Determining the Mass of a Submerged Cube in Water: A Practical Application of Archimedes' Principle

When a cubical block of wood floats in water, it does so in accordance with Archimedes' Principle. This principle states that the buoyant force on a submerged object is equal to the weight of the fluid displaced by the object. In this article, we will determine the mass of a cube of wood that floats one-fifth in water, using the concept of volume displacement and density.

Understanding the Problem

The problem describes a cubical block of wood with a cross-sectional area of 100 cm2 that floats one-fifth in water. We need to determine the mass of the block that is immersed in the water. Let's break down the problem step by step.

The Volume of the Block

We will assume the block is a perfect cube. The cross-sectional area of 100 square centimeters (cm2) means the side of the cube is 10 cm, as the area of a cube's face (100 cm2) can be expressed as the side length squared (10 cm × 10 cm).

The volume of the cube can therefore be calculated as:

V side3 10 cm × 10 cm × 10 cm 1000 cm3 0.001 m3

Volume of the Block Submerged in Water

According to the problem, one-fifth of the block is immersed in the water. Therefore, the volume of the block submerged in water can be calculated as:

v 1/5 V 1/5 × 0.001 m3 0.0002 m3

Calculating the Mass of the Block

The block floats by displacing an equal weight of water. The volume of water displaced is equal to the submerged volume of the block. The density of water is 1000 kg/m3. The mass of the water displaced can be calculated using the formula:

M dv

where:

? M is the mass of the water displaced ? d is the density of water (1000 kg/m3) ? v is the volume of the water displaced (0.0002 m3)

Substituting the values:

M 1000 kg/m3 × 0.0002 m3 0.2 kg

Thus, the mass of the block is 0.2 kg.

Further Explorations

The described scenario can be extended to explore the concept of density and buoyancy in different fluids or at different temperatures. For example, if the experiment were conducted in seawater, which has a density of 1025 kg/m3, the calculations would differ slightly:

1. The volume of the block remains 0.001 m3.

2. The volume of the water displaced is still 0.0002 m3.

3. The mass of the water displaced in seawater would be:

M 1025 kg/m3 × 0.0002 m3 0.205 kg

Hence, the mass of the block in seawater would be 0.205 kg.

This problem demonstrates the practical applications of Archimedes' Principle and how volume displacement can be used to determine the mass of an object. Understanding such principles is crucial in fields such as maritime engineering, buoyancy calculations, and fluid dynamics.