91Ó°ÊÓ

The Momentum Principle is a fundamental concept in dynamics that helps us understand how the motion of objects can be influenced by forces. It states that the change in momentum of an object is equal to the impulse applied to it. In mathematical terms, momentum ( p ) is defined as the product of mass ( m ) and velocity ( v ):

• Momentum: \[ p = m \cdot v \]
• Change in momentum: \[ \Delta p = F \cdot \Delta t \]

In the context of the exercise, we use the momentum principle to calculate the force ( F ) exerted by the airflow. By knowing the mass flow rate and the velocity of the air, we determine the momentum change, which helps us find the force acting on the door.
The force can be directly calculated with the formula:

• Force: \[ F = \dot{m} \times v \]

Where \( \dot{m} \) is the mass flow rate, and \( v \) is the velocity of the airflow.

Unit conversion is a crucial skill in physics and engineering, ensuring that all measurements are in compatible units for calculations. In this exercise, the velocity of the airflow is initially given in miles per hour (mi/hr). To use this value in our calculations, we need to convert it into feet per minute (ft/min), as the flow rate is given in cubic feet per minute.
Here is the conversion process:

• Convert miles to feet: \[ 1 \text{ mile} = 5280 \text{ feet} \]
• Convert hours to minutes: \[ 1 \text{ hour} = 60 \text{ minutes} \]

Using these conversion factors, velocity is calculated as:

• Velocity: \[ 130 \text{ mi/hr} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{60 \text{ min}} = 11440 \text{ ft/min} \]

The mass flow rate is an essential concept in fluid dynamics, quantifying the mass of fluid passing through a section per unit time. It is a vital step when calculating forces involving fluids. In this exercise, we calculate the mass flow rate using the given volumetric flow rate and the specific weight of air.
The formula for mass flow rate (\( \dot{m} \)) is straightforward:

• Mass flow rate: \[ \dot{m} = \text{Volumetric flow rate} \times \text{Specific weight} \]

Given:

• Flow rate: 410 \( \text{ft}^3/\text{min} \)
• Specific weight: 0.0753 \( \text{lb/ft}^3 \)

Hence,

• \[\dot{m} = 410 \text{ ft}^3/\text{min} \times 0.0753 \text{ lb/ft}^3 = 30.873 \text{ lb/min} \]

The specific weight of a fluid, such as air, is a significant parameter in many engineering calculations. It is defined as the weight per unit volume, representing how much a particular volume of the fluid weighs. For this exercise, the specific weight of air is given as 0.0753 lb/ft³. This value is crucial when computing the mass flow rate.
Specific weight ( γ ) can be calculated as follows:

• Specific weight formula: \[ \gamma = \frac{\text{Weight}}{\text{Volume}} \]

In fluid dynamics problems, knowing the specific weight helps in converting volumetric flow rates to mass flow rates since:

• Mass flow rate: \[ \dot{m} = \text{Flow rate} \times \gamma \]

This parameter allows for more accurate and meaningful calculations, particularly in determining the force exerted by moving air on surfaces like the garage door in the exercise scenario.