Volumetry

An additional factor that must be considered when considering whether or not the fat lady can float is her volume. Although the fat lady will not completely sink underwater, she will partially sink, especially depending on her volume.

For simplicity's sake, assume that the fat lady is comprised of 100% body fat. This would make the density of her body equal to 920kg/m^3. Make it also given that the fat lady's mass is 226.8 kilograms, her volume is .09m^3, and that her surface area is .15m/s^2. The first part of the problem is to determine how deep the fat lady will partially sink. The second part of the problem to determine how deep the fat lady will partially sink if her volume is increased to .11m^3 in comparison to original volume expressed in percents.

Here goes:

B=w p(fluid)Vg=p(object)Vg V/A=h V=Ah p(fluid)Ahg=p(object)Vg The g's cancel out and the equation left is:

p(fluid)Ah=p(object)V (1000kg/m^3)(.15m^2)h=(920kg/m^3)(.09m^3) h=(920)(.09)/(1000)(.15) h=.552m

From this, the conclusion can be made that the fat lady will sink .552m.

The same equation will be used, but the value of the volume will be changed. p(fluid)Ah=p(object)V (1000)(.15)h=(920)(.11) h=(920)(.11)/(1000)(.15) h=.675m

From this, the conclusion can be made that the fat lady will sink .675m.

In order to determine the percentage of the fat lady that is submerged in the water, her posterior to anterior height must be calculated under both conditions.

Her height will be equal to the quotient of her volume and surface area.

Originally, her height was (.09m^3)/(.15m^2)= .6m. Therefore, the percentage of her body that was submerged was .552m/.6m= 92%.

When her volume was increased, her height was (.11m^3)/(.15m^2)=.733m. Therefore, the percentage of her body that was submerged was .675m/.733m=92.1%

In conclusion, when the fat lady's volume was increased, she sunk a little bit further into the water percentage-wise.