[Physics] Wood in water and buoyant force

Question

Wood doesn't sink in water because its density is less than water density. So $ΣF=0$. But what is the force that counterbalances the gravitational force if the wood does not sink in water so that there is no water displaced and hence no buoyant force?

Answer 1

You are working with the false assumption that the wood does not displace any water. This is not the case. The wood will be somewhat in the water, so there is still a buoyant force. This is the force that counteracts the gravitational force.

To be more exact, let's assume a rectangular block of base area $A$ and density $\rho_b$ in a fluid of density $\rho_f$. Since the block is as rest, we can equate the buoyant force and the weight (i.e. $\sum F=0$, as you said in your question): $$mg=\rho_bV_Tg=F_b=\rho_fV_{sub}g$$ where $V_T$ is the entire volume of the block, and $V_{sub}$ is just the volume of the block that is submerged in the fluid.

Now, we can determine how much of the wood is submerged by defining the total height and submerged height respectively as $H$ and $h$. Therefore $$\rho_bAHg=\rho_fAhg$$ $$h=\frac{\rho_b}{\rho_f}H$$

Since $\rho_b<\rho_f$, we have that $0<h<H$. Therefore, the block must be submerged by some amount less than its total height (the block is not fully submerged).