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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 water bed weighs 8758.48 N, which exceeds the floor's limit.

Step by step solution

01

Understand the problem

We need to calculate the weight of the water in the bed and determine if it exceeds the tolerated weight of the floor. We'll use the dimensions of the bed to find the volume of water and then convert this volume to weight.
02

Calculate the volume of the water bed

The volume of a rectangular prism (like the water bed) is given by the formula \( V = l \times w \times h \), where \( l \), \( w \), and \( h \) are the length, width, and height, respectively. Plugging in the dimensions of the bed, we get:\[ V = 1.83 \text{ m} \times 2.13 \text{ m} \times 0.229 \text{ m} = 0.8929177 \text{ m}^3 \].
03

Convert volume to weight

The density of water is \( 1000 \text{ kg/m}^3 \). To find the weight, we use the formula \( ext{Weight} = ext{Volume} \times ext{Density} \times g \), where \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity). Plugging in the values, we get:\[ ext{Weight} = 0.8929177 \text{ m}^3 \times 1000 \text{ kg/m}^3 \times 9.81 \text{ m/s}^2 = 8758.48 \text{ N} \].
04

Determine if the floor can support the weight

The floor can tolerate an additional weight of no more than \( 6660 \text{ N} \). The weight of the water is \( 8758.48 \text{ N} \). Since \( 8758.48 \text{ N} \) is greater than \( 6660 \text{ N} \), the floor cannot support the additional weight of the water bed.

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Key Concepts

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

Density of Water
The density of water is a crucial concept in fluid mechanics. It measures how much mass exists within a specific volume of water. Water's density is typically 1000 kg/m鲁 at standard temperature and pressure.
Understanding this property helps us calculate the weight of water in various physical scenarios.
This constant density value means that for every cubic meter of water, there is 1000 kilograms of mass.
  • This measurement is often used in calculations involving fluids, like determining the weight from volume.
  • Density helps predict whether objects will float or sink in water, depending on their own densities.
Volume Calculation
Volume calculation is essential to determine the capacity of three-dimensional shapes. For a rectangular prism, like a water bed, finding volume is simple with the formula:
\[V = l \times w \times h\]
Where \( l \), \( w \), and \( h \) are the length, width, and height of the prism.
  • This formula helps us understand how much space is inside a shape.
  • It's applicable to various scenarios, from calculating shipping costs to filling containers.
In the problem example, the dimensions provided are 1.83 m, 2.13 m, and 0.229 m. Multiplying these gives the volume of the water bed:
0.8929177 m鲁.
This number indicates the amount of space the water takes up inside the bed.
Weight Calculation
Converting the volume of water into weight involves a straightforward formula that combines volume, density, and gravitational acceleration.
To calculate the weight:\[\text{Weight} = \text{Volume} \times \text{Density} \times g\]
Where \( g \) represents the acceleration due to gravity, approximately 9.81 m/s虏. Using the density of water (1000 kg/m鲁) and the volume of the bed, we calculate as:
0.8929177 m鲁 \( \times \) 1000 kg/m鲁 \( \times \) 9.81 m/s虏 = 8758.48 N.
  • This value shows the force exerted by the water due to gravity.
  • Understanding weight calculation is vital in assessing structural support and potential overloading.
Rectangular Prism
The rectangular prism is a fundamental shape in geometry and is prominent in everyday objects. It consists of six rectangular faces, with each pair of opposite faces being equal.
This makes it unique and straightforward to work with mathematically.
  • Real-world examples include boxes, rooms, and water tanks.
  • Knowing its volume calculation enables estimating capacity and material usage.
The dimensions of a rectangular prism allow us to calculate its volume using length, width, and height.
In practical applications, knowing the dimensions helps us solve problems related to space and capacity, like whether a water bed would fit in a bedroom or if the floor can support its weight.

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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?

The density of ice is \(917 \mathrm{kg} / \mathrm{m}^{3}\), and the density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3} .\) A swimming polar bear climbs onto a piece of floating ice that has a volume of \(5.2 \mathrm{m}^{3} .\) What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?

A pipe is horizontal and carries oil that has a viscosity of \(0.14\) \(\mathrm{Pa} \cdot \mathrm{s}\) The volume flow rate of the oil is \(5.3 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s} .\) The length of the pipe is \(37\) \(\mathrm{m},\) and its radius is \(0.60 \mathrm{cm} .\) At the output end of the pipe the pressure is atmospheric pressure. What is the absolute pressure at the input end?

If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure, while the external pressure increases due to the increased water depth. At sufficient depths, the difference between the external and internal pressures can rupture an eardrum. Eardrums can rupture when the pressure difference is as little as \(35\) \(\mathrm{kPa}\). What is the depth at which this pressure difference could occur? The density of seawater is \(1025\) \(\mathrm{kg} /\mathrm{m}^{3}\).

A hollow cubical box is \(0.30\) \(\mathrm{m}\) on an edge. This box is floating in a lake with one-third of its height beneath the surface. The walls of the box have a negligible thickness. Water from a hose is poured into the open top of the box. What is the depth of the water in the box just at the instant that water from the lake begins to pour into the box from the lake?
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