1. ## Archimedes' principle!

Hey guys, it's me again. I came across a pre-calculus problem in some homework and can't finish it. The problem gives you Archimedes' principle:
[IMG]file:///C:/Users/Andy/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
(π/3)(3rh2 – h3)dL = 4/3πr3ds
Where dL = 62.5lb/ft3

If r = 5 and ds= 20 lb/ft3

What is h = ?

I know the answer should come out approximate... If anyone could explain to me how to reach the answer, I would be very grateful!

2. Originally Posted by zacharyrod
Hey guys, it's me again. I came across a pre-calculus problem in some homework and can't finish it. The problem gives you Archimedes' principle:
[IMG]file:///C:/Users/Andy/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
(π/3)(3rh2 – h3)dL = 4/3πr3ds
Where dL = 62.5lb/ft3

If r = 5 and ds= 20 lb/ft3

What is h = ?

I know the answer should come out approximate... If anyone could explain to me how to reach the answer, I would be very grateful!
1. I assume that you want to calculate the draught of a sphere floating in a liquid.

2. Consequently ds means density of solid and
dL density of liquid.

3. If so: Plug in all given values:

$\frac \pi3 (3 r h^2 - h^3) \cdot d_L=\frac43 \pi r^3\cdot d_s$ $\implies$ $\frac \pi3 (3 \cdot 5 \cdot h^2 - h^3) \cdot 62.5=\frac43 \pi \cdot 5^3\cdot 20$

which simplifies to

$h^3 - 15 h^2+160=0$

Since the density of the liquid is nearly 3 times larger than the density of the solid the value of h must be smaller than the length of the radius.

4. I used Newton's method to get an approximate value for h:

$h_n=h_{n-1}-\dfrac{h_{n-1}^3 - 15h_{n-1}^2 + 160}{3h_{n-1}^2 - 30h_{n-1}}$

Start with $h_0 = 3$.

After 3 steps you'll get the value of h with an accuracy of 9 digits.