Problem 15 - Physics

Question

A suitcase (mass \(m=16 \mathrm{~kg}\) ) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures \(0.50 \mathrm{~m}\) by \(0.15 \mathrm{~m} .\) The elevator is moving upward, the magnitude of its acceleration being \(1.5 \mathrm{~m} / \mathrm{s}^{2}\). What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Step-by-Step Solution

Answer

The pressure applied to the floor beneath the suitcase is 2413.3 N/m² in excess of atmospheric pressure.

Step 1: Understand the Concept of Pressure

Pressure is defined as the force applied per unit area. In this problem, we are interested in the pressure exerted by the suitcase on the floor, specifically in excess of atmospheric pressure.

Step 2: Calculate the Force Exerted by the Suitcase

The force exerted by the suitcase is due to its weight and the additional force from the elevator's acceleration. Calculate this using the formula: \( F = m(g + a) \), where \( g = 9.81 \mathrm{~m/s^2} \) is the acceleration due to gravity, and \( a = 1.5 \mathrm{~m/s^2} \) is the elevator's acceleration.Substituting the values, we find:\[ F = 16 \times (9.81 + 1.5) = 181.0 \mathrm{~N} \]

Step 3: Calculate the Area of Contact

The area of the suitcase in contact with the floor is a rectangle. Calculate the area with the formula: \( A = ext{length} \times ext{width} \).Substituting the given dimensions:\[ A = 0.50 \times 0.15 = 0.075 \mathrm{~m^2} \]

Step 4: Calculate the Pressure Exerted by the Suitcase

Pressure is calculated using the formula: \( P = \frac{F}{A} \).Substituting the calculated force and area:\[ P = \frac{181.0}{0.075} = 2413.3 \mathrm{~N/m^2} \]

Step 5: Subtract Atmospheric Pressure to Find the Pressure in Excess

Standard atmospheric pressure is approximately \(101325 \mathrm{~N/m^2}\), but since we want the pressure in excess, we don't need to consider atmospheric pressure unless specified. Here, the problem asks for the internal pressure due to the suitcase contact which is already separate from atmospheric pressure effects. Thus, no subtraction is needed.

Key Concepts

Force Calculation

In physics, calculating force involves understanding the relationship between mass, acceleration, and gravity.

• The formula to find force is: \( F = m(g + a) \)
• Where \( F \) is the force in Newtons, \( m \) is the mass in kilograms, \( g \) is the gravitational acceleration (approximately \(9.81 \mathrm{~m/s^2}\), and \( a \) is any additional acceleration (e.g., from an elevator).

To illustrate, consider a suitcase with a mass of 16 kg placed on an elevator. The total acceleration is the sum of gravitational pull and the upward acceleration of the elevator, calculated as \( 9.81 \mathrm{~m/s^2} + 1.5 \mathrm{~m/s^2} \). The resulting force exerted by the suitcase is therefore \( 16 \times 11.31 = 181 \mathrm{~N}\). This calculation combines both the force of gravity acting on the suitcase and the force due to the lifting motion of the elevator. Breaking down each step helps in understanding how each factor contributes to the total force experienced.

Area of Contact

The area of contact underlines the importance of how weight is distributed over a surface. The pressure exerted depends on both the force applied and the area over which it is spread.

• The area of contact for a rectangular object is calculated using: \( A = \text{length} \times \text{width} \)

Given that the suitcase measures \(0.50 \mathrm{~m}\) by \(0.15 \mathrm{~m}\), its contact area is computed as \(0.50 \times 0.15 = 0.075 \mathrm{~m^2}\). Understanding this concept is useful, as a larger contact area results in lesser pressure exerted for the same force, akin to wearing snowshoes on ice. A solid grasp of this relation aids in addressing more complex physics problems involving pressure distribution across surfaces.

Elevator Acceleration

Elevator acceleration interestingly demonstrates how forces add up in a non-static system. When an elevator moves, an additional force due to acceleration is experienced.

• Acceleration is a vector quantity, having both magnitude and direction. It changes the net force acting on a body within the system.

In our scenario, the elevator accelerates upwards at \(1.5 \mathrm{~m/s^2}\). This alters the effective force acting on the suitcase, raising it from its static weight given by gravity alone, \(mg\), to \(m(g + a)\). By adding this acceleration to Earth's gravity, the force exerted by the suitcase grows, illustrating the practical implications such as increased pressure exerted during acceleration. While still subject to gravity, any additional upward pull requires calculus of the new force dynamics.

Pressure Exertion

Pressure relates directly to how force is applied over an area, making it a central topic in mechanics. It determines how objects interact with surfaces they engage with.

• Calculate pressure using \( P = \frac{F}{A} \), where \( P \) is pressure in pascals, \( F \) is force in newtons, and \( A \) is area in square meters.

To find the pressure exerted by a suitcase, which equals \( \frac{181}{0.075} = 2413.3 \mathrm{~N/m^2} \), helps understand how much force per unit area is applied to the elevator floor. In solving these problems, notice we focus on pressure excess of atmospheric, meaning we only consider the suitcase’s direct impact without subtracting standard atmospheric pressure, unless specified. This concept is vital in understanding how pressure changes with force distribution in different physical contexts.