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Suppose we have the following situation: A toy boat is placed in a tank of water that is perfectly balanced before. Which way, if at all, will the tank tip?

As the boat is placed in the water, by Archimedes' Law, it displaces a weight of water equal to its own weight, meaning there must be no net downwards force where the boat is added. I assume, however, that slightly more of the displaced water would be distributed to the side opposite the boat, as the boat itself obstructs the water: enter image description here However, the correct answer is that there is still no resultant torque. How is this the case?

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    $\begingroup$ I don't know how to answer your question, but the situation in picture (1) is unrealistic, because it is highly unstable. If the tank tilts, by even the most trivial amount, to either side, then the water will shift to that side and un-balance it. $\endgroup$ – Solomon Slow Oct 27 '19 at 17:26
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    $\begingroup$ A more realistic way to ask the question would be to draw the tank supported by two scales, with the scales positioned such that each one reads half of the combined weight of the water and the tank. Then, place the boat wherever, and ask, "how does the weight of the boat change the readings of the two scales?" $\endgroup$ – Solomon Slow Oct 27 '19 at 17:28
  • $\begingroup$ @SolomonSlow Agree, but this is theoretical anyway. Other possibility is a very light (low k) coil spring to indicate if there is a net torque created by the toy. $\endgroup$ – Bob D Oct 27 '19 at 17:58
  • $\begingroup$ I am not clear on the issue you describe. You perfectly well explain how the heavy object in one side causes more water to be displaced to the opposite side - more mass due to the block on the right-hand-side, more mass due to the water on the left-hand-side. In total, still the same mass and thus the same gravitational pull at either side. Just as you explained. Why would you then be confused about the correct answer confirming that? $\endgroup$ – Steeven Oct 27 '19 at 18:28
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    $\begingroup$ @Steeven The confusion was because, if I understand it correctly, the weight added due to the boat is exactly equal to the weight removed through the displacement of the water. So there is no extra torque because of that. But more water seemed to be displaced to the other side, which would make it tip. I think now it doesn't tip due to the displaced water being distributed equally, rather than a higher mass appearing on the right-hand side. (But please correct me if I'm wrong). Thanks! $\endgroup$ – Nick Maslov Oct 27 '19 at 18:54
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Your

by Archimedes' Law, it displaces a weight of water equal to its own weight

provides the answer: In your drawing you can replace the boat displacement with the equivalent amount of water. Then you can see that floating the boat is the same as simply adding the boat displacement's amount of water to the tank without the boat, which only increases the depth of the water (with no boat) which would not effect the balance of the tank.

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  • $\begingroup$ Thank you for your help, but I'm still a bit confused. I understand that the boat is equivalent to its weight of water. But when water levels rise, it seems they must rise everywhere except where the boat is (as it is in the way), which means there must be a tiny torque to the left in the image. What am I missing? $\endgroup$ – Nick Maslov Oct 27 '19 at 17:41
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    $\begingroup$ The water displaced by the boat only increases the height of the water column uniformly which the increases the water pressure at any given height. This water pressure will be uniform across the bottom of the tank so the balance will not change. $\endgroup$ – user45664 Oct 27 '19 at 17:57
  • $\begingroup$ Ah, I think I understand. I think my erroneous assumption was that the boat stays at the same height above the bottom of the tank. But the depth below it must increase by the same amount as the depth everywhere else. Thank you! $\endgroup$ – Nick Maslov Oct 27 '19 at 18:25
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I think it should be pointed out that the book answer is only correct if it assumes the toy floats or can otherwise remain suspended in the water (not sink to the bottom).

If the toy is heavy enough such that when submerged the downward force of gravity on the toy plus any water it may "contain" is greater than the upward buoyant force of the water (which equals the weight of the volume of water displaced), the toy will sink to the bottom and contribute a clockwise moment (torque) about the balance point.

I would also add the statement that the toy "displaces a weight of water equal to its own weight" can be misleading. If the toy is heavy enough to sink, adding more weight to the toy does not displace any more water.

Hope this helps.

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The answer lies in the comment givenby user45664 above.

The total mass in the tank is increased by the amount of mass of the boaat. However th boat floats, so we know that the amount of upward force on the boat is the same as the amount of downward force on it. It is held in place by the water which is, of course, a fluid.

The total mass of the water in the tank is transmitted evenly through the water to the base of the tank as the water pressure across the whole tank is the same.

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Add the boat slowly.

If you add the boat fast enough that it makes waves, the waves are an unbalancing force. Because you have added the boat faster than the water could adjust. That may or may not be enough to get the tank to tip over.

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