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Lets say a balloon with a certain amount of buoyant lift has a 2 kg mass strapped to it (including engines) and accelerates enough that it perceives a positive 2.5 g inertial force (2.5g * weight) acting downwards. Lets say initially the amount of buoyant force is equal to 50% of the perceived weight during no acceleration (1 kg total weight).

Because weight equals: mass * gravity (2.0 kg * 9.81)

And the g factor due to acceleration is applied to the weight to get g force: 2.5g * weight = Force

Then the g force would become 2.5g * 2.0 kg * 9.81 = 49.05 N

OR...

Is the g factor due to acceleration only applied to the resultant weight minus the helium buoyant force?

2.5g * 1kg (resultant weight) * 9.81 = 24.525 N

If the first case is true, the buoyancy becomes much less effective under an acceleration inertial load. If the second case is true, can someone show me mathematically or explain why the g factor is not multiplied by the entire mass?

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You assumed initially that the buoyant force is is equal to 50% of the perceived weight, that is for a 2kg mass you assumed $F_b = 2kg \cdot 9.8N/kg \cdot 0.5 = 9.8 N$. Then the acceleration would be $a = \frac{1}{2kg}(F_b - F_g) = -4.9N/kg$, so the mass will accelerate downwards at 0.5g, not 2.5g.

Your first calculation is correct, however your conclusion that the buoyant force is less effective under acceleration isn't. For a balloon of volume $V$ you can compute the buoyant force $F_b = gV(\rho_1 - \rho_2)$ where $\rho_1$ and $\rho_2$ are the densities of the air out of and in the balloon. As long as the densities remain constant, the buoyant force should remain constant, even if the balloon is accelerating.

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  • $\begingroup$ Sorry I should have re-worded but for theories sake can we assume there are engines strapped to this thing accelerating it to the point that it experiences 2.5g's? If your argument that the buoyant force remains constant regardless of external acceleration and only relies on density is true, then my argument that buoyant force becomes less effective under acceleration is also true. $\endgroup$ – Jason Oct 31 '18 at 0:24

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