Problem 31 - College Physics

Question

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?

Step-by-Step Solution

Answer

The volume of the boat is 5.75 m³. Throw 1150 kg overboard.

Step 1: Understand the Problem

We are given a floating boat on a freshwater lake with a total mass of 5750 kg. The water level is up to the top of its sides (gunwales), meaning the boat is at the brink of sinking. Our goal is to find the volume of the boat and determine how much weight needs to be removed for 20% of the boat's volume to be above the water.

Step 1: Determine the Volume of the Boat

By Archimedes’ principle, the volume of the water displaced by the boat is equal to the volume of the boat that is submerged. The weight of the boat equals the weight of the displaced water. The mass of the water displaced is 5750 kg, so using the density of water (1000 kg/m³), the volume of the boat (V) is:\[ V = \frac{5750 \, \text{kg}}{1000 \, \text{kg/m}^3} = 5.75 \, \text{m}^3 \]

Step 2: Calculate New Submerged Volume

Since the captain wants 20% of the boat's volume above water, 80% will be submerged. Calculate the new submerged volume:\[ V_{\text{sub}} = 0.8 \times 5.75 \, \text{m}^3 = 4.6 \, \text{m}^3 \]

Step 3: Determine New Mass and Mass to Remove

The new mass of the boat plus the cargo must displace exactly 4.6 m³ of water, so the mass of the displaced water should be equal to the new mass of the boat:\[ \text{New Mass} = 4.6 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 = 4600 \, \text{kg} \]The mass to be thrown overboard is:\[ 5750 \, \text{kg} - 4600 \, \text{kg} = 1150 \, \text{kg} \]

Key Concepts

Archimedes' Principle

Archimedes' Principle is a fundamental concept in physics that explains why objects float or sink when placed in a fluid, such as water. It states that any object, fully or partially immersed in a fluid, experiences an upward force known as buoyancy, which is equal to the weight of the fluid displaced by the object.
In the context of the exercise here, the boat floats because the buoyant force, which is equal to the weight of water displaced by the submerged part of the boat, perfectly counterbalances the weight of the boat including its passengers and cargo.
This concept is crucial in determining the volume of a floating object. When the boat is at the brink of sinking, this implies that the volume of water it displaces is equal to its own mass. Thus, by understanding Archimedes' Principle, we can calculate the boat's volume based on the water displaced, helping us solve the initial part of the exercise.

Density Calculation

Density is a measure of mass per unit volume and plays a pivotal role in understanding buoyancy and the behavior of objects in fluids. The formula for density is given by:- \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)For the boat floating in water, knowing the density of water (1000 kg/m³) allows us to find the volume of the displaced water, which should equal the boat's volume when it is on the verge of sinking.
In our case, with a total mass of 5750 kg, density helps us to calculate the boat's submerged volume by the formula:- \( V = \frac{\text{Mass of the boat}}{\text{Density of water}} \)
This calculation shows us that the boat displaces 5.75 m³ of water, confirming the boat's volume.

Floating and Sinking

The concepts of floating and sinking are determined by the balance between weight and buoyant force. An object will float if the buoyant force it experiences is equal to its weight. Conversely, it will sink if the weight is greater.
For the boat to float with 20% of its volume above water, the goal is to adjust the ballast or weight correctly. Initially, when the boat was entirely submerged, it was at risk of sinking as the weight was equal to the maximum buoyancy.
To increase the boat's freeboard, the captain must ensure only 80% of the boat's volume is submerged. This requires reducing the boat's mass to displace only 4.6 m³ of water (80% of 5.75 m³). The calculation shows that 1150 kg must be removed, reducing the weight so 20% of the volume is above the waterline, thus achieving a proper floating balance.