Depth to which the sphere is submerged

[haruspex]

It would be reasonable that the amount submerged should also be 2/5 th the radius

No, 2/5 of the volume, so nearly half the diameter.

As with most problems where you are given numerical values, it is better to set those aside and work purely algebraically. There are many advantages, including for other readers.
In the present case, you would get
$\frac 43\pi r^3\rho_m=\frac{\pi h^2}3(3r-h)\rho_w$
$4r^3\frac{\rho_m}{\rho_w}=3rh^2-h^3$
The cancellation will simplify the arithmetic and tend to improve the numerical accuracy of your answer.
At this point you can do the sanity check for the easy cases $\rho_m=\rho_w$ and $2\rho_m=\rho_w$.

 

[haruspex]

I think you are slightly evil...shall we put in viscosity next? This is a very nice problem actually. I always wanted to structure a course with one continually unfolding problem...for me it is a great way to learn.

A tougher set of problems concerns stable 'postures'. For a uniform solid cylinder radius r, length h, in which orientations can it float?

Edit: responses to that challenge moved to https://www.physicsforums.com/threads/stability-of-a-floating-cylinder.992553/#post-6380998

 

[Mike_A]

Homework Statement:: A solid sphere of volume 20 cm and uniform density $\rho_B$ = 400 $\text{kg m}^{-3}$ floats in a vessel containing pure water. Calculate the depth to which the sphere is submerged.
Relevant Equations:: 1. $\text{Law of floatation}$ : When a body floats in a liquid, the weight of the body is equal to that of the weight of the liquid displaced : $w_B = \Delta w_L$.
2. The volume of a spherical body of radius $r_B$ is given by : $V_B = \frac{4}{3}\pi r_B^3$.
3. The weight of a body is given by : $w_B = m_B g = \rho_B V_B g$ where $\rho_B$ is the density of the body and $V_B,m_B$ are the volume and mass of the body respectively.

View attachment 267778I draw a sketch of the diagram to the right.
The body has a radius of $r_B = 0.2$ m.
The volume of the body is given by $V_B = \frac{4}{3}\times \pi \times 0.2^3 = 0.034\;\text{m}^3$.
The mass of the body : $w_B = \rho_B V_B = 400\times 0.034 = 13.37\,\text{kg}$.
Hence the mass of the liquid displaced : $\Delta m_L = 13.37\, \text{kg}$ . (Law of floatation can also be "read off" as $m_B = \Delta m_L$)
The volume of liquid displaced : $\Delta V_L = \frac{\Delta m_L}{\rho_L} = \frac{13.37}{1000} = 1.34\times 10^{-3}\,\text{m}^3$.
This is the volume of the sphere below the blue line inside the water, as shown to the right.

Question is, how to connect the volume of liquid displaced $\mathbf{\Delta V_L}$ to the volume of the body $\mathbf{V_B}$ ?

I could proceed no further. Any help would be appreciated.
Here is the solution,

Homework Statement:: A solid sphere of volume 20 cm and uniform density $\rho_B$ = 400 $\text{kg m}^{-3}$ floats in a vessel containing pure water. Calculate the depth to which the sphere is submerged.
Relevant Equations:: 1. $\text{Law of floatation}$ : When a body floats in a liquid, the weight of the body is equal to that of the weight of the liquid displaced : $w_B = \Delta w_L$.
2. The volume of a spherical body of radius $r_B$ is given by : $V_B = \frac{4}{3}\pi r_B^3$.
3. The weight of a body is given by : $w_B = m_B g = \rho_B V_B g$ where $\rho_B$ is the density of the body and $V_B,m_B$ are the volume and mass of the body respectively.

View attachment 267778I draw a sketch of the diagram to the right.
The body has a radius of $r_B = 0.2$ m.
The volume of the body is given by $V_B = \frac{4}{3}\times \pi \times 0.2^3 = 0.034\;\text{m}^3$.
The mass of the body : $w_B = \rho_B V_B = 400\times 0.034 = 13.37\,\text{kg}$.
Hence the mass of the liquid displaced : $\Delta m_L = 13.37\, \text{kg}$ . (Law of floatation can also be "read off" as $m_B = \Delta m_L$)
The volume of liquid displaced : $\Delta V_L = \frac{\Delta m_L}{\rho_L} = \frac{13.37}{1000} = 1.34\times 10^{-3}\,\text{m}^3$.
This is the volume of the sphere below the blue line inside the water, as shown to the right.

Question is, how to connect the volume of liquid displaced $\mathbf{\Delta V_L}$ to the volume of the body $\mathbf{V_B}$ ?

I could proceed no further. Any help would be appreciated.
Here is the solution.

 

[Charles Link]

One thing that might be helpful is if you can follow the calculation of post 28. Then you need to subtract out the volume of a cone whose altitude is $h=R-d$, and whose $r^2$ of the base is $r^2=R^2-(R-d)^2$. You should know the volume of a cone as $V=\pi r^2 h/3$. This subtraction gives the volume of the submerged part as a function of $d$. See also post 46. This may be an easier approach than the solution you posted in post 53.

 

[haruspex]

@Mike_A :
Since the problem is already solved in the thread, I assume you have posted this as a model solution.
I am forever having to drum into students to resist the urge to plug in numbers straight away. Much better to keep everything symbolic as long as possible. There are many advantages.
In that spirit, I would write
$W_{sphere}=\rho_s\frac{4\pi r^3}3g$
$V_{cap}=\frac{\pi r^3}3(2+x)(1-x)^2$ where $x=\cos(\theta)$.
$\rho_s\frac{4\pi r^3}3g=\rho_w\frac{\pi r^3}3(2+x)(1-x)^2g$
Then cancel, etc.
Having obtained $3x=0.4+x^3$, N-R is overkill. It should be clear that x is small, so its cube very small. We can do one simpler iteration to be sure:
$x=0.4/3=0.13333...$
$x=(0.4+0.1333...^3)/3=0.1341...$. Time to stop already.
Finally, we don't need to find the angle, something else students routinely waste time, and maybe lose precision, doing. We only want its cosine, and we have that.

 

[Charles Link]

Just one comment is that they don't derive the formula they use for the submerged volume (in the solution posted in post 53). By substituting $d=R(1-x)$ where $x=\cos{\theta}$, the formula agrees with what was previously derived in this thread. Meanwhile, a very good analysis by @haruspex in post 55 of the solution posted in post 53. :)