Section 15.4 BuoyancyA 2.0 cm × 2.0 cm × 6.0 cm block

Step 1 of 2

We have to find the block’s mass density, if a 2.0 cm × 2.0 cm × 6.0 cm block floats in water with its long axis vertical and the length of the block above water is 2.0 cm.

The buoyant force on the cylinder is given by Archimedes principle. The cylinder is in static equilibrium so that the buoyant force is the equal to the weight of the displaced water.

\(F_{B}=F_{G}\)

Where,

\(F_{B} =\) Buoyant force \(=\rho_{w} V_{w} g\)

\(F_{G} =\) weight of the displaced water

\(=m g=\rho_{c y l} V_{c y l} g\)

\(\rho_{w} =\) density of water \(=1000 \mathrm{~kg} / \mathrm{m}^{3}\)

\(V_{w} =\) Volume of water displaced

\( \quad \text { in } \mathrm{m}^{3}\)

\(\rho_{c y} =\) density of block's mass in \(\mathrm{kg} / \mathrm{m}^{3}\)

\(V_{c y l} =\) Volume of immersed block in \(\mathrm{m}^{3}\)

\(g =9.80 \mathrm{~m} / \mathrm{s}^{2}\)