How can I calculate the true mass of an object in a fluid (f.e. air)?
Given:
- force measurement (F = 863000 N)
- air density (rhoA = 1,29 kg/m^3)
- object density (rhoO = 1100 kg/m^3)
- g (g = 9,807 m/s^2)
Solution: 88,1 kg.
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$\begingroup$ You compute the volume from the equilibrium and then the mass with volume and density. See the answer below $\endgroup$– basicsCommented Jan 14 at 14:02
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3 Answers
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When you're not given a volume, then that is your $x$. Find $x$ and $y$ will be revealed. ($y$ is mass).
So, I'm just going to think this out in real-time:
You measure a force: $F$
It is the weight of the object: $W = Mg$
($M$ is the mass of the object: $M = \rho_O V$)
minus the weight of the air: $w = \rho_A V$
so...
$$ F = W-w = \rho_O Vg - \rho_A Vg = Vg\Delta\rho$$
where delta rho is self-explanatory.
So then:
$$ V = \frac{F}{\Delta\rho g} $$
Sanity check: if the densities are the same, it diverges for non-zero $F$, and just doesn't work if $F=0$...good.
Back to mass:
$$M=\rho_O V = \frac{F\rho_O}{\Delta\rho g}$$
so there density units cancel and it's force divided by an acceleration...which Newtons say is mass. Done.
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Assuming $g$ acts downward, We would have to calculate the gravitational force $mg$ acting downwards.
Now we would have to calculate the buoyancy force acting upwards due to the mass of air displaced. The volume of the object is missing. you have probably forgotten to provide its value.
Now we can subtract both and we would get the apparent weight of the object in the fluid.
I would assume that you have defined a new unit system where the unit of force is "F = 863000N". Then you will have to convert the apparent weight from the unit system you have calculated into the new unit system.
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$\begingroup$ Down is defined by $g$, so no need to assume. Of course, when you do real geodesy, $\vec g(\phi, \lambda)$ doesn't point down (deflection of vertical), and $|\vec g(\phi, \lambda)| \ne g$ (the anomaly). $\endgroup$– JEBCommented Jan 14 at 14:37
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Equilibrium, i.e. sum of forces equals zero. Which forces act on the body?
- weight $\mathbf{W}$: pointing downwards
- buoyancy $\mathbf{F}_A$: Archimedes' tell you that it points upwards with magnitude equal to the weight of the displaced fluid
- reaction $\mathbf{R}$ of the measurement instrument, pointing upwards
\begin{equation} \mathbf{0} = \mathbf{W} + \mathbf{F}_A + \mathbf{R} \ . \end{equation}
From this equation, you have everything you need to compute the volume of the body. Then, given volume and density, you can compute the mass of the body, or the weight multiplying by $g$.
I won't complete the solution of the problem, but I stop here with a summary of the physical principles you should you to solve the problem, and provide you the final result in terms of weight
\begin{equation} W = \dfrac{1}{1-\frac{\rho_a}{\rho}} R \ , \end{equation} that relates the actual weight with the reading of the instrument $R$ as a function of the ratio of density $\rho_a / \rho$: for very small density ratio, the factor $\frac{1}{1 - \frac{\rho_a}{\rho}} \sim 1$, so that we could usually neglect buoyancy when we measure mass/weight of dense (much denser than air) objects.
You have everything you need to solve the problem
