/*! 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 31 When an open-faced boat has a ma... [FREE SOLUTION] | 91Ó°ÊÓ

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When an open-faced boat has a mass of \(5750 \mathrm{~kg}\), including its cargo and passengers, it floats with the water just up to the top of its gunwales (sides) on a freshwater lake. (a) What is the volume of this boat? (b) The captain decides that it is too dangerous to float with his boat on the verge of sinking, so he decides to throw some cargo overboard so that \(20 \%\) of the boat's volume will be above water. How much mass should he throw out?

Short Answer

Expert verified
(a) Volume of the boat: 5.75 m³. (b) Mass to throw out: 1150 kg.

Step by step solution

01

Identify the force acting on the boat

When the boat is floating, the buoyant force acting on it equals the gravitational force due to its weight. The buoyant force is given by the weight of the water displaced by the boat. Therefore, \( F_{buoyant} = F_{gravity} = m \cdot g \), where \( m = 5750 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \).
02

Calculate the volume of water displaced

The buoyant force is also given by \( F_{buoyant} = \rho \cdot V \cdot g \), where \( \rho = 1000 \text{ kg/m}^3 \) is the density of freshwater and \( V \) is the volume of water displaced (which is the volume of the boat when it's floating at the verge of sinking). Therefore, set \( 5750 \cdot 9.8 = 1000 \cdot V \cdot 9.8 \) and solve for \( V \).
03

Solve for the volume of the boat

From the equation \( 5750 \cdot 9.8 = 1000 \cdot V \cdot 9.8 \), simplify to find \( V = \frac{5750}{1000} = 5.75 \text{ m}^3 \). This is the total volume of the boat.
04

Find the new volume submerged when 20% is above water

To have 20% of the boat's volume above water, 80% will be submerged. Calculate the submerged volume as \( V_{sub} = 0.8 \cdot 5.75 \text{ m}^3 \).
05

Calculate the new submerged mass

Since the boat needs to float with only 80% volume submerged, the buoyant force must equal the weight of the new mass. Use \( F_{buoyant} = \rho \cdot V_{sub} \cdot g = m_{new} \cdot g \). Solve for \( m_{new} \).
06

Find the mass to throw out

Substitute in the equation \( 1000 \times (0.8 \times 5.75) \times 9.8 = m_{new} \times 9.8 \) and solve for \( m_{new} \). This gives \( m_{new} = 4600 \text{ kg} \). Therefore, the captain needs to throw out \( 5750 - 4600 = 1150 \text{ kg} \) of mass.

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Key Concepts

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

Archimedes' Principle
The concept of buoyancy, as described by Archimedes' Principle, is essential for understanding why objects float or sink. This principle states: An object submerged in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the object.

For the boat in our exercise, as it floats, it displaces water equivalent in weight to its own. Hence, the buoyant force equals the gravitational force acting on the boat. When the boat reaches the verge of sinking, the entire weight of the 5750 kg boat is balanced by the buoyant force produced by the displaced water.

In practical terms, this means:
  • A boat will float as long as the weight of the displaced water equals the weight of the boat.
  • An object sinks if its weight exceeds the weight of the fluid it displaces.
  • Archimedes' Principle applies to all fluids, including liquids and gases.
Understanding this principle helps in calculating the necessary adjustments, such as removing cargo, to prevent sinking.
Density of Water
In our scenario, we refer to freshwater with a density of 1000 kg/m³. Density is a measure of mass per unit volume and is critical in determining how much water must be displaced to support a floating object.

The density of water directly impacts the buoyant force. In the calculation, we use density as a factor to determine the volume of water the boat displaces. Using the formula for the buoyant force: \[ F_{buoyant} = \rho \cdot V \cdot g \] where:
  • \( \rho \) is the density of water,
  • \( V \) is the volume of water displaced, and
  • \( g \) is the acceleration due to gravity (9.8 m/s²).
For freshwater, which has a consistent density, this calculation remains straightforward. This consistency enables a predictable buoyant force for objects like the boat.
Gravitational Force
Gravitational force is the force that attracts any objects with mass toward the center of the earth. In the context of our exercise, it's the force acting downward on the boat, calculated by the formula:\[ F_{gravity} = m \cdot g \] where \( m \) is mass and \( g \) is gravitational acceleration (9.8 m/s²).

For our boat:
  • Mass: 5750 kg
  • Gravitational constant: 9.8 m/s²
  • Calculating: \( F_{gravity} = 5750 \cdot 9.8 \) Newtons
This gravitational force must be countered by an equal buoyant force in order to keep the boat afloat. If the buoyant force is less than the gravitational force, the boat will sink.
Volume Displacement
Volume displacement is the key to understanding how buoyancy works, especially in floating objects like boats. When a boat is placed in water, it pushes away a certain volume of water equivalent to the weight of the boat itself.

By displacing water:
  • The boat creates enough space for the buoyant force to act upon, which helps it to float.
  • The amount of water displaced equals the submerged volume of the boat.
For example, the volume of water displaced by our boat is calculated using the formula:\[ V = \frac{5750}{1000} = 5.75 \text{ m}^3 \] This represents the full volume when the boat is at the brink of sinking. By adjusting what is on board, such as throwing out excess cargo to change the displacement to a safer level with 20% of the boat above water, the captain can ensure safety.

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Most popular questions from this chapter

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