Problem 84 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(a) Water level rises by 0.212 m; (b) After metal is pushed overboard, level falls by 0.188 m.

Step 1: Determine the Volume Displacement by the Barge

The weight of the load placed on the barge is given as \(2.50 \times 10^6 \ \mathrm{N}\). Since the weight is equal to the product of the volume of the displaced water \( V \) and the gravitational force \( g \), which is \(9.81 \ \mathrm{m/s^2} \), we equate this as:\[ W = \rho_w \cdot g \cdot V \] where \( \rho_w = 1000 \ \mathrm{kg/m^3}\) (density of water) and solve for \( V \): \[ V = \frac{W}{\rho_w \cdot g} = \frac{2.50 \times 10^6}{1000 \times 9.81} \ \mathrm{m^3} \approx 254.34 \ \mathrm{m^3} \]

Step 2: Find the Rise in Water Level

Given the lock's base dimensions (length \(L = 60.0 \ \mathrm{m}\) and width \(W = 20.0 \ \mathrm{m}\)), the water level rise \(h\) can be found by equating the volume of displaced water to the volume of the cuboidal strip of water above its original level. Therefore, solve:\[ LW \cdot h = V \]Find \(h\):\[ h = \frac{V}{LW} = \frac{254.34}{60.0 \times 20.0} \ \mathrm{m} = 0.212 \ \mathrm{m} \]

Step 3: Effect of Metal Being Pushed Overboard

When the scrap metal is pushed overboard, the displaced volume is that of the metal itself since it now sinks. Calculate the volume of the scrap metal:\[ V_m = \frac{m}{\rho_m} = \frac{W/g}{\rho_m} = \frac{2.50 \times 10^6 / 9.81}{9000} \ \mathrm{m^3} \approx 28.41 \ \mathrm{m^3} \]This volume \( V_m\) is less than the water volume displaced when on the barge, hence the water level decreases.

Step 4: Calculate Change in Water Level with Metal in Water

Now that the volume while the scrap metal is in the water is \(V_m \approx 28.41 \ \mathrm{m^3}\), the change in water level is found as the difference in displaced volumes before and after.Calculate the new water level drop\[ \Delta h = \frac{V - V_m}{LW} = \frac{254.34 - 28.41}{60.0 \times 20.0} \ \mathrm{m} \approx 0.188 \ \mathrm{m} \]Thus, by pushing the metal into the water, the water level decreases.

Key Concepts

Archimedes' Principle

Understanding why objects float or sink in a fluid is crucial, and this is where Archimedes' Principle comes into play. It states that a body submerged in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the body. This is why heavy ships float; the water they displace weighs more than the weight of the ships themselves.
Archimedes' Principle can be visualized with the classic example of placing a solid object in water. The object pushes water aside, creating an upward buoyant force on itself. This principle helps us calculate how much of an object will be submerged and estimate water level changes in a contained environment like a barge in a lock.

Density of Materials

Density is a measure of how much mass is contained in a given volume, often expressed in kilograms per cubic meter (\(\mathrm{kg/m^3}\)). Understanding the density of materials is key in predicting their behavior when placed in a fluid.
For example, steel is denser than water, so it will sink, while wood is typically less dense and will float. In this problem, the scrap metal has a density of 9000 \(\mathrm{kg/m^3}\), meaning each tiny volume of it is quite heavy compared to water, which has a density of just 1000 \(\mathrm{kg/m^3}\). When identifying whether an object will float or sink, compare its density to that of the fluid it's in.

Fluid Displacement

Fluid displacement refers to the volume of fluid that is moved or pushed aside by an object. The displaced water's volume is crucial in calculating the buoyant force.
In our scenario, as the barge sits in the lock, any added load increases the water displacement. When scrap metal is loaded onto the barge, more water needs to be displaced to keep the barge afloat, reflecting in a raised water level. Conversely, if an object like our metal load enters the water directly, it will displace an amount of water equal to its volume if fully submerged.

Water Level Changes

Changes in water level are an immediate outcome once an object is placed in or removed from a body of water. These changes are determined by the volume of fluid displaced.
In a confined area like a lock, adding heavy items like scrap metal to a floating barge increases the total displacement, thus raising the water level. However, if that metal is moved from the barge into the water, it displaces less water due to its volume, causing the water level to drop.

• When on the barge: The load adds to the barge's weight, causing more displacement and a water rise.
• When in the water: The metal alone displaces less water volume than when on the barge, hence a drop in water level.

Such insights into water level dynamics can aid in planning and understanding real-life maritime and engineering projects.