Arquimedes Principle?

Question

You have an homogeneous spherical body with a radius of 5.5 cm and density of 0.6. Calculate (and please present a step-by-step solution) the dept of the part imersed in water. (Which will be smaller than de radius)

Please help

Answer 1

I assume that the density of water is 1 gr/cm^3 or w = 1

The volume of a sphere is Vol(sphere)= 4/3 * pi * r^3

The volume of a spherical cap is Vol(cap) = 1/3 * pi * h^2 * (3 * r - h)

where h is the "height" of the spherical cap.

Check: if h = r, or half a sphere Vol(cap) = 2/3 * pi * r^3

if d is the density of the sphere, and w is the density of water

then the sphere will be in equilibrium when

Vol(sphere) * d = 4/3 * pi * r^3 * d = w * (1/3 * pi * h^2 * (3 * r - h))

find the solution for h:

4 * r^3 * d / w =  h^2 * (3 * r - h)

h^3 - 3 * r * h^2 + 4 * r^3 * d /w = 0

with numerical values:

h^3 - 16.5 h^2 + 399. 3 = 0

h =  6.23775815137495 cm, the other roots are meaningless.

Note: x and h are the same thing.

Check: try with d = 1, then h = 11 or 2 * r which is correct since the sphere in completely under water.

Answer 2

The weight of the sphere will be balanced by the boyancy force

W = mg = ρVg = ρg(4/3)πr³

The boyance force is the weight of the displaced fluid, ρl is the density of the liquid.

F = ρlVg

you need to find the volume of the sphere under water:

V = (π/3)x²(3r - x)

so

F = ρlg(π/3)x²(3r - x)

F = W

ρlg(π/3)x²(3r - x) = ρg(4/3)πr³

ρl/ρ (3r - x)x² = 4r³

ρl/(0.6) (3*5.5 - x)x² = 4(5.5)³

ρl (16.5 - x)x² = 399.3

-x³ + 16.5x² - 399.3/ρl = 0

solve ofr x using the value of ρl