Q: Equation of motion for ice block in water I need to derive the equation of motion for an building block of ice that swims in water. The building block has an initial height that is looking out of the water called $h_A$ and is pushed into the water by a force $F$ such that it isn't completely under water, so $\Delta z < h_A$. Now I need to derive the equation of motion for the process when the force disappears and the building block goes up.  
What I tried to write down is the force that pushes the block up:  
$m \frac{d^2z}{dt^2} = F_G - F_A = mg - gA\big[l-(h_A-z)\big] \rho_w$  
with $ A $ the area of the block and $\rho_w$ the density of the water.  
To me this equation looks kind of hard to solve so I wanted to ask if that could be right. 

A: Named $z_0$ the equilibrium position of the block, since the force $F$ pushes down the block to $z_0 - \Delta z$. 
When the block is released it will raises 'til $z_0 + \Delta z$, then keeps on oscillatinig with a total amplitude of $2\Delta z$ (in abscence of drags).
So the block start oscillating around the equilibrium position, with a maximum in $z_0 + \Delta z$ and a minimum in $z_0 - \Delta z$, according to the simple harmonic oscillator equation .
In case you have drags the block will follow the damping oscillator equations.
The only problem remained is to calculate the pulsation $\omega$, but it is trivial calculate it from boundary conditions of your problem

A: So far you've got a second order Differential Equation (DE):
$$m \frac{d^2z}{dt^2} = F_G - F_A = mg - gA\big[l-(h_A-z)\big] \rho_w$$
But it looks worse than it is: make the following substitution:
$$u=mg - gA\big[l-(h_A-z)\big] \rho_w,$$
and because it's mostly constant terms, then:
$$du=-gA\rho_w dz$$
and:
$$\frac{d^2z}{dt^2}=-\frac{1}{gA\rho_w}\frac{d^2u}{dt^2}$$
Substitute back into the DE:
$$-\frac{m}{gA\rho_w}\frac{d^2u}{dt^2}-u=0$$
Or:
$$m\frac{d^2u}{dt^2}+gA\rho_wu=0$$
This is the classic DE of a Simple Harmonic Oscillator. Assuming at $t=0$, $u=u_0$, $\frac{du}{dt}=0$ then the solution is:
$$u=u_0\cos \omega t,$$
with:
$$\large{\omega=\sqrt{\frac{gA\rho_w}{m}}}$$
So ignoring friction the block will bob up and down with period $T=\frac{2\pi}{\omega}$.
You can transform $u$ back to $z$ and define a $z_0$ if you like.
