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As, Buoyant force is the net upward force on any object in any fluid. If the buoyant force is greater than the object's weight, the object will rise to the surface and float. If the buoyant force is less than the object's weight, the object will sink.

B(buoyant force) = -$V\rho g$

We all know that $V$ is volume of displaced fluid by an object.


$$Question$$


Suppose we have solid wooden cylinder which has iron ball attached to it on the TOP and I place this thing (wooden solid cylinder +small Iron ball system) inside water and we $observe$ that it is half submerged. So what is the buoyant force on the system by water.

A) Is it $ρALg/2$ where $ρ$ is density of water, $A$ is cross sectional area of solid cylinder, $L$ is the length of cylinder

B) or is it $ρVg/2$ where $V$ = Total volume of SOLID CYLINDER + SMALL COPPER BALL which isn't equal to $ρALg/2$ because total volume of the system is AL + V′ where V′ is volume of small copper ball

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  • $\begingroup$ Just to be clear, is your drawing intended to represent the condition where the iron block is on top of the wooden block, or before the iron block is placed on the wooden block. I ask because your statement "Now Let suppose there is a wooden block partially submerged in water and I placed another object (say iron block ) on the top of it of density $\sigma$" can be interpreted that the wooden block is submerged as shown BEFORE the iron block is placed on top. Please clarify $\endgroup$
    – Bob D
    Jun 9 at 14:39
  • $\begingroup$ Its system, ans not just wooden block.. $\endgroup$
    – 5 Dots
    Jun 9 at 15:55

2 Answers 2

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V is the volume of fluid displaced.

If the iron block has mass $M$ and we assume that the box does not sink after its introduction then there must be an additional $Mg$ buoyant force supplied, which can be linked to the additional volume of fluid displaced as

\begin{equation} \Delta V = \frac{M}{\rho_{\text{wood}}} \end{equation}

corresponding to an additional length of box, $\Delta l$, sinking into the water

\begin{equation} \Delta l = \frac{M}{A\rho_{\text{wood}}} \end{equation}

If $\Delta l$ is greater than the height of the box above the water prior to the iron block being placed on top then buoyancy alone can't support the weight of the block.

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  • $\begingroup$ Sir what is the buoyant force on the system(wooden + iron block) by the water $\endgroup$
    – 5 Dots
    Jun 9 at 16:05
  • $\begingroup$ Have you tried drawing a free body diagram for this? If the system is in equilibrium then what are the forces that need to be balanced? $\endgroup$
    – Niall
    Jun 9 at 17:02
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V would be equal to the volume of the liquid displaced by the submerged part. Now that you have added an iron block, the total mass of the iron block and the wooden block must have gone up and resulted in vertical disturbance since the buoyant force now has to change so as to be able to sustain the new weight. The mass has come a bit more submerged in the water, so now the buoyancy will increase. The V will get a bit large also. The thing to note is that V can be defined as the volume of the solid submerged under water as well as the volume of the liquid displaced. That's obvious though-if a box has volume of 100m^3, then upon fully submerging it into the water (forcefully or let it go naturally: how it goes isn't but it goes is important), only 100m^3 of water (any liquid or gas rather that you put it in-the volume of the displaced liquid doesn't depend on the density).

The volume of the displaced liquid will be equal to both

the liquid's volume which we can replace where the object meant to be inside liquid or is it the object's partial volume

if by "partial volume", you mean NOTHING ELSE but volume of the solid now submerged

For the second part, if the new submerged volume be V, then the buoyant force will be V*(rho)*g The negative sign that you have used will come if you take the usual down as positive.

EDIT:

I am not sure what we have misinterpreted. If in the present picture you have shown, you have let the system settle down for a while, then the buoyant force is:ρAg*l (small l). EDIT 2: I just saw that it is observed that the body sinks half. SO keep the half.

Now, there will be a DIFFERENCE IN BUOYANT FORCE if we consider that this setup as you have shown, is just at the instant that the iron block is placed. This is because iron has its own mass, adding to the weight of the total system, pushing it further down to displace a little more water so as to have a new submerged volume (or volume of displaced liquid: they are the same) so as to support the weight of the new "system". Therefore, we wouldn't be correct t use either of L or l-since the iron will force the wooden block surface to sink deeper. L submerging may be done when the iron block is way too heavy.

Say the new length of the box that is submerged becomes L' .

Then, L'-l is the "extra" distance that the iron block forced the wooden block to move. (since the wooden block has become more heavier with the iron block introduction, it becomes necessary for the block to sink just enough that there can be extra water displaced so as to compensate for the iron block mass)

Now, we need to assume one more thing for a simpler calculation. That is: the extra length L'-l isnt greater than L-l. The reason: so the iron block can always be kept at the top and we can take a value of "A" that is constant throughout with the height.

If you fancy though, then you can also discard the assumption. And say A(iron block)*(submerged length of iron block)+ A(wooden block)*100% of length (which is L from your picture), (because for the iron block to be wet, the wooden block has to first sink completely) for the calculation of volume submerged.

And L-l is exactly that buffer or "extra" length that can be submerged further before the iron block gets wet. (l=small L). If it is equal to L'-l, then the iron block surface will just touch the water surface, but not get wet fully. There's like 0.00000000000000000000000001% of length submerged, basically 0%

Else, assuming that L' is the new submerged length, and that L'>l, and that (L'-l)<=(L-l) {which ultimately is L'<=L, but I like to think of it in differences}, the after some time (required for the system to settle), new submerged length*area of wooden block (I assume L'=L, so iron block is yet to be submerged) is volume.

Mathematical part: New submerged volume= AL' Buoyant force=rhoA*L'*g If however, L'-l>L-l (i.e. the iron block gets submerged too)

Let's say iron block has total length of h, with x% submerged

New volume (submerged) =V'=AL + A(iron block)(hx%)

Now V'rhog will be the buoyant force on the system.

And to calculate the exact dip, I am afraid I find no success for a generalized formula. Maybe try to see some cases, derive some system of equations and add/subtract to get the difference in new submerged length and earlier submerged length. Then, you could maybe divide the result and the other side of the equation (i think it should involve mass) to get increase for the wood per added 1 kg?

EDIT: Whenever I use "length", just see to it that it is the measurement of that side that you have labelled in the cross section as l and L (height more apparently).

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  • $\begingroup$ Misinterpreted dear, ok ok ok let try another way around... Supposed we have solid cylinder(wooden) which has a small copper ball on the top of it by super duper glue. Now i place this thing (wooden block +small copper ball system) inside water and we observe that it is half submerged. So what is the buoyant force on the system by water. $\endgroup$
    – 5 Dots
    Jun 9 at 16:23
  • $\begingroup$ a) Is it $ \rho ALg/2$ where $\rho$ is density of water, A is cross sectional area of solid cylinder, L is the length of cylinder $\endgroup$
    – 5 Dots
    Jun 9 at 16:25
  • $\begingroup$ b) or is it $ \rho Vg/2$ where V = Total volume of SOLID CYLINDER + SMALL COPPER BALL which isn't equal to $ \rho ALg/2$ because total volume of the system is AL + $V'$ where $V'$ is volume of small copper ball $\endgroup$
    – 5 Dots
    Jun 9 at 16:30
  • $\begingroup$ see: V represents submerged volume. So, until and unless the iron ball (man you are changing the shapes-keep it as a iron box) is wet, it has no role in the displacement of waterr (since then it would be outside). If this is t=0, then that means you have just placed the iron box on top of the wooden box. So, if we allow it to settle, then more volume of the water is displaced, hence more buoyancy. For which volume to take in and how much, just refer to the EDIT I made. $\endgroup$
    – Aveer
    Jun 10 at 7:22

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