Question

Four cubes of the same volume are made of different materials: lead (density 11,300 kg/m3), aluminum (density 2700 kg/m3), wood (density 800 kg/m3), and Styrofoam (density 50 kg/m3). You place the cubes in a large container filled with water.Rank the buoyant forces that the water exerts on the cubes from largest to smallest.Rank from largest to smallest. To rank items as equivalent, overlap them.

Answer 1

Lead > Aluminium > Wood > Styrofoam

Step-by-step explanation:

Buoyant force is described by the Archimedes's Principle, which states that buoyant force is equal to the weight of the fluid displaced by the submerged object. By Newton's Laws, the buoyant force is represented by the following equation of equilibrium:

`\Sigma F = F_(D) - W_(cube) = 0`

`F_(D) = W_(cube)`

`F_(D) = \rho_(cube) \cdot g \cdot V_(cube)`

Where:

`\rho_(cube)` - Density of the cube, measured in kilograms per cubic meter.

`g` - Gravitational constant, measured in meters per square second.

`V_(cube)` - Volume of the cube, measured in cubic meters.

Let suppose that volume of the cube is known. Given that
`g = 9.807\,(m)/(s^(2))`, the buoyant force is computed for each material:

Lead (
`\rho_(cube) = 11,300\,(kg)/(m^(3))`)

`F_(D) = \left(11,300\,(kg)/(m^(3)) \right)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot V_(cube)`

`F_(D) = 110,819.1V_(cube)`

Aluminium (
`\rho_(cube) = 2,700\,(kg)/(m^(3))`)

`F_(D) = \left(2,700\,(kg)/(m^(3)) \right)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot V_(cube)`

`F_(D) = 26478.9V_(cube)`

Wood (
`\rho_(cube) = 800\,(kg)/(m^(3))`)

`F_(D) = \left(800\,(kg)/(m^(3)) \right)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot V_(cube)`

`F_(D) = 7845.6V_(cube)`

Styrofoam (
`\rho_(cube) = 50\,(kg)/(m^(3))`)

`F_(D) = \left(50\,(kg)/(m^(3)) \right)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot V_(cube)`

`F_(D) = 490.35V_(cube)`

Therefore, the buoyant forces that the water exerts on the cubes from largest to smallest corresponds to: Lead > Aluminium > Wood > Styrofoam.