# Confusion regarding a wooden block floating in water in a sealed container [closed]

Consider the following situation: a wooden block floats in water in a sealed container. When the container is at rest, $$\frac14^{th}$$ of the block is above the water.

When the air pressure inside the container above the water is increased, what fraction of the block will be submerged now?

My thinking is that $$P$$ = $$\frac{F}{A}$$, and since $$Pressure$$ increases and $$Area$$ remains constant, $$Force$$ which acts on the block from above would also increase and more than $$\frac34^{th}$$ of the block would be submerged.

But this logic is incorrect. Can someone tell me why this is incorrect and provide a hint/solution to the problem?

When the air pressure increases, the water pressure must increase as well to remain in equilibrium. It has to, otherwise there would be a net downward force on the layer of water at the surface. So you can't assume that the water pressure acting on the bottom of the block remains the same.

To actually solve this problem, consider a cylindrical block for simplicity. Remember that the forces acting on the block are the air pressure on the top face, the water pressure on the bottom face, and its own weight. (There are pressure forces on the sides as well, but they cancel.) Equilibrium will be reached when these forces cancel, or when the difference between the air pressure and the water pressure forces is equal to the weight of the block.

So you need to figure out how the difference in pressure between the top & bottom of the block changes (or if it does.)

I believe the degree to which the block floats above the water depends only on the density of the block relative to the density of the water. Increasing the air pressure above the water has little effect on the density of water (and virtually no effect on the density of the block, a solid) as water is relatively incompressible.

Therefore, increasing the air pressure above the water should have an insignificant effect on the buoyancy of the block, i.e., it should still float with about one fourth above the water.

Hope this helps.

Maybe the difference in pressure is miniscule? Isn't this analogous to exerting force on the block downwards with your own hands, and thus more than 3/4th is submerged? Unless, perhaps, since the pressure surrounding it increases equivalently, the force acting along the sides of the block should cancel each other but the pressure acting downwards still causes the block to be further submerged.

Why would this be wrong?

• No, the pressure change isn't miniscule, and is only experienced above the water, not from the sides of block or such. Thats the reason am confused. Jan 31 at 14:08
• @Haider Why would the pressure in the water not increase? Jan 31 at 15:18

The block would float slightly higher in the water. For normal pressures, it would be hard to measure.

As Michael Seifert says, the increase in air pressure would not matter. But the increase in air density does. At $$1$$ atmosphere, the density of air is about $$1/1000^{th}$$ that of water. So normally you disregard air, treating it like vacuum (unless you want to weigh helium balloons).

Suppose wood and water were truly incompressible and air followed the ideal gas law even at extreme pressure. At $$1000$$ atmospheres, the density of air would be the same as water. The block would float to the top of the chamber.

At lower pressures, you could treat it like a two fluid problem. Suppose the block was floating at the interface between water and a low density oil. How much of the wood would stick up into the oil?

The block would float so the force on the top and bottom surfaces cancelled the force of gravity. The pressure on the top is lowest. In oil, pressure increases relatively slowly with depth because of the low density of oil. It increases faster in water. So the total pressure difference depends on how much of the block is in each.