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Dropping Anchor. An iron anchor with mass 35.0 \(\mathrm{kg}\) and density 7860 \(\mathrm{kg} / \mathrm{m}^{3}\) lies on the deck of a small barge that has vertical sides and floats in a freshwater river. The area of the bottom of the barge is 8.00 \(\mathrm{m}^{2} .\) The anchor is thrown overboard but is suspended above the bottom of the river by a rope; the mass and volume of the rope are small enough to ignore. After the anchor is overboard and the barge has finally stopped bobbing up and down, has the barge risen or sunk down in the water? By what vertical distance?

Short Answer

Expert verified
The barge rises by approximately 4.375 mm.

Step by step solution

01

Calculate the Volume of the Anchor

To understand how much water the anchor displaces, we need to find its volume. The volume \( V \) can be found using the formula: \[ V = \frac{m}{\rho} \] where \( m \) is the mass of the anchor (35.0 kg) and \( \rho \) is the density of iron (7860 kg/m\(^3\)). Substitute these values to find \( V \): \[ V = \frac{35.0 \text{ kg}}{7860 \text{ kg/m}^3} \approx 0.004454 \text{ m}^3 \]
02

Calculate Initial Water Displacement

Initially, when the anchor is on the barge, it adds to the weight of the barge, and thus to the water it displaces. Since the total initial weight is the weight of the barge plus the weight of the anchor, the volume of water displaced is the mass of the anchor divided by the density of water (\(1000 \text{ kg/m}^3\)) at equilibrium. The initial weight contribution of the anchor is \(35.0 \text{ kg}\).
03

Calculate Final Water Displacement

Once the anchor is overboard, it does not contribute to the weight of the barge anymore. Therefore, the water displacement by the barge reduces by the volume that corresponds to the weight of the anchor (35.0 kg). Since no weight is on the barge, the volume of water displacement decreases by \(\frac{35.0 \text{ kg}}{1000 \text{ kg/m}^3} = 0.035 \text{ m}^3\).
04

Calculate the Change in Water Level

The decrease in volume displaced by the barge is \(0.035 \text{ m}^3\), and this volume is distributed across the area of the barge's bottom, \(8.00 \text{ m}^2\). Thus, the change in water level (change in displacement height) can be calculated as follows: \[ \Delta h = \frac{\Delta V}{A} = \frac{0.035 \text{ m}^3}{8.00 \text{ m}^2} \approx 0.004375 \text{ m} \] Therefore, the barge rises by approximately 0.004375 m (4.375 mm) after the anchor is thrown overboard.

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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 describes how objects experience buoyant forces when submerged in a fluid. It states that any object, completely or partially submerged in a fluid, is buoyed up by a force equal to the weight of the fluid it displaces. This force enables objects to float or at least weigh less in water than they do in air.
This principle can be applied in various contexts, such as determining whether an object will sink or float. When the buoyant force is greater than or equal to the object's weight, it floats; otherwise, it sinks. Understanding Archimedes' Principle is crucial in solving problems like the barge example, where the focus is on how the displacement of water affects the buoyancy of the barge. As the anchor is thrown overboard, it no longer displaces water around the barge, causing it to rise. The principle helps explain why the barge rises by the amount equal to the volume of water the anchor initially displaced, illustrating this principle in action.
Density and Volume Calculations
Density and Volume Calculations play a vital role in understanding how objects interact with their environments, especially within fluids. Density is the mass of an object divided by its volume, often expressed in units like kilograms per cubic meter (\( ext{kg/m}^3 \).When you calculate density, you use the formula: \[ \rho = \frac{m}{V} \]where \( m \) is mass and \( V \) is volume. This formula allows you to understand how tightly mass is packed within an object, which in turn determines how it behaves in a fluid.In the case of the anchor, knowing its density enabled us to find its volume. Using the values provided—mass (\( 35.0 ext{ kg} \)) and density (\( 7860 ext{ kg/m}^3 \))—we calculated that the anchor's volume is approximately \( 0.004454 ext{ m}^3 \).These calculations are essential for predicting the change in water displacement when the anchor is removed from the barge. Overall, understanding these calculations helps in anticipating the physical changes when objects are immersed or removed from fluids.
Fluid Displacement
Fluid Displacement is a concept closely linked with buoyancy and density. It refers to the process of an object causing a fluid to make space for it, leading to the displacement of some of the fluid's volume. When an object is placed into a fluid, it pushes aside a quantity of fluid equal to the object's volume if submerged, or part of its volume if floating. In the context of the anchor and the barge, fluid displacement helps explain how the barge's movement in water changes when the anchor is no longer aboard. Initially, the anchor's weight adds to the water the barge displaces, contributing to equilibrium. When the anchor is dropped into the water, the displaced water volume decreases by the volume corresponding to the anchor's weight divided by the density of water. The relationship here is straightforward: less weight means less water displaced, and the barge rises as a result. The decrease in displaced volume reflects a reduction in the draft of the barge—how deep it sits in water. By calculating this displaced volume accurately, one can determine by exactly how much the barge's position in the water changes, a clear example of fluid displacement at work.

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Most popular questions from this chapter

On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-m- high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m- deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 \(\mathrm{kg} / \mathrm{m}^{3} .\) If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (Hint: Consider the outward force on a circular ring of the tank wall of width \(d y\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

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