/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A suitcase (mass \(\mathrm{m}=16... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 15

A suitcase (mass \(\mathrm{m}=16 \mathrm{~kg}\) ) 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?

Short Answer

Expert verified
The excess pressure applied beneath the suitcase is 2410.67 Pa.

Step by step solution

01

Calculate the Force Exerted by the Suitcase

The suitcase experiences a normal force and gravity. The total force exerted by the suitcase is equal to the normal force required to support it plus the additional force due to the elevator's acceleration. The force due to gravity (weight) is calculated as \( F_g = m \cdot g \), where \( m = 16 \) kg and \( g = 9.8 \) m/s虏.Thus, \( F_g = 16 \times 9.8 = 156.8 \) N.The additional force due to acceleration is \( F_a = m \cdot a \), where \( a = 1.5 \) m/s虏.Thus, \( F_a = 16 \times 1.5 = 24 \) N.The total force \( F_t \) is the sum of these forces: \( F_t = F_g + F_a = 156.8 + 24 = 180.8 \) N.
02

Calculate the Area of Contact

The area \( A \) the suitcase has in contact with the floor is given by the product of its length and width. Given dimensions are \( 0.50 \) m by \( 0.15 \) m. Hence, the area is:\[ A = 0.50 \times 0.15 = 0.075 \text{ m}^2 \].
03

Calculate the Excess Pressure

Pressure \( P \) is determined by dividing the total force by the area of contact. Specifically, \( P = \frac{F_t}{A} \), where \( F_t = 180.8 \) N as previously calculated and \( A = 0.075 \) m虏.Thus:\[ P = \frac{180.8}{0.075} = 2410.67 \text{ Pa} \].
04

Interpret the Result

The calculated pressure represents the pressure in excess of atmospheric pressure applied by the suitcase to the floor. Atmospheric pressure is not factored into the result as we are considering the case of excess pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force of Gravity
When we talk about the force of gravity acting on an object, such as a suitcase, we are referring to the weight of the object. Weight is the force exerted by gravity on an object's mass.

To calculate this force, we use the formula: \( F_g = m \cdot g \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).
  • In the original exercise, the suitcase has a mass (\( m \)) of 16 kg.
  • Thus, the gravitational force is \( F_g = 16 \times 9.8 = 156.8 \text{ N} \).
This force acts downward, pulling the suitcase towards the center of the Earth. Understanding this concept is fundamental for calculating any additional forces and pressures acting on or exerted by the object when it's stationary or in motion.
Normal Force
The concept of the normal force is essential in physics, especially when dealing with objects on surfaces. This force is the support force exerted upon an object that is in contact with another stable surface. It is called "normal" because it acts perpendicular to the surface.

When the elevator is stationary or moving at a constant velocity, the normal force compensates for the force of gravity. However, in an accelerating elevator, the situation changes.
  • For an object at rest or moving at uniform velocity, normal force equals the gravitational force \( F_g \).
  • When the elevator accelerates upward, the additional force due to acceleration \( F_a = m \cdot a \) must be considered.
  • In our case, \( F_a = 16 \times 1.5 = 24 \text{ N} \).
So, the total force which normal force opposes upwards becomes: \( F_t = F_g + F_a = 156.8 + 24 = 180.8 \text{ N} \). The normal force here not only supports gravity but also the additional upward force.
Contact Area
The contact area is a fundamental concept when calculating pressure because pressure is defined as the force distributed over an area.

In our exercise, the suitcase is touching the elevator floor over a specific area which dictates how the exerted force is spread.
  • The suitcase's dimensions provide us with the contact area.
  • Here, the measurements are given as \( 0.50 \text{ m} \) by \( 0.15 \text{ m} \).
  • The contact area \( A = 0.50 \times 0.15 = 0.075 \text{ m}^2 \).
This small area means the total force will be concentrated over a relatively small surface, resulting in more significant pressure, which is critical when calculating how much force is applied at a specific spot.
Elevator Physics
Understanding elevator physics helps in recognizing how forces operate in practical scenarios. When an elevator accelerates, passengers and items in the elevator experience a change in normal force.

In physics terms, the elevator serves as a non-inertial frame of reference:
  • An upward acceleration increases the felt weight Force: \( F_t = F_g + F_a \).
  • This increase occurs because the elevator itself contributes to the upward force, adding to the existing gravitational force.
  • This scenario affects the pressure calculation as pressure \( P \) is the quotient of total force \( F_t \) and contact area \( A \): \( P = \frac{F_t}{A} = \frac{180.8}{0.075} = 2410.67 \text{ Pa} \).
In essence, moving elevators modify the forces acting on objects, which is crucial in contexts like elevator design or safety considerations.

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Most popular questions from this chapter

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of \(1.00 \mathrm{~m}\), one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance \(h\), as the drawing shows. It can then be squirted over the meat. (a) Is the absolute pressure in the bulb in the drawing greater than or less than atmospheric pressure? (b) In a second trial, the distance \(h\) is somewhat less than it is in the drawing. Is the absolute pressure in the bulb in the second trial greater or smaller than in the case shown in the drawing? Explain your answers. Using \(1.013 \times 10^{5} \mathrm{~Pa}\) for the atmospheric pressure and \(1200 \mathrm{~kg} / \mathrm{m}^{3}\) for the density of the sauce, find the absolute pressure in the bulb when the distance \(h\) is (a) \(0.15\) \(\mathrm{m}\) and (b) \(0.10 \mathrm{~m}\). Verify that your answers are consistent with your answers to the Concept Questions.

Interactive Solution \(\underline{11.73}\) at illustrates a model for solving this problem. A pressure difference of \(1.8 \times 10^{3} \mathrm{~Pa}\) is needed to drive water \(\left(\eta=1.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) through a pipe whose radius is \(5.1 \times 10^{-3} \mathrm{~m} .\) The volume flow rate of the water is \(2.8 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s} .\) What is the length of the pipe?

In a very large closed tank, the absolute pressure of the air above the water is \(6.01 \times 10^{5} \mathrm{~Pa}\). The water leaves the bottom of the tank through a nozzle that is directed straight upward. The opening of the nozzle is \(4.00 \mathrm{~m}\) below the surface of the water. (a) Find the speed at which the water leaves the nozzle. (b) Ignoring air resistance and viscous effects, determine the height to which the water rises.

A paperweight, when weighed in air, has a weight of \(\mathrm{W}=6.9 \mathrm{~N}\). When completely immersed in water, however, it has a weight of \(W_{\text {in water }}=4.3 \mathrm{~N}\). Find the volume of the paperweight.
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