Question

Someone with a force of 900 N does not float in a freshwater pool. To prevent them from sinking, 20 N must be applied in an upward fashion. What is their volume and average density?

Answer 1

Explanation:

The buoyant force $F_B$ is defined as

$F_B = \rho_wgV$

where $\rho_w$ is the density of the displaced fluid (freshwater), g is the acceleration due to gravity and V is the volume of the submerged object. In the case of freshwater, its density is $997\:\text{kg/m}^3.$ Since the buoyant force is 20 N, we can solve for the volume of the displaced fluid:

$F_B = \rho_wgV \Rightarrow V = \dfrac{F_B}{\rho_wg}$

Plugging in the values, we get

$V = \dfrac{20\:\text{N}}{(997\:\text{kg/m}^3)(9.8\:\text{m/s}^2)}$

$\:\:\:\:\:= 2.05×10^{-3}\:\text{m}^3$

Recall that the weight of an object in terms of its density and volume is given by

$W = \rho gV$

Using the value for the volume above, we can solve for the density of the object as follows:

$\rho = \dfrac{W}{gV} = \dfrac{900\:\text{N}}{(9.8\:\text{m/s}^2)(2.05×10^{-3}\:\text{m}^3)}$

$\:\:\:\:\:= 44,798\:\text{kg/m}^3$