B Using energy principles to find equilibrium depth of floating object

[brainpushups]

TL;DR Summary

I get an extra factor of 2 in the expected result

For fun I was trying to use energy considerations to determine the depth to which a solid object will sink in a fluid to reach equilibrium. The first approach that I tried was just to consider the change in potential energy of the block and the fluid as the block is lowered some unknown distance d into the fluid similar to what is shown in the answer to this post. Upon taking the limit as the vessel's cross sectional area approaches infinity I have an extra factor of 2 in the equilibrium depth.

That approach is a rather lengthy process, but I can get the same result when considering the work done on the block as explained below. I'm hoping the source of the error is the same so I'll use this shorter way to explain and hopefully someone can point out what I am overlooking.

Suppose a rectangular block with specific weight $\gamma_b$,cross section A and length L descends into an ideal fluid of specific weight $\gamma_f$ by a vertical distance d which is the equilibrium depth. The work done on the block is the difference between the work done by gravity and the work done by the buoyant force. For the former, this is just the product of weight and distance which we can write as $$W_g=\gamma_b A L d$$
The work done by the buoyant force must be integrated since it depends on the depth. After integration we obtain $$W_b=-\frac{1}{2} \gamma_f A d^2$$ which would also be the negative change in the potential energy of the fluid. The depth for which the change in the overall potential energy is zero can then be solved for which yields $$\frac{2L\gamma_b}{\gamma_f}$$

Without the factor of 2 we arrive at the expected result. Again, this is the same thing I find when going through the longer process of considering the change in the fluid height in a vessel and taking the limit as the cross sectional area of the vessel approaches infinity. Any insight about what I am missing here?

 

[Baluncore]

Any insight about what I am missing here?

As the body displaces fluid, the fluid surface will rise, so the reference from which d is measured is changing.

 

[brainpushups]

Thanks. I think I accounted for that in my potential energy approach, though I ignored it in my above example. To double check I calculated the total work again, this time including the area of the vessel. In the limit of a large cross sectional area for the vessel the factor of 2 persists.

To be clear, and to make sure I'm not making the same error over and over, let the cross sectional area of the block be $A_b$ and the cross sectional area of the vessel be $A_v$. As the block descends a distance $d$ the volume of water displaced is $A_b d$. Let the height that the fluid rises be $h$. Then, since the volume is constant, the height it rises is related to the depth the block sinks by $A_b d=(A_v-A_b)h$ and therefore $$h=\frac{A_b d}{A_v-A_b}$$
If the specific weight of the fluid is $\gamma_f$ then the work done by the buoyant force is $$- \gamma_f A_b \int y\, dy$$ with limits of integration from $-h$ as given by the expression above to $d$, the equilibrium depth.

If the length of the block is $L$ and its specific weight is $\gamma_b$ then the work done by gravity as the block descends a distance $d$ is just $\gamma_b A_b L d$.

To save some time I used Mathematica to evaluate the integral at the stated limits and then found the expression for the total work done on the block and, setting it equal to zero, solved for $d$. After taking the limit as $A_v$ gets very large the result is, yet again, that which was quoted in post #1.

I've either made an error here or I'm overlooking something else.

 

[brainpushups]

Wait a minute...for the work done by the buoyant force should I be considering the displacement of the center of buoyancy and not the surface itself? That would take care of the factor of 2.

 

[A.T.]

As the block descends a distance $d$ the volume of water displaced is $A_b d$.

If $d$ is measured relative to a fixed reference (e.g. mark on vessel wall) and "water displaced" is the change in volume of the object below the water level, then the above is false, unless the vessel was initially full and overflows or $A_v \gg A_b$ so $h = 0$.

You are computing the change in volume of the object below the original water level, not below the resulting water level.

 

[TSny]

If you release the block from rest at the position where it starts to submerge, then the only two forces doing work as the block submerges are gravity and buoyancy. But in this case, the object will bob up and down in oscillation. The net work is zero at the instant when the oscillating block is maximally submerged. The net force at this instant is not zero. The equilibrium position of the block is where the net force is zero. But this is not the position where the net work done (starting from the initial position) is zero. The maximum depth is twice the equilibrium depth.

You could lower the block by hand to the equilibrium depth to avoid oscillation. However, the net work would now include the work done by the force of the hand.

 

[brainpushups]

Of course! Now I'm feeling dense...