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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?

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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).

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if the wood doesn't sink in water therefore $A=ρgV=0$

As Aaron Stevens pointed out in their excellent answer, the assumption you made is not correct. The truth is that the piece of wood is not fully submerged in water, but rather it is partially submerged. Take a look at the following diagram:

enter image description here

One can find the exact relationship between the volume of wood submerged, $V_s$ and the whole volume of wood, $V$ by using Archimedes' principle:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid – Wikipedia (emphasis mine)

Assuming uniform cross-sectional area, one can consequently infer the relationship between the whole height of the piece of wood, $H$ and the height of the portion submerged in water, $h$.

$$\vec{F_B}+\vec{F_W}=0\implies \rho_l V_sg=mg\implies \rho_l V_s=\rho V$$

Where $\rho_l$ and $\rho$ are the densities of the liquid and of the wood respectively. Carrying on with the above reasoning:

$$\dfrac{V_s}{V}=\dfrac{\rho}{\rho_l}\stackrel{S=\text{ct.}}{\implies} \dfrac{Sh}{SH}=\dfrac{\rho}{\rho_l}\implies \boxed{\dfrac{h}{H}=\dfrac{\rho}{\rho_l}}\tag{1}$$

Conclusion. For wood in particular (ignoring surface tension forces), from eq. $(1)$ we can clearly see that $h=H\dfrac{\rho}{\rho_l}$. The density of most types of wood varies between $0.4\frac{g}{cm^3}$ and $0.8\frac{g}{cm^3}$ (there are exceptions), which is clearly less than the density of water, $1\frac{g}{cm^3}$. Consequently, $h$ may be even less than half of $H$. In reality, the surface tension forces also decrease $h$ further, plus there are some optical effects like refraction that, especially for thin pieces of wood, may have given you the false impression that the wood is fully outside the water. But it's not the case.

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As Archimedes discovered: every body immersed (partially or totally) in a fluid is acted with a (vertical) force that is equal to the weight of the volume of fluid that it displaces. This is the force you look for. It results from the difference in pressure on the body surface.

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  • $\begingroup$ yes but wood doesnt displace any volume of water because it doesnt sink. so which is the force if it is not the buoyant ? $\endgroup$ Oct 21 '18 at 1:56
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    $\begingroup$ For the problem you consider there is no other force than buoyancy force. If the wood do not sinks to the observer eyes it is because its density is too low; then the depth it sinks is ver small, but not zero. $\endgroup$
    – nodarkside
    Oct 21 '18 at 2:16
  • $\begingroup$ Still for the surface tension to have a component in the vertical direction it is necessary for the wood to sink. $\endgroup$
    – nodarkside
    Oct 21 '18 at 2:18
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For better understanding, you can consider the wood as two parts, one that totally sinked into the water and the one does not. In this case, the one which sink in the water will provide the overall force which is counter to the gravitational force.

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