Question

A 950-kg cylindrical can buoy floats vertically in seawater. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

Short answer

The buoy will sink an additional 0.126 m.

Step-by-step solution

Determine the Buoyant Force

The buoyant force is equal to the weight of the displaced water. Initially, the weight of the water displaced is equal to the weight of the buoy, which is given by \( W = mg = 950 \text{ kg} \times 9.81 \text{ m/s}^2 \).

Calculate Initial Buoyancy

The initial buoyancy is equal to the initial weight of the displaced water, which is \( 950 \times 9.81 = 9319.5 \text{ N} \).

Calculate Additional Weight

Find the additional force due to the man's weight, \( W_m = 80.0 \text{ kg} \times 9.81 \text{ m/s}^2 = 784.8 \text{ N} \).

Determine Total Buoyant Force Required

The total buoyant force needed for the buoy to remain floating is now \( 9319.5 \text{ N} + 784.8 \text{ N} = 10104.3 \text{ N} \).

Find Additional Volume of Displaced Water

The additional volume of water that needs to be displaced for the additional buoyancy is given by \( \frac{784.8}{1000 \text{ kg/m}^3 \times 9.81 \text{ m/s}^2} \). Calculating gives approximately \( 0.08 \text{ m}^3 \).

Calculate Additional Sinking Depth

The additional depth the buoy will sink is the additional volume of water displaced divided by the cross-sectional area of the buoy. The cross-sectional area is \( \pi \times (0.45)^2 \approx 0.636 \text{ m}^2 \). Dividing the additional volume by this area gives approximately \( \frac{0.08}{0.636} \approx 0.126 \text{ m} \).

Key concepts

Archimedes' Principle

Archimedes' Principle is a fundamental concept in fluid mechanics that explains why objects float or sink when placed in a fluid. It states that any object, completely or partially submerged in a fluid, experiences an upward force known as the buoyant force. This force is equal to the weight of the fluid that the object displaces.
Understanding Archimedes' Principle is critical for solving problems related to buoyancy, such as determining how much of an object will float above the fluid's surface or how much additional weight an object can support before sinking further.
Archimedes' Principle can be expressed mathematically as: \( F_b = \rho_{fluid} \times V_{displaced} \times g \), where:

• \( F_b \) is the buoyant force
• \( \rho_{fluid} \) is the fluid's density
• \( V_{displaced} \) is the volume of fluid displaced
• \( g \) is the acceleration due to gravity

This principle not only explains why ships made of steel can float but also details how submarines sink by displacing different volumes of water based on weight adjustments.

Cylindrical Buoy

A cylindrical buoy serves as a practical example of applying Archimedes’ Principle in real-world situations. Such buoys are often used for navigational markers on water bodies. They are designed to float vertically with a significant part submerged, maintaining stability.
The buoy discussed in the exercise has a diameter of 0.900 meters and a known weight, making it easy to calculate its buoyancy characteristics. For a cylindrical object, the key to calculating such characteristics lies in its geometry. The cross-sectional area, necessary to calculate the buoyancy, is determined using the formula for the area of a circle: \( A = \pi r^2 \), where \( r = \text{diameter}/2 \). Here, this helps us find how much water needs to be displaced when additional weight, like a person standing on it, is applied.
Understanding the dimensions and weight of a buoy allows us to assess how much of it will be submerged under different conditions, maintaining it as an efficient floating device.

Displacement of Water

Displacement of water is a key component of the buoyancy concept. When an object is placed in water, it pushes aside a volume of water equal to the volume of the portion of the object submerged. This displaced water supports the object via the buoyant force.
For example, when calculating how much further a buoy will sink if a weight is added, we need to determine how much more water needs to be displaced. If an 80 kg person stands on top of the buoy, the water displaced must weigh as much as the additional load (80 kg).
The volume of water displaced can be determined using the relation: \( V_{displaced} = \frac{F_{additional}}{\rho_{water} \times g} \), where:

• \( V_{displaced} \) is the additional volume of water displaced
• \( F_{additional} \) is the additional weight force
• \( \rho_{water} \) is the density of water, approximately \( 1000 \text{ kg/m}^3 \)
• \( g \) is the gravitational acceleration

The depth the buoy sinks can be calculated by dividing this volume by the buoy's cross-sectional area. This ensures the buoy remains afloat and helps in predicting its change in the position due to added weight.