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This problem has bothered me for quite some time and I can't solve it. I have even tried to make a construction, but it sometimes tips to the left and sometimes tips to the right :).

When we submerge the body in the water the water pushes it up. That is an action. At the same time the body pushes water down. That is a reaction. So the balance should be maintained.

But, since the water level rises the hydrostatic pressure on the bottom is greater so the right side should go down.

There is another similar problem but it's not the same.

Please help.

Qmechanic
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user3368512
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    This looks different, the bouyant mass is hanging from the scales not an external hook. Actually this makes all the difference to the answer. – JMLCarter Mar 16 '17 at 23:04
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    I have seen the problem you have suggested and it is not the same as JMLCarter already wrote – user3368512 Mar 16 '17 at 23:21
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    Are we assuming that the acceleration due to gravity is constant throughout the height difference for the purpose of this question? (I didn't think I had to ask this but maybe it should be cleared up) – JMac Mar 16 '17 at 23:47
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    @JMac Excellent question, it is completely relevant as we all know that objects weigh less at increased heights. This is what my answer is trying to address. Though I acknowledge that the difference between the two forces is very small it is certainly not negligible and (assuming the scales are accurate) it will be enough to make the scales rotate. – BLAZE Mar 16 '17 at 23:59
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    @BLAZE What you're describing is an impossible to balance scale in real life. By your logic any height difference in the objects will have a force great enough to make the scale start tipping. Any attempt to put two objects on and compare them would have picometres difference and start to tip. The reality is there is some contact surface are and imperfections and air resistance which also create small forces which on average dissipate the small differences in force due to all of these effects. For realistic analysis we either neglect all small forces or analyze them all... – JMac Mar 17 '17 at 00:05
  • ... if we want this to resemble real life in the slightest. You didn't ask the question originally though, which is why I want to clear up it's intent. – JMac Mar 17 '17 at 00:06
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    @JMac I agree with you that the question needs to be more restrictive, but in this particular problem there is an appreciable height difference. This is what the question is trying to show by the diagrams. But yes, clarification is needed. – BLAZE Mar 17 '17 at 00:13
  • @BLAZE I disagree with what you believe it is trying to show. I believe it's trying to show the force remains balanced even though there's a buoyant force acting on the mass, because there's an equal increased force on the weight, and it's all on the same scale. The difference between the two is like $0.00001 \frac {m}{s^2}$ for a $1m$ height difference. That's assuming a reasonable size setup in a lab. The other effects would likely outweigh that difference heavily. – JMac Mar 17 '17 at 00:33
  • I am aware that the question is not precise enough, but that is how it is given: Which way the scale will tip when we go from the first to the second figure? – user3368512 Mar 17 '17 at 00:52
  • @user3368512 But we can't answer that unless you state (or the question states) assumptions. It's very simple; if we neglect height difference of mass then the scales won't rotate. If we don't neglect height difference they will rotate. So if this is some kind of exam question I would have to give two answers one for each situation like I did in my answer. Everyone else neglected the height difference. – BLAZE Mar 17 '17 at 01:09
  • @BLAZE But my point is your assumptions are extremely unrealistic. In a realistic scenario the gravitational effects are one of many which essentially cancel out (and the scale would really be imperfect such that it could withstand a small amount of imbalance). There's no need to only analyze one of the small scale forces, it's not a realistic analysis, regardless of how the question is presented. There's some level of assumptions you have to be able to make regarding scales of magnitude, the effect of height on gravitational acceleration should be very low magnitude. – JMac Mar 17 '17 at 01:10
  • @JMac Sorry, I know what you're saying but I simply disagree; It wouldn't hurt to state assumptions no matter how unrealistic they are. – BLAZE Mar 17 '17 at 01:17
  • @user3368512, thanks for a different version of a demonstration that I have shown students. The scale will remain in balance for your stated problem, but it takes a bit of thinking to arrive at this conclusion. Now, all I have to do is think of a way to demonstrate this example in the class room. – David White Mar 17 '17 at 01:42
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    Again, this question is not a duplicate. I dont see how someone can say that. That means he/she have not read that or this question. Only thing that is the same is the title because the problems are similar, but certainly not the same. Cheers – user3368512 Mar 17 '17 at 08:07
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    This is not a duplicate. In the referred question some portion of the weight of the steel ball is carried by the string that is not connected to the scale. But in the current question, the string is connected to the scale. – lucas Mar 17 '17 at 08:39
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    The reopen review for this question has ended with 3 of 4 votes for "leave closed." Even though there's a difference in the setup between this one and the duplicate, the method for solving them is the same. Flaggers and commenters who feel differently might raise the issue on [meta] or [chat]. – rob Mar 17 '17 at 22:56

6 Answers6

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The scale will not move.

You don't need to think about buoyancy at all to answer this question. In the first picture, the scale is balanced, because the net force on each side (the weight) is equal. No mass is added to or removed from either side, so the net forces remain the same.

Nuclear Hoagie
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The balance will be maintained because there is no EXTERNAL force applied on left or right side. This is because of the same reason you can't push a car while sitting in it!

Alireza
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So various forces neglected (including gravity, air resistance, thermal forces, air pressure) - but your question suggest this is an inquiry about the effect of bouyancy.

That being the case, lowering the mass into the water brings into effect its bouyancy. This is a force lifting the mass and acting down on the water. However, as the mass is tethered to the scale the force acting down the string is reduced by the same amount as the bouyancy, and the net effect on the scale is zero.

(Also neglecting rotating frames of reference, light pressure, stress and young's modulus, magnetically induce currents etc)

JMLCarter
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If you consider everything put in the right side of balance scale as one system, only the internal forces are changed which will not affect the balance. No external force is applied and hence the balance will be maintained.

user42461
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Let $V$ be the volume of the hanging block, $A$ the area of the water surface, $\rho$ the density of water, and $g$ the acceleration due to gravity.

The extra hydrostatic force felt by the bottom of the container after lowering the block is $$\Delta F = \rho gA\Delta h,$$ where $\Delta h$ is the change in water height ($\rho g \Delta h$ would be the change in hydrostatic pressure). Because water is not compressible, $A\Delta h$ must be the volume of the submerged block. $$\Delta F = \rho g V.$$ Notice that this is exactly equal to the buoyant force on the block, so the two new forces created after the block is submerged negate each other.

Mark H
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EDIT:

Since this answer is not being very well received I will just say that if we neglect gravitational forces due to the height difference between the mass in the upper and lower pictures; then yes, the scales will be balanced as there is simply no net torque.

However, since the question did not stipulate these assumptions then the answer below is the correct one:

Look at the RHS of the scale and consider Newton's Law of Gravitation $$F=-\frac{G \cdot m\cdot m_E}{d^2}$$ On the RHS consider $m$ to be the mass of block suspended by the string and $m_E$ is the mass of the Earth $\approx 6\times 10^{24}$kg. $G$ is the gravitational constant and $d$ is the distance of the string between the block in the top picture and the one in the bottom plus the distance to the centre of the earth.

Now let $m_{s}$ be the mass of the submerged block in the container of the lower RHS and $d_E$ is the distance to the centre of the earth. $$F=-\frac{G \cdot m_{s}\cdot m_E}{d_E^2}$$

The gravitational force acting on the RHS of lower system is therefore greater by Newton's inverse square law.

Can you now understand why once the block is dropped into the fluid the RHS of the lower image will weigh more and hence tilt clockwise?

BLAZE
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  • I don't think Newtons Law of Gravity applies here... these forces are completely negligible compared to the force of Earth's gravity. – JMac Mar 16 '17 at 23:13
  • I'm saying this entire thing is being driven by acceleration due to gravity. You very rarely (I've never really seen it) use the force of gravity between two bodies when doing questions like this. In fact, draw FBD's for the gravitational force they will exert on each other. There's a gravitational force down on the mass and an equal force up on the water. Both the water and string are attached to the scale, the net force on the scale is still 0 due to the gravitational force between them. Instead we assume one of the masses is Earth then we know acceleration in freefall is $\approx$ g. – JMac Mar 16 '17 at 23:19
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    @JMac Yes, but they are at different heights with respect to the earth, so Newton's law of gravity is applicable here. – BLAZE Mar 16 '17 at 23:21
  • Use some realistic numbers here. The mass of Earth is $5.972 x 10^{24}$ kg, the radius is $6371$ km. Compare that to any values you would expect to see in this situation and you can recognize that even moving a 1 ton mass 100 meters won't have an appreciable effect on acceleration due to gravity. – JMac Mar 16 '17 at 23:23
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    I agree with JMac. Earth's gravity is way greater. However it is very interesting to take into the consideration changed gravity due to the changed distance between two bodies, although it is negligible. – user3368512 Mar 16 '17 at 23:25
  • These edits still don't clear it up. If you want to consider these effects, then we also have to consider other small forces like any potential imperfections in the system and the air around it and such. Generally when approaching a problem like this where we are assuming doing this on Earths surface, we assume acceleration due to gravity is constant. The extremely low magnitude different brought out by newtons law of gravitation would be completely unnoticed in a real scenario, so we ignore them in these contexts. You don't see people integrating $mg(h)dh$ for gravity potential usually. – JMac Mar 16 '17 at 23:36
  • I still think you are missing the point based on this new edit. You say "...if and only if we neglect gravitational forces due to the height difference between the mass in the upper and lower pictures; then yes, the scales will be balanced..." but that's not true. You can consider the effect of height on gravitational forces. But if you do, you also have to consider all the other forces of a similar magnitude that act in this scenario for the analysis to be remotely realistic. This analysis is nothing like what would happen in real life, and therefore isn't particularly helpful. – JMac Mar 17 '17 at 01:30