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Chapter 11: Problem 46

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 \(1025 \mathrm{kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The mass of the container is approximately 53582.5 kg.

Step by step solution

01

Calculate Volume of the Container

The volume of the shipping container can be found using the formula for the volume of a rectangular prism: \(V = \text{length} \times \text{width} \times \text{height}\). Substitute the given dimensions to get \(V = 6.1 \times 2.4 \times 2.6 = 38.064 \, \text{m}^3\).
02

Calculate Volume of the Balloon

The volume of a sphere is calculated using the formula \(V = \frac{4}{3} \pi r^3\). Substitute the radius of the balloon to find the volume: \(V = \frac{4}{3} \pi (1.5)^3 \approx 14.137 \, \text{m}^3\).
03

Calculate Total Buoyant Force

The buoyant force, which results from the displacement of water by both the balloon and the container, is given by \(F_{buoyant} = \rho_{water} \times g \times V_{total}\), where \(\rho_{water}\) is the density of seawater \(1025 \, \text{kg/m}^3\), \(g = 9.81 \, \text{m/s}^2\), and \(V_{total} = V_{container} + V_{balloon}\). Thus, \(V_{total} = 38.064 + 14.137 = 52.201 \, \text{m}^3\) and \(F_{buoyant} = 1025 \times 9.81 \times 52.201 \approx 525610.8 \, \text{N}\).
04

Equate Buoyant Force to Weight of the Container

When the container begins to rise, the buoyant force equals the gravitational force (weight) of the container: \(F_{buoyant} = m_{container} \times g\). Rearrange to solve for the mass of the container: \(m_{container} = \frac{F_{buoyant}}{g} = \frac{525610.8}{9.81} \approx 53582.5 \, \text{kg}\).

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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 objects is fundamental when dealing with problems involving buoyancy because the force of buoyancy depends on the volume of water displaced. For a shipping container shaped like a rectangular prism, the formula to calculate volume is straightforward: multiply the length, width, and height together.

In this particular case, the dimensions were given as 6.1 meters in length, 2.4 meters in width, and 2.6 meters in height. By substituting these values into the formula, we find the container's volume:
  • \( V = \text{Length} \times \text{Width} \times \text{Height} \)
  • \( V = 6.1 \times 2.4 \times 2.6 = 38.064 \, \text{m}^3 \)
Remember that each unit of volume, or cubic meter, will affect the amount of water displaced, thus impacting the buoyant force.
Buoyant Force
The buoyant force is the upward force exerted on objects submerged in a fluid, such as seawater. This force is a result of the pressure exerted by the fluid on the object. The concept of buoyancy is derived from Archimedes' principle, which states that the force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

To calculate the buoyant force, the equation is:
  • \( F_{\text{buoyant}} = \rho_{\text{water}} \times g \times V_{\text{total}} \)
where:
  • \( \rho_{\text{water}} \) is the density of the fluid (in this case, seawater)
  • \( g \) is the acceleration due to gravity
  • \( V_{\text{total}} \) is the total volume of water displaced, which includes both the container and the additional volume due to the balloon
In this exercise, we considered both the container's volume and the volume of an attached spherical balloon to find the total volume of displaced water.
Density of Seawater
Density is a measure of mass per unit volume and plays a crucial role in buoyancy calculations. Seawater is denser than fresh water, largely due to its salt content, which affects how objects float or sink in the ocean. In this problem, seawater density is given as 1025 kg/m鲁, which is typical for seawater.

Knowing the density is important because it enables us to compute the buoyant force exerted by the seawater on the submerged object.
  • The higher the density of the fluid, the greater the buoyant force.
  • This force determines whether the object will rise to the surface or remain submerged.
By multiplying the seawater density by the volume of displaced water and the acceleration due to gravity, we arrive at the buoyant force acting on the submerged container.

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Most popular questions from this chapter

A water line with an internal radius of \(6.5 \times 10^{-3} \mathrm{m}\) is connected to a shower head that has 12 holes. The speed of the water in the line is \(1.2 \mathrm{m} / \mathrm{s}\). (a) What is the volume flow rate in the line? (b) At what speed does the water leave one of the holes (effective hole radius \(=4.6 \times 10^{-4} \mathrm{m}\) ) in the head?

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is \(1.80 \times 10^{5}\) Pa. Assuming that the top and bottom surfaces of the cap each have an area of \(4.10 \times 10^{-4} \mathrm{m}^{2}\), obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.

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.

The Mariana trench is located in the floor of the Pacific Ocean at a depth of about \(11000 \mathrm{m}\) below the surface of the water. The density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3} .\) (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius \(=0.10 \mathrm{m}) ?\) (b) For comparison, determine the weight of a jetliner whose mass is \(1.2 \times 10^{5} \mathrm{kg}.\)

Water flows straight down from an open faucet. The crosssectional area of the faucet is \(1.8 \times 10^{-4} \mathrm{m}^{2},\) and the speed of the water is \(0.85 \mathrm{m} / \mathrm{s}\) as it leaves the faucet. Ignoring air resistance, find the crosssectional area of the water stream at a point \(0.10 \mathrm{m}\) below the faucet.
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