Problem 2 - Physics

Question

One of the concrete pillars that support a house is \(2.2 \mathrm{~m}\) tall and has a radius of \(0.50 \mathrm{~m}\). The density of concrete is about \(2.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Find the weight of this pillar in pounds \((1 \mathrm{~N}=0.2248 \mathrm{lb})\).

Step-by-Step Solution

Answer

The weight of the pillar is approximately 8368.06 pounds.

Step 1: Calculate the Volume of the Pillar

The pillar is in the shape of a cylinder. To find the volume of a cylinder, use the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Here, \( r = 0.50 \, \text{m} \) and \( h = 2.2 \, \text{m} \). Substitute these values into the formula: \( V = \pi \times (0.50)^2 \times 2.2 \). Calculate \( V \):\[ V = 3.14 \times 0.25 \times 2.2 = 1.727 \, \text{m}^3 \]

Step 2: Calculate the Mass of the Pillar

Use the formula for mass based on density: \( m = \rho \times V \), where \( \rho \) is the density and \( V \) is the volume. The density \( \rho \) is given as \( 2.2 \times 10^3 \, \text{kg/m}^3 \). Substitute the volume from Step 1:\[ m = 2.2 \times 10^3 \times 1.727 = 3799.4 \, \text{kg} \]

Step 3: Calculate the Weight of the Pillar in Newtons

Weight is the force due to gravity and can be calculated using \( W = m \times g \), where \( g = 9.81 \, \text{m/s}^2 \). Use the mass from Step 2:\[ W = 3799.4 \times 9.81 = 37243.014 \, \text{N} \]

Step 4: Convert the Weight from Newtons to Pounds

Given that \( 1 \text{ Newton} = 0.2248 \text{ pounds} \), convert the weight from newtons to pounds:\[ W = 37243.014 \times 0.2248 = 8368.06 \, \text{lb} \]

Key Concepts

Cylinder Volume

To find the volume of a cylinder, which is the shape of the concrete pillar, you need to apply the mathematical formula for the volume of a cylinder: \( V = \pi r^2 h \). Here, \( V \) represents the volume, \( r \) stands for the radius, and \( h \) is the height of the cylinder. The inputs here are the radius \( r = 0.50 \, \text{m} \) and the height \( h = 2.2 \, \text{m} \). This particular formula is derived from the fact that the area of the circular base is \( \pi r^2 \) and needs to be multiplied by the height to account for the whole cylinder.

• Substitute the given values into the formula: \( V = \pi \times (0.50)^2 \times 2.2 \).
• After simplifying, you find that \( V = 3.14 \times 0.25 \times 2.2 = 1.727 \, \text{m}^3 \).

This step of finding the volume is crucial because it forms the basis for determining the pillar's mass, using density, in the next step.

Density Calculation

Density is a key physical property that connects the mass of an object with its volume. The formula for density \( \rho \) is \( \rho = \frac{m}{V} \), which states that density is the mass \( m \) divided by the volume \( V \). In this problem, we work in the reverse direction to find mass from given density and calculated volume: \( m = \rho \times V \).

• The known density of concrete is \( 2.2 \times 10^3 \, \text{kg/m}^3 \).
• You previously calculated the volume as \( 1.727 \, \text{m}^3 \).
• Thus, the mass is calculated by multiplying density by volume: \[ m = 2.2 \times 10^3 \times 1.727 = 3799.4 \, \text{kg} \].

This step showcases the importance of understanding material properties and how they influence overall weight calculations of objects in physics.

Weight Conversion

Weight is the force due to gravity acting on an object's mass. Using the earth's gravitational acceleration \( g \approx 9.81 \, \text{m/s}^2 \), the weight \( W \) is given by \( W = m \times g \).

• Given the mass from the previous steps, \( m = 3799.4 \, \text{kg} \).
• Calculate the weight in newtons: \[ W = 3799.4 \times 9.81 = 37243.014 \, \text{N} \].

Since the problem requires the answer in pounds, we use conversion: \( 1 \, \text{N} = 0.2248 \, \text{lb} \).

• Convert the weight: \[ W = 37243.014 \times 0.2248 = 8368.06 \, \text{lb} \].

Weight conversion is a powerful tool in physics that allows us to express computations in different units of measurement to fit practical needs or regional standards.

Physics Problem Solving

In physics, problem-solving requires understanding how to apply mathematical formulas to real-world situations. Simplifying complex issues into step-by-step solutions is essential.

• Start by understanding the problem and identifying known variables, such as the dimensions of the cylinder and the density of the material.
• Use basic principles, like calculating volume and applying density formulas, to find intermediate values like mass.
• Translate these values into what is required, such as converting mass into weight using gravitational acceleration.
• Finally, convert units if necessary to satisfy the problem's output requirements, such as converting newtons to pounds.

Being able to break down complex situations into simple steps not only aids in solving the problems at hand but also helps in developing a deeper understanding of physical concepts and their applications.