/*! 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 76 A barge is in a rectangular lock... [FREE SOLUTION] | 91Ó°ÊÓ

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A barge is in a rectangular lock on a freshwater river. The lock is \(60.0 \mathrm{~m}\) long and \(20.0 \mathrm{~m}\) wide, and the steel doors on each end are closed. With the barge floating in the lock, a \(2.50 \times 10^{6} \mathrm{~N}\) load of scrap metal is put onto the barge. The metal has density \(7200 \mathrm{~kg} / \mathrm{m}^{3}\). (a) When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise? (b) The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same? If it rises or falls, by what vertical distance does it change?

Short Answer

Expert verified
Part (a): The water in the lock rises by a certain height 'h'. Part (b): The water level in the lock falls by the same height 'h'.

Step by step solution

01

Determine the Weight and Volume of the Scrap Metal

The weight of the scrap metal is given as \(2.50 \times 10^{6} \mathrm{N}\). From this, its volume can be calculated using its density. The formula to find the volume is mass divided by density. Since the weight is equivalent to mass times gravity, the volume (\(V_m\)) of the scrap metal can be calculated as \(V_m = \frac{Weight}{Density \times Gravity}\). Substituting the given values, we get: \(V_m = \frac{2.50 \times 10^{6} \mathrm{N}}{7200 \mathrm{~kg/m^{3}} \times 9.8 \mathrm{m/s^{2}}}\)
02

Calculate the Rise in Water Level for Part (a)

The volume of water displaced corresponds to the volume of the scrap metal. The rise in water level (\(h_1\)) can be calculated by dividing this volume by the area of the lock. The area of the lock (\(A_l\)) is given by its length times its width: \(A_l = 60.0 \mathrm{m} \times 20.0 \mathrm{m}\). Therefore, \(h_1 = \frac{V_m}{A_l}\)
03

Determine the Change in Water Level for Part (b)

Once the scrap metal is put overboard, it displaces water equal to its own volume. Since metal is denser than water, it displaces less water when sunk compared to when it was on the barge. Therefore, the water level falls. The fall in water level (\(h_2\)) can be found using the difference between the water displacements when the scrap metal was on the barge and after it is sunk. That is, \(h_2 = h_1 - \frac{V_m}{A_l}\), where \(V_s\) is the volume of scrap metal after it is put overboard. Since weight, density, and gravity remain the same, \(V_s\) is equal to \(V_m\). Therefore, \(h_2 = h_1\)

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Key Concepts

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

Buoyancy
Buoyancy is the force exerted by a fluid that opposes the weight of an object immersed in it. It's the reason why objects such as boats and barges can float. According to Archimedes' principle, the buoyant force on a submerged object is equal to the weight of the fluid it displaces.

For example, in the problem where a load of scrap metal is placed onto a barge, the barge sinks further into the water, displacing more water and thus experiencing a greater buoyant force. This increase in buoyant force continues until it balances the increased weight of the barge with the metal load. If the buoyant force is less than the weight of the object, the object will sink until it rests on the bottom or is fully submerged.
Density
Density is a property that measures how much mass is contained in a given volume. It's mathematically defined as mass per unit volume and typically expressed in units such as kilograms per cubic meter (\text{kg/m}^{3}).

The concept of density is crucial in understanding why certain objects float or sink. For instance, the problem's reference to the density of the metal at 7200 \(\mathrm{kg/m}^{3}\) is much greater than the density of freshwater, which is about 1000 \(\mathrm{kg/m}^{3}\). An object will sink if its density is greater than the fluid it displaces, which is exactly what happens to the metal when pushed overboard into the water.
Displacement
Displacement in physics usually refers to the volume of fluid that an object pushes aside when it is placed in the fluid. This concept is directly related to the calculation of buoyancy. The displaced fluid's volume is equivalent to the volume of the part of the object that is submerged.

In the situation where the scrap metal is added to the barge, it displaces an amount of water equal to the volume of metal. If the metal is later pushed overboard, it will displace an amount of water equivalent to its own submerged volume, which affects the water level in the lock. Displacement helps us understand changes in water level, as seen in this problem.
Fluid Mechanics
Fluid mechanics is the study of fluids (liquids, gases, and plasmas) and the forces on them. It's divided into fluid statics (studying fluids at rest) and fluid dynamics (studying fluids in motion).

This area of physics provides the theoretical foundation for understanding the problem's scenario, where the barge's interaction with the water is a fluid statics problem. Fundamental principles from fluid mechanics, such as Archimedes’ principle and the concept of pressure, directly contribute to solving problems involving buoyancy, density, and displacement.

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Most popular questions from this chapter

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 2200.355 -L cans per minute. At point 2 in the pipe, the gauge pressure is \(152 \mathrm{kPa}\) and the cross-sectional area is \(8.00 \mathrm{~cm}^{2}\). At point \(1,1.35 \mathrm{~m}\) above point \(2,\) the cross- sectional area is \(2.00 \mathrm{~cm}^{2}\). Find the (a) mass flow rate; (b) volume flow rate; (c) flow speeds at points 1 and \(2 ;\) (d) gauge pressure at point 1 .

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