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I'm not able to comprehend the meaning of buoyant force. As per Archimedes Principle, it's the force equal to the weight of the volume of water displaced by the body. But wouldn't this be only true when an object is partially submerged?

As just for an example,I have a tub of water and a cube(of around same density as of water so that it doesn't sink or float but stay at a certain level throughout the time, say 1g/1cm³) of dimensions 1m, which is completely submerged inside at a certain depth, say the top surface of the cube is 5m from the surface of the water and the bottom part of the cube is 6m.

I find myself with two options to calculate it:

  1. To calculate the difference of forces experienced by the bottom and top surface of the cube which will give me the net force experienced by the cube.
  2. As the cube is at a constant position at the depth, the vertical forces should cancel each other out, thus the net force at the bottom part of the cube would be the same as the buoyant force it experiences,which seems logical to do.

1st Method:

Force on the bottom of the cube(F2): F(atmospheric pressure) + F(weight of cube) + F(weight of the fluid column just above the cube).

Force on top of the cube(F1): F(atmospheric pressure) + F(weight of fluid column above the cube)

---> Net force: F2 - F1 = F(due to the weight of cube) = m(cube) *g = 1000kg/m³ * 1m³ * 10m/sec² (to make calculation easy) = 10⁴N

2nd Method: Total force experienced at the bottom part of the cube = Buoyant force:

F2 = F(atmospheric) + F(weight of cube) + F(weight of fluid column above)

----> 10⁵N(approx.) + 10000N + 50000(5m*1m² * 10000kg/m³ * 10m/sec²) = (10⁵ + 60000)N

Here I got 2 different values , although I believe the first method seems more related to what the principle says, but wouldn't the total force experienced at the bottom point of the cube be the actual buoyant force?

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2 Answers 2

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You are just making some mistakes in what we are trying to say.

Let the cube have side length $\ell$ and the top surface be submerged to depth $h$ from the top surface of the water of density $\rho$, exposed to atmospheric pressure $P_\text{air}$. The forces on the top and bottom surfaces of the cube are $F_\text{top}$ and $F_\text{bot}$ respectively. Then, they are $$ \begin{align} \tag1F_\text{top}&=\ell^2\left[P_\text{air}+\rho g h \right]\qquad\quad\ \text{downwards}\\ \tag2F_\text{bot}&=\ell^2\left[P_\text{air}+\rho g(h+\ell)\right]\qquad\text{upwards}\\ \tag3F_\text{bouyancy}&=F_\text{bot}-F_\text{top}=\ell^2\rho g\ell=\rho g \ell^3 \end {align} $$ which is independent of the atmospheric pressure. It is only dependent upon the water's density $\rho$ and the cube's volume $\ell^3$, i.e. the volume of the water that was displaced by the cube. i.e. $F_\text{bouyancy}=W_\text{displaced water}$

For the cube to stay afloat in the middle, this means that $F_\text{bouyancy}=W_\text{cube}=mg=\rho_\text{cube}\ell^3g$ and just by comparing with Equation (3), we see that it is necessary for $\rho_\text{cube}=\rho_\text{water}$ for that to happen.

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  • $\begingroup$ I'm able to understand the derivation and approach. But I still got one doubt , the buoyant force is just the force required to push an object/body upwards as what the principle says. But shouldn't it be the net force acting on the bottom point of the cube? Because not only the force(Buoyant ) is pushing the cube upwards but also the water column just above the cube, as it's also applying a significant pressure(and thus a force) on the cube which ultimately be the actual buoyant force? $\endgroup$
    – Aryan
    Sep 30, 2023 at 5:02
  • $\begingroup$ The bouyant force is a convenient mathematical consequence of the imbalance of the pressures pushing down from on top and pushing up from below. What particular net force are you trying to talk about? The bottom force is indeed strong enough to prop up the water column above the cube; that is simply subtracted off, as you saw in the derivation. $\endgroup$ Sep 30, 2023 at 6:25
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There are two problems with your logic:

  1. You count the weight of the cube pushing down on the bottom surface, but not the weight of the cube pulling down on the top surface. Though I suggest just not worrying about internal forces to begin with.

  2. The external force on the bottom of the cube is not the vertical column above the top of the cube. The external force on the bottom can only come from the water molecules hitting it. Those water molecules are at a pressure $P$ so they exert a force $F = PA$, and that force must be in the direction normal to the bottom surface.

Like wise, the only external force on the top of the cube is $F = PA$; but the water pressure is slightly different here since the pressure depends on the depth of the water $d$; with $P = \rho gd $.

So the difference in force on the top vs. bottom of the cube is simply $A * (P_T - P_B) = \rho gd_T - \rho gd_B = \rho gL$ for a cube of length $L$. That is the buoyancy force.

You can then also factor in gravity, which acts on the entire body so that the net force on the block is $F_{net} = F_{gravity} - \rho gL$

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  • $\begingroup$ I'm not able to understand a part of the first point where you say " but not the weight of cube pulling down on top surface", could you please elaborate? $\endgroup$
    – Aryan
    Sep 30, 2023 at 4:52

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