Problem 25 - University Physics with Modern Physics

Question

A 950 -kg cylindrical can buoy floats vertically in salt water. The diameter of the buoy is 0.900 \(\mathrm{m} .\) Calculate the additional distance the buoy will sink when a 70.0 -kg man stands on top of it.

Step-by-Step Solution

Answer

The buoy sinks an additional 10.8 cm when the man stands on it.

Step 1: Understand the forces involved

The buoy is initially in equilibrium with the buoyant force equal to its weight. When a person stands on it, the combined weight must be supported by a new buoyant force. Therefore, we need to find how much deeper the buoy sinks to create this additional buoyant force.

Step 2: Apply Archimedes' Principle

Archimedes' Principle states that the buoyant force is equal to the weight of the displaced fluid. The additional weight of the man requires additional water displacement, which results in the buoy sinking further.

Step 3: Calculate initial and additional forces

The initial force is the weight of the buoy: \[ F_{ ext{buoy initial}} = m_{ ext{buoy}} \times g = 950 \times 9.8 \, \text{N}. \]The additional force when the man stands on it is:\[ F_{ ext{man}} = m_{ ext{man}} \times g = 70 \times 9.8 \, \text{N}. \]

Step 4: Determine cross-sectional area

The buoy has a diameter of 0.900 m, so its radius is 0.450 m. The cross-sectional area \( A \) is:\[ A = \pi \times (0.450)^2 = 0.63617 \, ext{m}^2. \]

Step 5: Calculate additional displacement needed

The total additional force needed is the force due to the man, calculated as before. To balance this force, additional volume of water must be displaced. Therefore, use:\[ F_{ ext{additional}} = \rho_{w} \times V_{w} \times g, \]where \( \rho_{w} \) is density of salt water (approximately \( 1025 \, \text{kg/m}^3 \)), and \( V_{w} = A \times h \), where \( h \) is the depth increase.

Step 6: Solve for additional distance the buoy sinks, \( h \)

Re-arrange and substitute for \( h \):\[ 70 \times 9.8 = 1025 \times 0.63617 \times h \times 9.8, \]Cancel \( 9.8 \) and solve for \( h \):\[ h = \frac{70}{1025 \times 0.63617} = 0.108 \, ext{m} = 10.8 \, ext{cm}. \]

Step 7: Conclusion

Thus, the additional distance the buoy will sink is 10.8 cm when the man stands on it.

Key Concepts

Buoyant Force

The buoyant force is a fascinating phenomenon that keeps objects afloat in a fluid. It is an upward force exerted by the fluid, countering the object’s weight.
Understanding buoyant force is essential in explaining why some objects float while others sink. According to Archimedes' Principle, the buoyant force on a submerged object is equal to the weight of the fluid displaced by the object.
In the context of our problem, the weight of the buoy originally matched the buoyant force of the displaced water, keeping it afloat. When the man stood on the buoy, his weight added to the system required an increased buoyant force to maintain floatation. To achieve this, the buoy had to displace more water, causing it to sink further into the fluid. This sinking increased the volume of displaced water, generating the necessary additional buoyant force to balance the combined weight.

Displacement

Displacement refers to the amount of fluid displaced by an object when it is submerged. In simpler terms, it's the volume of water that is "pushed aside" by the object. Displacement is what governs the buoyant force, as per Archimedes' Principle.
For the buoy in our exercise, the original volume of water displaced was enough to support the buoy's weight. However, when the man stepped onto the buoy, additional displacement was required to generate the extra buoyant force needed to support his weight.

• This additional displacement is represented by the increased distance the buoy sinks into the water.
• The deeper the buoy goes, the more water it displaces, consequently creating a larger buoyant force.

The relationship between displacement and the buoyant force is key to understanding how objects interact with fluids.

Salt Water Density

Salt water density is a crucial component in understanding buoyancy in marine environments. Density is defined as mass per unit volume and significantly influences the buoyant force.
For salt water, the typical density is around 1025 kg/m³. This is slightly higher than fresh water, due to the dissolved salts, making objects slightly more buoyant in salt water than in fresh water.

• In our problem, this higher density means that less displacement is required to provide the additional buoyant force needed for the buoy and the man.
• Slight differences in fluid density can significantly affect the calculations involved in buoyancy problems.

Therefore, knowing the density of the fluid is essential to accurately predict how deep an object will sink when additional weight is applied. Understanding these principles helps comprehend why boats float and how they can carry additional loads on different bodies of water.