the cube’s height is submerged in oil.

### Problem 12:

What is the apparent weight of a 50 kg block of wood with a density of \(700 kg/m^3\) when submerged in freshwater with a density of \(1000 kg/m^3\)?

### Solution 12:

Calculate the volume of the wood block:

\[ Volume_{wood} = \frac{Mass_{wood}}{Density_{wood}} \]

\[ Volume_{wood} = \frac{50 kg}{700 kg/m^3} \]

\[ Volume_{wood} \approx 0.07143 m^3 \]

Calculate the buoyant force:

\[ F_b = \rho_{water} \cdot g \cdot V_{wood} \]

\[ F_b = 1000 kg/m^3 \times 9.8 m/s^2 \times 0.07143 m^3 \]

\[ F_b \approx 700 N \]

The weight of the wood block is:

\[ Weight_{wood} = Mass_{wood} \times g \]

\[ Weight_{wood} = 50 kg \times 9.8 m/s^2 \]

\[ Weight_{wood} \approx 490 N \]

The apparent weight is the actual weight minus the buoyant force:

\[ Apparent weight = Weight_{wood} – F_b \]

\[ Apparent weight = 490 N – 700 N \]

\[ Apparent weight = -210 N \]

Since the buoyant force exceeds the actual weight, the block is pushed upwards, and the apparent weight would be negative, implying that the block will float.

### Problem 13:

A cube of side 0.5 m is made from a material with a density of \(1200 kg/m^3\) and is fully immersed in a fluid with a density of \(800 kg/m^