A 60-kg woman with a density of 980 kg/m^3 stands on a bathroom scale. Determine the reduction of the scale reading due to air. (The answer is 0.774 N)
The woman displaces a volume of air equivalent to her own volume; VV; V=\dfrac{m}{\rho_{\text{woman}}}=\dfrac{60}{980}=0.0612\mathrm{\ m^3}V= ρ woman m = 98060 =0.0612 m 3
