Problem 45 - College Physics

Question

An object has a weight of \(8.0 \mathrm{~N}\) in air. However, it apparently weighs only \(4.0 \mathrm{~N}\) when it is completely submerged in water. What is the density of the object?

Step-by-Step Solution

Answer

The density of the object is approximately 2000 kg/m³.

Step 1: Understanding the Given

We know the weight of the object in air is given as 8.0 N, and its apparent weight in water is 4.0 N. The difference in weight is due to the buoyant force exerted by water.

Step 2: Calculating Buoyant Force

The buoyant force can be calculated by taking the difference between the weight of the object in air and its apparent weight in water. This can be expressed as: \[ F_b = W_{air} - W_{water} = 8.0 \text{ N} - 4.0 \text{ N} = 4.0 \text{ N} \]

Step 3: Applying Archimedes' Principle

According to Archimedes' principle, the buoyant force is equal to the weight of the displaced fluid. The volume of the displaced water equals the volume of the object, and the weight of displaced water is the buoyant force, which is 4.0 N.

Step 4: Finding the Volume of Water Displaced

The weight of the displaced water can be expressed as \( \rho_{water} \cdot V \cdot g \), where \( \rho_{water} \) is the density of water (approximately 1000 kg/m³), \( V \) is the volume displaced, and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²). Solving for \( V \) from the buoyant force gives: \[ \rho_{water} \cdot V \cdot g = 4.0 \text{ N} \] \[ 1000 \cdot V \cdot 9.8 = 4.0 \text{ N} \] \[ V = \frac{4.0}{1000 \times 9.8} \] \[ V = 4.08 \times 10^{-4} \text{ m}^3 \]

Step 5: Calculating Density of the Object

The density of the object \( \rho_{object} \) can be found using the formula \( \rho = \frac{m}{V} \), where \( m \) is the mass of the object. The mass can be found from its weight in air \( m \cdot g = 8.0 \text{ N} \), so \( m = \frac{8.0}{9.8} \approx 0.816 \text{ kg} \). Substituting into the density equation gives: \[ \rho_{object} = \frac{0.816}{4.08 \times 10^{-4}} \approx 2000 \text{ kg/m}^3 \]

Key Concepts

Buoyant Force

Buoyant force is a fascinating concept in physics, particularly in fluid mechanics. It is the upward force exerted by a fluid on an object submerged in it.
What makes it special is that it acts opposite to gravity. This opposing force is why an object seems lighter when underwater.
In the original exercise, we calculated the buoyant force by finding the difference between the object's weight in air and its apparent weight in water. Mathematically, it's expressed as:

• Original weight in air: 8.0 N
• Weight in water: 4.0 N
• Buoyant force: 8.0 N - 4.0 N = 4.0 N

The buoyant force here is 4.0 N, which means the object experiences a 4.0 N upward push while submerged.
This is aligned with Archimedes' principle, which states that this force is equal to the weight of the fluid displaced by the object.

Density Calculation

Density is a measure of how much mass is contained in a given volume. It's an important property in fluid mechanics to understand how objects interact within a fluid.
The formula for density \( \rho \) is given by:\[ \rho = \frac{m}{V} \]where \( m \) is the mass and \( V \) is the volume. The problem asks for the density of the object.
In step 5 of the original solution, we first found the object's mass from its weight in air:

• Weight in air: 8.0 N, which corresponds to a mass \( m = \frac{8.0}{9.8} \approx 0.816 \) kg

Next, using the volume calculated from the displaced water, the density was computed as:\[ \rho_{object} = \frac{0.816}{4.08 \times 10^{-4}} \approx 2000 \text{ kg/m}^3 \]
This result indicates that the object is denser than water, which is typically about 1000 kg/m³, explaining why it does not float.

Volume Displacement

Volume displacement is key to understanding how buoyancy operates. When an object is submerged in a fluid, it displaces a volume of that fluid equal to its own volume.
This concept is crucial in the calculation of buoyant force, as the weight of the displaced fluid equates to this force.
During the step-by-step solution, we used Archimedes' principle to equate the buoyant force to the weight of the displaced water. Using the given buoyant force of 4.0 N, we derived the volume of displaced water using:\[ \rho_{water} \cdot V \cdot g = 4.0 \text{ N} \]where \( \rho_{water} \approx 1000 \text{ kg/m}^3 \) and \( g \approx 9.8 \text{ m/s}^2 \).By rearranging, we obtained:\[ V = \frac{4.0}{1000 \times 9.8} = 4.08 \times 10^{-4} \text{ m}^3 \]
This calculated volume shows how much water was displaced by our object, reinforcing the basis of Archimedes' principle.

Fluid Mechanics

Fluid mechanics studies fluids (liquids and gases) and the forces acting upon them.
It is a vital field of physics, providing insights into phenomena such as buoyancy, forces, energy transfer, and much more.
One fundamental principle within fluid mechanics is Archimedes' principle. It connects the forces experienced by submerged objects to the properties of the fluid itself. In the exercise, the object in water displayed a reduced apparent weight due to the buoyant force provided by the water.
Fluid mechanics also helps in understanding the behavior of different substances, such as the way density influences buoyancy. In our example:

• The object has a density greater than water, meaning it sinks.
• If it had a lower density, it would float, showcasing how fluid mechanics principles dictate its position within the fluid.

These concepts showcase the significance of fluid mechanics in explaining real-world physical interactions involving liquids and gases.