Question

A rectangular box is immersed in water. Its upper surface is at a depth of 2 m, and the lower surface is at a depth of 4 m. The area of the lower surface is 10 m². You need to calculate the upthrust (buoyant force) due to the liquid acting on the box.

Answer 1

To calculate the upthrust (buoyant force) acting on the rectangular box, we can use Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the submerged part of the object.

The buoyant force \( F_b \) is given by the formula:

\[
F_b = \rho \cdot g \cdot V
\]

where:
- \( \rho \) is the density of the fluid (for water, this is approximately \( 1000 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( V \) is the volume of the fluid displaced by the submerged part of the object.

To find the volume \( V \), we need to determine the height of the box that is submerged:

- The upper surface of the box is at a depth of 2 m.
- The lower surface of the box is at a depth of 4 m.

The vertical height of the box that is submerged (the height difference between the upper and lower surfaces) is:

\[
\text{Submerged height} = 4 \, \text{m} - 2 \, \text{m} = 2 \, \text{m}
\]

Now, we can calculate the volume of the submerged part of the box, \( V \), using the area of the lower surface:

\[
V = \text{Area} \times \text{Submerged height} = 10 \, \text{m}^2 \times 2 \, \text{m} = 20 \, \text{m}^3
\]

Now we can substitute the known values into the buoyancy formula:

- Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \)
- Volume \( V = 20 \, \text{m}^3 \)

Now we can calculate the buoyant force:

\[
F_b = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 20 \, \text{m}^3
\]

Calculating this:

\[
F_b = 1000 \times 9.81 \times 20 = 196200 \, \text{N}
\]

Thus, the upthrust (buoyant force) acting on the box is approximately:

\[
\boxed{196200 \, \text{N}}
\]