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If I got it right, the buoyancy force is a result of pressure changes exerted by the gravitational force. Archimedes’ law then generally predicts the buoyant force to be $\rho_\text{fluid} Vg$. What I fail to understand is why when I calculate the total force due to pressure around a sphere of radius $R$ for example, I don't get exactly this force, but the force factored by some constant. Following is an example of what I'm talking about for the sake of making myself clear, I'm not concerned about specific examples:

$$F_\text{buoyancy}=\int_{0}^{2\pi}\int_{0}^{\pi}\rho_\text{fluid} gh\,R^{2}\sin\theta d\theta d\varphi$$ when putting $h = -R(cos\theta-1)$ for the depth of some infinitesimal surface area, we get:

$$F_\text{buoyancy}=-2\pi g\rho_\text{fluid} gR^{3}\int_{0}^{\pi}(\cos\theta-1)\sin\theta d\theta=4\pi R^{3}\rho_\text{fluid}g$$

In this calculation for example, I'm missing a $\frac{1}{3}$ factor in order to get Archimedes law right. Does that mean the buoyancy force isn't generally just the pressure exerted on a body? Or am I missing something? Seems too close to be a coincidence.

BioPhysicist
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Darkenin
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1 Answers1

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You are missing a $-\cos\theta$ factor: this will account for the fact that you are integrating the vertical (upward) components of the force distribution on the sphere. Without this, you are just integrating the magnitude of this force, which isn't very useful.

Puk
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  • Well, it does work out this way. I'm not sure I understand why I should generally only count the upward component though. Aren't all the pressure components contributing to the general buoyant force? What do all the other components physically do? – Darkenin Oct 23 '20 at 21:45
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    The pressure does not have components. Forces do. And they are vectors and add like vectors. The sum of the magnitudes is not the same as magnitude of the sum. – nasu Oct 23 '20 at 21:53
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    @Darkenin The buoyant force is always upward. The other components end up canceling out, so you don't need to include them in your calculation. If you do want to show that they cancel out, you need to integrate the normal force vector per unit area, breaking it down to its components, rather than integrating the (scalar) pressure. In this example, symmetry considerations will reveal immediately that horizontal components cancel (this happens in the $\varphi$ integral). – Puk Oct 23 '20 at 21:54
  • @Frobenius For the sake of clarity, in this case just multiplying the integrand by $\sin \theta$ gives the magnitude of the horizontal force per unit area, but the horizontal component is still a vector with a direction that depends on the position along the surface (always pointing toward the vertical axis of symmetry of the sphere). So that won't actually integrate to zero. You can further decompose it into Cartesian components, which will contribute $\cos \varphi$ and $\sin \varphi$ factors that will cause the $\varphi$ integral to yield $0$. – Puk Oct 23 '20 at 22:34
  • You are absolutely right. I post a new comment. – Frobenius Oct 23 '20 at 22:38
  • @Darkenin : The horizontal hydrostatic force is always zero for any body (not for spherical ones) for any immersion depth. If not, then a boat in a calm lake would automatically move horizontally to a random direction, see my answer as "user82794" here Proof of Archimedes Principle. – Frobenius Oct 24 '20 at 00:42