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Chapter 14: Problem 82

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 \(9000 \mathrm{kg} / \mathrm{m}^{3} .\) (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
(a) Water rises by 0.2125 m. (b) Water falls by 0.1888 m.

Step by step solution

01

Determine the Initial Displacement

The first step is to calculate the volume of water displaced by the barge when the scrap metal is placed on it. Using the buoyant force, which equals the weight of the scrap metal (\(2.50 \times 10^6 \text{ N}\)), we can find the volume displaced. Since the density of water is \(1000 \text{ kg/m}^3\), the volume of displacement \(V_d\) is given by: \[ V_d = \frac{F}{\text{density} \times g} = \frac{2.50 \times 10^6}{1000 \times 9.81} \approx 255 \text{ m}^3. \]
02

Calculate the Rise in Water Level

The water level rises because the displacement volume causes the water in the lock to rise. Since the lock is a rectangular prism with a base area of \(60.0 \text{ m} \times 20.0 \text{ m}\), the rise in water level \(h\) is found by dividing the displaced volume by the area:\[ h = \frac{V_d}{A} = \frac{255}{60 \times 20} \approx 0.2125 \text{ m}. \] Thus, the water in the lock rises by approximately 0.2125 meters.
03

Analyze the Impact of Discarding Scrap Metal

When the scrap metal is discarded into the water, it displaces its own volume of water. The volume \(V_m\) of the scrap metal is given by its mass divided by its density:\[ V_m = \frac{m}{\text{density}} = \frac{2.50 \times 10^6 / 9.81}{9000} \approx 28.4 \text{ m}^3. \]
04

Determine the Change in Water Level

When the scrap metal goes overboard, water displacement changes. Initially, water displacement equaled the volume corresponding to weight force. Upon immersion, displaced volume equals the metal's physical volume, which is smaller. So, water level reduces as decrease in displaced volume \(\Delta V\) is \[ \Delta V = 255 - 28.4 \approx 226.6 \text{ m}^3. \]Water level change \(\Delta h\) is then:\[ \Delta h = \frac{\Delta V}{A} = \frac{226.6}{60 \times 20} \approx 0.1888 \text{ m}. \]Thus, the water level drops by approximately 0.1888 meters.

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Key Concepts

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

Archimedes' Principle
Archimedes' Principle is a fundamental concept in physics that helps us understand buoyancy. It states that an object completely or partially submerged in a fluid experiences an upward force equal to the weight of the fluid displaced by the object.

This principle is the key to solving many problems involving floating and submerged objects. When you place an object in a fluid, like water, the fluid gets pushed aside, or displaced. As a result, the object experiences a buoyant force that makes it feel lighter. In the case of our problem, the barge floats because the water in the lock exerts this buoyant force. When metal is added, its weight is supported by this force.

By using Archimedes' Principle, you can calculate how much water the object will displace and, consequently, how much the water level will rise or fall in a confined area like a lock.
Volume Displacement
Volume displacement is about understanding how much space an object takes up when submerged in a fluid. In our exercise, when the scrap metal is put onto the barge, it causes the barge to sink further into the water, displacing additional water.

The displaced volume can be calculated by measuring the buoyant force which equals the weight of the submerged object. This is because the displaced fluid needs to balance the weight of the object for it to float. This principle allows us to determine how much the water level rises, as we know both the volume of the water displaced and the area of the lock.

It's important to note that volume displacement is directly linked to the shape and size of the object and the density of the fluid. Understanding this concept helps us visualize how objects interact with fluids in various real-world scenarios.
Density Calculations
Density calculations are crucial for understanding how different materials interact with fluids, particularly in relation to buoyancy. Density is defined as mass per unit volume: \( \rho = \frac{m}{V} \). For solids like scrap metal, this property determines how they will behave when submerged.

In the exercise, the metal's density is given, which allows us to calculate its volume using the relationship between mass, volume, and density. Essentially, we determine how much an object will sink by comparing its density to that of the fluid. If it's denser, as in our exercise, it will sink until sufficiently buoyed by the displaced fluid.

These calculations help determine the change in water level when the metal is placed on the barge and subsequently when it's discarded into the water. Understanding density allows you to predict movements in fluids accurately, which is an essential skill in both educational and practical settings.

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Most popular questions from this chapter

A hunk of aluminum is completely covered with a gold shell to form an ingot of weight 45.0 \(\mathrm{N}\) . When you suspend the ingot from a spring balance and submerge the ingot in water, the balance reads 39.0 \(\mathrm{N}\) . What is the weight of the gold in the shell?

A hollow plastic sphere is held below the surface of a fresh-water lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 \(\mathrm{m}^{3}\) and the tension in the cord is 900 \(\mathrm{N}\) . (a) Calculate the buoyant force exerted by the water on the sphere. (b) What is the mass of the sphere? (c) The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

A cubical block of wood 0.100 \(\mathrm{m}\) on a side and with a density of 550 \(\mathrm{kg} / \mathrm{m}^{3}\) floats in a jar of water. Oil with a density of 750 \(\mathrm{kg} / \mathrm{m}^{3}\) is poured on the water until the top of the oil layer is 0.035 \(\mathrm{m}\) below the top of the block. (a) How deep is the oil layer? (b) What is the gauge pressure at the block's lower face?

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 \(\mathrm{cm}^{3} / \mathrm{s}\) . At one point in the pipe, where the radius is 4.00 \(\mathrm{cm}\) , the water's absolute pressure is \(2.40 \times 10^{5} \mathrm{Pa}\) . At a second point in the pipe, the water passes through a constriction where the radius is \(2.00 \mathrm{cm} .\) What is the water's absolute pressure as it flows through this constriction?

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 \(\mathrm{kg}\) .
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