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Chapter 11: Problem 89

A water bed for sale has dimensions of \(1.83 \mathrm{m} \times 2.13 \mathrm{m} \times 0.229 \mathrm{m}\) The floor of the bedroom will tolerate an additional weight of no more than \(6660 \mathrm{N}\). Find the weight of the water in the bed and determine whether the bed should be purchased.

Short Answer

Expert verified
The bed should not be purchased; its weight exceeds the floor's limit.

Step by step solution

01

Calculate the Volume of the Water Bed

To find the volume of the water bed, use the formula for the volume of a rectangular prism: Volume = length 脳 width 脳 height. Substitute the given dimensions into the formula, so the volume is \(1.83 \, \text{m} \times 2.13 \, \text{m} \times 0.229 \, \text{m}\).
02

Calculate the Volume

Perform the multiplication: \(1.83 \times 2.13 \times 0.229 = 0.8911367 \, \text{m}^3\). So, the volume of the water bed is approximately \(0.8911 \, \text{m}^3\).
03

Calculate the Weight of the Water

The weight of the water can be found by multiplying its volume by the density of water and gravity. Use the formula: Weight = volume \(\times\) density \(\times\) gravity. The density of water is \(1000 \, \text{kg/m}^3\), and gravity is \(9.81 \, \text{m/s}^2\). Thus, the weight of the water is \(0.8911 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2\).
04

Calculate the Weight

Perform the calculation: \(0.8911 \times 1000 \times 9.81 = 8743.791 \, \text{N}\). Therefore, the weight of the water is approximately \(8744 \, \text{N}\).
05

Compare the Weight to the Floor's Limit

The calculated weight of the water is \(8744 \, \text{N}\). Compare this to the floor's tolerance of \(6660 \, \text{N}\). Since \(8744 \, \text{N} > 6660 \, \text{N}\), the weight of the water exceeds the floor's weight tolerance.

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Key Concepts

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

Volume Calculation
Calculating the volume of an object is an important first step in many physics problems, especially those involving fluids. In this particular problem, the object in question is a water bed, which can be modeled as a three-dimensional rectangular prism. To calculate its volume, use the formula for the volume of a rectangular prism:
  • Formula: Volume = length 脳 width 脳 height
  • Example Calculation: Given dimensions are 1.83 m, 2.13 m, and 0.229 m.
  • Substituting these dimensions: Volume = 1.83 m 脳 2.13 m 脳 0.229 m = 0.8911 m鲁.
This calculated volume tells us how much space the water inside the bed occupies.
Density of Water
Density is a core concept in understanding how weight relates to volume. The density of a substance is defined as its mass per unit volume. When considering water, the density is a constant that is essential for calculations involving water weight.
For water, the density is typically 1000 kg/m鲁 under standard conditions, which is a useful number to commit to memory. This figure allows us to easily transition from thinking about the amount of space the water occupies (volume) to thinking about its "heaviness" or weight.
When combining volume and density, remember:
  • Densities help convert volume to mass.
  • Mass can then contribute to calculating weight, when considering gravity.
Weight and Gravity
Understanding the concept of weight in physics requires recognizing its dependence on gravity. Weight is effectively the force exerted by the mass of an object when placed in a gravitational field like Earth's.
To find the weight of the water in the waterbed, we utilize:
  • Formula: Weight = mass 脳 gravity
  • In our calculation: Weight = volume 脳 density 脳 gravity
  • Using known values: Weight = 0.8911 m鲁 脳 1000 kg/m鲁 脳 9.81 m/s虏
  • Result: 8744 N.
This value represents the force with which the waterbed would press down on its supporting surface.
Structural Load Capacity
A critical aspect of selecting household furnishings like a water bed is ensuring compatibility with the existing structural load capacity of a building. Every structure can bear only a certain amount of weight safely. For example, the bed's intended placement floor can tolerate an additional load up to 6660 N.
  • Comparison: The waterbed weighs 8744 N, surpassing the floor's weight tolerance of 6660 N.
  • Conclusion: If the weight of the load exceeds the capacity, as in this situation, it may lead to structural damage.
Evaluating these factors ensures that the installation of such items does not pose a risk to the integrity of building structures.

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Most popular questions from this chapter

A liquid is flowing through a horizontal pipe whose radius is \(0.0200 \mathrm{m}\). The pipe bends straight upward through a height of \(10.0 \mathrm{m}\) and joins another horizontal pipe whose radius is \(0.0400 \mathrm{m} .\) What volume flow rate will keep the pressures in the two horizontal pipes the same?

A lost shipping container is found resting on the ocean floor and completely submerged. The container is \(6.1 \mathrm{m}\) long, \(2.4 \mathrm{m}\) wide, and \(2.6 \mathrm{m}\) high. Salvage experts attach a spherical balloon to the top of the container and inflate it with air pumped down from the surface. When the balloon's radius is \(1.5 \mathrm{m},\) the shipping container just begins to rise toward the surface. What is the mass of the container? Ignore the mass of the balloon and the air within it. Do not neglect the buoyant force exerted on the shipping container by the water. The density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3}\)

The human lungs can function satisfactorily up to a limit where the pressure difference between the outside and inside of the lungs is one-twentieth of an atmosphere. If a diver uses a snorkel for breathing, how far below the water can she swim? Assume the diver is in salt water whose density is \(1025 \mathrm{kg} / \mathrm{m}^{3}.\)

A solid concrete block weighs 169 N and is resting on the ground. Its dimensions are \(0.400 \mathrm{m} \times 0.200 \mathrm{m} \times 0.100 \mathrm{m} .\) A number of identical blocks are stacked on top of this one. What is the smallest number of whole blocks (including the one on the ground) that can be stacked so that their weight creates a pressure of at least two atmospheres on the ground beneath the first block?

An airplane wing is designed so that the speed of the air across the top of the wing is \(251 \mathrm{m} / \mathrm{s}\) when the speed of the air below the wing is \(225 \mathrm{m} / \mathrm{s} .\) The density of the air is \(1.29 \mathrm{kg} / \mathrm{m}^{3} .\) What is the lifting force on a wing of area \(24.0 \mathrm{m}^{2} ?\)
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