Question

The diameter of the Styrox ball is 38 cm and the density is 84 kg / m³. How much power
needed to keep the ball completely under the water (1000 kg / m³)?

Answer 1

Given:

$d = 38~\text{cm} = 0.38~\text{m}; \rho = 84~\frac{\text{kg}}{\text{m}^3}; {\rho}_w = 1000~\frac{\text{kg}}{\text{m}^3}.$

In order to keep the ball completely under the water, the applied force should compensate the expelling force of the water (according to the Archimedes' principle), taking into account the gravity weight of the ball:

$F = {{\rho}_w}Vg - mg = {{\rho}_w}Vg - {\rho}Vg = ({{\rho}_w} - {\rho})Vg = ({{\rho}_w} - {\rho})\frac{4}{3}{\pi}{r^3}g = ({{\rho}_w} - {\rho})\frac{1}{6}{\pi}{d^3}g = \frac{1}{6}~*~(1000 - 84)~\frac{\text{kg}}{\text{m}^3}~*~{\pi}~*~(0.38~\text{m})^3~*~9.81~\frac{\text{m}}{\text{s}^2} \approx 258.17~\text{N}.$