Deducing the Relative Density of an Object via Buoyancy

Understanding the buoyant forces and how they relate to an object's relative density can be a fundamental and practical application of basic physics principles. In this article, we will explore how to calculate the relative density of an object given its weights in air, water, and another liquid.

Introduction to Buoyancy and Relative Density

When an object is submerged in a fluid (such as water or another liquid), it experiences an upward force called the buoyant force. This force is equal to the weight of the fluid displaced by the object. Relative density, also known as specific gravity, is the ratio of the density of the object to the density of the reference fluid (typically water).

Problem Setup

Consider an object weighing 14 N in air, 10 N in water, and 6 N in a liquid of unknown density. Our goal is to find the relative density of the object.

Step-by-Step Solution

1. Buoyant Force in Water:

The buoyant force in water can be calculated as the difference between the weight of the object in air and its weight in water.

Weight in air (W_{air} 14 , text{N})

Weight in water (W_{water} 10 , text{N})

Buoyant force in water ( W_{air} - W_{water} 14 , text{N} - 10 , text{N} 4 , text{N})

2. Volume of the Object:

The volume of the object can be found using the buoyant force and the properties of water. The buoyant force equals the weight of the water displaced, given by:

$$text{Buoyant force} text{density of water} times text{volume} times g$$

Assuming the density of water is (1000 , text{kg/m}^3) and (g approx 9.81 , text{m/s}^2):

$$4 , text{N} 1000 , text{kg/m}^3 times text{volume} times 9.81 , text{m/s}^2$$

Solving for the volume:

$$text{volume} frac{4 , text{N}}{1000 , text{kg/m}^3 times 9.81 , text{m/s}^2} approx 0.000408 , text{m}^3$$

3. Mass of the Object:

The mass of the object can be found from its weight in air:

$$text{mass} frac{W_{air}}{g} frac{14 , text{N}}{9.81 , text{m/s}^2} approx 1.43 , text{kg}$$

4. Density of the Object:

Now, we can find the density of the object using its mass and volume:

$$text{density of object} frac{text{mass}}{text{volume}} frac{1.43 , text{kg}}{0.000408 , text{m}^3} approx 3500 , text{kg/m}^3$$

5. Relative Density:

Finally, the relative density (specific gravity) of the object is the ratio of the density of the object to the density of water:

$$text{Relative density} frac{text{density of object}}{text{density of water}} frac{3500 , text{kg/m}^3}{1000 , text{kg/m}^3} 3.5$$

Conclusion

Thus, the relative density of the object is approximately 3.5. This method can be applied to find the relative density of any object given its weights in air and water or any other liquid. Understanding these principles can be useful in various scientific and industrial applications.

Re-explanation Simplified

Let the volume of the object A be (v) units. A weighs 14 N in air and 10 N in water. A displaces (v) units of water. Since the relative density of water is 1, we can say the weight displaced is 4 N (weight equivalent to (v) units of water). The relative density of the object is the object's weight divided by the weight of the displaced water: (14/4 3.5).

The same object, when immersed in a liquid, weighs 6 N. This means the volume of the liquid displaced by the object is also (v) units. The liquid has an upward thrust on the object of 14 - 6 8 N. As 4 units of the liquid weigh 8 N compared to the same volume of water that weighs 4 N, the relative density of the liquid to water is (8N/4N 2).