/*! 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 82 Assume that crude oil from a sup... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Assume that crude oil from a supertanker has density 750 \(\mathrm{kg} / \mathrm{m}^{3} .\) The tanker runs aground on a sandbar. To refloat the tanker, its oil cargo is pumped out into steel barrels, each of which has a mass of 15.0 \(\mathrm{kg}\) when empty and holds 0.120 \(\mathrm{m}^{3}\) of oil. You can ignore the volume occupied by the steel from which the barrel is made. (a) If a salvage worker accidentally drops a filled, sealed barrel overboard, will it float or sink in the seawater? (b) If the barrel floats, what fraction of its volume will be above the water surface? If it sinks, what minimum tension would have to be exerted by a rope to haul the barrel up from the ocean floor? (c) Repeat parts (a) and (b) if the density of the oil is 910 \(\mathrm{kg} / \mathrm{m}^{3}\) and the mass of each empty barrel is 32.0 \(\mathrm{kg}\) .

Short Answer

Expert verified
(a) The first barrel floats, with 14.59% above water; the second barrel sinks. (b) Minimum tension for the sunken barrel: 178.96 N.

Step by step solution

01

Calculate Mass of Oil in Barrel

Determine the mass of oil contained in a full barrel. Use the formula for mass: \( m = \rho \times V \), where \( \rho = 750 \, \mathrm{kg/m^3} \) is the density of oil, and \( V = 0.120 \, \mathrm{m^3} \) is the volume of the barrel. \[m = 750 \, \mathrm{kg/m^3} \times 0.120 \, \mathrm{m^3} = 90 \, \mathrm{kg}\]
02

Calculate Total Mass of Filled Barrel

Add the mass of the oil to the mass of the empty barrel to find the total mass. The mass of the empty barrel is \( 15.0 \, \mathrm{kg} \) for the first case.\[m_{\text{total}} = m_{\text{oil}} + m_{\text{barrel}} = 90 \, \mathrm{kg} + 15 \, \mathrm{kg} = 105 \, \mathrm{kg}\]
03

Determine Buoyant Force

Calculate the buoyant force acting on the barrel using Archimedes’ Principle. The buoyant force is equal to the weight of the displaced seawater. Assume the density of seawater \( \rho_{\text{sea}} = 1025 \, \mathrm{kg/m^3} \).\[F_{\text{buoyant}} = \rho_{\text{sea}} \times V \times g = 1025 \, \mathrm{kg/m^3} \times 0.120 \, \mathrm{m^3} \times 9.8 \, \mathrm{m/s^2} = 1204.8 \, \mathrm{N}\]
04

Compare Gravitational and Buoyant Forces

Compare the gravitational force on the filled barrel with the buoyant force to determine if it will float. Calculate the gravitational force:\[F_{\text{gravity}} = m_{\text{total}} \times g = 105 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 1029 \, \mathrm{N}\]Since \( F_{\text{buoyant} = 1204.8 \, \mathrm{N} } > F_{\text{gravity} = 1029 \, \mathrm{N} } \), the barrel will float.
05

Calculate the Fraction of Volume Above Water

Calculate the fraction of the barrel's volume that will be above the water by using the volume ratio principle, as follows:\[\text{fraction above water} = 1 - \left(\frac{F_{\text{gravity}}}{F_{\text{buoyant}}}\right)\]\[\text{fraction above water} = 1 - \left(\frac{1029 \, \mathrm{N}}{1204.8 \, \mathrm{N}}\right) = 0.1459 \]Approximately 14.59% of the barrel's volume will be above water.
06

Repeat with Different Oil Density and Barrel Mass

Now consider oil density \( \rho = 910 \, \mathrm{kg/m^3} \) and barrel mass \( 32.0 \, \mathrm{kg} \).1. Calculate mass of the new oil:\[m = 910 \, \mathrm{kg/m^3} \times 0.120 \, \mathrm{m^3} = 109.2 \, \mathrm{kg}\]2. Total mass of the filled barrel:\[m_{\text{total}} = 109.2 \, \mathrm{kg} + 32.0 \, \mathrm{kg} = 141.2 \, \mathrm{kg}\]3. Gravitational force:\[F_{\text{gravity}} = 141.2 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 1383.76 \, \mathrm{N}\]Since \( F_{\text{gravity} = 1383.76 \, \mathrm{N} } > F_{\text{buoyant} = 1204.8 \, \mathrm{N} } \), the barrel will sink.
07

Calculate Minimum Tension in Rope (Sinking Case)

For the barrel to be hauled up from the ocean floor, calculate the difference between the gravitational force and the buoyant force, which will be the tension in the rope.\[T = F_{\text{gravity}} - F_{\text{buoyant}} = 1383.76 \, \mathrm{N} - 1204.8 \, \mathrm{N} = 178.96 \, \mathrm{N}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Buoyancy
Buoyancy is the upward force exerted by a fluid that opposes the weight of an object immersed in it. This force is what makes objects like ships and barrels float in water. The principle of buoyancy is described by Archimedes' Principle, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object.
In our example, the sealed barrel filled with oil experiences a buoyant force from the seawater. The magnitude of this force can be calculated using the density of the seawater, the volume of the displaced water, and the acceleration due to gravity. By comparing this buoyant force with the gravitational force acting on the barrel, we can predict whether the barrel will float or sink.
Understanding buoyancy is crucial in many fields, including ship-building and engineering, as it directly affects how objects behave in fluids.
Density
Density is defined as mass per unit volume. It is a key property of materials that influences how they interact with fluids. Higher density objects are typically heavier and may sink, while lower density objects can float.
In our exercise, we consider two density values of oil - 750 kg/m³ and 910 kg/m³. These densities, along with the volume of oil each barrel holds, determine the mass of the oil in each barrel.
  • The formula used is: \[ \text{mass} = \text{density} \times \text{volume} \]
With a lower density oil, a barrel is lighter and more likely to float, whereas a higher density oil increases the barrel's total mass, making it more likely to sink.
Understanding the concept of density helps us predict the buoyant behavior of objects in fluids.
Gravitational Force
Gravitational force is the downward force exerted by gravity on an object. It is calculated as the product of an object's mass and the acceleration due to gravity (9.8 m/s² on Earth).
For the oil barrels, we calculate the gravitational force acting on them by summing the masses of the oil and the barrel structure.
  • Formula: \[ F_{\text{gravity}} = m_{\text{total}} \times g \]
Where \( m_{\text{total}} \) is the sum of the mass of the oil and the empty barrel. This gravitational force affects whether the barrel sinks or floats. If the gravitational force is less than the buoyant force, the barrel floats. If it's greater, the barrel sinks.
This balance of forces is an essential consideration in all applications involving gravity and buoyancy.
Floating and Sinking
Floating and sinking are determined by the relative magnitudes of buoyant force and gravitational force acting on an object. Whether an object floats or sinks depends on the comparison between these two forces.
In the given exercise, after calculating both forces for the different scenarios, it's evident that a barrel filled with lower density oil (750 kg/m³) floats, since the buoyant force exceeds the gravitational force. Conversely, with the denser oil (910 kg/m³), the gravitational force surpasses the buoyant force, causing the barrel to sink.
This basic principle is pervasive in understanding how different objects behave when placed in fluids and helps in designing vessels that need to remain buoyant.
Fluid Mechanics
Fluid mechanics is the branch of physics dealing with the behavior of fluids (liquids and gases) in motion and rest. It encompasses concepts such as buoyancy, pressure, and viscosity.
In the context of our exercise, fluid mechanics principles help explain why barrels filled with differing densities of oil behave differently in seawater. It also helps in calculating the forces at play, for example:
  • Buoyant force (Archimedes' Principle)
  • Gravitational force calculation based on mass and gravity
These calculations illustrate how objects can be designed and configured to float efficiently or retrieve sunken items from underwater. Understanding fluid mechanics is vital for creating technology and structures that must function in aquatic environments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density 1050 \(\mathrm{kg} / \mathrm{m}^{3} )\) located at height \(h\) above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is 5980 \(\mathrm{Pat}\) is the minimum value of \(h\) that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity (see Section 12.6\()\) of the fluid.

The densities of air, helium, and hydrogen (at \(p=1.0\) atm and \(T=20^{\circ} \mathrm{C}\) ) are \(1.20 \mathrm{kg} / \mathrm{m}^{3}, 0.166 \mathrm{kg} / \mathrm{m}^{3},\) and \(0.0899 \mathrm{kg} / \mathrm{m}^{3},\) respectively. (a) What is the volume in cubic meters displaced by a hydrogen-filled airship that has a total "lift" of 90.0 \(\mathrm{kN}\) ? (The "lift" is the amount by which the buoyant force exceeds the weight of the gas that fills the airship.) (b) What would be the "lift" if helium were used instead of hydrogen? In view of your answer, why is helium used in modern airships like advertising blimps?

An incompressible fluid with density \(\rho\) is in a horizontal test tube of inner cross-sectional area \(A .\) The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed \omega. Gravitational forces are negligible. Consider a volume element of the fluid of area \(A\) and thickness \(d r^{\prime}\) a distance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho \omega^{2} r^{\prime} d r^{\prime}\) . (b) If the surface of the fluid is at a radius \(r_{0}\) where the pressure is \(p_{0},\) show that the pressure \(p\) at a distance \(r \geq r_{0}\) is \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .\) (c) An object of volume \(V\) and density \(\rho_{\mathrm{ob}}\) has its center of mass at a distance \(R_{\mathrm{cmob}}\) from the axis. Show that the net horizontal force on the object is \(\rho V \omega^{2} R_{\mathrm{cm}},\) where \(R_{\mathrm{cm}}\) is the distance from the axis to the center of mass of the displaced fluid. (d) Explain why the object will move inward if \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{ob}} R_{\mathrm{cmob}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}} .\) (e) For small objects of uniform density, \(R_{\mathrm{cm}}=R_{\mathrm{cmob}}\) . What happens to a mixture of small objects of this kind with different densities in an ultracentrifuge?

A hydrometer consists of a spherical bulb and a cylindrical stem with a cross- sectional area of 0.400 \(\mathrm{cm}^{2}\) (see Fig. 12.12 \(\mathrm{a} ) .\) The total volume of bulb and stem is 13.2 \(\mathrm{cm}^{3} .\) When immersed in water, the hydrometer floats with 8.00 cm of the stem above the water surface. When the hydrometer is immersed in an organic fluid, 3.20 cm of the stem is above the surface. Find the density of the organic fluid. (Note: This illustrates the precision of such a hydrometer. Relatively small density differences give rise to relatively large differences in hydrometer readings.)

A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.