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Please read the following. A cube submerged in water exerts a force on the water. The cube exerts a force equal to its weight on the water below it, and by Newton's third law it should feel a reaction force to this weight. This means that there are four forces on it: one due to the pressure and weight of the water on top, one due to the pressure of the water below, one due to gravity and one due to the reaction force to its weight.

But the pressure from below is due to the weight of the water above it pushing downwards, but there is no water because it was displaced, so the pressure from below can only be due to random collisions. The pressure from above is due to random collisions and the weight of all the water. The displaced water also moves up, so its weight adds to the pressure on the upper surface of the cube.

The weight of the object is balanced by the reaction force of the water below, like when you place something on a table and it does not fall. So the sum of all the forces on the object is simply the weight of the water above, plus the weight of the water displaced. Therefore no object can ever sink because the net force is downwards.

What is wrong with my line of thought? I can see that it is obviously wrong but I have no idea why.

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  • $\begingroup$ Reductionism is an excellent way to tackle a problem, but I think you have broken down the problem way too much! It is enough to think in terms of fluid pressure and weight of body. Read Fluid mechanics by F.M. White. $\endgroup$ – Deep Dec 25 '16 at 4:46
  • $\begingroup$ Think carefully , Your concepts are not clear . You are counting forces multiple times $\endgroup$ – InquisitiveMind Dec 25 '16 at 4:49
  • $\begingroup$ It would give me peace of mind this way. I mean, something is wrong with my understanding because this is incorrect, and I want to know what and why. $\endgroup$ – Albertrichard Dec 25 '16 at 4:55
  • $\begingroup$ Which forces am I counting more than once? The weight of the displaced water? $\endgroup$ – Albertrichard Dec 25 '16 at 4:57
  • $\begingroup$ Related : Proof of Archimedes Principle $\endgroup$ – Frobenius Dec 25 '16 at 6:46
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The cube exerts a force equal to its weight on the water below it,

This is incorrect, and leads to many other errors in your thinking.

The force on a submerged object in a gravitational field is easily calculated by considering the pressure in the fluid at each vertical position. That being the case, all the horizontal forces add to zero leaving only the forces on the top and the bottom which are different.

Now lets consider a submerged cube oriented with vertical and horizontal faces. The downward force on top ($\rho_w g d_{\mathrm{top}}A$) is less than the upward force on the bottom ($\rho_w g d_{\mathrm{bottom}}A$), where $A$ is the area of the face of the cube, $\rho_w$ is the density of water, and the $d$s are the depths. There is also the weight force on the cube which is totally independent of the liquid, so there are 3 important forces.

Now, the sinking or floating of the cube depends on the density of the cube. Summing the forces, taking down as positive, we have $$F_{\mathrm{down}}=\rho_w g d_{\mathrm{top}}A-\rho_w g d_{\mathrm{bottom}}A +(\rho_{\mathrm{cube}}Vg).$$

Now the length of one side is $A^{1/2}$, so the volume is $A^{3/2}$ and $d_{\mathrm{bottom}}-d_{\mathrm{top}}=A^{1/2}.$ Substituting gives us $$F_{\mathrm{down}}=\rho_{\mathrm{cube}}gA^{3/2}-\rho_wgA^{3/2}=gA^{3/2}(\rho_{\mathrm{cube}}-\rho_w).$$

So, if the density of the cube is greater than the density of water, there is a net downward force. If it's less, there is a net upward force.

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  • $\begingroup$ This. You can even measure this force by placing a beaker on a balance and lower a mass in on a string so that it is submerged but does not touch the sides. The force depends on the volume of the object but not on the mass beyond the requirement that you be able to get it underwater. $\endgroup$ – dmckee Dec 25 '16 at 17:48
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Okay! If you want me to point out the essential error that you make, it is this:

The normal reaction of the surface of a table is NOT the reaction to the weight of the object. (In the Newton's 3rd law sense). And let me explain:

When you place an object on a table this is what happens:
1)The earth is pulling the object towards itself(weight) and the object is pulling the earth towards itself. This is the action-reaction pair.
2)The object pushes on the table surface and the table surface pushes it back. Both are applying normal reactions to each other because the molecules of a solid don't give in. This is the second action-reaction pair.

Do you now see your error? The law is this: If object A applies a force on object B then object B applies an equal and opposite force on object A.

You can apply it to your particular example now. The forces are:
1)The earth is pulling the object towards itself(weight) and the object is pulling the earth towards itself...you get it right?
2)The molecules of water impinging on the solid's surface due to random collisions and the pushing back of the solid's molecules because they don't give in.

Therefore, on the object there are only two forces: weight and upthrust.

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  • $\begingroup$ I meant that the force with which the object pushes into a table is equal to the weight. $\endgroup$ – Albertrichard Dec 25 '16 at 5:04
  • $\begingroup$ Are you assuming that the pressure on the upper surface of the object is equal to the pressure on the lower surface? Because that would be wrong. In fact, it is the difference betweeen the pressures that gives rise to upthrust. $\endgroup$ – Abhijeet Melkani Dec 25 '16 at 5:08
  • $\begingroup$ Are you assuming that because the force with which the table pushes the object is equal to the object's weight, that the upward push of the water must also be equal to the weight of the object? Because that is also not true. They do not form an action-reaction pair. $\endgroup$ – Abhijeet Melkani Dec 25 '16 at 5:11
  • $\begingroup$ That was what I was assuming. And I pressed enter prematurely on the comment. I thought it would space it out, but it added it and I can't edit apparently. The object's push into the fluid is equal to the pressure difference right? $\endgroup$ – Albertrichard Dec 25 '16 at 5:13
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    $\begingroup$ I see. So a fluid would not balance the weight because it flows. $\endgroup$ – Albertrichard Dec 25 '16 at 5:31

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