Calculate Average Density of a Person Hanging in Mid-Air

[faoltaem]

Homework Statement

a person sits in a harness hamging in mid-air from a set of scales. the scales display her weight as 550.0 N. she then empties her lungs as much as possible and is completely immersed in water. the scales now give her weight as 21.2 N. what is her average density?

Homework Equations

$$\rho$$ = $$\frac{m}{v}$$
W = mg
$$\rho$$$$_{water}$$ = 1000 kg/m$$^{3}$$

The Attempt at a Solution

m$$_{above}$$ = $$\frac{550}{9.81}$$ = 56kg
i'm not really sure what to do from here:
m$$_{below}$$ = $$\frac{21.2}{9.81}$$ = 2.16kg
v = $$\frac{m_{below}}{\rho_{water}}$$ = $$\frac{2.16}{1000}$$ = 0.00216 m$$^{3}$$
$$\rho$$$$_{person}$$ = $$\frac{m_{above}}{v}$$ = $$\frac{56}{0.00210}$$ = 26666 kg/m$$^{3}$$

But i know this is wrong cause the answer is 1040 kg/m$$^{3}$$

 

[Kurdt]

Her mass under water is her mass above minus the mass of the displaced water. So you need to use the mass of displaced water to work out her volume not her under water weight.

 

[faoltaem]

thanks, that was really helpful