Chapter Notes on Buoyancy Concepts and Problem Solving

Class focus on the topic of buoyancy.
Previous session featured a brief video explaining buoyancy and hydrostatic force.
Important to solve problems related to buoyancy in the current session.
Definition of Buoyancy:
- Buoyancy is a force experienced by an object submerged in a fluid, affecting its position based on visible portion above the fluid's surface and total submersion.
- Two scenarios: an object is fully submerged or partially floats at the surface.
Buoyant force is defined as the force acting vertically upward through the centroid of the displaced volume of fluid.
Key formula for buoyancy:
- The summation of forces in the vertical direction (y-axis) equals zero (
- This assumes the object is at rest in the fluid with no external forces involved.
Neutral Buoyancy:
- Occurs when a body's average specific weight equals that of the surrounding fluid, allowing it to remain submerged without rising or sinking.

Example discussed using ice as a demonstration of neutral buoyancy in a glass of water.
An object that shares density with water will demonstrate neutral buoyancy.
Problem: Container's dimensions are 600 mm x 350 mm x 900 mm, with unit weight of steel as 77 kN/m³.
Weight of the container is 500 N, and weight of a steel block is 200 N.

Analysis of forces acting on the system to determine buoyant force:
- Sum of forces in y-direction is set to zero: (buoyant force) - (weight of container) - (weight of steel block) = 0.
- Known values: weight of container = 500 N; weight of steel = 200 N.
Buoyancy Force Formula:
- $F_b = ext{(unit weight of fluid)} imes V_d$ where $V_d$ is the volume displaced.
- Volume displaced can be calculated as: $V_d = ext{Area} imes ext{Height}$ .
- Given measures: $0.6 ext{m} imes 0.9 ext{m} imes d$ (d as depth).

Calculation with specific weights:
- Unit weight of water is $9.81 ext{ kN/m}^3$ .
- Substitute values to find d in meters: $d = 0.13214 ext{ m}$ , which converts to mm as 132.14 mm.

For letter b, block positioned directly beneath the container affects the forces:
- Re-evaluation of sum of forces along y for the system:
- Buoyant force due to both container and steel block is considered in calculation.
- Maintaining upward force as positive; initial forces reiterated: upward buoyant force - weight of container - weight of steel block = 0.
Calculate total buoyant forces acting on the container and the steel block respectively.

Volume of steel block determined by:
- ext{Volume} = rac{ ext{Weight of steel block}}{ ext{Unit weight of steel}} ,
- Use known values for computations leading to results in meters.
- Final displaced height results calculated to be 127.33 mm.

Next example begins involving a glass of 50 mm in diameter, filled to a specified water level:
- The weight of ice placed within the glass introduces new calculations.
- Key relationships discussed: buoyant force equals weight of the object, where buoyant force equals unit weight of liquid times volume displaced, aligning with previous principles.
- Unit weight formula for water: $ext{Unit weight} = ext{density of water} imes g$ ,
- Reiterate the density of water = 1000 kg/m³ and density of ice = 920 kg/m³.

Calculation of area as related to volume of displaced liquid: use circle area formula, employing diameter measurements.
Introduce final displacement values leading to d being calculated as 7.32 mm, leading to a total water height of 107.32 mm.

Activity presented to the class: Solve for a wood cube with dimensions given, alongside a specific weight parameter.
Floatation principle reiterated concerning exposed area and resultant calculations.

Instructor availability for queries via messaging platforms or group chat for collective clarification.
Expressed gratitude towards students for participation and attention.