# Confusion about the buoyant force applied by fluids

EDIT

I found that a similar question was asked by user Muno in this question in a follow up comment which is as follows (in short this is what I'm asking for)

to make my question more succinct: if buoyant force depends on a difference in pressure, and pressure at a particular depth depends upon the weight above it, why isn't a submerged object's weight factored into pressure?

I have heard of the reason that water applies buoyant force due to a gradient of pressure. But why does it arise?

Consider the following (the cause of my problem)

When analyzing the situation it's said that the object feels a force which is equal to the weight of water that it displaces. But I'm a bit (or say too much) confused on this too. It's as follows :

The water above the object (say at a depth $$h_a$$) is applying a force equal to its weight which is $$\pi r^2 h \rho _{water}$$. Now consider the lower portion, the object and the water column above are applying a force equal to

$$\pi r^2 h \rho _\text{water} g+W_\text{object}$$

but the water column below is applying a force equals to

$$-(\pi r^2 h \rho _\text{water} g +W_\text{object})$$ (via Newton's third law)

therefore the net force on the object is $$-W_\text{object}$$. So why is this not the case?

• "itis said" Archimedes said over 2000 years ago. Nov 22 '19 at 8:38

The water above applies pressure according to it's depth, the water below always has greater depth, so applies greater pressure. If the cylinder's weight is less than the difference in the upper and lower pressures, then it will be pushed upwards. In other words if the cylinder weighs less than the same volume of water, it's displacement, it will float up out of the water, until the portion of it below the water level displaces it's weight.

but the water column below is applying a force equals to $$-(\pi r^2 h \rho _{water} g +W_{object})$$ (via Newton's third law)

This isn't true. Unlike a solid surface which would have applied an equal and opposite reaction under equilibrium, for incompressible fluids the situation is different as they displace and deform.* The displaced liquid tends to push back. To account for the fluid nature, we use pressure to evaluate forces on submerged surfaces. The pressure at bottom surface is $$\rho_{water}g(h+h_{cylinder})$$. This gives a net force on object $$-\rho_{water} g \pi r^2 h_{cylinder}$$ which is the weight of the displaced water i.e. the Archimedes principle.

Buoyant force is a contact or "reaction" force. Microscopically it has the same origin as all contact forces

• the repulsion between molecules which are squashed together.

A solid cannot get out of the way because it does not flow and hence when a force is applied on it it gets compressed and a restoring force is generated, which increases in magnitude till it becomes equal to the force applied by the object. Whereas for the fluid they simply get out of the way rather than getting compressed (as their molecules are free to move) hence you cannot just say that the force applied by fluids equals the weight above it.

When you lower an object into a liquid, the reaction force increase because as the depth increases so does the speed with which the molecules bump into the object. But after the object is fully submerged molecules of water start pushing the object downward and in due process cancelling the effect of increasing upward force. So after that the buoyant force remains constant.

At each stage the reaction force from the fluid on the block ("upthrust") equals the force which the block exerts on the fluid. If the "upthrust" at the current depth or deformation does not equal the weight of the block, there is an unbalanced force on the block, which moves further down.

$$\underline{\text{An Intuitive Way to Understand Archimedes' Principle}}$$ Consider the following figure.

It is just a container filled with water (density $$\rho_w$$) and I have marked a closed surface $$S$$ inside it of arbitrary shape enclosing volume $$V$$. I haven't introduced any external object yet. Now, we know that the water outside the enclosed volume exerts pressure on the water inside. The exact pressure distribution on the surface is complicated. But, we do know one thing for sure : the water inside is at equilibrium. This implies the following ($$d\mathbf{A}=dA \hat{n}$$, where $$\hat{n}$$ is the inward pointing normal unit vector),

$$\oint_S p d\mathbf{A}=-\rho_w Vg \hat{k} \tag{1}$$

Upon considering the special case of a cylinder to be the closed surface $$S$$ (so that the pressure distribution on the sides of the cylinder are symmetrical and cancel out), we can find that the pressure distribution varies linearly with height.

Now, replace the water enclosed by $$S$$ by an object of the exact size and shape. Will the water molecules outside the enclosed volume change their behavior and apply a different pressure distribution simply because you changed the water inside with a different material? No.

Therefore, the force exerted by water on the object will still be equal to $$-\rho_w V g \hat{k}$$.

$$\underline{\text{Where you went wrong (possibly)}}$$ I believe you're assuming a cylinder as your object. The force exerted on the bottom of the cylinder is not $$\pi r^2 h \rho_w + W_{object}$$, but instead $$\pi r^2 (h+h_{cyl}) \rho_w$$. This is also the force exerted by the cylinder on the bottom portion of the water ($$\pi r^2 (h+h_{cyl}) \rho_w$$ is the action-reaction pair). In short, I believe you have made a mistake in identifying the action-reaction pair.

This is kind of similar to what happens when you place a block of mass $$M$$ on top of a vertical spring (with spring constant $$k$$) attached to the ground that's unstretched/uncompressed. When you place it, does the reaction force on the block due to the spring immediately become $$Mg$$ and the block just sits still? No, it oscillates, right? The action-reaction pair in this case is $$kx$$.

The part where you have first made a mistake is when you said that the force the body exerts on the water below is the pressure on the upper part of the object plus the weight of the object. This is not true, because a rigid solid object does not build a pressure gradient similarly as a liquid does. Therefore the pressure at the bottom part of the object is determined actually just by the height of the water (because pressure disperses also horizontally, so the pressure must be the same as next to the solid object, where only a water column exists above the level of interest) and the solid object does not affect the pressure.