Question

What is the density (in kg/m3) of a woman who floats in freshwater with 4.32% of her volume above the surface? (This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water.) (Give your answer to at least three significant figures.) What percent of her volume is above the surface when she floats in seawater?

Answer 1

In equilibrium Fnet = 0

Fb - Fg = 0

Fb = buoyancy force = Dwater*Vimmersed*g

Fg = m*g = Dwoman*Vwoman*g

Vimmersed = (1-0.0432)*vwoman = 0.9568*Vwoman

1000*0.9568*Vwoman*g - Dwoman*Vwoman*g = 0

Dwoma = 1000*0.9568 = 956.8 kg/m^3

=================

In sea water

Dwater = 1027 kg /m^3

1027*Vimmeresed*g - 956.8*Vwoman*g = 0

Vimmersed = 956.8/1027*Vwoman = 0.9316*voman m^3

Vabove = Vwoman - Vimmersed = (1 - 0.9316)*voman = 0.0684*voman

percent = (Vabove/Vwoman)*100 = 6.84%