# Work done when submerging an object into fluid [closed]

How does one calculate the work needed to submerge, e.g. wooden, object into water?

For example, you want to submerge wooden cube, density 800 $\mathrm {kg/m^3}$ and side's length a=0.2 m.

The way I was doing it is I calculated force of gravity $G$ and buoyancy $U$, then I got total force $F=U-G$. Then, to get length of path along which one must apply that force, I calculated percentage of cube's volume that is inside the fluid when in equilibrium $(G=U)$. For that I got 0.8, so length of path is $s=0.2a$, and work needed would be $W=F\times s$.

But I can't get the result right. What did I miss?

## closed as off-topic by John Rennie, AccidentalFourierTransform, Kyle Kanos, Jon Custer, heatherJan 6 '17 at 16:07

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• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, AccidentalFourierTransform, Kyle Kanos, Jon Custer, heather
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• Since I'm not allowed to help you with homework based questions, here's how I can help you with hints. As you push your body into the fluid, you displace more volume of the fluid and hence result in a buoyant force which increases with the volume pushed into the fluid. Assuming you push the body perpendicularly in to the fluid. Can you find a relation between the force $F(x)$ and distance the body has been pushed in $x$?? Once you do that, I'll help you further. – ubuntu_noob Jan 4 '17 at 17:49
• Also, don't work with units, assume the density of the fluid is $\rho_f$, the density of wood is $\rho_w$ and the side length of the cube is $a$. – ubuntu_noob Jan 4 '17 at 17:53
• Firstly, this si an exercise, not homework. I thought about that, too. Buoyant force is not always the same, U(x)=ρfgAx=ρfg*(a^2)*x. So so to get total U I integrate (ρfga^2*dx) from 0.8*a to a, and add to that starting buoyancy, 0.8*Vgρf, but then I'm left with regular Vgρf. – DoroEser Jan 4 '17 at 18:10
• The tag is "homework-and-exercises". The site policy makes no distinction and applies to both. – sammy gerbil Jan 4 '17 at 18:15

An alternative approach, avoiding integration, is to calculate the increase in gravitational potential energy when the block descends from the floating position (left diagram) to the just-submerged position (right diagram) and the water moves up.

As the block is submerged, the water level rises from $h_2$ to $h_3$ measured from the base of the container, while the bottom of the block moves down from $h_1$ to the base. For convenience I assume the block just touches the base of the container when it is just submerged (right diagram). The top portion of the block (grey) moves to replace the water below the block (dark blue), while this water moves up to cause the increase in water level. All other portions of block or water remain in the same position, so they can be ignored. The grey and dark blue volumes are equal.

The condition for the block to float is
$\frac{h_2-h_1}{h_3}=\frac{\rho}{\rho_w}$
where $\rho, \rho_w$ are the densities of the block and water respectively.

The CG of the water which moves up is initially $\frac12h_1$ and finally $\frac12(h_3+h_2)$ above the base. The volume of this water is the same in both positions, so
$(h_3-h_2)(A-a)=h_1a$
where $A, a$ are the cross-sectional areas of the container and block respectively.

The CG of the grey portion of the block moves down by distance $h_3$. The volume of this portion is $h_1a$.

The above equations should be sufficient for you to calculate the overall increase in GPE when the block is submerged. This equals the work required to submerge the block.

If the block is being submerged in a large body of water instead of a container, then you should apply the limit $\frac{a}{A} \to 0$.

• but how do you calculate the work done?? By the difference in potential energies??? – ubuntu_noob Jan 4 '17 at 19:42
• By Total force * Distance – Zwolf Jan 4 '17 at 20:25
• But have you checked whether the results are the same? Ithink calculating the difference between the sum of the potential energies of the entire system water+block should work – ubuntu_noob Jan 4 '17 at 20:55
• I mean the difference between the initial and final potential energies. See there's so many approaches that can be discussed but I can't post them because of the homework restriction rules. I think they should allow alternative solutions to be posted at least. – ubuntu_noob Jan 4 '17 at 20:59

I can think of at least four things you would need to clear up/think about before you can answer this question:

1. Does "submerging" start when the block first touches the water, and end when it is just submerged?
2. Can we assume the block is oriented so that the surface of the cube is parallel to the surface of the water?
3. Do we expect the water level to rise as the cube is submerged (see Sammy Gerbil's answer)
4. How does the buoyancy force change as more of the cube is submerged?
5. What is the role of gravity in all this - as the cube moves down, gravity will do some of the work. Does that count in the "work done"?

When you have thought about all these things, you may be able to get the right answer yourself.

Don't be upset, I want to help you too, but I'm restricted by the rules of this site. See what you need to do is to balance forces, good that you developed it thus far. This allows me to help you further. The work done is done by the resultant force and not just by the buoyant force, since you apply the resultant force since the weight of the object is already helping you. $F_{resultant}=\rho_f ga^2x-\rho_wga^3$. Now try to get the work done by using $\int F_{resultant}\mathrm{d}x$ within the limits you specified.

I hope, this does not qualify as an answer. I'm just clearing a part of a misconception.

Disclaimer: I am adding this part after the OP showed his work.

The work done is calculated as $$\int\mathrm{d}W=\int F_{resultant}\mathrm{d}x=\int_{0.8a}^{a}(\rho_f g a^2x-\rho_wga^3)\mathrm{d}x$$

• I'm not upset at all, and sorry if I made that impression. I respect the rules of the site. Problem is that our collage system doesn't teach us calculus in math by the time we have to apply it in physics. Can you then just point out if the problem is in integration? I thought I could just integrate buoyant force seperatly from resultant force, then use the result in calculatin resultant force. – DoroEser Jan 4 '17 at 18:27
• could you show me your work and post it here, it'll be more clear as to where you are faltering. – ubuntu_noob Jan 4 '17 at 18:30
• @ubuntu_noob : If this does not qualify as an answer then it shouldn't be posted as one! The rules only discourage posting complete answers to homework-like questions. – sammy gerbil Jan 4 '17 at 18:40
• @sammygerbil: I wanted to be able to read the latex math I was typing in, see I even urged him to post his own answer... – ubuntu_noob Jan 4 '17 at 18:42
• @ubuntu_noob : I am not criticizing your decision to post an answer (instead of answering-in-comment), but your inconsistency of logic (posting an answer but hoping that it does not qualify as one). – sammy gerbil Jan 4 '17 at 18:51

An object in general describes a cross-sectional area $A(\ell)$ where $\ell$ is the distance that it has been pushed into the water; a good rule of thumb is $\int_0^L d\ell~A(\ell) = V,$ the total volume of the object, when it is fully submerged at depth $L$. Of course in your case $A(\ell)$ is just a constant $A$.

In order to calculate the work needed to submerge a floating object into the water, you indeed need to solve first for the equilibrium floating depth $h$, $$\rho_\text{obj}~V= \rho_\text{fluid} ~\int_0^hd\ell ~A(\ell),$$and then you need to submerge it further. As you submerge it further the force needed will increase with the amount of water displaced,$$F(x) = \rho_\text{fluid} ~g~\int_h^{h+x} d\ell ~A(\ell).$$Because this force is not constant to get the work you must integrate over the distance you push the item again,$$W=\int dx~F(x) = \rho_\text{fluid} ~g~ \int_0^{L-h}dx~\int_h^{h+x}d\ell~A(\ell).$$

For a constant area you should just get $A~x$ for the first integral and therefore $\frac12 A (L-h)^2$ for the second. You seem to have gotten this but not realized that it was a work, not a force.

A more realistic scenario for approximating various boats is to assume $A(\ell) = \gamma~\ell$ for some side-steepness factor $\gamma$, this corresponds to a boat which is a triangular prism. Or you could try to model one as a prism made by a parabola if you like.