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Suppose there is a wooden hemisphere on a horizontal plane, apex touching the plane. A small wooden cube is slowly placed on the base of the hemisphere anywhere other than the centre. As a result of extra torque provided by the weight of the cube, the hemisphere will lean on one side. Assume that there is enough friction to prevent cube from sliding.

solid

Now replace the wooden hemisphere with a bowl of similar shape filled with water (This time its edge is higher than before, as water should not run off. But the volume of water is the same as the volume of the hemisphere in the previous case). Now slowly place the small wooden cube on the water again as before. Will the bowl lean as before?

liquid

So far, my idea was: No, because when we place the cube on the water, a same amount of water equal to the submerged volume will be displaced. (My guess is) That will keep the centre of mass fixed, so there will be no extra torque about the contact point.

But the answers I got to these questions made me confused over my opinion: Can the fish topple the bowl and What is the rower actually doing? Pushing the water or pushing the lake?. ( Read them only if you are interested. I expect an answer for this question.)

Is the cube still able to change the CoM of the system?

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  • $\begingroup$ I'm confused. Do you place the cube in the water? Do you keep it fixed in that position somehow or does it sink? $\endgroup$
    – Prallax
    Aug 7 at 8:00
  • $\begingroup$ @Prallax, Place the cube on the water in the way how it was placed on the wooden hemisphere. I don't understand what you mean by in the water. The cube should float on the water, partially submerged. $\endgroup$
    – ACB
    Aug 7 at 10:40
  • $\begingroup$ Ok, thank you, now I understand $\endgroup$
    – Prallax
    Aug 7 at 12:55
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You are right, the bowl doesn't lean. From Archimedes' principle we know that the mass of the displaced fluid is the same as the mass of the cube. This means that the weight of the cube is the same as the weight of some water that would occupy the submerged part of the cube. In other words, if you substitute the cube with a volume of water equal to the volume of the submerged part, the force is the same.

But of course, if there is only water, the bowl is in equilibrium. Therefore, since the situation is dynamically equivalent, the bowl is in equilibrium also with the cube.

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  • $\begingroup$ Your answer seems correct to me. And it is the answer that was in my mind as well. Though this should imply that if we move the cube horizontally on the water, it will not create a motion. But if you read the answers to the second question I linked above, you can see a different answer. I'm in a tough situation. By the way, thanks for your answer. I am not asking you to expand your answer. Only if you are interested....... $\endgroup$
    – ACB
    Aug 7 at 18:06
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    $\begingroup$ The answer I gave was only about the static case of the cube in the water. If the cube had rows attached and somehow started rowing, the situation would change, with ripples, waves and water currents. By considering that the center of mass (in absence of friction with the ground) stays fixed and that the total torque is null, it is reasonable to think that the bowl would oscillate and slide, don't you agree? $\endgroup$
    – Prallax
    Aug 7 at 20:48
  • $\begingroup$ If instead the cube rowed very slowly, in order not to disturb the water much, by the same principles of conservation the bowl wouldn't move, passing continuously from one equilibrium state to the other $\endgroup$
    – Prallax
    Aug 7 at 20:56

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