Problem 40 - University Physics with Modern Physics

Question

A supertanker filled with oil has a total mass of \(10.2 \cdot 10^{8} \mathrm{~kg}\). If the dimensions of the ship are those of a rectangular box \(250 . \mathrm{m}\) long, \(80.0 \mathrm{~m}\) wide, and \(80.0 \mathrm{~m}\) high, determine how far the bottom of the ship is below sea level \(\left(\rho_{\mathrm{sea}}=1020 \mathrm{~kg} / \mathrm{m}^{3}\right)\)

Step-by-Step Solution

Answer

Answer: 50 meters

Step 1: Write down the Archimedes' principle formula for buoyant force

The formula for the buoyant force is given as: \(F_B = \rho_{sea} Vg\), where \(F_B\) is the buoyant force, \(\rho_{sea}\) is the density of sea water, \(V\) is the submerged volume of the ship and \(g\) is the acceleration due to gravity.

Step 2: Use the Archimedes' principle to relate the buoyant force to the ship's mass

We know that the ship is in equilibrium when the buoyant force equals the weight of the submerged ship: \(F_B = mg\). Therefore, we can write: \(\rho_{sea} Vg = mg\)

Step 3: Solve for the submerged volume, V

We can solve for the submerged volume by dividing both sides of the equation by \(\rho_{sea}g\): \(V = \frac{mg}{\rho_{sea}g}\). Now we can plug in the given values: \(V = \frac{(10.2 \cdot 10^{8}\,\text{kg})(9.81\,\text{m/s}^2)}{1020\,\text{kg/m}^3(9.81\,\text{m/s}^2)}\)

Step 4: Calculate the submerged volume, V

Performing the calculation, we get: \(V = \frac{(10.2 \cdot 10^{8}\,\text{kg})(9.81\,\text{m/s}^2)}{1020\,\text{kg/m}^3(9.81\,\text{m/s}^2)} = 10^7\,\text{m}^3\)

Step 5: Find the depth of submersion (h) using the submerged volume

Since the ship's shape is a rectangular box, we can express the submerged volume as \(V = lwh\), where \(l\) is length, \(w\) is width and \(h\) is the submerged depth. Plug in the given dimensions and calculated volume: \(10^7\,\text{m}^3 = (250\,\text{m})(80\,\text{m})h\)

Step 6: Solve for the submerged depth (h)

Divide both sides by \((250\,\text{m})(80\,\text{m})\) to get the submerged depth: \(h = \frac{10^7\,\text{m}^3}{(250\,\text{m})(80\,\text{m})}\)

Step 7: Calculate the submerged depth (h)

Performing the calculation, we get: \(h = \frac{10^7\,\text{m}^3}{(250\,\text{m})(80\,\text{m})} = 50\,\text{m}\) Thus, the bottom of the ship is 50 meters below sea level.

Key Concepts

Buoyant Force

When an object is submerged in a fluid, it experiences an upward force known as the buoyant force. This force is crucial in understanding why objects float or sink. According to Archimedes' principle, the amount of the buoyant force is equal to the weight of the fluid that the object displaces.

This means for our supertanker, the buoyant force works against gravity to support the tanker's weight in water. If the tanker's weight is greater than the buoyant force, it will sink until it displaces enough water to equal its own mass, achieving equilibrium. In the case of the supertanker, the equilibrium occurs when the buoyant force balances the weight of the ship, allowing us to solve for the submerged volume using the relationship, \( F_B = \rho_{sea} Vg \).

Density

Density is defined as mass per unit volume and is a key concept when discussing buoyancy and fluid mechanics. The density of the fluid into which the object is submerged plays a critical role in determining the buoyant force acting on the object.

In our textbook solution, the sea water's density \( \rho_{sea} = 1020 \text{kg/m}^3 \) is an important factor as it determines the weight of water displaced by the supertanker's submerged volume. A fluid with a higher density would exert a greater buoyant force, influencing how much of the supertanker is submerged.

Equilibrium in Fluids

Equilibrium in fluids is the point at which the upward buoyant force is equal to the downward gravitational force on an object. When an object is fully or partially submerged, it will move until these forces are balanced.

For our supertanker, it reached equilibrium when the force due to gravity, \( mg \), on the tanker matched the buoyant force, \( \rho_{sea} Vg \). It's essential in this case to understand that the ship does not need to be fully submerged to be in equilibrium; it only needs to displace a volume of water whose weight equals its own.

Submerged Volume Calculation

Calculating the submerged volume of an object involves determining the volume of fluid displaced by the object. For regularly shaped objects like our supertanker, modeled as a rectangular box, this calculation is straightforward once you understand the principles of buoyancy.

In the textbook solution, we used the equation \( V = \frac{mg}{\rho_{sea}g} \) to find that the submerged volume of the tanker was \( 10^7 \text{m}^3 \). Then, recognizing that the tanker is box-shaped, we derive the depth of submersion, \( h \), by dividing this volume by the product of the tanker's length and width. This mathematical approach elegantly combines principles of buoyancy with basic geometry to provide a clear solution.