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I was inspired by the question Which way does the scale tip?, so I have conducted a branch of simplified experiment. I have weighted on the scales glass of water without and with stone immersed inside water and suspended by a thread kept with my hand. Results that I've got are:

enter image description here

And with stone:

enter image description here

So scales shows additional $30~g$ mass, which exactly is mass of water expelled by stone due to Buoyancy force, i.e. $B/g = 1 g/cm^3 \times 30~cm^3~~\{\text{stone volume}\}.$ Stone mass itself is about $70~g$, it's calculated density is $\approx 2.3~g/cm^3$.

So net force acting down on scales is $W_{glass}+B_0.$, i.e. Can we say that exact additional force which is pressing scales down in case stone is immersed inside glass ,- is third Newton force pair reactive Buoyancy force induced from a stone to water ?

In short, can we assume that $B_{stone\to~water} = - B_{water \to stone} $ (i.e. apply third Newton law to Buoyancy) and that $B_{stone\to~water}$ is the exact force which scales measures in a second case ?

I never heard of reactive Buoyancy forces, so here's my doubts and confusion on true nature of this additional force generated.

Qmechanic
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Agnius Vasiliauskas
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2 Answers2

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You're not missing anything. Newton's 3rd law says that since the water exerts a surface force (buoyant force) upward on the stone, the stone exerts a surface force downward on the water of the same magnitude. (I reject the characterization of either one of these forces as "reactive.") This increases the force that the glass must exert on the water to maintain equilibrium, which in turn increases the force the scale must exert on the glass. This latter force is what the scale reads.

d_b
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In a straightforward manner, it's just Newton's third law applied to buoyant force. But, the result can be explained from the origin of buoyant force i.e. the pressure of water.

Consider a simple case of a cylinder of length $l$ and cross-section area $A$ submerged in water vertically $h$ distance below it. What is the pressure at the two ends?

On the upper end, it's:

$$ P_{u} = \rho gh \implies \vec{F_{u}} = \rho hA \vec{g}, $$

where, $\rho$ is the density of water. And for the lower end: $$ P_{l} = \rho g(h+l) \implies \vec{F_{l}} = -\rho (h+l)A\vec{g}. $$

Also, $\vec{F_u}$ points downward on the cylinder and $\vec{F_l}$ points upward on it. Here Newton's third law comes and we say that the cylinder applies an upward force of magnitude, $|\vec{F_u}|$ and a downward force of magnitude, $|\vec{F_l}|$ on the water. So, net force on water(due to the cylinder) is:

$$ \vec{F} = (-\vec{F_l})+(-\vec{F_u}) =[\rho (h+l)A-\rho hA ]\vec{g} $$ $$ \implies\vec{F}=\rho lA \vec{g} = \rho V\vec{g}, $$

in the downward direction. Here, $V$ is the volume of the submerged object(cylinder). And, this is ,if you will, the reactive buoyant force on the water in downward direction.

This is what extra force is measured by the scale on somewhat fundamental level (if we ignore the origin of pressure, i.e. the jingling of water molecules).

CP of Physics
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