3
$\begingroup$

Suppose I have a wooden block that floats in water and I put it in an empty glass beaker. Then I poured water in the beaker so carefully that no water goes under the block. In that case, would the block float or it will remain under the water?

As there is no water under the block there is no upward thrust on the base. Therefore it seems that the block would not float.

On contrary, there is tangential upward thrust on the sides of the block. Because of this thrust the block will float.

Which is the correct answer?

EDIT

I think the right answer is : the block would not float. Thanks to sammygerbil, anna v, Kieran Moynihan.

This reminds me another question.

A wooden block is floating on water in a air tight container. Is it going to float more or float less in the following cases?

  1. All the air has been pulled out from the container.
  2. We compress the air above the water.

My Thought

Case 1:

When all the air has been pulled out from the container there is no buoyant force due to air, so total downward force acting on the block would increase and the block would float less.

OR,

As there is no air below the block no buyout force due to air. When all the air has been pulled out from the container there is no air pressure, so the block would float more.

Case 2:

When we compress the air, the density of air would increase. So the weight of the displaced air as well as the buoyant force due to air would also increase and the block would float more.

OR,

As there is no air below the block no buyout force due to air. When we compress the air, so the block would float less as air puts more downward pressure on the block.

Now which is correct or incorrect and why?

$\endgroup$
  • 1
    $\begingroup$ What is tangential upward thrust? $\endgroup$ – sammy gerbil Oct 30 '17 at 15:46
  • $\begingroup$ @sammygerbil yes. I realise there is no tangential upward thrust. Initially I thought that on each point of the surface of the sides of the block pressure acts along all possible direction giving a tangential upward thrust. $\endgroup$ – ddas Nov 1 '17 at 7:00
  • $\begingroup$ What about a tangential downward thrust? Is there any reason for it to be only upward? If you have ruled out the tangential upward thrust then there is only one option left in your question. $\endgroup$ – sammy gerbil Nov 1 '17 at 7:20
  • $\begingroup$ yes, you are right. This reminds me another problem. A wooden block is floating on water in a air tight container. Is it going to float more or float less in the following cases? 1. All the air has been pulled out from the container. 2. We compress the air above the water. $\endgroup$ – ddas Nov 1 '17 at 7:30
  • $\begingroup$ In case 1 wouldn't the water evaporate assuming there is a vacuum pump sucking the air out? $\endgroup$ – Communisty Nov 1 '17 at 7:51
2
$\begingroup$

IN ANSWER TO THE ORIGINAL QUESTION

Yes, there must be fluid vertically below some part of the object for it to float, but there need not be fluid below all parts of it. The object must have some surface area which has an inward normal with an upward component, so that the water pressure on it has an upward component. For example, the block could float if the sides slope inwards toward the base, as Kieran Moynihan suggests.

In his answer to No buoyancy inside liquid Chester Miller shows that the resultant buoyant force of fluid pressure ("upthrust") on an object which is touching the bottom of a container is $$B=(V-hA)\rho_wg$$ where $V$ is the volume of the submerged part of the object, $h$ is the depth of fluid and $A$ is the area of contact with the bottom of the container.

If the sides of the block are vertical then $hA \ge V$ for all values of $V$. (The equality applies if the object is not fully submerged.) The formula confirms anna v's observation that no amount of water will make the block float, even if it is far less dense than water.

This formula also shows that even if $A$ is very small you can make $B$ negative (ie the "buoyant" force becomes a downward force) by increasing the depth $h$ of fluid. The surprising consequence is that an object which would otherwise float when free of the bottom of the container (ie when there is fluid below all parts of it) can be made to stay on the bottom if a small part of it is already touching the bottom. For example, a large helium balloon can be tethered to the ground using a light suction cup which is much smaller in area than the cross-section of the balloon.


IN ANSWER TO THE QUESTION IN YOUR EDIT

Intuitively you would think that increasing the pressure of the air increases the downward force on the block, making it sink lower in the water, but this is not the case. In both cases what happens to the block does not depend on the pressure of the air but the pressure gradient in the air. If the pressure is uniform throughout the air space, an increase or decrease has no effect on the depth at which the object floats in the water. But if there is a pressure gradient (which necessarily increases downwards) then an increase in the average pressure makes the object rise up in the water, and a decrease makes it sink lower.

Explanation

The explanation is similar to that in Why does a helium filled ballon move forward in a car when the car is accelerating?

The forces on the block are initially balanced. The vertical forces are the weight $W$ of the block and the pressure-forces $F_1$ of the air on the upper face and $F_2$ of the water on the lower face of the block : $$F_2=W+F_1$$ To avoid complications I assume that the block is cuboid so that the areas of upper and lower faces are equal.

Suppose the air pressure is constant throughout the upper part of the container. Then an increase in air pressure increases the forces $F_1, F_2$ equally, so the depth at which the block floats in the water does not change. The increase in pressure at the upper face is transmitted through the air and water to the lower face, increasing it by the same amount.

The air pressure would be approximately constant throughout its volume if the air is only slightly compressible and its density is low compared with that of the water. Both these conditions usually apply at typical atmospheric pressures.

However, if there is a significant pressure gradient in the air then the pressure at the surface of the water will be greater than at the upper face of the block. It is the pressure at the water surface which is transmitted to the lower face of the block, so the increase in force on the lower face would be greater than that on the upper face, and the block would rise up in the water.

Another way of seeing this is to imagine that the air becomes as dense as the water. Then since the block floats in water it will also float upwards into the dense air.

There will be a significant pressure gradient in the air if either it is compressible or its density is comparable with that of the water.

$\endgroup$
  • 1
    $\begingroup$ Well ... actually an increase in pressure in the air makes the water very, very slightly denser, which means the object will float a very small amount higher, assuming that the water is more compressible than the object. This is a tiny effect. (But so is the pressure gradient in the air.) $\endgroup$ – Peter Shor Nov 5 '17 at 3:57
  • $\begingroup$ @PeterShor True, but this effect is smaller for water than for air. At STP air is 20,000 times more compressible than water, but only 784 times less dense. $\endgroup$ – sammy gerbil Nov 6 '17 at 15:33
  • $\begingroup$ But the effect in water depends on the absolute pressure, and the effect in air depends on the pressure gradient. So for water, the pressure difference might be on the order of an atmosphere of pressure, which is equivalent to the weight of several kilometers of air, while for the other effect you only get the weight of a few centimeters of air. $\endgroup$ – Peter Shor Nov 6 '17 at 15:59
3
$\begingroup$

It is the "no fluid" condition that makes a difference.

If the block is glued to the bottom, i.e. there is no possibility for the fluid to seep under it, it will stay put because of the cohesive forces of solids, it has become one with the floor.

Even if only microscopic paths allow for seepage of the fluid under the block, it stays put until the weight of the displaced water reaches the weight of the block and then it slowly starts floating. I tried it with a small block of wood in a plate filling it up with water slowly.

$\endgroup$
0
$\begingroup$

This is by no means a definitive answer.

Assuming a solid impermeable by the fluid, and no gaps between the solid and container, I would suggest that it probably depends on the density of the fluid and the malleability, height, and density of the solid.

In the example you provided of the wooden block in a container with water, it seems doubtful that the wooden block would rise, given how low the density of the water is (experimentally false due to imperfections in the surface of the materials—as anna v stated). However, if you were to use a much, much more dense fluid, it is possible that the gradient of compression pressure between the top and bottom of the block is sufficient to cause the walls of the block to become inclined inwards. If this effect were to happen to a large enough degree, there could be enough water underneath the leading edges of the block to get it off of the bottom.

You can see from the above example where the malleability and density of the solid comes into play. It might, for example, be much harder to raise a piece of titanium than a piece of gold by this method, despite gold being more dense than titanium, because gold is so much more malleable. Additionally, the solid block must be tall enough that a sufficient pressure gradient can exist between the top and bottom of the block.


In short, if the density of the fluid is great enough, it may be able to deform the sides of a malleable enough block that is sufficiently tall to be effected by the gradient of the fluid's compressive stress acting on its walls such that some amount of water can flow underneath the leading edges of the block.

$\endgroup$
  • $\begingroup$ The shape of the wooden block is not specified. If it need not be a cuboid shape then simpler than deformation of a malleable block is a rigid block which is already deformed. $\endgroup$ – sammy gerbil Nov 1 '17 at 7:37
  • $\begingroup$ If the block is already shaped such that the walls slope outwards as measured from the base to the top of the block, then it may be raised by the water underneath the slope $\endgroup$ – Kieran Moynihan Nov 1 '17 at 15:08
  • $\begingroup$ Yes, in the same way that it would be raised if the extreme density deformed the block into this shape. It is raised because of the shape, not the extreme density. So it is not necessary to postulate extreme density. $\endgroup$ – sammy gerbil Nov 1 '17 at 15:46
  • $\begingroup$ I assumed, when writing my answer, that the block would not be deformed such that it would rise anyway, because then there would be no problem to solve. $\endgroup$ – Kieran Moynihan Nov 1 '17 at 15:48
  • $\begingroup$ You assumed that it was not initially deformed, but that it was deformable. So you might as well have assumed it was deformed at the start. $\endgroup$ – sammy gerbil Nov 1 '17 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.