Question

A piece of metal has a weight of 6.5 N when it is in air and 3.0 N when it is submerged into water. What is the buoyant force on the piece of iron?

Answer 1

Answer:

The buoyant force on the piece of iron is 3.5 newtons.

Explanation:

According to the Archimedes' Principle, the piece of metal experiments a buoyant force equal to the weight of the displaced fluid. By Newton's Laws, we have the following formula:

$F_{B} - W_{P}+W_{F} = 0$ (1)

Where:

$F_{B}$ - Buoyant force on the piece of iron, in newtons.

$W_{P}$ - Weight of the piece of metal, in newtons.

$W_{F}$ - Weight of the fluid displaced by the piece of metal, in newtons.

$F_{B} = W_{P}-W_{F}$

If we know that $W_{P} = 6.5\,N$ and $W_{F} = 3\,N$, then the buoyant force on the piece of iron is:

$F_{B} = 6.5\,N - 3\,N$

$F_{B} = 3.5\,N$

The buoyant force on the piece of iron is 3.5 newtons.