Problem 33 - Fundamentals of Physics

Question

An iron anchor of density \(7870 \mathrm{~kg} / \mathrm{m}^{3}\) appears \(200 \mathrm{~N}\) lighter in water than in air. (a) What is the volume of the anchor? (b) How much does it weigh in air?

Step-by-Step Solution

Answer

The anchor's volume is \(0.0204 \, \mathrm{m^3}\), and it weighs approximately \(1580 \, \mathrm{N}\) in air.

Step 1: Understand the Problem

The anchor weighs 200 N less in water due to buoyant force. We need to find the anchor's volume and its weight in air, knowing its density is \(7870\, \mathrm{kg/m^3}\).

Step 2: Apply the Buoyant Force Formula

The buoyant force is equal to the weight loss in water. This can be given as \(F_b = \rho_w \cdot V \cdot g\), where \(\rho_w = 1000 \, \mathrm{kg/m^3}\) is the density of water, \(V\) is the volume of the anchor, and \(g = 9.81 \, \mathrm{m/s^2}\) is the acceleration due to gravity. Here, \(F_b = 200 \, \mathrm{N}\).

Step 3: Solve for Volume

Using the equation from Step 2: \[F_b = \rho_w \cdot V \cdot g \]Solve for \(V\):\[V = \frac{F_b}{\rho_w \cdot g} = \frac{200 \, \mathrm{N}}{1000 \, \mathrm{kg/m^3} \cdot 9.81 \, \mathrm{m/s^2}} = 0.0204 \, \mathrm{m^3}\]

Step 4: Calculate Weight in Air

The weight of the anchor in air can be found using the density and volume from earlier steps. The formula is: \[W_{air} = \rho \cdot V \cdot g\]where \(\rho = 7870 \, \mathrm{kg/m^3}\).Substituting:\[W_{air} = 7870 \, \mathrm{kg/m^3} \cdot 0.0204 \, \mathrm{m^3} \cdot 9.81 \, \mathrm{m/s^2} \approx 1580 \, \mathrm{N}\]

Step 5: Verify and Conclude the Answers

We found the volume to be \(0.0204 \, \mathrm{m^3}\) and the weight in air to be approximately \(1580 \, \mathrm{N}\). These answers are consistent with the given conditions about buoyancy and the anchor's density.

Key Concepts

Density

Density is an important concept that describes how much mass is contained in a unit volume of a substance. If you imagine how packed a material is in a specific space, you've got the right idea about density. It's calculated with the formula \[ \rho = \frac{m}{V} \] where:

• \( \rho \) is the density,
• \( m \) is the mass, and
• \( V \) is the volume.

In the case of the anchor, its density is given as \( 7870 \, \mathrm{kg/m^3} \). This high density indicates that iron is a dense material, meaning it has a lot of mass in a small volume.
This density is crucial to solve the problem since it helps calculate the anchor's volume and eventually its weight when in air.

Volume Calculation

Volume calculation is necessary to figure out the size of the space that an object occupies. With the anchor problem, volume tells us how much space the anchor takes up under water. By using buoyant force information, we determine the anchor's volume.
The buoyant force in water, a central part to solving this problem, equals the "apparent weight" loss of the anchor. The formula we use here is: \[ F_b = \rho_w \cdot V \cdot g \] where:

• \(F_b = 200 \, \mathrm{N}\) is the buoyant force,
• \( \rho_w = 1000 \, \mathrm{kg/m^3} \) is the density of water,
• \( g = 9.81 \, \mathrm{m/s^2} \) is gravitational acceleration, and
• \( V \) is the volume of the anchor.

Rearranging the formula, we find \( V = 0.0204 \, \mathrm{m^3} \). This calculation shows that the anchor displaces a specific volume of water, which in turn is responsible for the buoyant force making the anchor appear lighter in water.

Weight Measurement

Weight is the force exerted by the gravitational pull on an object's mass. In the anchor exercise, understanding weight both in air and water is crucial.
To find the anchor's weight in air, use the formula: \[ W_{air} = \rho \cdot V \cdot g \] where:

• \( \rho = 7870 \, \mathrm{kg/m^3} \) is the density of iron,
• \( V = 0.0204 \, \mathrm{m^3} \) is the volume, and
• \( g = 9.81 \, \mathrm{m/s^2} \) is gravity.

Using these values, the weight in air is calculated as approximately \( 1580 \, \mathrm{N} \). This weight is greater than in water due to the absence of buoyant force. By knowing this weight, students can see how much more a dense object like iron weighs out of water compared to submersed in it, illustrating the effectiveness of buoyant force.