Problem 69 - University Physics with Modern Physics

Question

When an open-faced boat has a mass of 5750 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

(a) 5.75 m³; (b) Throw out 1150 kg.

Step 1: Understand the Concept of Buoyancy

The boat floats due to the buoyant force, which is equal to the weight of the water displaced by the boat. For a floating object, its weight is equal to the buoyant force.

Step 2: Calculate the Volume of the Boat

The weight of the boat is given as 5750 kg. Since it floats, the weight of the boat is equal to the weight of the water displaced. In fresh water, the density is approximately 1000 kg/m³. Therefore, the volume of the boat can be calculated using the formula:\[ V = \frac{m}{\rho} \]where \( m = 5750 \text{ kg} \) and \( \rho = 1000 \text{ kg/m}^3 \).Thus, the volume \( V \) is:\[ V = \frac{5750}{1000} = 5.75 \text{ m}^3 \].

Step 3: Calculate the New Required Submerged Volume

The captain requires 20% of the boat's volume to be above water, meaning 80% of the boat should be submerged. Thus, the submerged volume is:\[ V_{\text{submerged}} = 0.8 \times V = 0.8 \times 5.75 = 4.6 \text{ m}^3 \].

Step 4: Determine the Mass of Water Displaced with New Volume

The mass of water displaced by the new submerged volume is equivalent to the weight that the boat can support in this condition. Calculate using:\[ m_{\text{water}} = \rho \times V_{\text{submerged}} = 1000 \times 4.6 = 4600 \text{ kg} \].

Step 5: Calculate Mass to Be Thrown Overboard

Initially, the total mass is 5750 kg. With a submerged volume corresponding to 4600 kg, the reduction in mass is:\[ m_{\text{throw}} = 5750 - 4600 = 1150 \text{ kg} \].Therefore, the captain should throw overboard 1150 kg of mass.

Key Concepts

Floating Objects

Floating objects have a unique property where they rest on the surface of a liquid without sinking. This happens because the buoyant force pushing upwards against the object is equal to the object's weight. This principle is known as Archimedes' principle. When any object is submerged in a fluid, it experiences an upward force called the buoyant force.
The buoyant force's strength depends on the volume of the object that is submerged and the density of the fluid. This is why a boat, despite being heavy, can float when placed in water. The boat displaces enough water to create a buoyant force equal to its weight,
keeping it afloat.

Density of Freshwater

The density of a fluid is a measure of its mass per unit volume. For freshwater, this is usually around 1000 kg/m³. Density plays a critical role in determining whether an object floats or sinks.
When an object is in freshwater, it will displace some of the water based on its own weight. The volume of water displaced correlates to the object's weight divided by the water's density, as seen in the formula:

• \[ V = \frac{m}{\rho} \]

where \( V \) is the volume of water displaced, \( m \) is the mass of the object, and \( \rho \) is the density of freshwater.
This calculation is critical for determining the volume of the part of an object that remains submerged when floating.

Displacement

Displacement refers to the volume of fluid that is moved out of place by an object that is submerged. In the context of floating objects, like a boat, the amount of water displaced is equal to the volume of the object submerged.
For a boat floating on water, the volume of water displaced by the boat equals the volume of the submerged part of the boat.
This means, if a boat's total volume is known and the volume submerged is required, the submerged volume can be determined using the total volume and the percentage submerged. For instance, if a boat floats at 80% of its volume (leaving 20% above water), it means 80% of the boat's total volume is pushing away water.
This principle is utilized to determine how much of a boat needs to be unloaded to adjust how high it sits on the water's surface.